The Free High School Science Texts: Textbooks for High School Students Chemistry

The Free High School Science Texts: Textbooks for High School Students Chemistry
FHSST Authors
The Free High School Science Texts:
Textbooks for High School Students
Studying the Sciences
Chemistry
Grades 10 - 12
Version 0
November 9, 2008
ii
Copyright 2007 “Free High School Science Texts”
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FHSST Core Team
Mark Horner ; Samuel Halliday ; Sarah Blyth ; Rory Adams ; Spencer Wheaton
FHSST Editors
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Whitfield
FHSST Contributors
Rory Adams ; Prashant Arora ; Richard Baxter ; Dr. Sarah Blyth ; Sebastian Bodenstein ;
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Daniels ; Sean Dobbs ; Fernando Durrell ; Dr. Dan Dwyer ; Frans van Eeden ; Giovanni
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Andrew Kubik ; Dr. Marco van Leeuwen ; Dr. Anton Machacek ; Dr. Komal Maheshwari ;
Kosma von Maltitz ; Nicole Masureik ; John Mathew ; JoEllen McBride ; Nikolai Meures ;
Riana Meyer ; Jenny Miller ; Abdul Mirza ; Asogan Moodaly ; Jothi Moodley ; Nolene Naidu ;
Tyrone Negus ; Thomas O’Donnell ; Dr. Markus Oldenburg ; Dr. Jaynie Padayachee ;
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iii
iv
Contents
I
II
Introduction
1
Matter and Materials
3
1 Classification of Matter - Grade 10
1.1
1.2
5
Mixtures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
1.1.1
Heterogeneous mixtures . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
1.1.2
Homogeneous mixtures . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
1.1.3
Separating mixtures . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
Pure Substances: Elements and Compounds . . . . . . . . . . . . . . . . . . . .
9
1.2.1
Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
1.2.2
Compounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
1.3
Giving names and formulae to substances . . . . . . . . . . . . . . . . . . . . . 10
1.4
Metals, Semi-metals and Non-metals . . . . . . . . . . . . . . . . . . . . . . . . 13
1.4.1
Metals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
1.4.2
Non-metals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
1.4.3
Semi-metals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
1.5
Electrical conductors, semi-conductors and insulators . . . . . . . . . . . . . . . 14
1.6
Thermal Conductors and Insulators . . . . . . . . . . . . . . . . . . . . . . . . . 15
1.7
Magnetic and Non-magnetic Materials . . . . . . . . . . . . . . . . . . . . . . . 17
1.8
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
2 What are the objects around us made of? - Grade 10
21
2.1
Introduction: The atom as the building block of matter . . . . . . . . . . . . . . 21
2.2
Molecules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
2.2.1
Representing molecules . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
2.3
Intramolecular and intermolecular forces . . . . . . . . . . . . . . . . . . . . . . 25
2.4
The Kinetic Theory of Matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
2.5
The Properties of Matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
2.6
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
3 The Atom - Grade 10
3.1
35
Models of the Atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
3.1.1
The Plum Pudding Model . . . . . . . . . . . . . . . . . . . . . . . . . . 35
3.1.2
Rutherford’s model of the atom
v
. . . . . . . . . . . . . . . . . . . . . . 36
CONTENTS
3.1.3
3.2
3.3
CONTENTS
The Bohr Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
How big is an atom? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
3.2.1
How heavy is an atom? . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
3.2.2
How big is an atom? . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
Atomic structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
3.3.1
The Electron . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
3.3.2
The Nucleus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
3.4
Atomic number and atomic mass number . . . . . . . . . . . . . . . . . . . . . 40
3.5
Isotopes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
3.6
3.7
3.8
3.9
3.5.1
What is an isotope? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
3.5.2
Relative atomic mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
Energy quantisation and electron configuration . . . . . . . . . . . . . . . . . . 46
3.6.1
The energy of electrons . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
3.6.2
Energy quantisation and line emission spectra . . . . . . . . . . . . . . . 47
3.6.3
Electron configuration . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
3.6.4
Core and valence electrons . . . . . . . . . . . . . . . . . . . . . . . . . 51
3.6.5
The importance of understanding electron configuration . . . . . . . . . 51
Ionisation Energy and the Periodic Table . . . . . . . . . . . . . . . . . . . . . . 53
3.7.1
Ions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
3.7.2
Ionisation Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
The Arrangement of Atoms in the Periodic Table . . . . . . . . . . . . . . . . . 56
3.8.1
Groups in the periodic table
. . . . . . . . . . . . . . . . . . . . . . . . 56
3.8.2
Periods in the periodic table . . . . . . . . . . . . . . . . . . . . . . . . 58
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
4 Atomic Combinations - Grade 11
63
4.1
Why do atoms bond? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
4.2
Energy and bonding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
4.3
What happens when atoms bond? . . . . . . . . . . . . . . . . . . . . . . . . . 65
4.4
Covalent Bonding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
4.4.1
The nature of the covalent bond . . . . . . . . . . . . . . . . . . . . . . 65
4.5
Lewis notation and molecular structure . . . . . . . . . . . . . . . . . . . . . . . 69
4.6
Electronegativity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
4.7
4.8
4.6.1
Non-polar and polar covalent bonds . . . . . . . . . . . . . . . . . . . . 73
4.6.2
Polar molecules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
Ionic Bonding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
4.7.1
The nature of the ionic bond . . . . . . . . . . . . . . . . . . . . . . . . 74
4.7.2
The crystal lattice structure of ionic compounds . . . . . . . . . . . . . . 76
4.7.3
Properties of Ionic Compounds . . . . . . . . . . . . . . . . . . . . . . . 76
Metallic bonds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
4.8.1
The nature of the metallic bond . . . . . . . . . . . . . . . . . . . . . . 76
4.8.2
The properties of metals . . . . . . . . . . . . . . . . . . . . . . . . . . 77
vi
CONTENTS
4.9
CONTENTS
Writing chemical formulae
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
4.9.1
The formulae of covalent compounds . . . . . . . . . . . . . . . . . . . . 78
4.9.2
The formulae of ionic compounds . . . . . . . . . . . . . . . . . . . . . 80
4.10 The Shape of Molecules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
4.10.1 Valence Shell Electron Pair Repulsion (VSEPR) theory . . . . . . . . . . 82
4.10.2 Determining the shape of a molecule . . . . . . . . . . . . . . . . . . . . 82
4.11 Oxidation numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
4.12 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
5 Intermolecular Forces - Grade 11
91
5.1
Types of Intermolecular Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
5.2
Understanding intermolecular forces . . . . . . . . . . . . . . . . . . . . . . . . 94
5.3
Intermolecular forces in liquids . . . . . . . . . . . . . . . . . . . . . . . . . . . 96
5.4
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
6 Solutions and solubility - Grade 11
101
6.1
Types of solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
6.2
Forces and solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
6.3
Solubility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
6.4
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106
7 Atomic Nuclei - Grade 11
107
7.1
Nuclear structure and stability . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
7.2
The Discovery of Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
7.3
Radioactivity and Types of Radiation . . . . . . . . . . . . . . . . . . . . . . . . 108
7.4
7.3.1
Alpha (α) particles and alpha decay . . . . . . . . . . . . . . . . . . . . 109
7.3.2
Beta (β) particles and beta decay . . . . . . . . . . . . . . . . . . . . . 109
7.3.3
Gamma (γ) rays and gamma decay . . . . . . . . . . . . . . . . . . . . . 110
Sources of radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112
7.4.1
Natural background radiation . . . . . . . . . . . . . . . . . . . . . . . . 112
7.4.2
Man-made sources of radiation . . . . . . . . . . . . . . . . . . . . . . . 113
7.5
The ’half-life’ of an element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
7.6
The Dangers of Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116
7.7
The Uses of Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117
7.8
Nuclear Fission
7.9
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
7.8.1
The Atomic bomb - an abuse of nuclear fission . . . . . . . . . . . . . . 119
7.8.2
Nuclear power - harnessing energy . . . . . . . . . . . . . . . . . . . . . 120
Nuclear Fusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120
7.10 Nucleosynthesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121
7.10.1 Age of Nucleosynthesis (225 s - 103 s) . . . . . . . . . . . . . . . . . . . 121
7.10.2 Age of Ions (103 s - 1013 s) . . . . . . . . . . . . . . . . . . . . . . . . . 122
7.10.3 Age of Atoms (1013 s - 1015 s) . . . . . . . . . . . . . . . . . . . . . . . 122
7.10.4 Age of Stars and Galaxies (the universe today) . . . . . . . . . . . . . . 122
7.11 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
vii
CONTENTS
CONTENTS
8 Thermal Properties and Ideal Gases - Grade 11
125
8.1
A review of the kinetic theory of matter . . . . . . . . . . . . . . . . . . . . . . 125
8.2
Boyle’s Law: Pressure and volume of an enclosed gas . . . . . . . . . . . . . . . 126
8.3
Charles’s Law: Volume and Temperature of an enclosed gas . . . . . . . . . . . 132
8.4
The relationship between temperature and pressure . . . . . . . . . . . . . . . . 136
8.5
The general gas equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137
8.6
The ideal gas equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140
8.7
Molar volume of gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145
8.8
Ideal gases and non-ideal gas behaviour . . . . . . . . . . . . . . . . . . . . . . 146
8.9
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147
9 Organic Molecules - Grade 12
151
9.1
What is organic chemistry? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151
9.2
Sources of carbon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151
9.3
Unique properties of carbon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152
9.4
Representing organic compounds . . . . . . . . . . . . . . . . . . . . . . . . . . 152
9.4.1
Molecular formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152
9.4.2
Structural formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153
9.4.3
Condensed structural formula . . . . . . . . . . . . . . . . . . . . . . . . 153
9.5
Isomerism in organic compounds . . . . . . . . . . . . . . . . . . . . . . . . . . 154
9.6
Functional groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155
9.7
The Hydrocarbons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155
9.7.1
The Alkanes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158
9.7.2
Naming the alkanes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159
9.7.3
Properties of the alkanes . . . . . . . . . . . . . . . . . . . . . . . . . . 163
9.7.4
Reactions of the alkanes . . . . . . . . . . . . . . . . . . . . . . . . . . 163
9.7.5
The alkenes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166
9.7.6
Naming the alkenes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166
9.7.7
The properties of the alkenes . . . . . . . . . . . . . . . . . . . . . . . . 169
9.7.8
Reactions of the alkenes
9.7.9
The Alkynes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171
. . . . . . . . . . . . . . . . . . . . . . . . . . 169
9.7.10 Naming the alkynes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171
9.8
9.9
The Alcohols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172
9.8.1
Naming the alcohols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173
9.8.2
Physical and chemical properties of the alcohols . . . . . . . . . . . . . . 175
Carboxylic Acids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176
9.9.1
Physical Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177
9.9.2
Derivatives of carboxylic acids: The esters . . . . . . . . . . . . . . . . . 178
9.10 The Amino Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178
9.11 The Carbonyl Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178
9.12 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179
viii
CONTENTS
CONTENTS
10 Organic Macromolecules - Grade 12
185
10.1 Polymers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185
10.2 How do polymers form? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186
10.2.1 Addition polymerisation . . . . . . . . . . . . . . . . . . . . . . . . . . . 186
10.2.2 Condensation polymerisation . . . . . . . . . . . . . . . . . . . . . . . . 188
10.3 The chemical properties of polymers . . . . . . . . . . . . . . . . . . . . . . . . 190
10.4 Types of polymers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191
10.5 Plastics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191
10.5.1 The uses of plastics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192
10.5.2 Thermoplastics and thermosetting plastics . . . . . . . . . . . . . . . . . 194
10.5.3 Plastics and the environment . . . . . . . . . . . . . . . . . . . . . . . . 195
10.6 Biological Macromolecules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196
10.6.1 Carbohydrates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197
10.6.2 Proteins . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199
10.6.3 Nucleic Acids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202
10.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204
III
Chemical Change
209
11 Physical and Chemical Change - Grade 10
211
11.1 Physical changes in matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211
11.2 Chemical Changes in Matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212
11.2.1 Decomposition reactions . . . . . . . . . . . . . . . . . . . . . . . . . . 213
11.2.2 Synthesis reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214
11.3 Energy changes in chemical reactions . . . . . . . . . . . . . . . . . . . . . . . . 217
11.4 Conservation of atoms and mass in reactions . . . . . . . . . . . . . . . . . . . . 217
11.5 Law of constant composition . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219
11.6 Volume relationships in gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219
11.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220
12 Representing Chemical Change - Grade 10
223
12.1 Chemical symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223
12.2 Writing chemical formulae
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224
12.3 Balancing chemical equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224
12.3.1 The law of conservation of mass . . . . . . . . . . . . . . . . . . . . . . 224
12.3.2 Steps to balance a chemical equation
. . . . . . . . . . . . . . . . . . . 226
12.4 State symbols and other information . . . . . . . . . . . . . . . . . . . . . . . . 230
12.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232
13 Quantitative Aspects of Chemical Change - Grade 11
233
13.1 The Mole . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233
13.2 Molar Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235
13.3 An equation to calculate moles and mass in chemical reactions . . . . . . . . . . 237
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CONTENTS
13.4 Molecules and compounds
CONTENTS
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239
13.5 The Composition of Substances . . . . . . . . . . . . . . . . . . . . . . . . . . . 242
13.6 Molar Volumes of Gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246
13.7 Molar concentrations in liquids . . . . . . . . . . . . . . . . . . . . . . . . . . . 247
13.8 Stoichiometric calculations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249
13.9 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252
14 Energy Changes In Chemical Reactions - Grade 11
255
14.1 What causes the energy changes in chemical reactions? . . . . . . . . . . . . . . 255
14.2 Exothermic and endothermic reactions . . . . . . . . . . . . . . . . . . . . . . . 255
14.3 The heat of reaction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257
14.4 Examples of endothermic and exothermic reactions . . . . . . . . . . . . . . . . 259
14.5 Spontaneous and non-spontaneous reactions . . . . . . . . . . . . . . . . . . . . 260
14.6 Activation energy and the activated complex . . . . . . . . . . . . . . . . . . . . 261
14.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264
15 Types of Reactions - Grade 11
267
15.1 Acid-base reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267
15.1.1 What are acids and bases? . . . . . . . . . . . . . . . . . . . . . . . . . 267
15.1.2 Defining acids and bases . . . . . . . . . . . . . . . . . . . . . . . . . . 267
15.1.3 Conjugate acid-base pairs . . . . . . . . . . . . . . . . . . . . . . . . . . 269
15.1.4 Acid-base reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270
15.1.5 Acid-carbonate reactions . . . . . . . . . . . . . . . . . . . . . . . . . . 274
15.2 Redox reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276
15.2.1 Oxidation and reduction
. . . . . . . . . . . . . . . . . . . . . . . . . . 277
15.2.2 Redox reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278
15.3 Addition, substitution and elimination reactions . . . . . . . . . . . . . . . . . . 280
15.3.1 Addition reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 280
15.3.2 Elimination reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281
15.3.3 Substitution reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282
15.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283
16 Reaction Rates - Grade 12
287
16.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287
16.2 Factors affecting reaction rates . . . . . . . . . . . . . . . . . . . . . . . . . . . 289
16.3 Reaction rates and collision theory . . . . . . . . . . . . . . . . . . . . . . . . . 293
16.4 Measuring Rates of Reaction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295
16.5 Mechanism of reaction and catalysis . . . . . . . . . . . . . . . . . . . . . . . . 297
16.6 Chemical equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 300
16.6.1 Open and closed systems . . . . . . . . . . . . . . . . . . . . . . . . . . 302
16.6.2 Reversible reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302
16.6.3 Chemical equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303
16.7 The equilibrium constant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304
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CONTENTS
CONTENTS
16.7.1 Calculating the equilibrium constant . . . . . . . . . . . . . . . . . . . . 305
16.7.2 The meaning of kc values . . . . . . . . . . . . . . . . . . . . . . . . . . 306
16.8 Le Chatelier’s principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 310
16.8.1 The effect of concentration on equilibrium . . . . . . . . . . . . . . . . . 310
16.8.2 The effect of temperature on equilibrium . . . . . . . . . . . . . . . . . . 310
16.8.3 The effect of pressure on equilibrium . . . . . . . . . . . . . . . . . . . . 312
16.9 Industrial applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315
16.10Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 316
17 Electrochemical Reactions - Grade 12
319
17.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319
17.2 The Galvanic Cell . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 320
17.2.1 Half-cell reactions in the Zn-Cu cell . . . . . . . . . . . . . . . . . . . . 321
17.2.2 Components of the Zn-Cu cell . . . . . . . . . . . . . . . . . . . . . . . 322
17.2.3 The Galvanic cell . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323
17.2.4 Uses and applications of the galvanic cell . . . . . . . . . . . . . . . . . 324
17.3 The Electrolytic cell . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325
17.3.1 The electrolysis of copper sulphate . . . . . . . . . . . . . . . . . . . . . 326
17.3.2 The electrolysis of water . . . . . . . . . . . . . . . . . . . . . . . . . . 327
17.3.3 A comparison of galvanic and electrolytic cells . . . . . . . . . . . . . . . 328
17.4 Standard Electrode Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . 328
17.4.1 The different reactivities of metals . . . . . . . . . . . . . . . . . . . . . 329
17.4.2 Equilibrium reactions in half cells . . . . . . . . . . . . . . . . . . . . . . 329
17.4.3 Measuring electrode potential . . . . . . . . . . . . . . . . . . . . . . . . 330
17.4.4 The standard hydrogen electrode . . . . . . . . . . . . . . . . . . . . . . 330
17.4.5 Standard electrode potentials . . . . . . . . . . . . . . . . . . . . . . . . 333
17.4.6 Combining half cells . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337
17.4.7 Uses of standard electrode potential . . . . . . . . . . . . . . . . . . . . 338
17.5 Balancing redox reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342
17.6 Applications of electrochemistry . . . . . . . . . . . . . . . . . . . . . . . . . . 347
17.6.1 Electroplating . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347
17.6.2 The production of chlorine . . . . . . . . . . . . . . . . . . . . . . . . . 348
17.6.3 Extraction of aluminium
. . . . . . . . . . . . . . . . . . . . . . . . . . 349
17.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 349
IV
Chemical Systems
353
18 The Water Cycle - Grade 10
355
18.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355
18.2 The importance of water . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355
18.3 The movement of water through the water cycle . . . . . . . . . . . . . . . . . . 356
18.4 The microscopic structure of water . . . . . . . . . . . . . . . . . . . . . . . . . 359
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18.4.1 The polar nature of water . . . . . . . . . . . . . . . . . . . . . . . . . . 359
18.4.2 Hydrogen bonding in water molecules . . . . . . . . . . . . . . . . . . . 359
18.5 The unique properties of water . . . . . . . . . . . . . . . . . . . . . . . . . . . 360
18.6 Water conservation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363
18.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 366
19 Global Cycles: The Nitrogen Cycle - Grade 10
369
19.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 369
19.2 Nitrogen fixation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 369
19.3 Nitrification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371
19.4 Denitrification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372
19.5 Human Influences on the Nitrogen Cycle . . . . . . . . . . . . . . . . . . . . . . 372
19.6 The industrial fixation of nitrogen . . . . . . . . . . . . . . . . . . . . . . . . . 373
19.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374
20 The Hydrosphere - Grade 10
377
20.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 377
20.2 Interactions of the hydrosphere . . . . . . . . . . . . . . . . . . . . . . . . . . . 377
20.3 Exploring the Hydrosphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 378
20.4 The Importance of the Hydrosphere . . . . . . . . . . . . . . . . . . . . . . . . 379
20.5 Ions in aqueous solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 379
20.5.1 Dissociation in water . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380
20.5.2 Ions and water hardness . . . . . . . . . . . . . . . . . . . . . . . . . . . 382
20.5.3 The pH scale . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382
20.5.4 Acid rain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384
20.6 Electrolytes, ionisation and conductivity . . . . . . . . . . . . . . . . . . . . . . 386
20.6.1 Electrolytes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 386
20.6.2 Non-electrolytes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 387
20.6.3 Factors that affect the conductivity of water . . . . . . . . . . . . . . . . 387
20.7 Precipitation reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 389
20.8 Testing for common anions in solution . . . . . . . . . . . . . . . . . . . . . . . 391
20.8.1 Test for a chloride . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 391
20.8.2 Test for a sulphate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 391
20.8.3 Test for a carbonate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392
20.8.4 Test for bromides and iodides . . . . . . . . . . . . . . . . . . . . . . . . 392
20.9 Threats to the Hydrosphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393
20.10Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394
21 The Lithosphere - Grade 11
397
21.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 397
21.2 The chemistry of the earth’s crust . . . . . . . . . . . . . . . . . . . . . . . . . 398
21.3 A brief history of mineral use . . . . . . . . . . . . . . . . . . . . . . . . . . . . 399
21.4 Energy resources and their uses . . . . . . . . . . . . . . . . . . . . . . . . . . . 400
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CONTENTS
21.5 Mining and Mineral Processing: Gold . . . . . . . . . . . . . . . . . . . . . . . . 401
21.5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401
21.5.2 Mining the Gold . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401
21.5.3 Processing the gold ore . . . . . . . . . . . . . . . . . . . . . . . . . . . 401
21.5.4 Characteristics and uses of gold . . . . . . . . . . . . . . . . . . . . . . . 402
21.5.5 Environmental impacts of gold mining . . . . . . . . . . . . . . . . . . . 404
21.6 Mining and mineral processing: Iron . . . . . . . . . . . . . . . . . . . . . . . . 406
21.6.1 Iron mining and iron ore processing . . . . . . . . . . . . . . . . . . . . . 406
21.6.2 Types of iron . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 407
21.6.3 Iron in South Africa . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 408
21.7 Mining and mineral processing: Phosphates . . . . . . . . . . . . . . . . . . . . 409
21.7.1 Mining phosphates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409
21.7.2 Uses of phosphates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409
21.8 Energy resources and their uses: Coal . . . . . . . . . . . . . . . . . . . . . . . 411
21.8.1 The formation of coal . . . . . . . . . . . . . . . . . . . . . . . . . . . . 411
21.8.2 How coal is removed from the ground . . . . . . . . . . . . . . . . . . . 411
21.8.3 The uses of coal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412
21.8.4 Coal and the South African economy . . . . . . . . . . . . . . . . . . . . 412
21.8.5 The environmental impacts of coal mining . . . . . . . . . . . . . . . . . 413
21.9 Energy resources and their uses: Oil . . . . . . . . . . . . . . . . . . . . . . . . 414
21.9.1 How oil is formed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 414
21.9.2 Extracting oil . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 414
21.9.3 Other oil products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415
21.9.4 The environmental impacts of oil extraction and use . . . . . . . . . . . 415
21.10Alternative energy resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415
21.11Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 417
22 The Atmosphere - Grade 11
421
22.1 The composition of the atmosphere . . . . . . . . . . . . . . . . . . . . . . . . 421
22.2 The structure of the atmosphere . . . . . . . . . . . . . . . . . . . . . . . . . . 422
22.2.1 The troposphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 422
22.2.2 The stratosphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 422
22.2.3 The mesosphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 424
22.2.4 The thermosphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 424
22.3 Greenhouse gases and global warming . . . . . . . . . . . . . . . . . . . . . . . 426
22.3.1 The heating of the atmosphere . . . . . . . . . . . . . . . . . . . . . . . 426
22.3.2 The greenhouse gases and global warming . . . . . . . . . . . . . . . . . 426
22.3.3 The consequences of global warming . . . . . . . . . . . . . . . . . . . . 429
22.3.4 Taking action to combat global warming . . . . . . . . . . . . . . . . . . 430
22.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 431
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23 The Chemical Industry - Grade 12
435
23.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 435
23.2 Sasol . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 435
23.2.1 Sasol today: Technology and production . . . . . . . . . . . . . . . . . . 436
23.2.2 Sasol and the environment . . . . . . . . . . . . . . . . . . . . . . . . . 440
23.3 The Chloralkali Industry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 442
23.3.1 The Industrial Production of Chlorine and Sodium Hydroxide . . . . . . . 442
23.3.2 Soaps and Detergents . . . . . . . . . . . . . . . . . . . . . . . . . . . . 446
23.4 The Fertiliser Industry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 450
23.4.1 The value of nutrients . . . . . . . . . . . . . . . . . . . . . . . . . . . . 450
23.4.2 The Role of fertilisers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 450
23.4.3 The Industrial Production of Fertilisers . . . . . . . . . . . . . . . . . . . 451
23.4.4 Fertilisers and the Environment: Eutrophication . . . . . . . . . . . . . . 454
23.5 Electrochemistry and batteries . . . . . . . . . . . . . . . . . . . . . . . . . . . 456
23.5.1 How batteries work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 456
23.5.2 Battery capacity and energy . . . . . . . . . . . . . . . . . . . . . . . . 457
23.5.3 Lead-acid batteries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 457
23.5.4 The zinc-carbon dry cell . . . . . . . . . . . . . . . . . . . . . . . . . . . 459
23.5.5 Environmental considerations . . . . . . . . . . . . . . . . . . . . . . . . 460
23.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 461
A GNU Free Documentation License
467
xiv
Part I
Introduction
1
Part II
Matter and Materials
3
Chapter 1
Classification of Matter - Grade 10
All the objects that we see in the world around us, are made of matter. Matter makes up the
air we breathe, the ground we walk on, the food we eat and the animals and plants that live
around us. Even our own human bodies are made of matter!
Different objects can be made of different types of matter, or materials. For example, a cupboard (an object) is made of wood, nails and hinges (the materials). The properties of the
materials will affect the properties of the object. In the example of the cupboard, the strength
of the wood and metals make the cupboard strong and durable. In the same way, the raincoats
that you wear during bad weather, are made of a material that is waterproof. The electrical wires
in your home are made of metal because metals are a type of material that is able to conduct
electricity. It is very important to understand the properties of materials, so that we can use
them in our homes, in industry and in other applications. In this chapter, we will be looking at
different types of materials and their properties.
The diagram below shows one way in which matter can be classified (grouped) according to
its different properties. As you read further in this chapter, you will see that there are also
other ways of classifying materials, for example according to whether they are good electrical
conductors.
MATTER
MIXTURES
Homogeneous
PURE SUBSTANCES
Heterogeneous
Elements
Metals
Magnetic
Compounds
Non-metals
Non-magnetic
Figure 1.1: The classification of matter
1.1
Mixtures
We see mixtures all the time in our everyday lives. A stew, for example, is a mixture of different
foods such as meat and vegetables; sea water is a mixture of water, salt and other substances,
and air is a mixture of gases such as carbon dioxide, oxygen and nitrogen.
5
1.1
CHAPTER 1. CLASSIFICATION OF MATTER - GRADE 10
Definition: Mixture
A mixture is a combination of more than one substance, where these substances are not
bonded to each other.
In a mixture, the substances that make up the mixture:
• are not in a fixed ratio
Imagine, for example, that you have a 250 ml beaker of water. It doesn’t matter whether
you add 20 g, 40 g, 100 g or any other mass of sand to the water; it will still be called a
mixture of sand and water.
• keep their physical properties
In the example we used of the sand and water, neither of these substances has changed in
any way when they are mixed together. Even though the sand is in water, it still has the
same properties as when it was out of the water.
• can be separated by mechanical means
To separate something by ’mechanical means’, means that there is no chemical process
involved. In our sand and water example, it is possible to separate the mixture by simply
pouring the water through a filter. Something physical is done to the mixture, rather than
something chemical.
Some other examples of mixtures include blood (a mixture of blood cells, platelets and plasma),
steel (a mixture of iron and other materials) and the gold that is used to make jewellery. The
gold in jewellery is not pure gold but is a mixture of metals. The carat of the gold gives an idea
of how much gold is in the item.
We can group mixtures further by dividing them into those that are heterogeneous and those
that are homogeneous.
1.1.1
Heterogeneous mixtures
A heterogeneous mixture does not have a definite composition. Think of a pizza, that is a
mixture of cheese, tomato, mushrooms and peppers. Each slice will probably be slightly different
from the next because the toppings like the mushrooms and peppers are not evenly distributed.
Another example would be granite, a type of rock. Granite is made up of lots of different mineral
substances including quartz and feldspar. But these minerals are not spread evenly through the
rock and so some parts of the rock may have more quartz than others. Another example is
a mixture of oil and water. Although you may add one substance to the other, they will stay
separate in the mixture. We say that these heterogeneous mixtures are non-uniform, in other
words they are not exactly the same throughout.
Definition: Heterogeneous mixture
A heterogeneous mixture is one that is non-uniform, and where the different components
of the mixture can be seen.
1.1.2
Homogeneous mixtures
A homogeneous mixture has a definite composition, and specific properties. In a homogeneous
mixture, the different parts cannot be seen. A solution of salt dissolved in water is an example
of a homogeneous mixture. When the salt dissolves, it will spread evenly through the water so
that all parts of the solution are the same, and you can no longer see the salt as being separate
from the water. Think also of a powdered drink that you mix with water. Provided you give the
container a good shake after you have added the powder to the water, the drink will have the
same sweet taste for anyone who drinks it, it won’t matter whether they take a sip from the top
6
CHAPTER 1. CLASSIFICATION OF MATTER - GRADE 10
1.1
or from the bottom. The air we breathe is another example of a homogeneous mixture since it is
made up of different gases which are in a constant ratio, and which can’t be distinguished from
each other.
Definition: Homogeneous mixture
A homogeneous mixture is one that is uniform, and where the different components of the
mixture cannot be seen.
An alloy is a homogeneous mixture of two or more elements, at least one of which is a metal,
where the resulting material has metallic properties. Alloys are usually made to improve on the
properties of the elements that make them up. Steel for example, is much stronger than iron,
which is its main component.
1.1.3
Separating mixtures
Sometimes it is important to be able to separate a mixture. There are lots of different ways to
do this. These are some examples:
• Filtration
A piece of filter paper in a funnel can be used to separate a mixture of sand and water.
• Heating / evaporation
Sometimes, heating a solution causes the water to evaporate, leaving the other part of the
mixture behind. You can try this using a salt solution.
• Centrifugation
This is a laboratory process which uses the centrifugal force of spinning objects to separate
out the heavier substances from a mixture. This process is used to separate the cells and
plasma in blood. When the test tubes that hold the blood are spun round in the machine,
the heavier cells sink to the bottom of the test tube. Can you think of a reason why it
might be important to have a way of separating blood in this way?
• Dialysis
This is an interesting way of separating a mixture because it can be used in some important
applications. Dialysis works using a process called diffusion. Diffusion takes place when
one substance in a mixture moves from an area where it has a high concentration to an
area where its concentration is lower. This movement takes place across a semi-permeable
membrane. A semi-permeable membrane is a barrier that lets some things move across it,
but not others. This process is very important for people whose kidneys are not functioning
properly, an illness called renal failure.
teresting Normally, healthy kidneys remove waste products from the blood. When a person
Interesting
Fact
Fact
has renal failure, their kidneys cannot do this any more, and this can be lifethreatening. Using dialysis, the blood of the patient flows on one side of a
semi-permeable membrane. On the other side there will be a fluid that has no
waste products but lots of other important substances such as potassium ions
(K + ) that the person will need. Waste products from the blood diffuse from
where their concentration is high (i.e. in the person’s blood) into the ’clean’
fluid on the other side of the membrane. The potassium ions will move in the
opposite direction from the fluid into the blood. Through this process, waste
products are taken out of the blood so that the person stays healthy.
7
1.1
CHAPTER 1. CLASSIFICATION OF MATTER - GRADE 10
Activity :: Investigation : The separation of a salt solution
Aim:
To demonstrate that a homogeneous salt solution can be separated using physical
methods.
Apparatus:
glass beaker, salt, water, retort stand, bunsen burner.
Method:
1. Pour a small amount of water (about 20 ml) into a beaker.
2. Measure a teaspoon of salt and pour this into the water.
3. Stir until the salt dissolves completely. This is now called a salt solution. This
salt solution is a homogeneous mixture.
4. Place the beaker on a retort stand over a bunsen burner and heat gently. You
should increase the heat until the water almost boils.
5. Watch the beaker until all the water has evaporated. What do you see in the
beaker?
H2 O
salt
solution
water evaporates
when the solution
is heated
salt crystals
remain at the
bottom of the beaker
stand
bunsen
burner
Results:
The water evaporates from the beaker and tiny grains of salt remain at the
bottom.
Conclusion:
The sodium chloride solution, which was a homogeneous mixture of salt and
water, has been separated using heating and evaporation.
Activity :: Discussion : Separating mixtures
Work in groups of 3-4
Imagine that you have been given a container which holds a mixture of sand,
iron filings (small pieces of iron metal), salt and small stones of different sizes. Is
this a homogeneous or a heterogeneous mixture? In your group, discuss how you
would go about separating this mixture into the four materials that it contains.
8
CHAPTER 1. CLASSIFICATION OF MATTER - GRADE 10
1.2
Exercise: Mixtures
1. Which of the following subtances are mixtures?
(a)
(b)
(c)
(d)
(e)
(f)
tap water
brass (an alloy of copper and zinc)
concrete
aluminium
Coca cola
distilled water
2. In each of the examples above, say whether the mixture is homogeneous or
heterogeneous
1.2
Pure Substances: Elements and Compounds
Any material that is not a mixture, is called a pure substance. Pure substances include elements
and compounds. It is much more difficult to break down pure substances into their parts, and
complex chemical methods are needed to do this.
1.2.1
Elements
An element is a chemical substance that can’t be divided or changed into other chemical
substances by any ordinary chemical means. The smallest unit of an element is the atom.
Definition: Element
An element is a substance that cannot be broken down into other substances through
chemical means.
There are 109 known elements. Most of these are natural, but some are man-made. The
elements we know are represented in the Periodic Table of the Elements, where each element
is abbreviated to a chemical symbol. Examples of elements are magnesium (Mg), hydrogen (H),
oxygen (O) and carbon (C). On the Periodic Table you will notice that some of the abbreviations
do not seem to match the elements they represent. The element iron, for example, has the
chemical formula Fe. This is because the elements were originally given Latin names. Iron has
the abbreviation Fe because its Latin name is ’ferrum’. In the same way, sodium’s Latin name
is ’natrium’ (Na) and gold’s is ’aurum’ (Au).
1.2.2
Compounds
A compound is a chemical substance that forms when two or more elements combine in a fixed
ratio. Water (H2 O), for example, is a compound that is made up of two hydrogen atoms for
every one oxygen atom. Sodium chloride (NaCl) is a compound made up of one sodium atom
for every chlorine atom. An important characteristic of a compound is that it has a chemical
formula, which describes the ratio in which the atoms of each element in the compound occur.
Definition: Compound
A substance made up of two or more elements that are joined together in a fixed ratio.
Diagram 1.2 might help you to understand the difference between the terms element, mixture
and compound. Iron (Fe) and sulfur (S) are two elements. When they are added together, they
9
CHAPTER 1. CLASSIFICATION OF MATTER - GRADE 10
Fe
A mixture of iron and sulfur
S
Fe S
S
An atom
of the element iron
(Fe)
S
S
S
Fe
Fe
Fe
Fe
Fe
Fe
S
S
Fe
S
S
Fe
Fe
S
An atom
of the element sulfur (S)
Fe S
1.3
The compound iron sulfide
(FeS)
Figure 1.2: Understanding the difference between a mixture and a compound
form a mixture or iron and sulfur. The iron and sulfur are not joined together. However, if
the mixture is heated, a new compound is formed, which is called iron sulfide (FeS). In this
compound, the iron and sulfur are joined to each other in a ratio of 1:1. In other words, one
atom of iron is joined to one atom of sulfur in the compound iron sulfide.
Exercise: Elements, mixtures and compounds
1. In the following table, tick whether each of the substances listed is a mixture
or a pure substance. If it is a mixture, also say whether it is a homogeneous or
heterogeneous mixture.
Substance
Mixture or pure
Homogeneous
heterogeneous
mixture
or
fizzy colddrink
steel
oxygen
iron filings
smoke
limestone (CaCO3 )
2. In each of the following cases, say whether the substance is an element, a
mixture or a compound.
(a)
(b)
(c)
(d)
(e)
1.3
Cu
iron and sulfur
Al
H2 SO4
SO3
Giving names and formulae to substances
It is easy to describe elements and mixtures. But how are compounds named? In the example
of iron sulfide that was used earlier, which element is named first, and which ’ending’ is given
to the compound name (in this case, the ending is -ide)?
The following are some guidelines for naming compounds:
10
CHAPTER 1. CLASSIFICATION OF MATTER - GRADE 10
1.3
1. The compound name will always include the names of the elements that are part of it.
• A compound of iron (Fe) and sulfur (S) is iron sulf ide (FeS)
• A compound of potassium (K) and bromine (S) is potassium bromide (KBr)
• A compound of sodium (Na) and chlorine (Cl) is sodium chlor ide (NaCl)
2. In a compound, the element that is to the left and lower down on the Periodic Table,
is used first when naming the compound. In the example of NaCl, sodium is a group 1
element on the left hand side of the table, while chlorine is in group 7 on the right of the
table. Sodium therefore comes first in the compound name. The same is true for FeS and
KBr.
3. The symbols of the elements can be used to represent compounds e.g. FeS, NaCl and
KBr. These are called chemical formulae. In these three examples, the ratio of the
elements in each compound is 1:1. So, for FeS, there is one atom of iron for every atom
of sulfur in the compound.
4. A compound may contain compound ions. Some of the more common compound ions
and their names are shown below.
Name of compound ion
Carbonate
sulphate
Hydroxide
Ammonium
Nitrate
Hydrogen carbonate
Phosphate
Chlorate
Cyanide
Chromate
Permanganate
formula
CO3 2−
SO4 2−
OH−
NH4 +
NO3 −
HCO3 −
PO4 3−
ClO3 −
CN−
CrO4 2−
MnO4 −
5. When there are only two elements in the compound, the compound is often given a suffix
(ending) of -ide. You would have seen this in some of the examples we have used so far.
When a non-metal is combined with oxygen to form a negative ion (anion) which then
combines with a positive ion (cation) from hydrogen or a metal, then the suffix of the
name will be ...ate or ...ite. NO−
3 for example, is a negative ion, which may combine with
a cation such as hydrogen (HNO3 ) or a metal like potassium (KNO3 ). The NO−
3 anion
has the name nitrate. SO3 in a formula is sulphite, e.g. sodium sulphite (Na2 SO3 ). SO4
is sulphate and PO4 is phosphate.
6. Prefixes can be used to describe the ratio of the elements that are in the compound. You
should know the following prefixes: ’mono’ (one), ’di’ (two) and ’tri’ (three).
• CO (carbon monoxide) - There is one atom of oxygen for every one atom of carbon
• N O2 (nitrogen dioxide) - There are two atoms of oxygen for every one atom of
nitrogen
• SO3 (sulfur trioxide) - There are three atoms of oxygen for every one atom of sulfur
Important:
When numbers are written as ’subscripts’ in compounds (i.e. they are written below the
element symbol), this tells us how many atoms of that element there are in relation to other
elements in the compound. For example in nitrogen dioxide (NO2 ) there are two oxygen
atoms for every one atom of nitrogen. In sulfur trioxide (SO3 ), there are three oxygen atoms
for every one atom of sulfur in the compound. Later, when we start looking at chemical
equations, you will notice that sometimes there are numbers before the compound name.
For example, 2H2 O means that there are two molecules of water, and that in each molecule
there are two hydrogen atoms for every one oxygen atom.
11
1.3
CHAPTER 1. CLASSIFICATION OF MATTER - GRADE 10
Exercise: Naming compounds
1. The formula for calcium carbonate is CaCO3 .
(a) Is calcium carbonate a mixture or a compound? Give a reason for your
answer.
(b) What is the ratio of Ca:C:O atoms in the formula?
2. Give the name of each of the following substances.
(a)
(b)
(c)
(d)
(e)
(f)
KBr
HCl
KMnO4
NO2
NH4 OH
Na2 SO4
3. Give the chemical formula for each of the following compounds.
(a)
(b)
(c)
(d)
(e)
potassium nitrate
sodium iodide
barium sulphate
nitrogen dioxide
sodium monosulphate
4. Refer to the diagram below, showing sodium chloride and water, and then
answer the questions that follow.
(a)
(b)
(c)
(d)
What is the chemical formula for water?
What is the chemical formula for sodium chloride?
Label the water and sodium chloride in the diagram.
Which of the following statements most accurately describes the picture?
i. The picture shows a mixture of an element and a compound
ii. The picture shows a mixture of two compounds
iii. The picture shows two compounds that have been chemically bonded
to each other
5. What is the formula of this molecule?
H
H
H
A
B
C
D
C
C
H
H
C6 H2 O
C2 H6 O
2C6HO
2 CH6 O
12
O
H
CHAPTER 1. CLASSIFICATION OF MATTER - GRADE 10
1.4
1.4
Metals, Semi-metals and Non-metals
The elements in the Periodic Table can also be divided according to whether they are metals,
semi-metals or non-metals. On the right hand side of the Periodic Table is a dark ’zigzag’ line.
This line separates all the elements that are metals from those that are non-metals. Metals are
found on the left of the line, and non-metals are those on the right. Metals, semi-metals and
non-metals all have their own specific properties.
1.4.1
Metals
Examples of metals include copper (Cu), zinc (Zn), gold (Au) and silver (Ag). On the Periodic
Table, the metals are on the left of the zig-zag line. There are a large number of elements that
are metals. The following are some of the properties of metals:
• Thermal conductors
Metals are good conductors of heat and are therefore used in cooking utensils such as pots
and pans.
• Electrical conductors
Metals are good conductors of electricity, and are therefore used in electrical conducting
wires.
• Shiny metallic lustre
Metals have a characteristic shiny appearance and are often used to make jewellery.
• Malleable
This means that they can be bent into shape without breaking.
• Ductile
Metals can stretched into thin wires such as copper, which can then be used to conduct
electricity.
• Melting point
Metals usually have a high melting point and can therefore be used to make cooking pots
and other equipment that needs to become very hot, without being damaged.
You can see how the properties of metals make them very useful in certain applications.
Activity :: Group Work : Looking at metals
1. Collect a number of metal items from your home or school. Some examples
are listed below:
•
•
•
•
•
•
hammer
electrical wiring
cooking pots
jewellery
burglar bars
coins
2. In groups of 3-4, combine your collection of metal objects.
3. What is the function of each of these objects?
4. Discuss why you think metal was used to make each object. You should consider
the properties of metals when you answer this question.
13
1.5
CHAPTER 1. CLASSIFICATION OF MATTER - GRADE 10
1.4.2
Non-metals
In contrast to metals, non-metals are poor thermal conductors, good electrical insulators (meaning that they do not conduct electrical charge) and are neither malleable nor ductile. The
non-metals are found on the right hand side of the Periodic Table, and include elements such as
sulfur (S), phosphorus (P), nitrogen (N) and oxygen (O).
1.4.3
Semi-metals
Semi-metals have mostly non-metallic properties. One of their distinguishing characteristics is
that their conductivity increases as their temperature increases. This is the opposite of what
happens in metals. The semi-metals include elements such as silicon (Si) and germanium (Ge).
Notice where these elements are positioned in the Periodic Table.
1.5
Electrical conductors, semi-conductors and insulators
An electrical conductor is a substance that allows an electrical current to pass through it.
Many electrical conductors are metals, but non-metals can also be good conductors. Copper is
one of the best electrical conductors, and this is why it is used to make conducting wire. In
reality, silver actually has an even higher electrical conductivity than copper, but because silver
is so expensive, it is not practical to use it for electrical wiring because such large amounts are
needed. In the overhead power lines that we see above us, aluminium is used. The aluminium
usually surrounds a steel core which adds tensile strength to the metal so that it doesn’t break
when it is stretched across distances. Occasionally gold is used to make wire, not because it is
a particularly good conductor, but because it is very resistant to surface corrosion. Corrosion is
when a material starts to deteriorate at the surface because of its reactions with the surroundings, for example oxygen and water in the air.
An insulator is a non-conducting material that does not carry any charge. Examples of insulators
would be plastic and wood. Do you understand now why electrical wires are normally covered
with plastic insulation? Semi-conductors behave like insulators when they are cold, and like
conductors when they are hot. The elements silicon and germanium are examples of semiconductors.
Definition: Conductors and insulators
A conductor allows the easy movement or flow of something such as heat or electrical charge
through it. Insulators are the opposite to conductors because they inhibit or reduce the flow
of heat, electrical charge, sound etc through them.
Activity :: Experiment : Electrical conductivity
Aim:
To investigate the electrical conductivity of a number of substances
Apparatus:
• two or three cells
• light bulb
• crocodile clips
• wire leads
• a selection of test substances (e.g. a piece of plastic, aluminium can, metal
pencil sharpener, metal magnet, wood, chalk).
14
CHAPTER 1. CLASSIFICATION OF MATTER - GRADE 10
1.6
light bulb
battery
test substance
X
crocodile clip
Method:
1. Set up the circuit as shown above, so that the test substance is held between
the two crocodile clips. The wire leads should be connected to the cells and
the light bulb should also be connected into the circuit.
2. Place the test substances one by one between the crocodile clips and see what
happens to the light bulb.
Results:
Record your results in the table below:
Test substance
Metal/non-metal
Does
glow?
bulb
Conductor or
insulator
Conclusions:
In the substances that were tested, the metals were able to conduct electricity
and the non-metals were not. Metals are good electrical conductors and non-metals
are not.
1.6
Thermal Conductors and Insulators
A thermal conductor is a material that allows energy in the form of heat, to be transferred
within the material, without any movement of the material itself. An easy way to understand
this concept is through a simple demonstration.
Activity :: Demonstration : Thermal conductivity
Aim:
To demonstrate the ability of different substances to conduct heat.
Apparatus:
15
1.6
CHAPTER 1. CLASSIFICATION OF MATTER - GRADE 10
You will need two cups (made from the same material e.g. plastic); a metal
spoon and a plastic spoon.
Method:
• Pour boiling water into the two cups so that they are about half full.
• At the same time, place a metal spoon into one cup and a plastic spoon in the
other.
• Note which spoon heats up more quickly
Results:
The metal spoon heats up more quickly than the plastic spoon. In other words,
the metal conducts heat well, but the plastic does not.
Conclusion:
Metal is a good thermal conductor, while plastic is a poor thermal conductor.
This explains why cooking pots are metal, but their handles are often plastic or
wooden. The pot itself must be metal so that heat from the cooking surface can
heat up the pot to cook the food inside it, but the handle is made from a poor
thermal conductor so that the heat does not burn the hand of the person who is
cooking.
An insulator is a material that does not allow a transfer of electricity or energy. Materials that
are poor thermal conductors can also be described as being good insulators.
teresting Water is a better thermal conductor than air and conducts heat away from the
Interesting
Fact
Fact
body about 20 times more efficiently than air. A person who is not wearing
a wetsuit, will lose heat very quickly to the water around them and can be
vulnerable to hypothermia. Wetsuits help to preserve body heat by trapping a
layer of water against the skin. This water is then warmed by body heat and acts
as an insulator. Wetsuits are made out of closed-cell, foam neoprene. Neoprene
is a synthetic rubber that contains small bubbles of nitrogen gas when made for
use as wetsuit material. Nitrogen gas has very low thermal conductivity, so it
does not allow heat from the body (or the water trapped between the body and
the wetsuit) to be lost to the water outside of the wetsuit. In this way a person
in a wetsuit is able to keep their body temperature much higher than they would
otherwise.
Activity :: Investigation : A closer look at thermal conductivity
Look at the table below, which shows the thermal conductivity of a number
of different materials, and then answer the questions that follow. The higher the
number in the second column, the better the material is at conducting heat (i.e. it is
a good thermal conductor). Remember that a material that conducts heat efficiently,
will also lose heat more quickly than an insulating material.
16
CHAPTER 1. CLASSIFICATION OF MATTER - GRADE 10
Material
Silver
Stainless steel
Standard glass
Concrete
Red brick
Water
Snow
Wood
Polystyrene
Air
1.7
Thermal Conductivity (W/m/K)
429
16
1.05
0.9 - 2
0.69
0.58
0.5 - 0.25
0.04 - 0.12
0.03
0.024
Use this information to answer the following questions:
1. Name two materials that are good thermal conductors.
2. Name two materials that are good insulators.
3. Explain why:
(a) cooler boxes are often made of polystyrene
(b) homes that are made from wood need less internal heating during the
winter months.
(c) igloos (homes made from snow) are so good at maintaining warm temperatures, even in freezing conditions.
teresting It is a known fact that well-insulated buildings need less energy for heating than
Interesting
Fact
Fact
do buildings that have no insulation. Two building materials that are being used
more and more worldwide, are mineral wool and polystyrene. Mineral wool
is a good insulator because it holds air still in the matrix of the wool so that
heat is not lost. Since air is a poor conductor and a good insulator, this helps
to keep energy within the building. Polystyrene is also a good insulator and is
able to keep cool things cool and hot things hot! It has the added advantage of
being resistant to moisture, mould and mildew.
Remember that concepts such as conductivity and insulation are not only relevant in the building,
industrial and home environments. Think for example of the layer of blubber or fat that we find
in animals. In very cold environments, fat and blubber not only provide protection, but also act
as an insulator to help the animal to keep its body temperature at the right level. This is known
as thermoregulation.
1.7
Magnetic and Non-magnetic Materials
We have now looked at a number of ways in which matter can be grouped, such as into metals,
semi-metals and non-metals; electrical conductors and insulators, and thermal conductors and
insulators. One way in which we can further group metals, is to divide them into those that are
magnetic and those that are non-magnetic.
Definition: Magnetism
Magnetism is one of the phenomena by which materials exert attractive or repulsive forces
on other materials.
17
1.8
CHAPTER 1. CLASSIFICATION OF MATTER - GRADE 10
A metal is said to be ferromagnetic if it can be magnetised (i.e. made into a magnet). If you
hold a magnet very close to a metal object, it may happen that its own electrical field will be
induced and the object becomes magnetic. Some metals keep their magnetism for longer than
others. Look at iron and steel for example. Iron loses its magnetism quite quickly if it is taken
away from the magnet. Steel on the other hand will stay magnetic for a longer time. Steel is
often used to make permanent magnets that can be used for a variety of purposes.
Magnets are used to sort the metals in a scrap yard, in compasses to find direction, in the magnetic strips of video tapes and ATM cards where information must be stored, in computers and
TV’s, as well as in generators and electric motors.
Activity :: Investigation : Magnetism
You can test whether an object is magnetic or not by holding another magnet
close to it. If the object is attracted to the magnet, then it too is magnetic.
Find some objects in your classroom or your home and test whether they are
magnetic or not. Then complete the table below:
Object
Magnetic
magnetic
or
non-
Activity :: Group Discussion : Properties of materials
In groups of 4-5, discuss how our knowledge of the properties of materials has
allowed society to:
• develop advanced computer technology
• provide homes with electricity
• find ways to conserve energy
1.8
Summary
• All the objects and substances that we see in the world are made of matter.
• This matter can be classified according to whether it is a mixture or a pure substance.
• A mixture is a combination of one or more substances that are not chemically bonded
to each other. Examples of mixtures are air (a mixture of different gases) and blood (a
mixture of cells, platelets and plasma).
• The main characteristics of mixtures are that the substances that make them up are not
in a fixed ratio, they keep their individual properties and they can be separated from each
other using mechanical means.
18
CHAPTER 1. CLASSIFICATION OF MATTER - GRADE 10
1.8
• A heterogeneous mixture is non-uniform and the different parts of the mixture can be
seen. An example would be a mixture of sand and salt.
• A homogeneous mixture is uniform, and the different components of the mixture can’t
be seen. An example would be a salt solution. A salt solution is a mixture of salt and
water. The salt dissolves in the water, meaning that you can’t see the individual salt
particles. They are interspersed between the water molecules. Another example is a metal
alloy such as steel.
• Mixtures can be separated using a number of methods such as filtration, heating, evaporation, centrifugation and dialysis.
• Pure substances can be further divided into elements and compounds.
• An element is a substance that can’t be broken down into simpler substances through
chemical means.
• All the elements are recorded in the Periodic Table of the Elements. Each element has
its own chemical symbol. Examples are iron (Fe), sulfur (S), calcium (Ca), magnesium
(Mg) and fluorine (F).
• A compound is a substance that is made up of two or more elements that are chemically
bonded to each other in a fixed ratio. Examples of compounds are sodium chloride (NaCl),
iron sulfide (FeS), calcium carbonate (CaCO3 ) and water (H2 O).
• When naming compounds and writing their chemical formula, it is important to know
the elements that are in the compound, how many atoms of each of these elements will
combine in the compound and where the elements are in the Periodic Table. A number of
rules can then be followed to name the compound.
• Another way of classifying matter is into metals (e.g. iron, gold, copper), semi-metals
(e.g. silicon and germanium) and non-metals (e.g. sulfur, phosphorus and nitrogen).
• Metals are good electrical and thermal conductors, they have a shiny lustre, they are
malleable and ductile, and they have a high melting point. These properties make metals
very useful in electrical wires, cooking utensils, jewellery and many other applications.
• A further way of classifying matter is into electrical conductors, semi-conductors and
insulators.
• An electrical conductor allows an electrical current to pass through it. Most metals are
good electrical conductors.
• An electrical insulator is not able to carry an electrical current. Examples are plastic,
wood, cotton material and ceramic.
• Materials may also be classified as thermal conductors or thermal insulators depending
on whether or not they are able to conduct heat.
• Materials may also be either magnetic or non-magnetic.
Exercise: Summary
1. For each of the following multiple choice questions, choose one correct answer
from the list provided.
A Which of the following can be classified as a mixture:
i. sugar
ii. table salt
iii. air
iv. Iron
B An element can be defined as:
19
1.8
CHAPTER 1. CLASSIFICATION OF MATTER - GRADE 10
i. A substance that cannot be separated into two or more substances by
ordinary chemical (or physical) means
ii. A substance with constant composition
iii. A substance that contains two or more substances, in definite proportion by weight
iv. A uniform substance
2. Classify each of the following substances as an element, a compound, a solution(homogeneous mixture), or a heterogeneous mixture: salt, pure water, soil,
salt water, pure air, carbon dioxide, gold and bronze
3. Look at the table below. In the first column (A) is a list of substances. In the
second column (B) is a description of the group that each of these substances
belongs in. Match up the substance in Column A with the description in
Column B.
Column A
iron
H2 S
sugar solution
sand and stones
steel
Column B
a compound containing 2 elements
a heterogeneous mixture
a metal alloy
an element
a homogeneous mixture
4. You are given a test tube that contains a mixture of iron filings and sulfur. You
are asked to weigh the amount of iron in the sample.
a Suggest one method that you could use to separate the iron filings from
the sulfur.
b What property of metals allows you to do this?
5. Given the following descriptions, write the chemical formula for each of the
following substances:
a silver metal
b a compound that contains only potassium and bromine
c a gas that contains the elements carbon and oxygen in a ratio of 1:2
6. Give the names of each of the following compounds:
a NaBr
b BaSO4
c SO2
7. For each of the following materials, say what properties of the material make
it important in carrying out its particular function.
a
b
c
d
e
f
tar on roads
iron burglar bars
plastic furniture
metal jewellery
clay for building
cotton clothing
20
Chapter 2
What are the objects around us
made of? - Grade 10
2.1
Introduction: The atom as the building block of matter
We have now seen that different materials have different properties. Some materials are metals
and some are non-metals; some are electrical or thermal conductors, while others are not. Depending on the properties of these materials, they can be used in lots of useful applications. But
what is it exactly that makes up these materials? In other words, if we were to break down a
material into the parts that make it up, what would we find? And how is it that a material’s
microscopic structure is able to give it all these different properties?
The answer lies in the smallest building block of matter: the atom. It is the type of atoms, and
the way in which they are arranged in a material, that affects the properties of that substance.
It is not often that substances are found in atomic form. Normally, atoms are bonded to other
atoms to form compounds or molecules. It is only in the noble gases (e.g. helium, neon and
argon) that atoms are found individually and are not bonded to other atoms. We will look at
the reasons for this in a later chapter.
2.2
Molecules
Definition: Molecule
A molecule is a group of two or more atoms that are attracted to each other by relatively
strong forces or bonds
Almost everything around us is made up of molecules. Water is made up of molecules, each of
which has two hydrogen atoms joined to one oxygen atom. Oxygen is a molecule that is made
up of two oxygen atoms that are joined to one another. Even the food that we eat is made
up of molecules that contain atoms of elements such as carbon, hydrogen and oxygen that are
joined to one another in different ways. All of these are known as small molecules because
there are only a few atoms in each molecule. Giant molecules are those where there may be
millions of atoms per molecule. Examples of giant molecules are diamonds, which are made up
of millions of carbon atoms bonded to each other, and metals, which are made up of millions of
metal atoms bonded to each other.
2.2.1
Representing molecules
The structure of a molecule can be shown in many different ways. Sometimes it is easiest to
show what a molecule looks like by using different types of diagrams, but at other times, we
may decide to simply represent a molecule using its chemical formula or its written name.
21
2.2
CHAPTER 2. WHAT ARE THE OBJECTS AROUND US MADE OF? - GRADE 10
1. Using formulae to show the structure of a molecule
A chemical formula is an abbreviated (shortened) way of describing a molecule, or some
other chemical substance. In chapter 1, we saw how chemical compounds can be represented using element symbols from the Periodic Table. A chemical formula can also tell
us the number of atoms of each element that are in a molecule, and their ratio in that
molecule.
For example, the chemical formula for a molecule of carbon dioxide is:
CO2
The formula above is called the molecular formula of that compound. The formula tells
us that in one molecule of carbon dioxide, there is one atom of carbon and two atoms of
oxygen. The ratio of carbon atoms to oxygen atoms is 1:2.
Definition: Molecular formula
A concise way of expressing information about the atoms that make up a particular chemical
compound. The molecular formula gives the exact number of each type of atom in the
molecule.
A molecule of glucose has the molecular formula:
C6 H12 O6
In each glucose molecule, there are six carbon atoms, twelve hydrogen atoms and six oxygen atoms. The ratio of carbon:hydrogen:oxygen is 6:12:6. We can simplify this ratio to
write 1:2:1, or if we were to use the element symbols, the formula would be written as
CH2 O. This is called the empirical formula of the molecule.
Definition: Empirical formula
This is a way of expressing the relative number of each type of atom in a chemical compound.
In most cases, the empirical formula does not show the exact number of atoms, but rather
the simplest ratio of the atoms in the compound.
The empirical formula is useful when we want to write the formula for a giant molecule.
Since giant molecules may consist of millions of atoms, it is impossible to say exactly how
many atoms are in each molecule. It makes sense then to represent these molecules using
their empirical formula. So, in the case of a metal such as copper, we would simply write
Cu, or if we were to represent a molecule of sodium chloride, we would simply write NaCl.
Chemical formulae therefore tell us something about the types of atoms that are in a
molecule and the ratio in which these atoms occur in the molecule, but they don’t give us
any idea of what the molecule actually looks like, in other words its shape. Another useful
way of representing molecules is to use diagrams.
Another type of formula that can be used to describe a molecule is its structural formula.
A structural formula uses a graphical representation to show a molecule’s structure (figure
2.1).
2. Using diagrams to show the structure of a molecule
Diagrams of molecules are very useful because they give us an idea of the space that is
occupied by the molecule, and they also help us to picture how the atoms are arranged in
the molecule. There are two types of diagrams that are commonly used:
22
CHAPTER 2. WHAT ARE THE OBJECTS AROUND US MADE OF? - GRADE 10
CH3
(a) C4 H10
(b) C2 H5
2.2
CH
CH3
CH3
Figure 2.1: Diagram showing (a) the molecular, (b) the empirical and (c) the structural formula
of isobutane
• Ball and stick models
This is a 3-dimensional molecular model that uses ’balls’ to represent atoms and
’sticks’ to represent the bonds between them. The centres of the atoms (the balls)
are connected by straight lines which represent the bonds between them. A simplified
example is shown in figure 2.2.
oxygen atom
hydrogen atom
Figure 2.2: A ball and stick model of a water molecule
• Space-filling model
This is also a 3-dimensional molecular model. The atoms are represented by multicoloured spheres. Space-filling models of water and ammonia are shown in figures
2.3 and 2.4.
Figures 2.3 and 2.4 are some examples of simple molecules that are represented in different ways.
oxygen atom
O
hydrogen atoms
H
H
Figure 2.3: A space-filling model and structural formula of a water molecule. Each molecule
is made up of two hydrogen atoms that are attached to one oxygen atom. This is a simple
molecule.
Figure 2.5 shows the bonds between the carbon atoms in diamond, which is a giant
molecule. Each carbon atom is joined to four others, and this pattern repeats itself until
a complex lattice structure is formed. Each black ball in the diagram represents a carbon
atom, and each line represents the bond between two carbon atoms.
teresting Diamonds are most often thought of in terms of their use in the jewellery industry.
Interesting
Fact
Fact
However, about 80% of mined diamonds are unsuitable for use as gemstones and
are therefore used in industry because of their strength and hardness. These
23
2.2
CHAPTER 2. WHAT ARE THE OBJECTS AROUND US MADE OF? - GRADE 10
nitrogen atom
hydrogen atom
N
H
H
H
Figure 2.4: A space-filling model and structural formula of a molecule of ammonia. Each
molecule is made up of one nitrogen atom and three hydrogen atoms. This is a simple molecule.
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
Figure 2.5: Diagrams showing the microscopic structure of diamond. The diagram on the left
shows part of a diamond lattice, made up of numerous carbon atoms. The diagram on the right
shows how each carbon atom in the lattice is joined to four others. This forms the basis of the
lattice structure. Diamond is a giant molecule.
properties of diamonds are due to the strong covalent bonds betwene the carbon
atoms in diamond. The most common uses for diamonds in industry are in
cutting, drilling, grinding, and polishing.
Exercise: Atoms and molecules
1. In each of the following, say whether the chemical substance is made up of
single atoms, simple molecules or giant molecules.
(a) ammonia gas (NH3 )
(b) zinc metal (Zn)
(c) graphite (C)
(d) nitric acid (HNO3 )
(e) neon gas (Ne2 )
2. Refer to the diagram below and then answer the questions that follow:
24
CHAPTER 2. WHAT ARE THE OBJECTS AROUND US MADE OF? - GRADE 10
O
C
2.3
O
(a) Identify the molecule.
(b) Write the molecular formula for the molecule.
(c) Is the molecule a simple or giant molecule?
3. Represent each of the following molecules using its chemical formula, structural
formula and ball and stick model.
(a) H2
(b) NH3
(c) sulfur dioxide
2.3
Intramolecular and intermolecular forces
When atoms join to form molecules, they are held together by chemical bonds. The type of
bond, and the strength of the bond, depends on the atoms that are involved. These bonds are
called intramolecular forces because they are bonding forces inside a molecule (’intra’ means
’within’ or ’inside’). Sometimes we simply call these intramolecular forces chemical bonds.
Definition: Intramolecular force
The force between the atoms of a molecule, which holds them together.
Examples of the types of chemical bonds that can exist between atoms inside a molecule are
shown below. These will be looked at in more detail in chapter 4.
• Covalent bond
Covalent bonds exist between non-metal atoms e.g. There are covalent bonds between
the carbon and oxygen atoms in a molecule of carbon dioxide.
• Ionic bond
Ionic bonds occur between non-metal and metal atoms e.g. There are ionic bonds between
the sodium and chlorine atoms in a molecule of sodium chloride.
• Metallic bond
Metallic bonds join metal atoms e.g. There are metallic bonds between copper atoms in
a piece of copper metal.
Intermolecular forces are those bonds that hold molecules together. A glass of water for
example, contains many molecules of water. These molecules are held together by intermolecular
forces. The strength of the intermolecular forces is important because they affect properties such
as melting point and boiling point. For example, the stronger the intermolecular forces, the higher
the melting point and boiling point for that substance. The strength of the intermolecular forces
increases as the size of the molecule increases.
25
2.4
CHAPTER 2. WHAT ARE THE OBJECTS AROUND US MADE OF? - GRADE 10
Definition: Intermolecular force
A force between molecules, which holds them together.
Diagram 2.6 may help you to understand the difference between intramolecular forces and intermolecular forces.
intermolecular forces
intramolecular forces
H
O
O
H
O
O
H
O
O
Figure 2.6: Two representations showing the intermolecular and intramolecular forces in water:
space-filling model and structural formula.
It should be clearer now that there are two types of forces that hold matter together. In the case
of water, there are intramolecular forces that hold the two hydrogen atoms to the oxygen atom
in each molecule of water. There are also intramolecular forces between each of these water
molecules. As mentioned earlier, these forces are very important because they affect many of
the properties of matter such as boiling point, melting point and a number of other properties.
Before we go on to look at some of these examples, it is important that we first take a look at
the Kinetic Theory of Matter.
Exercise: Intramolecular and intermolecular forces
1. Using ammonia gas as an example...
(a) Explain what is meant by an intramolecular force or chemical bond.
(b) Explain what is meant by an intermolecular force.
2. Draw a diagram showing three molecules of carbon dioxide. On the diagram,
show where the intramolecular and intermolecular forces are.
3. Why is it important to understand the types of forces that exist between atoms
and between molecules? Try to use some practical examples in your answer.
2.4
The Kinetic Theory of Matter
The kinetic theory of matter is used to explain why matter exists in different phases (i.e. solid,
liquid and gas), and how matter can change from one phase to the next. The kinetic theory of
matter also helps us to understand other properties of matter. It is important to realise that
what we will go on to describe is only a theory. It cannot be proved beyond doubt, but the fact
that it helps us to explain our observations of changes in phase, and other properties of matter,
suggests that it probably is more than just a theory.
Broadly, the Kinetic Theory of Matter says that:
26
CHAPTER 2. WHAT ARE THE OBJECTS AROUND US MADE OF? - GRADE 10
2.4
• Matter is made up of particles that are constantly moving.
• All particles have energy, but the energy varies depending on whether the substance is a
solid, liquid or gas. Solid particles have the least energy and gas particles have the most
amount of energy.
• The temperature of a substance is a measure of the average kinetic energy of the particles.
• A change in phase may occur when the energy of the particles is changed.
• There are spaces between the particles of matter.
• There are attractive forces between particles and these become stronger as the particles
move closer together. These attractive forces will either be intramolecular forces (if the
particles are atoms) or intermolecular forces (if the particles are molecules). When the
particles are extremely close, repulsive forces start to act.
Table 2.1 summarises the characteristics of the particles that are in each phase of matter.
Table 2.1: Table summarising the general features of solids, liquids and gases.
Property of matter Gas
Liquid
Gas
Particles
Atoms or molecules
Atoms or molecules
Atoms or molecules
Energy and move- Particles have high Particles have less Low energy - partiment of particles
energy and are con- energy than in the cles vibrate around a
stantly moving
gas phase
fixed point
Spaces between par- Large spaces be- Smaller spaces than Very little space
ticles
cause of high energy in gases
between particles.
Particles are tightly
packed together
Attractive forces be- Weak forces because Stronger forces than Very strong forces.
tween particles
of the large distance in gas. Liquids can Solids have a fixed
between particles
be poured.
volume.
Changes in phase
In general a gas A liquid becomes a Solids become liqbecomes a liquid gas if its tempera- uids or gases if their
or solid when it is ture is increased. It temperature is incooled.
Particles becomes a solid if creased.
have less energy its temperature deand therefore move creases.
closer together so
that the attractive forces become
stronger, and the
gas becomes a liquid
or a solid
Let’s look at an example that involves the three phases of water: ice (solid), water (liquid) and
water vapour (gas).
solid
liquid
gas
Figure 2.7: The three phases of matter
In a solid (e.g. ice), the water molecules have very little energy and can’t move away from each
other. The molecules are held close together in a regular pattern called a lattice. If the ice is
27
2.5
CHAPTER 2. WHAT ARE THE OBJECTS AROUND US MADE OF? - GRADE 10
heated, the energy of the molecules increases. This means that some of the water molecules are
able to overcome the intermolecular forces that are holding them together, and the molecules
move further apart to form liquid water. This is why liquid water is able to flow, because the
molecules are more free to move than they were in the solid lattice. If the molecules are heated
further, the liquid water will become water vapour, which is a gas. Gas particles have lots of
energy and are far away from each other. That is why it is difficult to keep a gas in a specific
area! The attractive forces between the particles are very weak and they are only loosely held
together. Figure 2.8 shows the changes in phase that may occur in matter, and the names that
describe these processes.
Gas
co n
eva dens
p o a ti o
rat n
io n
on
a ti
lim o n
s u b a ti
re- blim
su
Liquid
Solid
freezing
melting
Figure 2.8: Changes in phase
2.5
The Properties of Matter
Let us now look at what we have learned about chemical bonds, intermolecular forces and the
kinetic theory of matter, and see whether this can help us to understand some of the macroscopic
properties of materials.
1. Melting point
Definition: Melting point
The temperature at which a solid changes its phase or state to become a liquid. The reverse
process (change in phase from liquid to solid) is called freezing.
In order for a solid to melt, the energy of the particles must increase enough to overcome
the bonds that are holding the particles together. It makes sense then that a solid which is
held together by strong bonds will have a higher melting point than one where the bonds
are weak, because more energy (heat) is needed to break the bonds. In the examples we
have looked at, metals, ionic solids and some atomic lattices (e.g. diamond) have high
melting points, whereas the melting points for molecular solids and other atomic lattices
(e.g. graphite) are much lower. Generally, the intermolecular forces between molecular
solids are weaker than those between ionic and metallic solids.
2. Boiling point
Definition: Boiling point
The temperature at which a liquid changes its phase to become a gas.
28
CHAPTER 2. WHAT ARE THE OBJECTS AROUND US MADE OF? - GRADE 10
2.5
When the temperature of a liquid increases, the average kinetic energy of the particles also
increases, and they are able to overcome the bonding forces that are holding them in the
liquid. When boiling point is reached, evaporation takes place and some particles in the
liquid become a gas. In other words, the energy of the particles is too great for them to
be held in a liquid anymore. The stronger the bonds within a liquid, the higher the boiling
point needs to be in order to break these bonds. Metallic and ionic compounds have high
boiling points while the boiling point for molecular liquids is lower.
The data in table 2.2 below may help you to understand some of the concepts we have
explained. Not all of the substances in the table are solids at room temperature, so for
now, let’s just focus on the boiling points for each of these substances. Of the substances
listed, ethanol has the weakest intermolecular forces, and sodium chloride and mercury
have the strongest. What do you notice?
Substance
Ethanol (C2 H6 O)
Water
Mercury
Sodium chloride
Melting point (0 C)
-114,3
0
-38,83
801
Boiling point (0 C)
78,4
100
356,73
1465
Table 2.2: The melting and boiling points for a number of substances
You will have seen that substances such as ethanol, with relatively weak intermolecular
forces, have the lowest boiling point, while substances with stronger intermolecular forces
such as sodium chloride and mercury, must be heated much more if the particles are to
have enough energy to overcome the forces that are holding them together in the liquid
or solid phase.
Exercise: Forces and boiling point
The table below gives the molecular formula and the boiling point for a
number of organic compounds called alkanes. Refer to the table and then
answer the questions that follow.
Organic compound Molecular formula Boiling point (0 C)
Methane
CH2
-161.6
Ethane
C2 H6
-88.6
Propane
C3 H8
-45
Butane
C4 H10
-0.5
Pentane
C5 H12
36.1
Hexane
C6 H14
69
Heptane
C7 H16
98.42
Octane
C8 H18
125.52
Data from: http://www.wikipedia.com
(a) Draw a graph to show the relationship between the number of carbon atoms
in each alkane, and its boiling point (Number of carbon atoms will go on
the x-axis and boiling point on the y-axis).
(b) Describe what you see.
(c) Suggest a reason for what you have observed.
(d) Why was it enough for us to use ’number of carbon atoms’ as a measure
of the molecular weight of the molecules?
3. Density and viscosity
Density is a measure of the mass of a substance per unit volume. The density of a solid
is generally higher than that of a liquid because the particles are hold much more closely
29
2.5
CHAPTER 2. WHAT ARE THE OBJECTS AROUND US MADE OF? - GRADE 10
together and therefore there are more particles packed together in a particular volume. In
other words, there is a greater mass of the substance in a particular volume. In general,
density increases as the strength of the intermolecular forces increases. Viscosity is a
measure of how resistant a liquid is to changing its form. Viscosity is also sometimes
described as the ’thickness’ of a fluid. Think for example of syrup and how slowly it pours
from one container into another. Now compare this to how easy it is to pour water. The
viscosity of syrup is greater than the viscosity of water. Once again, the stronger the
intermolecular forces in the liquid, the greater its viscosity.
It should be clear now that we can explain a lot of the macroscopic properties of matter (i.e.
the characteristics we can see or observe) by understanding their microscopic structure and
the way in which the atoms and molecules that make up matter are held together.
Activity :: Investigation : Determining the density of liquids:
Density is a very important property because it helps us to identify different
materials. Every material, depending on the elements that make it up, and the
arrangement of its atoms, will have a different density.
The equation for density is:
Density = Mass/Volume
Discussion questions:
To calculate the density of liquids and solids, we need to be able to first determine
their mass and volume. As a group, think about the following questions:
• How would you determine the mass of a liquid?
• How would you determine the volume of an irregular solid?
Apparatus:
Laboratory mass balance, 10 ml and 100 ml graduated cylinders, thread, distilled
water, two different liquids.
Method:
Determine the density of the distilled water and two liquids as follows:
1. Measure and record the mass of a 10 ml graduated cyclinder.
2. Pour an amount of distilled water into the cylinder.
3. Measure and record the combined mass of the water and cylinder.
4. Record the volume of distilled water in the cylinder
5. Empty, clean and dry the graduated cylinder.
6. Repeat the above steps for the other two liquids you have.
7. Complete the table below.
Liquid
Distilled water
Liquid 1
Liquid 2
Mass (g)
Volume (ml)
Density (g/ml)
Activity :: Investigation : Determining the density of irregular solids:
Apparatus:
Use the same materials and equpiment as before (for the liquids). Also find a
number of solids that have an irregular shape.
Method:
Determine the density of irregular solids as follows:
30
CHAPTER 2. WHAT ARE THE OBJECTS AROUND US MADE OF? - GRADE 10
2.6
1. Measure and record the mass of one of the irregular solids.
2. Tie a piece of thread around the solid.
3. Pour some water into a 100 ml graduated cylinder and record the volume.
4. Gently lower the solid into the water, keeping hold of the thread. Record the
combined volume of the solid and the water.
5. Dtermine the volume of the solid by subtracting the combined volume from the
original volume of the water only.
6. Repeat these steps for the second object.
7. Complete the table below.
Solid
Solid 1
Solid 2
Solid 3
2.6
Mass (g)
Volume (ml)
Density (g/ml)
Summary
• The smallest unit of matter is the atom. Atoms can combine to form molecules.
• A molecule is a group of two or more atoms that are attracted to each other by chemical
bonds.
• A small molecule consists of a few atoms per molecule. A giant molecule consists of
millions of atoms per molecule, for example metals and diamonds.
• The structure of a molecule can be represented in a number of ways.
• The chemical formula of a molecule is an abbreviated way of showing a molecule, using
the symbols for the elements in the molecule. There are two types of chemical formulae:
molecular and empirical formula.
• The molecular formula of a molecule gives the exact number of atoms of each element
that are in the molecule.
• The empirical formula of a molecule gives the relative number of atoms of each element
in the molecule.
• Molecules can also be represented using diagrams.
• A ball and stick diagram is a 3-dimensional molecular model that uses ’balls’ to represent
atoms and ’sticks’ to represent the bonds between them.
• A space-filling model is also a 3-dimensional molecular model. The atoms are represented
by multi-coloured spheres.
• In a molecule, atoms are held together by chemical bonds or intramolecular forces.
Covalent bonds, ionic bonds and metallic bonds are examples of chemical bonds.
• A covalent bond exists between non-metal atoms. An ionic bond exists between nonmetal and metal atoms, and a metallic bond exists between metal atoms.
• Intermolecular forces are the bonds that hold molecules together.
• The kinetic theory of matter attempts to explain the behaviour of matter in different
phases.
• The theory says that all matter is composed of particles which have a certain amount
of energy which allows them to move at different speeds depending on the temperature
(energy). There are spaces between the particles, and also attractive forces between
particles when they come close together.
31
2.6
CHAPTER 2. WHAT ARE THE OBJECTS AROUND US MADE OF? - GRADE 10
• Understanding chemical bonds, intermolecular forces and the kinetic theory of matter, can
help to explain many of the macroscopic properties of matter.
• Melting point is the temperature at which a solid changes its phase to become a liquid.
The reverse process (change in phase from liquid to solid) is called freezing. The stronger
the chemical bonds and intermolecular forces in a substance, the higher the melting point
will be.
• Boiling point is the temperature at which a liquid changes phase to become a gas. The
stronger the chemical bonds and intermolecular forces in a substance, the higher the boiling
point will be.
• Density is a measure of the mass of a substance per unit volume.
• Viscosity is a measure of how resistant a liquid is to changing its form.
Exercise: Summary exercise
1. Give one word or term for each of the following descriptions.
(a) The property that determines how easily a liquid flows.
(b) The change in phase from liquid to gas.
(c) A composition of two or more atoms that act as a unit.
(d) Chemical formula that gives the relative number of atoms of each element
that are in a molecule.
2. For each of the following questions, choose the one correct answer from the
list provided.
A Ammonia, an ingredient in household cleaners, can be broken down to
form one part nitrogen (N) and three parts hydrogen (H). This means that
ammonia...
i. is a colourless gas
ii. is not a compound
iii. cannot be an element
iv. has the formula N3 H
B If one substance A has a melting point that is lower than the melting point
of substance B, this suggests that...
i. A will be a liquid at room temperature.
ii. The chemical bonds in substance A are weaker than those in substance
B.
iii. The chemical bonds in substance A are stronger than those in substance B.
iv. B will be a gas at room temperature.
3. Boiling point is an important concept to understand.
a Define ’boiling point’.
b What change in phase takes place when a liquid reaches its boiling point?
c What is the boiling point of water?
d Use the kinetic theory of matter and your knowledge of intermolecular
forces, to explain why water changes phase at this temperature.
4. Refer to the table below which gives the melting and boiling points of a
number of elements, and then answer the questions that follow. (Data from
http://www.chemicalelements.com)
Element
copper
magnesium
oxygen
carbon
helium
sulfur
Melting point
1083
650
-218.4
3500
-272
112.8
32
Boiling point (0 C)
2567
1107
-183
4827
-268.6
444.6
CHAPTER 2. WHAT ARE THE OBJECTS AROUND US MADE OF? - GRADE 10
a What state of matter (i.e. solid, liquid or gas) will each of these elements
be in at room temperature?
b Which of these elements has the strongest forces between its atoms? Give
a reason for your answer.
c Which of these elements has the weakest forces between its atoms? Give
a reason for your answer.
33
2.6
2.6
CHAPTER 2. WHAT ARE THE OBJECTS AROUND US MADE OF? - GRADE 10
34
Chapter 3
The Atom - Grade 10
We have now looked at many examples of the types of matter and materials that exist around
us, and we have investigated some of the ways that materials are classified. But what is it that
makes up these materials? And what makes one material different from another? In order to
understand this, we need to take a closer look at the building block of matter, the atom. Atoms
are the basis of all the structures and organisms in the universe. The planets, the sun, grass and
trees, the air we breathe, and people are all made up of different combinations of atoms.
3.1
Models of the Atom
It is important to realise that a lot of what we know about the structure of atoms has been
developed over a long period of time. This is often how scientific knowledge develops, with
one person building on the ideas of someone else. We are going to look at how our modern
understanding of the atom has evolved over time.
The idea of atoms was invented by two Greek philosophers, Democritus and Leucippus in the
fifth century BC. The Greek word ατ oµoν (atom) means indivisible because they believed that
atoms could not be broken into smaller pieces.
Nowadays, we know that atoms are made up of a positively charged nucleus in the centre
surrounded by negatively charged electrons. However, in the past, before the structure of the
atom was properly understood, scientists came up with lots of different models or pictures to
describe what atoms look like.
Definition: Model
A model is a representation of a system in the real world. Models help us to understand
systems and their properties. For example, an atomic model represents what the structure
of an atom could look like, based on what we know about how atoms behave. It is not
necessarily a true picture of the exact structure of an atom.
3.1.1
The Plum Pudding Model
After the electron was discovered by J.J. Thomson in 1897, people realised that atoms were made
up of even smaller particles than they had previously thought. However, the atomic nucleus had
not been discovered yet, and so the ’plum pudding model’ was put forward in 1904. In this
model, the atom is made up of negative electrons that float in a soup of positive charge, much
like plums in a pudding or raisins in a fruit cake (figure 3.1). In 1906, Thomson was awarded
the Nobel Prize for his work in this field. However, even with the Plum Pudding Model, there
was still no understanding of how these electrons in the atom were arranged.
35
3.1
CHAPTER 3. THE ATOM - GRADE 10
-
-
-
electrons
-
-
-
’soup’ of positive charge
-
Figure 3.1: A schematic diagram to show what the atom looked like according to the Plum
Pudding model
The discovery of radiation was the next step along the path to building an accurate picture of
atomic structure. In the early twentieth century, Marie Curie and her husband discovered that
some elements (the radioactive elements) emit particles, which are able to pass through matter
in a similar way to X-rays (read more about this in chapter 7). It was Ernest Rutherford who, in
1911, used this discovery to revise the model of the atom.
3.1.2
Rutherford’s model of the atom
Radioactive elements emit different types of particles. Some of these are positively charged alpha
(α) particles. Rutherford carried out a series of experiments where he bombarded sheets of gold
foil with these particles, to try to get a better understanding of where the positive charge in the
atom was. A simplified diagram of his experiment is shown in figure 3.2.
C
B
b
gold sheet
b
radioactive
substance
A
α particles
A
α particles
b
b
b
b
b
b
b
C
(a)
B
C
screen
b
b
b
b
b
b
b
b
b
b
b
b
A
b
b
b
b
B
b
b
b
b
nucleus of
gold atom
(b)
Figure 3.2: Rutherford’s gold foil experiment. Figure (a) shows the path of the α particles after
they hit the gold sheet. Figure (b) shows the arrangement of atoms in the gold sheets, and the
path of the α particles in relation to this.
Rutherford set up his experiment so that a beam of alpha particles was directed at the gold
sheets. Behind the gold sheets, was a screen made of zinc sulfide. This screen allowed Rutherford to see where the alpha particles were landing. Rutherford knew that the electrons in the gold
atoms would not really affect the path of the alpha particles, because the mass of an electron is
so much smaller than that of a proton. He reasoned that the positively charged protons would
be the ones to repel the positively charged alpha particles and alter their path.
36
CHAPTER 3. THE ATOM - GRADE 10
3.1
What he discovered was that most of the alpha particles passed through the foil undisturbed,
and could be detected on the screen directly behind the foil (A). Some of the particles ended up
being slightly deflected onto other parts of the screen (B). But what was even more interesting
was that some of the particles were deflected straight back in the direction from where they
had come (C)! These were the particles that had been repelled by the positive protons in the
gold atoms. If the Plum Pudding model of the atom were true, then Rutherford would have
expected much more repulsion since the positive charge, according to that model, is distributed
throughout the atom. But this was not the case. The fact that most particles passed straight
through suggested that the positive charge was concentrated in one part of the atom only.
Rutherford’s work led to a change in ideas around the atom. His new model described the
atom as a tiny, dense, positively charged core called a nucleus, surrounded by lighter, negatively
charged electrons. Another way of thinking about this model was that the atom was seen to be
like a mini solar system where the electrons orbit the nucleus like planets orbiting around the
sun. A simplified picture of this is shown in figure 3.3.
b
b
b
electron orbiting the nucleus
b
b
b
nucleus (containing protons and neutrons)
Figure 3.3: Rutherford’s model of the atom
3.1.3
The Bohr Model
There were, however, some problems with this model: for example it could not explain the very
interesting observation that atoms only emit light at certain wavelengths or frequencies. Niels
Bohr solved this problem by proposing that the electrons could only orbit the nucleus in certain
special orbits at different energy levels around the nucleus. The exact energies of the orbitals in
each energy level depends on the type of atom. Helium for example, has different energy levels
to Carbon. If an electron jumps down from a higher energy level to a lower energy level, then
light is emitted from the atom. The energy of the light emitted is the same as the gap in the
energy between the two energy levels. You can read more about this in section 3.6. The distance
between the nucleus and the electron in the lowest energy level of a hydrogen atom is known as
the Bohr radius.
teresting Light has the properties of both a particle and a wave! Einstein discovered that
Interesting
Fact
Fact
light comes in energy packets which are called photons. When an electron in
an atom changes energy levels, a photon of light is emitted. This photon has the
same energy as the difference between the two electron energy levels.
37
3.2
3.2
CHAPTER 3. THE ATOM - GRADE 10
How big is an atom?
It is difficult sometimes to imagine the size of an atom, or its mass, because we cannot see them,
and also because we are not used to working with such small measurements.
3.2.1
How heavy is an atom?
It is possible to determine the mass of a single atom in kilograms. But to do this, you would
need very modern mass spectrometers, and the values you would get would be very clumsy and
difficult to use. The mass of a carbon atom, for example, is about 1.99 x 10−26 kg, while the
mass of an atom of hydrogen is about 1.67 x 10−27 kg. Looking at these very small numbers
makes it difficult to compare how much bigger the mass of one atom is when compared to another.
To make the situation simpler, scientists use a different unit of mass when they are describing
the mass of an atom. This unit is called the atomic mass unit (amu). We can abbreviate
(shorten) this unit to just ’u’. If we give carbon an atomic mass of 12 u, then the mass of an
atom of hydrogen will be 1 u. You can check this by dividing the mass of a carbon atom in
kilograms (see above) by the mass of a hydrogen atom in kilograms (you will need to use a
calculator for this!). If you do this calculation, you will see that the mass of a carbon atom is
twelve times greater than the mass of a hydrogen atom. When we use atomic mass units instead
of kilograms, it becomes easier to see this. Atomic mass units are therefore not giving us the
actual mass of an atom, but rather its mass relative to the mass of other atoms in the Periodic
Table. The atomic masses of some elements are shown in table 3.1 below.
Table 3.1: The atomic mass of a number of elements
Element
Atomic mass (u)
Nitrogen (N)
14
Bromine (Br)
80
Magnesium (Mg)
24
Potassium (K)
39
Calcium (Ca)
40
Oxygen (O)
16
The actual value of 1 atomic mass unit is 1.67 x 10−24 g or 1.67 x 10−27 kg. This is a very tiny
mass!
3.2.2
pm stands for
picometres. 1
pm = 10−12
m
How big is an atom?
Atomic diameter also varies depending on the element. On average, the diameter of an atom
ranges from 100 pm (Helium) to 670 pm (Caesium). Using different units, 100 pm = 1 Angstrom,
and 1 Angstrom = 10−10 m. That is the same as saying that 1 Angstrom = 0,0000000010 m
or that 100 pm = 0,0000000010 m! In other words, the diameter of an atom ranges from
0.0000000010 m to 0.0000000067 m. This is very small indeed.
3.3
Atomic structure
As a result of the models that we discussed in section 3.1, scientists now have a good idea of
what an atom looks like. This knowledge is important because it helps us to understand things
like why materials have different properties and why some materials bond with others. Let us
now take a closer look at the microscopic structure of the atom.
So far, we have discussed that atoms are made up of a positively charged nucleus surrounded
by one or more negatively charged electrons. These electrons orbit the nucleus.
38
CHAPTER 3. THE ATOM - GRADE 10
3.3.1
3.3
The Electron
The electron is a very light particle. It has a mass of 9.11 x 10−31 kg. Scientists believe that the
electron can be treated as a point particle or elementary particle meaning that it can’t be broken
down into anything smaller. The electron also carries one unit of negative electric charge which
is the same as 1.6 x 10−19 C (Coulombs).
3.3.2
The Nucleus
Unlike the electron, the nucleus can be broken up into smaller building blocks called protons
and neutrons. Together, the protons and neutrons are called nucleons.
The Proton
Each proton carries one unit of positive electric charge. Since we know that atoms are electrically
neutral, i.e. do not carry any extra charge, then the number of protons in an atom has to be the
same as the number of electrons to balance out the positive and negative charge to zero. The
total positive charge of a nucleus is equal to the number of protons in the nucleus. The proton
is much heavier than the electron (10 000 times heavier!) and has a mass of 1.6726 x 10−27 kg.
When we talk about the atomic mass of an atom, we are mostly referring to the combined mass
of the protons and neutrons, i.e. the nucleons.
The Neutron
The neutron is electrically neutral i.e. it carries no charge at all. Like the proton, it is much
heavier than the electron and its mass is 1.6749 x 10−27 kg (slightly heavier than the proton).
teresting Rutherford predicted (in 1920) that another kind of particle must be present in
Interesting
Fact
Fact
the nucleus along with the proton. He predicted this because if there were only
positively charged protons in the nucleus, then it should break into bits because
of the repulsive forces between the like-charged protons! Also, if protons were
the only particles in the nucleus, then a helium nucleus (atomic number 2) would
have two protons and therefore only twice the mass of hydrogen. However, it is
actually four times heavier than hydrogen. This suggested that there must be
something else inside the nucleus as well as the protons. To make sure that the
atom stays electrically neutral, this particle would have to be neutral itself. In
1932 James Chadwick discovered the neutron and measured its mass.
Mass (kg)
Units of charge
Charge (C)
proton
1.6726 x 10−27
+1
1.6 x 10−19
neutron
1.6749 x 10−27
0
0
electron
9.11 x 10−31
-1
-1.6 x 10−19
Table 3.2: Summary of the particles inside the atom
teresting Unlike the electron which is thought to be a point particle and unable to be
Interesting
Fact
Fact
broken up into smaller pieces, the proton and neutron can be divided. Protons
and neutrons are built up of smaller particles called quarks. The proton and
neutron are made up of 3 quarks each.
39
3.4
3.4
CHAPTER 3. THE ATOM - GRADE 10
Atomic number and atomic mass number
The chemical properties of an element are determined by the charge of its nucleus, i.e. by the
number of protons. This number is called the atomic number and is denoted by the letter Z.
Definition: Atomic number (Z)
The number of protons in an atom
The mass of an atom depends on how many nucleons its nucleus contains. The number of
nucleons, i.e. the total number of protons plus neutrons, is called the atomic mass number
and is denoted by the letter A.
Definition: Atomic mass number (A)
The number of protons and neutrons in the nucleus of an atom
Standard notation shows the chemical symbol, the atomic mass number and the atomic number
of an element as follows:
number of nucleons
A
ZX
chemical symbol
number of protons
For example, the iron nucleus which has 26 protons and 30 neutrons, is denoted as
56
26 Fe
,
where the total nuclear charge is Z = 26 and the mass number A = 56. The number of neutrons
is simply the difference N = A − Z.
40
CHAPTER 3. THE ATOM - GRADE 10
3.4
Important:
Don’t confuse the notation we have used above, with the way this information appears
on the Periodic Table. On the Periodic Table, the atomic number usually appears in the
top lefthand corner of the block or immediately above the element’s symbol. The number
below the element’s symbol is its relative atomic mass. This is not exactly the same as
the atomic mass number. This will be explained in section 3.5. The example of iron is used
again below.
26
Fe
55.85
You will notice in the example of iron that the atomic mass number is more or less the same as
its atomic mass. Generally, an atom that contains n protons and neutrons (i.e. Z = n), will have
a mass approximately equal to n u. The reason is that a C-12 atom has 6 protons, 6 neutrons
and 6 electrons, with the protons and neutrons having about the same mass and the electron
mass being negligible in comparison.
Exercise: The structure of the atom
1. Explain the meaning of each of the following terms:
(a) nucleus
(b) electron
(c) atomic mass
2. Complete the following table: (Note: You will see that the atomic masses on
the Periodic Table are not whole numbers. This will be explained later. For
now, you can round off to the nearest whole number.)
Element Atomic
Atomic
Number
Number
Number
mass
number
of pro- of elec- of neutons
trons
trons
Mg
24
12
O
8
17
Ni
28
40
20
Zn
0
C
12
6
3. Use standard notation to represent the following elements:
(a) potassium
(b) copper
(c) chlorine
4. For the element
35
17 Cl,
give the number of ...
(a) protons
(b) neutrons
(c) electrons
... in the atom.
41
3.5
CHAPTER 3. THE ATOM - GRADE 10
5. Which of the following atoms has 7 electrons?
(a) 52 He
(b) 13
6 C
(c) 73 Li
(d) 15
7 N
6. In each of the following cases, give the number or the element symbol represented by ’X’.
(a) 40
18 X
(b) x20 Ca
(c) 31
x P
7. Complete the following table:
A
Z
N
235
92 U
238
92 U
In these two different forms of Uranium...
(a) What is the same?
(b) What is different?
Uranium can occur in different forms, called isotopes. You will learn more
about isotopes in section 3.5.
3.5
3.5.1
Isotopes
What is an isotope?
If a few neutrons are added to or removed from a nucleus, the chemical properties of the atom
will stay the same because its charge is still the same. Therefore, the chemical properties of an
element depend on the number of protons inside the atom. This means that such an atom should
remain in the same place in the Periodic table. For example, no matter how many neutrons we
add or subtract from a nucleus with 6 protons, that element will always be called carbon and
have the element symbol C (see the Table of Elements). Atoms which have the same number
of protons, but a different number of neutrons, are called isotopes.
Definition: Isotope
The isotope of a particular element, is made up of atoms which have the same number of
protons as the atoms in the orginal element, but a different number of neutrons.
The different isotopes of an element have the same atomic number Z but different mass numbers
A because they have a different number of neutrons N . The chemical properties of the different
isotopes of an element are the same, but they might vary in how stable their nucleus is. Note
that if an element is written for example as C-12, the ’12’ is the atomic mass of that atom. So,
Cl-35 has an atomic mass of 35 u, while Cl-37 has an atomic mass of 37 u.
teresting In Greek, “same place” reads as ὶσoς τ òπoς (isos topos). This is why atoms
Interesting
Fact
Fact
which have the same number of protons, but different numbers of neutrons, are
called isotopes. They are in the same place on the Periodic Table!
The following worked examples will help you to understand the concept of an isotope better.
42
CHAPTER 3. THE ATOM - GRADE 10
3.5
Worked Example 1: Isotopes
Question: For the element
234
92 U
(uranium), use standard notation to describe:
1. the isotope with 2 fewer neutrons
2. the isotope with 4 more neutrons
Answer
Step 1 : Go over the definition of isotope
We know that isotopes of any element have the same number of protons (same
atomic number) in each atom which means that they have the same chemical symbol. However, they have a different number of neutrons, and therefore a different
mass number.
Step 2 : Rewrite the notation for the isotopes
Therefore, any isotope of uranium will have the symbol:
U
Also, since the number of protons in uranium isotopes is always the same, we can
write down the atomic number:
92 U
Now, if the isotope we want has 2 fewer neutrons than
original mass number and subtract 2, which gives:
234
92 U,
then we take the
232
92 U
Following the steps above, we can write the isotope with 4 more neutrons as:
238
92 U
Worked Example 2: Isotopes
Question: Which of the following are isotopes of
•
•
•
40
20 Ca?
40
19 K
42
20 Ca
40
18 Ar
Answer
Step 1 : Go over the definition of isotope:
We know that isotopes have the same atomic number but different mass numbers.
Step 2 : Determine which of the elements listed fits the definition of an
isotope.
You need to look for the element that has the same atomic number but a different
atomic mass number. The only element is 12
20 Ca. What is different is that there are
2 more neutrons than in the original element.
43
3.5
CHAPTER 3. THE ATOM - GRADE 10
Worked Example 3: Isotopes
Question: For the sulfur isotope
1.
2.
3.
4.
33
16 S,
give the number of...
protons
nucleons
electrons
neutrons
Answer
Step 1 : Determine the number of protons by looking at the atomic number,
Z.
Z = 16, therefore the number of protons is 16 (answer to (a)).
Step 2 : Determine the number of nucleons by looking at the atomic mass
number, A.
A = 33, therefore the number of nucleons is 33 (answer to (b)).
Step 3 : Determine the number of electrons
The atom is neutral, and therefore the number of electrons is the same as the number of protons. The number of electrons is 16 (answer to (c)).
Step 4 : Calculate the number of neutrons
N = A − Z = 33 − 16 = 17
The number of neutrons is 17 (answer to (d)).
Exercise: Isotopes
1. Atom A has 5 protons and 5 neutrons, and atom B has 6 protons and 5 neutrons.
These atoms are...
(a) allotropes
(b) isotopes
(c) isomers
(d) atoms of different elements
34
2. For the sulfur isotopes, 32
16 S and 16 S, give the number of...
(a) protons
(b) nucleons
(c) electrons
(d) neutrons
3. Which of the following are isotopes of Cl35 ?
(a) 17
35 Cl
(b) 35
17 Cl
(c) 37
17 Cl
4. Which of the following are isotopes of U-235? (X represents an element symbol)
(a) 238
92 X
(b) 238
90 X
(c) 235
92 X
44
CHAPTER 3. THE ATOM - GRADE 10
3.5.2
3.5
Relative atomic mass
It is important to realise that the atomic mass of isotopes of the same element will be different
because they have a different number of nucleons. Chlorine, for example, has two common
isotopes which are chlorine-35 and chlorine-37. Chlorine-35 has an atomic mass of 35 u, while
chlorine-37 has an atomic mass of 37 u. In the world around us, both of these isotopes occur
naturally. It doesn’t make sense to say that the element chlorine has an atomic mass of 35
u, or that it has an atomic mass of 37 u. Neither of these are absolutely true since the mass
varies depending on the form in which the element occurs. We need to look at how much more
common one is than the other in order to calculate the relative atomic mass for the element
chlorine. This is then the number that will appear on the Periodic Table.
Definition: Relative atomic mass
Relative atomic mass is the average mass of one atom of all the naturally occurring isotopes
of a particular chemical element, expressed in atomic mass units.
Worked Example 4: The relative atomic mass of an isotopic element
Question: The element chlorine has two isotopes, chlorine-35 and chlorine-37. The
abundance of these isotopes when they occur naturally is 75% chlorine-35 and 25%
chlorine-37. Calculate the average relative atomic mass for chlorine.
Answer
Step 1 : Calculate the mass contribution of chlorine-35 to the average relative atomic mass
Contribution of Cl-35 = (75/100 x 35) = 26.25 u
Step 2 : Calculate the contribution of chlorine-37 to the average relative
atomic mass
Contribution of Cl-37 = (25/100 x 37) = 9.25 u
Step 3 : Add the two values to arrive at the average relative atomic mass of
chlorine
Relative atomic mass of chlorine = 26.25 u + 9.25 u = 35.5 u.
If you look on the periodic table, the average relative atomic mass for chlorine is 35,5
u. You will notice that for many elements, the relative atomic mass that is shown
is not a whole number. You should now understand that this number is the average
relative atomic mass for those elements that have naturally occurring isotopes.
Exercise: Isotopes
You are given a sample that contains carbon-12 and carbon-14.
1. Complete the table below:
45
3.6
CHAPTER 3. THE ATOM - GRADE 10
Isotope
Carbon-12
Carbon-14
Chlorine-35
Chlorine-37
Z
A
Protons
Neutrons
Electrons
2. If the sample you have contains 90% carbon-12 and 10% carbon-14, calculate
the relative atomic mass of an atom in that sample.
3. In another sample, you have 22.5% Cl-37 and 77.5% Cl-35. Calculate the
relative atomic mass of an atom in that sample.
Activity :: Group Discussion : The changing nature of scientific knowledge
Scientific knowledge is not static: it changes and evolves over time as scientists
build on the ideas of others to come up with revised (and often improved) theories and
ideas. In this chapter for example, we saw how peoples’ understanding of atomic
structure changed as more information was gathered about the atom. There are
many more examples like this one in the field of science. Think for example, about
our knowledge of the solar system and the origin of the universe, or about the particle
and wave nature of light.
Often, these changes in scientific thinking can be very controversial because they
disturb what people have come to know and accept. It is important that we realise
that what we know now about science may also change. An important part of
being a scientist is to be a critical thinker. This means that you need to question
information that you are given and decide whether it is accurate and whether it can
be accepted as true. At the same time, you need to learn to be open to new ideas
and not to become stuck in what you believe is right... there might just be something
new waiting around the corner that you have not thought about!
In groups of 4-5, discuss the following questions:
• Think about some other examples where scientific knowledge has changed because of new ideas and discoveries:
–
–
–
–
What were these new ideas?
Were they controversial? If so, why?
What role (if any) did technology play in developing these new ideas?
How have these ideas affected the way we understand the world?
• Many people come up with their own ideas about how the world works. The
same is true in science. So how do we, and other scientists, know what to
believe and what not to? How do we know when new ideas are ’good’ science
or ’bad’ science? In your groups, discuss some of the things that would need
to be done to check whether a new idea or theory was worth listening to, or
whether it was not.
• Present your ideas to the rest of the class.
3.6
3.6.1
Energy quantisation and electron configuration
The energy of electrons
You will remember from our earlier discussions, that an atom is made up of a central nucleus,
which contains protons and neutrons, and that this nucleus is surrounded by electrons. Although
46
CHAPTER 3. THE ATOM - GRADE 10
3.6
these electrons all have the same charge and the same mass, each electron in an atom has a
different amount of energy. Electrons that have the lowest energy are found closest to the nucleus
where the attractive force of the positively charged nucleus is the greatest. Those electrons that
have higher energy, and which are able to overcome the attractive force of the nucleus, are found
further away.
3.6.2
Energy quantisation and line emission spectra
If the energy of an atom is increased (for example when a substance is heated), the energy of the
electrons inside the atom can be increased (when an electron has a higher energy than normal
it is said to be ”excited”). For the excited electron to go back to its original energy (called the
ground state), it needs to release energy. It releases energy by emitting light. If one heats up
different elements, one will see that for each element, light is emitted only at certain frequencies
(or wavelengths). Instead of a smooth continuum of frequencies, we see lines (called emission
lines) at particular frequencies. These frequencies correspond to the energy of the emitted light.
If electrons could be excited to any energy and lose any amount of energy, there would be a
continuous spread of light frequencies emitted. However, the sharp lines we see mean that there
are only certain particular energies that an electron can be excited to, or can lose, for each
element.
You can think of this like going up a flight of steps: you can’t lift your foot by any amount to
go from the ground to the first step. If you lift your foot too low you’ll bump into the step and
be stuck on the ground level. You have to lift your foot just the right amount (the height of the
step) to go to the next step, and so on. The same goes for electrons and the amount of energy
they can have. This is called quantisation of energy because there are only certain quantities
of energy that an electron can have in an atom. Like steps, we can think of these quantities as
energy levels in the atom. The energy of the light released when an electron drops down from
a higher energy level to a lower energy level is the same as the difference in energy between the
two levels.
3.6.3
Electron configuration
Electrons are arranged in energy levels around the nucleus of an atom. Electrons that are in the
energy level that is closest to the nucleus, will have the lowest energy and those further away
will have a higher energy. Each energy level can only hold a certain number of electrons, and
an electron will only be found in the second energy level once the first energy level is full. The
same rule applies for the higher energy levels. You will need to learn the following rules:
• The 1st energy level can hold a maximum of 2 electrons
• The 2nd energy level can hold a maximum of 8 electrons
• The 3rd energy level can hold a maximum of 8 electrons
• If the number of electrons in the atom is greater than 18, they will need to move to the
4th energy level.
In the following examples, the energy levels are shown as concentric circles around the central
nucleus.
1. Lithium
Lithium (Li) has an atomic number of 3, meaning that in a neutral atom, the number of
electrons will also be 3. The first two electrons are found in the first energy level, while
the third electron is found in the second energy level (figure 3.11).
2. Fluorine
Fluorine (F) has an atomic number of 9, meaning that a neutral atom also has 9 electrons.
The first 2 electrons are found in the first energy level, while the other 7 are found in the
second energy level (figure 3.12).
47
3.6
CHAPTER 3. THE ATOM - GRADE 10
second energy level
electrons
first energy level
nucleus, containing 3 protons and 4 neutrons
Figure 3.4: The arrangement of electrons in a lithium atom.
Figure 3.5: The arrangement of electrons in a fluorine atom.
3. Argon
Argon has an atomic number of 18, meaning that a neutral atom also has 18 electrons.
The first 2 electrons are found in the first energy level, the next 8 are found in the second
energy level, and the last 8 are found in the third energy level (figure 3.6).
Figure 3.6: The arrangement of electrons in an argon atom.
But the situation is slightly more complicated than this. Within each energy level, the electrons
move in orbitals. An orbital defines the spaces or regions where electrons move.
Definition: Atomic orbital
An atomic orbital is the region in which an electron may be found around a single atom.
There are different orbital shapes, but we will be dealing with only two. These are the ’s’ and ’p’
orbitals (there are also ’d’ and ’f’ orbitals). The first energy level contains only one ’s’ orbital,
the second energy level contains one ’s’ orbital and three ’p’ orbitals and the third energy level
also contains one ’s’ orbital and three ’p’ orbitals. Within each energy level, the ’s’ orbital is at
a lower energy than the ’p’ orbitals. This arrangement is shown in figure 3.7.
When we want to show how electrons are arranged in an atom, we need to remember the
following principles:
48
CHAPTER 3. THE ATOM - GRADE 10
3.6
4s
3p
Third main
energy level
2p
Second main
energy level
3s
E
N
E
R
G
Y
2s
First main
energy level
1s
Figure 3.7: The positions of the first ten orbits of an atom on an energy diagram. Note that
each block is able to hold two electrons.
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3.6
CHAPTER 3. THE ATOM - GRADE 10
• Each orbital can only hold two electrons. Electrons that occur together in an orbital are
called an electron pair. These electrons spin in opposite directions around their own axes.
• An electron will always try to enter an orbital with the lowest possible energy.
• An electron will occupy an orbital on its own, rather than share an orbital with another
electron. An electron would also rather occupy a lower energy orbital with another electron,
before occupying a higher energy orbital. In other words, within one energy level, electrons
will fill an ’s’ orbital before starting to fill ’p’ orbitals.
The way that electrons are arranged in an atom is called its electron configuration.
Definition: Electron configuration
Electron configuration is the arrangement of electrons in an atom, molecule, or other physical
structure.
An element’s electron configuration can be represented using Aufbau diagrams or energy level
diagrams. An Aufbau diagram uses arrows to represent electrons. You can use the following
steps to help you to draw an Aufbau diagram:
1. Determine the number of electrons that the atom has.
2. Fill the ’s’ orbital in the first energy level (the 1s orbital) with the first two electrons.
3. Fill the ’s’ orbital in the second energy level (the 2s orbital) with the second two electrons.
4. Put one electron in each of the three ’p’ orbitals in the second energy level (the 2p orbitals),
and then if there are still electrons remaining, go back and place a second electron in each
of the 2p orbitals to complete the electron pairs.
5. Carry on in this way through each of the successive energy levels until all the electrons
have been drawn.
Important:
When there are two electrons in an orbital, the electrons are called an electron pair. If the
orbital only has one electron, this electron is said to be an unpaired electron. Electron
pairs are shown with arrows in opposite directions. This is because when two electrons
occupy the same orbital, they spin in opposite directions on their axes.
An Aufbau diagram for the element Lithium is shown in figure 3.8.
2s
1s
Figure 3.8: The electron configuration of Lithium, shown on an Aufbau diagram
A special type of notation is used to show an atom’s electron configuration. The notation describes the energy levels, orbitals and the number of electrons in each. For example, the electron
configuration of lithium is 1s2 2s1 . The number and letter describe the energy level and orbital,
and the number above the orbital shows how many electrons are in that orbital.
Aufbau diagrams for the elements fluorine and argon are shown in figures 3.9 and 3.10 respectively. Using standard notation, the electron configuration of fluorine is 1s2 2s2 2p5 and the
electron configuration of argon is 1s2 2s2 2p6 3s2 3p6 .
50
CHAPTER 3. THE ATOM - GRADE 10
3.6
4s
3p
3s
2p
E
N
E
R
G
Y
2s
1s
Fluorine
Figure 3.9: An Aufbau diagram showing the electron configuration of fluorine
3.6.4
Core and valence electrons
Electrons in the outermost energy level of an atom are called valence electrons. The electrons
that are in the energy shells closer to the nucleus are called core electrons. Core electrons are
all the electrons in an atom, excluding the valence electrons. An element that has its valence
energy level full is more stable and less likely to react than other elements with a valence energy
level that is not full.
Definition: Valence electrons
The electrons in the outer energy level of an atom
Definition: Core electrons
All the electrons in an atom, excluding the valence electrons
3.6.5
The importance of understanding electron configuration
By this stage, you may well be wondering why it is important for you to understand how electrons
are arranged around the nucleus of an atom. Remember that during chemical reactions, when
atoms come into contact with one another, it is the electrons of these atoms that will interact
first. More specifically, it is the valence electrons of the atoms that will determine how they
51
3.6
CHAPTER 3. THE ATOM - GRADE 10
4s
3p
3s
2p
2s
1s
Argon
Figure 3.10: An Aufbau diagram showing the electron configuration of argon
react with one another.
To take this a step further, an atom is at its most stable (and therefore unreactive) when all
its orbitals are full. On the other hand, an atom is least stable (and therefore most reactive)
when its valence electron orbitals are not full. This will make more sense when we go on to
look at chemical bonding in a later chapter. To put it simply, the valence electrons are largely
responsible for an element’s chemical behaviour, and elements that have the same number of
valence electrons often have similar chemical properties.
Exercise: Energy diagrams and electrons
1. Draw Aufbau diagrams to show the electron configuration of each of the following elements:
(a) magnesium
(b) potassium
(c) sulfur
(d) neon
(e) nitrogen
2. Use the Aufbau diagrams you drew to help you complete the following table:
52
CHAPTER 3. THE ATOM - GRADE 10
Element
No.
of
energy
levels
3.7
No.
of
core electrons
No.
of
valence
electrons
Electron
configuration
(standard
notation)
Mg
K
S
Ne
N
3. Rank the elements used above in order of increasing reactivity. Give reasons
for the order you give.
Activity :: Group work : Building a model of an atom
Earlier in this chapter, we talked about different ’models’ of the atom. In science,
one of the uses of models is that they can help us to understand the structure of
something that we can’t see. In the case of the atom, models help us to build a
picture in our heads of what the atom looks like.
Models are often simplified. The small toy cars that you may have played with as
a child are models. They give you a good idea of what a real car looks like, but they
are much smaller and much simpler. A model cannot always be absolutely accurate
and it is important that we realise this so that we don’t build up a false idea about
something.
In groups of 4-5, you are going to build a model of an atom. Before you start,
think about these questions:
• What information do I know about the structure of the atom? (e.g. what parts
make it up? how big is it?)
• What materials can I use to represent these parts of the atom as accurately as
I can?
• How will I put all these different parts together in my model?
As a group, share your ideas and then plan how you will build your model. Once
you have built your model, discuss the following questions:
• Does our model give a good idea of what the atom actually looks like?
• In what ways is our model inaccurate? For example, we know that electrons
move around the atom’s nucleus, but in your model, it might not have been
possible for you to show this.
• Are there any ways in which our model could be improved?
Now look at what other groups have done. Discuss the same questions for each
of the models you see and record your answers.
3.7
3.7.1
Ionisation Energy and the Periodic Table
Ions
In the previous section, we focused our attention on the electron configuration of neutral atoms.
In a neutral atom, the number of protons is the same as the number of electrons. But what
53
3.7
CHAPTER 3. THE ATOM - GRADE 10
happens if an atom gains or loses electrons? Does it mean that the atom will still be part of the
same element?
A change in the number of electrons of an atom does not change the type of atom that it is.
However, the charge of the atom will change. If electrons are added, then the atom will become
more negative. If electrons are taken away, then the atom will become more positive. The atom
that is formed in either of these cases is called an ion. Put simply, an ion is a charged atom.
Definition: Ion
An ion is a charged atom. A positively charged ion is called a cation e.g. N a+ , and a
negatively charged ion is called an anion e.g. F − . The charge on an ion depends on the
number of electrons that have been lost or gained.
Look at the following examples. Notice the number of valence electrons in the neutral atom,
the number of electrons that are lost or gained, and the final charge of the ion that is formed.
Lithium
A lithium atoms loses one electrons to form a positive ion (figure 3.11).
1 electron lost
Li
Li+
Li atom with 3 electrons
Li+ ion with only 2 electrons
Figure 3.11: The arrangement of electrons in a lithium ion.
In this example, the lithium atom loses an electron to form the cation Li+ .
Fluorine
A fluorine atom gains one electron to form a negative ion (figure 3.12).
1 electron gained
F
F−
Figure 3.12: The arrangement of electrons in a fluorine ion.
Activity :: Investigation : The formation of ions
54
CHAPTER 3. THE ATOM - GRADE 10
3.7
1. Use the diagram for lithium as a guide and draw similar diagrams to show how
each of the following ions is formed:
(a) Mg2+
(b) Na+
(c) Cl−
(d) O2−
2. Do you notice anything interesting about the charge on each of these ions?
Hint: Look at the number of valence electrons in the neutral atom and the
charge on the final ion.
Observations:
Once you have completed the activity, you should notice that:
• In each case the number of electrons that is either gained or lost, is the same as the number
of electrons that are needed for the atoms to achieve a full or an empty valence energy
level.
• If you look at an energy level diagram for sodium (Na), you will see that in a neutral
atom, there is only one valence electron. In order to achieve an empty valence level, and
therefore a more stable state for the atom, this electron will be lost.
• In the case of oxygen (O), there are six valence electrons. To fill the valence energy level,
it makes more sense for this atom to gain two electrons. A negative ion is formed.
3.7.2
Ionisation Energy
Ionisation energy is the energy that is needed to remove one electron from an atom. The ionisation energy will be different for different atoms.
The second ionisation energy is the energy that is needed to remove a second electron from an
atom, and so on. As an energy level becomes more full, it becomes more and more difficult to
remove an electron and the ionisation energy increases. On the Periodic Table of the Elements,
a group is a vertical column of the elements, and a period is a horizontal row. In the periodic
table, ionisation energy increases across a period, but decreases as you move down a group.
The lower the ionisation energy, the more reactive the element will be because there is a greater
chance of electrons being involved in chemical reactions. We will look at this in more detail in
the next section.
Exercise: The formation of ions
Match the information in column A with the information in column B by writing
only the letter (A to I) next to the question number (1 to 7)
55
3.8
CHAPTER 3. THE ATOM - GRADE 10
1. A positive ion that has 3 less electrons
than its neutral atom
2. An ion that has 1 more electron than
its neutral atom
3.
The anion that is formed when
bromine gains an electron
4. The cation that is formed from a magnesium atom
5. An example of a compound ion
6. A positive ion with the electron configuration of argon
7. A negative ion with the electron configuration of neon
A. Mg2+
B. Cl−
C. CO2−
3
D. Al3+
E. Br2−
F. K+
G. Mg+
H. O2−
I. Br−
3.8
The Arrangement of Atoms in the Periodic Table
The periodic table of the elements is a tabular method of showing the chemical elements.
Most of the work that was done to arrive at the periodic table that we know, can be attributed to
a man called Dmitri Mendeleev in 1869. Mendeleev was a Russian chemist who designed the
table in such a way that recurring (”periodic”) trends in the properties of the elements could be
shown. Using the trends he observed, he even left gaps for those elements that he thought were
’missing’. He even predicted the properties that he thought the missing elements would have
when they were discovered. Many of these elements were indeed discovered and Mendeleev’s
predictions were proved to be correct.
To show the recurring properties that he had observed, Mendeleev began new rows in his table
so that elements with similar properties were in the same vertical columns, called groups. Each
row was referred to as a period. One important feature to note in the periodic table is that all
the non-metals are to the right of the zig-zag line drawn under the element boron. The rest of
the elements are metals, with the exception of hydrogen which occurs in the first block of the
table despite being a non-metal.
Group
group number
1
8
H
2
3
4
5
6
7
He
Li
Be
B
C
N
O
F
Ne
Na Mg
Al
Si
P
S
Cl
Ar
Cr Mn Fe Co Ni Cu Zn Ga Ge As Se
Br
Kr
K
Ca Sc
Ti
V
Period
Figure 3.13: A simplified diagram showing part of the Periodic Table
3.8.1
Groups in the periodic table
A group is a vertical column in the periodic table, and is considered to be the most important
way of classifying the elements. If you look at a periodic table, you will see the groups numbered
56
CHAPTER 3. THE ATOM - GRADE 10
3.8
at the top of each column. The groups are numbered from left to right as follows: 1, 2, then an
open space which contains the transition elements, followed by groups 3 to 8. These numbers
are normally represented using roman numerals. In some periodic tables, all the groups are
numbered from left to right from number 1 to number 18. In some groups, the elements display
very similar chemical properties, and the groups are even given separate names to identify them.
The characteristics of each group are mostly determined by the electron configuration of the
atoms of the element.
• Group 1: These elements are known as the alkali metals and they are very reactive.
Hydrogen
Lithium
Sodium
Potassium
Figure 3.14: Electron diagrams for some of the Group 1 elements
Activity :: Investigation : The properties of elements
Refer to figure 3.14.
1. Use a Periodic Table to help you to complete the last two diagrams for
sodium (Na) and potassium (K).
2. What do you notice about the number of electrons in the valence energy
level in each case?
3. Explain why elements from group 1 are more reactive than elements from
group 2 on the periodic table (Hint: Think back to ’ionisation energy’).
• Group 2: These elements are known as the alkali earth metals. Each element only has
two valence electrons and so in chemical reactions, the group 2 elements tend to lose these
electrons so that the energy shells are complete. These elements are less reactive than
those in group 1 because it is more difficult to lose two electrons than it is to lose one.
Group 3 elements have three valence electrons.
Important: The number of valence electrons of an element corresponds to its group number
on the periodic table.
• Group 7: These elements are known as the halogens. Each element is missing just one
electron from its outer energy shell. These elements tend to gain electrons to fill this shell,
rather than losing them.
• Group 8: These elements are the noble gases. All of the energy shells of the halogens are
full, and so these elements are very unreactive.
• Transition metals: The differences between groups in the transition metals are not usually
dramatic.
57
3.8
CHAPTER 3. THE ATOM - GRADE 10
Helium
Lithium
Figure 3.15: Electron diagrams for two of the noble gases, helium (He) and neon (Ne).
It is worth noting that in each of the groups described above, the atomic diameter of the
elements increases as you move down the group. This is because, while the number of valence
electrons is the same in each element, the number of core electrons increases as one moves down
the group.
3.8.2
Periods in the periodic table
A period is a horizontal row in the periodic table of the elements. Some of the trends that can
be observed within a period are highlighted below:
• As you move from one group to the next within a period, the number of valence electrons
increases by one each time.
• Within a single period, all the valence electrons occur in the same energy shell. If the
period increases, so does the energy shell in which the valence electrons occur.
• In general, the diameter of atoms decreases as one moves from left to right across a period.
Consider the attractive force between the positively charged nucleus and the negatively
charged electrons in an atom. As you move across a period, the number of protons in
each atom increases. The number of electrons also increases, but these electrons will still
be in the same energy shell. As the number of protons increases, the force of attraction
between the nucleus and the electrons will increase and the atomic diameter will decrease.
• Ionisation energy increases as one moves from left to right across a period. As the valence
electron shell moves closer to being full, it becomes more difficult to remove electrons. The
opposite is true when you move down a group in the table because more energy shells are
being added. The electrons that are closer to the nucleus ’shield’ the outer electrons from
the attractive force of the positive nucleus. Because these electrons are not being held
to the nucleus as strongly, it is easier for them to be removed and the ionisation energy
decreases.
• In general, the reactivity of the elements decreases from left to right across a period.
Exercise: Trends in ionisation energy
Refer to the data table below which gives the ionisation energy (in kJ/mol) and
atomic number (Z) for a number of elements in the periodic table:
58
CHAPTER 3. THE ATOM - GRADE 10
Z
1
2
3
4
5
6
7
8
9
Ionisation energy
1310
2360
517
895
797
1087
1397
1307
1673
3.9
Z
10
11
12
13
14
15
16
17
18
Ionisation energy
2072
494
734
575
783
1051
994
1250
1540
1. Draw a line graph to show the relationship between atomic number (on the
x-axis) and ionisation energy (y-axis).
2. Describe any trends that you observe.
3. Explain why...
(a) the ionisation energy for Z=2 is higher than for Z=1
(b) the ionisation energy for Z=3 is lower than for Z=2
(c) the ionisation energy increases between Z=5 and Z=7
Exercise: Elements in the Periodic Table
Refer to the elements listed below:
Lithium (Li); Chlorine (Cl); Magnesium (Mg); Neon (Ne); Oxygen (O); Calcium
(Ca); Carbon (C)
Which of the elements listed above:
1. belongs to Group 1
2. is a halogen
3. is a noble gas
4. is an alkali metal
5. has an atomic number of 12
6. has 4 neutrons in the nucleus of its atoms
7. contains electrons in the 4th energy level
8. has only one valence electron
9. has all its energy orbitals full
10. will have chemical properties that are most similar
11. will form positive ions
3.9
Summary
• Much of what we know today about the atom, has been the result of the work of a
number of scientists who have added to each other’s work to give us a good understanding
of atomic structure.
59
3.9
CHAPTER 3. THE ATOM - GRADE 10
• Some of the important scientific contributors include J.J.Thomson (discovery of the electron, which led to the Plum Pudding Model of the atom), Ernest Rutherford (discovery
that positive charge is concentrated in the centre of the atom) and Niels Bohr (the
arrangement of electrons around the nucleus in energy levels).
• Because of the very small mass of atoms, their mass in measured in atomic mass units
(u). 1 u = 1.67 × 10−24 g.
• An atom is made up of a central nucleus (containing protons and neutrons), surrounded
by electrons.
• The atomic number (Z) is the number of protons in an atom.
• The atomic mass number (A) is the number of protons and neutrons in the nucleus of
an atom.
• The standard notation that is used to write an element, is A
Z X, where X is the element
symbol, A is the atomic mass number and Z is the atomic number.
• The isotope of a particular element is made up of atoms which have the same number
of protons as the atoms in the original element, but a different number of neutrons. This
means that not all atoms of an element will have the same atomic mass.
• The relative atomic mass of an element is the average mass of one atom of all the
naturally occurring isotopes of a particular chemical element, expressed in atomic mass
units. The relative atomic mass is written under the elements’ symbol on the Periodic
Table.
• The energy of electrons in an atom is quantised. Electrons occur in specific energy levels
around an atom’s nucleus.
• Within each energy level, an electron may move within a particular shape of orbital. An
orbital defines the space in which an electron is most likely to be found. There are different
orbital shapes, including s, p, d and f orbitals.
• Energy diagrams such as Aufbau diagrams are used to show the electron configuration
of atoms.
• The electrons in the outermost energy level are called valence electrons.
• The electrons that are not valence electrons are called core electrons.
• Atoms whose outermost energy level is full, are less chemically reactive and therefore more
stable, than those atoms whose outer energy level is not full.
• An ion is a charged atom. A cation is a positively charged ion and an anion is a negatively
charged ion.
• When forming an ion, an atom will lose or gain the number of electrons that will make its
valence energy level full.
• An element’s ionisation energy is the energy that is needed to remove one electron from
an atom.
• Ionisation energy increases across a period in the Periodic Table.
• Ionisation energy decreases down a group in the Periodic Table.
Exercise: Summary
1. Write down only the word/term for each of the following descriptions.
(a) The sum of the number of protons and neutrons in an atom
60
CHAPTER 3. THE ATOM - GRADE 10
3.9
(b) The defined space around an atom’s nucleus, where an electron is most
likely to be found
2. For each of the following, say whether the statement is True or False. If it is
False, re-write the statement correctly.
22
(a) 20
10 Ne and 10 Ne each have 10 protons, 12 electrons and 12 neutrons.
(b) The atomic mass of any atom of a particular element is always the same.
(c) It is safer to use helium gas rather than hydrogen gas in balloons.
(d) Group 1 elements readily form negative ions.
3. Multiple choice questions: In each of the following, choose the one correct
answer.
(a) The three basic components of an atom are:
i. protons, neutrons, and ions
ii. protons, neutrons, and electrons
iii. protons, neutrinos, and ions
iv. protium, deuterium, and tritium
(b) The charge of an atom is...
i. positive
ii. neutral
iii. negative
(c) If Rutherford had used neutrons instead of alpha particles in his scattering
experiment, the neutrons would...
i. not deflect because they have no charge
ii. have deflected more often
iii. have been attracted to the nucleus easily
iv. have given the same results
(d) Consider the isotope 234
92 U. Which of the following statements is true?
i. The element is an isotope of 234
94 Pu
ii. The element contains 234 neutrons
iii. The element has the same electron configuration as 238
92 U
iv. The element has an atomic mass number of 92
(e) The electron configuration of an atom of chlorine can be represented using
the following notation:
i. 1s2 2s8 3s7
ii. 1s2 2s2 2p6 3s2 3p5
iii. 1s2 2s2 2p6 3s2 3p6
iv. 1s2 2s2 2p5
4. The following table shows the first ionisation energies for the elements of period
1 and 2.
Period
Element
First ionisation energy (kJ.mol−1 )
1
H
He
Li
Be
B
C
N
O
F
Ne
1312
2372
520
899
801
1086
1402
1314
1681
2081
2
(a) What is the meaning of the term first ionisation energy ?
(b) Identify the pattern of first ionisation energies in a period.
(c) Which TWO elements exert the strongest attractive forces on their electrons? Use the data in the table to give a reason for your answer.
61
3.9
CHAPTER 3. THE ATOM - GRADE 10
(d) Draw Aufbau diagrams for the TWO elements you listed in the previous
question and explain why these elements are so stable.
(e) It is safer to use helium gas than hydrogen gas in balloons. Which property
of helium makes it a safer option?
(f) ’Group 1 elements readily form positive ions’.
Is this statement correct? Explain your answer by referring to the table.
62
Chapter 4
Atomic Combinations - Grade 11
When you look at the matter around you, you will realise that atoms seldom exist on their own.
More often, the things around us are made up of different atoms that have been joined together.
This is called chemical bonding. Chemical bonding is one of the most important processes in
chemistry because it allows all sorts of different molecules and combinations of atoms to form,
which then make up the objects in the complex world around us. There are, however, some
atoms that do exist on their own, and which do not bond with others. The noble gases in
Group 8 of the Periodic Table, behave in this way. They include elements like neon (Ne), helium
(He) and argon (Ar). The important question then is, why do some atoms bond but others do
not?
4.1
Why do atoms bond?
As we begin this section, it’s important to remember that what we will go on to discuss is a model
of bonding, that is based on a particular model of the atom. You will remember from section
3.1 that a model is a representation of what is happening in reality. In the model of the atom
that has been used so far, the atom is made up of a central nucleus, surrounded by electrons
that are arranged in fixed energy levels (also sometimes called shells). Within each energy level,
electrons move in orbitals of different shapes. The electrons in the outermost energy level of an
atom are called the valence electrons. This model of the atom is useful in trying to understand
how different types of bonding take place between atoms.
You will remember from these earlier discussions of electrons and energy levels in the atom,
that electrons always try to occupy the lowest possible energy level. In the same way, an atom
also prefers to exist at the lowest possible energy state so that it is most stable. An atom is
most stable when all its valence electron orbitals are full. In other words, the outer energy level
of the atom contains the maximum number of electrons that it can. A stable atom is also an
unreactive one, and is unlikely to bond with other atoms. This explains why the noble gases
are unreactive and why they exist as atoms, rather than as molecules. Look for example at the
electron configuration of neon (1s2 2s2 3p6 ). Neon has eight valence electrons in its valence
energy shell. This is the maximum that it can hold and so neon is very stable and unreactive,
and will not form new bonds. Other atoms, whose valence energy levels are not full, are more
likely to bond in order to become more stable. We are going to look a bit more closely at some
of the energy changes that take place when atoms bond.
4.2
Energy and bonding
Let’s start by imagining that there are two hydrogen atoms approaching one another. As they
move closer together, there are three forces that act on the atoms at the same time. These
forces are shown in figure 4.1 and are described below:
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4.2
CHAPTER 4. ATOMIC COMBINATIONS - GRADE 11
(1)
(2)
+
(3)
+
Figure 4.1: Forces acting on two approaching atoms: (1) repulsion between electrons, (2)
attraction between protons and electrons and (3) repulsion between protons.
1. repulsive force between the electrons of the atoms, since like charges repel
2. attractive force between the nucleus of one atom and the electrons of another
3. repulsive force between the two positively-charged nuclei
Now look at figure 4.2 to understand the energy changes that take place when the two atoms
move towards each other.
+
Distance between atomic nuclei
Energy
P
A
0
Q
-
X
Figure 4.2: Graph showing the change in energy that takes place as atoms move closer together
In the example of the two hydrogen atoms, where the resultant force between them is attraction,
the energy of the system is zero when the atoms are far apart (point A), because there is no
interaction between the atoms. When the atoms are closer together, attractive forces dominate
and the atoms are pulled towards each other. As this happens, the potential energy of the
system decreases because energy would now need to be supplied to the system in order to move
the atoms apart. However, as the atoms move closer together (i.e. left along the horizontal
axis of the graph), repulsive forces start to dominate and this causes the potential energy of the
system to rise again. At some point, the attractive and repulsive effects are balanced, and the
energy of the system is at its minimum (point X). It is at this point, when the energy is at a
minimum, that bonding takes place.
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CHAPTER 4. ATOMIC COMBINATIONS - GRADE 11
4.3
The distance marked ’P’ is the bond length, i.e. the distance between the nuclei of the atoms
when they bond. ’Q’ represents the bond energy i.e. the amount of energy that must be added
to the system to break the bonds that have formed. Bond strength means how strongly one
atom attracts and is held to another. The strength of a bond is related to the bond length, the
size of the bonded atoms and the number of bonds between the atoms. In general, the shorter
the bond length, the stronger the bond between the atoms, and the smaller the atoms involved,
the stronger the bond. The greater the number of bonds between atoms, the greater will be the
bond strength.
4.3
What happens when atoms bond?
A chemical bond is formed when atoms are held together by attractive forces. This attraction
occurs when electrons are shared between atoms, or when electrons are exchanged between the
atoms that are involved in the bond. The sharing or exchange of electrons takes place so that the
outer energy levels of the atoms involved are filled and the atoms are more stable. If an electron
is shared, it means that it will spend its time moving in the electron orbitals around both atoms.
If an electron is exchanged it means that it is transferred from one atom to another, in other
words one atom gains an electron while the other loses an electron.
Definition: Chemical bond
A chemical bond is the physical process that causes atoms and molecules to be attracted
to each other, and held together in more stable chemical compounds.
The type of bond that is formed depends on the elements that are involved. In this section, we
will be looking at three types of chemical bonding: covalent, ionic and metallic bonding.
You need to remember that it is the valence electrons that are involved in bonding and that
atoms will try to fill their outer energy levels so that they are more stable.
4.4
4.4.1
Covalent Bonding
The nature of the covalent bond
Covalent bonding occurs between the atoms of non-metals. The outermost orbitals of the atoms
overlap so that unpaired electrons in each of the bonding atoms can be shared. By overlapping
orbitals, the outer energy shells of all the bonding atoms are filled. The shared electrons move in
the orbitals around both atoms. As they move, there is an attraction between these negatively
charged electrons and the positively charged nuclei, and this force holds the atoms together in a
covalent bond.
Definition: Covalent bond
Covalent bonding is a form of chemical bonding where pairs of electrons are shared between
atoms.
Below are a few examples. Remember that it is only the valence electrons that are involved in
bonding, and so when diagrams are drawn to show what is happening during bonding, it is only
these electrons that are shown. Circles and crosses represent electrons in different atoms.
Worked Example 5: Covalent bonding
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4.4
CHAPTER 4. ATOMIC COMBINATIONS - GRADE 11
Question: How do hydrogen and chlorine atoms bond covalently in a molecule of
hydrogen chloride?
Answer
Step 1 : Determine the electron configuration of each of the bonding atoms.
A chlorine atom has 17 electrons, and an electron configuration of 1s2 2s2 2p6 3s2
3p5 . A hydrogen atom has only 1 electron, and an electron configuration of 1s1 .
Step 2 : Determine the number of valence electrons for each atom, and how
many of the electrons are paired or unpaired.
Chlorine has 7 valence electrons. One of these electrons is unpaired. Hydrogen has
1 valence electron and it is unpaired.
Step 3 : Look to see how the electrons can be shared between the atoms so
that the outermost energy levels of both atoms are full.
The hydrogen atom needs one more electron to complete its valence shell. The
chlorine atom also needs one more electron to complete its shell. Therefore one pair
of electrons must be shared between the two atoms. In other words, one electron
from the chlorine atom will spend some of its time orbiting the hydrogen atom so
that hydrogen’s valence shell is full. The hydrogen electron will spend some of its
time orbiting the chlorine atom so that chlorine’s valence shell is also full. A molecule
of hydrogen chloride is formed (figure 4.3). Notice the shared electron pair in the
overlapping orbitals.
paired electrons in valence energy level
xx
H
+
x
Cl
xx
unpaired electrons
xx
x
x
H
x
Cl
x
x
xx
overlap of electron orbitals and
sharing of electron pair
Figure 4.3: Covalent bonding in a molecule of hydrogen chloride
Worked Example 6: Covalent bonding involving multiple bonds
Question: How do nitrogen and hydrogen atoms bond to form a molecule of ammonia (NH3 )?
Answer
Step 1 : Determine the electron configuration of each of the bonding atoms.
A nitrogen atom has 7 electrons, and an electron configuration of 1s2 2s2 2p3 . A
hydrogen atom has only 1 electron, and an electron configuration of 1s1 .
Step 2 : Determine the number of valence electrons for each atom, and how
many of the electrons are paired or unpaired.
Nitrogen has 5 valence electrons meaning that 3 electrons are unpaired. Hydrogen
has 1 valence electron and it is unpaired.
Step 3 : Look to see how the electrons can be shared between the atoms so
that the outer energy shells of all atoms are full.
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CHAPTER 4. ATOMIC COMBINATIONS - GRADE 11
4.4
Each hydrogen atom needs one more electron to complete its valence energy shell.
The nitrogen atom needs three more electrons to complete its valence energy shell.
Therefore three pairs of electrons must be shared between the four atoms involved.
The nitrogen atom will share three of its electrons so that each of the hydrogen
atoms now has a complete valence shell. Each of the hydrogen atoms will share its
electron with the nitrogen atom to complete its valence shell (figure 4.4).
xx
3
H
+
x
N
x
xx
x
H
x
N
x
H
x
H
Figure 4.4: Covalent bonding in a molecule of ammonia
The above examples all show single covalent bonds, where only one pair of electrons is shared
between the same two atoms. If two pairs of electrons are shared between the same two atoms,
this is called a double bond. A triple bond is formed if three pairs of electrons are shared.
Worked Example 7: Covalent bonding involving a double bond
Question: How do oxygen atoms bond covalently to form an oxygen molecule?
Answer
Step 1 : Determine the electron configuration of the bonding atoms.
Each oxygen atom has 8 electrons, and their electron configuration is 1s2 2s2 2p4 .
Step 2 : Determine the number of valence electrons for each atom and how
many of these electrons are paired and unpaired.
Each oxygen atom has 6 valence electrons, meaning that each atom has 2 unpaired
electrons.
Step 3 : Look to see how the electrons can be shared between atoms so that
the outer energy shells of all the atoms are full.
Each oxygen atom needs two more electrons to complete its valence energy shell.
Therefore two pairs of electrons must be shared between the two oxygen atoms so
that both valence shells are full. Notice that the two electron pairs are being shared
between the same two atoms, and so we call this a double bond (figure 4.5).
You will have noticed in the above examples that the number of electrons that are involved in
bonding varies between atoms. We say that the valency of the atoms is different.
Definition: Valency
The number of electrons in an atom which are used to form a bond.
67
4.4
CHAPTER 4. ATOMIC COMBINATIONS - GRADE 11
xx
x
x
O
xx
x+
x
x
O
O
x
x
O
x
Figure 4.5: A double covalent bond in an oxygen molecule
In the first example, the valency of both hydrogen and chlorine is one, therefore there is a single
covalent bond between these two atoms. In the second example, nitrogen has a valency of three
and hydrogen has a valency of one. This means that three hydrogen atoms will need to bond
with a single nitrogen atom. There are three single covalent bonds in a molecule of ammonia.
In the third example, the valency of oxygen is two. This means that each oxygen atom will form
two bonds with another atom. Since there is only one other atom in a molecule of O2 , a double
covalent bond is formed between these two atoms.
Important: There is a relationship between the valency of an element and its position on
the Periodic Table. For the elements in groups 1 to 4, the valency is the same as the group
number. For elements in groups 5 to 7, the valency is calculated by subtracting the group
number from 8. For example, the valency of fluorine (group 7) is 8-7=1, while the valency
of calcium (group 2) is 2. Some elements have more than one possible valency, so you
always need to be careful when you are writing a chemical formula. Often, the valency will
be written in a bracket after the element symbol e.g. carbon (iv) oxide, means that in this
molecule carbon has a valency of 4.
Exercise: Covalent bonding and valency
1. Explain the difference between the valence electrons and the valency of an
element.
2. Complete the table below by filling in the number of valence electrons and the
valency for each of the elements shown:
Element
No. of valence
electrons
No.
of electrons needed to
fill outer shell
Valency
Mg
K
F
Ar
C
N
O
3. Draw simple diagrams to show how electrons are arranged in the following
covalent molecules:
(a) Calcium oxide (CaO)
(b) Water (H2 O)
(c) Chlorine (Cl2 )
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CHAPTER 4. ATOMIC COMBINATIONS - GRADE 11
4.5
4.5
Lewis notation and molecular structure
Although we have used diagrams to show the structure of molecules, there are other forms of
notation that can be used, such as Lewis notation and Couper notation. Lewis notation uses
dots and crosses to represent the valence electrons on different atoms. The chemical symbol
of the element is used to represent the nucleus and the core electrons of the atom.
So, for example, a hydrogen atom would be represented like this:
H
•
A chlorine atom would look like this:
××
××
×
Cl
××
A molecule of hydrogen chloride would be shown like this:
Cl
××
××
ו
H
××
The dot and cross inbetween the two atoms, represent the pair of electrons that are shared in
the covalent bond.
Worked Example 8: Lewis notation: Simple molecules
Question: Represent the molecule H2 O using Lewis notation
Answer
Step 1 : For each atom, determine the number of valence electrons in the
atom, and represent these using dots and crosses.
The electron configuration of hydrogen is 1s1 and the electron configuration for
oxygen is 1s2 2s2 2p4 . Each hydrogen atom has one valence electron, which is
unpaired, and the oxygen atom has six valence electrons with two unpaired.
O
×
××
××
×
2H•
Step 2 : Arrange the electrons so that the outermost energy level of each
atom is full.
The water molecule is represented below.
××
××
ו
H O
ו
H
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CHAPTER 4. ATOMIC COMBINATIONS - GRADE 11
Worked Example 9: Lewis notation: Molecules with multiple bonds
Question: Represent the molecule HCN using Lewis notation
Answer
Step 1 : For each atom, determine the number of valence electrons that the
atom has from its electron configuration.
The electron configuration of hydrogen is 1s1 , the electron configuration of nitrogen
is 1s2 2s2 2p3 and for carbon is 1s2 2s2 2p2 . This means that hydrogen has one
valence electron which is unpaired, carbon has four valence electrons, all of which
are unpaired, and nitrogen has five valence electrons, three of which are unpaired.
•
N
×
×
••
C
•
×
H•
•
×
N
••
C
•••
ו
H
×××
Step 2 : Arrange the electrons in the HCN molecule so that the outermost
energy level in each atom is full.
The HCN molecule is represented below. Notice the three electron pairs between
the nitrogen and carbon atom. Because these three covalent bonds are between the
same two atoms, this is a triple bond.
Worked Example 10: Lewis notation: Atoms with variable valencies
Question: Represent the molecule H2 S using Lewis notation
Answer
Step 1 : Determine the number of valence electrons for each atom.
Hydrogen has an electron configuration of 1s1 and sulfur has an electron configuration of 1s2 2s2 2p6 3s2 3p4 . Each hydrogen atom has one valence electron which is
unpaired, and sulfur has six valence electrons. Although sulfur has a variable valency,
we know that the sulfur will be able to form 2 bonds with the hydrogen atoms. In
this case, the valency of sulfur must be two.
S
×
××
××
×
2H•
Step 2 : Arrange the atoms in the molecule so that the outermost energy
level in each atom is full.
The H2 S molecule is represented below.
S
××
××
ו
H
ו
H
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CHAPTER 4. ATOMIC COMBINATIONS - GRADE 11
4.5
Another way of representing molecules is using Couper notation. In this case, only the electrons
that are involved in the bond between the atoms are shown. A line is used for each covalent
bond. Using Couper notation, a molecule of water and a molecule of HCN would be represented
as shown in figures 4.6 and 4.7 below.
H O
H
Figure 4.6: A water molecule represented using Couper notation
H
C
N
Figure 4.7: A molecule of HCN represented using Couper notation
Extension: Dative covalent bonds
A dative covalent bond (also known as a coordinate covalent bond) is a description of covalent bonding between two atoms in which both electrons shared in
the bond come from the same atom. This happens when a Lewis base (an electron
donor) donates a pair of electrons to a Lewis acid (an electron acceptor). Lewis
acids and bases will be discussed in section 15.1 in chapter 15.
One example of a molecule that contains a dative covalent bond is the ammonium
+
ion (NH+
4 ) shown in the figure below. The hydrogen ion H does not contain any
electrons, and therefore the electrons that are in the bond that forms between this
ion and the nitrogen atom, come only from the nitrogen.
H
××
××
××
H
××
ו
ו
ו
H N H + [H]+
H N H
ו
H
Exercise: Atomic bonding and Lewis notation
1. Represent each of the following atoms using Lewis notation:
(a) beryllium
(b) calcium
(c) lithium
2. Represent each of the following molecules using Lewis notation:
(a) bromine gas (Br2 )
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4.6
CHAPTER 4. ATOMIC COMBINATIONS - GRADE 11
(b) calcium chloride (CaCl2 )
(c) carbon dioxide (CO2 )
3. Which of the three molecules listed above contains a double bond?
4. Two chemical reactions are described below.
• nitrogen and hydrogen react to form ammonia
• carbon and hydrogen bond to form a molecule of methane (CH4 )
For each reaction, give:
(a)
(b)
(c)
(d)
the
the
the
the
valency of each of the atoms involved in the reaction
Lewis structure of the product that is formed
chemical formula of the product
name of the product
5. A chemical compound has the following Lewis notation:
Y
××
××
ו
X
ו
H
(a)
(b)
(c)
(d)
(e)
4.6
How many valence electrons does element Y have?
What is the valency of element Y?
What is the valency of element X?
How many covalent bonds are in the molecule?
Suggest a name for the elements X and Y.
Electronegativity
Electronegativity is a measure of how strongly an atom pulls a shared electron pair towards it.
The table below shows the electronegativities of a number of elements:
Table 4.1: Table of electronegativities for selected elements
Element
Electronegativity
Hydrogen (H)
2.1
Sodium (Na)
0.9
Magnesium (Mg)
1.2
Calcium (Ca)
1.0
Chlorine (Cl)
3.0
Bromine (Br)
2.8
Definition: Electronegativity
Electronegativity is a chemical property which describes the power of an atom to attract
electrons towards itself.
The greater the electronegativity of an element, the stronger its attractive pull on electrons.
For example, in a molecule of hydrogen bromide (HBr), the electronegativity of bromine (2.8)
is higher than that of hydrogen (2.1), and so the shared electrons will spend more of their time
closer to the bromine atom. Bromine will have a slightly negative charge, and hydrogen will have
a slightly positive charge. In a molecule like hydrogen (H2 ) where the electronegativities of the
atoms in the molecule are the same, both atoms have a neutral charge.
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CHAPTER 4. ATOMIC COMBINATIONS - GRADE 11
4.6
teresting The concept of electronegativity was introduced by Linus Pauling in 1932,
Interesting
Fact
Fact
and this became very useful in predicting the nature of bonds between atoms
in molecules. In 1939, he published a book called ’The Nature of the Chemical
Bond’, which became one of the most influential chemistry books ever published.
For this work, Pauling was awarded the Nobel Prize in Chemistry in 1954. He
also received the Nobel Peace Prize in 1962 for his campaign against aboveground nuclear testing.
4.6.1
Non-polar and polar covalent bonds
Electronegativity can be used to explain the difference between two types of covalent bonds.
Non-polar covalent bonds occur between two identical non-metal atoms, e.g. H2 , Cl2 and O2 .
Because the two atoms have the same electronegativity, the electron pair in the covalent bond is
shared equally between them. However, if two different non-metal atoms bond then the shared
electron pair will be pulled more strongly by the atom with the highest electronegativity. As a
result, a polar covalent bond is formed where one atom will have a slightly negative charge and
the other a slightly positive charge. This is represented using the symbols δ + (slightly positive)
+
and δ − (slightly negative). So, in a molecule such as hydrogen chloride (HCl), hydrogen is H δ
−
and chlorine is Clδ .
4.6.2
Polar molecules
Some molecules with polar covalent bonds are polar molecules, e.g. H2 O. But not all molecules
with polar covalent bonds are polar. An example is CO2 . Although CO2 has two polar covalent
bonds (between C+ atom and the two O− atoms), the molecule itself is not polar. The reason
is that CO2 is a linear molecule and is therefore symmetrical. So there is no difference in charge
between the two ends of the molecule. The polarity of molecules affects properties such as
solubility, melting points and boiling points.
Definition: Polar and non-polar molecules
A polar molecule is one that has one end with a slightly positive charge, and one end with
a slightly negative charge. A non-polar molecule is one where the charge is equally spread
across the molecule.
Exercise: Electronegativity
1. In a molecule of hydrogen chloride (HCl),
(a) What is the electronegativity of hydrogen
(b) What is the electronegativity of chlorine?
(c) Which atom will have a slightly positive charge and which will have a
slightly negative charge in the molecule?
(d) Is the bond a non-polar or polar covalent bond?
(e) Is the molecule polar or non-polar?
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CHAPTER 4. ATOMIC COMBINATIONS - GRADE 11
2. Complete the table below:
Molecule Difference
in
electronegativity
between atoms
H2 O
HBr
NO2
F2
CH4
4.7
4.7.1
Non-polar/polar
covalent bond
Polar/non-polar
molecule
Ionic Bonding
The nature of the ionic bond
You will remember that when atoms bond, electrons are either shared or they are transferred
between the atoms that are bonding. In covalent bonding, electrons are shared between the
atoms. There is another type of bonding, where electrons are transferred from one atom to
another. This is called ionic bonding.
Ionic bonding takes place when the difference in electronegativity between the two atoms is more
than 1.7. This usually happens when a metal atom bonds with a non-metal atom. When the
difference in electronegativity is large, one atom will attract the shared electron pair much more
strongly than the other, causing electrons to be transferred from one atom to the other.
Definition: Ionic bond
An ionic bond is a type of chemical bond based on the electrostatic forces between two
oppositely-charged ions. When ionic bonds form, a metal donates an electron, due to a
low electronegativity, to form a positive ion or cation. The non-metal atom has a high
electronegativity, and therefore readily gains electrons to form negative ions or anions. The
two or more ions are then attracted to each other by electrostatic forces.
Example 1:
In the case of NaCl, the difference in electronegativity is 2.1. Sodium has only one valence
electron, while chlorine has seven. Because the electronegativity of chlorine is higher than the
electronegativity of sodium, chlorine will attract the valence electron in the sodium atom very
strongly. This electron from sodium is transferred to chlorine. Sodium has lost an electron and
forms a N a+ ion. Chlorine gains an electron and forms a Cl− ion. The attractive force between
the positive and negative ion is what holds the molecule together.
The balanced equation for the reaction is:
N a + Cl → N aCl
This can be represented using Lewis notation:
Example 2:
Another example of ionic bonding takes place between magnesium (Mg) and oxygen (O) to
form magnesium oxide (MgO). Magnesium has two valence electrons and an electronegativity
of 1.2, while oxygen has six valence electrons and an electronegativity of 3.5. Since oxygen has
74
CHAPTER 4. ATOMIC COMBINATIONS - GRADE 11
Na
• +×
××
Cl
××
4.7
+
×
×
××
[Na] [ Cl ××]−
×
•
electron transer from
sodium to chlorine
××
Figure 4.8: Ionic bonding in sodium chloride
a higher electronegativity, it attracts the two valence electrons from the magnesium atom and
these electrons are transferred from the magnesium atom to the oxygen atom. Magnesium loses
two electrons to form M g 2+ , and oxygen gains two electrons to form O2− . The attractive force
between the oppositely charged ions is what holds the molecule together.
The balanced equation for the reaction is:
2M g + O2 → 2M gO
Because oxygen is a diatomic molecule, two magnesium atoms will be needed to combine with
two oxygen atoms to produce two molecules of magnesium oxide (MgO).
•
Mg •
×
××
O×
2+
×
×
[Mg]
two electrons transferred
from Mg to O
××
[ O ×× ]2−
×
•
• ×
Figure 4.9: Ionic bonding in magnesium oxide
Important: Notice that the number of electrons that is either lost or gained by an atom
during ionic bonding, is the same as the valency of that element
Exercise: Ionic compounds
1. Explain the difference between a covalent and an ionic bond.
2. Magnesium and chlorine react to form magnesium chloride.
(a) What is the difference in electronegativity between these two elements?
(b) Give the chemical formula for:
• a magnesium ion
• a choride ion
• the ionic compound that is produced during this reaction
(c) Write a balanced chemical equation for the reaction that takes place.
3. Draw Lewis diagrams to represent the following ionic compounds:
(a) sodium iodide (NaI)
(b) calcium bromide (CaBr2 )
(c) potassium chloride (KCl)
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CHAPTER 4. ATOMIC COMBINATIONS - GRADE 11
4.7.2
The crystal lattice structure of ionic compounds
Ionic substances are actually a combination of lots of ions bonded together into a giant molecule.
The arrangement of ions in a regular, geometric structure is called a crystal lattice. So in fact
NaCl does not contain one Na and one Cl ion, but rather a lot of these two ions arranged in a
crystal lattice where the ratio of Na to Cl ions is 1:1. The structure of a crystal lattice is shown
in figure 4.10.
b
c
b
b
c
b
b
c
b
b
c
b
b
bc
b
bc
b
bc
b
bc
b
bc
b
bc
b
bc
b
bc
b
atom of element 1 (e.g. Na)
bc
atom of element 2 (e.g. Cl)
b
ionic bonds hold atoms together
in the lattice structure
Figure 4.10: The crystal lattice arrangement in an ionic compound (e.g. NaCl)
4.7.3
Properties of Ionic Compounds
Ionic compounds have a number of properties:
• Ions are arranged in a lattice structure
• Ionic solids are crystalline at room temperature
• The ionic bond is a strong electrical attraction. This means that ionic compounds are
often hard and have high melting and boiling points
• Ionic compounds are brittle, and bonds are broken along planes when the compound is
stressed
• Solid crystals don’t conduct electricity, but ionic solutions do
4.8
4.8.1
Metallic bonds
The nature of the metallic bond
The structure of a metallic bond is quite different from covalent and ionic bonds. In a metal
bond, the valence electrons are delocalised, meaning that an atom’s electrons do not stay around
that one nucleus. In a metallic bond, the positive atomic nuclei (sometimes called the ’atomic
kernels’) are surrounded by a sea of delocalised electrons which are attracted to the nuclei (figure
4.11).
Definition: Metallic bond
Metallic bonding is the electrostatic attraction between the positively charged atomic nuclei
of metal atoms and the delocalised electrons in the metal.
76
CHAPTER 4. ATOMIC COMBINATIONS - GRADE 11
bc
bc
bc
b
bc
+
b
+
b
c
+b
b
+
+
b
bc
+
bc
+
b
c
b
b
bc
+
b
bc
+
b
b
b
b
b
b
b
+
b
b
c
+b
c
+b
b
b
c
+b
c
+b
b
bc
+
bc
4.8
+
b
b
bc
b
bc
+
b
b
b
b
b
bc
b
c
+b
b
+
b
+
b
b
c
b
c
b
b
bc
c
b
+
+
b
+
b
c
b
+
c
b
c
b
+
+
+
Figure 4.11: Positive atomic nuclei (+) surrounded by delocalised electrons (•)
4.8.2
The properties of metals
Metals have several unique properties as a result of this arrangement:
• Thermal conductors
Metals are good conductors of heat and are therefore used in cooking utensils such as
pots and pans. Because the electrons are loosely bound and are able to move, they can
transport heat energy from one part of the material to another.
• Electrical conductors
Metals are good conductors of electricity, and are therefore used in electrical conducting
wires. The loosely bound electrons are able to move easily and to transfer charge from
one part of the material to another.
• Shiny metallic lustre
Metals have a characteristic shiny appearance and are often used to make jewellery. The
loosely bound electrons are able to absorb and reflect light at all frequencies, making metals
look polished and shiny.
• Malleable and ductile
This means that they can be bent into shape without breaking (malleable) and can be
stretched into thin wires (ductile) such as copper, which can then be used to conduct
electricity. Because the bonds are not fixed in a particular direction, atoms can slide easily
over one another, making metals easy to shape, mould or draw into threads.
• Melting point
Metals usually have a high melting point and can therefore be used to make cooking pots
and other equipment that needs to become very hot, without being damaged. The high
melting point is due to the high strength of metallic bonds.
• Density
Metals have a high density because their atoms are packed closely together.
Exercise: Chemical bonding
1. Give two examples of everyday objects that contain..
(a) covalent bonds
(b) ionic bonds
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4.9
CHAPTER 4. ATOMIC COMBINATIONS - GRADE 11
(c) metallic bonds
2. Complete the table which compares the different types of bonding:
Covalent
Ionic
Metallic
Types of atoms involved
Nature of bond between atoms
Melting Point (high/low)
Conducts electricity? (yes/no)
Other properties
3. Complete the table below by identifying the type of bond (covalent, ionic or
metallic) in each of the compounds:
Molecular formula
H2 SO4
FeS
NaI
MgCl2
Zn
Type of bond
4. Which of these substances will conduct electricity most effectively? Give a
reason for your answer.
5. Use your knowledge of the different types of bonding to explain the following
statements:
(a) Swimming during an electric thunderstorm (i.e. where there is lightning)
can be very dangerous.
(b) Most jewellery items are made from metals.
(c) Plastics are good insulators.
4.9
Writing chemical formulae
4.9.1
The formulae of covalent compounds
To work out the formulae of covalent compounds, we need to use the valency of the atoms in the
compound. This is because the valency tells us how many bonds each atom can form. This in
turn can help to work out how many atoms of each element are in the compound, and therefore
what its formula is. The following are some examples where this information is used to write the
chemical formula of a compound.
Worked Example 11: Formulae of covalent compounds
Question: Write the chemical formula for water
Answer
Step 1 : Write down the elements that make up the compound.
A molecule of water contains the elements hydrogen and oxygen.
Step 2 : Determine the valency of each element
The valency of hydrogen is 1 and the valency of oxygen is 2. This means that oxygen
can form two bonds with other elements and each of the hydrogen atoms can form
one.
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CHAPTER 4. ATOMIC COMBINATIONS - GRADE 11
Step 3 : Write the chemical formula
Using the valencies of hydrogen and oxygen, we know that in a single water molecule,
two hydrogen atoms will combine with one oxygen atom. The chemical formula for
water is therefore:
H2 O.
Worked Example 12: Formulae of covalent compounds
Question: Write the chemical formula for magnesium oxide
Answer
Step 1 : Write down the elements that make up the compound.
A molecule of magnesium oxide contains the elements magnesium and oxygen.
Step 2 : Determine the valency of each element
The valency of magnesium is 2, while the valency of oxygen is also 2. In a molecule
of magnesium oxide, one atom of magnesium will combine with one atom of oxygen.
Step 3 : Write the chemical formula
The chemical formula for magnesium oxide is therefore:
MgO
Worked Example 13: Formulae of covalent compounds
Question: Write the chemical formula for copper (II) chloride.
Answer
Step 1 : Write down the elements that make up the compound.
A molecule of copper (II) chloride contains the elements copper and chlorine.
Step 2 : Determine the valency of each element
The valency of copper is 2, while the valency of chlorine is 1. In a molecule of copper
(II) chloride, two atoms of chlorine will combine with one atom of copper.
Step 3 : Write the chemical formula
The chemical formula for copper (II) chloride is therefore:
CuCl2
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4.9
4.9
CHAPTER 4. ATOMIC COMBINATIONS - GRADE 11
4.9.2
The formulae of ionic compounds
The overall charge of an ionic compound will always be zero and so the negative and positive
charge must be the same size. We can use this information to work out what the chemical
formula of an ionic compound is if we know the charge on the individual ions. In the case of
NaCl for example, the charge on the sodium is +1 and the charge on the chlorine is -1. The
charges balance (+1-1=0) and therefore the ionic compound is neutral. In MgO, magnesium has
a charge of +2 and oxygen has a charge of -2. Again, the charges balance and the compound is
neutral. Positive ions are called cations and negative ions are called anions.
Some ions are made up of groups of atoms, and these are called compound ions. It is a good
idea to learn the compound ions that are shown in table 4.2
Table 4.2: Table showing common compound ions and their formulae
Name of compound ion
Carbonate
sulphate
Hydroxide
Ammonium
Nitrate
Hydrogen carbonate
Phosphate
Chlorate
Cyanide
Chromate
Permanganate
formula
CO3 2−
SO4 2−
OH−
NH4 +
NO3 −
HCO3 −
PO4 3−
ClO3 −
CN−
CrO4 2−
MnO4 −
In the case of ionic compounds, the valency of an ion is the same as its charge (Note: valency
is always expressed as a positive number e.g. valency of the chloride ion is 1 and not -1). Since
an ionic compound is always neutral, the positive charges in the compound must balance out
the negative. The following are some examples:
Worked Example 14: Formulae of ionic compounds
Question: Write the chemical formula for potassium iodide.
Answer
Step 1 : Write down the ions that make up the compound.
Potassium iodide contains potassium and iodide ions.
Step 2 : Determine the valency and charge of each ion.
Potassium iodide contains the ions K+ (valency = 1; charge = +1) and I− (valency
= 1; charge = -1). In order to balance the charge in a single molecule, one atom of
potassium will be needed for every one atom of iodine.
Step 3 : Write the chemical formula
The chemical formula for potassium iodide is therefore:
KI
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CHAPTER 4. ATOMIC COMBINATIONS - GRADE 11
4.9
Worked Example 15: Formulae of ionic compounds
Question: Write the chemical formula for sodium sulphate.
Answer
Step 1 : Write down the ions that make up the compound.
Sodium sulphate contains sodium ions and sulphate ions.
Step 2 : Determine the valency and charge of each ion.
Na+ (valency = 1; charge = +1) and SO4 2− (valency = 2; charge = -2).
Step 3 : Write the chemical formula.
Two sodium ions will be needed to balance the charge of the sulphate ion. The
chemical formula for sodium sulphate is therefore:
Na2 SO4 2−
Worked Example 16: Formulae of ionic compounds
Question: Write the chemical formula for calcium hydroxide.
Answer
Step 1 : Write down the ions that make up the compound.
Calcium hydroxide contains calcium ions and hydroxide ions.
Step 2 : Determine the valency and charge of each ion.
Calcium hydroxide contains the ions Ca2+ (charge = +2) and OH− (charge = -1).
In order to balance the charge in a single molecule, two hydroxide ions will be needed
for every ion of calcium.
Step 3 : Write the chemical formula.
The chemical formula for calcium hydroxide is therefore:
Ca(OH)2
Exercise: Chemical formulae
1. Copy and complete the table below:
Compound
Cation
Na+
potassium bromide
NH+
4
Anion
Cl−
Br−
Cl−
Formula
potassium chromate
PbI
potassium permanganate
calcium phosphate
81
4.10
CHAPTER 4. ATOMIC COMBINATIONS - GRADE 11
2. Write the chemical formula for each of the following compounds:
(a)
(b)
(c)
(d)
(e)
(f)
hydrogen cyanide
carbon dioxide
sodium carbonate
ammonium hydroxide
barium sulphate
potassium permanganate
4.10
The Shape of Molecules
4.10.1
Valence Shell Electron Pair Repulsion (VSEPR) theory
The shape of a covalent molecule can be predicted using the Valence Shell Electron Pair Repulsion (VSEPR) theory. This is a model in chemistry that tries to predict the shapes of molecules.
Very simply, VSEPR theory says that the electron pairs in a molecule will arrange themselves
around the central atom of the molecule so that the repulsion between their negative charges is
as small as possible. In other words, the electron pairs arrange themselves so that they are as far
apart as they can be. Depending on the number of electron pairs in the molecule, it will have a
different shape.
Definition: Valence Shell Electron Pair Repulsion Theory
Valence shell electron pair repulsion theory (VSEPR) is a model in chemistry, which is used
to predict the shape of individual molecules, based upon the extent of their electron-pair
repulsion.
VSEPR theory is based on the idea that the geometry of a molecule is mostly determined by
repulsion among the pairs of electrons around a central atom. The pairs of electrons may
be bonding or non-bonding (also called lone pairs). Only valence electrons of the central
atom influence the molecular shape in a meaningful way.
4.10.2
Determining the shape of a molecule
To predict the shape of a covalent molecule, follow these steps:
Step 1:
Draw the molecule using Lewis notation. Make sure that you draw all the electrons around the
molecule’s central atom.
Step 2:
Count the number of electron pairs around the central atom.
Step 3:
Determine the basic geometry of the molecule using the table below. For example, a molecule
with two electron pairs around the central atom has a linear shape, and one with four electron
pairs around the central atom would have a tetrahedral shape. The situation is actually more
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CHAPTER 4. ATOMIC COMBINATIONS - GRADE 11
4.10
Table 4.3: The effect of electron pairs in determining the shape of molecules
Number of electron pairs Geometry
2
linear
3
trigonal planar
4
tetrahedral
5
trigonal bipyramidal
6
octahedral
complicated than this, but this will be discussed later in this section.
Figure 4.12 shows each of these shapes. Remember that the shapes are 3-dimensional, and so
you need to try to imagine them in this way. In the diagrams, the shaded part represents those
parts of the molecule that are ’in front’, while the dashed lines represent those parts that are ’at
the back’ of the molecule.
180◦
120◦
109◦
linear
trigonal planar
tetrahedral
90◦
90◦
octahedral
trigonal hipyramidal
Figure 4.12: Some common molecular shapes
Worked Example 17: The shapes of molecules
Question: Determine the shape of a molecule of O2
Answer
Step 1 : Draw the molecule using Lewis notation
O
••
××
O
••
••
××
××
Step 2 : Count the number of electron pairs around the central atom
There are two electron pairs.
Step 3 : Determine the basic geometry of the molecule
Since there are two electron pairs, the molecule must be linear.
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CHAPTER 4. ATOMIC COMBINATIONS - GRADE 11
Worked Example 18: The shapes of molecules
Question: Determine the shape of a molecule of BF3
Answer
Step 1 : Draw the molecule using Lewis notation
••
••
B
••
••
F
F
•×
ו
••
••
ו
••
••
F
••
Step 2 : Count the number of electron pairs around the central atom
There are three electron pairs.
Step 3 : Determine the basic geometry of the molecule
Since there are three electron pairs, the molecule must be trigonal planar.
Extension: More about molecular shapes
Determining the shape of a molecule can be a bit more complicated. In the
examples we have used above, we looked only at the number of bonding electron
pairs when we were trying to decide on the molecules’ shape. But there are also
other electron pairs in the molecules. These electrons, which are not involved in
bonding but which are also around the central atom, are called lone pairs. The
worked example below will give you an indea of how these lone pairs can affect the
shape of the molecule.
Worked Example 19: Advanced
Question: Determine the shape of a molecule of N H3
Answer
Step 1 : Draw the molecule using Lewis notation
lone pair of electrons
ו
××
ו
4.10
H N H
ו
H
Step 2 : Count the number of electron pairs around the central atom
There are four electron pairs.
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CHAPTER 4. ATOMIC COMBINATIONS - GRADE 11
4.11
Step 3 : Determine the basic geometry of the molecule
Since there are four electron pairs, the molecule must be tetrahedral.
Step 4 : Determine how many lone pairs are around the central atom
There is one lone pair of electrons and this will affect the shape of the molecule.
Step 5 : Determine the final shape of the molecule
The lone pair needs more space than the bonding pairs, and therefore pushes the
three hydrogen atoms together a little more. The bond angles between the hydrogen
and nitrogen atoms in the molecule become 106 degrees, rather than the usual 109
degrees of a tetrahedral molecule. The shape of the molecule is trigonal pyramidal.
Activity :: Group work : Building molecular models
In groups, you are going to build a number of molecules using marshmallows to
represent the atoms in the molecule, and toothpicks to represent the bonds between
the atoms. In other words, the toothpicks will hold the atoms (marshmallows) in the
molecule together. Try to use different coloured marshmallows to represent different
elements.
You will build models of the following molecules:
HCl, CH4 , H2 O, HBr and NH3
For each molecule, you need to:
• Determine the basic geometry of the molecule
• Build your model so that the atoms are as far apart from each other as possible
(remember that the electrons around the central atom will try to avoid the
repulsions between them).
• Decide whether this shape is accurate for that molecule or whether there are
any lone pairs that may influence it.
• Adjust the position of the atoms so that the bonding pairs are further away
from the lone pairs.
• How has the shape of the molecule changed?
• Draw a simple diagram to show the shape of the molecule. It doesn’t matter
if it is not 100% accurate. This exercise is only to help you to visualise the
3-dimensional shapes of molecules.
Do the models help you to have a clearer picture of what the molecules look like?
Try to build some more models for other molecules you can think of.
4.11
Oxidation numbers
When reactions occur, an exchange of electrons takes place. Oxidation is the loss of electrons
from an atom, while reduction is the gain of electrons by an atom. By giving elements an
oxidation number, it is possible to keep track of whether that element is losing or gaining
electrons during a chemical reaction. The loss of electrons in one part of the reaction, must be
balanced by a gain of electrons in another part of the reaction.
Definition: Oxidation number
A simplified way of understanding an oxidation number is to say that it is the charge an
atom would have if it was in a compound composed of ions.
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4.11
CHAPTER 4. ATOMIC COMBINATIONS - GRADE 11
There are a number of rules that you need to know about oxidation numbers, and these are listed
below. These will probably not make much sense at first, but once you have worked through
some examples, you will soon start to understand!
1. Rule 1: An element always has an oxidation number of zero, since it is neutral.
In the reaction H2 + Br2 → 2HBr, the oxidation numbers of hydrogen and bromine
on the left hand side of the equation are both zero.
2. Rule 2: In most cases, an atom that is part of a molecule will have an oxidation number
that has the same numerical value as its valency.
3. Rule 3: Monatomic ions have an oxidation number that is equal to the charge on the ion.
The chloride ion Cl− has an oxidation number of -1, and the magnesium ion M g 2+ has
an oxidation number of +2.
4. Rule 4: In a molecule, the oxidation number for the whole molecule will be zero, unless
the molecule has a charge, in which case the oxidation number is equal to the charge.
5. Rule 5: Use a table of electronegativities to determine whether an atom has a positive or
a negative oxidation number. For example, in a molecule of water, oxygen has a higher
electronegativity so it must be negative because it attracts electrons more strongly. It will
have a negative oxidation number (-2). Hydrogen will have a positive oxidation number
(+1).
6. Rule 6: An oxygen atom usually has an oxidation number of -2, although there are some
cases where its oxidation number is -1.
7. Rule 7: The oxidation number of hydrogen is usually +1. There are some exceptions
where its oxidation number is -1.
8. Rule 8: In most compounds, the oxidation number of the halogens is -1.
Important: You will notice that the oxidation number of an atom is the same as its valency.
Whether an oxidation number os positive or negative, is determined by the electronegativities
of the atoms involved.
Worked Example 20: Oxidation numbers
Question: Give the oxidation numbers for all the atoms in the reaction between
sodium and chlorine to form sodium chloride.
N a + Cl → N aCl
Answer
Step 1 : Determine which atom will have a positive or negative oxidation
number
Sodium will have a positive oxidation number and chlorine will have a negative oxidation number.
Step 2 : Determine the oxidation number for each atom
Sodium (group 1) will have an oxidation number of +1. Chlorine (group 7) will have
an oxidation number of -1.
Step 3 : Check whether the oxidation numbers add up to the charge on the
molecule
In the equation N a + Cl → N aCl, the overall charge on the NaCl molecule is
+1-1=0. This is correct since NaCl is neutral. This means that, in a molecule
of NaCl, sodium has an oxidation number of +1 and chlorine has an oxidation
number of -1.
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CHAPTER 4. ATOMIC COMBINATIONS - GRADE 11
Worked Example 21: Oxidation numbers
Question: Give the oxidation numbers for all the atoms in the reaction between
hydrogen and oxygen to produce water. The unbalanced equation is shown below:
H2 + O2 → H2 O
Answer
Step 1 : Determine which atom will have a positive or negative oxidation
number
Hydrogen will have a positive oxidation number and oxygen will have a negative
oxidation number.
Step 2 : Determine the oxidation number for each atom
Hydrogen (group 1) will have an oxidation number of +1. Oxygen (group 6) will
have an oxidation number of -2.
Step 3 : Check whether the oxidation numbers add up to the charge on the
molecule
In the reaction H2 + O2 → H2 O, the oxidation numbers for hydrogen and oxygen
(on the left hand side of the equation) are zero since these are elements. In the
water molecule, the sum of the oxidation numbers is 2(+1)-2=0. This is correct
since the oxidation number of water is zero. Therefore, in water, hydrogen has
an oxidation number of +1 and oxygen has an oxidation number of -2.
Worked Example 22: Oxidation numbers
Question: Give the oxidation number of sulfur in a sulphate (SO42− ) ion
Answer
Step 1 : Determine which atom will have a positive or negative oxidation
number
Sulfur has a positive oxidation number and oxygen will have a negative oxidation
number.
Step 2 : Determine the oxidation number for each atom
Oxygen (group 6) will have an oxidation number of -2. The oxidation number of
sulfur at this stage is uncertain.
Step 3 : Determine the oxidation number of sulfur by using the fact that the
oxidation numbers of the atoms must add up to the charge on the molecule
In the polyatomic SO42− ion, the sum of the oxidation numbers must be -2. Since
there are four oxygen atoms in the ion, the total charge of the oxygen is -8. If the
overall charge of the ion is -2, then the oxidation number of sulfur must be +6.
87
4.11
4.12
CHAPTER 4. ATOMIC COMBINATIONS - GRADE 11
Exercise: Oxidation numbers
1. Give the oxidation numbers for each element in the following chemical compounds:
(a) NO2
(b) BaCl2
(c) H2 SO4
2. Give the oxidation numbers for the reactants and products in each of the following reactions:
(a) C + O2 → CO2
(b) N2 + 3H2 → 2NH3
(c) Magnesium metal burns in oxygen
4.12
Summary
• A chemical bond is the physical process that causes atoms and molecules to be attracted
together and to be bound in new compounds.
• Atoms are more reactive, and therefore more likely to bond, when their outer electron
orbitals are not full. Atoms are less reactive when these outer orbitals contain the maximum
number of electrons. This explains why the noble gases do not combine to form molecules.
• There are a number of forces that act between atoms: attractive forces between the
positive nucleus of one atom and the negative electrons of another; repulsive forces between
like-charged electrons, and repulsion between like-charged nuclei.
• Chemical bonding occurs when the energy of the system is at its lowest.
• Bond length is the distance between the nuclei of the atoms when they bond.
• Bond energy is the energy that must be added to the system for the bonds to break.
• When atoms bond, electrons are either shared or exchanged.
• Covalent bonding occurs between the atoms of non-metals and involves a sharing of
electrons so that the orbitals of the outermost energy levels in the atoms are filled.
• The valency of an atom is the number of bonds that it can form with other atoms.
• A double or triple bond occurs if there are two or three electron pairs that are shared
between the same two atoms.
• A dative covalent bond is a bond between two atoms in which both the electrons that
are shared in the bond come from the same atom.
• Lewis and Couper notation are two ways of representing molecular structure. In Lewis
notation, dots and crosses are used to represent the valence electrons around the central
atom. In Couper notation, lines are used to represent the bonds between atoms.
• Electronegativity measures how strongly an atom draws electrons to it.
• Electronegativity can be used to explain the difference between two types of covalent
bonds: polar covalent bonds (between non-identical atoms) and non-polar covalent
bonds (between identical atoms).
• An ionic bond occurs between atoms where the difference in electronegativity is greater
than 2.1. An exchange of electrons takes place and the atoms are held together by the
electrostatic force of attraction between oppositely-charged ions.
• Ionic solids are arranged in a crystal lattice structure.
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CHAPTER 4. ATOMIC COMBINATIONS - GRADE 11
4.12
• Ionic compounds have a number of specific properties, including their high melting and
boiling points, brittle nature, the arrangement of solids in a lattice structure and the ability
of ionic solutions to conduct electricity.
• A metallic bond is the electrostatic attraction between the positively charged nuclei of
metal atoms and the delocalised electrons in the metal.
• Metals also have a number of properties, including their ability to conduct heat and electricity, their metallic lustre, the fact that they are both malleable and ductile, and their
high melting point and density.
• The valency of atoms, and the way they bond, can be used to determine the chemical
formulae of compounds.
• The shape of molecules can be predicted using the VSEPR theory, which uses the arrangement of electrons around the central atom to determine the most likely shape of the
molecule.
• Oxidation numbers are used to determine whether an atom has gained or lost electrons
during a chemical reaction.
Exercise: Summary exercise
1. Give one word/term for each of the following descriptions.
(a) The distance between two atoms in a molecule
(b) A type of chemical bond that involves the transfer of electrons from one
atom to another.
(c) A measure of an atom’s ability to attract electrons to it.
2. Which ONE of the following best describes the bond formed between an H+
ion and the NH3 molecule?
(a)
(b)
(c)
(d)
Covalent bond
Dative covalent (coordinate covalent) bond
Ionic Bond
Hydrogen Bond
3. Explain the meaning of each of the following terms:
(a) valency
(b) bond energy
(c) covalent bond
4. Which of the following reactions will not take place? Explain your answer.
(a) H + H → H2
(b) Ne + Ne → Ne2
(c) I + I → I2
5. Draw the Lewis structure for each of the following:
(a)
(b)
(c)
(d)
calcium
iodine
hydrogen bromide (HBr)
nitrogen dioxide (NO2 )
6. Given the following Lewis structure, where X and Y each represent a different
element...
X
Y
ו
ו
××
ו
X
(a) What is the valency of X?
89
X
4.12
CHAPTER 4. ATOMIC COMBINATIONS - GRADE 11
(b) What is the valency of Y?
(c) Which elements could X and Y represent?
7. A molecule of ethane has the formula C2 H6 . Which of the following diagrams
(Couper notation) accurately represents this molecule?
(a)
H
H
H
C
C
H
H
(c)
H
(b)
H
H
H
H
H
C
C
H
H
C
C
H
H
H
H
8. Potassium dichromate is dissolved in water.
(a) Give the name and chemical formula for each of the ions in solution.
(b) What is the chemical formula for potassium dichromate?
(c) Give the oxidation number for each element in potassium dichromate.
90
Chapter 5
Intermolecular Forces - Grade 11
In the previous chapter, we discussed the different forces that exist between atoms (intramolecular
forces). When atoms are joined to one another they form molecules, and these molecules in turn
have forces that bind them together. These forces are known as intermolecular forces, and we
are going to look at them in more detail in this next section.
Definition: Intermolecular forces
Intermolecular forces are forces that act between stable molecules.
You will also remember from the previous chapter, that we can describe molecules as being either
polar or non-polar. A polar molecule is one in which there is a difference in electronegativity
between the atoms in the molecule, such that the shared electron pair spends more time close
to the atom that attracts it more strongly. The result is that one end of the molecule will have
a slightly positive charge (δ + ), and the other end will have a slightly negative charge (δ + ). The
molecule is said to be a dipole. However, it is important to remember that just because the
bonds within a molecule are polar, the molecule itself may not necessarily be polar. The shape
of the molecule may also affect its polarity. A few examples are shown in table 5.1 to refresh
your memory!
5.1
Types of Intermolecular Forces
It is important to be able to recognise whether the molecules in a substance are polar or nonpolar because this will determine what type of inermolecular forces there are. This is important
in explaining the properties of the substance.
1. Van der Waals forces
These intermolecular forces are named after a Dutch physicist called Johannes van der
Waals (1837 -1923), who recognised that there were weak attractive and repulsive forces
between the molecules of a gas, and that these forces caused gases to deviate from ’ideal
gas’ behaviour. Van der Waals forces are weak intermolecular forces, and can be divided
into three types:
(a) Dipole-dipole forces
Figure 5.1 shows a simplified dipole molecule, with one end slightly positive and the
other slightly negative.
When one dipole molecule comes into contact with another dipole molecule, the
positive pole of the one molecule will be attracted to the negative pole of the other,
and the molecules will be held together in this way (figure 5.2). Examples of molecules
that are held together by dipole-dipole forces are HCl, FeS, KBr, SO2 and NO2 .
(b) Ion-dipole forces
As the name suggests, this type of intermolecular force exists between an ion and
a dipole molecule. You will remember that an ion is a charged atom, and this will
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5.1
CHAPTER 5. INTERMOLECULAR FORCES - GRADE 11
Table 5.1: Polarity in molecules with different atomic bonds and molecular shapes
Molecule
Chemical
formula
Hydrogen
H2
Hydrogen
ride
chlo-
Carbon tetrafluoromethane
HCl
CF4
Bond
between
atoms
Covalent
Shape of molecule
Polar covalent
Linear molecule
H
H
Non-polar
Linear molecule
Polar
Hδ
Polar covalent
Polarity of
molecule
+
Clδ
−
Non-polar
Tetrahedral molecule
Fδ
δ+
−
Fδ
−
Cδ
+
Fδ
−
Fδ
−
δ−
Figure 5.1: A simplified diagram of a dipole molecule
δ+
δ−
δ+
δ−
Figure 5.2: Two dipole molecules are held together by the attractive force between their oppositely charged poles
be attracted to one of the charged ends of the polar molecule. A positive ion will
be attracted to the negative pole of the polar molecule, while a negative ion will be
attracted to the positive pole of the polar molecule. This can be seen when sodium
chloride (NaCl) dissolves in water. The positive sodium ion (Na+ ) will be attracted to
the slightly negative oxygen atoms in the water molecule, while the negative chloride
ion (Cl− ) is attracted to the slightly positive hydrogen atom. These intermolecular
forces weaken the ionic bonds between the sodium and chloride ions so that the
sodium chloride dissolves in the water (figure 5.3).
(c) London forces
These intermolecular forces are also sometimes called ’dipole- induced dipole’ or
’momentary dipole’ forces. Not all molecules are polar, and yet we know that there
are also intermolecular forces between non-polar molecules such as carbon dioxide. In
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CHAPTER 5. INTERMOLECULAR FORCES - GRADE 11
5.1
Cl−
δ+
H2 O
δ−
Cl− δ + H2 O δ − N a+ δ − H2 O δ + Cl−
δ−
H2 O
δ+
Cl−
Figure 5.3: Ion-dipole forces in a sodium chloride solution
non-polar molecules the electronic charge is evenly distributed but it is possible that
at a particular moment in time, the electrons might not be evenly distributed. The
molecule will have a temporary dipole. In other words, each end of the molecules has
a slight charge, either positive or negative. When this happens, molecules that are
next to each other attract each other very weakly. These London forces are found
in the halogens (e.g. F2 and I2 ), the noble gases (e.g. Ne and Ar) and in other
non-polar molecules such as carbon dioxide and methane.
2. Hydrogen bonds
As the name implies, this type of intermolecular bond involves a hydrogen atom. The
hydrogen must be attached to another atom that is strongly electronegative, such as
oxygen, nitrogen or fluorine. Water molecules for example, are held together by hydrogen
bonds between the hydrogen atom of one molecule and the oxygen atom of another (figure
5.4). Hydrogen bonds are stronger than van der Waals forces. It is important to note
however, that both van der Waals forces and hydrogen bonds are weaker than the covalent
and ionic bonds that exist between atoms.
hydrogen bonds
atomic bonds
H
O
O
H
O
O
H
O
O
Figure 5.4: Two representations showing the hydrogen bonds between water molecules: spacefilling model and structural formula.
93
5.2
CHAPTER 5. INTERMOLECULAR FORCES - GRADE 11
Exercise: Types of intermolecular forces
1. Complete the following table by placing a tick to show which type of intermolecular force occurs in each substance:
Formula DipoleMomentary Ion-dipole
hydrogen
dipole
dipole
bond
HCl
CO2
I2
H2 O
KI(aq)
NH3
2. In which of the substances above are the intermolecular forces...
(a) strongest
(b) weakest
5.2
Understanding intermolecular forces
The types of intermolecular forces that occur in a substance will affect its properties, such as
its phase, melting point and boiling point. You should remember, if you think back to the
kinetic theory of matter, that the phase of a substance is determined by how strong the forces
are between its particles. The weaker the forces, the more likely the substance is to exist as
a gas because the particles are far apart. If the forces are very strong, the particles are held
closely together in a solid structure. Remember also that the temperature of a material affects
the energy of its particles. The more energy the particles have, the more likely they are to be
able to overcome the forces that are holding them together. This can cause a change in phase.
Definition: Boiling point
The temperature at which a material will change from being a liquid to being a gas.
Definition: Melting point
The temperature at which a material will change from being a solid to being a liquid.
Now look at the data in table 5.2.
Formula
He
Ne
Ar
F2
Cl2
Br2
NH3
H2 O
HF
Formula mass
4
20
40
38
71
160
17
18
20
Melting point (0 C)
-270
-249
-189
-220
-101
-7
-78
0
-83
Boiling point (0 C) at 1 atm
-269
-246
-186
-188
-35
58
-33
100
20
Table 5.2: Melting point and boiling point of a number of chemical substances
The melting point and boiling point of a substance, give us information about the phase of the
substance at room temperature, and the strength of the intermolecular forces. The examples
below will help to explain this.
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CHAPTER 5. INTERMOLECULAR FORCES - GRADE 11
5.2
Example 1: Fluorine (F2 )
Phase at room temperature
Fluorine (F2 ) has a melting point of -2200 C and a boiling point of -1880C. This means that
for any temperature that is greater than -1880C, fluorine will be a gas. At temperatures below
-2200C, fluorine would be a solid, and at any temperature inbetween these two, fluorine will be
a liquid. So, at room temperature, fluorine exists as a gas.
Strength of intermolecular forces
What does this information tell us about the intermolecular forces in fluorine? In fluorine, these
forces must be very weak for it to exist as a gas at room temperature. Only at temperatures
below -1880 C will the molecules have a low enough energy that they will come close enough to
each other for forces of attraction to act between the molecules. The intermolecular forces in
fluorine are very weak van der Waals forces because the molecules are non-polar.
Example 2: Hydrogen fluoride (HF)
Phase at room temperature
For temperatures below -830 C, hydrogen fluoride is a solid. Between -830 C and 200 C, it exists
as a liquid, and if the temperature is increased above 200 C, it will become a gas.
Strength of intermolecular forces
What does this tell us about the intermolecular forces in hydrogen fluoride? The forces are
stronger than those in fluorine, because more energy is needed for fluorine to change into the
gaseous phase. In other words, more energy is needed for the intermolecular forces to be overcome so that the molecules can move further apart. Intermolecular forces will exist between the
hydrogen atom of one molecule and the fluorine atom of another. These are hydrogen bonds,
which are stronger than van der Waals forces.
What do you notice about water? Luckily for us, water behaves quite differently from the rest
of the halides. Imagine if water were like ammonia (NH3 ), which is a gas above a temperature
of -330 C! There would be no liquid water on the planet, and that would mean that no life would
be able to survive here. The hydrogen bonds in water are particularly strong and this gives water
unique qualities when compared to other molecules with hydrogen bonds. This will be discussed
more in chapter ?? deals with this in more detail. You should also note that the strength of
the intermolecular forces increases with an increase in formula mass. This can be seen by the
increasing melting and boiling points of substances as formula mass increases.
Exercise: Applying your knowledge of intermolecular forces
Refer to the data in table 5.2 and then use your knowledge of different types of
intermolecular forces to explain the following statements:
•
•
•
•
The boiling point of F2 is much lower than the boiling point of NH3
At room temperature, many elements exist naturally as gases
The boiling point of HF is higher than the boiling point of Cl2
The boiling point of water is much higher than HF, even though they both
contain hydrogen bonds
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5.3
CHAPTER 5. INTERMOLECULAR FORCES - GRADE 11
5.3
Intermolecular forces in liquids
Intermolecular forces affect a number of properties in liquids:
• Surface tension
You may have noticed how some insects are able to walk across a water surface, and how
leaves float in water. This is because of surface tension. In water, each molecule is held to
the surrounding molecules by strong hydrogen bonds. Molecules in the centre of the liquid
are completely surrounded by other molecules, so these forces are exerted in all directions.
However, molecules at the surface do not have any water molecules above them to pull
them upwards. Because they are only pulled sideways and downwards, the water molecules
at the surface are held more closely together. This is called surface tension.
For molecules at the
surface there are no
upward forces, so
the molecules are
closer together.
For molecules in the
centre of the liquid, the intermolecular forces act in all
directions.
Figure 5.5: Surface tension in a liquid
• Capillarity
Activity :: Investigation : Capillarity
Half fill a beaker with water and hold a hollow glass tube in the centre as
shown below. Mark the level of the water in the glass tube, and look carefully
at the shape of the air-water interface in the tube. What do you notice?
meniscus
water
At the air-water interface, you will notice a meniscus, where the water
appears to dip in the centre. In the glass tube, the attractive forces between
the glass and the water are stronger than the intermolecular forces between the
water molecules. This causes the water to be held more closely to the glass,
and a meniscus forms. The forces between the glass and the water also mean
that the water can be ’pulled up’ higher when it is in the tube than when it is
in teh beaker. Capillarity is the surface tension that occurs in liquids that are
inside tubes.
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CHAPTER 5. INTERMOLECULAR FORCES - GRADE 11
5.4
• Evaporation
In a liquid, each particle has kinetic energy, but some particles will have more energy than
others. We therefore refer to the average kinetic energy of the molecules when we describe
the liquid. When the liquid is heated, those particles which have the highest energy will
be able to overcome the intermolecular forces holding them in the liquid phase, and will
become a gas. This is called evaporation. Evaporation occurs when a liquid changes to a
gas. The stronger the intermolecular forces in a liquid, the higher the temperature of the
molecules will have to be for it to become a gas. You should note that a liquid doesn’t
necessarily have to reach boiling point before evaporation can occur. Evaporation takes
place all the time. You will see this if you leave a glass of water outside in the sun. Slowly
the water level will drop over a period of time.
What happens then to the molecules of water that remain in the liquid? Remember that it
was the molecules with the highest energy that left the liquid. This means that the average
kinetic energy of the remaining molecules will decrease, and so will the temperature of the
liquid.
A similar process takes place when a person sweats during exercise. When you exercise,
your body temperature increases and you begin to release moisture (sweat) through the
pores in your skin. The sweat quickly evaporates and causes the temperature of your skin
to drop. This helps to keep your body temperature at a level that is suitable forit to
function properly.
teresting Transpiration in plants - Did you know that plants also ’sweat’ ? In plants,
Interesting
Fact
Fact
this is called transpiration, and a plant will lose water through spaces in the leaf
surface called stomata. Although this water loss is important in the survival of a
plant, if a plant loses too much water, it will die. Plants that live in very hot, dry
places such as deserts, must be specially adapted to reduce the amount of water
that transpires (evaporates) from their leaf surface. Desert plants have some
amazing adaptations to deal with this problem! Some have hairs on their leaves,
which reflect sunlight so that the temperature is not as high as it would be,
while others have a thin waxy layer covering their leaves, which reduces water
loss. Some plants are even able to close their stomata during the day when
temperatures (and therefore transpiration) are highest.
Important: In the same way that intermolecular forces affect the properties of liquids, they
also affect the properties of solids. For example, the stronger the intermolecular forces
between the particles that make up the solid, the harder the solid is likely to be, and the
higher its melting point is likely to be.
5.4
Summary
• Intermolecular forces are the forces that act between stable molecules.
• The type of intermolecular force in a substance, will depend on the nature of the
molecules.
• Polar molecules have an unequal distribution of charge, meaning that one part of the
molecule is slightly positive and the other part is slightly negative. Non-polar molecules
have an equal distribution of charge.
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5.4
CHAPTER 5. INTERMOLECULAR FORCES - GRADE 11
• There are three types of Van der Waal’s forces. These are dipole-dipole, ion-dipole and
London forces (momentary dipole).
• Dipole-dipole forces exist between two polar molecules, for example between two molecules
of hydrogen chloride.
• Ion-dipole forces exist between ions and dipole molecules. The ion is attracted to the
part of the molecule that has an opposite charge to its own. An example of this is when
an ionic solid such as sodium chloride dissolves in water.
• Momentary dipole forces occur between two non-polar molecules, where at some point
there is an uequal distribution of charge in the molecule. For example, there are London
forces between two molecules of carbon dioxide.
• Hydrogen bonds occur between hydrogen atoms and other atoms that have a high
electronegativity such as oxygen, nitrogen and fluorine. The hydrogen atom in one
molecule will be attracted to the nitrogen atom in another molecule, for example. There
are hydrogen bonds between water molecules and between ammonia molecules.
• Intermolecular forces affect the properties of substances. For example, the stronger the
intermolecular forces, the higher the melting point of that substance, and the more likely
that substance is to exist as a solid or liquid. Its boiling point will also be higher.
• In liquids, properties such as surface tension, capillarity and evaporation are the result
of intermolecular forces.
Exercise: Summary Exercise
1. Give one word or term for each of the following descriptions:
(a) The tendency of an atom in a molecule to attract bonding electrons.
(b) A molecule that has an unequal distribution of charge.
(c) A charged atom.
2. For each of the following questions, choose the one correct answer from the
list provided.
(a) The following table gives the melting points of various hydrides:
Hydride Melting point (0 C)
HI
-50
NH3
-78
H2 S
-83
CH4
-184
In which of these hydrides does hydrogen bonding occur?
i. HI only
ii. NH3 only
iii. HI and NH3 only
iv. HI, NH3 and H2 S
(IEB Paper 2, 2003)
(b) Refer to the list of substances below:
HCl, Cl2 , H2 O, NH3 , N2 , HF
Select the true statement from the list below:
i. NH3 is a non-polar molecule
ii. The melting point of NH3 will be higher than for Cl2
iii. Ion-dipole forces exist between molecules of HF
iv. At room temperature N2 is usually a liquid
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CHAPTER 5. INTERMOLECULAR FORCES - GRADE 11
3. The respective boiling points for four chemical substances are given below:
Hydrogen sulphide -600 C
Ammonia -330 C
Hydrogen fluoride 200 C
Water 1000 C
(a) Which one of the substances exhibits the strongest forces of attraction
between its molecules in the liquid state?
(b) Give the name of the force responsible for the relatively high boiling points
of ammonia and water and explain how this force originates.
(c) The shapes of the molecules of hydrogen sulphide and water are similar,
yet their boiling points differ. Explain.
(IEB Paper 2, 2002)
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5.4
5.4
CHAPTER 5. INTERMOLECULAR FORCES - GRADE 11
100
Chapter 6
Solutions and solubility - Grade 11
We are surrounded by different types of solutions in our daily lives. Any solution is made up of
a solute and a solvent. A solute is a substance that dissolves in a solvent. In the case of a salt
(NaCl) solution, the salt crystals are the solute. A solvent is the substance in which the solute
dissolves. In the case of the NaCl solution, the solvent would be the water. In most cases, there
is always more of the solvent than there is of the solute in a solution.
Definition: Solutes and solvents
A solute is a substance that is dissolved in another substance. A solute can be a solid,
liquid or gas. A solvent is the liquid that dissolves a solid, liquid, or gaseous solute.
6.1
Types of solutions
When a solute is mixed with a solvent, a mixture is formed, and this may be either heterogeneous or homogeneous. If you mix sand and water for example, the sand does not dissolve in the
water. This is a heterogeneous mixture. When you mix salt and water, the resulting mixture
is homogeneous because the solute has dissolved in the solvent.
Definition: Solution
In chemistry, a solution is a homogeneous mixture that consists of a solute that has been
dissolved in a solvent.
A solution then is a homogeneous mixture of a solute and a solvent. Examples of solutions are:
• A solid solute dissolved in a liquid solvent e.g. sodium chloride dissolved in water.
• A gas solute dissolved in a liquid solvent e.g. carbon dioxide dissolved in water (fizzy
drinks) or oxygen dissolved in water (aquatic ecosystems).
• A liquid solute dissolved in a liquid solvent e.g. ethanol in water.
• A solid solute in a solid solvent e.g. metal alloys.
• A gas solute in a gas solvent e.g. the homogeneous mixture of gases in the air that we
breathe.
While there are many different types of solutions, most of those we will be discussing are liquids.
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CHAPTER 6. SOLUTIONS AND SOLUBILITY - GRADE 11
6.2
Forces and solutions
An important question to ask is why some solutes dissolve in certain solvents and not in others.
The answer lies in understanding the interaction between the intramolecular and intermolecular
forces between the solute and solvent particles.
Activity :: Experiment : Solubility
Aim:
To investigate the solubility of solutes in different solvents.
Apparatus:
Salt, vinegar, iodine, ethanol
Method:
1. Mix half a teaspoon of salt in 100cm3 of water
2. Mix half a teaspoon of vinegar (acetic acid) in 100cm3 of water
3. Mix a few grains of iodine in ethanol
4. Mix a few grains of iodine in 100cm3 of water
Results:
Record your observations in the table below:
Solute
Iodine
Iodine
Vinegar
Salt
Polar, non-polar or
ionic solute
Solvent
Polar, non-polar or
ionic solvent
Does solute dissolve?
Ethanol
Water
Water
Water
You should have noticed that in some cases, the solute dissolves in the solvent,
while in other cases it does not.
Conclusions:
In general, polar and ionic solutes dissolve well in polar solvents, while non-polar
solutes dissolve well in non-polar solvents. An easy way to remember this is that ’like
dissolves like’, in other words, if the solute and the solvent have similar intermolecular
forces, there is a high possibility that dissolution will occur. This will be explained
in more detail below.
• Non-polar solutes and non-polar solvents (e.g. iodine and ether)
Iodine molecules are non-polar, and the forces between the molecules are weak van der
Waals forces. There are also weak van der Waals forces between ether molecules. Because the intermolecular forces in both the solute and the solvent are similar, it is easy for
these to be broken in the solute, allowing the solute to move into the spaces between the
molecules of the solvent. The solute dissolves in the solvent.
• Polar solutes and polar solvents (e.g. salt and water)
There are strong electrostatic forces between the ions of a salt such as sodium chloride.
There are also strong hydrogen bonds between water molecules. Because the strength of
the intermolecular forces in the solute and solvent are similar, the solute will dissolve in
the solvent.
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CHAPTER 6. SOLUTIONS AND SOLUBILITY - GRADE 11
6.3
6.3
Solubility
You may have noticed sometimes that, if you try to dissolve salt (or some other solute) in a
small amount of water, it will initially dissolve, but then appears not to be able to dissolve any
further when you keep adding more solute to the solvent. This is called the solubility of the
solution. Solubility refers to the maximum amount of solute that will dissolve in a solvent under
certain conditions.
Definition: Solubility
Solubility is the ability of a given substance, the solute, to dissolve in a solvent. If a substance
has a high solubility, it means that lots of the solute is able to dissolve in the solvent.
So what factors affect solubility? Below are some of the factors that affect solubility:
• the quantity of solute and solvent in the solution
• the temperature of the solution
• other compounds in the solvent affect solubility because they take up some of the spaces
between molecules of the solvent, that could otherwise be taken by the solute itself
• the strength of the forces between particles of the solute, and the strength of forces between
particles of the solvent
Activity :: Experiment : Factors affecting solubility
Aim:
To determine the effect of temperature on solubility
Method:
1. Measure 100cm3 of water into a beaker
2. Measure 100 g of salt and place into another beaker
3. Slowly pour the salt into the beaker with the water, stirring it as you add. Keep
adding salt until you notice that the salt is not dissolving anymore.
4. Record the amount of salt that has been added to the water and the temperature of the solution.
5. Now increase the temperature of the water by heating it over a bunsen burner.
6. Repeat the steps above so that you obtain the solubility limit of salt at this
higher temperature. You will need to start again with new salt and water!
7. Continue to increase the temperature as many times as possible and record
your results.
Results:
Record your results in the table below:
Temp (0 C)
Amount of solute that dissolves
in 100 cm3 of water (g)
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6.3
CHAPTER 6. SOLUTIONS AND SOLUBILITY - GRADE 11
As you increase the temperature of the water, are you able to dissolve more or
less salt?
Conclusions:
As the temperature of the solution increases, so does the amount of salt that will
dissolve. The solubility of sodium chloride increases as the temperature increases.
Exercise: Investigating the solubility of salts
The data table below gives the solubility (measured in grams of salt per 100 g
water) of a number of different salts at various temperatures. Look at the data and
then answer the questions that follow.
Temp (0 C)
0
10
20
30
40
50
60
KNO3
13.9
21.2
31.6
45.3
61.4
83.5
106.0
Solubility (g salt per 100 g H2 O)
K2 SO4
NaCl
7.4
35.7
9.3
35.8
11.1
36.0
13.0
36.2
14.8
36.5
16.5
36.8
18.2
37.3
1. On the same set of axes, draw line graphs to show how the solubility of the
three salts changes with an increase in temperature.
2. Describe what happens to salt solubility as temperature increases. Suggest a
reason why this happens.
3. Write an equation to show how each of the following salts ionises in water:
(a) KNO3
(b) K2 SO4
4. You are given three beakers, each containing the same amount of water. 5 g
KNO3 is added to beaker 1, 5 g K2 SO4 is added to beaker 2 and 5 g NaCl
is added to beaker 3. The beakers are heated over a bunsen burner until the
temperature of their solutions is 600 C.
(a) Which salt solution will have the highest conductivity under these conditions?
(b) Explain your answer.
Exercise: Experiments and solubility
Two grade 10 learners, Siphiwe and Ann, wish to separately investigate the solubility of potassium chloride at room temperature. They follow the list of instructions
shown below, using the apparatus that has been given to them:
Method:
1. Determine the mass of an empty, dry evaporating basin using an electronic
balance and record the mass.
2. Pour 50 ml water into a 250 ml beaker.
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CHAPTER 6. SOLUTIONS AND SOLUBILITY - GRADE 11
6.3
3. Add potassium chloride crystals to the water in the beaker in small portions.
4. Stir the solution until the salt dissolves.
5. Repeat the addition of potassium chloride (steps a and b) until no more salt
dissolves and some salt remains undissolved.
6. Record the temperature of the potassium chloride solution.
7. Filter the solution into the evaporating basin.
8. Determine the mass of the evaporating basin containing the solution that has
passed through the filter (the filtrate) on the electronic balance and record the
mass.
9. Ignite the Bunsen burner.
10. Carefully heat the filtrate in the evaporating basin until the salt is dry.
11. Place the evaporating basin in the desiccator (a large glass container in which
there is a dehydrating agent like calcium sulphate that absorbs water) until it
reaches room temperature.
12. Determine the mass of the evaporating basin containing the dry cool salt on
the electronic balance and record the mass.
On completion of the experiment, their results were as follows:
Siphiwe’s
results
Temperature (0 C)
15
Mass of evaporating basin (g)
65.32
Mass of evaporating basin + salt solution (g) 125.32
Mass of evaporating basin + salt (g)
81.32
Ann’s
results
26
67.55
137.55
85.75
1. Calculate the solubility of potassium chloride, using the data recorded by
(a) Siphiwe
(b) Ann
A reference book lists the solubility of potassium chloride as 35.0 g per
100 ml of water at 250 C.
(c) Give a reason why you think each obtained results different from each other
and the value in the reference book.
2. Siphiwe and Ann now expand their investigation and work together. They
investigate the solubility of potassium chloride at different temperatures but
also the solubility of copper (II) sulphate at these same temperatures. They
collect and write up their results as follows:
In each experiment we used 50 ml of water in the beaker. We found the
following masses of substance dissolved in the 50 ml of water. At 00 C, mass
of potassium chloride is 14.0 g and copper sulphate is 14.3 g. At 100 C, 15.6
g and 17.4 g respectively. At 200 C, 17.3 g and 20.7 g respectively. At 400 C,
potassium chloride mass is 20.2 g and copper sulphate is 28.5 g, at 600 C, 23.1
g and 40.0 g and lastly at 800 C, the masses were 26.4 g and 55.0 g respectively.
(a) From the record of data provided above, draw up a neat table to record
Siphiwe and Ann’s results.
(b) Identify the dependent and independent variables in their investigation.
(c) Choose an appropriate scale and plot a graph of these results.
(d) From the graph, determine:
i. the temperature at which the solubility of copper sulphate is 50 g per
50 ml of water.
ii. the maximum number of grams of potassium chloride which will dissolve in 100 ml of water at 700 C.
(IEB Exemplar Paper 2, 2006)
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6.4
CHAPTER 6. SOLUTIONS AND SOLUBILITY - GRADE 11
6.4
Summary
• In chemistry, a solution is a homogenous mixture of a solute in a solvent.
• A solute is a substance that dissolves in a solute. A solute can be a solid, liquid or gas.
• A solvent is a substance in which a solute dissolves. A solvent can also be a solid, liquid
or gas.
• Examples of solutions include salt solutions, metal alloys, the air we breathe and gases
such as oxygen and carbon dioxide dissolved in water.
• Not all solutes will dissolve in all solvents. A general rule is the like dissolves like. Solutes
and solvents that have similar intermolecular forces are more likely to dissolve.
• Polar and ionic solutes will be more likely to dissolve in polar solvents, while non-polar
solutes will be more likely to dissolve in polar solvents.
• Solubility is the extent to which a solute is able to dissolve in a solvent under certain
conditions.
• Factors that affect solubility are the quantity of solute and solvent, temperature, the
intermolecular forces in the solute and solvent and other substances that may be in
the solvent.
Exercise: Summary Exercise
1. Give one word or term for each of the following descriptions:
(a) A type of mixture where the solute has completely dissolved in the solvent.
(b) A measure of how much solute is dissolved in a solution.
(c) Forces between the molecules in a substance.
2. For each of the following questions, choose the one correct answer from the
list provided.
A Which one of the following will readily dissolve in water?
i. I2 (s)
ii. NaI(s)
iii. CCl4 (l)
iv. BaSO4 (s)
(IEB Paper 2, 2005)
b In which of the following pairs of substances will the dissolving process
happen most readily?
Solute Solvent
A
S8
H2 O
B
KCl
CCl4
C KNO3
H2 O
D NH4 Cl
CCl4
(IEB Paper 2, 2004)
3. Which one of the following three substances is the most soluble in pure water
at room temperature?
Hydrogen sulphide, ammonia and hydrogen fluoride
4. Briefly explain in terms of intermolecular forces why solid iodine does not dissolve in pure water, yet it dissolves in xylene, an organic liquid at room temperature.
(IEB Paper 2, 2002)
106
Chapter 7
Atomic Nuclei - Grade 11
Nuclear physics is the branch of physics which deals with the nucleus of the atom. Within
this field, some scientists focus their attention on looking at the particles inside the nucleus
and understanding how they interact, while others classify and interpret the properties of nuclei.
This detailed knowledge of the nucleus makes it possible for technological advances to be made.
In this next chapter, we are going to touch on each of these different areas within the field of
nuclear physics.
7.1
Nuclear structure and stability
You will remember from an earlier chapter that an atom is made up of different types of particles:
protons (positive charge) neutrons (neutral) and electrons (negative charge). The nucleus is the
part of the atom that contains the protons and the neutrons, while the electrons are found in
energy orbitals around the nucleus. The protons and neutrons together are called nucleons. It
is the nucleus that makes up most of an atom’s atomic mass, because an electron has a very
small mass when compared with a proton or a neutron.
Within the nucleus, there are different forces which act between the particles. The strong
nuclear force is the force between two or more nucleons, and this force binds protons and
neutrons together inside the nucleus. This force is most powerful when the nucleus is small,
and the nucleons are close together. The electromagnetic force causes the repulsion between
like-charged (positive) protons. In a way then, these forces are trying to produce opposite effects
in the nucleus. The strong nuclear force acts to hold all the protons and neutrons close together,
while the electromagnetic force acts to push protons further apart. In atoms where the nuclei are
small, the strong nuclear force overpowers the electromagnetic force. However, as the nucleus
gets bigger (in elements with a higher number of nucleons), the electromagnetic force becomes
greater than the strong nuclear force. In these nuclei, it becomes possible for particles and energy
to be ejected from the nucleus. These nuclei are called unstable. The particles and energy that
a nucleus releases are referred to as radiation, and the atom is said to be radioactive. We are
going to look at these concepts in more detail in the next few sections.
7.2
The Discovery of Radiation
Radioactivity was first discovered in 1896 by a French scientist called Henri Becquerel while he
was working on phosphorescent materials. He happened to place some uranium crystals on black
paper that he had used to cover a piece of film. When he looked more carefully, he noticed that
the film had lots of patches on it, and that this did not happen when other elements were placed
on the paper. He eventually concluded that some rays must be coming out of the uranium
crystals to produce this effect.
His observations were taken further by the Polish scientist Marie Curie and her husband Pierre,
who increased our knowledge of radioactive elements. In 1903, Henri, Marie and Pierre were
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7.3
CHAPTER 7. ATOMIC NUCLEI - GRADE 11
awarded the Nobel Prize in Physics for their work on radioactive elements. This award made
Marie the first woman ever to receive a Nobel Prize. Marie Curie and her husband went on to
discover two new elements, which they named polonium (Po) after Marie’s home country, and
radium (Ra) after its highly radioactive characteristics. For these dicoveries, Marie was awarded a
Nobel Prize in Chemistry in 1911, making her one of very few people to receive two Nobel Prizes.
teresting Marie Curie died in 1934 from aplastic anemia, which was almost certainly partly
Interesting
Fact
Fact
due to her massive exposure to radiation during her lifetime. Most of her work
was carried out in a shed without safety measures, and she was known to carry
test tubes full of radioactive isotopes in her pockets and to store them in her
desk drawers. By the end of her life, not only was she very ill, but her hands
had become badly deformed due to their constant exposure to radiation. Unfortunately it was only later in her life that the full dangers of radiation were
realised.
7.3
Radioactivity and Types of Radiation
In section 7.1, we discussed that when a nucleus is unstable it can emit particles and energy.
This is called radioactive decay.
Definition: Radioactive decay
Radioactive decay is the process in which an unstable atomic nucleus loses energy by emitting
particles or electromagnetic waves. Radiation is the name for the emitted particles or
electromagnetic waves.
When a nucleus undergoes radioactive decay, it emits radiation and the nucleus is called radioactive. We are exposed to small amounts of radiation all the time. Even the rocks around us emit
radiation! However some elements are far more radioactive than others. Isotopes tend to be
less stable because they contain a larger number of nucleons than ’non-isotopes’ of the same
element. These radioactive isotopes are called radioisotopes.
Radiation can be emitted in different forms. There are three main types of radiation: alpha,
beta and gamma radiation. These are shown in figure 7.1, and are described below.
alpha (α)
beta (β)
gamma (γ)
paper
aluminium
Figure 7.1: Types of radiation
108
lead
CHAPTER 7. ATOMIC NUCLEI - GRADE 11
7.3.1
7.3
Alpha (α) particles and alpha decay
An alpha particle is made up of two protons and two neutrons bound together. This type of
radiation has a positive charge. An alpha particle is sometimes represented using the chemical
symbol He2+ , because it has the same structure as a Helium atom (two neutrons and two
protons) which is missing its two electrons, hence the overall charge of +2. Alpha particles
have very low penetration power. Penetration power describes how easily the particles can pass
through another material. Because alpha particles have a low penetration power, it means that
even something as thin as a piece of paper or the outside layer of the human skin, will absorb
these particles so that they can’t go any further.
Alpha decay occurs because the nucleus has too many protons, and this causes a lot of repulsion
between these like charges. To try to reduce this repulsion, the nucleus emits an α particle. This
can be seen in the decay of Americium (Am) to Neptunium (Np).
Example:
241
95 Am
→237
93 Np + αparticle
Let’s take a closer look at what has happened during this reaction. Americium (Z = 95; A =
241) undergoes α decay and releases one alpha particle (i.e. 2 protons and 2 neutrons). The
atom now has only 93 protons (Z = 93). On the periodic table, the element which has 93 protons
(Z = 93) is called Neptunium. Therefore, the Americium atom has become a Neptunium atom.
The atomic mass of the neptunium atom is 237 (A = 237) because 4 nucleons (2 protons and
2 neutrons) were emitted from the atom of Americium.
7.3.2
Beta (β) particles and beta decay
In certain types of radioactive nuclei that have too many neutrons, a neutron may be converted
into a proton, an electron and another particle (called a neutrino). The high energy electrons
that are released in this way are called beta particles. Beta particles have a higher penetration
power than alpha particles and are able to pass through thicker materials such as paper.
The diagram below shows what happens during β decay:
electron (β particle)
One of the neutrons from H-3
is converted to
a proton
Atomic nucleus
neutrino (ν̄)
Hydrogen-3
= one proton
Helium-3
An electron and a
neutrino are released
= one neutron
Figure 7.2: β decay in a hydrogen atom
During beta decay, the number of neutrons in the atom decreases by one, and the number of
protons increases by one. Since the number of protons before and after the decay is different,
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CHAPTER 7. ATOMIC NUCLEI - GRADE 11
the atom has changed into a different element. In figure 7.2, Hydrogen has become Helium.
The beta decay of the Hydrogen-3 atom can be represented as follows:
3
1H
→32 He + βparticle + ν̄
teresting When scientists added up all the energy from the neutrons, protons and electrons
Interesting
Fact
Fact
involved in β-decays, they noticed that there was always some energy missing.
We know that energy is always conserved, which led Wolfgang Pauli in 1930 to
come up with the idea that another particle, which was not detected yet, also
had to be involved in the decay. He called this particle the neutrino (Italian
for ”little neutral one”), because he knew it had to be neutral, have little or no
mass, and interact only very weakly, making it very hard to find experimentally!
The neutrino was finally identified experimentally about 25 years after Pauli first
thought of it.
Due to the radioactive processes inside the sun, each 1 cm2 patch of the earth
receives 70 billion (70×109) neutrinos each second! Luckily neutrinos only interact very weakly so they do not harm our bodies when billions of them pass
through us every second.
7.3.3
Gamma (γ) rays and gamma decay
When particles inside the nucleus collide during radioactive decay, energy is released. This energy can leave the nucleus in the form of waves of electromagnetic energy called gamma rays.
Gamma radiation is part of the electromagnetic spectrum, just like visible light. However, unlike
visible light, humans cannot see gamma rays because they are at a higher frequency and a higher
energy. Gamma radiation has no mass or charge. This type of radiation is able to penetrate
most common substances, including metals. The only substances that can absorb this radiation
are thick lead and concrete.
Gamma decay occurs if the nucleus is at too high an energy level. Since gamma rays are part
of the electromagnetic spectrum, they can be thought of as waves or particles. Therefore in
gamma decay, we can think of a ray or a particle (called a photon) being released. The atomic
number and atomic mass remain unchanged.
Table 7.1 summarises and compares the three types of radioactive decay that have been discussed.
Table 7.1: A comparison of alpha, beta and gamma decay
Type of decay Particle/ray released
Change in
element
Alpha (α)
α particle (2 protons and 2 neutrons) Yes
Beta (β)
β particle (electron)
Yes
Gamma (γ)
γ ray (electromagnetic energy)
No
Penetration
power
Low
Medium
High
Worked Example 23: Radioactive decay
Question: The isotope
particles.
241
95 Pb
undergoes radioactive decay and loses two alpha
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7.3
photon (γ particle)
Helium-3
Helium-3
Figure 7.3: γ decay in a helium atom
1. Write the chemical formula of the element that is produced as a result of the
decay.
2. Write an equation for this decay process.
Answer
Step 1 : Work out the number of protons and/or neutrons that the radioisotope loses during radioactive decay
One α particle consists of two protons and two neutrons. Since two α particles are
released, the total number of protons lost is four and the total number of neutrons
lost is also four.
Step 2 : Calculate the atomic number (Z) and atomic mass number (A) of
the element that is formed.
Z = 95 − 4 = 91
A = 241 − 4 = 237
Step 3 : Refer to the periodic table to see which element has the atomic
number that you have calculated.
The element that has Z = 91 is Protactinium (Pa).
Step 4 : Write the symbol for the element that has formed as a result of
radioactive decay.
237
91 Pa
Step 5 : Write an equation for the decay process.
241
95 P b
→237
91 P a + 2 protons + 2 neutrons
Activity :: Discussion : Radiation
In groups of 3-4, discuss the following questions:
• Which of the three types of radiation is most dangerous to living creatures
(including humans!)
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• What can happen to people if they are exposed to high levels of radiation?
• What can be done to protect yourself from radiation (Hint: Think of what the
radiologist does when you go for an X-ray)?
Exercise: Radiation and radioactive elements
1. There are two main forces inside an atomic nucleus:
(a) Name these two forces.
(b) Explain why atoms that contain a greater number of nucleons are more
likely to be radioactive.
2. The isotope 241
95 Pb undergoes radioactive decay and loses three alpha particles.
(a) Write the chemical formula of the element that is produced as a result of
the decay.
(b) How many nucleons does this element contain?
3. Complete the following equation:
210
82 Am
→ (alpha decay)
4. Radium-228 decays by emitting a beta particle. Write an equation for this
decay process.
5. Describe how gamma decay differs from alpha and beta decay.
7.4
Sources of radiation
The sources of radiation can be either natural or man-made.
7.4.1
Natural background radiation
• Cosmic radiation
The Earth, and all living things on it, are constantly bombarded by radiation from space.
Charged particles from the sun and stars interact with the Earth’s atmosphere and magnetic
field to produce a shower of radiation, mostly beta and gamma radiation. The amount of
cosmic radiation varies in different parts of the world because of differences in elevation
and also the effects of the Earth’s magnetic field.
• Terrestrial Radiation
Radioactive material is found throughout nature. It occurs naturally in the soil, water, and
vegetation. The major isotopes that are of concern are uranium and the decay products of
uranium, such as thorium, radium, and radon. Low levels of uranium, thorium, and their
decay products are found everywhere. Some of these materials are ingested (taken in)
with food and water, while others are breathed in. The dose of radiation from terrestrial
sources varies in different parts of the world.
teresting Cosmic and terrestrial radiation are not the only natural sources. All people
Interesting
Fact
Fact
have radioactive potassium-40, carbon-14, lead-210 and other isotopes inside
their bodies from birth.
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7.4.2
7.5
Man-made sources of radiation
Although all living things are exposed to natural background radiation, there are other sources
of radiation. Some of these will affect most members of the public, while others will only affect
those people who are exposed to radiation through their work.
• Members of the Public
Man-made radiation sources that affect members of the public include televisions, tobacco (polonium-210), combustible fuels, smoke detectors (americium), luminous watches
(tritium) and building materials. By far, the most significant source of man-made radiation exposure to the public is from medical procedures, such as diagnostic x-rays, nuclear
medicine, and radiation therapy. Some of the major isotopes involved are I-131, Tc-99m,
Co-60, Ir-192, and Cs-137. The production of nuclear fuel using uranium is also a source
of radiation for the public, as is fallout from nuclear weapons testing or use.
• Individuals who are exposed through their work
Any people who work in the following environments are exposed to radiation at some time:
radiology (X-ray) departments, nuclear power plants, nuclear medicine departments and
radiation oncology (the study of cancer) departments. Some of the isotopes that are of
concern are cobalt-60, cesium-137, americium-241, and others.
teresting Radiation therapy (or radiotherapy) uses ionising radiation as part of cancer
Interesting
Fact
Fact
treatment to control malignant cells. In cancer, a malignant cell is one that
divides very rapidly to produce many more cells. These groups of dividing cells
can form a growth or tumour. The malignant cells in the tumour can take
nutrition away from other healthy body cells, causing them to die, or can increase the pressure in parts of the body because of the space that they take up.
Radiation therapy uses radiation to try to target these malignant cells and kill
them. However, the radiation can also damage other, healthy cells in the body.
To stop this from happening, shaped radiation beams are aimed from several
angles to intersect at the tumour, so that the radiation dose here is much higher
than in the surrounding, healthy tissue. But even doing this doesn’t protect all
the healthy cells, and that is why people have side-effects to this treatment.
Note that radiation therapy is different from chemotherapy, which uses chemicals,
rather than radiation, to destroy malignant cells. Generally, the side effects of
chemotherapy are greater because the treatment is not as localised as it is with
radiation therapy. The chemicals travel throughout the body, affecting many
healthy cells.
7.5
The ’half-life’ of an element
Definition: Half-life
The half-life of an element is the time it takes for half the atoms of a radioisotope to decay
into other atoms.
Table 7.2 gives some examples of the half-lives of different elements.
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Table 7.2: Table showing the half-life of
Radioisotope Chemical symbol
Polonium-212 Po-212
Sodium-24
Na-24
Strontium-90
Sr-90
Cobalt-60
Co-60
Caesium-137
Cs-137
Carbon-14
C-14
Calcium
Ca
Beryllium
Be
Uranium-235
U-235
a number of elements
Half-life
0.16 seconds
15 hours
28 days
5.3 years
30 years
5 760 years
100 000 years
2 700 000 years
7.1 billion years
So, in the case of Sr-90, it will take 28 days for half of the atoms to decay into other atoms. It
will take another 28 days for half of the remaining atoms to decay. Let’s assume that we have
a sample of strontium that weighs 8g. After the first 28 days there will be:
1/2 x 8 = 4 g Sr-90 left
After 56 days, there will be:
1/2 x 4 g = 2 g Sr-90 left
After 84 days, there will be:
1/2 x 2 g = 1 g Sr-90 left
If we convert these amounts to a fraction of the original sample, then after 28 days 1/2 of the
sample remains undecayed. After 56 days 1/4 is undecayed and after 84 days, 1/8 and so on.
Activity :: Group work : Understanding half-life
Work in groups of 4-5
You will need:
16 sheets of A4 paper per group, scissors, 2 boxes per group, a marking pen and
timer/stopwatch.
What to do:
• Your group should have two boxes. Label one ’decayed’ and the other ’radioactive’.
• Take the A4 pages and cut each into 4 pieces of the same size. You should now
have 64 pieces of paper. Stack these neatly and place them in the ’radioactive’
box. The paper is going to represent some radioactive material.
• Set the timer for one minute. After one minute, remove half the sheets of
paper from the radioactive box and put them in the ’decayed’ box.
• Set the timer for another minute and repeat the previous step, again removing
half the pieces of paper that are left in the radioactive box and putting them
in the decayed box.
• Repeat this process until 8 minutes have passed. You may need to start cutting
your pieces of paper into even smaller pieces as you progress.
Questions:
1. How many pages were left in the radioactive box after...
(a) 1 minute
(b) 3 minutes
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(c) 5 minutes
2. What percentage (%) of the pages were left in the radioactive box after...
(a) 2 minutes
(b) 4 minutes
3. After how many minutes is there 1/128 of radioactive material remaining?
4. What is the half-life of the ’radioactive’ material in this exercise?
Worked Example 24: Half-life 1
Question: A 100 g sample of Cs-137 is allowed to decay. Calculate the mass of
Cs-137 that will be left after 90 years
Answer
Step 1 : You need to know the half-life of Cs-137
The half-life of Cs-137 is 30 years.
Step 2 : Determine how many times the quantity of sample will be halved in
90 years.
If the half-life of Cs-137 is 30 years, and the sample is left to decay for 90 years,
then the number of times the quantity of sample will be halved is 90/30 = 3.
Step 3 : Calculate the quantity that will be left by halving the mass of Cs-137
three times
1. After 30 years, the mass left is 100 g × 1/2 = 50 g
2. After 60 years, the mass left is 50 g × 1/2 = 25 g
3. After 90 years, the mass left is 25 g × 1/2 = 12.5 g
Note that a quicker way to do this calculation is as follows:
Mass left after 90 years = (1/2)3 × 100 g = 12.5 g (The exponent is the number
of times the quantity is halved)
Worked Example 25: Half-life 2
Question: An 80 g sample of Po-212 decays until only 10 g is left. How long did it
take for this decay to take place?
Answer
Step 1 : Calculate the fraction of the original sample that is left after decay
Fraction remaining = 10 g/80 g = 1/8
Step 2 : Calculate how many half-life periods of decay (x) must have taken
place for 1/8 of the original sample to be left
1
1
( )x =
2
8
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CHAPTER 7. ATOMIC NUCLEI - GRADE 11
Therefore, x = 3
Step 3 : Use the half-life of Po-212 to calculate how long the sample was
left to decay
The half-life of Po-212 is 0.16 seconds. Therefore if there were three periods of
decay, then the total time is 0.16 × 3. The time that the sample was left to decay
is 0.48 seconds.
Exercise: Looking at half life
1. Imagine that you have 100 g of Na-24.
(a) What is the half life of Na-24?
(b) How much of this isotope will be left after 45 hours?
(c) What percentage of the original sample will be left after 60 hours?
2. A sample of Sr-90 is allowed to decay. After 84 days, 10 g of the sample
remains.
(a) What is the half life of Sr-90?
(b) How much Sr-90 was in the original sample?
(c) How much Sr-90 will be left after 112 days?
7.6
The Dangers of Radiation
Natural radiation comes from a variety of sources such as the rocks, sun and from space. However, when we are exposed to large amounts of radiation, this can cause damage to cells. γ
radiation is particularly dangerous because it is able to penetrate the body, unlike α and β
particles whose penetration power is less. Some of the dangers of radiation are listed below:
• Damage to cells
Radiation is able to penetrate the body, and also to penetrate the membranes of the
cells within our bodies, causing massive damage. Radiation poisoning occurs when a
person is exposed to large amounts of this type of radiation. Radiation poisoning damages
tissues within the body, causing symptoms such as diarrhoea, vomiting, loss of hair and
convulsions.
• Genetic abnormalities
When radiation penetrates cell membranes, it can damage chromosomes within the nucleus
of the cell. The chromosomes contain all the genetic information for that person. If the
chromosomes are changed, this may lead to genetic abnormalities in any children that are
born to the person who has been exposed to radiation. Long after the nuclear disaster
of Chernobyl in Russia in 1986, babies were born with defects such as missing limbs and
abnormal growths.
• Cancer
Small amounts of radiation can cause cancers such as leukemia (cancer of the blood)
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7.7
7.7
The Uses of Radiation
However, despite the many dangers of radiation, it does have many powerful uses, some of which
are listed below:
• Medical Field
Radioactive chemical tracers emitting γ rays can give information about a person’s internal
anatomy and the functioning of specific organs. The radioactive material may be injected
into the patient, from where it will target specific areas such as bones or tumours. As
the material decays and releases radiation, this can be seen using a special type of camera
or other instrument. The radioactive material that is used for this purpose must have a
short half-life so that the radiation can be detected quickly and also so that the material
is quickly removed from the patient’s body. Using radioactive materials for this purpose
can mean that a tumour or cancer may be diagnosed long before these would have been
detected using other methods such as X-rays.
Radiation may also be used to sterilise medical equipment.
Activity :: Research Project : The medical uses of radioisotopes
Carry out your own research to find out more about the radioisotopes that
are used to diagnose diseases in the following parts of the body:
– thyroid gland
– kidneys
– brain
In each case, try to find out...
1. which radioisotope is used
2. what the sources of this radioisotope are
3. how the radioisotope enters the patient’s body and how it is monitored
• Biochemistry and Genetics
Radioisotopes may be used as tracers to label molecules so that chemical processes such
as DNA replication or amino acid transport can be traced.
• Food preservation
Irradiation of food can stop vegetables or plants from sprouting after they have been
harvested. It also kills bacteria and parasites, and controls the ripening of fruits.
• Environment
Radioisotopes can be used to trace and analyse pollutants.
• Archaeology and Carbon dating
Natural radioisotopes such as C-14 can be used to determine the age of organic remains.
All living organisms (e.g. trees, humans) contain carbon. Carbon is taken in by plants
and trees through the process of photosynthesis in the form of carbon dioxide and is then
converted into organic molecules. When animals feed on plants, they also obtain carbon
through these organic compounds. Some of the carbon in carbon dioxide is the radioactive
C-14, while the rest is a non-radioactive form of carbon. When an organism dies, no more
carbon is taken in and there is a limited amount of C-14 in the body. From this point
onwards, C-14 begins its radioactive decay. When scientists uncover remains, they are able
to estimate the age of the remains by seeing how much C-14 is left in the body relative
to the amount of non-radioactive carbon. The less C-14 there is, the older the remains
because radioactive decay must have been taking place for a long time. Because scientists
know the exact rate of decay of C-14, they can calculate a very accurate estimate of the
age of the remains. Carbon dating has been a very important tool in building up accurate
historical records.
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Activity :: Case Study : Using radiocarbon dating
Radiocarbon dating has played an important role in uncovering many aspects of South Africa’s history. Read the following extract from an article that
appeared in Afrol news on 10th February 2007 and then answer the questions
that follow.
The world famous rock art in South Africa’s uKhahlamba-Drakensberg,
a World Heritage Site, is three times older than previously thought, archaeologists conclude in a new study. The more than 40,000 paintings
were made by the San people some 3000 years ago, a new analysis had
shown.
Previous work on the age of the rock art in uKhahlamba-Drakensberg
concluded it is less than 1,000 years old. But the new study - headed
by a South African archaeologist leading a team from the University
of Newcastle upon Tyne (UK) and Australian National University in
Canberra - estimates the panels were created up to 3,000 years ago.
They used the latest radio-carbon dating technology.
The findings, published in the current edition of the academic journal
’South African Humanities’, have ”major implications for our understanding of how the rock artists lived and the social changes that were
taking place over the last three millennia,” according to a press release
from the British university.
Questions:
1. What is the half-life of carbon-14?
2. In the news article, what role did radiocarbon dating play in increasing our
knowledge of South Africa’s history?
3. Radiocarbon dating can also be used to analyse the remains of once-living
organisms. Imagine that a set of bones are found between layers of sediment
and rock in a remote area. A group of archaeologists carries out a series
of tests to try to estimate the age of the bones. They calculate that the
bones are approximately 23 040 years old.
What percentage of the original carbon-14 must have been left in the bones
for them to arrive at this estimate?
7.8
Nuclear Fission
Nuclear fission is a process where the nucleus of an atom is split into two or more smaller
nuclei, known as fission products. The fission of heavy elements is an exothermic reaction and
huge amounts of energy are released in the process. This energy can be used to produce nuclear
power or to make nuclear weapons, both of which we will discuss a little later.
Definition: Nuclear fission
The splitting of an atomic nucleus
Below is a diagram showing the nuclear fission of Uranium-235. An atom of Uranium-235 is
bombarded with a neutron to initiate the fission process. This neutron is absorbed by Uranium235, to become Uranium-236. Uranium-236 is highly unstable and breaks down into a number
of lighter elements, releasing energy in the process. Free neutrons are also produced during this
process, and these are then available to bombard other fissionable elements. This process is
known as a fission chain reaction, and occurs when one nuclear reaction starts off another,
which then also starts off another one so that there is a rapid increase in the number of nuclear
reactions that are taking place.
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7.8
Massive release of energy during nuclear
fission
Neutron is absorbed
by the nucleus of the
U-235 atom to form
U-236
U-235
U-236
b
neutron
The elements and
number of neutrons
produced
in
the
process, is random.
U-236
splits
into
lighter elements called
fission products and
free neutrons
7.8.1
The Atomic bomb - an abuse of nuclear fission
A nuclear chain reaction can happen very quickly, releasing vast amounts of energy in the process.
In 1939, it was discovered that Uranium could undergo nuclear fission. In fact, it was uranium
that was used in the first atomic bomb. The bomb contained huge amounts of Uranium-235,
enough to start a runaway nuclear fission chain reaction. Because the process was uncontrolled,
the energy from the fission reactions was released in a matter of seconds, resulting in the massive
explosion of that first bomb. Since then, more atomic bombs have been dropped, causing massive
destruction and loss of life.
Activity :: Discussion : Nuclear weapons testing - an ongoing issue
Read the article below which has been adapted from one that appeared in ’The
Globe’ in Washington on 10th October 2006, and then answer the questions that
follow.
US officials and arms control specialists warned yesterday that North Korea’s test of a small nuclear device could start an arms race in the region
and threaten the landmark global treaty designed nearly four decades ago
to halt the spread of nuclear weapons. US officials expressed concern
that North Korea’s neighbors, including Japan, Taiwan, and South Korea,
could eventually decide to develop weapons of their own. They also fear
that North Korea’s moves could embolden Iran, and that this in turn could
encourage Saudi Arabia or other neighbours in the volatile Middle East to
one day seek nuclear deterrents, analysts say.
North Korea is the first country to conduct a nuclear test after pulling
out of the Nuclear Nonproliferation Treaty. The treaty, which was created
in 1968, now includes 185 nations (nearly every country in the world).
Under the treaty, the five declared nuclear powers at the time (United
States, the Soviet Union, France, China, and Great Britain) agreed to
reduce their supplies of nuclear weapons. The treaty has also helped to
limit the number of new nuclear weapons nations.
But there have also been serious setbacks. India and Pakistan, which
never signed the treaty, became new nuclear powers, shocking the world
with test explosions in 1998. The current issue of nuclear weapons testing
in North Korea, is another such setback and a blow to the treaty.
Group discussion questions:
1. Discuss what is meant by an ’arms race’ and a ’treaty’.
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2. Do you think it is important to have such treaties in place to control the testing
and use of nuclear weapons? Explain your answer.
3. Discuss some of the reasons why countries might not agree to be part of a
nuclear weapons treaty.
4. How would you feel if South Africa decided to develop its own nuclear weapons?
7.8.2
Nuclear power - harnessing energy
However, nuclear fission can also be carried out in a controlled way in a nuclear reactor. A nuclear
reactor is a piece of equpiment where nuclear chain reactions can be started in a controlled and
sustained way. This is different from a nuclear explosion where the chain reaction occurs in
seconds. The most important use of nuclear reactors at the moment is to produce electrical
power, and most of these nuclear reactors use nuclear fission. A nuclear fuel is a chemical
isotope that can keep a fission chain reaction going. The most common isotopes that are used
are Uranium-235 and Plutonium-239. The amount of free energy that is in nuclear fuels is far
greater than the energy in a similar amount of other fuels such as gasoline. In many countries,
nuclear power is seen as a relatively environmentally friendly alternative to fossil fuels, which
release large amounts of greenhouse gases, and are also non-renewable resources. However,
one of the concerns around the use of nuclear power, is the production of nuclear waste which
contains radioactive chemical elements.
Activity :: Debate : Nuclear Power
The use of nuclear power as a source of energy has been a subject of much
debate. There are many advantages of nuclear power over other energy sources.
These include the large amount of energy that can be produced at a small plant,
little atmospheric pollution and the small quantity of waste. However there are also
disadvantages. These include the expense of maintaining nuclear power stations, the
huge impact that an accident could have as well as the disposal of dangerous nuclear
waste.
Use these ideas as a starting point for a class debate.
Nuclear power - An energy alternative or environmental hazard?
Your teacher will divide the class into teams. Some of the teams will be ’pro’
nuclear power while the others will be ’anti’ nuclear power.
7.9
Nuclear Fusion
Nuclear fusion is the joining together of the nuclei of two atoms to form a heavier nucleus.
If the atoms involved are small, this process is accompanied by the release of energy. It is the
nuclear fusion of elements that causes stars to shine and hydrogen bombs to explode. As with
nuclear fission then, there are both positive and negative uses of nuclear fusion.
Definition: Nuclear fusion
The joining together of the nuclei of two atoms.
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7.10
You will remember that nuclei naturally repel one another because of the electrostatic force between their positively charged protons. So, in order to bring two nuclei together, a lot of energy
must be supplied if fusion is to take place. If two nuclei can be brought close enough together
however, the electrostatic force is overwhelmed by the more powerful strong nuclear force which
only operates over short distances. If this happens, nuclear fusion can take place. Inside the
cores of stars, the temperature is high enough for hydrogen fusion to take place but scientists
have so far been unsuccessful in making it work in the laboratory. One of the huge advantages
of nuclear fusion, if it could be made to happen, is that it is a relatively environmentally friendly
source of energy. The helium that is produced is not radioactive or poisonous and does not carry
the dangers of nuclear fission.
7.10
Nucleosynthesis
An astronomer named Edwin Hubble discovered in the 1920’s that the universe is expanding. He
measured that far-away galaxies are moving away from the earth at great speed, and the further
away they are, the faster they are moving.
Extension: What are galaxies?
Galaxies are huge clusters of stars and matter in the universe. The earth is part
of the Milky Way galaxy which is shaped like a very large spiral. Astronomers can
measure the light coming from distant galaxies using telescopes. Edwin Hubble was
also able to measure the velocities of galaxies.
These observations led people to see that the universe is expanding. It also led to the Big Bang
hypothesis. The ’Big Bang’ hypothesis is an idea about how the universe may have started.
According to this theory, the universe started off at the beginning of time as a point which then
exploded and expanded into the universe we live in today. This happened between 10 and 14
billion years ago.
Just after the Big Bang, when the universe was only 10−43 s old, it was very hot and was
made up of quarks and leptons (an example of a lepton is the electron). As the universe
expanded, (∼ 10−2 s) and cooled, the quarks started binding together to form protons and
neutrons (together called nucleons).
7.10.1
Age of Nucleosynthesis (225 s - 103 s)
About 225 s after the Big Bang, the protons and neutrons started binding together to form simple
nuclei. The process of forming nuclei is called nucleosynthesis. When a proton and a neutron
bind together, they form the deuteron. The deuteron is like a hydrogen nucleus (which is just a
proton) with a neutron added to it so it can be written as 2 H. Using protons and neutrons as
building blocks, more nuclei can be formed as shown below. For example, the Helium-4 nucleus
(also called an alpha particle) can be formed in the following ways:
2
H + n → 3H
deuteron + neutron → triton
then:
3
H+p →
4
He
triton + proton → Helium4 (alpha particle)
or
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2
H+p →
3
He
deuteron + proton → Helium3
then:
3
He + n →
4
He
Helium3 + neutron → Helium4 (alpha particle)
Some 7 Li nuclei could also have been formed by the fusion of 4 He and 3 H.
7.10.2
Age of Ions (103 s - 1013 s)
However, at this time the universe was still very hot and the electrons still had too much energy
to become bound to the alpha particles to form helium atoms. Also, the nuclei with mass
numbers greater than 4 (i.e. greater than 4 He) are very short-lived and would have decayed
almost immediately after being formed. Therefore, the universe moved through a stage called
the Age of Ions when it consisted of free positively charged H+ ions and 4 He ions, and negatively
charged electrons not yet bound into atoms.
7.10.3
Age of Atoms (1013 s - 1015 s)
As the universe expanded further, it cooled down until the electrons were able to bind to the
hydrogen and helium nuclei to form hydrogen and helium atoms. Earlier, during the Age of Ions,
both the hydrogen and helium ions were positively charged which meant that they repelled each
other (electrostatically). During the Age of Atoms, the hydrogen and helium along with the
electrons, were in the form of atoms which are electrically neutral and so they no longer repelled
each other and instead pulled together under gravity to form clouds of gas, which evetually
formed stars.
7.10.4
Age of Stars and Galaxies (the universe today)
Inside the core of stars, the densities and temperatures are high enough for fusion reactions to
occur. Most of the heavier nuclei that exist today were formed inside stars from thermonuclear reactions! (It’s interesting to think that the atoms that we are made of were actually
manufactured inside stars!). Since stars are mostly composed of hydrogen, the first stage of
thermonuclear reactions inside stars involves hydrogen and is called hydrogen burning. The
process has three steps and results in four hydrogen atoms being formed into a helium atom
with (among other things) two photons (light!) being released.
The next stage is helium burning which results in the formation of carbon. All these reactions
release a large amount of energy and heat the star which causes heavier and heavier nuclei to
fuse into nuclei with higher and higher atomic numbers. The process stops with the formation
of 56 Fe, which is the most strongly bound nucleus. To make heavier nuclei, even higher energies
are needed than is possible inside normal stars. These nuclei are most likely formed when huge
amounts of energy are released, for example when stars explode (an exploding star is called a
supernova). This is also how all the nuclei formed inside stars get ”recycled” in the universe to
become part of new stars and planets.
7.11
Summary
• Nuclear physics is the branch of physics that deals with the nucleus of an atom.
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7.11
• There are two forces between the particles of the nucleus. The strong nuclear force is
an attractive force between the neutrons and the electromagnetic force is the repulsive
force between like-charged protons.
• In atoms with large nuclei, the electromagnetic force becomes greater than the strong
nuclear force and particles or energy may be released from the nucleus.
• Radioactive decay occurs when an unstable atomic nucleus loses energy by emitting
particles or electromagnetic waves.
• The particles and energy released are called radiation and the atom is said to be radioactive.
• Radioactive isotopes are called radioisotopes.
• Radioactivity was first discovered by Marie Curie and her husband Pierre.
• There are three types of radiation from radioactive decay: alpha (α), beta (β) and
gamma (γ) radiation.
• During alpha decay, an alpha particle is released. An alpha particle consists of two protons
and two neutrons bound together. Alpha radiation has low penetration power.
• During beta decay, a beta particle is released. During beta decay, a neutron is converted
to a proton, an electron and a neutrino. A beta particle is the electron that is released.
Beta radiation has greater penetration power than alpha radiation.
• During gamma decay, electromagnetic energy is released as gamma rays. Gamma radiation has the highest penetration power of the three radiation types.
• There are many sources of radiation. Some of natural and others are man-made.
• Natural sources of radiation include cosmic and terrestrial radiation.
• Man-made sources of radiation include televisions, smoke detectors, X-rays and radiation
therapy.
• The half-life of an element is the time it takes for half the atoms of a radioisotope to
decay into other atoms.
• Radiation can be very damaging. Some of the negative impacts of radiation exposure
include damage to cells, genetic abnormalities and cancer.
• However, radiation can also have many positive uses. These include use in the medical
field (e.g. chemical tracers), biochemistry and genetics, use in food preservation, the
environment and in archaeology.
• Nuclear fission is the splitting of an atomic nucleus into smaller fission products. Nuclear
fission produces large amounts of energy, which can be used to produce nuclear power,
and to make nuclear weapons.
• Nuclear fusion is the joining together of the nuclei of two atoms to form a heavier nucleus.
In stars, fusion reactions involve the joining of hydrogen atoms to form helium atoms.
• Nucleosynthesis is the process of forming nuclei. This was very important in helping to
form the universe as we know it.
Exercise: Summary exercise
1. Explain each of the following terms:
(a) electromagnetic force
(b) radioactive decay
(c) radiocarbon dating
123
7.11
CHAPTER 7. ATOMIC NUCLEI - GRADE 11
2. For each of the following questions, choose the one correct answer:
(a) The part of the atom that undergoes radioactive decay is the...
i. neutrons
ii. nucleus
iii. electrons
iv. entire atom
(b) The radio-isotope Po-212 undergoes alpha decay. Which of the following
statements is true?
i. The number of protons in the element remains unchanged.
ii. The number of nucleons after decay is 212.
iii. The number of protons in the element after decay is 82.
iv. The end product after decay is Po-208.
3. 20 g of sodium-24 undergoes radoactive decay. Calculate the percentage of the
original sample that remains after 60 hours.
4. Nuclear physics can be controversial. Many people argue that studying the
nucleus has led to devastation and huge loss of life. Others would argue that
the benefits of nuclear physics far outweigh the negative things that have come
from it.
(a) Outline some of the ways in which nuclear physics has been used in negative
ways.
(b) Outline some of the benefits that have come from nuclear physics.
124
Chapter 8
Thermal Properties and Ideal
Gases - Grade 11
We are surrounded by gases in our atmosphere which support and protect life on this planet.
In this chapter, we are going to try to understand more about gases, and learn how to predict
how they will behave under different conditions. The kinetic theory of matter was discussed in
chapter 2. This theory is very important in understanding how gases behave.
8.1
A review of the kinetic theory of matter
The main assumptions of the kinetic theory of matter are as follows:
• Matter is made up of particles (e.g. atoms or molecules)
• These particles are constantly moving because they have kinetic energy. The space in
which the particles move is the volume of the gas.
• There are spaces between the particles
• There are attractive forces between particles and these become stronger as the particles
move closer together.
• All particles have energy. The temperature of a substance is a measure of the average
kinetic energy of the particles.
• A change in phase may occur when the energy of the particles is changed.
The kinetic theory applies to all matter, including gases. In a gas, the particles are far apart and
have a high kinetic energy. They move around freely, colliding with each other or with the sides
of the container if the gas is enclosed. The pressure of a gas is a measure of the frequency
of collisions of the gas particles with each other and with the sides of the container that they
are in. If the gas is heated, the average kinetic energy of the gas particles will increase and
if the temperature is decreased, so does their energy. If the energy of the particles decreases
significantly, the gas liquifies. An ideal gas is one that obeys all the assumptions of the kinetic
theory of matter. A real gas behaves like an ideal gas, except at high pressures and low
temperatures. This will be discussed in more detail later in this chapter.
Definition: Ideal gas
An ideal gas or perfect gas is a hypothetical gas that obeys all the assumptions of the kinetic
theory of matter. In other words, an ideal gas would have identical particles of zero volume,
with no intermolecular forces between them. The atoms or molecules in an ideal gas would
also undergo elastic collisions with the walls of their container.
125
8.2
CHAPTER 8. THERMAL PROPERTIES AND IDEAL GASES - GRADE 11
Definition: Real gas
Real gases behave more or less like ideal gases except under certain conditions e.g. high
pressures and low temperatures.
There are a number of laws that describe how gases behave. It will be easy to make sense of
these laws if you understand the kinetic theory of gases that was discussed above.
8.2
Boyle’s Law: Pressure and volume of an enclosed gas
Activity :: Demonstration : Boyle’s Law
If you have ever tried to force in the plunger of a syringe or a bicycle pump while
sealing the opening with a finger, you will have seen Boyle’s Law in action! This will
now be demonstrated using a 10 ml syringe.
Aim:
To demonstrate Boyle’s law.
Apparatus:
You will only need a syringe for this demonstration.
5
10
mℓ
Method:
1. Hold the syringe in one hand, and with the other pull the plunger out towards
you so that the syringe is now full of air.
2. Seal the opening of the syringe with your finger so that no air can escape the
syringe.
3. Slowly push the plunger in, and notice whether it becomes more or less difficult
to push the plunger in.
Results:
What did you notice when you pushed the plunger in? What happens to the
volume of air inside the syringe? Did it become more or less difficult to push the
plunger in as the volume of the air in the syringe decreased? In other words, did
you have to apply more or less pressure to the plunger as the volume of air in the
syringe decreased?
As the volume of air in the syringe decreases, you have to apply more pressure
to the plunger to keep forcing it down. The pressure of the gas inside the syringe
pushing back on the plunger is greater. Another way of saying this is that as the
volume of the gas in the syringe decreases, the pressure of that gas increases.
Conclusion:
If the volume of the gas decreases, the pressure of the gas increases. If the
volume of the gas increases, the pressure decreases. These results support Boyle’s
law.
In the previous demonstration, the volume of the gas decreased when the pressure increased,
and the volume increased when the pressure decreased. This is called an inverse relationship.
The inverse relationship between pressure and volume is shown in figure 8.1.
126
8.2
Pressure
CHAPTER 8. THERMAL PROPERTIES AND IDEAL GASES - GRADE 11
Volume
Figure 8.1: Graph showing the inverse relationship between pressure and volume
Can you use the kinetic theory of gases to explain this inverse relationship between the pressure
and volume of a gas? Let’s think about it. If you decrease the volume of a gas, this means that
the same number of gas particles are now going to come into contact with each other and with
the sides of the container much more often. You may remember from earlier that we said that
pressure is a measure of the frequency of collisions of gas particles with each other and with
the sides of the container they are in. So, if the volume decreases, the pressure will naturally
increase. The opposite is true if the volume of the gas is increased. Now, the gas particles collide
less frequently and the pressure will decrease.
Pressure
It was an Englishman named Robert Boyle who was able to take very accurate measurements of
gas pressures and volumes using excellent vacuum pumps. He discovered the startlingly simple
fact that the pressure and volume of a gas are not just vaguely inversely related, but are exactly
inversely proportional. This can be seen when a graph of pressure against the inverse of volume
is plotted. When the values are plotted, the graph is a straight line. This relationship is shown
in figure 8.2.
1/Volume
Figure 8.2: The graph of pressure plotted against the inverse of volume, produces a straight line.
This shows that pressure and volume are exactly inversely proportional.
Definition: Boyle’s Law
The pressure of a fixed quantity of gas is inversely proportional to the volume it occupies
so long as the temperature remains constant.
127
8.2
CHAPTER 8. THERMAL PROPERTIES AND IDEAL GASES - GRADE 11
Important: Proportionality
During this chapter, the terms directly proportional and inversely proportional will be
used a lot, and it is important that you understand their meaning. Two quantities are
said to be proportional if they vary in such a way that one of the quantities is a constant
multiple of the other, or if they have a constant ratio. We will look at two examples to
show the difference between directly proportional and inversely proportional.
1. Directly proportional
A car travels at a constant speed of 120 km/h. The time and the distance covered
are shown in the table below.
Time (mins)
10
20
30
40
Distance (km)
20
40
60
80
What you will notice is that the two quantities shown are constant multiples of each
other. If you divide each distance value by the time the car has been driving, you
will always get 2. This shows that the values are proportional to each other. They
are directly proportional because both values are increasing. In other words, as
the driving time increases, so does the distance covered. The same is true if the
values decrease. The shorter the driving time, the smaller the distance covered. This
relationship can be described mathematically as:
y = kx
where y is distance, x is time and k is the proportionality constant, which in this case
is 2. Note that this is the equation for a straight line graph! The symbol ∝ is also
used to show a directly proportional relationship.
2. Inversely proportional
Two variables are inversely proportional if one of the variables is directly proportional
to the multiplicative inverse of the other. In other words,
y∝
1
x
y=
k
x
or
This means that as one value gets bigger, the other value will get smaller. For example,
the time taken for a journey is inversely proportional to the speed of travel. Look at
the table below to check this for yourself. For this example, assume that the distance
of the journey is 100 km.
Speed (km/h)
100
80
60
40
Time (mins)
60
75
100
150
According to our definition, the two variables are inversely proportional is one variable
is directly proportional to the inverse of the other. In other words, if we divide one
of the variables by the inverse of the other, we should always get the same number.
For example,
100
= 6000
1/60
128you will find that the answer is always 6000.
If you repeat this using the other values,
The variables are inversely proportional to each other.
CHAPTER 8. THERMAL PROPERTIES AND IDEAL GASES - GRADE 11
8.2
We know now that the pressure of a gas is inversely proportional to the volume of the gas,
provided the temperature stays the same. We can write this relationship symbolically as
p∝
1
V
This equation can also be written as follows:
p=
k
V
where k is a proportionality constant. If we rearrange this equation, we can say that:
pV = k
This equation means that, assuming the temperature is constant, multiplying any pressure and
volume values for a fixed amount of gas will always give the same value. So, for example, p1 V1
= k and p2 V2 = k, where the subscripts 1 and 2 refer to two pairs of pressure and volume
readings for the same mass of gas at the same temperature.
From this, we can then say that:
p1 V1 = p2 V2
Important: Remember that Boyle’s Law requires two conditions. First, the amount of
gas must stay constant. Clearly, if you let a little of the air escape from the container
in which it is enclosed, the pressure of the gas will decrease along with the volume, and
the inverse proportion relationship is broken. Second, the temperature must stay constant.
Cooling or heating matter generally causes it to contract or expand. In our original syringe
demonstration, if you were to heat up the gas in the syringe, it would expand and force you
to apply a greater force to keep the plunger at a given position. Again, the proportionality
would be broken.
Activity :: Investigation : Boyle’s Law
Here are some of Boyle’s original data. Note that pressure would originally have
been measured using a mercury manometer and the units for pressure would have
been millimetres mercury or mm Hg. However, to make things a bit easier for you,
the pressure data have been converted to a unit that is more familiar. Note that the
volume is given in terms of arbitrary marks (evenly made).
Volume
(graduation
mark)
12
14
16
18
20
22
24
26
Pressure
(kPa)
398
340
298
264
239
217
199
184
Volume
(graduation
mark)
28
30
32
34
36
38
40
Pressure
(kPa)
170
159
150
141
133
125
120
1. Plot a graph of pressure (p) against volume (V). Volume will be on the x-axis
and pressure on the y-axis. Describe the relationship that you see.
129
In the gas
equations, k
is a ”variable
constant”.
This means
that k is
constant in a
particular set
of situations,
but in two
different sets
of situations
it has different constant
values.
8.2
CHAPTER 8. THERMAL PROPERTIES AND IDEAL GASES - GRADE 11
2. Plot a graph of p against 1/V . Describe the relationship that you see.
3. Do your results support Boyle’s Law? Explain your answer.
teresting Did you know that the mechanisms involved in breathing also relate to Boyle’s
Interesting
Fact
Fact
Law? Just below the lungs is a muscle called the diaphragm. When a person
breathes in, the diaphragm moves down and becomes more ’flattened’ so that the
volume of the lungs can increase. When the lung volume increases, the pressure
in the lungs decreases (Boyle’s law). Since air always moves from areas of high
pressure to areas of lower pressure, air will now be drawn into the lungs because
the air pressure outside the body is higher than the pressure in the lungs. The
opposite process happens when a person breathes out. Now, the diaphragm
moves upwards and causes the volume of the lungs to decrease. The pressure
in the lungs will increase, and the air that was in the lungs will be forced out
towards the lower air pressure outside the body.
Worked Example 26: Boyle’s Law 1
Question: A sample of helium occupies a volume of 160 cm3 at 100 kPa and 25 ◦ C.
What volume will it occupy if the pressure is adjusted to 80 kPa and if the temperature remains unchanged?
Answer
Step 4 : Write down all the information that you know about the gas.
V1 = 160 cm3 and V2 = ?
p1 = 100 kPa and p2 = 80 kPa
Step 1 : Use an appropriate gas law equation to calculate the unknown
variable.
Because the temperature of the gas stays the same, the following equation can be
used:
p1 V1 = p2 V2
If the equation is rearranged, then
V2 =
p1 V1
p2
Step 2 : Substitute the known values into the equation, making sure that
the units for each variable are the same. Calculate the unknown variable.
V2 =
100 × 160
= 200cm3
80
The volume occupied by the gas at a pressure of 80kPa, is 200 cm3
130
CHAPTER 8. THERMAL PROPERTIES AND IDEAL GASES - GRADE 11
8.2
Worked Example 27: Boyle’s Law 2
Question: The pressure on a 2.5 l volume of gas is increased from 695 Pa to 755 Pa
while a constant temperature is maintained. What is the volume of the gas under
these pressure conditions?
Answer
Step 1 : Write down all the information that you know about the gas.
V1 = 2.5 l and V2 = ?
p1 = 695 Pa and p2 = 755 Pa
Step 2 : Choose a relevant gas law equation to calculate the unknown variable.
At constant temperature,
p1 V1 = p2 V2
Therefore,
V2 =
p1 V1
p2
Step 3 : Substitute the known values into the equation, making sure that
the units for each variable are the same. Calculate the unknown variable.
V2 =
695 × 2.5
= 2.3l
755
Important:
It is not necessary to convert to Standard International (SI) units in the examples we have
used above. Changing pressure and volume into different units involves multiplication. If
you were to change the units in the above equation, this would involve multiplication on
both sides of the equation, and so the conversions cancel each other out. However, although
SI units don’t have to be used, you must make sure that for each variable you use the same
units throughout the equation. This is not true for some of the calculations we will do at a
later stage, where SI units must be used.
Exercise: Boyle’s Law
1. An unknown gas has an initial pressure of 150 kPa and a volume of 1 L. If the
volume is increased to 1.5 L, what will the pressure now be?
2. A bicycle pump contains 250 cm3 of air at a pressure of 90 kPa. If the air is
compressed, the volume is reduced to 200 cm3 . What is the pressure of the air
inside the pump?
3. The air inside a syringe occupies a volume of 10 cm3 and exerts a pressure of
100 kPa. If the end of the syringe is sealed and the plunger is pushed down, the
pressure increases to 120 kPa. What is the volume of the air in the syringe?
4. During an investigation to find the relationship between the pressure and volume
of an enclosed gas at constant temperature, the following results were obtained.
131
8.3
CHAPTER 8. THERMAL PROPERTIES AND IDEAL GASES - GRADE 11
Volume (cm3 )
40
30
25
Pressure (kPa)
125.0
166.7
200.0
(a) For the results given in the above table, plot a graph of pressure (y-axis)
against the inverse of volume (x-axis).
(b) From the graph, deduce the relationship between the pressure and volume
of an enclosed gas at constant temperature.
(c) Use the graph to predict what the volume of the gas would be at a pressure
of 40 kPa. Show on your graph how you arrived at your answer.
(IEB 2004 Paper 2 )
8.3
Charles’s Law: Volume and Temperature of an enclosed
gas
Charles’s law describes the relationship between the volume and temperature of a gas. The
law was first published by Joseph Louis Gay-Lussac in 1802, but he referenced unpublished work
by Jacques Charles from around 1787. This law states that at constant pressure, the volume of
a given mass of an ideal gas increases or decreases by the same factor as its temperature (in
kelvin) increases or decreases. Another way of saying this is that temperature and volume are
directly proportional (figure ??).
Definition: Charles’s Law
The volume of an enclosed sample of gas is directly proportional to its absolute temperature
provided the pressure is kept constant.
teresting Charles’s Law is also known as Gay-Lussac’s Law. This is because Charles
Interesting
Fact
Fact
did not publish his discovery, and it was rediscovered independently by another
French Chemist Joseph Louis Gay-Lussac some years later.
Activity :: Demonstration : Charles’s Law
Aim:
To demonstrate Charles’s Law using simple materials.
Apparatus:
glass bottle (e.g. empty glass coke bottle), balloon, bunsen burner, retort stand
Method:
1. Place the balloon over the opening of the empty bottle.
2. Place the bottle on the retort stand over the bunsen burner and allow it to
heat up. Observe what happens to the balloon. WARNING: Be careful when
handling the heated bottle. You may need to wear gloves for protection.
132
CHAPTER 8. THERMAL PROPERTIES AND IDEAL GASES - GRADE 11
8.3
Results:
You should see that the balloon starts to expand. As the air inside the bottle
is heated, the pressure also increases, causing the volume to increase. Since the
volume of the glass bottle can’t increase, the air moves into the balloon, causing it
to expand.
Conclusion:
The temperature and volume of the gas are directly related to each other. As
one increases, so does the other.
Mathematically, the relationship between temperature and pressure can be represented as follows:
V ∝T
or
V = kT
If the equation is rearranged, then...
V
=k
T
and, following the same logic that was used for Boyle’s law:
V2
V1
=
T1
T2
Volume
The equation relating volume and temperature produces a straight line graph (refer back to the
notes on proportionality if this is unclear). This relationship is shown in figure 8.3.
0
Temperature (K)
Figure 8.3: The volume of a gas is directly proportional to its temperature, provided the pressure
of the gas is constant.
However, if this graph is plotted on a celsius temperature scale, the zero point of temperature
doesn’t correspond to the zero point of volume. When the volume is zero, the temperature is
actually -273.150C (figure 8.4.
A new temperature scale, the Kelvin scale must be used instead. Since zero on the Celsius
scale corresponds with a Kelvin temperature of -273.150C, it can be said that:
Kelvin temperature (T) = Celsius temperature (t) + 273.15
133
CHAPTER 8. THERMAL PROPERTIES AND IDEAL GASES - GRADE 11
Volume (kPa)
8.3
-273◦ C
0K
0◦ C
273 K
Temperature
Figure 8.4: The relationship between volume and temperature, shown on a Celsius temperature
scale.
At school level, you can simplify this slightly and convert between the two temperature scales as
follows:
T = t + 273
or
t = T - 273
Can you explain Charles’s law in terms of the kinetic theory of gases? When the temperature
of a gas increases, so does the average speed of its molecules. The molecules collide with the
walls of the container more often and with greater impact. These collisions will push back the
walls, so that the gas occupies a greater volume than it did at the start. We saw this in the
first demonstration. Because the glass bottle couldn’t expand, the gas pushed out the balloon
instead.
Exercise: Charles’s law
The table below gives the temperature (in 0 C) of a number of gases under
different volumes at a constant pressure.
Volume (l)
0
0.25
0.5
0.75
1.0
1.5
2
2.5
3.0
3.5
He
-272.4
-245.5
-218.6
-191.8
-164.9
-111.1
-57.4
-3.6
50.2
103.9
H2
-271.8
-192.4
-113.1
-33.7
45.7
204.4
363.1
521.8
680.6
839.3
N2 O
-275.0
-123.5
28.1
179.6
331.1
634.1
937.2
1240.2
1543.2
1846.2
1. On the same set of axes, draw graphs to show the relationship between temperature and volume for each of the gases.
2. Describe the relationship you observe.
3. If you extrapolate the graphs (in other words, extend the graph line even though
you may not have the exact data points), at what temperature do they intersect?
4. What is significant about this temperature?
134
CHAPTER 8. THERMAL PROPERTIES AND IDEAL GASES - GRADE 11
8.3
Worked Example 28: Charles’s Law 1
Question: Ammonium chloride and calcium hydroxide are allowed to react. The
ammonia that is released in the reaction is collected in a gas syringe and sealed in.
This gas is allowed to come to room temperature which is 32◦ C. The volume of the
ammonia is found to be 122 ml. It is now placed in a water bath set at 7◦ C. What
will be the volume reading after the syringe has been left in the bath for a good
while (assume the plunger moves completely freely)?
Answer
Step 1 : Write down all the information that you know about the gas.
V1 = 122 ml and V2 = ?
T1 = 320 C and T2 = 70 C
Step 2 : Convert the known values to SI units if necessary.
Here, temperature must be converted into Kelvin, therefore:
T1 = 32 + 273 = 305 K
T2 = 7 + 273 = 280 K
Step 3 : Choose a relevant gas law equation that will allow you to calculate
the unknown variable.
V2
V1
=
T1
T2
Therefore,
V2 =
V1 × T2
T1
Step 4 : Substitute the known values into the equation.
unknown variable.
V2 =
Calculate the
122 × 280
= 112ml
305
Important:
Note that here the temperature must be converted to Kelvin (SI) since the change from
degrees Celcius involves addition, not multiplication by a fixed conversion ratio (as is the
case with pressure and volume.)
Worked Example 29: Charles’s Law 2
Question: At a temperature of 298 K, a certain amount of CO2 gas occupies a
volume of 6 l. What volume will the gas occupy if its temperature is reduced to 273
K?
Answer
135
8.4
CHAPTER 8. THERMAL PROPERTIES AND IDEAL GASES - GRADE 11
Step 1 : Write down all the information that you know about the gas.
V1 = 6 l and V2 = ?
T1 = 298 K and T2 = 273 K
Step 2 : Convert the known values to SI units if necessary.
Temperature data is already in Kelvin, and so no conversions are necessary.
Step 3 : Choose a relevant gas law equation that will allow you to calculate
the unknown variable.
V2
V1
=
T1
T2
Therefore,
V2 =
V1 × T2
T1
Step 4 : Substitute the known values into the equation.
unknown variable.
V2 =
8.4
Calculate the
6 × 273
= 5.5l
298
The relationship between temperature and pressure
Pressure
The pressure of a gas is directly proportional to its temperature, if the volume is kept constant
(figure 8.5). When the temperature of a gas increases, so does the energy of the particles.
This causes them to move more rapidly and to collide with each other and with the side of the
container more often. Since pressure is a measure of these collisions, the pressure of the gas
increases with an increase in temperature. The pressure of the gas will decrease if its temperature
decreases.
0
Temperature (K)
Figure 8.5: The relationship between the temperature and pressure of a gas
In the same way that we have done for the other gas laws, we can describe the relationship
between temperature and pressure using symbols, as follows:
T ∝ p, therefore p = kT
We can also say that:
p
=k
T
136
CHAPTER 8. THERMAL PROPERTIES AND IDEAL GASES - GRADE 11
8.5
and that, provided the amount of gas stays the same...
p1
p2
=
T1
T2
Exercise: More gas laws
1. A gas of unknown volume has a temperature of 14◦ C. When the temperature
of the gas is increased to 100◦ C, the volume is found to be 5.5 L. What was
the initial volume of the gas?
2. A gas has an initial volume of 2600 mL and a temperature of 350 K.
(a) If the volume is reduced to 1500 mL, what will the temperature of the gas
be in Kelvin?
(b) Has the temperature increased or decreased?
(c) Explain this change, using the kinetic theory of matter.
3. A cylinder of propane gas at a temperature of 20◦ C exerts a pressure of 8 atm.
When a cylinder has been placed in sunlight, its temperature increases to 25◦ C.
What is the pressure of the gas inside the cylinder at this temperature?
8.5
The general gas equation
All the gas laws we have described so far rely on the fact that at least one variable (T, p or V)
remains constant. Since this is unlikely to be the case most times, it is useful to combine the
relationships into one equation. These relationships are as follows:
Boyle’s law: p ∝
1
V
(constant T)
Relationship between p and T: p ∝ T (constant V)
If we combine these relationships, we get p ∝
T
V
If we introduce the proportionality constant k, we get p = k VT
or, rearranging the equation...
pV = kT
We can also rewrite this relationship as follows:
pV
=k
T
Provided the mass of the gas stays the same, we can also say that:
p2 V2
p1 V1
=
T1
T2
In the above equation, the subscripts 1 and 2 refer to two pressure and volume readings for
the same mass of gas under different conditions. This is known as the general gas equation.
137
8.5
CHAPTER 8. THERMAL PROPERTIES AND IDEAL GASES - GRADE 11
Temperature is always in kelvin and the units used for pressure and volume must be the same
on both sides of the equation.
Important:
Remember that the general gas equation only applies if the mass of the gas is fixed.
Worked Example 30: General Gas Equation 1
Question: At the beginning of a journey, a truck tyre has a volume of 30 dm3 and
an internal pressure of 170 kPa. The temperature of the tyre is 160 C. By the end
of the trip, the volume of the tyre has increased to 32 dm3 and the temperature
of the air inside the tyre is 350 C. What is the tyre pressure at the end of the journey?
Answer
Step 1 : Write down all the information that you know about the gas.
p1 = 170 kPa and p2 = ?
V1 = 30 dm3 and V2 = 32 dm3
T1 = 160 C and T2 = 400 C
Step 2 : Convert the known values to SI units if necessary.
Here, temperature must be converted into Kelvin, therefore:
T1 = 16 + 273 = 289 K
T2 = 40 + 273 = 313 K
Step 3 : Choose a relevant gas law equation that will allow you to calculate
the unknown variable.
Use the general gas equation to solve this problem:
p1 × V1
p2 × V2
=
T1
T2
Therefore,
p2 =
p1 × V1 × T2
T1 × V2
Step 4 : Substitute the known values into the equation.
unknown variable.
Calculate the
170 × 30 × 313
= 173kP a
289 × 32
The pressure of the tyre at the end of the journey is 173 kPa.
p2 =
Worked Example 31: General Gas Equation 2
Question: A cylinder that contains methane gas is kept at a temperature of 150 C
and exerts a pressure of 7 atm. If the temperature of the cylinder increases to 250 C,
what pressure does the gas now exert? (Refer to table 8.1 to see what an ’atm’ is.
Answer
Step 1 : Write down all the information that you know about the gas.
p1 = 7 atm and p2 = ?
138
CHAPTER 8. THERMAL PROPERTIES AND IDEAL GASES - GRADE 11
T1 = 150 C and T2 = 250 C
Step 2 : Convert the known values to SI units if necessary.
Here, temperature must be converted into Kelvin, therefore:
T1 = 15 + 273 = 288 K
T2 = 25 + 273 = 298 K
Step 3 : Choose a relevant gas law equation that will allow you to calculate
the unknown variable.
Since the volume of the cylinder is constant, we can write:
p1
p2
=
T1
T2
Therefore,
p2 =
p1 × T 2
T1
Step 4 : Substitute the known values into the equation.
unknown variable.
Calculate the
7 × 298
= 7.24atm
288
The pressure of the gas is 7.24 atm.
p2 =
Worked Example 32: General Gas Equation 3
Question: A gas container can withstand a pressure of 130 kPa before it will start
to leak. Assuming that the volume of the gas in the container stays the same, at
what temperature will the container start to leak if the gas exerts a pressure of 100
kPa at 150 C?
Answer
Step 1 : Write down all the information that you know about the gas.
p1 = 100 kPa and p2 = 130 kPa
T1 = 150 C and T2 = ?
Step 2 : Convert the known values to SI units if necessary.
Here, temperature must be converted into Kelvin, therefore:
T1 = 15 + 273 = 288 K
Step 3 : Choose a relevant gas law equation that will allow you to calculate
the unknown variable.
Since the volume of the container is constant, we can write:
p1
p2
=
T1
T2
Therefore,
Therefore,
p1
1
=
T2
T 1 × p2
T2 =
T 1 × p2
p1
139
8.5
8.6
CHAPTER 8. THERMAL PROPERTIES AND IDEAL GASES - GRADE 11
Step 4 : Substitute the known values into the equation.
unknown variable.
T2 =
Calculate the
288 × 130
= 374.4K = 101.40 C
100
Exercise: The general gas equation
1. A closed gas system initially has a volume of 8 L and a temperature of 100◦ C.
The pressure of the gas is unknown. If the temperature of the gas decreases to
50◦ C, the gas occupies a volume of 5 L. If the pressure of the gas under these
conditions is 1.2 atm, what was the initial pressure of the gas?
2. A balloon is filled with helium gas at 27◦ C and a pressure of 1.0 atm. As the
balloon rises, the volume of the balloon increases by a factor of 1.6 and the
temperature decreases to 15◦ C. What is the final pressure of the gas (assuming
none has escaped)?
3. 25 cm3 of gas at 1 atm has a temperature of 20◦ C. When the gas is compressed
to 20 cm3 , the temperature of the gas increases to 28◦ C. Calculate the final
pressure of the gas.
8.6
The ideal gas equation
In the early 1800’s, Amedeo Avogadro hypothesised that if you have samples of different gases,
of the same volume, at a fixed temperature and pressure, then the samples must contain the
same number of freely moving particles (i.e. atoms or molecules).
Definition: Avogadro’s Law
Equal volumes of gases, at the same temperature and pressure, contain the same number
of molecules.
You will remember from an earlier section, that we combined different gas law equations to get
one that included temperature, volume and pressure. In this equation, pV = kT, the value of k
is different for different masses of gas. If we were to measure the amount of gas in moles, then
k = nR, where n is the number of moles of gas and R is the universal gas constant. The value
of R is 8.3143 J.K−1 , or for most calculations, 8.3 J.K−1 . So, if we replace k in the general gas
equation, we get the following ideal gas equation.
pV = nRT
Important:
1. The value of R is the same for all gases
2. All quantities in the equation pV = nRT must be in the same units as the value of
R. In other words, SI units must be used throughout the equation.
The following table may help you when you convert to SI units.
140
CHAPTER 8. THERMAL PROPERTIES AND IDEAL GASES - GRADE 11
8.6
Table 8.1: Conversion table showing different units of measurement for volume, pressure and
temperature.
Variable
Pressure (p)
Volume (V)
moles (n)
SI unit
Other units
and conversions
Pascals (Pa)
760 mm Hg
= 1 atm =
101325 Pa =
101.325 kPa
m3
1
m3
=
1000000 cm3
= 1000 dm3
= 1000 litres
mol
universal gas
constant (R)
J.mol.K−1
temperature
(K)
kelvin (K)
K = 0C +
273
Worked Example 33: Ideal gas equation 1
Question: Two moles of oxygen (O2 ) gas occupy a volume of 25 dm3 at a temperature of 400 C. Calculate the pressure of the gas under these conditions.
Answer
Step 1 : Write down all the information that you know about the gas.
p=?
V = 25 dm3
n=2
T = 400 C
Step 2 : Convert the known values to SI units if necessary.
V =
25
= 0.025m3
1000
T = 40 + 273 = 313K
Step 3 : Choose a relevant gas law equation that will allow you to calculate
the unknown variable.
pV = nRT
Therefore,
p=
nRT
V
Step 4 : Substitute the known values into the equation.
unknown variable.
Calculate the
p = 2 × 8.3 × 3130.025 = 207832P a = 207.8kP a
Worked Example 34: Ideal gas equation 2
Question: Carbon dioxide (CO2 ) gas is produced as a result of the reaction between
calcium carbonate and hydrochloric acid. The gas that is produced is collected in
a 20 dm3 container. The pressure of the gas is 105 kPa at a temperature of 200 C.
141
8.6
CHAPTER 8. THERMAL PROPERTIES AND IDEAL GASES - GRADE 11
What mass of carbon dioxide was produced?
Answer
Step 1 : Write down all the information that you know about the gas.
p = 105 kPa
V = 20 dm3
T = 200 C
Step 2 : Convert the known values to SI units if necessary.
p = 105 × 1000 = 105000P a
T = 20 + 273 = 293K
V =
20
= 0.02m3
1000
Step 3 : Choose a relevant gas law equation that will allow you to calculate
the unknown variable.
pV = nRT
Therefore,
n=
pV
RT
Step 4 : Substitute the known values into the equation.
unknown variable.
n=
Calculate the
105000 × 0.02
= 0.86moles
8.3 × 293
Step 5 : Calculate mass from moles
n=
m
M
Therefore,
m=n×M
The molar mass of CO2 is calculated as follows:
M = 12 + (2 × 16) = 44g.mol−1
Therefore,
m = 0.86 × 44 = 37.84g
Worked Example 35: Ideal gas equation 3
Question: 1 mole of nitrogen (N2 ) reacts with hydrogen (H2 ) according to the
following equation:
N2 + 3H2 → 2N H3
142
CHAPTER 8. THERMAL PROPERTIES AND IDEAL GASES - GRADE 11
The ammonia (NH3 ) gas is collected in a separate gas cylinder which has a volume
of 25 dm3 . The temperature of the gas is 220 C. Calculate the pressure of the gas
inside the cylinder.
Answer
Step 1 : Write down all the information that you know about the gas.
V = 25 dm3
n = 2 (Calculate this by looking at the mole ratio of nitrogen to ammonia, which is
1:2)
T = 220 C
Step 2 : Convert the known values to SI units if necessary.
V =
25
= 0.025m3
1000
T = 22 + 273 = 295K
Step 3 : Choose a relevant gas law equation that will allow you to calculate
the unknown variable.
pV = nRT
Therefore,
p=
nRT
V
Step 4 : Substitute the known values into the equation.
unknown variable.
p=
Calculate the
2 × 8.3 × 295
= 195880P a = 195.89kP a
0.025
Worked Example 36: Ideal gas equation 4
Question: Calculate the number of air particles in a 10 m by 7 m by 2 m classroom
on a day when the temperature is 23◦ C and the air pressure is 98 kPa.
Answer
Step 1 : Write down all the information that you know about the gas.
V = 10 m × 7 m × 2m = 140 m3
p = 98 kPa
T = 230 C
Step 2 : Convert the known values to SI units if necessary.
p = 98 × 1000 = 98000P a
T = 23 + 273 = 296K
Step 3 : Choose a relevant gas law equation that will allow you to calculate
the unknown variable.
pV = nRT
143
8.6
8.6
CHAPTER 8. THERMAL PROPERTIES AND IDEAL GASES - GRADE 11
Therefore,
n=
pV
RT
Step 4 : Substitute the known values into the equation.
unknown variable.
n=
Calculate the
98000 × 140
= 5584.5mol
8.3 × 296
Worked Example 37: Applying the gas laws
Question: Most modern cars are equipped with airbags for both the driver and the
passenger. An airbag will completely inflate in 0,05 s. This is important because a
typical car collision lasts about 0,125 s. The following reaction of sodium azide (a
compound found in airbags) is activated by an electrical signal:
2N aN3 (s) → 2N a(s) + 3N2 (g)
1. Calculate the mass of N2 (g) needed to inflate a sample airbag to a volume of
65 dm3 at 25 ◦ C and 99,3 kPa. Assume the gas temperature remains constant
during the reaction.
2. In reality the above reaction is exothermic. Describe, in terms of the kinetic
molecular theory, how the pressure in the sample airbag will change, if at all,
as the gas temperature returns to 25 ◦ C.
Answer
Step 1 : Look at the information you have been given, and the information
you still need.
Here you are given the volume, temperature and pressure. You are required to work
out the mass of N2 .
Step 2 : Check that all the units are S.I. units
Pressure: 93,3 × 103 Pa
Volume: 65 × 10−3 m3
Temperature: (273 + 25) K
Gas Constant: 8,31
Step 3 : Write out the Ideal Gas formula
pV = nRT
Step 4 : Solve for the required quantity using symbols
n=
pV
RT
Step 5 : Solve by substituting numbers into the equation to solve for ’n’.
n=
99,3 × 103 × 65 × 10−3
8,31 × (273 + 25)
Step 6 : Convert the number of moles to number of grams
144
CHAPTER 8. THERMAL PROPERTIES AND IDEAL GASES - GRADE 11
m
=
m
m
=
=
8.7
n×M
2,61 × 28
73,0g
Step 7 : Theory Question
When the temperature decreases the intensity of collisions with the walls of the
airbag and between particles decreases. Therefore pressure decreases.
Exercise: The ideal gas equation
1. An unknown gas has pressure, volume and temperature of 0.9 atm, 8 L and
120◦ C respectively. How many moles of gas are present?
2. 6 g of chlorine (Cl2 ) occupies a volume of 0.002 m3 at a temperature of 26◦ C.
What is the pressure of the gas under these conditions?
3. An average pair of human lungs contains about 3.5 L of air after inhalation
and about 3.0 L after exhalation. Assuming that air in your lungs is at 37◦ C
and 1.0 atm, determine the number of moles of air in a typical breath.
4. A learner is asked to calculate the answer to the problem below:
Calculate the pressure exerted by 1.5 moles of nitrogen gas in a container with
a volume of 20 dm3 at a temperature of 37◦ C.
The learner writes the solution as follows:
V = 20 dm3
n = 1.5 mol
R = 8.3 J.K−1 .mol−1
T = 37 + 273 = 310 K
p = nRT, therefore
p=
=
nRV
T
1.5 × 8.3 × 20
310
= 0.8 kPa
(a) Identify 2 mistakes the learner has made in the calculation.
(b) Are the units of the final answer correct?
(c) Rewrite the solution, correcting the mistakes to arrive at the right answer.
8.7
Molar volume of gases
It is possible to calculate the volume of a mole of gas at STP using what we now know about
gases.
1. Write down the ideal gas equation
145
8.8
CHAPTER 8. THERMAL PROPERTIES AND IDEAL GASES - GRADE 11
pV = nRT, therefore V =
nRT
p
2. Record the values that you know, making sure that they are in SI units
You know that the gas is under STP conditions. These are as follows:
p = 101.3 kPa = 101300 Pa
n = 1 mole
R = 8.3 J.K−1 .mol−1
T = 273 K
3. Substitute these values into the original equation.
V =
V =
nRT
p
1mol × 8.3J.K −1 .mol−1 × 273K
101300P a
4. Calculate the volume of 1 mole of gas under these conditions
The volume of 1 mole of gas at STP is 22.4 × 10−3 m3 = 22.4 dm3 .
8.8
Ideal gases and non-ideal gas behaviour
In looking at the behaviour of gases to arrive at the Ideal Gas Law, we have limited our examination to a small range of temperature and pressure. Most gases do obey these laws most
of the time, and are called ideal gases, but there are deviations at high pressures and low
temperatures. So what is happening at these two extremes?
Earlier when we discussed the kinetic theory of gases, we made a number of assumptions about
the behaviour of gases. We now need to look at two of these again because they affect how
gases behave either when pressures are high or when temperatures are low.
1. Molecules do occupy volume
Volume
This means that when pressures are very high and the molecules are compressed, their volume becomes significant. This means that the total volume available for the gas molecules
to move is reduced and collisions become more frequent. This causes the pressure of the
gas to be higher than what would normally have been predicted by Boyle’s law (figure
8.6).
rea
i de
al
ga
lg
as
s
Pressure
Figure 8.6: Gases deviate from ideal gas behaviour at high pressure.
146
CHAPTER 8. THERMAL PROPERTIES AND IDEAL GASES - GRADE 11
8.9
2. Forces of attraction do exist between molecules
as
lg
lg
as
ea
id
re
a
Pressure
At low temperatures, when the speed of the molecules decreases and they move closer
together, the intermolecular forces become more apparent. As the attraction between
molecules increases, their movement decreases and there are fewer collisions between them.
The pressure of the gas at low temperatures is therefore lower than what would have been
expected for an ideal gas (figure 8.7. If the temperature is low enough or the pressure high
enough, a real gas will liquify.
Temperature
Figure 8.7: Gases deviate from ideal gas behaviour at low temperatures
8.9
Summary
• The kinetic theory of matter helps to explain the behaviour of gases under different
conditions.
• An ideal gas is one that obeys all the assumptions of the kinetic theory.
• A real gas behaves like an ideal gas, except at high pressures and low temperatures. Under
these conditions, the forces between molecules become significant and the gas will liquify.
• Boyle’s law states that the pressure of a fixed quantity of gas is inversely proportional to
its volume, as long as the temperature stays the same. In other words, pV = k or
p1 V1 = p2 V2 .
• Charles’s law states that the volume of an enclosed sample of gas is directly proportional
to its temperature, as long as the pressure stays the same. In other words,
V1
V2
=
T1
T2
• The temperature of a fixed mass of gas is directly proportional to its pressure, if the
volume is constant. In other words,
p1
p2
=
T1
T2
• In the above equations, temperature must be written in Kelvin. Temperature in degrees
Celsius (temperature = t) can be converted to temperature in Kelvin (temperature = T)
using the following equation:
T = t + 273
147
8.9
CHAPTER 8. THERMAL PROPERTIES AND IDEAL GASES - GRADE 11
• Combining Boyle’s law and the relationship between the temperature and pressure of a
gas, gives the general gas equation, which applies as long as the amount of gas remains
constant. The general gas equation is pV = kT, or
p1 V1
p2 V2
=
T1
T2
• Because the mass of gas is not always constant, another equation is needed for these
situations. The ideal gas equation can be written as
pV = nRT
where n is the number of moles of gas and R is the universal gas constant, which is 8.3
J.K−1 .mol−1 . In this equation, SI units must be used. Volume (m3 ), pressure (Pa) and
temperature (K).
• The volume of one mole of gas under STP is 22.4 dm3 . This is called the molar gas
volume.
s
Exercise: Summary exercise
1. For each of the following, say whether the statement is true or false. If the
statement is false, rewrite the statement correctly.
(a) Real gases behave like ideal gases, except at low pressures and low temperatures.
(b) The volume of a given mass of gas is inversely proportional to the pressure
it exerts.
(c) The temperature of a fixed mass of gas is directly proportional to its pressure, regardless of the volume of the gas.
2. For each of the following multiple choice questions, choose the one correct
answer.
(a) Which one of the following properties of a fixed quantity of a gas must be
kept constant during a Boyle’s law investigation?
i. density
ii. pressure
iii. temperature
iv. volume
(IEB 2003 Paper 2 )
(b) Three containers of EQUAL VOLUME are filled with EQUAL MASSES
of helium, nitrogen and carbon dioxide gas respectively. The gases in the
three containers are all at the same TEMPERATURE. Which one of the
following statements is correct regarding the pressure of the gases?
i. All three gases will be at the same pressure
ii. The helium will be at the greatest pressure
iii. The nitrogen will be at the greatest pressure
iv. The carbon dioxide will be at the greatest pressure
(IEB 2004 Paper 2 )
(c) One mole of an ideal gas is stored at a temperature T (in Kelvin) in a rigid
gas tank. If the average speed of the gas particles is doubled, what is the
new Kelvin temperature of the gas?
i. 4T
ii. 2T
√
iii. 2T
148
CHAPTER 8. THERMAL PROPERTIES AND IDEAL GASES - GRADE 11
iv. 0.5 T
(IEB 2002 Paper 2 )
(d) The ideal gas equation is given by pV = nRT. Which one of the following
conditions is true according to Avogadro’s hypothesis?
a p ∝ 1/V
(T = constant)
b
V∝T
(p = constant)
c
V∝n
(p, T = constant)
d
p∝T
(n = constant)
(DoE Exemplar paper 2, 2007 )
3. Use your knowledge of the gas laws to explain the following statements.
(a) It is dangerous to put an aerosol can near heat.
(b) A pressure vessel that is poorly designed and made can be a serious safety
hazard (a pressure vessel is a closed, rigi container that is used to hold
gases at a pressure that is higher than the normal air pressure).
(c) The volume of a car tyre increases after a trip on a hot road.
4. Copy the following set of labelled axes and answer the questions that follow:
Volume (m3 )
0
Temperature (K)
(a) On the axes, using a solid line, draw the graph that would be obtained
for a fixed mass of an ideal gas if the pressure is kept constant.
(b) If the gradient of the above graph is measured to be 0.008 m3 .K−1 , calculate the pressure that 0.3 mol of this gas would exert.
(IEB 2002 Paper 2)
5. Two gas cylinders, A and B, have a volume of 0.15 m3 and 0.20 m3 respectively.
Cylinder A contains 1.25 mol He gas at pressure p and cylinder B contains 2.45
mol He gas at standard pressure. The ratio of the Kelvin temperatures A:B is
1.80:1.00. Calculate the pressure of the gas (in kPa) in cylinder A.
(IEB 2002 Paper 2 )
6. A learner investigates the relationship between the Celsius temperature and the
pressure of a fixed amount of helium gas in a 500 cm3 closed container. From
the results of the investigation, she draws the graph below:
pressure
(kPa)
300
10
20 temperature (0 C)
(a) Under the conditions of this investigation, helium gas behaves like an ideal
gas. Explain briefly why this is so.
149
8.9
8.9
CHAPTER 8. THERMAL PROPERTIES AND IDEAL GASES - GRADE 11
(b) From the shape of the graph, the learner concludes that the pressure of
the helium gas is directly proportional to the Celcius temperature. Is her
conclusion correct? Briefly explain your answer.
(c) Calculate the pressure of the helium gas at 0 ◦ C.
(d) Calculate the mass of helium gas in the container.
(IEB 2003 Paper 2 )
7. One of the cylinders of a motor car engine, before compression contains 450
cm3 of a mixture of air and petrol in the gaseous phase, at a temperature of
30◦ C and a pressure of 100 kPa. If the volume of the cylinder after compression
decreases to one tenth of the original volume, and the temperature of the gas
mixture rises to 140◦C, calculate the pressure now exerted by the gas mixture.
8. In an experiment to determine the relationship between pressure and temperature of a fixed mass of gas, a group of learners obtained the following results:
Pressure (kPa)
Temperature (0 C)
Total gas volume (cm3 )
101
0
250
120
50
250
130.5
80
250
138
100
250
(a) Draw a straight-line graph of pressure (on the dependent, y-axis) versus
temperature (on the independent, x-axis) on a piece of graph paper. Plot
the points. Give your graph a suitable heading.
A straight-line graph passing through the origin is essential to obtain a
mathematical relationship between pressure and temperature.
(b) Extrapolate (extend) your graph and determine the temperature (in 0 C)
at which the graph will pass through the temperature axis.
(c) Write down, in words, the relationship between pressure and Kelvin temperature.
(d) From your graph, determine the pressure (in kPa) at 173 K. Indicate on
your graph how you obtained this value.
(e) How would the gradient of the graph be affected (if at all) if a larger mass
of the gas is used? Write down ONLY increases, decreases or stays the
same.
(DoE Exemplar Paper 2, 2007 )
150
Chapter 9
Organic Molecules - Grade 12
9.1
What is organic chemistry?
Organic chemistry is the branch of chemistry that deals with organic molecules. An organic molecule is one which contains carbon, and these molecules can range in size from simple
molecules to complex structures containing thousands of atoms! Although the main element in
organic compounds is carbon, other elements such as hydrogen (H), oxygen (O), nitrogen (N),
sulfur (S) and phosphorus (P) are also common in these molecules.
Until the early nineteenth century, chemists had managed to make many simple compounds
in the laboratory, but were still unable to produce the complex molecules that they found in
living organisms. It was around this time that a Swedish chemist called Jons Jakob Berzelius
suggested that compounds found only in living organisms (the organic compounds) should be
grouped separately from those found in the non-living world (the inorganic compounds). He also
suggested that the laws that governed how organic compounds formed, were different from those
for inorganic compounds. From this, the idea developed that there was a ’vital force’ in organic
compounds. In other words, scientists believed that organic compounds would not follow the
normal physical and chemical laws that applied to other inorganic compounds because the very
’force of life’ made them different.
This idea of a mystical ’vital force’ in organic compounds was weakened when scientists began to
manufacture organic compounds in the laboratory from non-living materials. One of the first to
do this was Friedrich Wohler in 1828, who successfully prepared urea, an organic compound in
the urine of animals which, until that point, had only been found in animals. A few years later a
student of Wohler’s, Hermann Kolbe, made the organic compound acetic acid from inorganic
compounds. By this stage it was acknowledged that organic compounds are governed by exactly
the same laws that apply to inorganic compounds. The properties of organic compounds are not
due to a ’vital force’ but to the unique properties of the carbon atom itself.
Organic compounds are very important in daily life. They make up a big part of our own bodies,
they are in the food we eat and in the clothes we wear. Organic compounds are also used to
make products such as medicines, plastics, washing powders, dyes, along with a list of other
items.
9.2
Sources of carbon
The main source of the carbon in organic compounds is carbon dioxide in the air. Plants use
sunlight to convert carbon dioxide into organic compounds through the process of photosynthesis. Plants are therefore able to make their own organic compounds through photosynthesis,
while animals feed on plants or plant products so that they gain the organic compounds that
they need to survive.
151
9.3
CHAPTER 9. ORGANIC MOLECULES - GRADE 12
Another important source of carbon is fossil fuels such as coal, petroleum and natural gas. This
is because fossil fuels are themselves formed from the decaying remains of dead organisms (refer
to chapter 21 for more information on fossil fuels).
9.3
Unique properties of carbon
Carbon has a number of unique properties which influence how it behaves and how it bonds with
other atoms:
• Carbon has four valence electrons which means that each carbon atom can form bonds
with four other atoms. Because of this, long chain structures can form. These chains
can either be unbranched (figure 9.1) or branched (figure 9.2). Because of the number of
bonds that carbon can form with other atoms, organic compounds can be very complex.
C
C
C
C
Figure 9.1: An unbranched carbon chain
C
C
C
C
C
C
C
Figure 9.2: A branched carbon chain
• Because of its position on the Periodic Table, most of the bonds that carbon forms with
other atoms are covalent. Think for example of a C-C bond. The difference in electronegativity between the two atoms is zero, so this is a pure covalent bond. In the case of a
C-H bond, the difference in electronegativity between carbon (2.5) and hydrogen (2.1) is
so small that C-H bonds are almost purely covalent. The result of this is that most organic
compounds are non-polar. This affects some of the properties of organic compounds.
9.4
Representing organic compounds
There are a number of ways to represent organic compounds. It is useful to know all of these so
that you can recognise a molecule however it is shown. There are three main ways of representing
a compound. We will use the example of a molecule called 2-methylpropane to help explain the
difference between each.
9.4.1
Molecular formula
The molecular formula of a compound shows how many atoms of each type are in a molecule.
The number of each atom is written as a subscript after the atomic symbol. The molecular
formula of 2-methylpropane is:
152
CHAPTER 9. ORGANIC MOLECULES - GRADE 12
9.4
C4 H10
9.4.2
Structural formula
The structural formula of an organic compound shows every bond between every atom in the
molecule. Each bond is represented by a line. The structural formula of 2-methylpropane is
shown in figure 9.3.
H
H
H
C
H
H
H
C
C
C
H
H
H
H
Figure 9.3: The structural formula of 2-methylpropane
9.4.3
Condensed structural formula
When a compound is represented using its condensed structural formula, each carbon atom and
the hydrogen atoms that are bonded directly to it are listed as a molecular formula, followed
by a similar molecular formula for the neighbouring carbon atom. Branched groups are shown
in brackets after the carbon atom to which they are bonded. The condensed structural formula
below shows that in 2-methylpropane, there is a branched chain attached to the second carbon
atom of the main chain. You can check this by looking at the structural formula in figure ??.
CH3 CH(CH3 )CH3
Exercise: Representing organic compounds
1. For each of the following organic compounds, give the condensed structural
formula and the molecular formula.
H
H
H
H
H
C
C
C
C
H
H
(a)
153
H
9.5
CHAPTER 9. ORGANIC MOLECULES - GRADE 12
H
H
H
H
C
H
C
H
H
C
C
C
H
(b)
H
H
2. For each of the following, give the structural formula and the molecular
formula.
(a) CH3 CH2 CH3
(b) CH3 CH2 CH(CH3 )CH3
(c) C2 H6
3. Give two possible structural formulae for the compound with a molecular formula of C4 H10 .
9.5
Isomerism in organic compounds
It is possible for two organic compounds to have the same molecular formula but a different
structural formula. Look for example at the two organic compounds that are shown in figure
9.4.
H
H
H
H
H
H
H
C
C
C
C
H
H
H
H
H
H
C
H
H
H
H
C
C
C
H
H
H
Figure 9.4: Isomers of a 4-carbon organic compound
If you were to count the number of carbon and hydrogen atoms in each compound, you would
find that they are the same. They both have the same molecular formula (C4 H10 ), but their
structure is different and so are their properties. Such compounds are called isomers.
Definition: Isomer
In chemistry, isomers are molecules with the same molecular formula and often with the same
kinds of chemical bonds between atoms, but in which the atoms are arranged differently.
Exercise: Isomers
Match the organic compound in Column A with its isomer Column B:
154
CHAPTER 9. ORGANIC MOLECULES - GRADE 12
9.6
Column A
CH3 CH(CH3 )OH
H
H
9.6
Column B
CH3 CH(CH3 )CH3
H
H
H
H
C
C
C
C
H
H
H
H
CH3
H
H
C
C
C
H
H
H
H H
H
H
CH3
C
C
C
H
H
H
H
H
C3 H7 OH
Functional groups
All organic compounds have a particular bond or group of atoms which we call its functional
group. This group is important in determining how a compound will react.
Definition: Functional group
In organic chemistry, a functional group is a specific group of atoms within molecules,
that are responsible for the characteristic chemical reactions of those molecules. The same
functional group will undergo the same or similar chemical reaction(s) regardless of the size
of the molecule it is a part of.
In one group of organic compounds called the hydrocarbons, the single, double and triple bonds
of the alkanes, elkenes and alkynes are examples of functional groups. In another group, the
alcohols, an oxygen and a hydrogen atom that are bonded to each other form the functional
group for those compounds. All alcohols will contain an oxygen and a hydrogen atom bonded
together in some part of the molecule.
Table 9.1 summarises some of the common functional groups. We will look at these in more
detail later in this chapter.
9.7
The Hydrocarbons
Let us first look at a group of organic compounds known as the hydrocarbons. These molecules
only contain carbon and hydrogen. The hydrocarbons that we are going to look at are called
aliphatic compounds. The aliphatic compounds are divided into acyclic compounds (chain
structures) and cyclic compounds (ring structures). The chain structures are further divided into
structures that contain only single bonds (alkanes), those that contain double bonds (alkenes)
and those that contain triple bonds (alkynes). Cyclic compounds include structures such as the
benzene ring. Figure 9.5 summarises the classification of the hydrocarbons.
Hydrocarbons that contain only single bonds are called saturated hydrocarbons because each
carbon atom is bonded to as many hydrogen atoms as possible. Figure 9.6 shows a molecule of
ethane which is a saturated hydrocarbon.
155
9.7
Name of group
CHAPTER 9. ORGANIC MOLECULES - GRADE 12
Functional group
C
Example
Diagram
C
H
Alkane
H
H
C
C
H
H
H
Ethane
H
C
H
C
C
C
H
Alkene
H
Ethene
C
C
H
Alkyne
C
C
H
Ethyne (acetylene)
H
C
CH3
X
X
C
H
(X=F,Cl,Br,I)
Halo-alkane
Chloroethane
C
H
OH
C
C
H
H
OH
H
Alcohol/ alkanol
H
Ethanol
O
O
CH3
C
OH
Carboxylic acid
ethanoic acid
H
O
C
H
R
Amine
OH
N
H
CH3
C
C
H
N
H
Glycine
Table 9.1: Some functional groups of organic compounds
Hydrocarbons that contain double or triple bonds are called unsaturated hydrocarbons because
they don’t contain as many hydrogen atoms as possible. Figure 9.7 shows a molecule of ethene
which is an unsaturated hydrocarbon. If you compare the number of carbon and hydrogen atoms
156
H
CHAPTER 9. ORGANIC MOLECULES - GRADE 12
9.7
Aliphatic hydrocarbons
Acyclic compounds
(chain structures)
Alkanes (single bonds)
Cyclic compounds
(ring structures e.g. benzene ring)
Alkenes (contain double bonds)
Alkynes (contain triple bonds)
Figure 9.5: The classification of the aliphatic hydrocarbons
H
H
H
C
C
H
H
H
Figure 9.6: A saturated hydrocarbon
in a molecule of ethane and a molecule of ethene, you will see that the number of hydrogen
atoms in ethene is less than the number of hydrogen atoms in ethane despite the fact that they
both contain two carbon atoms. In order for an unsaturated compound to become saturated,
a double bond has to be broken, and another two hydrogen atoms added for each double bond
that is replaced by a single bond.
H
H
C
C
H
H
Figure 9.7: An unsaturated hydrocarbon
teresting Fat that occurs naturally in living matter such as animals and plants is used as
Interesting
Fact
Fact
food for human consumption and contains varying proportions of saturated and
unsaturated fat. Foods that contain a high proportion of saturated fat are butter,
ghee, suet, tallow, lard, coconut oil, cottonseed oil, and palm kernel oil, dairy
products (especially cream and cheese), meat, and some prepared foods. Diets
high in saturated fat are correlated with an increased incidence of atherosclerosis
and coronary heart disease according to a number of studies. Vegetable oils
contain unsaturated fats and can be hardened to form margarine by adding
hydrogen on to some of the carbon=carbon double bonds using a nickel catalyst.
The process is called hydrogenation
157
9.7
CHAPTER 9. ORGANIC MOLECULES - GRADE 12
We will now go on to look at each of the hydrocarbon groups in more detail. These groups are
the alkanes, the alkenes and the alkynes.
9.7.1
The Alkanes
The alkanes are hydrocarbons that only contain single covalent bonds between their carbon
atoms. This means that they are saturated compounds and are quite unreactive. The simplest
alkane has only one carbon atom and is called methane. This molecule is shown in figure 9.8.
H
(a)
H
C
H
(b)
CH4
H
Figure 9.8: The structural (a) and molecular formula (b) for methane
The second alkane in the series has two carbon atoms and is called ethane. This is shown in
figure 9.9.
(a)
H
H
H
C
C
H
H
(b) C2 H6
H
Figure 9.9: The structural (a) and molecular formula (b) for ethane
The third alkane in the series has three carbon atoms and is called propane (Figure 9.10).
(a)
H
H
H
H
C
C
C
H
H
H
H
(b) C3 H8
Figure 9.10: The structural (a) and molecular formula (b) for propane
When you look at the molecular formula for each of the alkanes, you should notice a pattern
developing. For each carbon atom that is added to the molecule, two hydrogen atoms are added.
In other words, each molecule differs from the one before it by CH2 . This is called a homologous
series. The alkanes have the general formula Cn H2n+2 .
The alkanes are the most important source of fuel in the world and are used extensively in the
chemical industry. Some are gases (e.g. methane and ethane), while others are liquid fuels (e.g.
octane, an important component of petrol).
teresting Some fungi use alkanes as a source of carbon and energy. One fungus AmorInteresting
Fact
Fact
photheca resinae prefers the alkanes used in aviation fuel, and this can cause
problems for aircraft in tropical areas!
158
CHAPTER 9. ORGANIC MOLECULES - GRADE 12
9.7.2
9.7
Naming the alkanes
In order to give compounds a name, certain rules must be followed. When naming organic
compounds, the IUPAC (International Union of Pure and Applied Chemistry) nomenclature is
used. We will first look at some of the steps that need to be followed when naming a compound,
and then try to apply these rules to some specific examples.
1. STEP 1: Recognise the functional group in the compound. This will determine the suffix
(the ’end’) of the name. For example, if the compound is an alkane, the suffix will be
-ane; if the compound is an alkene the suffix will be -ene; if the compound is an alcohol
the suffix will be -ol, and so on.
2. STEP 2: Find the longest continuous carbon chain (it won’t always be a straight chain)
and count the number of carbon atoms in this chain. This number will determine the prefix
(the ’beginning’) of the compound’s name. These prefixes are shown in table 9.2. So, for
example, an alkane that has 3 carbon atoms will have the suffix prop and the compound’s
name will be propane.
Carbon atoms
1
2
3
4
5
6
7
8
9
10
prefix
meth(ane)
eth(ane)
prop(ane)
but(ane)
pent(ane)
hex(ane)
hept(ane)
oct(ane)
non(ane)
dec(ane)
Table 9.2: The prefix of a compound’s name is determined by the number of carbon atoms in
the longest chain
3. STEP 3: Number the carbons in the longest carbon chain (Important: If there is a double
or triple bond, you need to start numbering so that the bond is at the carbon with the
lowest number.
4. STEP 4: Look for any branched groups and name them. Also give them a number to
show their position on the carbon chain. If there are no branched groups, this step can be
left out.
5. STEP 5: Combine the elements of the name into a single word in the following order:
branched groups; prefix; name ending according to the functional group and its position
along the longest carbon chain.
Worked Example 38: Naming the alkanes
Question: Give the IUPAC name for the following compound:
Note: The numbers attached to the carbon atoms would not normally be shown.
The atoms have been numbered to help you to name the compound.
Answer
Step 1 : Identify the functional group
The compound is a hydrocarbon with single bonds between the carbon atoms. It is
an alkane and will have a suffix of -ane.
Step 2 : Find the longest carbon chain
159
9.7
CHAPTER 9. ORGANIC MOLECULES - GRADE 12
H
H
H
H
H
C(1)
C(2)
C(3)
C(4)
H
H
H
H
H
There are four carbon atoms in the longest chain. The prefix of the compound will
be ’but’.
Step 3 : Number the carbons in the longest chain
In this case, it is easy. The carbons are numbered from left to right, from one to four.
Step 4 : Look for any branched groups, name them and give their position
on the carbon chain
There are no branched groups in this compound.
Step 5 : Combine the elements of the name into a single word
The name of the compound is butane.
Worked Example 39: Naming the alkanes
Question: Give the IUPAC name for the following compound:
H
H
H
H
C
C
C
H
H
H
H
C
H
H
Answer
Step 1 : Identify the functional group
The compound is an alkane and will have the suffix -ane.
Step 2 : Find the longest carbon chain
There are three carbons in the longest chain. The prefix for this compound is -prop.
Step 3 : Number the carbons in the carbon chain
If we start at the carbon on the left, we can number the atoms as shown below:
Step 4 : Look for any branched groups, name them and give their position
on the carbon chain
There is a branched group attached to the second carbon atom. This group has the
formula CH3 which is methane. However, because it is not part of the main chain,
it is given the suffix -yl (i.e. methyl). The position of the methyl group comes just
160
CHAPTER 9. ORGANIC MOLECULES - GRADE 12
H
9.7
H
H
H
C(1)
C(2)
C(3)
H
H
H
H
C
H
H
before its name (see next step).
Step 5 : Combine the elements of the compound’s name into a single word in
the order of branched groups; prefix; name ending according to the functional
group.
The compound’s name is 2-methylpropane.
Worked Example 40: Naming the alkanes
Question: Give the IUPAC name for the following compound:
CH3 CH(CH3 )CH(CH3 )CH3
(Remember that the side groups are shown in brackets after the carbon atom to
which they are attached.)
Answer
Step 1 : Draw the compound from its condensed structural formula
The structural formula of the compound is:
H
H
CH3 CH3
H
C(1)
C(2)
C(3)
C(4)
H
H
H
H
H
Step 2 : Identify the functional group
The compound is an alkane and will have the suffix -ane.
Step 3 : Find the longest carbon chain
There are four carbons in the longest chain. The prefix for this compound is -but.
Step 4 : Number the carbons in the carbon chain
If we start at the carbon on the left, carbon atoms are numbered as shown in the
diagram above. A second way that the carbons could be numbered is:
CH3 CH3(4)
H
C(1)
C(2)
C(3)
C
H
H
H
H
H
H
161
H
9.7
CHAPTER 9. ORGANIC MOLECULES - GRADE 12
Step 5 : Look for any branched groups, name them and give their position
on the carbon chain
There are two methyl groups attached to the main chain. The first one is attached
to the second carbon atom and the second methyl group is attached to the third
carbon atom. Notice that in this example it does not matter how you have chosen
to number the carbons in the main chain; the methyl groups are still attached to the
second and third carbons and so the naming of the compound is not affected.
Step 6 : Combine the elements of the compound’s name into a single word in
the order of branched groups; prefix; name ending according to the functional
group.
The compound’s name is 2,3-dimethyl-butane.
Worked Example 41: Naming the alkanes
Question: Give the IUPAC name for the following compound:
H
CH3
H
H
H
C
C
C
C
H
H
CH2
H
H
CH3
Answer
Step 1 : Identify the functional group
The compound is an alkane and will have the suffix -ane.
Step 2 : Find the longest carbon chain and number the carbons in the longest
chain.
There are five carbons in the longest chain if they are numbered as shown below.
The prefix for the compound is -pent.
H
CH3
H
H
H
C(1)
C(2)
C(3)
C
H
H CH2(4)
H
H
CH3(5)
Step 3 : Look for any branched groups, name them and give their position
on the carbon chain
There are two methyl groups attached to the main chain. The first one is attached
to the first carbon atom and the second methyl group is attached to the third carbon
atom.
Step 4 : Combine the elements of the compound’s name into a single word in
the order of branched groups; prefix; name ending according to the functional
group.
The compound’s name is 1,3-dimethyl-pentane.
162
CHAPTER 9. ORGANIC MOLECULES - GRADE 12
9.7
Exercise: Naming the alkanes
1. Give the structural formula for each of the following:
(a)
(b)
(c)
(d)
Octane
CH3 CH2 CH3
CH3 CH(CH3 )CH3
3-ethyl-pentane
2. Give the IUPAC name for each of the following organic compounds.
H
H
H
CH3
C
C
C
H
H
H
H
(a)
(b) CH3 CH2 CH(CH3 )CH2 CH3
(c) CH3 CH(CH3 )CH2 CH(CH3 )CH3
9.7.3
Properties of the alkanes
We have already mentioned that the alkanes are relatively unreactive because of their stable
C-C and C-H bonds. The boiling point and melting point of these molecules is determined by
their molecular structure, and their surface area. The more carbon atoms there are in an alkane,
the greater the surface area and therefore the higher the boiling point. The melting point also
increases as the number of carbon atoms in the molecule increases. This can be seen in the data
in table 9.3.
Formula
CH4
C2 H6
C3 H8
C4 H10
C5 H12
C6 H14
C17 H36
Name
methane
ethane
propane
butane
pentane
hexane
heptadecane
Melting point (0 C)
-183
-182
-187
-138
-130
-95
22
Boiling point (0 C)
-162
-88
-45
-0.5
36
69
302
Phase at room temperature
gas
gas
gas
gas
liquid
liquid
solid
Table 9.3: Properties of some of the alkanes
You will also notice that, when the molecular mass of the alkanes is low (i.e. there are few
carbon atoms), the organic compounds are gases because the intermolecular forces are weak. As
the number of carbon atoms and the molecular mass increases, the compounds are more likely
to be liquids or solids because the intermolecular forces are stronger.
9.7.4
Reactions of the alkanes
There are three types of reactions that can occur in saturated compounds such as the alkanes.
1. Substitution reactions
Substitution reactions involve the removal of a hydrogen atom which is replaced by an
atom of another element, such as a halogen (F, Cl, Br or I) (figure 9.11). The product is
called a halo-alkane. Since alkanes are not very reactive, heat or light are needed for this
163
9.7
CHAPTER 9. ORGANIC MOLECULES - GRADE 12
H
H
H
C
C
+
H
HBr
H
H
H
C
C
H
H
Br
Figure 9.11: A substitution reaction
reaction to take place.
e.g. CH2 =CH2 + HBr → CH3 -CH2 -Br (halo-alkane)
Halo-alkanes (also sometimes called alkyl halides) that contain methane and chlorine
are substances that can be used as anaesthetics during operations. One example is
trichloromethane, also known as ’chloroform’ (figure 9.12).
H
Cl
CHCl3
C
Cl
Cl
Figure 9.12: Trichloromethane
2. Elimination reactions
Saturated compounds can also undergo elimination reactions to become unsaturated (figure 9.13). In the example below, an atom of hydrogen and chlorine are eliminated from
the original compound to form an unsaturated halo-alkene.
e.g. CH2 Cl − CH2 Cl → CH2 = CHCl + HCl
H
H
H
C
C
Cl
Cl
H
H
H
H
C
C
Cl
+
HCl
Figure 9.13: An elimination reaction
3. Oxidation reactions
When alkanes are burnt in air, they react with the oxygen in air and heat is produced. This
is called an oxidation or combustion reaction. Carbon dioxide and water are given off as
products. Heat is also released during the reaction. The burning of alkanes provides most
of the energy that is used by man.
e.g. CH4 + 2O2 → CO2 + 2H2 O + heat
Exercise: The Alkanes
164
CHAPTER 9. ORGANIC MOLECULES - GRADE 12
1. Give the IUPAC name for each of the following alkanes:
(a) C6 H14
H
H
C
H
H
C
H
(b)
(c) CH3 CH3
H
H
H
C
C
C
H
H
H
H
2. Give the structural formula for each of the following compounds:
(a) octane
(b) 3-methyl-hexane
3. Methane is one of the simplest alkanes and yet it is an important fuel source.
Methane occurs naturally in wetlands, natural gas and permafrost. However,
methane can also be produced when organic wastes (e.g. animal manure and
decaying material) are broken down by bacteria under conditions that are anaerobic (there is no oxygen). The simplified reaction is shown below:
Organic matter → Simple organic acids → Biogas
The organic matter could be carbohydrates, proteins or fats which are broken
down by acid-forming bacteria into simple organic acids such as acetic acid or
formic acid. Methane-forming bacteria then convert these acids into biogases
such as methane and ammonia.
The production of methane in this way is very important because methane can
be used as a fuel source. One of the advantages of methane over other fuels like
coal, is that it produces more energy but with lower carbon dioxide emissions.
The problem however, is that methane itself is a greenhouse gas and has a much
higher global warming potential than carbon dioxide. So, producing methane
may in fact have an even more dangerous impact on the environment.
(a) What is the structural formula of methane?
(b) Write an equation to show the reaction that takes place when methane is
burned as a fuel.
(c) Explain what is meant by the statement that methane ’has a greater global
warming potential than carbon dioxide’.
4. Chlorine and ethane react to form chloroethane and hydrogen chloride.
(a) Write a balanced chemical equation for this reaction, using molecular formulae.
(b) Give the structural formula of chloroethane.
(c) What type of reaction has taken place in this example?
5. Petrol (C8 H18 ) is in fact not pure C8 H18 but a mixture of various alkanes. The
’octane rating’ of petrol refers to the percentage of the petrol which is C8 H18 .
For example, 93 octane fuel contains 93% C8 H18 and 7% other alkanes. The
isomer of C8 H18 referred to in the ’octane rating’ is in fact not octane but
2,2,4-trimethylpentane.
(a) Write an unbalanced equation for the chemical reaction which takes place
when petrol (C8 H18 ) burns in excess oxygen.
(b) Write the general formula of the alkanes.
(c) Define the term structural isomer.
(d) Use the information given in this question and your knowledge of naming
organic compounds to deduce and draw the full structural formula for
2,2,4-trimethylpentane. (IEB pg 25)
165
9.7
9.7
CHAPTER 9. ORGANIC MOLECULES - GRADE 12
9.7.5
The alkenes
In the alkenes, there is at least one double bond between two carbon atoms. This means that
they are unsaturated and are more reactive than the alkanes. The simplest alkene is ethene
(also known as ethylene), which is shown in figure 9.14.
H
(a)
H
C
(b)
C
H
CH2 CH2
(c)
C2 H4
H
Figure 9.14: The (a) structural, (b) condensed structural and (c) molecular structure representations of ethene
As with the alkanes, the elkenes also form a homologous series. They have the general formula
Cn H2n . The second alkene in the series would therefore be C3 H6 . This molecule is known as
propene (figure 9.15). Note that if an alkene has two double bonds, it is called a diene and if
it has three double bonds it is called a triene.
(a)
H
H
H
C
C
H
H
(b) CH3 CHCH2
C
(c)
C3 H6
H
Figure 9.15: The (a) structural, (b) condensed structural and (c) molecular structure representations of propene
The elkenes have a variety of uses. Ethylene for example is a hormone in plants that stimulates
the ripening of fruits and the opening of flowers. Propene is an important compound in the
petrochemicals industry. It is used as a monomer to make polypropylene and is also used as a
fuel gas for other industrial processes.
9.7.6
Naming the alkenes
Similar rules will apply in naming the alkenes, as for the alkanes.
Worked Example 42: Naming the alkenes
Question: Give the IUPAC name for the following compound:
166
CHAPTER 9. ORGANIC MOLECULES - GRADE 12
H
9.7
H
H
H
H
C(1)
C(2)
C(3)
C(4)
H
H
H
Answer
Step 1 : Identify the functional group
The compound is an alkene and will have the suffix -ene.
Step 2 : Find the longest carbon chain
There are four carbon atoms in the longest chain and so the prefix for this compound
will be ’but’.
Step 3 : Number the carbon atoms
Remember that when there is a double or triple bond, the carbon atoms must be
numbered so that the double or triple bond is at the lowest numbered carbon. In
this case, it doesn’t matter whether we number the carbons from the left to right,
or from the right to left. The double bond will still fall between C2 and C3 . The
position of the bond will come just before the suffix in the compound’s name.
Step 4 : Look for any branched groups, name them and give their position
on the carbon chain
There are no branched groups in this molecule.
Step 5 : Name the compound
The name of this compound is but-2-ene.
Worked Example 43: Naming the alkenes
Question: Draw the structural formula for the organic compound 3-methyl-butene
Answer
Step 1 : Identify the functional group
The suffix -ene means that this compound is an alkene and there must be a double
bond in the molecule. There is no number immediately before the suffix which means
that the double bond must be at the first carbon in the chain.
Step 2 : Determine the number of carbons in the longest chain
The prefix for the compound is ’but’ so there must be four carbons in the longest
chain.
Step 3 : Look for any branched groups
There is a methyl group at the third carbon atom in the chain.
Step 4 : Combine this information to draw the structural formula for this
molecule.
167
9.7
CHAPTER 9. ORGANIC MOLECULES - GRADE 12
H
H
C
H
H
C
H
H
C
H
C
C
H
H
H
Worked Example 44: Naming the alkenes
Question: Give the IUPAC name for the following compound:
CH3
H
H
CH2
H
H
C(1)
C(2)
C(3)
C(4)
H
Answer
Step 1 : Identify the functional group
The compound is an alkene and will have the suffix -ene. There is a double bond
between the first and second carbons and also between the third and forth carbons.
The organic compound is therefore a ’diene’.
Step 2 : Find the longest carbon chain and number the carbon atoms
There are four carbon atoms in the longest chain and so the prefix for this compound will be ’but’. The carbon atoms are numbered 1 to 4 in the diagram above.
Remember that the main carbon chain must contain both the double bonds.
Step 3 : Look for any branched groups, name them and give their position
on the carbon chain
There is a methyl group on the first carbon and an ethyl group on the second carbon.
Step 4 : Name the compound
The name of this compound is 1-methyl,2-ethyl-1,3 diene.
Exercise: Naming the alkenes
Give the IUPAC name for each of the following alkenes:
1. C5 H10
2. CH3 CHCHCH3
168
CHAPTER 9. ORGANIC MOLECULES - GRADE 12
H
H
C
C
H
H
C
C
H
H
3.
9.7.7
9.7
The properties of the alkenes
The properties of the alkenes are very similar to those of the alkanes, except that the alkenes are
more reactive because they are unsaturated. As with the alkanes, compounds that have four or
less carbon atoms are gases at room temperature, while those with five or more carbon atoms
are liquids.
9.7.8
Reactions of the alkenes
Alkenes can undergo addition reactions because they are unsaturated. They readily react with
hydrogen, water and the halogens. The double bond is broken and a single, saturated bond is
formed. A new group is then added to one or both of the carbon atoms that previously made
up the double bond. The following are some examples:
1. Hydrogenation reactions
A catalyst such as platinum is normally needed for these reactions
CH2 = CH2 + H2 → CH3 − CH3 (figure 9.16)
H
H
H
C
C
H
+
H2
H
H
H
C
C
H
H
H
H
C
C
H
H
H
Figure 9.16: A hydrogenation reaction
2. Halogenation reactions
CH2 = CH2 + HBr → CH3 − CH2 − Br (figure 9.17)
H
H
H
C
C
H
+
HBr
H
Figure 9.17: A halogenation reaction
3. The formation of alcohols
CH2 = CH2 + H2 O → CH3 − CH2 − OH (figure 9.18)
169
Br
9.7
CHAPTER 9. ORGANIC MOLECULES - GRADE 12
H
H
H
C
C
+
H
H2 O
H
H
H
C
C
H
H
OH
Figure 9.18: The formation of an alcohol
Exercise: The Alkenes
1. Give the IUPAC name for each of the following organic compounds:
H
H
H
C
C
C
H
H
H
H
(a)
(b) CH3 CHCH2
C
H
C
H
H
2. Refer to the data table below which shows the melting point and boiling point
for a number of different organic compounds.
Formula
C4 H10
C5 H12
C6 H14
C4 H8
C5 H10
C6 H12
Name
Butane
Pentane
Hexane
Butene
Pentene
Hexene
Melting point (0 C)
-138
-130
-95
-185
-138
-140
Boiling point (0 C)
-0.5
36
69
-6
30
63
(a) At room temperature (approx. 250 C), which of the organic compounds in
the table are:
i. gases
ii. liquids
(b) In the alkanes...
i. Describe what happens to the melting point and boiling point as the
number of carbon atoms in the compound increases.
ii. Explain why this is the case.
(c) If you look at an alkane and an alkene that have the same number of
carbon atoms...
i. How do their melting points and boiling points compare?
ii. Can you explain why their melting points and boiling points are different?
(d) Which of the compounds, hexane or hexene, is more reactive? Explain
your answer.
3. The following reaction takes place:
CH3 CHCH2 + H2 → CH3 CH2 CH3
(a) Give the name of the organic compound in the reactants.
(b) What is the name of the product?
(c) What type of reaction is this?
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CHAPTER 9. ORGANIC MOLECULES - GRADE 12
9.7
(d) Which compound in the reaction is a saturated hydrocarbon?
9.7.9
The Alkynes
In the alkynes, there is at least one triple bond between two of the carbon atoms. They are
unsaturated compounds and are therefore highly reactive. Their general formula is Cn H2n−2 .
The simplest alkyne is ethyne (figure 9.19), also known as acetylene. Many of the alkynes are
used to synthesise other chemical products.
H
C
C
H
Figure 9.19: Ethyne (acetylene)
teresting The raw materials that are needed to make acetylene are calcium carbonate and
Interesting
Fact
Fact
coal. Acetylene can be produced through the following reactions:
CaCO3 → CaO
CaO + 3C → CaC2 + CO
CaC2 + 2H2 O → Ca(OH)2 + C2 H2
An important use of acetylene is in oxyacetylene gas welding. The fuel gas burns
with oxygen in a torch. An incredibly high heat is produced, and this is enough
to melt metal.
9.7.10
Naming the alkynes
The same rules will apply as for the alkanes and alkenes, except that the suffix of the name will
now be -yne.
Worked Example 45: Naming the alkynes
Question: Give the IUPAC name for the following compound:
CH3
CH
CH2
C
CH3
Answer
Step 1 : Identify the functional group
171
C
CH3
9.8
CHAPTER 9. ORGANIC MOLECULES - GRADE 12
There is a triple bond between two of the carbon atoms, so this compound is an
alkyne. The suffix will be -yne. The triple bond is at the second carbon, so the suffix
will in fact be 2-yne.
Step 2 : Find the longest carbon chain and give the compound the correct
prefix
If we count the carbons in a straight line, there are six. The prefix of the compound’s
name will be ’hex’.
Step 3 : Number the carbons in the longest chain
In this example, you will need to number the carbons from right to left so that the
triple bond is between carbon atoms with the lowest numbers.
Step 4 : Look for any branched groups, name them and show the number of
the carbon atom to which the group is attached
There is a methyl (CH3 ) group attached to the fifth carbon (remember we have
numbered the carbon atoms from right to left).
Step 5 : Combine the elements of the name into a single word in the following
order: branched groups; prefix; name ending according to the functional
group and its position along the longest carbon chain.
If we follow this order, the name of the compound is 5-methyl-hex-2-yne.
Exercise: The alkynes
Give the IUPAC name for each of the following organic compounds.
H
H
CH3
C
C
H
H
C
C
H
1.
2. C2 H2
3. CH3 CH2 CCH
9.8
The Alcohols
An alcohol is any organic compound where there is a hydroxyl functional group (-OH) bound to
a carbon atom. The general formula for a simple alcohol is Cn H2n+1 OH.
The simplest and most commonly used alcohols are methanol and ethanol (figure 9.20).
The alcohols have a number of different uses:
• methylated spirits (surgical spirits) is a form of ethanol where methanol has been added
• ethanol is used in alcoholic drinks
• ethanol is used as an industrial solvent
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CHAPTER 9. ORGANIC MOLECULES - GRADE 12
(a)
OH
H
C
H
H
9.8
(b)
H
OH
H
C
C
H
H
H
Figure 9.20: (a) methanol and (b) ethanol
• methanol and ethanol can both be used as a fuel and they burn more cleanly than gasoline
or diesel (refer to chapter 21 for more information on biofuels as an alternative energy
resource.)
• ethanol is used as a solvent in medical drugs, perfumes and vegetable essences
• ethanol is an antiseptic
teresting
Interesting
Fact
Fact
’Fermentation’ refers to the conversion of sugar to alcohol using yeast (a fungus). The
process of fermentation produces items such as wine, beer and yoghurt. To make wine,
grape juice is fermented to produce alcohol. This reaction is shown below:
C6 H12 O6 → 2CO2 + 2C2 H5 OH + energy
teresting Ethanol is a diuretic. In humans, ethanol reduces the secretion of a hormone
Interesting
Fact
Fact
called antidiuretic hormone (ADH). The role of ADH is to control the amount
of water that the body retains. When this hormone is not secreted in the right
quantities, it can cause dehyration because too much water is lost from the body
in the urine. This is why people who drink too much alcohol can become dehydrated, and experience symptoms such as headaches, dry mouth, and lethargy.
Part of the reason for the headaches is that dehydration causes the brain to
shrink away from the skull slightly. The effects of drinking too much alcohol can
be reduced by drinking lots of water.
9.8.1
Naming the alcohols
The rules used to name the alcohols are similar to those already discussed for the other compounds, except that the suffix of the name will be different because the compound is an alcohol.
173
9.8
CHAPTER 9. ORGANIC MOLECULES - GRADE 12
H
H
OH
H
C1
C2
C3
H
H
H
H
Worked Example 46: Naming alcohols 1
Question: Give the IUPAC name for the following organic compound
Answer
Step 1 : Identify the functional group
The compound has an -OH (hydroxyl) functional group and is therefore an alcohol.
The compound will have the suffix -ol.
Step 2 : Find the longest carbon chain
There are three carbons in the longest chain. The prefix for this compound will
be ’prop’. Since there are only single bonds between the carbon atoms, the suffix
becomes ’propan’ (similar to the alkane ’propane’).
Step 3 : Number the carbons in the carbon chain
In this case, it doesn’t matter whether you start numbering from the left or right.
The hydroxyl group will still be attached to the middle carbon atom, numbered ’2’.
Step 4 : Look for any branched groups, name them and give their position
on the carbon chain.
There are no branched groups in this compound, but you still need to indicate the
position of the hydroxyl group on the second carbon. The suffix will be -2-ol because
the hydroxyl group is attached to the second carbon.
Step 5 : Combine the elements of the compound’s name into a single word in
the order of branched groups; prefix; name ending according to the functional
group.
The compound’s name is propan-2-ol.
Worked Example 47: Naming alcohols 2
Question: Give the IUPAC name for the following compound:
H
OH
OH
H
H
C1
C2
C3
C4
H
H
H
H
174
H
CHAPTER 9. ORGANIC MOLECULES - GRADE 12
9.8
Answer
Step 1 : Identify the functional group
The compound has an -OH (hydroxyl) functional group and is therefore an alcohol.
There are two hydroxyl groups in the compound, so the suffix will be -diol.
Step 2 : Find the longest carbon chain
There are four carbons in the longest chain. The prefix for this compound will be
’butan’.
Step 3 : Number the carbons in the carbon chain
The carbons will be numbered from left to right so that the two hydroxyl groups are
attached to carbon atoms with the lowest numbers.
Step 4 : Look for any branched groups, name them and give their position
on the carbon chain.
There are no branched groups in this compound, but you still need to indicate the
position of the hydroxyl groups on the first and second carbon atoms. The suffix
will therefore become 1,2-diol.
Step 5 : Combine the elements of the compound’s name into a single word in
the order of branched groups; prefix; name ending according to the functional
group.
The compound’s name is butan-1,2-diol.
Exercise: Naming the alcohols
1. Give the structural formula of each of the following organic compounds:
(a) pentan-3-ol
(b) butan-2,3-diol
(c) 2-methyl-propanol
2. Give the IUPAC name for each of the following:
(a) CH3 CH2 CH(OH)CH3
H
CH3
(b)
9.8.2
C
CH2
CH2
CH3
OH
Physical and chemical properties of the alcohols
The hydroxyl group affects the solubility of the alcohols. The hydroxyl group generally makes
the alcohol molecule polar and therefore more likely to be soluble in water. However, the carbon
chain resists solubility, so there are two opposing trends in the alcohols. Alcohols with shorter
carbon chains are usually more soluble than those with longer carbon chains.
Alcohols tend to have higher boiling points than the hydrocarbons because of the strong hydrogen bond between hydrogen and oxygen atoms.
175
9.9
CHAPTER 9. ORGANIC MOLECULES - GRADE 12
Alcohols can show either acidic or basic properties because of the hydroxyl group. They also
undergo oxidation reactions to form aldehydes, ketones and carboxylic acids.
Activity :: Case Study : The uses of the alcohols
Read the extract below and then answer the questions that follow:
The alcohols are a very important group of organic compounds, and they have
a variety of uses. Our most common use of the word ’alcohol’ is with reference
to alcoholic drinks. The alcohol in drinks is in fact ethanol. But ethanol has
many more uses apart from alcoholic drinks! When ethanol burns in air, it produces
carbon dioxide, water and energy and can therefore be used as a fuel on its own,
or in mixtures with petrol. Because ethanol can be produced through fermentation,
this is a useful way for countries without an oil industry to reduce imports of petrol.
Ethanol is also used as a solvent in many perfumes and cosmetics.
Methanol can also be used as a fuel, or as a petrol additive to improve combustion. Most methanol is used as an industrial feedstock, in other words it is used
to make other things such as methanal (formaldehyde), ethanoic acid and methyl
esters. In most cases, these are then turned into other products.
Propan-2-ol is another important alcohol, which is used in a variety of applications as a solvent.
Questions
1. Give the structural formula for propan-2-ol.
2. Write a balanced chemical equation for the combustion reaction of ethanol.
3. Explain why the alcohols are good solvents.
9.9
Carboxylic Acids
Carboxylic acids are organic acids that are characterised by having a carboxyl group, which has
the formula -(C=O)-OH, or more commonly written as -COOH. In a carboxyl group, an oxygen
atom is double-bonded to a carbon atom, which is also bonded to a hydroxyl group. The simplest
carboxylic acid, methanoic acid, is shown in figure 9.21.
O
C
H
OH
Figure 9.21: Methanoic acid
Carboxylic acids are widespread in nature. Methanoic acid (also known as formic acid) has the
formula HCOOH and is found in insect stings. Ethanoic acid (CH3 COOH), or acetic acid, is
the main component of vinegar. More complex organic acids also have a variety of different
functions. Benzoic acid (C6 H5 COOH) for example, is used as a food preservative.
teresting
Interesting
Fact
Fact
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CHAPTER 9. ORGANIC MOLECULES - GRADE 12
9.9
A certain type of ant, called formicine ants, manufacture and secrete formic acid, which is
used to defend themselves against other organisms that might try to eat them.
9.9.1
Physical Properties
Carboxylic acids are weak acids, in other words they only dissociate partially. Why does the
carboxyl group have acidic properties? In the carboxyl group, the hydrogen tends to separate
itself from the oxygen atom. In other words, the carboxyl group becomes a source of positivelycharged hydrogen ions (H+ ). This is shown in figure 9.22.
H
H
C
H
O
C
H
Acetic acid
H
OH
C
O
+
C
H+
O−
H
Acetate ion
Hydrogen ion
Figure 9.22: The dissociation of a carboxylic acid
Exercise: Carboxylic acids
1. Refer to the table below which gives information about a number of carboxylic
acids, and then answer the questions that follow.
Formula
Common Source
name
formic
acid
boiling
point
(0 C)
101
butter
melting
point
(0 C)
methanoic 8.4
acid
ethanoic 16.6
acid
propanoic -20.8
acid
-5.5
valerian
root
goats
pentanoic -34.5
acid
-4
186
ants
CH3 CO2 H
vinegar
propionic
acid
CH3 (CH2 )2 CO2 H butyric
acid
valeric
acid
CH3 (CH2 )4 CO2 H caproic
acid
enanthic
acid
CH3 (CH2 )6 CO2 H caprylic
acid
milk
IUPAC
name
118
141
164
205
vines
-7.5
223
goats
16.3
239
(a) Fill in the missing spaces in the table by writing the formula, common
name or IUPAC name.
(b) Draw the structural formula for butyric acid.
(c) Give the molecular formula for caprylic acid.
(d) Draw a graph to show the relationship between molecular mass (on the
x-axis) and boiling point (on the y-axis)
177
9.10
CHAPTER 9. ORGANIC MOLECULES - GRADE 12
i. Describe the trend you see.
ii. Suggest a reason for this trend.
9.9.2
Derivatives of carboxylic acids: The esters
When an alcohol reacts with a carboxylic acid, an ester is formed. Most esters have a characteristic and pleasant smell. In the reaction, the hydrogen atom from the hydroxyl group, and
an OH from the carboxlic acid, form a molecule of water. A new bond is formed between what
remains of the alcohol and acid. The name of the ester is a combination of the names of the
alcohol and carboxylic acid. The suffix for an ester is -oate. An example is shown in figure 9.23.
H
H
C
H
O
O
H + H
O
C
H
H
H
C
O
O
C
H + H2 O
H
methanol
methyl methanoate
methanoic acid
Figure 9.23: The formation of an ester from an alcohol and carboxylic acid
9.10
The Amino Group
The amino group has the formula -NH2 and consists of a nitrogen atom that is bonded to
two hydrogen atoms, and to the carbon skeleton. Organic compounds that contain this functional group are called amines. One example is glycine. Glycine belongs to a group of organic
compounds called amino acids, which are the building blocks of proteins.
H
O
C
OH
C
H
H
N
H
Figure 9.24: A molecule of glycine
9.11
The Carbonyl Group
The carbonyl group (-CO) consists of a carbon atom that is joined to an oxygen by a double
bond. If the functional group is on the end of the carbon chain, the organic compound is called
a ketone. The simplest ketone is acetone, which contains three carbon atoms. A ketone has
the ending ’one’ in its IUPAC name.
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CHAPTER 9. ORGANIC MOLECULES - GRADE 12
9.12
Exercise: Carboxylic acids, esters, amines and ketones
1. Look at the list of organic compounds in the table below:
Organic compound
CH3 CH2 CH2 COOH
NH2 CH2 COOH
propyl ethanoate
CH3 CHO
Type of compound
(a) Complete the table by identifying each compound as either a carboxylic
acid, ester, amine or ketone.
(b) Give the name of the compounds that have been written as condensed
structural formulae.
2. A chemical reaction takes place and ethyl methanoate is formed.
(a) What type of organic compound is ethyl methanoate?
(b) Name the two reactants in this chemical reaction.
(c) Give the structural formula of ethyl methanoate.
9.12
Summary
• Organic chemistry is the branch of chemistry that deals with organic molecules. An
organic molecule is one that contains carbon.
• All living organisms contain carbon. Plants use sunlight to convert carbon dioxide in the
air into organic compounds through the process of photosynthesis. Animals and other
organisms then feed on plants to obtain their own organic compounds. Fossil fuels are
another important source of carbon.
• It is the unique properties of the carbon atom that give organic compounds certain
properties.
• The carbon atom has four valence electrons, so it can bond with many other atoms,
often resulting in long chain structures. It also forms mostly covalent bonds with the
atoms that it bonds to, meaning that most organic molecules are non-polar.
• An organic compound can be represented in different ways, using its molecular formula,
structural formula or condensed structural formula.
• If two compounds are isomers, it means that they have the same molecular formulae but
different structural formulae.
• A functional group is a particular group of atoms within a molecule, which give it certain
reaction characteristics. Organic compounds can be grouped according to their functional
group.
• The hydrocarbons are organic compounds that contain only carbon and hydrogen. They
can be further divided into the alkanes, alkenes and alkynes, based on the type of bonds
between the carbon atoms.
• The alkanes have only single bonds between their carbon atoms and are unreactive.
• The alkenes have at least one double bond between two of their carbon atoms. They
are more reactive than the alkanes.
• The alkynes have at least one triple bond between two of their carbon atoms. They are
the most reactive of the three groups.
• A hydrocarbon is said to be saturated if it contains the maximum possible number of
hydrogen atoms for that molecule. The alkanes are all saturated compounds.
179
9.12
CHAPTER 9. ORGANIC MOLECULES - GRADE 12
• A hydrocarbon is unsaturated if it does not contain the maximum number of hydrogen
atoms for that molecule. The alkenes and alkynes are examples of unsaturated molecules.
If a double or triple bond is broken, more hydrogen atoms can be added to the molecule.
• There are three types of reactions that occur in the alkanes: substitution, elimination
and oxidation reactions.
• The alkenes undergo addition reactions because they are unsaturated.
• Organic compounds are named according to their functional group and its position in the
molecule, the number of carbon atoms in the molecule and the position of any double
and triple bonds. The IUPAC rules for nomenclature are used in the naming of organic
molecules.
• Many of the properties of the hydrocarbons are determined by their molecular structure,
the bonds between atoms and molecules, and their surface area.
• The melting point and boiling point of the hydrocarbons increases as their number of
carbon atoms increases.
• The molecular mass of the hydrocarbons determines whether they will be in the gaseous,
liquid or solid phase at certain temperatures.
• An alcohol is an organic compound that contains a hydroxyl group (OH).
• The alcohols have a number of different uses including their use as a solvent, for medicinal
purposes and in alcoholic drinks.
• The alcohols share a number of properties because of the hydroxyl group. The hydroxyl
group affects the solubility of the alcohols. Those with shorter carbon chains are generally
more soluble, and those with longer chains are less soluble. The strong hydrogen bond
between the hydrogen and oxygen atoms in the hydroxyl group gives alcohols a higher
melting point and boiling point than other organic compounds. The hydroxyl group also
gives the alcohols both acidic and basic properties.
• The carboxylic acids are organic acids that contain a carboxyl group with the formula
COOH. In a carboxyl group, an oxygen atom is double-bonded to a carbon atom, which is
also bonded to a hydroxyl group.
• The carboxylic acids have weak acidic properties because the hydrogen atom is able to
dissociate from the carboxyl group.
• An ester is formed when an alcohol reacts with a carboxylic acid.
• The amines are organic compounds that contain an amino functional group, which has
the formula NH2 . Some amines belong to the amino acid group, which are the building
blocks of proteins.
• The ketones are a group of compounds that contain a carbonyl group, which consists of
an oxygen atom that is double-bonded to a carbon atom. In a ketone, the carbonyl group
is on the end of the carbon chain.
Exercise: Summary exercise
1. Give one word for each of the following descriptions:
(a)
(b)
(c)
(d)
(e)
The
The
The
The
The
group of hydrocarbons to which 2-methyl-propene belongs.
name of the functional group that gives alcohols their properties.
group of organic compounds that have acidic properties.
name of the organic compound that is found in vinegar.
name of the organic compound that is found in alcoholic beverages.
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CHAPTER 9. ORGANIC MOLECULES - GRADE 12
9.12
2. In each of the following questions, choose the one correct answer from the
list provided.
(a) When 1-propanol is oxidised by acidified potassium permanganate, the
possible product formed is...
i. propane
ii. propanoic acid
iii. methyl propanol
iv. propyl methanoate
(IEB 2004)
(b) What is the IUPAC name for the compound represented by the following
structural formula?
H
Cl
Cl
H
C
C
C
H
H
Cl
C
H
H
H
i.
ii.
iii.
iv.
1,2,2-trichlorobutane
1-chloro-2,2-dichlorobutane
1,2,2-trichloro-3-methylpropane
1-chloro-2,2-dichloro-3-methylpropane
(IEB 2003)
3. Write balanced equations for the following reactions:
(a) Ethene reacts with bromine
(b) Ethyne gas burns in an excess of oxygen
(c) Ethanoic acid ionises in water
4. The table below gives the boiling point of ten organic compounds.
1
2
3
4
5
6
7
8
9
10
Compound
methane
ethane
propane
butane
pentane
methanol
ethanol
propan-1-ol
propan-1,2-diol
propan-1,2,3-triol
Formula
CH4
C2 H6
C3 H8
C4 H10
C5 H12
CH3 OH
C2 H5 OH
C3 H7 OH
CH3 CHOHCH2 OH
CH2 OHCHOHCH2 OH
Boiling Point (0 C)
-164
-88
-42
0
36
65
78
98
189
290
The following questions refer to the compounds shown in the above table.
(a) To which homologous series do the following compounds belong?
i. Compounds 1,2 and 3
ii. Compounds 6,7 and 8
(b) Which of the above compounds are gases at room temperature?
(c) What causes the trend of increasing boiling points of compounds 1 to 5?
(d) Despite the fact that the length of the carbon chain in compounds 8,9
and 10 is the same, the boiling point of propan-1,2,3-triol is much higher
than the boiling point of propan-1-ol. What is responsible for this large
difference in boiling point?
(e) Give the IUPAC name and the structural formula of an isomer of butane.
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9.12
CHAPTER 9. ORGANIC MOLECULES - GRADE 12
(f) Which one of the above substances is used as a reactant in the preparation
of the ester ethylmethanoate?
(g) Using structural formulae, write an equation for the reaction which produces ethylmethanoate.
(IEB 2004 )
5. Refer to the numbered diagrams below and then answer the questions that
follow.
1
O
H
H
H
HO
C
C
C
C
H
H
H
H
Br
C
C
H
Br
3
H
C
C
H
H
2
H
H
HO
C
C
H
H
4
H
O
H
C
C
H
H
H
O
C
H
H
(a) Which one of the above compounds is produced from the fermentation of
starches and sugars in plant matter?
i. compound 1
ii. compound 2
iii. compound 3
iv. compound 4
(b) To which one of the following homologous series does compound 1 belong?
i. esters
ii. alcohols
iii. aldehydes
iv. carboxylic acids
(c) The correct IUPAC name for compound 3 is...
i. 1,1-dibromo-3-butyne
ii. 4,4-dibromo-1-butyne
iii. 2,4-dibromo-1-butyne
iv. 4,4-dibromo-1-propyne
(d) What is the correct IUPAC name for compound 4?
i. propanoic acid
ii. ethylmethanoate
iii. methylethanoate
iv. methylpropanoate
IEB 2005
6. Answer the following questions:
(a) What is a homologous series?
(b) A mixture of ethanoic acid and methanol is warmed in the presence of
concentrated sulphuric acid.
i. Using structural formulae, give an equation for the reaction which takes
place.
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CHAPTER 9. ORGANIC MOLECULES - GRADE 12
9.12
ii. What is the IUPAC name of the organic compound formed in this
reaction?
(c) Consider the following unsaturated hydrocarbon:
H
H
H
H
C
C
C
C
H
H
H
i. Give the IUPAC name for this compound.
ii. Give the balanced equation for the combustion of this compound in
excess oxygen.
(IEB Paper 2, 2003)
7. Consider the organic compounds labelled A to E.
A. CH3 CH2 CH2 CH2 CH2 CH3
B. C6 H6
C. CH3 -Cl
D. Methylamine
E
H
H
C
H
H
C
C
H
(a)
(b)
(c)
(d)
H
H
O
H
C
C
C
H
H
C
H
H
C
H
H
H
H
Write a balanced chemical equation for the preparation of compound C
using an alkane as one of the reactants.
Write down the IUPAC name for compound E.
Write down the structural formula of an isomer of compound A that has
only FOUR carbon atoms in the longest chain.
Write down the structural formula for compound B.
183
9.12
CHAPTER 9. ORGANIC MOLECULES - GRADE 12
184
Chapter 10
Organic Macromolecules - Grade
12
As its name suggests, a macromolecule is a large molecule that forms when lots of smaller
molecules are joined together. In this chapter, we will be taking a closer look at the structure
and properties of different macromolecules, and at how they form.
10.1
Polymers
Some macromolecules are made up of lots of repeating structural units called monomers. To
put it more simply, a monomer is like a building block. When lots of similar monomers are joined
together by covalent bonds, they form a polymer. In an organic polymer, the monomers would
be joined by the carbon atoms of the polymer ’backbone’. A polymer can also be inorganic, in
which case there may be atoms such as silicon in the place of carbon atoms. The key feature
that makes a polymer different from other macromolecules, is the repetition of identical or similar
monomers in the polymer chain. The examples shown below will help to make these concepts
clearer.
Definition: Polymer
Polymer is a term used to describe large molecules consisting of repeating structural units,
or monomers, connected by covalent chemical bonds.
1. Polyethene
Chapter 9 looked at the structure of a group of hydrocarbons called the alkenes. One
example is the molecule ethene. The structural formula of ethene is is shown in figure
10.1. When lots of ethene molecules bond together, a polymer called polyethene is
formed. Ethene is the monomer which, when joined to other ethene molecules, forms the
polymer polyethene. Polyethene is a cheap plastic that is used to make plastic bags and
bottles.
(a)
H
H
C
H
(b)
C
H
H
H
H
H
C
C
C
C
H
H
H
H
Figure 10.1: (a) Ethene monomer and (b) polyethene polymer
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CHAPTER 10. ORGANIC MACROMOLECULES - GRADE 12
A polymer may be a chain of thousands of monomers, and so it is impossible to draw
the entire polymer. Rather, the structure of a polymer can be condensed and represented
as shown in figure 10.2. The monomer is enclosed in brackets and the ’n’ represents the
number of ethene molecules in the polymer, where ’n’ is any whole number. What this
shows is that the ethene monomer is repeated an indefinite number of times in a molecule
of polyethene.
n
H
H
C
C
H
H
Figure 10.2: A simplified representation of a polyethene molecule
2. Polypropene
Another example of a polymer is polypropene (fig 10.3). Polypropene is also a plastic, but
is stronger than polyethene and is used to make crates, fibres and ropes. In this polymer,
the monomer is the alkene called propene.
(a)
CH3
H
C
H
C
H
(b)
CH3
H
CH3
H
C
C
C
C
H
H
H
H
CH3
H
C
C
H
H
or n
Figure 10.3: (a) Propene monomer and (b) polypropene polymer
10.2
How do polymers form?
Polymers are formed through a process called polymerisation, where monomer molecules react together to form a polymer chain. Two types of polymerisation reactions are addition
polymerisation and condensation polymerisation.
Definition: Polymerisation
In chemistry, polymerisation is a process of bonding monomers, or single units together
through a variety of reaction mechanisms to form longer chains called polymers.
10.2.1
Addition polymerisation
In this type of reaction, monomer molecules are added to a growing polymer chain one at a time.
No small molecules are eliminated in the process. An example of this type of reaction is the
formation of polyethene from ethene (fig 10.1). When molecules of ethene are joined to each
other, the only thing that changes is that the double bond between the carbon atoms in each
ethene monomer is replaced by a single bond so that a new carbon-carbon bond can be formed
with the next monomer in the chain. In other words, the monomer is an unsaturated compound
which, after an addition reaction, becomes a saturated compound.
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Extension: Initiation, propagation and termination
There are three stages in the process of addition polymerisation. Initiation refers
to a chemical reaction that triggers off another reaction. In other words, initiation
is the starting point of the polymerisation reaction. Chain propagation is the part
where monomers are continually added to form a longer and longer polymer chain.
During chain propagation, it is the reactive end groups of the polymer chain that
react in each propagation step, to add a new monomer to the chain. Once a monomer
has been added, the reactive part of the polymer is now in this last monomer unit
so that propagation will continue. Termination refers to a chemical reaction that
destroys the reactive part of the polymer chain so that propagation stops.
Worked Example 48: Polymerisation reactions
Question: A polymerisation reaction takes place and the following polymer is
formed:
n
W
X
C
C
Y
Z
Note: W, X, Y and Z could represent a number of different atoms or combinations
of atoms e.g. H, F, Cl or CH3 .
1. Give the structural formula of the monomer of this polymer.
2. To what group of organic compounds does this monomer belong?
3. What type of polymerisation reaction has taken place to join these monomers
to form the polymer?
Answer
Step 1 : Look at the structure of the repeating unit in the polymer to
determine the monomer.
The monomer is:
W
X
C
C
Y
Z
Step 2 : Look at the atoms and bonds in the monomer to determine which
group of organic compounds it belongs to.
The monomer has a double bond between two carbon atoms. The monomer must
be an alkene.
Step 3 : Determine the type of polymerisation reaction.
In this example, unsaturated monomers combine to form a saturated polymer. No
atoms are lost or gained for the bonds between monomers to form. They are simply
added to each other. This is an addition reaction.
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10.2
10.2
CHAPTER 10. ORGANIC MACROMOLECULES - GRADE 12
10.2.2
Condensation polymerisation
In this type of reaction, two monomer molecules form a covalent bond and a small molecule such
as water is lost in the bonding process. Nearly all biological reactions are of this type. Polyester
and nylon are examples of polymers that form through condensation polymerisation.
1. Polyester
Polyesters are a group of polymers that contain the ester functional group in their main
chain. Although there are many forms of polyesters, the term polyester usually refers to
polyethylene terephthalate (PET). PET is made from ethylene glycol (an alcohol) and
terephthalic acid (an acid). In the reaction, a hydrogen atom is lost from the alcohol, and
a hydroxyl group is lost from the carboxylic acid. Together these form one water molecule
which is lost during condensation reactions. A new bond is formed between an oxygen
and a carbon atom. This bond is called an ester linkage. The reaction is shown in figure
10.4.
(a)
CH2 CH2
HO
+
OH
HO
ethylene glycol
O
O
C
C
terephthalic acid
H2 O molecule lost
(b)
HO
CH2 CH2
O
O
O
C
C
OH
ester linkage
Figure 10.4: An acid and an alcohol monomer react (a) to form a molecule of the polyester
’polyethylene terephthalate’ (b).
Polyesters have a number of characteristics which make them very useful. They are resistant to stretching and shrinking, they are easily washed and dry quickly, and they are
resistant to mildew. It is for these reasons that polyesters are being used more and more
in textiles. Polyesters are stretched out into fibres and can then be made into fabric and
articles of clothing. In the home, polyesters are used to make clothing, carpets, curtains,
sheets, pillows and upholstery.
teresting Polyester is not just a textile. Polyethylene terephthalate is in fact a plastic
Interesting
Fact
Fact
which can also be used to make plastic drink bottles. Many drink bottles
are recycled by being reheated and turned into polyester fibres. This type
of recycling helps to reduce disposal problems.
2. Nylon
Nylon was the first polymer to be commercially successful. Nylon replaced silk, and was
used to make parachutes during World War 2. Nylon is very strong and resistant, and is
used in fishing line, shoes, toothbrush bristles, guitar strings and machine parts to name
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CHAPTER 10. ORGANIC MACROMOLECULES - GRADE 12
10.2
just a few. Nylon is formed from the reaction of an amine (1,6-diaminohexane) and an acid
monomer (adipic acid) (figure 10.5). The bond that forms between the two monomers is
called an amide linkage. An amide linkage forms between a nitrogen atom in the amine
monomer and the carbonyl group in the carboxylic acid.
(a)
H
H
N
H
(CH2 )4
N
O
+
H
HO
C
O
(CH2 )4
C
OH
H2 O molecule is lost
(b)
H
H
N
(CH2 )4
H
O
N
C
O
(CH2 )4
C
OH
amide linkage
Figure 10.5: An amine and an acid monomer (a) combine to form a section of a nylon polymer
(b).
teresting Nylon was first introduced around 1939 and was in high demand to make stockInteresting
Fact
Fact
ings. However, as World War 2 progressed, nylon was used more and more to
make parachutes, and so stockings became more difficult to buy. After the war,
when manufacturers were able to shift their focus from parachutes back to stockings, a number of riots took place as women queued to get stockings. In one of
the worst disturbances, 40 000 women queued up for 13 000 pairs of stockings,
which led to fights breaking out!
Exercise: Polymers
1. The following monomer is a reactant in a polymerisation reaction:
H
CH3
C
C
H
CH3
(a) What is the IUPAC name of this monomer?
(b) Give the structural formula of the polymer that is formed in this polymerisation reaction.
(c) Is the reaction an addition or condensation reaction?
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CHAPTER 10. ORGANIC MACROMOLECULES - GRADE 12
2. The polymer below is the product of a polymerisation reaction.
(a)
(b)
(c)
(d)
H
Cl
H
Cl
H
Cl
C
C
C
C
C
C
H
H
H
H
H
H
Give the structural formula of the monomer in this polymer.
What is the name of the monomer?
Draw the abbreviated structural formula for the polymer.
Has this polymer been formed through an addition or condensation polymerisation reaction?
3. A condensation reaction takes place between methanol and methanoic acid.
(a) Give the structural formula for...
i. methanol
ii. methanoic acid
iii. the product of the reaction
(b) What is the name of the product? (Hint: The product is an ester)
10.3
The chemical properties of polymers
The attractive forces between polymer chains play a large part in determining a polymer’s properties. Because polymer chains are so long, these interchain forces are very important. It is usually
the side groups on the polymer that determine what types of intermolecular forces will exist.
The greater the strength of the intermolecular forces, the greater will be the tensile strength and
melting point of the polymer. Below are some examples:
• Hydrogen bonds between adjacent chains
Polymers that contain amide or carbonyl groups can form hydrogen bonds between adjacent chains. The positive hydrogen atoms in the N-H groups of one chain are strongly
attracted to the oxygen atoms in the C=O groups on another. Polymers that contain urea
linkages would fall into this category. The structural formula for urea is shown in figure
10.6. Polymers that contain urea linkages have high tensile strength and a high melting
point.
O
C
H2 N
NH2
Figure 10.6: The structural formula for urea
• Dipole-dipole bonds between adjacent chains
Polyesters have dipole-dipole bonding between their polymer chains. Dipole bonding is
not as strong as hydrogen bonding, so a polyester’s melting point and strength are lower
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10.4
than those of the polymers where there are hydrogen bonds between the chains. However,
the weaker bonds between the chains means that polyesters have greater flexibility. The
greater the flexibility of a polymer, the more likely it is to be moulded or stretched into
fibres.
• Weak van der Waal’s forces
Other molecules such as ethene do not have a permanent dipole and so the attractive forces
between polyethene chains arise from weak van der Waals forces. Polyethene therefore has
a lower melting point than many other polymers.
10.4
Types of polymers
There are many different types of polymers. Some are organic, while others are inorganic. Organic
polymers can be broadly grouped into either synthetic/semi-synthetic (artificial) or biological
(natural) polymers. We are going to take a look at two groups of organic polymers: plastics,
which are usually synthetic or semi-synthetic and biological macromolecules which are natural
polymers. Both of these groups of polymers play a very important role in our lives.
10.5
Plastics
In today’s world, we can hardly imagine life without plastic. From cellphones to food packaging,
fishing line to plumbing pipes, compact discs to electronic equipment, plastics have become a
very important part of our daily lives. ”Plastics” cover a range of synthetic and semi-synthetic
organic polymers. Their name comes from the fact that they are ’malleable’, in other words their
shape can be changed and moulded.
Definition: Plastic
Plastic covers a range of synthetic or semisynthetic organic polymers. Plastics may contain
other substances to improve their performance. Their name comes from the fact that many
of them are malleable, in other words they have the property of plasticity.
It was only in the nineteenth century that it was discovered that plastics could be made by
chemically changing natural polymers. For centuries before this, only natural organic polymers
had been used. Examples of natural organic polymers include waxes from plants, cellulose (a
plant polymer used in fibres and ropes) and natural rubber from rubber trees. But in many
cases, these natural organic polymers didn’t have the characteristics that were needed for them
to be used in specific ways. Natural rubber for example, is sensitive to temperature and becomes
sticky and smelly in hot weather and brittle in cold weather.
In 1834 two inventors, Friedrich Ludersdorf of Germany and Nathaniel Hayward of the US,
independently discovered that adding sulfur to raw rubber helped to stop the material from
becoming sticky. After this, Charles Goodyear discovered that heating this modified rubber
made it more resistant to abrasion, more elastic and much less sensitive to temperature. What
these inventors had done was to improve the properties of a natural polymer so that it could be
used in new ways. An important use of rubber now is in vehicle tyres, where these properties of
rubber are critically important.
teresting The first true plastic (i.e. one that was not based on any material found in
Interesting
Fact
Fact
nature) was Bakelite, a cheap, strong and durable plastic. Some of these plastics
are still used for example in electronic circuit boards, where their properties of
insulation and heat resistance are very important.
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CHAPTER 10. ORGANIC MACROMOLECULES - GRADE 12
10.5.1
The uses of plastics
There is such a variety of different plastics available, each having their own specific properties
and uses. The following are just a few examples.
• Polystyrene
Polystyrene (figure 15.2) is a common plastic that is used in model kits, disposable eating
utensils and a variety of other products. In the polystyrene polymer, the monomer is
styrene, a liquid hydrocarbon that is manufactured from petroleum.
CH2
CH2
CH
CH2
CH
CH
CH2
CH
polymerisation
etc
Figure 10.7: The polymerisation of a styrene monomer to form a polystyrene polymer
• Polyvinylchloride (PVC)
Polyvinyl chloride (PVC) (figure 10.8) is used in plumbing, gutters, electronic equipment,
wires and food packaging. The side chains of PVC contain chlorine atoms, which give it
its particular characteristics.
H
n
Cl
C
H
C
H
Figure 10.8: Polyvinyl chloride
teresting
Interesting
Fact
Fact
Many vinyl products have other chemicals added to them to give them particular properties. Some of these chemicals, called additives, can leach out
of the vinyl products. In PVC, plasticizers are used to make PVC more
flexible. Because many baby toys are made from PVC, there is concern that
some of these products may leach into the mouths of the babies that are
chewing on them. In the USA, most companies have stopped making PVC
toys. There are also concerns that some of the plasticizers added to PVC
may cause a number of health conditions including cancer.
• Synthetic rubber
Another plastic that was critical to the World War 2 effort was synthetic rubber, which was
produced in a variety of forms. Not only were worldwide natural rubber supplies limited,
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CHAPTER 10. ORGANIC MACROMOLECULES - GRADE 12
10.5
but most rubber-producing areas were under Japanese control. Rubber was needed for
tyres and parts of war machinery. After the war, synthetic rubber also played an important
part in the space race and nuclear arms race.
• Polyethene/polyethylene (PE)
Polyethylene (figure 10.1) was discovered in 1933. It is a cheap, flexible and durable plastic
and is used to make films and packaging materials, containers and car fittings. One of
the most well known polyethylene products is ’Tupperware’, the sealable food containers
designed by Earl Tupper and promoted through a network of housewives!
• Polytetrafluoroethylene (PTFE)
Polytetrafluoroethylene (figure 10.9) is more commonly known as ’Teflon’ and is most well
known for its use in non-stick frying pans. Teflon is also used to make the breathable fabric
Gore-Tex.
F
F
C
C
F
F
n
F
F
C
C
F
F
Figure 10.9: A tetra fluoroethylene monomer and polytetrafluoroethylene polymer
Table 10.1 summarises the formulae, properties and uses of some of the most common plastics.
Table 10.1: A summary of the formulae, properties and uses of some common plastics
Name
Polyethene (low density)
Polyethene (high density)
Polypropene
Formula
-(CH2 -CH2 )n -
Monomer
CH2 =CH2
Properties
soft, waxy solid
-(CH2 -CH2 )n -
CH2 =CH2
rigid
-[CH2 -CH(CH3 )]n -
CH2 =CHCH3
different
grades:
some are soft and
others hard
strong, rigid
Polyvinylchloride
-(CH2 -CHCl)n (PVC)
Polystyrene
-[CH2 -CH(C6 H5 )]n
Polytetrafluoroethylene -(CF2 -CF2 )n -
CH2 =CHCl
CH2 =CHC6 H5 hard, rigid
CF2 =CF2
resistant, smooth,
solid
Uses
film wrap and plastic
bags
electrical insulation,
bottles and toys
carpets and upholstery
pipes, flooring
toys, packaging
non-stick surfaces,
electrical insulation
Exercise: Plastics
1. It is possible for macromolecules to be composed of more than one type of
repeating monomer. The resulting polymer is called a copolymer. Varying
the monomers that combine to form a polymer, is one way of controlling the
properties of the resulting material. Refer to the table below which shows a
number of different copolymers of rubber, and answer the questions that follow:
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10.5
CHAPTER 10. ORGANIC MACROMOLECULES - GRADE 12
Monomer A
H2 C=CHCl
Monomer B
H2 C=CCl2
Copolymer
Saran
H2 C=CHC6 H5
H2 C=C-CH=CH2
H2 C=CHCN
H2 C=C-CH=CH2
SBR (styrene
butadiene
rubber)
Nitrile rubber
H2 C=C(CH3 )2
F2 C=CF(CF3 )
H2 C=C-CH=CH2
H2 C=CHF
Butyl rubber
Viton
Uses
films and fibres
tyres
adhesives and
hoses
inner tubes
gaskets
(a) Give the structural formula for each of the monomers of nitrile rubber.
(b) Give the structural formula of the copolymer viton.
(c) In what ways would you expect the properties of SBR to be different from
nitrile rubber?
(d) Suggest a reason why the properties of these polymers are different.
2. In your home, find as many examples of different types of plastics that you
can. Bring them to school and show them to your group. Together, use your
examples to complete the following table:
Object
10.5.2
Type of plastic
Properties
Uses
Thermoplastics and thermosetting plastics
A thermoplastic is a plastic that can be melted to a liquid when it is heated and freezes to
a brittle, glassy state when it is cooled enough. These properties of thermoplastics are mostly
due to the fact that the forces between chains are weak. This also means that these plastics
can be easily stretched or moulded into any shape. Examples of thermoplastics include nylon,
polystyrene, polyethylene, polypropylene and PVC. Thermoplastics are more easily recyclable
than some other plastics.
Thermosetting plastics differ from thermoplastics because once they have been formed, they
cannot be remelted or remoulded. Examples include bakelite, vulcanised rubber, melanine (used
to make furniture), and many glues. Thermosetting plastics are generally stronger than thermoplastics and are better suited to being used in situations where there are high temperatures.
They are not able to be recycled. Thermosetting plastics have strong covalent bonds between
chains and this makes them very strong.
Activity :: Case Study : Biodegradable plastics
Read the article below and then answer the questions that follow.
Our whole world seems to be wrapped in plastic. Almost every product we
buy, most of the food we eat and many of the liquids we drink come encased in
plastic. Plastic packaging provides excellent protection for the product, it is cheap
to manufacture and seems to last forever. Lasting forever, however, is proving to
be a major environmental problem. Another problem is that traditional plastics are
manufactured from non-renewable resources - oil, coal and natural gas. In an effort
to overcome these problems, researchers and engineers have been trying to develop
biodegradable plastics that are made from renewable resources, such as plants.
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CHAPTER 10. ORGANIC MACROMOLECULES - GRADE 12
10.5
The term biodegradable means that a substance can be broken down into simpler
substances by the activities of living organisms, and therefore is unlikely to remain
in the environment. The reason most plastics are not biodegradable is because their
long polymer molecules are too large and too tightly bonded together to be broken
apart and used by decomposer organisms. However, plastics based on natural plant
polymers that come from wheat or corn starch have molecules that can be more
easily broken down by microbes.
Starch is a natural polymer. It is a white, granular carbohydrate produced by
plants during photosynthesis and it serves as the plant’s energy store. Many plants
contain large amounts of starch. Starch can be processed directly into a bioplastic
but, because it is soluble in water, articles made from starch will swell and deform
when exposed to moisture, and this limits its use. This problem can be overcome
by changing starch into a different polymer. First, starch is harvested from corn,
wheat or potatoes, then microorganisms transform it into lactic acid, a monomer.
Finally, the lactic acid is chemically treated to cause the molecules of lactic acid to
link up into long chains or polymers, which bond together to form a plastic called
polylactide (PLA).
PLA can be used for products such as plant pots and disposable nappies. It has
been commercially available in some countries since 1990, and certain blends have
proved successful in medical implants, sutures and drug delivery systems because
they are able to dissolve away over time. However, because PLA is much more
expensive than normal plastics, it has not become as popular as one would have
hoped.
Questions
1. In your own words, explain what is meant by a ’biodegradable plastic’.
2. Using your knowledge of chemical bonding, explain why some polymers are
biodegradable and others are not.
3. Explain why lactic acid is a more useful monomer than starch, when making a
biodegradable plastic.
4. If you were a consumer (shopper), would you choose to buy a biodegradable
plastic rather than another? Explain your answer.
5. What do you think could be done to make biodegradable plastics more popular
with consumers?
10.5.3
Plastics and the environment
Although plastics have had a huge impact globally, there is also an environmental price that has
to be paid for their use. The following are just some of the ways in which plastics can cause
damage to the environment.
1. Waste disposal
Plastics are not easily broken down by micro-organisms and therefore most are not easily
biodegradeable. This leads to waste dispoal problems.
2. Air pollution
When plastics burn, they can produce toxic gases such as carbon monoxide, hydrogen
cyanide and hydrogen chloride (particularly from PVC and other plastics that contain
chlorine and nitrogen).
3. Recycling
It is very difficult to recycle plastics because each type of plastic has different properties
and so different recycling methods may be needed for each plastic. However, attempts
are being made to find ways of recycling plastics more effectively. Some plastics can
be remelted and re-used, while others can be ground up and used as a filler. However,
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CHAPTER 10. ORGANIC MACROMOLECULES - GRADE 12
one of the problems with recycling plastics is that they have to be sorted according to
plastic type. This process is difficult to do with machinery, and therefore needs a lot
of labour. Alternatively, plastics should be re-used. In many countries, including South
Africa, shoppers must now pay for plastic bags. This encourages people to collect and
re-use the bags they already have.
Activity :: Case Study : Plastic pollution in South Africa
Read the following extract, taken from ’Planet Ark’ (September 2003), and then
answer the questions that follow.
South Africa launches a programme this week to exterminate its ”national
flower” - the millions of used plastic bags that litter the landscape.
Beginning on Friday, plastic shopping bags used in the country must be
both thicker and more recyclable, a move officials hope will stop people from simply tossing them away. ”Government has targeted plastic
bags because they are the most visible kind of waste,” said Phindile Makwakwa, spokeswoman for the Department of Environmental Affairs and
Tourism. ”But this is mostly about changing people’s mindsets about the
environment.”
South Africa is awash in plastic pollution. Plastic bags are such a common
eyesore that they are dubbed ”roadside daisies” and referred to as the
national flower. Bill Naude of the Plastics Federation of South Africa said
the country used about eight billion plastic bags annually, a figure which
could drop by 50 percent if the new law works.
It is difficult sometimes to imagine exactly how much waste is produced in our
country every year. Where does all of this go to? You are going to do some simple
calculations to try to estimate the volume of plastic packets that is produced in
South Africa every year.
1. Take a plastic shopping packet and squash it into a tight ball.
(a) Measure the approximate length, breadth and depth of your squashed plastic bag.
(b) Calculate the approximate volume that is occupied by the packet.
(c) Now calculate the approximate volume of your classroom by measuring its
length, breadth and height.
(d) Calculate the number of squashed plastic packets that would fit into a
classroom of this volume.
(e) If South Africa produces an average of 8 billion plastic bags each year, how
many clasrooms would be filled if all of these bags were thrown away and
not re-used?
2. What has South Africa done to try to reduce the number of plastic bags that
are produced?
3. Do you think this has helped the situation?
4. What can you do to reduce the amount of plastic that you throw away?
10.6
Biological Macromolecules
A biological macromolecule is one that is found in living organisms. Biological macromolecules
include molecules such as carbohydrates, proteins and nucleic acids. Lipids are also biological
macromolecules. They are essential for all forms of life to survive.
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CHAPTER 10. ORGANIC MACROMOLECULES - GRADE 12
10.6
Definition: Biological macromolecule
A biological macromolecule is a polymer that occurs naturally in living organisms. These
molecules are essential to the survival of life.
10.6.1
Carbohydrates
Carbohydrates include the sugars and their polymers. One key characteristic of the carbohydrates is that they contain only the elements carbon, hydrogen and oxygen. In the carbohydrate
monomers, every carbon except one has a hydroxyl group attached to it, and the remaining
carbon atom is double bonded to an oxygen atom to form a carbonyl group. One of the most
important monomers in the carbohydrates is glucose (figure 10.10). The glucose molecule can
exist in an open-chain (acyclic) and ring (cyclic) form.
(a)
O
C1
OH
H
OH
OH
OH
C2
C3
C4
C5
C6
H
OH
H
H
H
H
H
CH2 OH
(b)
C5
O
H
C4 OH
H
C3
C2
H
OH
H
OH
H
C1
OH
Figure 10.10: The open chain (a) and cyclic (b) structure of a glucose molecule
Glucose is produced during photosynthesis, which takes place in plants. During photosynthesis,
sunlight (solar energy), water and carbon dioxide are involved in a chemical reaction that produces
glucose and oxygen. This glucose is stored in various ways in the plant.
The photosynthesis reaction is as follows:
6CO2 + 6H2 O + sunlight → C6 H12 O6 + 6O2
Glucose is an important source of energy for both the plant itself, and also for the other animals
and organisms that may feed on it. Glucose plays a critical role in cellular respiration, which is
a chemical reaction that occurs in the cells of all living organisms. During this reaction, glucose
and oxygen react to produce carbon dioxide, water and ATP energy. ATP is a type of energy
that can be used by the body’s cells so that they can function normally. The purpose of eating
then, is to obtain glucose which the body can then convert into the ATP energy it needs to be
able to survive.
The reaction for cellular respiration is as follows:
6C6 H12 O6 + 602 → 6CO2 + 6H2 O + ATP (cell energy)
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We don’t often eat glucose in its simple form. More often, we eat complex carbohydrates that
our bodies have to break down into individual glucose molecules before they can be used in cellular respiration. These complex carbohydrates are polymers, which form through condensation
polymerisation reactions (figure 10.11). Starch and cellulose are two example of carbohydrates
that are polymers composed of glucose monomers.
CH2 OH
(a)
CH2 OH
C5
O
H
C4 OH
H
C3
C2
H
OH
H
OH
H
+
C
OH
C5
H
C4
OH
O
H
C4 OH
H
C3
C2
H
OH
OH
CH2 OH
(b)
C5
H
H
C
OH
CH2 OH
O
H
C5
O
H
C4 OH
H
C3
C2
H
OH
H
H
C
C3
C2
H
H
OH
O
H
C
+ H2 O
OH
Figure 10.11: Two glucose monomers (a) undergo a condensation reaction to produce a section
of a carbohydrate polymer (b). One molecule of water is produced for every two monomers that
react.
• Starch
Starch is used by plants to store excess glucose, and consists of long chains of glucose
monomers. Potatoes are made up almost entirely of starch. This is why potatoes are such
a good source of energy. Animals are also able to store glucose, but in this case it is stored
as a compound called glycogen, rather than as starch.
• Cellulose
Cellulose is also made up of chains of glucose molecules, but the bonding between the
polymers is slightly different from that in starch. Cellulose is found in the cell walls of
plants and is used by plants as a building material.
teresting It is very difficult for animals to digest the cellulose in plants that they
Interesting
Fact
Fact
may have been feeding on. However, fungi and some protozoa are able to
break down cellulose. Many animals, including termites and cows, use these
organisms to break cellulose down into glucose, which they can then use
more easily.
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10.6.2
10.6
Proteins
Proteins are an incredibly important part of any cell, and they carry out a number of functions
such as support, storage and transport within the body. The monomers of proteins are called
amino acids. An amino acid is an organic molecule that contains a carboxyl and an amino
group, as well as a carbon side chain. The carbon side chain varies from one amino acid to the
next, and is sometimes simply represented by the letter ’R’ in a molecule’s structural formula.
Figure 10.12 shows some examples of different amino acids.
Carboxyl group
H
H2 N
Amino group
C
H
O
H
H2 N
C
C
O
C
CH3
OH
OH
Side chain (’R’)
glycine
alanine
H
H2 N
C
O
C
CH2
OH
OH
serine
Figure 10.12: Three amino acids: glycine, alanine and serine
Although each of these amino acids has the same basic structure, their side chains (’R’ groups)
are different. In the amino acid glycine, the side chain consists only of a hydrogen atom, while
alanine has a methyl side chain. The ’R’ group in serine is CH2 - OH. Amongst other things,
the side chains affect whether the amino acid is hydrophilic (attracted to water) or hydrophobic
(repelled by water). If the side chain is polar, then the amino acid is hydrophilic, but if the side
chain is non-polar then the amino acid is hydrophobic. Glycine and alanine both have non-polar
side chains, while serine has a polar side chain.
Extension: Charged regions in an amino acid
In an amino acid, the amino group acts as a base because the nitrogen atom has a
pair of unpaired electrons which it can use to bond to a hydrogen ion. The amino
group therefore attracts the hydrogen ion from the carboxyl group, and ends up having a charge of +1. The carboxyl group from which the hydrogen ion has been taken
then has a charge of -1. The amino acid glycine can therefore also be represented
as shown in the figure below.
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CHAPTER 10. ORGANIC MACROMOLECULES - GRADE 12
H
H3 N+
O
C
C
H
O−
glycine
When two amino acid monomers are close together, they may be joined to each other by peptide
bonds (figure 10.13) to form a polypeptide chain. . The reaction is a condensation reaction.
Polypeptides can vary in length from a few amino acids to a thousand or more. The polpeptide
chains are then joined to each other in different ways to form a protein. It is the sequence of
the amino acids in the polymer that gives a protein its particular properties.
The sequence of the amino acids in the chain is known as the protein’s primary structure. As
the chain grows in size, it begins to twist, curl and fold upon itself. The different parts of the
polypeptide are held together by hydrogen bonds, which form between hydrogen atoms in one
part of the chain and oxygen or nitrogen atoms in another part of the chain. This is known as
the secondary structure of the protein. Sometimes, in this coiled helical structure, bonds may
form between the side chains (R groups) of the amino acids. This results in even more irregular
contortions of the protein. This is called the tertiary structure of the protein.
(a)
H2 N
H
C
O
H
+
C
H
H2 N
C
CH3
OH
O
C
OH
Peptide bond
(b)
H2 N
H
O
C
C
H
H
N
C
H
CH3
O
C
+ H2 O
OH
Figure 10.13: Two amino acids (glycine and alanine) combine to form part of a polypeptide
chain. The amino acids are joined by a peptide bond between a carbon atom of one amino acid
and a nitrogen atom of the other amino acid.
teresting There are twenty different amino acids that exist. All cells, both plant and
Interesting
Fact
Fact
animal, build their proteins from only twenty amino acids. At first, this seems
like a very small number, especially considering the huge number of different
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CHAPTER 10. ORGANIC MACROMOLECULES - GRADE 12
10.6
proteins that exist. However, if you consider that most proteins are made up of
polypeptide chains that contain at least 100 amino acids, you will start to realise
the endless possible combinations of amino acids that are available.
The functions of proteins
Proteins have a number of functions in living organisms.
• Structural proteins such as collagen in animal connective tissue and keratin in hair, horns
and feather quills, all provide support.
• Storage proteins such as albumin in egg white provide a source of energy. Plants store
proteins in their seeds to provide energy for the new growing plant.
• Transport proteins transport other substances in the body. Haemoglobin in the blood
for example, is a protein that contains iron. Haemoglobin has an affinity (attraction) for
oxygen and so this is how oxygen is transported around the body in the blood.
• Hormonal proteins coordinate the body’s activities. Insulin for example, is a hormonal
protein that controls the sugar levels in the blood.
• Enzymes are chemical catalysts and speed up chemical reactions. Digestive enzymes such
as salivary amylase in your saliva, help to break down polymers in food. Enzymes play an
important role in all cellular reactions such as respiration, photosynthesis and many others.
Activity :: Research Project : Macromolecules in our daily diet
1. In order to keep our bodies healthy, it is important that we eat a balanced
diet with the right amounts of carbohydrates, proteins and fats. Fats are an
important source of energy, they provide insulation for the body, and they also
provide a protective layer around many vital organs. Our bodies also need
certain essential vitamins and minerals. Most food packaging has a label that
provides this information.
Choose a number of different food items that you eat. Look at the food label
for each, and then complete the following table:
Food
Carbohydrates Proteins (%)
Fats (%)
(%)
(a) Which food type contains the largest proportion of protein?
(b) Which food type contains the largest proportion of carbohydrates?
(c) Which of the food types you have listed would you consider to be the
’healthiest’ ? Give a reason for your answer.
2. In an effort to lose weight, many people choose to diet. There are many diets
on offer, each of which is based on particular theories about how to lose weight
most effectively. Look at the list of diets below:
• Vegetarian diet
• Low fat diet
• Atkin’s diet
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CHAPTER 10. ORGANIC MACROMOLECULES - GRADE 12
• Weight Watchers
For each of these diets, answer the following questions:
(a) What theory of weight loss does each type of diet propose?
(b) What are the benefits of the diet?
(c) What are the potential problems with the diet?
Exercise: Carbohydrates and proteins
1. Give the structural formula for each of the following:
(a) A polymer chain, consisting of three glucose molecules.
(b) A polypeptide chain, consisting of two molecules of alanine and one molecule
of serine.
2. Write balanced equations to show the polymerisation reactions that produce
the polymers described above.
3. The following polypeptide is the end product of a polymerisation reaction:
H2 N
H
O
C
C
CH3
H
O
N
C
C
H
CH2
H
N
C
H
H
O
C
OH
SH
(a) Give the structural formula of the monomers that make up the polypeptide.
(b) On the structural formula of the first monomer, label the amino group and
the carboxyl group.
(c) What is the chemical formula for the carbon side chain in the second
monomer?
(d) Name the bond that forms between the monomers of the polypeptide.
10.6.3
Nucleic Acids
You will remember that we mentioned earlier that each protein is different because of its unique
sequence of amino acids. But what controls how the amino acids arrange themselves to form
the specific proteins that are needed by an organism? This task is for the gene. A gene contains
DNA (deoxyribonucleic acid) which is a polymer that belongs to a class of compounds called the
nucleic acids. DNA is the genetic material that organisms inherit from their parents. It is DNA
that provides the genetic coding that is needed to form the specific proteins that an organism
needs. Another nucleic acid is RNA (ribonucleic acid).
The DNA polymer is made up of monomers called nucleotides. Each nucleotide has three
parts: a sugar, a phosphate and a nitrogenous base. The diagram in figure 10.14 may help you
to understand this better.
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CHAPTER 10. ORGANIC MACROMOLECULES - GRADE 12
phosphate sugar
10.6
nitrogenous base
DNA polymer made up of
four nucleotides
nucleotide
Figure 10.14: Nucleotide monomers make up the DNA polymer
There are five different nitrogenous bases: adenine (A), guanine (G), cytosine (C), thymine (T)
and uracil (U). It is the sequence of the nitrogenous bases in a DNA polymer that will determine
the genetic code for that organism. Three consecutive nitrogenous bases provide the coding
for one amino acid. So, for example, if the nitrogenous bases on three nucleotides are uracil,
cytosine and uracil (in that order), one serine amino acid will become part of the polypeptide
chain. The polypeptide chain is built up in this way until it is long enough (and with the right
amino acid sequence) to be a protein. Since proteins control much of what happens in living
organisms, it is easy to see how important nucleic acids are as the starting point of this process.
teresting A single defect in even one nucleotide, can be devastating to an organism.
Interesting
Fact
Fact
One example of this is a disease called sickle cell anaemia. Because of one
wrong nucletide in the genetic code, the body produces a protein called sickle
haemoglobin. Haemoglobin is the protein in red blood cells that helps to
transport oxygen around the body. When sickle haemoglobin is produced, the
red blood cells change shape. This process damages the red blood cell membrane,
and can cause the cells to become stuck in blood vessels. This then means that
the red blood cells, whcih are carrying oxygen, can’t get to the tissues where
they are needed. This can cause serious organ damage. Individuals who have
sickle cell anaemia generally have a lower life expectancy.
Table 10.2 shows some other examples of genetic coding for different amino acids.
Exercise: Nucleic acids
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10.7
CHAPTER 10. ORGANIC MACROMOLECULES - GRADE 12
Table 10.2: Nitrogenouse base sequences and the corresponding amino acid
Nitrogenous base sequence
Amino acid
UUU
Phenylalanine
CUU
Leucine
UCU
Serine
UAU
Tyrosine
UGU
Cysteine
GUU
Valine
GCU
Alanine
GGU
Glycine
1. For each of the following, say whether the statement is true or false. If the
statement is false, give a reason for your answer.
(a) Deoxyribonucleic acid (DNA) is an example of a polymer and a nucleotide
is an example of a monomer.
(b) Thymine and uracil are examples of nucleotides.
(c) A person’s DNA will determine what proteins their body will produce, and
therefore what characteristics they will have.
(d) An amino acid is a protein monomer.
(e) A polypeptide that consists of five amino acids, will also contain five nucleotides.
2. For each of the following sequences of nitrogenous bases, write down the amino
acid/s that will be part of the polypeptide chain.
(a) UUU
(b) UCUUUU
(c) GGUUAUGUUGGU
3. A polypeptide chain consists of three amino acids. The sequence of nitrogenous
bases in the nucleotides of the DNA is GCUGGUGCU. Give the structural
formula of the polypeptide.
10.7
Summary
• A polymer is a macromolecule that is made up of many repeating structural units called
monomers which are joined by covalent bonds.
• Polymers that contain carbon atoms in the main chain are called organic polymers.
• Organic polymers can be divided into natural organic polymers (e.g. natural rubber) or
synthetic organic polymers (e.g. polystyrene).
• The polymer polyethene for example, is made up of many ethene monomers that have
been joined into a polymer chain.
• Polymers form through a process called polymerisation.
• Two examples of polymerisation reactions are addition and condensation reactions.
• An addition reaction occurs when unsaturated monomers (e.g. alkenes) are added to
each other one by one. The breaking of a double bond between carbon atoms in the
monomer, means that a bond can form with the next monomer. The polymer polyethene
is formed through an addition reaction.
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10.7
• In a condensation reaction, a molecule of water is released as a product of the reaction.
The water molecule is made up of atoms that have been lost from each of the monomers.
Polyesters and nylon are polymers that are produced through a condensation reaction.
• The chemical properties of polymers (e.g. tensile strength and melting point) are determined by the types of atoms in the polymer, and by the strength of the bonds between
adjacent polymer chains. The stronger the bonds, the greater the strength of the polymer,
and the higher its melting point.
• One group of synthetic organic polymers, are the plastics.
• Polystyrene is a plastic that is made up of styrene monomers. Polystyrene is used a lot
in packaging.
• Polyvinyl chloride (PVC) consists of vinyl chloride monomers. PVC is used to make pipes
and flooring.
• Polyethene, or polyethylene, is made from ethene monomers. Polyethene is used to
make film wrapping, plastic bags, electrical insulation and bottles.
• Polytetrafluoroethylene is used in non-stick frying pans and electrical insulation.
• A thermoplastic can be heated and melted to a liquid. It freezes to a brittle, glassy state
when cooled very quickly. Examples of thermoplastics are polyethene and PVC.
• A thermoset plastic cannot be melted or re-shaped once formed. Examples of thermoset
plastics are vulcanised rubber and melanine.
• It is not easy to recycle all plastics, and so they create environmental problems.
• Some of these environmental problems include issues of waste disposal, air pollution and
recycling.
• A biological macromolecule is a polymer that occurs naturally in living organisms.
• Examples of biological macromolecules include carbohydrates and proteins, both of which
are essential for life to survive.
• Carbohydrates include the sugars and their polymers, and are an important source of
energy in living organisms.
• Glucose is a carbohydrate monomer. Glucose is the molecule that is needed for photosynthesis in plants.
• The glucose monomer is also a building block for carbohydrate polymers such as starch,
glycogen and cellulose.
• Proteins have a number of important functions. These include their roles in structures,
transport, storage, hormonal proteins and enzymes.
• A protein consists of monomers called amino acids, which are joined by peptide bonds.
• A protein has a primary, secondary and tertiary structure.
• An amino acid is an organic molecule, made up of a carboxyl and an amino group, as
well as a carbon side chain of varying lengths.
• It is the sequence of amino acids that determines the nature of the protein.
• It is the DNA of an organism that determines the order in which amino acids combine to
make a protein.
• DNA is a nucleic acid. DNA is a polymer, and is made up of monomers called nucleotides.
• Each nucleotide consists of a sugar, a phosphate and a nitrogenous base. It is the
sequence of the nitrogenous bases that provides the ’code’ for the arrangement of the
amino acids in a protein.
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10.7
CHAPTER 10. ORGANIC MACROMOLECULES - GRADE 12
Exercise: Summary exercise
1. Give one word for each of the following descriptions:
(a) A chain of monomers joined by covalent bonds.
(b) A polymerisation reaction that produces a molecule of water for every two
monomers that bond.
(c) The bond that forms between an alcohol and a carboxylic acid monomer
during a polymerisation reaction.
(d) The name given to a protein monomer.
(e) A six-carbon sugar monomer.
(f) The monomer of DNA, which determines the sequence of amino acids that
will make up a protein.
2. For each of the following questions, choose the one correct answer from the
list provided.
(a) A polymer is made up of monomers, each of which has the formula CH2 =CHCN.
The formula of the polymer is:
i. -(CH2 =CHCN)n ii. -(CH2 -CHCN)n iii. -(CH-CHCN)n iv. -(CH3 -CHCN)n (b) A polymer has the formula -[CO(CH2 )4 CO-NH(CH2 )6NH]n -. Which of
the following statements is true?
i. The polymer is the product of an addition reaction.
ii. The polymer is a polyester.
iii. The polymer contains an amide linkage.
iv. The polymer contains an ester linkage.
(c) Glucose...
i. is a monomer that is produced during cellular respiration
ii. is a sugar polymer
iii. is the monomer of starch
iv. is a polymer produced during photosynthesis
3. The following monomers are involved in a polymerisation reaction:
H2 N
H
O
C
C
OH
+
H
(a)
(b)
(c)
(d)
(e)
H2 N
H
O
C
C
OH
H
Give the structural formula of the polymer that is produced.
Is the reaction an addition or condensation reaction?
To what group of organic compounds do the two monomers belong?
What is the name of the monomers?
What type of bond forms between the monomers in the final polymer?
4. The table below shows the melting point for three plastics. Suggest a reason
why the melting point of PVC is higher than the melting point for polyethene,
but lower than that for polyester.
Plastic
Polyethene
PVC
Polyester
Melting point (0 C)
105 - 115
212
260
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CHAPTER 10. ORGANIC MACROMOLECULES - GRADE 12
5. An amino acid has the formula H2 NCH(CH2 CH2 SCH3 )COOH.
(a) Give the structural formula of this amino acid.
(b) What is the chemical formula of the carbon side chain in this molecule?
(c) Are there any peptide bonds in this molecule? Give a reason for your
answer.
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10.7
10.7
CHAPTER 10. ORGANIC MACROMOLECULES - GRADE 12
208
Part III
Chemical Change
209
Chapter 11
Physical and Chemical Change Grade 10
Matter is all around us. The desks we sit at, the air we breathe and the water we drink, are all
examples of matter. But matter doesn’t always stay the same. It can change in many different
ways. In this chapter, we are going to take a closer look at physical and chemical changes that
occur in matter.
11.1
Physical changes in matter
A physical change is one where the particles of the substances that are involved in the change
are not broken up in any way. When water is heated for example, the temperature and energy
of the water molecules increases and the liquid water evaporates to form water vapour. When
this happens, some kind of change has taken place, but the molecular structure of the water has
not changed. This is an example of a physical change.
H2 O(l) → H2 O(g)
Conduction (the transfer of energy through a material) is another example of a physical change.
As energy is transferred from one material to another, the energy of each material is changed,
but not its chemical makeup. Dissolving one substance in another is also a physical change.
Definition: Physical change
A change that can be seen or felt, but that doesn’t involve the break up of the particles in
the reaction. During a physical change, the form of matter may change, but not its identity.
A change in temperature is an example of a physical change.
There are some important things to remember about physical changes in matter:
• Arrangement of particles
When a physical change occurs, the particles (e.g. atoms, molecules) may re-arrange
themselves without actually breaking up in any way. In the example of evaporation that
we used earlier, the water molecules move further apart as their temperature (and therefore
energy) increases. The same would be true if ice were to melt. In the solid phase, water
molecules are packed close together in a very ordered way, but when the ice is heated, the
molecules overcome the forces holding them together and they move apart. Once again,
the particles have re-arranged themselves, but have not broken up.
H2 O(s) → H2 O(l)
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11.2
CHAPTER 11. PHYSICAL AND CHEMICAL CHANGE - GRADE 10
solid
gas
liquid
Figure 11.1: The arrangement of water molecules in the three phases of matter
Figure 11.1 shows this more clearly. In each phase of water, the water molecule itself stays
the same, but the way the molecules are arranged has changed.
In a physical change, the total mass, the number of atoms and the number of molecules
will always stay the same.
• Energy changes
Energy changes may take place when there is a physical change in matter, but these energy
changes are normally smaller than the energy changes that take place during a chemical
change.
• Reversibility
Physical changes in matter are usually easier to reverse than chemical changes. Water
vapour for example, can be changed back to liquid water if the temperature is lowered.
Liquid water can be changed into ice by simply increasing the temperature, and so on.
11.2
Chemical Changes in Matter
When a chemical change takes place, new substances are formed in a chemical reaction. These
new products may have very different properties from the substances that were there at the start
of the reaction.
The breakdown of copper(II) chloride to form copper and chlorine is an example of chemical
change. A simplified diagram of this reaction is shown in figure 11.2. In this reaction, the initial
substance is copper(II) chloride but, once the reaction is complete, the products are copper and
chlorine.
Cl
Cu
Cl
Cu
+
Cl
Cl
CuCl2 → Cu + Cl2
Figure 11.2: The decomposition of copper(II) chloride to form copper and chlorine
Definition: Chemical change
The formation of new substances in a chemical reaction. One type of matter is changed
into something different.
There are some important things to remember about chemical changes:
• Arrangement of particles
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CHAPTER 11. PHYSICAL AND CHEMICAL CHANGE - GRADE 10
11.2
During a chemical change, the particles themselves are changed in some way. In the
example of copper (II) chloride that was used earlier, the CuCl2 molecules were split
up into their component atoms. The number of particles will change because each one
CuCl2 molecule breaks down into one copper atom (Cu) and one chlorine molecule (Cl2 ).
However, what you should have noticed, is that the number of atoms of each element
stays the same, as does the total mass of the atoms. This will be discussed in more detail
in a later section.
• Energy changes
The energy changes that take place during a chemical reaction are much greater than those
that take place during a physical change in matter. During a chemical reaction, energy
is used up in order to break bonds, and then energy is released when the new product is
formed. This will be discussed in more detail in section ??.
• Reversibility
Chemical changes are far more difficult to reverse than physical changes.
Two types of chemical reactions are decomposition reactions and synthesis reactions.
11.2.1
Decomposition reactions
A decomposition reaction occurs when a chemical compound is broken down into elements or
smaller compounds. The generalised equation for a decomposition reaction is:
AB → A + B
One example of such a reaction is the decomposition of hydrogen peroxide (figure 11.3) to form
hydrogen and oxygen according to the following equation:
2H2 O2 → 2H2 O + O2
H
H
O
O
H
H
H
O
H
O
H
H
+
O
O
Figure 11.3: The decomposition of H2 O2 to form H2 O and O2
The decomposition of mercury (II) oxide is another example.
Activity :: Experiment : The decomposition of mercury (II) oxide
Aim:
To observe the decomposition of mercury (II) oxide when it is heated.
Note: Because this experiment involves mercury, which is a poisonous substance,
it should be done in a fume cupboard, and all the products of the reaction must be
very carefully disposed of.
Apparatus:
Mercury (II) oxide (an orange-red product); two test tubes; a large beaker; stopper and delivery tube; Bunsen burner; wooden splinter.
213
O
O
11.2
CHAPTER 11. PHYSICAL AND CHEMICAL CHANGE - GRADE 10
delivery
tube
c cb
b
c
b
rubber
stopper
b bc
c
c
b
bc bc
c b
c
cb
b
bc bc
c bc
b
cbc
b
bubbles of
oxygen gas
collecting in
second test
tube
mercury
(II) oxide
water
bunsen
burner
Method:
1. Put a small amount of mercury (II) oxide in a test tube and heat it gently over
a Bunsen burner. Then allow it to cool. What do you notice about the colour
of the mercury (II) oxide?
2. Heat the test tube again, and note what happens. Do you notice anything on
the walls of the test tube? Record these observations.
3. Test for the presence of oxygen using a glowing splinter.
Results:
• During the first heating of mercury (II) oxide, the only change that took place
was a change in colour from orange-red to black and then back to its original
colour.
• When the test tube was heated again, deposits of mercury formed on the inner
surface of the test tube. What colour is this mercury?
• The glowing splinter burst into flame when it was placed in the test tube,
meaning that oxygen is present.
Conclusions:
When mercury is heated, it decomposes to form mercury and oxygen. The
chemical decomposition reaction that takes place can be written as follows:
2HgO → 2Hg + O2
11.2.2
Synthesis reactions
During a synthesis reaction, a new product is formed from smaller elements or compounds.
The generalised equation for a synthesis reaction is as follows:
A + B → AB
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CHAPTER 11. PHYSICAL AND CHEMICAL CHANGE - GRADE 10
11.2
One example of a synthesis reaction is the burning of magnesium in oxygen to form magnesium
oxide. The equation for the reaction is:
2M g + O2 → 2M gO
Figure 11.4 shows the chemical changes that take place at a microscopic level during this chemical
reaction.
Mg
Mg
+
O
O
Mg
O
Mg
Figure 11.4: The synthesis of magnesium oxide (MgO) from magnesium and oxygen
Activity :: Experiment : Chemical reactions involving iron and sulfur
Aim:
To demonstrate the synthesis of iron sulfide from iron and sulfur.
Apparatus:
5.6 g iron filings and 3.2 g powdered sulfur; porcelain dish; test tube; bunsen
burner
Method:
1. Before you carry out the experiment, write a balanced equation for the reaction
you expect will take place.
2. Measure the quantity of iron and sulfur that you need and mix them in a
porcelain dish.
3. Take some of this mixture and place it in the test tube. The test tube should
be about 1/3 full.
4. This reaction should ideally take place in a fume cupboard. Heat the test tube
containing the mixture over the Bunsen burner. Increase the heat if no reaction
takes place. Once the reaction begins, you will need to remove the test tube
from the flame. Record your observations.
215
O
11.2
CHAPTER 11. PHYSICAL AND CHEMICAL CHANGE - GRADE 10
5. Wait for the product to cool before breaking the test tube with a hammer.
Make sure that the test tube is rolled in paper before you do this, otherwise
the glass will shatter everywhere and you may be hurt.
6. What does the product look like? Does it look anything like the original reactants? Does it have any of the properties of the reactants (e.g. the magnetism
of iron)?
Results:
• After you removed the test tube from the flame, the mixture glowed a bright
red colour. The reaction is exothermic and produces energy.
• The product, iron sulfide, is a dark colour and does not share any of the
properties of the original reactants. It is an entirely new product.
Conclusions:
A synthesis reaction has taken place. The equation for the reaction is:
F e + S → F eS
Activity :: Investigation : Physical or chemical change?
Apparatus:
Bunsen burner, 4 test tubes, a test tube rack and a test tube holder, small
spatula, pipette, magnet, a birthday candle, NaCl (table salt), 0.1M AgNO3 , 6M
HCl, magnesium ribbon, iron filings, sulfur.
Method:
1. Place a small amount of wax from a birthday candle into a test tube and heat
it over the bunsen burner until it melts. Leave it to cool.
2. Add a small spatula of NaCl to 5 ml water in a test tube and shake. Then use
the pipette to add 10 drops of AgNO3 to the sodium chloride solution.
3. Take a 5 cm piece of magnesium ribbon and tear it into 1 cm pieces. Place
two of these pieces into a test tube and add a few drops of 6M HCl. NOTE:
Be very careful when you handle this acid because it can cause major burns.
4. Take about 0.5 g iron filings and 0.5 g sulfur. Test each substance with a
magnet. Mix the two samples in a test tube, and run a magnet alongside the
outside of the test tube.
5. Now heat the test tube that contains the iron and sulfur. What changes do
you see? What happens now, if you run a magnet along the outside of the test
tube?
6. In each of the above cases, record your observations.
Questions:
Decide whether each of the following changes are physical or chemical and give
a reason for your answer in each case. Record your answers in the table below:
Description
Physical
chemical
change
melting candle wax
dissolving NaCl
mixing NaCl with AgNO3
tearing magnesium ribbon
adding HCl to magnesium ribbon
mixing iron and sulfur
heating iron and sulfur
216
or
Reason
CHAPTER 11. PHYSICAL AND CHEMICAL CHANGE - GRADE 10
11.3
11.3
Energy changes in chemical reactions
All reactions involve some change in energy. During a physical change in matter, such as the
evaporation of liquid water to water vapour, the energy of the water molecules increases. However, the change in energy is much smaller than in chemical reactions.
When a chemical reaction occurs, some bonds will break, while new bonds may form. Energy
changes in chemical reactions result from the breaking and forming of bonds. For bonds to
break, energy must be absorbed. When new bonds form, energy will be released because the
new product has a lower energy than the ’inbetween’ stage of the reaction when the bonds in
the reactants have just been broken.
In some reactions, the energy that must be absorbed to break the bonds in the reactants, is less
than the total energy that is released when new bonds are formed. This means that in the overall
reaction, energy is released. This type of reaction is known as an exothermic reaction. In other
reactions, the energy that must be absorbed to break the bonds in the reactants, is more than
the total energy that is released when new bonds are formed. This means that in the overall
reaction, energy must be absorbed from the surroundings. This type of reaction is known as an
endothermic reaction. In the earlier part of this chapter, most decomposition reactions were
endothermic, and heating was needed for the reaction to occur. Most of the synthesis reactions
were exothermic, meaning that energy was given off in the form of heat or light.
More simply, we can describe the energy changes that take place during a chemical reaction as:
Total energy absorbed to break bonds - Total energy released when new bonds form
So, for example, in the reaction...
2M g + O2 → 2M gO
Energy is needed to break the O-O bonds in the oxygen molecule so that new Mg-O bonds can
be formed, and energy is released when the product (MgO) forms.
Despite all the energy changes that seem to take place during reactions, it is important to
remember that energy cannot be created or destroyed. Energy that enters a system will have
come from the surrounding environment, and energy that leaves a system will again become part
of that environment. This principle is known as the principle of conservation of energy.
Definition: Conservation of energy principle
Energy cannot be created or destroyed. It can only be changed from one form to another.
Chemical reactions may produce some very visible, and often violent, changes. An explosion,
for example, is a sudden increase in volume and release of energy when high temperatures are
generated and gases are released. For example, NH4 NO3 can be heated to generate nitrous oxide.
Under these conditions, it is highly sensitive and can detonate easily in an explosive exothermic
reaction.
11.4
Conservation of atoms and mass in reactions
The total mass of all the substances taking part in a chemical reaction is conserved during a
chemical reaction. This is known as the law of conservation of mass. The total number of
atoms of each element also remains the same during a reaction, although these may be arranged
differently in the products.
We will use two of our earlier examples of chemical reactions to demonstrate this:
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11.4
CHAPTER 11. PHYSICAL AND CHEMICAL CHANGE - GRADE 10
• The decomposition of hydrogen peroxide into water and oxygen
2H2 O2 → 2H2 O + O2
H
H
O
H
O
H
H
O
H
O
H
H
+
O
O
O
Left hand side of the equation
Total atomic mass = (4 × 1) + (4 × 16) = 68 u
Number of atoms of each element = (4 × H) + (4 × O)
Right hand side of the equation
Total atomic mass = (4 × 1) + (2 × 16) + (2 × 16) = 68 u
Number of atoms of each element = (4 × H) + (4 × O)
Both the atomic mass and the number of atoms of each element are conserved in the
reaction.
• The synthesis of magnesium and oxygen to form magnesium oxide
2M g + O2 → 2M gO
Mg
Mg
+
O
O
Mg
O
Mg
O
Left hand side of the equation
Total atomic mass = (2 × 24.3) + (2 × 16) = 80.6 u
Number of atoms of each element = (2 × Mg) + (2 × O)
Right hand side of the equation
Total atomic mass = (2 × 24.3) + (2 × 16) = 80.6 u
Number of atoms of each element = (2 × Mg) + (2 × O)
Both the atomic mass and the number of atoms of each element are conserved in the
reaction.
Activity :: Demonstration : The conservation of atoms in chemical reactions
Materials:
• Coloured marbles or small balls to represent atoms. Each colour will represent
a different element.
218
O
CHAPTER 11. PHYSICAL AND CHEMICAL CHANGE - GRADE 10
11.5
• Prestik
Method:
1. Choose a reaction from any that have been used in this chapter or any other
balanced chemical reaction that you can think of. To help to explain this
activity, we will use the decomposition reaction of calcium carbonate to produce
carbon dioxide and calcium oxide.
CaCO3 → CO2 + CaO
2. Stick marbles together to represent the reactants and put these on one side of
your table. In this example you may for example join one red marble (calcium),
one green marble (carbon) and three yellow marbles (oxygen) together to form
the molecule calcium carbonate (CaCO3 ).
3. Leaving your reactants on the table, use marbles to make the product molecules
and place these on the other side of the table.
4. Now count the number of atoms on each side of the table. What do you notice?
5. Observe whether there is any difference between the molecules in the reactants
and the molecules in the products.
Discussion
You should have noticed that the number of atoms in the reactants is the same
as the number of atoms in the product. The number of atoms is conserved during
the reaction. However, you will also see that the molecules in the reactants and
products is not the same. The arrangement of atoms is not conserved during the
reaction.
11.5
Law of constant composition
In any given chemical compound, the elements always combine in the same proportion with each
other. This is the law of constant proportions.
The law of constant composition says that, in any particular chemical compound, all samples
of that compound will be made up of the same elements in the same proportion or ratio. For
example, any water molecule is always made up of two hydrogen atoms and one oxygen atom in
a 2:1 ratio. If we look at the relative masses of oxygen and hydrogen in a water molecule, we
see that 94% of the mass of a water molecule is accounted for by oxygen, and the remaining 6%
is the mass of hydrogen. This mass proportion will be the same for any water molecule.
This does not mean that hydrogen and oxygen always combine in a 2:1 ratio to form H2 O.
Multiple proportions are possible. For example, hydrogen and oxygen may combine in different proportions to form H2 O2 rather than H2 O. In H2 O2 , the H:O ratio is 1:1 and the mass
ratio of hydrogen to oxygen is 1:16. This will be the same for any molecule of hydrogen peroxide.
11.6
Volume relationships in gases
In a chemical reaction between gases, the relative volumes of the gases in the reaction are present
in a ratio of small whole numbers if all the gases are at the same temperature and pressure. This
relationship is also known as Gay-Lussac’s Law.
For example, in the reaction between hydrogen and oxygen to produce water, two volumes of
H2 react with 1 volume of O2 to produce 2 volumes of H2 O.
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CHAPTER 11. PHYSICAL AND CHEMICAL CHANGE - GRADE 10
2H2 + O2 → 2H2 O
In the reaction to produce ammonia, one volume of nitrogen gas reacts with three volumes of
hydrogen gas to produce two volumes of ammonia gas.
N2 + 3H2 → 2N H3
This relationship will also be true for all other chemical reactions.
11.7
Summary
• Matter does not stay the same. It may undergo physical or chemical changes
• A physical change means that the form of matter may change, but not its identity. For
example, when water evaporates, the energy and the arrangement of water molecules will
change, but not the structure of the water molecule itself.
• During a physical change, the arrangement of particles may change but the mass, number
of atoms and number of molecules will stay the same.
• Physical changes involve small changes in energy, and are easily reversible.
• A chemical change occurs when one form of matter changes into something else. A
chemical reaction involves the formation of new substances with different properties.
For example, carbon dioxide reacts with water to form carbonic acid.
CO2 + H2 O → H2 CO3
• A chemical change may involve a decomposition or synthesis reaction. During chemical
change, the mass and number of atoms is conserved, but the number of molecules is not
always the same.
• Chemical reactions involve larger changes in energy. During a reaction, energy is needed
to break bonds in the reactants, and energy is released when new products form. If the
energy released is greater than the energy absorbed, then the reaction is exothermic. If the
energy released is less than the energy absorbed, then the reaction is endothermic. These
chemical reactions are not easily reversible.
• Decomposition reactions are usually endothermic and synthesis reactions are usually
exothermic.
• The law of conservation of mass states that the total mass of all the substances taking
part in a chemical reaction is conserved and the number of atoms of each element in the
reaction does not change when a new product is formed.
• The conservation of energy principle states that energy cannot be created or destroyed,
it can only change from one form to another.
• The law of constant composition states that in any particular compound, all samples of
that compound will be made up of the same elements in the same proportion or ratio.
• Gay-Lussac’s Law states that in a chemical reaction between gases, the relative volumes
of the gases in the reaction are present in a ratio of small whole numbers if all the gases
are at the same temperature and pressure.
Exercise: Summary exercise
1. Complete the following table by saying whether each of the descriptions is an
example of a physical or chemical change:
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CHAPTER 11. PHYSICAL AND CHEMICAL CHANGE - GRADE 10
Description
Physical
chemical
11.7
or
hot and cold water mix together
milk turns sour
a car starts to rust
food digests in the stomach
alcohol disappears when it is placed on your skin
warming food in a microwave
separating sand and gravel
fireworks exploding
2. For each of the following reactions, say whether it is an example of a synthesis
or decomposition reaction:
(a)
(b)
(c)
(d)
(N H4 )2 CO3 → 2N H3 + CO2 + H2 O
4F e + 3O2 → 2F e2 O3
N2 (g) + 3H2 (g) → 2N H3
CaCO3 (s) → CaO + CO2
3. For the following equation:
CaCO3 → CO2 + CaO
Show that the ’law of conservation of mass’ applies.
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11.7
CHAPTER 11. PHYSICAL AND CHEMICAL CHANGE - GRADE 10
222
Chapter 12
Representing Chemical Change Grade 10
As we have already mentioned, a number of changes can occur when elements react with one
another. These changes may either be physical or chemical. One way of representing these
changes is through balanced chemical equations. A chemical equation describes a chemical
reaction by using symbols for the elements involved. For example, if we look at the reaction
between iron (Fe) and sulfur (S) to form iron sulfide (FeS), we could represent these changes
either in words or using chemical symbols:
iron + sulfur → iron sulfide
or
F e + S → F eS
Another example would be:
ammonia + oxygen → nitric oxide + water
or
4N H3 + 5O2 → 4N O + 6H2 O
Compounds on the left of the arrow are called the reactants and these are needed for the reaction to take place. In this equation, the reactants are ammonia and oxygen. The compounds on
the right are called the products and these are what is formed from the reaction.
In order to be able to write a balanced chemical equation, there are a number of important
things that need to be done:
1. Know the chemical symbols for the elements involved in the reaction
2. Be able to write the chemical formulae for different reactants and products
3. Balance chemical equations by understanding the laws that govern chemical change
4. Know the state symbols for the equation
We will look at each of these steps separately in the next sections.
12.1
Chemical symbols
It is very important to know the chemical symbols for common elements in the Periodic Table
so that you are able to write chemical equations and to recognise different compounds.
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CHAPTER 12. REPRESENTING CHEMICAL CHANGE - GRADE 10
Exercise: Revising common chemical symbols
• Write down the chemical symbols and names of all the elements that you know.
• Compare your list with another learner and add any symbols and names that
you don’t have.
• Spend some time, either in class or at home, learning the symbols for at least the
first twenty elements in the periodic table. You should also learn the symbols
for other common elements that are not in the first twenty.
• Write a short test for someone else in the class and then exchange tests with
them so that you each have the chance to answer one.
12.2
Writing chemical formulae
A chemical formula is a concise way of giving information about the atoms that make up a
particular chemical compound. A chemical formula shows each element by its symbol, and also
shows how many atoms of each element are found in that compound. The number of atoms (if
greater than one) is shown as a subscript.
Examples:
CH4 (methane)
Number of atoms: (1 x carbon) + (4 x hydrogen) = 5 atoms in one methane molecule
H2 SO4 (sulfuric acid)
Number of atoms: (2 x hydrogen) + (1 x sulfur) + (4 x oxygen) = 7 atoms in one molecule of
sulfuric acid
A chemical formula may also give information about how the atoms are arranged in a molecule
if it is written in a particular way. A molecule of ethane, for example, has the chemical formula
C2 H6 . This formula tells us how many atoms of each element are in the molecule, but doesn’t
tell us anything about how these atoms are arranged. In fact, each carbon atom in the ethane
molecule is bonded to three hydrogen atoms. Another way of writing the formula for ethane is
CH3 CH3 . The number of atoms of each element has not changed, but this formula gives us
more information about how the atoms are arranged in relation to each other.
The slightly tricky part of writing chemical formulae comes when you have to work out the ratio
in which the elements combine. For example, you may know that sodium (Na) and chlorine (Cl)
react to form sodium chloride, but how do you know that in each molecule of sodium chloride
there is only one atom of sodium for every one atom of chlorine? It all comes down to the
valency of an atom or group of atoms. Valency is the number of bonds that an element can
form with another element. Working out the chemical formulae of chemical compounds using
their valency, will be covered in chapter 4. For now, we will use formulae that you already know.
12.3
Balancing chemical equations
12.3.1
The law of conservation of mass
In order to balance a chemical equation, it is important to understand the law of conservation
of mass.
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CHAPTER 12. REPRESENTING CHEMICAL CHANGE - GRADE 10
12.3
Definition: The law of conservation of mass
The mass of a closed system of substances will remain constant, regardless of the processes
acting inside the system. Matter can change form, but cannot be created or destroyed. For
any chemical process in a closed system, the mass of the reactants must equal the mass of
the products.
In a chemical equation then, the mass of the reactants must be equal to the mass of the products. In order to make sure that this is the case, the number of atoms of each element in the
reactants must be equal to the number of atoms of those same elements in the products. Some
examples are shown below:
Example 1:
F e + S → F eS
Fe
+
S
Fe
S
Reactants
Atomic mass of reactants = 55.8 u + 32.1 u = 87.9 u
Number of atoms of each element in the reactants: (1 × Fe) and (1 × S)
Products
Atomic mass of product = 55.8 u + 32.1 u = 87.9 u
Number of atoms of each element in the products: (1 × Fe) and (1 × S)
Since the number of atoms of each element is the same in the reactants and in the products, we
say that the equation is balanced.
Example 2:
H2 + O2 → H2 O
H
H
+
O
O
H
O
H
Reactants
Atomic mass of reactants = (1 + 1) + (16 + 16) = 34 u
Number of atoms of each element in the reactants: (2 × H) and (2 × O)
Product
Atomic mass of product = (1 + 1 + 16) = 18 u
Number of atoms of each element in the products: (2 × H) and (1 × O)
Since the total atomic mass of the reactants and the products is not the same, and since there are
more oxygen atoms in the reactants than there are in the product, the equation is not balanced.
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12.3
CHAPTER 12. REPRESENTING CHEMICAL CHANGE - GRADE 10
Example 3:
N aOH + HCl → N aCl + H2 O
Na
O
H
+
H
Cl
Na
Cl
+
H
O
H
Reactants
Atomic mass of reactants = (23 + 16 + 1) + (1 + 35.4) = 76.4 u
Number of atoms of each element in the reactants: (1 × Na) + (1 × O) + (2 × H) + (1 × Cl)
Products
Atomic mass of products = (23 + 35.4) + (1 + 1 + 16) = 76.4 u
Number of atoms of each element in the products: (1 × Na) + (1 × O) + (2 × H) + (1 × Cl)
Since the number of atoms of each element is the same in the reactants and in the products, we
say that the equation is balanced.
We now need to find a way to balance those equations that are not balanced so that the number
of atoms of each element in the reactants is the same as that for the products. This can be
done by changing the coefficients of the molecules until the atoms on each side of the arrow
are balanced. You will see later in chapter 13 that these coefficients tell us something about the
mole ratio in which substances react. They also tell us about the volume relationship between
gases in the reactants and products.
Important: Coefficients
Remember that if you put a number in front of a molecule, that number applies to the whole
molecule. For example, if you write 2H2 O, this means that there are 2 molecules of water. In
other words, there are 4 hydrogen atoms and 2 oxygen atoms. If we write 3HCl, this means that
there are 3 molecules of HCl. In other words there are 3 hydrogen atoms and 3 chlorine atoms
in total. In the first example, 2 is the coefficient and in the second example, 3 is the coefficient.
12.3.2
Steps to balance a chemical equation
When balancing a chemical equation, there are a number of steps that need to be followed.
• STEP 1: Identify the reactants and the products in the reaction, and write their chemical
formulae.
• STEP 2: Write the equation by putting the reactants on the left of the arrow, and the
products on the right.
• STEP 3: Count the number of atoms of each element in the reactants and the number of
atoms of each element in the products.
• STEP 4: If the equation is not balanced, change the coefficients of the molecules until the
number of atoms of each element on either side of the equation balance.
• STEP 5: Check that the atoms are in fact balanced.
• STEP 6 (we will look at this a little later): Add any extra details to the equation e.g.
phase.
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CHAPTER 12. REPRESENTING CHEMICAL CHANGE - GRADE 10
Worked Example 49: Balancing chemical equations 1
Question: Balance the following equation:
M g + HCl → M gCl2 + H2
Answer
Step 1 : Because the equation has been written for you, you can move
straight on to counting the number of atoms of each element in the reactants
and products
Reactants: Mg = 1 atom; H = 1 atom and Cl = 1 atom
Products: Mg = 1 atom; H = 2 atoms and Cl = 2 atoms
Step 2 : Balance the equation
The equation is not balanced since there are 2 chlorine atoms in the product and only
1 in the reactants. If we add a coefficient of 2 to the HCl to increase the number of
H and Cl atoms in the reactants, the equation will look like this:
M g + 2HCl → M gCl2 + H2
Step 3 : Check that the atoms are balanced
If we count the atoms on each side of the equation, we find the following:
Reactants: Mg = 1; H = 2; Cl = 2
Products: Mg = 1; H = 2; Cl = 2
The equation is balanced. The final equation is:
M g + 2HCl → M gCl2 + H2
Worked Example 50: Balancing chemical equations 2
Question: Balance the following equation:
CH4 + O2 → CO2 + H2 O
Answer
Step 1 : Count the number of atoms of each element in the reactants and
products
Reactants: C = 1; H = 4; O = 2
Products: C = 1; H = 2; O = 3
Step 2 : Balance the equation
If we add a coefficient of 2 to H2 O, then the number of hydrogen atoms in the
reactants will be 4, which is the same as for the reactants. The equation will be:
CH4 + O2 → CO2 + 2H2 O
Step 3 : Check that the atoms balance
Reactants: C = 1; H = 4; O = 2
Products: C = 1; H = 4; O = 4
You will see that, although the number of hydrogen atoms now balances, there are
more oxygen atoms in the products. You now need to repeat the previous step. If
we put a coefficient of 2 in front of O2 , then we will increase the number of oxygen
atoms in the reactants by 2. The new equation is:
CH4 + 2O2 → CO2 + 2H2 O
When we check the number of atoms again, we find that the number of atoms
of each element in the reactants is the same as the number in the products. The
equation is now balanced.
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12.3
12.3
CHAPTER 12. REPRESENTING CHEMICAL CHANGE - GRADE 10
Worked Example 51: Balancing chemical equations 3
Question: Nitrogen gas reacts with hydrogen gas to form ammonia. Write a balanced chemical equation for this reaction.
Answer
Step 1 : Identify the reactants and the products, and write their chemical
formulae
The reactants are nitrogen (N2 ) and hydrogen (H2 ), and the product is ammonia
(NH3 ).
Step 2 : Write the equation so that the reactants are on the left and products
on the right of the arrow
The equation is as follows:
N 2 + H2 → N H 3
Step 3 : Count the atoms of each element in the reactants and products
Reactants: N = 2; H = 2
Products: N = 1; H = 3
Step 4 : Balance the equation
In order to balance the number of nitrogen atoms, we could rewrite the equation as:
N2 + H2 → 2N H3
Step 5 : Check that the atoms are balanced
In the above equation, the nitrogen atoms now balance, but the hydrogen atoms
don’t (there are 2 hydrogen atoms in the reactants and 6 in the product). If we put
a coefficient of 3 in front of the hydrogen (H2 ), then the hydrogen atoms and the
nitrogen atoms balance. The final equation is:
N2 + 3H2 → 2N H3
Worked Example 52: Balancing chemical equations 4
Question: In our bodies, sugar (C6 H12 O6 ) reacts with the oxygen we breathe in
to produce carbon dioxide, water and energy. Write the balanced equation for this
reaction.
Answer
Step 1 : Identify the reactants and products in the reaction, and write their
chemical formulae.
Reactants: sugar (C6 H12 O6 ) and oxygen (O2 )
Products: carbon dioxide (CO2 ) and water (H2 O)
Step 2 : Write the equation by putting the reactants on the left of the arrow,
and the products on the right
C6 H12 O6 + O2 → CO2 + H2 O
Step 3 : Count the number of atoms of each element in the reactants and
the number of atoms of each element in the products
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CHAPTER 12. REPRESENTING CHEMICAL CHANGE - GRADE 10
Reactants: C=6; H=12; O=8;
Products: C=1; H=2; O=3;
Step 4 : Change the coefficents of the molecules until the number of atoms
of each element on either side of the equation balance.
It is easier to start with carbon as it only appears once on each side. If we add a 6
in front of CO2 , the equation looks like this:
C6 H12 O6 + O2 → 6CO2 + H2 O
Reactants: C=6; H=12; O=8;
Products: C=6; H=2; O=13;
Step 5 : Change the coefficients again to try to balance the equation.
Let’s try to get the number of hydrogens the same this time.
C6 H12 O6 + O2 → 6CO2 + 6H2 O
Reactants: C=6; H=12; O=8;
Products: C=6; H=12; O=18;
Step 6 : Now we just need to balance the oxygen atoms.
C6 H12 O6 + 6O2 → 6CO2 + 6H2 O
Reactants: C=6; H=12; O=18;
Products: C=6; H=12; O=18;
Exercise: Balancing simple chemical equations
Balance the following equations:
1. Hydrogen fuel cells are extremely important in the development of alternative
energy sources. Many of these cells work by reacting hydrogen and oxygen gases
together to form water, a reaction which also produces electricity. Balance the
following equation:
H2 (g) + O2 (g) → H2 O(l)
2. The synthesis of ammonia (NH3 ), made famous by the German chemist Fritz
Haber in the early 20th century, is one of the most important reactions in the
chemical industry. Balance the following equation used to produce ammonia:
3.
4.
5.
6.
7.
8.
N2 (g) + H2 (g) → N H3 (g)
M g + P4 → M g3 P2
Ca + H2 O → Ca(OH)2 + H2
CuCO3 + H2 SO4 → CuSO4 + H2 O + CO2
CaCl2 + N a2 CO3 → CaCO3 + N aCl
C12 H22 O11 + O2 → CO2 + H2 O
Barium chloride reacts with sulphuric acid to produce barium sulphate and
hydrochloric acid.
9. Ethane (C2 H6 ) reacts with oxygen to form carbon dioxide and steam.
10. Ammonium carbonate is often used as a smelling salt. Balance the following
reaction for the decomposition of ammonium carbonate:
(N H4 )2 CO3 (s) → N H3 (aq) + CO2 (g) + H2 O(l)
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12.3
12.4
CHAPTER 12. REPRESENTING CHEMICAL CHANGE - GRADE 10
12.4
State symbols and other information
The state (phase) of the compounds can be expressed in the chemical equation. This is done by
placing the correct label on the right hand side of the formula. There are only four labels that
can be used:
1. (g) for gaseous compounds
2. (l) for liquids
3. (s) for solid compounds
4. (aq) for an aqueous (water) solution
Occasionally, a catalyst is added to the reaction. A catalyst is a substance that speeds up the
reaction without undergoing any change to itself. In a chemical equation, this is shown by using
the symbol of the catalyst above the arrow in the equation.
To show that heat was needed for the reaction, a Greek delta (∆) is placed above the arrow in
the same way as the catalyst.
Important: You may remember from chapter 11 that energy cannot be created or destroyed
during a chemical reaction but it may change form. In an exothermic reaction, ∆H is less
than zero, and in an endothermic reaction, ∆H is greater than zero. This value is often
written at the end of a chemical equation.
Worked Example 53: Balancing chemical equations 4
Question: Solid zinc metal reacts with aqueous hydrochloric acid to form an aqueous solution of zinc chloride (ZnCl2 )and hydrogen gas. Write a balanced equation
for this reaction.
Answer
Step 1 : Identify the reactants and products and their chemical formulae
The reactants are zinc (Zn) and hydrochloric acid (HCl). The products are zinc
chloride (ZnCl2 ) and hydrogen (H2 ).
Step 2 : Place the reactants on the left of the equation and the products on
the right hand side of the arrow.
Zn + HCl → ZnCl2 + H2
Step 3 : Balance the equation
You will notice that the zinc atoms balance but the chlorine and hydrogen atoms
don’t. Since there are two chlorine atoms on the right and only one on the left, we
will give HCl a coefficient of 2 so that there will be two chlorine atoms on each side
of the equation.
Zn + 2HCl → ZnCl2 + H2
Step 4 : Check that all the atoms balance
When you look at the equation again, you will see that all the atoms are now balanced.
Step 5 : Ensure all details (e.g. state symbols) are added
In the initial description, you were told that zinc was a metal, hydrochloric acid and
zinc chloride were in aqueous solutions and hydrogen was a gas.
Zn(s) + 2HCl(aq) → ZnCl2 (aq) + H2 (g)
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CHAPTER 12. REPRESENTING CHEMICAL CHANGE - GRADE 10
Worked Example 54: Balancing chemical equations 5 (advanced)
Question: Balance the following equation:
(N H4 )2 SO4 + N aOH → N H3 + H2 O + N a2 SO4
In this example, the first two steps are not necessary because the reactants and
products have already been given.
Answer
Step 1 : Balance the equation
With a complex equation, it is always best to start with atoms that appear only once
on each side i.e. Na, N and S atoms. Since the S atoms already balance, we will
start with Na and N atoms. There are two Na atoms on the right and one on the
left. We will add a second Na atom by giving NaOH a coefficient of two. There are
two N atoms on the left and one on the right. To balance the N atoms, NH3 will
be given a coefficient of two. The equation now looks as follows:
(N H4 )2 SO4 + 2N aOH → 2N H3 + H2 O + N a2 SO4
Step 2 : Check that all atoms balance
N, Na and S atoms balance, but O and H atoms do not. There are six O atoms and
ten H atoms on the left, and five O atoms and eight H atoms on the right. We need
to add one O atom and two H atoms on the right to balance the equation. This
is done by adding another H2 O molecule on the right hand side. We now need to
check the equation again:
(N H4 )2 SO4 + 2N aOH → 2N H3 + 2H2 O + N a2 SO4
The equation is now balanced.
Exercise: Balancing more advanced chemical equations
Write balanced equations for each of the following reactions:
1. Al2 O3 (s) + H2 SO4 (aq) → Al2 (SO4 )3 (aq) + 3H2 O(l)
2. M g(OH)2 (aq) + HN O3 (aq) → M g(N O3 )2 (aq) + 2H2 O(l)
3. Lead(ll)nitrate solution reacts with potassium iodide solution.
4. When heated, aluminium reacts with solid copper oxide to produce copper
metal and aluminium oxide (Al2 O3 ).
5. When calcium chloride solution is mixed with silver nitrate solution, a white
precipitate (solid) of silver chloride appears. Calcium nitrate (Ca(NO3 )2 ) is
also produced in the solution.
231
12.4
12.5
12.5
CHAPTER 12. REPRESENTING CHEMICAL CHANGE - GRADE 10
Summary
• A chemical equation uses symbols to describe a chemical reaction.
• In a chemical equation, reactants are written on the left hand side of the equation, and
the products on the right. The arrow is used to show the direction of the reaction.
• When representing chemical change, it is important to be able to write the chemical
formula of a compound.
• In any chemical reaction, the law of conservation of mass applies. This means that
the total atomic mass of the reactants must be the same as the total atomic mass of the
products. This also means that the number of atoms of each element in the reactants
must be the same as the number of atoms of each element in the product.
• If the number of atoms of each element in the reactants is the same as the number of
atoms of each element in the product, then the equation is balanced.
• If the number of atoms of each element in the reactants is not the same as the number of
atoms of each element in the product, then the equation is not balanced.
• In order to balance an equation, coefficients can be placed in front of the reactants and
products until the number of atoms of each element is the same on both sides of the
equation.
Exercise: Summary exercise
Balance each of the following chemical equations:
1. N H4 + H2 O → N H4 OH
2. Sodium chloride and water react to form sodium hydroxide, chlorine and hydrogen.
3. Propane is a fuel that is commonly used as a heat source for engines and homes.
Balance the following equation for the combustion of propane:
C3 H8 (l) + O2 (g) → CO2 (g) + H2 O(l)
4. Aspartame, an artificial sweetener, has the formula C14 H18 N2 O5 . Write the
balanced equation for its combustion (reaction with O2 ) to form CO2 gas,
liquid H2 O, and N2 gas.
5. F e2 (SO4 )3 + K(SCN ) → K3 F e(SCN )6 + K2 SO4
6. Chemical weapons were banned by the Geneva Protocol in 1925. According
to this protocol, all chemicals that release suffocating and poisonous gases
are not to be used as weapons. White phosphorus, a very reactive allotrope
of phosphorus, was recently used during a military attack. Phosphorus burns
vigorously in oxygen. Many people got severe burns and some died as a result.
The equation for this spontaneous reaction is:
P4 (s) + O2 (g) → P2 O5 (s)
(a) Balance the chemical equation.
(b) Prove that the law of conservation of mass is obeyed during this chemical
reaction.
(c) Name the product formed during this reaction.
(d) Classify the reaction as endothermic or exothermic. Give a reason for your
answer.
(e) Classify the reaction as a sythesis or decomposition reaction. Give a reason
for your answer.
(DoE Exemplar Paper 2 2007)
232
Chapter 13
Quantitative Aspects of Chemical
Change - Grade 11
An equation for a chemical reaction can provide us with a lot of useful information. It tells us
what the reactants and the products are in the reaction, and it also tells us the ratio in which
the reactants combine to form products. Look at the equation below:
F e + S → F eS
In this reaction, every atom of iron (Fe) will react with a single atom of sulfur (S) to form one
molecule of iron sulfide (FeS). However, what the equation doesn’t tell us, is the quantities or
the amount of each substance that is involved. You may for example be given a small sample
of iron for the reaction. How will you know how many atoms of iron are in this sample? And
how many atoms of sulfur will you need for the reaction to use up all the iron you have? Is
there a way of knowing what mass of iron sulfide will be produced at the end of the reaction?
These are all very important questions, especially when the reaction is an industrial one, where
it is important to know the quantities of reactants that are needed, and the quantity of product
that will be formed. This chapter will look at how to quantify the changes that take place in
chemical reactions.
13.1
The Mole
Sometimes it is important to know exactly how many particles (e.g. atoms or molecules) are in
a sample of a substance, or what quantity of a substance is needed for a chemical reaction to
take place.
You will remember from chapter 3 that the relative atomic mass of an element, describes the
mass of an atom of that element relative to the mass of an atom of carbon-12. So the mass of
an atom of carbon (relative atomic mass is 12 u) for example, is twelve times greater than the
mass of an atom of hydrogen, which has a relative atomic mass of 1 u. How can this information
be used to help us to know what mass of each element will be needed if we want to end up with
the same number of atoms of carbon and hydrogen?
Let’s say for example, that we have a sample of 12g carbon. What mass of hydrogen will contain
the same number of atoms as 12 g carbon? We know that each atom of carbon weighs twelve
times more than an atom of hydrogen. Surely then, we will only need 1g of hydrogen for the
number of atoms in the two samples to be the same? You will notice that the number of particles
(in this case, atoms) in the two substances is the same when the ratio of their sample masses
(12g carbon: 1g hydrogen = 12:1) is the same as the ratio of their relative atomic masses (12
u: 1 u = 12:1).
233
13.1
CHAPTER 13. QUANTITATIVE ASPECTS OF CHEMICAL CHANGE - GRADE 11
To take this a step further, if you were to weigh out samples of a number of elements so that the
mass of the sample was the same as the relative atomic mass of that element, you would find
that the number of particles in each sample is 6.023 x 1023 . These results are shown in table
13.1 below for a number of different elements. So, 24.31 g of magnesium (relative atomic mass
= 24.31 u) for example, has the same number of atoms as 40.08 g of calcium (relative atomic
mass = 40.08 u).
Table 13.1: Table showing the relationship between the sample mass, the relative atomic mass
and the number of atoms in a sample, for a number of elements.
Element
Relative atomic mass (u) Sample mass (g) Atoms in sample
Hydrogen (H)
1.01
1.01
6.023 x 1023
Carbon (C)
12.01
12.01
6.023 x 1023
Magnesium (Mg)
24.31
24.31
6.023 x 1023
Sulfur (S)
32.07
32.07
6.023 x 1023
Calcium (Ca)
40.08
40.08
6.023 x 1023
This result is so important that scientists decided to use a special unit of measurement to define
this quantity: the mole or ’mol’. A mole is defined as being an amount of a substance which
contains the same number of particles as there are atoms in 12 g of carbon. In the examples
that were used earlier, 24.31 g magnesium is one mole of magnesium, while 40.08 g of calcium
is one mole of calcium. A mole of any substance always contains the same number of particles.
Definition: Mole
The mole (abbreviation ’n’) is the SI (Standard International) unit for ’amount of substance’.
It is defined as an amount of substance that contains the same number of particles (atoms,
molecules or other particle units) as there are atoms in 12 g carbon.
In one mole of any substance, there are 6.023 x 1023 particles. This is known as Avogadro’s
number.
Definition: Avogadro constant
The number of particles in a mole, equal to 6.023 x 1023 . It is also sometimes referred to
as the number of atoms in 12 g of carbon-12.
teresting The original hypothesis that was proposed by Amadeo Avogadro was that ’equal
Interesting
Fact
Fact
volumes of gases, at the same temperature and pressure, contain the same number
of molecules’. His ideas were not accepted by the scientific community and it
was only four years after his death, that his original hypothesis was accepted
and that it became known as ’Avogadro’s Law’. In honour of his contribution
to science, the number of particles in one mole was named Avogadro’s number.
Exercise: Moles and mass
1. Complete the following table:
234
CHAPTER 13. QUANTITATIVE ASPECTS OF CHEMICAL CHANGE - GRADE 11
Element
Hydrogen
Magnesium
Carbon
Chlorine
Nitrogen
Relative
atomic mass
(u)
1.01
24.31
12.01
35.45
Sample mass
(g)
13.2
Number
of
moles in the
sample
1.01
24.31
24.02
70.9
42.08
2. How many atoms are there in...
(a) 1 mole of a substance
(b) 2 moles of calcium
(c) 5 moles of phosphorus
(d) 24.31 g of magnesium
(e) 24.02 g of carbon
13.2
Molar Mass
Definition: Molar mass
Molar mass (M) is the mass of 1 mole of a chemical substance. The unit for molar mass is
grams per mole or g.mol−1 .
Refer to table 13.1. You will remember that when the mass, in grams, of an element is equal to
its relative atomic mass, the sample contains one mole of that element. This mass is called the
molar mass of that element.
It is worth remembering the following: On the Periodic Table, the relative atomic mass that is
shown can be interpreted in two ways.
1. The mass of a single, average atom of that element relative to the mass of an atom of
carbon.
2. The mass of one mole of the element. This second use is the molar mass of the element.
Table 13.2: The relationship between relative atomic mass, molar mass and the mass of one
mole for a number of elements.
Element
Magnesium
Lithium
Oxygen
Nitrogen
Iron
Relative
atomic mass
(u)
24.31
6.94
16
14.01
55.85
Molar mass
(g.mol−1 )
24.31
6.94
16
14.01
55.85
235
Mass of one
mole of the
element (g)
24.31
6.94
16
14.01
55.85
13.2
CHAPTER 13. QUANTITATIVE ASPECTS OF CHEMICAL CHANGE - GRADE 11
Worked Example 55: Calculating the number of moles from mass
Question: Calculate the number of moles of iron (Fe) in a 111.7 g sample.
Answer
Step 1 : Find the molar mass of iron
If we look at the periodic table, we see that the molar mass of iron is 55.85 g.mol−1 .
This means that 1 mole of iron will have a mass of 55.85 g.
Step 2 : Use the molar mass and sample mass to calculate the number of
moles of iron
If 1 mole of iron has a mass of 55.85 g, then: the number of moles of iron in 111.7
g must be:
111.7g
= 2mol
55.85g.mol−1
There are 2 moles of iron in the sample.
Worked Example 56: Calculating mass from moles
Question: You have a sample that contains 5 moles of zinc.
1. What is the mass of the zinc in the sample?
2. How many atoms of zinc are in the sample?
Answer
Step 1 : Find the molar mass of zinc
Molar mass of zinc is 65.38 g.mol−1 , meaning that 1 mole of zinc has a mass of
65.38 g.
Step 2 : Calculate the mass of zinc, using moles and molar mass.
If 1 mole of zinc has a mass of 65.38 g, then 5 moles of zinc has a mass of:
65.38 g x 5 mol = 326.9 g (answer to a)
Step 3 : Use the number of moles of zinc and Avogadro’s number to calculate
the number of zinc atoms in the sample.
5 × 6.023 × 1023 = 30.115 × 1023
Exercise: Moles and molar mass
1. Give the molar mass of each of the following elements:
(a) hydrogen
(b) nitrogen
(c) bromine
2. Calculate the number of moles in each of the following samples:
(a) 21.62 g of boron (B)
236
CHAPTER 13. QUANTITATIVE ASPECTS OF CHEMICAL CHANGE - GRADE 11
13.3
(b) 54.94 g of manganese (Mn)
(c) 100.3 g of mercury (Hg)
(d) 50 g of barium (Ba)
(e) 40 g of lead (Pb)
13.3
An equation to calculate moles and mass in chemical
reactions
The calculations that have been used so far, can be made much simpler by using the following
equation:
n (number of moles) =
m (mass of substance in g)
M (molar mass of substance in g · mol−1 )
Important: Remember that when you use the equation n = m/M, the mass is always in
grams (g) and molar mass is in grams per mol (g.mol−1 ).
The equation can also be used to calculate mass and molar mass, using the following equations:
m=n×M
and
M=
m
n
The following diagram may help to remember the relationship between these three variables.
You need to imagine that the horizontal line is like a ’division’ sign and that the vertical line is
like a ’multiplication’ sign. So, for example, if you want to calculate ’M’, then the remaining two
letters in the triangle are ’m’ and ’n’ and ’m’ is above ’n’ with a division sign between them. In
your calculation then, ’m’ will be the numerator and ’n’ will be the denominator.
m
n
M
237
13.3
CHAPTER 13. QUANTITATIVE ASPECTS OF CHEMICAL CHANGE - GRADE 11
Worked Example 57: Calculating moles from mass
Question: Calculate the number of moles of copper there are in a sample that
weighs 127 g.
Answer
Step 1 : Write the equation to calculate the number of moles
n=
m
M
Step 2 : Substitute numbers into the equation
127
=2
63.55
There are 2 moles of copper in the sample.
n=
Worked Example 58: Calculating mass from moles
Question: You are given a 5 mol sample of sodium. What mass of sodium is in the
sample?
Answer
Step 1 : Write the equation to calculate the sample mass.
m=n×M
Step 2 : Substitute values into the equation.
MN a = 22.99 g.mol−1
Therefore,
m = 5 × 22.99 = 114.95g
The sample of sodium has a mass of 114.95 g.
Worked Example 59: Calculating atoms from mass
Question: Calculate the number of atoms there are in a sample of aluminium that
weighs 80.94 g.
Answer
Step 1 : Calculate the number of moles of aluminium in the sample.
n=
m
80.94
=
= 3moles
M
26.98
Step 2 : Use Avogadro’s number to calculate the number of atoms in the
sample.
Number of atoms in 3 mol aluminium = 3 × 6.023 × 1023
There are 18.069 × 1023 aluminium atoms in a sample of 80.94 g.
238
CHAPTER 13. QUANTITATIVE ASPECTS OF CHEMICAL CHANGE - GRADE 11
13.4
Exercise: Some simple calculations
1. Calculate the number of moles in each of the following samples:
(a) 5.6 g of calcium
(b) 0.02 g of manganese
(c) 40 g of aluminium
2. A lead sinker has a mass of 5 g.
(a) Calculate the number of moles of lead the sinker contains.
(b) How many lead atoms are in the sinker?
3. Calculate the mass of each of the following samples:
(a) 2.5 mol magnesium
(b) 12 g lithium
(c) 4.5 × 1025 atoms of silica
13.4
Molecules and compounds
So far, we have only discussed moles, mass and molar mass in relation to elements. But what
happens if we are dealing with a molecule or some other chemical compound? Do the same
concepts and rules apply? The answer is ’yes’. However, you need to remember that all your calculations will apply to the whole molecule. So, when you calculate the molar mass of a molecule,
you will need to add the molar mass of each atom in that compound. Also, the number of moles
will also apply to the whole molecule. For example, if you have one mole of nitric acid (HNO3 ),
it means you have 6.023 x 1023 molecules of nitric acid in the sample. This also means that
there are 6.023 × 1023 atoms of hydrogen, 6.023 × 1023 atoms of nitrogen and (3 × 6.023 ×
1023 ) atoms of oxygen in the sample.
In a balanced chemical equation, the number that is written in front of the element or compound,
shows the mole ratio in which the reactants combine to form a product. If there are no numbers
in front of the element symbol, this means the number is ’1’.
e.g. N2 + 3H2 → 2N H3
In this reaction, 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of
ammonia.
Worked Example 60: Calculating molar mass
Question: Calculate the molar mass of H2 SO4 .
Answer
Step 1 : Use the periodic table to find the molar mass for each element in
the molecule.
Hydrogen = 1.008 g.mol−1 ; Sulfur = 32.07 g.mol−1 ; Oxygen = 16 g.mol−1
239
13.4
CHAPTER 13. QUANTITATIVE ASPECTS OF CHEMICAL CHANGE - GRADE 11
Step 2 : Add the molar masses of each atom in the molecule
M(H2 SO4 ) = (2 × 1.008) + (32.07) + (4 × 16) = 98.09g.mol−1
Worked Example 61: Calculating moles from mass
Question: Calculate the number of moles there are in 1kg of MgCl2 .
Answer
Step 1 : Write the equation for calculating the number of moles in the
sample.
n=
m
M
Step 2 : Calculate the values that you will need, to substitute into the
equation
1. Convert mass into grams
m = 1kg × 1000 = 1000g
2. Calculate the molar mass of MgCl2 .
M(MgCl2 ) = 24.31 + (2 × 35.45) = 95.21g.mol−1
Step 3 : Substitute values into the equation
1000
= 10.5mol
95.21
There are 10.5 moles of magnesium chloride in a 1 kg sample.
n=
Worked Example 62: Calculating the mass of reactants and products
Question: Barium chloride and sulfuric acid react according to the following equation to produce barium sulphate and hydrochloric acid.
BaCl2 + H2 SO4 → BaSO4 + 2HCl
If you have 2 g of BaCl2 ...
1. What quantity (in g) of H2 SO4 will you need for the reaction so that all the
barium chloride is used up?
2. What mass of HCl is produced during the reaction?
Answer
Step 1 : Calculate the number of moles of BaCl2 that react.
n=
2
m
=
= 0.0096mol
M
208.24
240
CHAPTER 13. QUANTITATIVE ASPECTS OF CHEMICAL CHANGE - GRADE 11
Step 2 : Determine how many moles of H2 SO4 are needed for the reaction
According to the balanced equation, 1 mole of BaCl2 will react with 1 mole of
H2 SO4 . Therefore, if 0.0096 moles of BaCl2 react, then there must be the same
number of moles of H2 SO4 that react because their mole ratio is 1:1.
Step 3 : Calculate the mass of H2 SO4 that is needed.
m = n × M = 0.0096 × 98.086 = 0.94g
(answer to 1)
Step 4 : Determine the number of moles of HCl produced.
According to the balanced equation, 2 moles of HCl are produced for every 1 mole of
the two reactants. Therefore the number of moles of HCl produced is (2 × 0.0096),
which equals 0.0192 moles.
Step 5 : Calculate the mass of HCl.
m = n × M = 0.0192 × 35.73 = 0.69g
(answer to 2)
Activity :: Group work : Understanding moles, molecules and Avogadro’s
number
Divide into groups of three and spend about 20 minutes answering the following
questions together:
1. What are the units of the mole? Hint: Check the definition of the mole.
2. You have a 56 g sample of iron sulfide (FeS)
(a) How many moles of FeS are there in the sample?
(b) How many molecules of FeS are there in the sample?
(c) What is the difference between a mole and a molecule?
3. The exact size of Avogadro’s number is sometimes difficult to imagine.
(a) Write down Avogadro’s number without using scientific notation.
(b) How long would it take to count to Avogadro’s number? You can assume
that you can count two numbers in each second.
Exercise: More advanced calculations
1. Calculate the molar mass of the following chemical compounds:
(a) KOH
(b) FeCl3
(c) Mg(OH)2
2. How many moles are present in:
(a) 10 g of Na2 SO4
(b) 34 g of Ca(OH)2
(c) 2.45 x 1023 molecules of CH4 ?
241
13.4
13.5
CHAPTER 13. QUANTITATIVE ASPECTS OF CHEMICAL CHANGE - GRADE 11
3. For a sample of 0.2 moles of potassium bromide (KBr), calculate...
(a) the number of moles of K+ ions
(b) the number of moles of Br− ions
4. You have a sample containing 3 moles of calcium chloride.
(a) What is the chemical formula of calcium chloride?
(b) How many calcium atoms are in the sample?
5. Calculate the mass of:
(a) 3 moles of NH4 OH
(b) 4.2 moles of Ca(NO3 )2
6. 96.2 g sulfur reacts with an unknown quantity of zinc according to the following
equation:
Zn + S → ZnS
(a) What mass of zinc will you need for the reaction, if all the sulfur is to be
used up?
(b) What mass of zinc sulfide will this reaction produce?
7. Calcium chloride reacts with carbonic acid to produce calcium carbonate and
hydrochloric acid according to the following equation:
CaCl2 + H2 CO3 → CaCO3 + 2HCl
If you want to produce 10 g of calcium carbonate through this chemical reaction,
what quantity (in g) of calcium chloride will you need at the start of the
reaction?
13.5
The Composition of Substances
The empirical formula of a chemical compound is a simple expression of the relative number
of each type of atom in it. In contrast, the molecular formula of a chemical compound gives
the actual number of atoms of each element found in a molecule of that compound.
Definition: Empirical formula
The empirical formula of a chemical compound gives the relative number of each type of
atom in it.
Definition: Molecular formula
The molecular formula of a chemical compound gives the exact number of atoms of each
element in one molecule of that compound.
The compound ethanoic acid for example, has the molecular formula CH3 COOH or simply
C2 H4 O2 . In one molecule of this acid, there are two carbon atoms, four hydrogen atoms and
two oxygen atoms. The ratio of atoms in the compound is 2:4:2, which can be simplified to 1:2:1.
Therefore, the empirical formula for this compound is CH2 O. The empirical formula contains the
smallest whole number ratio of the elements that make up a compound.
Knowing either the empirical or molecular formula of a compound, can help to determine its
composition in more detail. The opposite is also true. Knowing the composition of a substance
can help you to determine its formula. There are three different types of composition problems
that you might come across:
242
CHAPTER 13. QUANTITATIVE ASPECTS OF CHEMICAL CHANGE - GRADE 11
13.5
1. Problems where you will be given the formula of the substance and asked to calculate the
percentage by mass of each element in the substance.
2. Problems where you will be given the percentage composition and asked to calculate the
formula.
3. Problems where you will be given the products of a chemical reaction and asked to calculate
the formula of one of the reactants. These are usually referred to as combustion analysis
problems.
Worked Example 63: Calculating the percentage by mass of elements in a
compound
Question: Calculate the percentage that each element contributes to the overall
mass of sulfuric acid (H2 SO4 ).
Answer
Step 1 : Write down the relative atomic mass of each element in the compound.
Hydrogen = 1.008 × 2 = 2.016 u
Sulfur = 32.07 u
Oxygen = 4 × 16 = 64 u
Step 2 : Calculate the molecular mass of sulfuric acid.
Use the calculations in the previous step to calculate the molecular mass of sulfuric
acid.
M ass = 2.016 + 32.07 + 64 = 98.09u
Step 3 : Convert the mass of each element to a percentage of the total mass
of the compound
Use the equation:
Percentage by mass = atomic mass / molecular mass of H2 SO4 × 100%
Hydrogen
2.016
× 100% = 2.06%
98.09
Sulfur
32.07
× 100% = 32.69%
98.09
Oxygen
64
× 100% = 65.25%
98.09
(You should check at the end that these percentages add up to 100%!)
In other words, in one molecule of sulfuric acid, hydrogen makes up 2.06% of the
mass of the compound, sulfur makes up 32.69% and oxygen makes up 65.25%.
243
13.5
CHAPTER 13. QUANTITATIVE ASPECTS OF CHEMICAL CHANGE - GRADE 11
Worked Example 64: Determining the empirical formula of a compound
Question: A compound contains 52.2% carbon (C), 13.0% hydrogen (H) and
34.8% oxygen (O). Determine its empirical formula.
Answer
Step 1 : If we assume that we have 100 g of this substance, then we can
convert each element percentage into a mass in grams.
Carbon = 52.2 g, hydrogen = 13 g and oxygen = 34.8 g
Step 2 : Convert the mass of each element into number of moles
n=
m
M
Therefore,
52.2
= 4.35mol
12.01
n(carbon) =
n(hydrogen) =
13
= 12.90mol
1.008
34.8
= 2.18mol
16
Step 3 : Convert these numbers to the simplest mole ratio by dividing by the
smallest number of moles
In this case, the smallest number of moles is 2.18. Therefore...
Carbon
n(oxygen) =
4.35
=2
2.18
Hydrogen
12.90
=6
2.18
Oxygen
2.18
=1
2.18
Therefore the empirical formula of this substance is: C2 H6 O. Do you recognise this
compound?
Worked Example 65: Determining the formula of a compound
Question: 207 g of lead combines with oxygen to form 239 g of a lead oxide. Use
this information to work out the formula of the lead oxide (Relative atomic masses:
Pb = 207 u and O = 16 u).
Answer
Step 1 : Calculate the mass of oxygen in the reactants
239 − 207 = 32g
244
CHAPTER 13. QUANTITATIVE ASPECTS OF CHEMICAL CHANGE - GRADE 11
Step 2 : Calculate the number of moles of lead and oxygen in the reactants.
n=
m
M
Lead
207
= 1mol
207
Oxygen
32
= 2mol
16
Step 3 : Deduce the formula of the compound
The mole ratio of Pb:O in the product is 1:2, which means that for every atom of
lead, there will be two atoms of oxygen. The formula of the compound is PbO2 .
Worked Example 66: Empirical and molecular formula
Question: Vinegar, which is used in our homes, is a dilute form of acetic acid. A
sample of acetic acid has the following percentage composition: 39.9% carbon, 6.7%
hyrogen and 53.4% oxygen.
1. Determine the empirical formula of acetic acid.
2. Determine the molecular formula of acetic acid if the molar mass of acetic acid
is 60g/mol.
Answer
Step 1 : Calculate the mass of each element in 100 g of acetic acid.
In 100g of acetic acid, there is 39.9 g C, 6.7 g H and 53.4 g O
Step 2 : Calculate the number of moles of each element in 100 g of acetic
acid.
m
n= M
nC
=
nH
=
nO
=
39.9
= 3.33 mol
12
6.7
= 6.7 mol
1
53.4
= 3.34 mol
16
Step 3 : Divide the number of moles of each element by the lowest number
to get the simplest mole ratio of the elements (i.e. the empirical formula) in
acetic acid.
Empirical formula is CH2 O
Step 4 : Calculate the molecular formula, using the molar mass of acetic
acid.
The molar mass of acetic acid using the empirical formula is 30 g/mol. Therefore the
actual number of moles of each element must be double what it is in the emprical
formula.
The molecular formula is therefore C2 H4 O2 or CH3 COOH
245
13.5
13.6
CHAPTER 13. QUANTITATIVE ASPECTS OF CHEMICAL CHANGE - GRADE 11
Exercise: Moles and empirical formulae
1. Calcium chloride is produced as the product of a chemical reaction.
(a) What is the formula of calcium chloride?
(b) What percentage does each of the elements contribute to the mass of a
molecule of calcium chloride?
(c) If the sample contains 5 g of calcium chloride, what is the mass of calcium
in the sample?
(d) How many moles of calcium chloride are in the sample?
2. 13g of zinc combines with 6.4g of sulfur.What is the empirical formula of zinc
sulfide?
(a) What mass of zinc sulfide will be produced?
(b) What percentage does each of the elements in zinc sulfide contribute to
its mass?
(c) Determine the formula of zinc sulfide.
3. A calcium mineral consisted of 29.4% calcium, 23.5% sulphur and 47.1% oxygen
by mass. Calculate the empirical formula of the mineral.
4. A chlorinated hydrocarbon compound when analysed, consisted of 24.24% carbon, 4.04% hydrogen, 71.72% chlorine. The molecular mass was found to be
99 from another experiment. Deduce the empirical and molecular formula.
13.6
Molar Volumes of Gases
It is possible to calculate the volume of a mole of gas at STP using what we now know about
gases.
1. Write down the ideal gas equation
pV = nRT, therefore V =
nRT
p
2. Record the values that you know, making sure that they are in SI units
You know that the gas is under STP conditions. These are as follows:
p = 101.3 kPa = 101300 Pa
n = 1 mole
R = 8.3 J.K−1 .mol−1
T = 273 K
3. Substitute these values into the original equation.
V =
V =
nRT
p
1mol × 8.3J.K −1 .mol−1 × 273K
101300P a
4. Calculate the volume of 1 mole of gas under these conditions
The volume of 1 mole of gas at STP is 22.4 × 10−3 m3 = 22.4 dm3 .
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CHAPTER 13. QUANTITATIVE ASPECTS OF CHEMICAL CHANGE - GRADE 11
13.7
Important: The standard units used for this equation are P in Pa, V in m3 and T in K.
Remember also that 1000cm3 = 1dm3 and 1000dm3 = 1m3 .
Worked Example 67: Ideal Gas
Question: A sample of gas occupies a volume of 20 dm3 , has a temperature of
280 K and has a pressure of 105 Pa. Calculate the number of moles of gas that are
present in the sample.
Answer
Step 1 : Convert all values into SI units
The only value that is not in SI units is volume. V = 0.02 m3 .
Step 2 : Write the equation for calculating the number of moles in a gas.
We know that pV = nRT
Therefore,
n=
pV
RT
Step 3 : Substitute values into the equation to calculate the number of moles
of the gas.
n=
105 × 0.02
2.1
=
= 0.0009moles
8.31 × 280
2326.8
Exercise: Using the combined gas law
1. An enclosed gas has a volume of 300 cm3 and a temperature of 300 K. The
pressure of the gas is 50 kPa. Calculate the number of moles of gas that are
present in the container.
2. What pressure will 3 mol gaseous nitrogen exert if it is pumped into a container
that has a volume of 25 dm3 at a temperature of 29 0 C?
3. The volume of air inside a tyre is 19 litres and the temperature is 290 K. You
check the pressure of your tyres and find that the pressure is 190 kPa. How
many moles of air are present in the tyre?
4. Compressed carbon dioxide is contained within a gas cylinder at a pressure of
700 kPa. The temperature of the gas in the cylinder is 310 K and the number
of moles of gas is 13 moles carbon dioxide. What is the volume of the gas
inside?
13.7
Molar concentrations in liquids
A typical solution is made by dissolving some solid substance in a liquid. The amount of substance
that is dissolved in a given volume of liquid is known as the concentration of the liquid.
Mathematically, concentration (C) is defined as moles of solute (n) per unit volume (V) of
solution.
247
13.7
CHAPTER 13. QUANTITATIVE ASPECTS OF CHEMICAL CHANGE - GRADE 11
C=
n
V
For this equation, the units for volume are dm3 . Therefore, the unit of concentration is mol.dm−3 .
When concentration is expressed in mol.dm−3 it is known as the molarity (M) of the solution.
Molarity is the most common expression for concentration.
Definition: Concentration
Concentration is a measure of the amount of solute that is dissolved in a given volume of
liquid. It is measured in mol.dm−3 . Another term that is used for concentration is molarity
(M)
Worked Example 68: Concentration Calculations 1
Question: If 3.5 g of sodium hydroxide (NaOH) is dissolved in 2.5 dm3 of water,
what is the concentration of the solution in mol.dm−3 ?
Answer
Step 1 : Convert the mass of NaOH into moles
m
3.5
=
= 0.0875mol
M
40
Step 2 : Calculate the concentration of the solution.
n=
C=
n
0.0875
=
= 0.035
V
2.5
The concentration of the solution is 0.035 mol.dm−3 or 0.035 M
Worked Example 69: Concentration Calculations 2
Question: You have a 1 dm3 container in which to prepare a solution of potassium
permanganate (KMnO4 ). What mass of KMnO4 is needed to make a solution with
a concentration of 0.2 M?
Answer
Step 1 : Calculate the number of moles of KMnO4 needed.
C=
n
V
therefore
n = C × V = 0.2 × 1 = 0.2mol
Step 2 : Convert the number of moles of KMnO4 to mass.
m = n × M = 0.2 × 158.04 = 31.61g
The mass of KMnO4 that is needed is 31.61 g.
248
CHAPTER 13. QUANTITATIVE ASPECTS OF CHEMICAL CHANGE - GRADE 11
13.8
Worked Example 70: Concentration Calculations 3
Question: How much sodium chloride (in g) will one need to prepare 500 cm3 of
solution with a concentration of 0.01 M?
Answer
Step 1 : Convert all quantities into the correct units for this equation.
V =
500
= 0.5dm3
1000
Step 2 : Calculate the number of moles of sodium chloride needed.
n = C × V = 0.01 × 0.5 = 0.005mol
Step 3 : Convert moles of KMnO4 to mass.
m = n × M = 0.005 × 58.45 = 0.29g
The mass of sodium chloride needed is 0.29 g
Exercise: Molarity and the concentration of solutions
1. 5.95g of potassium bromide was dissolved in 400cm3 of water. Calculate its
molarity.
2. 100 g of sodium chloride (NaCl) is dissolved in 450 cm3 of water.
(a)
(b)
(c)
(d)
How many moles of NaCl are present in solution?
What is the volume of water (in dm3 )?
Calculate the concentration of the solution.
What mass of sodium chloride would need to be added for the concentration to become 5.7 mol.dm−3 ?
3. What is the molarity of the solution formed by dissolving 80 g of sodium hydroxide (NaOH) in 500 cm3 of water?
4. What mass (g) of hydrogen chloride (HCl) is needed to make up 1000 cm3 of
a solution of concentration 1 mol.dm−3 ?
5. How many moles of H2 SO4 are there in 250 cm3 of a 0.8M sulphuric acid
solution? What mass of acid is in this solution?
13.8
Stoichiometric calculations
Stoichiometry is the study and calculation of relationships between reactants and products of
chemical reactions. Chapter 12 showed how to write balanced chemical equations. By knowing
the ratios of substances in a reaction, it is possible to use stoichiometry to calculate the amount
of reactants and products that are involved in the reaction. Some examples are shown below.
249
13.8
CHAPTER 13. QUANTITATIVE ASPECTS OF CHEMICAL CHANGE - GRADE 11
Worked Example 71: Stoichiometric calculation 1
Question: What volume of oxygen at S.T.P. is needed for the complete combustion
of 2dm3 of propane (C3 H8 )? (Hint: CO2 and H2 O are the products in this reaction)
Answer
Step 1 : Write a balanced equation for the reaction.
C3 H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2 O(g)
Step 2 : Determine the ratio of oxygen to propane that is needed for the
reaction.
From the balanced equation, the ratio of oxygen to propane in the reactants is 5:1.
Step 3 : Determine the volume of oxygen needed for the reaction.
1 volume of propane needs 5 volumes of oxygen, therefore 2 dm3 of propane will
need 10 dm3 of oxygen for the reaction to proceed to completion.
Worked Example 72: Stoichiometric calculation 2
Question: What mass of iron (II) sulphide is formed when 5.6 g of iron is completely
reacted with sulfur?
Answer
Step 1 : Write a balanced chemical equation for the reaction.
F e(s) + S(s) → F eS(s)
Step 2 : Calculate the number of moles of iron that react.
n=
5.6
m
=
= 0.1mol
M
55.85
Step 3 : Determine the number of moles of FeS produced.
From the equation 1 mole of Fe gives 1 mole of FeS. Therefore, 0.1 moles of iron in
the reactants will give 0.1 moles of iron sulfide in the product.
Step 4 : Calculate the mass of iron sulfide formed
m = n × M = 0.1 × 87.911 = 8.79g
The mass of iron (II) sulfide that is produced during this reaction is 8.79 g.
Important:
A closer look at the previous worked example shows that 5.6 g of iron is needed to produce
8.79 g of iron (II) sulphide. The amount of sulfur that is needed in the reactants is 3.2
g. What would happen if the amount of sulfur in the reactants was increased to 6.4 g but
the amount of iron was still 5.6 g? Would more FeS be produced? In fact, the amount
of iron(II) sulfide produced remains the same. No matter how much sulfur is added to the
system, the amount of iron (II) sulfide will not increase because there is not enough iron
to react with the additional sulfur in the reactants to produce more FeS. When all the iron
is used up the reaction stops. In this example, the iron is called the limiting reagent.
Because there is more sulfur than can be used up in the reaction, it is called the excess
reagent.
250
CHAPTER 13. QUANTITATIVE ASPECTS OF CHEMICAL CHANGE - GRADE 11
Worked Example 73: Industrial reaction to produce fertiliser
Question: Sulfuric acid (H2 SO4 ) reacts with ammonia (NH3 ) to produce the fertiliser ammonium sulphate ((NH4 )2 SO4 ) according to the following equation:
H2 SO4 (aq) + 2N H3 (g) → (N H4 )2 SO4 (aq)
What is the maximum mass of ammonium sulphate that can be obtained from 2.0
kg of sulfuric acid and 1.0 kg of ammonia?
Answer
Step 1 : Convert the mass of sulfuric acid and ammonia into moles
n(H2 SO4 ) =
2000g
m
=
= 20.39mol
M
98.078g/mol
n(N H3 ) =
1000g
= 58.72mol
17.03g/mol
Step 2 : Use the balanced equation to determine which of the reactants is
limiting.
From the balanced chemical equation, 1 mole of H2 SO4 reacts with 2 moles of NH3
to give 1 mole of (NH4 )2 SO4 . Therefore 20.39 moles of H2 SO4 need to react with
40.78 moles of NH3 . In this example, NH3 is in excess and H2 SO4 is the limiting
reagent.
Step 3 : Calculate the maximum amount of ammonium sulphate that can be
produced
Again from the equation, the mole ratio of H2 SO4 in the reactants to (NH4 )2 SO4
in the product is 1:1. Therefore, 20.39 moles of H2 SO4 will produce 20.39 moles of
(NH4 )2 SO4 .
The maximum mass of ammonium sulphate that can be produced is calculated as
follows:
m = n × M = 20.41mol × 132g/mol = 2694g
The maximum amount of ammonium sulphate that can be produced is 2.694 kg.
Exercise: Stoichiometry
1. Diborane, B2 H6 , was once considered for use as a rocket fuel. The combustion
reaction for diborane is:
B2 H6 (g) + 3O2 (l) → 2HBO2 (g) + 2H2 O(l)
If we react 2.37 grams of diborane, how many grams of water would we expect
to produce?
2. Sodium azide is a commonly used compound in airbags. When triggered, it
has the following reaction:
2N aN3 (s) → 2N a(s) + 3N2 (g)
If 23.4 grams of sodium azide are reacted, how many moles of nitrogen gas
would we expect to produce?
251
13.8
13.9
CHAPTER 13. QUANTITATIVE ASPECTS OF CHEMICAL CHANGE - GRADE 11
3. Photosynthesis is a chemical reaction that is vital to the existence of life on
Earth. During photosynthesis, plants and bacteria convert carbon dioxide gas,
liquid water, and light into glucose (C6 H12 O6 ) and oxygen gas.
(a) Write down the equation for the photosynthesis reaction.
(b) Balance the equation.
(c) If 3 moles of carbon dioxide are used up in the photosynthesis reaction,
what mass of glucose will be produced?
13.9
Summary
• It is important to be able to quantify the changes that take place during a chemical
reaction.
• The mole (n) is a SI unit that is used to describe an amount of substance that contains
the same number of particles as there are atoms in 12 g of carbon.
• The number of particles in a mole is called the Avogadro constant and its value is 6.023
× 1023 . These particles could be atoms, molecules or other particle units, depending on
the substance.
• The molar mass (M) is the mass of one mole of a substance and is measured in grams
per mole or g.mol−1 . The numerical value of an element’s molar mass is the same as its
relative atomic mass. For a compound, the molar mass has the same numerical value as
the molecular mass of that compound.
• The relationship between moles (n), mass in grams (m) and molar mass (M) is defined by
the following equation:
m
n=
M
• In a balanced chemical equation, the number in front of the chemical symbols describes
the mole ratio of the reactants and products.
• The empirical formula of a compound is an expression of the relative number of each
type of atom in the compound.
• The molecular formula of a compound describes the actual number of atoms of each
element in a molecule of the compound.
• The formula of a substance can be used to calculate the percentage by mass that each
element contributes to the compound.
• The percentage composition of a substance can be used to deduce its chemical formula.
• One mole of gas occupies a volume of 22.4 dm3 .
• The concentration of a solution can be calculated using the following equation,
C=
n
V
where C is the concentration (in mol.dm−3 ), n is the number of moles of solute dissolved
in the solution and V is the volume of the solution (in dm3 ).
• Molarity is a measure of the concentration of a solution, and its units are mol.dm−3 .
• Stoichiometry, the study of the relationships between reactants and products, can be
used to determine the quantities of reactants and products that are involved in chemical
reactions.
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CHAPTER 13. QUANTITATIVE ASPECTS OF CHEMICAL CHANGE - GRADE 11
13.9
• A limiting reagent is the chemical that is used up first in a reaction, and which therefore
determines how far the reaction will go before it has to stop.
• An excess reagent is a chemical that is in greater quantity than the limiting reagent in
the reaction. Once the reaction is complete, there will still be some of this chemical that
has not been used up.
Exercise: Summary Exercise
1. Write only the word/term for each of the following descriptions:
(a) the mass of one mole of a substance
(b) the number of particles in one mole of a substance
2. Multiple choice: Choose the one correct answer from those given.
A 5 g of magnesium chloride is formed as the product of a chemical reaction.
Select the true statement from the answers below:
i. 0.08 moles of magnesium chloride are formed in the reaction
ii. the number of atoms of Cl in the product is approximately 0.6023 ×
1023
iii. the number of atoms of Mg is 0.05
iv. the atomic ratio of Mg atoms to Cl atoms in the product is 1:1
B 2 moles of oxygen gas react with hydrogen. What is the mass of oxygen
in the reactants?
i. 32 g
ii. 0.125 g
iii. 64 g
iv. 0.063 g
C In the compound potassium sulphate (K2 SO4 ), oxygen makes up x% of
the mass of the compound. x = ...
i. 36.8
ii. 9.2
iii. 4
iv. 18.3
D The molarity of a 150 cm3 solution, containing 5 g of NaCl is...
i. 0.09 M
ii. 5.7 × 10−4 M
iii. 0.57 M
iv. 0.03 M
3. 300 cm3 of a 0.1 mol.dm−3 solution of sulfuric acid is added to 200 cm3 of a
0.5 mol.dm−3 solution of sodium hydroxide.
a Write down a balanced equation for the reaction which takes place when
these two solutions are mixed.
b Calculate the number of moles of sulfuric acid which were added to the
sodium hydroxide solution.
c Is the number of moles of sulfuric acid enough to fully neutralise the sodium
hydroxide solution? Support your answer by showing all relevant calculations.
(IEB Paper 2 2004)
4. Ozone (O3 ) reacts with nitrogen monoxide gas (NO) to produce NO2 gas. The
NO gas forms largely as a result of emissions from the exhausts of motor vehicles
and from certain jet planes. The NO2 gas also causes the brown smog (smoke
and fog), which is seen over most urban areas. This gas is also harmful to
humans, as it causes breathing (respiratory) problems. The following equation
indicates the reaction between ozone and nitrogen monoxide:
253
13.9
CHAPTER 13. QUANTITATIVE ASPECTS OF CHEMICAL CHANGE - GRADE 11
O3 (g) + N O(g) → O2 (g) + N O2 (g)
In one such reaction 0.74 g of O3 reacts with 0.67 g NO.
a Calculate the number of moles of O3 and of NO present at the start of the
reaction.
b Identify the limiting reagent in the reaction and justify your answer.
c Calculate the mass of NO2 produced from the reaction.
(DoE Exemplar Paper 2, 2007)
5. A learner is asked to make 200 cm3 of sodium hydroxide (NaOH) solution of
concentration 0.5 mol.dm−3 .
a Determine the mass of sodium hydroxide pellets he needs to use to do this.
b Using an accurate balance the learner accurately measures the correct mass
of the NaOH pellets. To the pellets he now adds exactly 200 cm3 of pure
water. Will his solution have the correct concentration? Explain your
answer.
300 cm3 of a 0.1 mol.dm−3 solution of sulfuric acid (H2 SO4 ) is added to
200 cm3 of a 0.5 mol.dm−3 solution of NaOH at 250 C.
c Write down a balanced equation for the reaction which takes place when
these two solutions are mixed.
d Calculate the number of moles of H2 SO4 which were added to the NaOH
solution.
e Is the number of moles of H2 SO4 calculated in the previous question
enough to fully neutralise the NaOH solution? Support your answer by
showing all the relevant calculations.
(IEB Paper 2, 2004)
254
Chapter 14
Energy Changes In Chemical
Reactions - Grade 11
All chemical reactions involve energy changes. In some reactions, we are able to see these energy
changes by either an increase or a decrease in the overall energy of the system.
14.1
What causes the energy changes in chemical reactions?
When a chemical reaction occurs, bonds in the reactants break, while new bonds form in the
product. The following example may help to explain this.
Hydrogen reacts with oxygen to form water, according to the following equation:
2H2 + O2 → 2H2 O
In this reaction, the bond between the two hydrogen atoms in the H2 molecule will break, as will
the bond between the oxygen atoms in the O2 molecule. New bonds will form between the two
hydrogen atoms and the single oxygen atom in the water molecule that is formed as the product.
For bonds to break, energy must be absorbed. When new bonds form, energy is released. The
energy that is needed to break a bond is called the bond energy or bond dissociation energy.
Bond energies are measured in units of kJ.mol−1 .
Definition: Bond energy
Bond energy is a measure of bond strength in a chemical bond. It is the amount of energy
(in kJ.mol−1 ) that is needed to break the chemical bond between two atoms.
14.2
Exothermic and endothermic reactions
In some reactions, the energy that must be absorbed to break the bonds in the reactants, is less
than the total energy that is released when new bonds are formed. This means that in the overall
reaction, energy is released as either heat or light. This type of reaction is called an exothermic
reaction. Another way of describing an exothermic reaction is that it is one in which the energy
of the product is less than the energy of the reactants, because energy has been released during
the reaction. We can represent this using the following general formula:
Reactants → Product + Energy
255
14.2
CHAPTER 14. ENERGY CHANGES IN CHEMICAL REACTIONS - GRADE 11
Definition: Exothermic reaction
An exothermic reaction is one that releases energy in the form of heat or light.
In other reactions,the energy that must be absorbed to break the bonds in the reactants, is more
than the total energy that is released when new bonds are formed. This means that in the overall
reaction, energy must be absorbed from the surroundings. This type of reaction is known as an
endothermic reaction. Another way of describing an endothermic reaction is that it is one in
which the energy of the product is greater than the energy of the reactants, because energy has
been absorbed during the reaction. This can be represented by the following formula:
Reactants + Energy → Product
Definition: Endothermic reaction
An endothermic reaction is one that absorbs energy in the form of heat.
The difference in energy (E) between the reactants and the products is known as the heat of
the reaction. It is also sometimes referred to as the enthalpy change of the system.
Activity :: Demonstration : Endothermic and exothermic reactions 1
Apparatus and materials:
You will need citric acid, sodium bicarbonate, a glass beaker, the lid of an icecream container, thermometer, glass stirring rod and a pair of scissors. Note that
citric acid is found in citrus fruits such as lemons. Sodium bicarbonate is actually
bicarbonate of soda (baking soda), the baking ingredient that helps cakes to rise.
Method:
1. Cut a piece of plastic from the ice-cream container lid that will be big enough
to cover the top of the beaker. Cut a small hole in the centre of this piece of
plastic and place the thermometer through it.
2. Pour some citric acid (H3 C6 H5 O7 ) into the glass beaker, cover the beaker with
its ’lid’ and record the temperature of the solution.
3. Stir in the sodium bicarbonate (NaHCO3 ), then cover the beaker again.
4. Immediately record the temperature, and then take a temperature reading every
two minutes after that. Record your results in a table like the one below.
Time (mins)
Temperature (0 C)
0
2
4
6
The equation for the reaction that takes place is:
H3 C6 H5 O7 (aq) + 3N aHCO3 (s) → 3CO2 (g) + 3H2 O(l) + N aC6 H5 O7 (aq)
Results:
• Plot your temperature results on a graph of temperature against time. What
happens to the temperature during this reaction?
• Is this an exothermic or an endothermic reaction?
• Why was it important to keep the beaker covered with a lid?
• Do you think a glass beaker is the best thing to use for this experiment? Explain
your answer.
• Suggest another container that could have been used and give reasons for your
choice. It might help you to look back to chapter ?? for some ideas!
256
CHAPTER 14. ENERGY CHANGES IN CHEMICAL REACTIONS - GRADE 11
14.3
Activity :: Demonstration : Endothermic and exothermic reactions 2
Apparatus and materials:
Vinegar, steel wool, thermometer, glass beaker and plastic lid (from previous
demonstration).
Method:
1. Put the thermometer through the plastic lid, cover the beaker and record the
temperature in the empty beaker. You will need to leave the thermometer in
the beaker for about 5 minutes in order to get an accurate reading.
2. Take the thermometer out of the jar.
3. Soak a piece of steel wool in vinegar for about a minute. The vinegar removes
the protective coating from the steel wool so that the metal is exposed to
oxygen.
4. After the steel wool has been in the vinegar, remove it and squeeze out any
vinegar that is still on the wool. Wrap the steel wool around the thermometer
and place it (still wrapped round the thermometer) back into the jar. The jar
is automatically sealed when you do this because the thermometer is through
the top of the lid.
5. Leave the steel wool in the beaker for about 5 minutes and then record the
temperature. Record your observations.
Results:
You should notice that the temperature increases when the steel wool is wrapped
around the thermometer.
Conclusion:
The reaction between oxygen and the exposed metal in the steel wool, is exothermic, which means that energy is released and the temperature increases.
14.3
The heat of reaction
The heat of the reaction is represented by the symbol ∆H, where:
∆H = Eprod − Ereact
• In an exothermic reaction, ∆H is less than zero because the energy of the reactants is
greater than the energy of the product. For example,
H2 + Cl2 → 2HCl ∆H = -183 kJ
• In an endothermic reaction, ∆H is greater than zero because the energy of the reactants
is less than the energy of the product. For example,
C + H2 O → CO + H2 ∆H = +131 kJ
Some of the information relating to exothermic and endothermic reactions is summarised in table
14.1.
257
14.3
CHAPTER 14. ENERGY CHANGES IN CHEMICAL REACTIONS - GRADE 11
Table 14.1: A comparison of exothermic and endothermic reactions
Type of reaction
Exothermic
Endothermic
Energy absorbed or re- Released
Absorbed
leased
Relative energy of reac- Energy of reactants greater Energy of reactants less
tants and products
than energy of product
than energy of product
Sign of ∆H
Negative
Positive
Definition: Enthalpy
Enthalpy is the heat content of a chemical system, and is given the symbol ’H’.
Important: Writing equations using ∆H
There are two ways to write the heat of the reaction in an equation
For the exothermic reaction C(s) + O2 (g) → CO2 (g), we can write:
C(s) + O2 (g) → CO2 (g) ∆H = -393 kJ.mol−1 or
C(s) + O2 (g) → CO2 (g) + 393 kJ.mol−1
For the endothermic reaction, C(s) + H2 O(g) → H2 (g) + CO(g), we can write:
C(s) + H2 O(g) → H2 (g) + CO(g) ∆H = +131 kJ.mol−1 or
C(s) + H2 O(g) + 131 kJ.mol−1 → CO + H2
The units for ∆H are kJ.mol−1 . In other words, the ∆H value gives the amount of
energy that is absorbed or released per mole of product that is formed. Units can
also be written as kJ, which then gives the total amount of energy that is released or
absorbed when the product forms.
Activity :: Investigation : Endothermic and exothermic reactions
Apparatus and materials:
Approximately 2 g each of calcium chloride (CaCl2 ), sodium hydroxide (NaOH),
potassium nitrate (KNO3 ) and barium chloride (BaCl2 ); concentrated sulfuric acid
(H2 SO4 ); 5 test tubes; thermometer.
Method:
1. Dissolve about 1 g of each of the following substances in 5-10 cm3 of water in
a test tube: CaCl2 , NaOH, KNO3 and BaCl2 .
2. Observe whether the reaction is endothermic or exothermic, either by feeling
whether the side of the test tube gets hot or cold, or using a thermometer.
3. Dilute 3 cm3 of concentrated H2 SO4 in 10 cm3 of water in the fifth test tube
and observe whether the temperature changes.
4. Wait a few minutes and then add NaOH to the H2 SO4 . Observe any energy
changes.
5. Record which of the above reactions are endothermic and which are exothermic.
Results:
• When BaCl2 and KNO3 dissolve in water, they take in heat from the surroundings. The dissolution of these salts is endothermic.
• When CaCl2 and NaOH dissolve in water, heat is released. The process is
exothermic.
258
CHAPTER 14. ENERGY CHANGES IN CHEMICAL REACTIONS - GRADE 11
14.4
• The reaction of H2 SO4 and NaOH is also exothermic.
14.4
Examples of endothermic and exothermic reactions
There are many examples of endothermic and exothermic reactions that occur around us all the
time. The following are just a few examples.
1. Endothermic reactions
• Photosynthesis
Photosynthesis is the chemical reaction that takes place in plants, which uses energy
from the sun to change carbon dioxide and water into food that the plant needs to
survive, and which other organisms (such as humans and other animals) can eat so
that they too can survive. The equation for this reaction is:
6CO2 + 12H2 O + energy → C6 H12 O6 + 6O2 + 6H2 O
Photosynthesis is an endothermic reaction because it will not happen without an external source of energy, which in this case is sunlight.
• The thermal decomposition of limestone
In industry, the breakdown of limestone into quicklime and carbon dioxide is very
important. Quicklime can be used to make steel from iron and also to neutralise soils
that are too acid. However, the limestone must be heated in a kiln at a temperature
of over 9000 C before the decomposition reaction will take place. The equation for
the reaction is shown below:
CaCO3 → CaO + CO2
2. Exothermic reactions
• Combustion reactions - The burning of fuel is an example of a combustion reaction, and we as humans rely heavily on this process for our energy requirements.
The following equations describe the combustion of a hydrocarbon such as methane
(CH4 ):
Fuel + Oxygen → Heat + Water + CarbonDioxide
CH4 + 2O2 → Heat + H2 O + CO2
This is why we burn fuels for energy, because the chemical changes that take place
during the reaction release huge amounts of energy, which we then use for things like
power and electricity. You should also note that carbon dioxide is produced during
this reaction. Later we will discuss some of the negative impacts of CO2 on the
environment. The chemical reaction that takes place when fuels burn therefore has
both positive and negative consequences.
• Respiration
Respiration is the chemical reaction that happens in our bodies to produce energy for
our cells. The equation below describes what happens during this reaction:
C6 H12 O6 + 6O2 → 6CO2 + 6H2 O + energy
In the reaction above, glucose (a type of carbohydrate in the food we eat) reacts with
oxygen from the air that we breathe in, to form carbon dioxide (which we breathe
out), water and energy. The energy that is produced allows the cell to carry out its
functions efficiently. Can you see now why you are always told that you must eat
food to get energy? It is not the food itself that provides you with energy, but the
exothermic reaction that takes place when compounds within the food react with the
oxygen you have breathed in!
259
14.5
CHAPTER 14. ENERGY CHANGES IN CHEMICAL REACTIONS - GRADE 11
teresting Lightsticks or glowsticks are used by divers, campers, and for decoration and
Interesting
Fact
Fact
fun. A lightstick is a plastic tube with a glass vial inside it. To activate a
lightstick, you bend the plastic stick, which breaks the glass vial. This allows
the chemicals that are inside the glass to mix with the chemicals in the plastic
tube. These two chemicals react and release energy. Another part of a lightstick
is a fluorescent dye which changes this energy into light, causing the lightstick
to glow!
Exercise: Endothermic and exothermic reactions
1. In each of the following reactions, say whether the reaction is endothermic or
exothermic, and give a reason for your answer.
(a) H2 + I2 → 2HI + 21kJ
(b) CH4 + 2O2 → CO2 + 2H2 O ∆ H = -802 kJ
(c) The following reaction takes place in a flask:
Ba(OH)2 .8H2 O + 2N H4 N O3 → Ba(N O3 )2 + 2N H3 + 10H2 O
Within a few minutes, the temperature of the flask drops by approximately 20C.
(d) N a + Cl2 → 2N aCl ∆H = -411 kJ
(e) C + O2 → CO2
2. For each of the following descriptions, say whether the process is endothermic
or exothermic and give a reason for your answer.
(a) evaporation
(b) the combustion reaction in a car engine
(c) bomb explosions
(d) melting ice
(e) digestion of food
(f) condensation
14.5
Spontaneous and non-spontaneous reactions
Activity :: Demonstration : Spontaneous and non-spontaneous reactions
Apparatus and materials:
A length of magnesium ribbon, thick copper wire and a bunsen burner
260
CHAPTER 14. ENERGY CHANGES IN CHEMICAL REACTIONS - GRADE 11
14.6
magnesium
ribbon
Method:
1. Scrape the length of magnesium ribbon and copper wire clean.
2. Heat each piece of metal over the bunsen burner, in a non-luminous flame.
Observe whether any chemical reaction takes place.
3. Remove the metals from the flame and observe whether the reaction stops. If
the reaction stops, return the metal to the bunsen flame and continue to heat
it.
Results:
• Did any reaction take place before the metals were heated?
• Did either of the reactions continue after they were removed from the flame?
• Write a balanced equation for each of the chemical reactions that takes place.
In the demonstration above, the reaction between magnesium and oxygen, and the reaction
between copper and oxygen are both non-spontaneous. Before the metals were held over the
bunsen burner, no reaction was observed. They need energy to initiate the reaction. After
the reaction has started, it may then carry on spontaneously. This is what happened when the
magnesium reacted with oxygen. Even after the magnesium was removed from the flame, the
reaction continued. Other reactions will not carry on unless there is a constant addition of energy. This was the case when copper reacted with oxygen. As soon as the copper was removed
from the flame, the reaction stopped.
Now try adding a solution of dilute sulfuric acid with a solution of sodium hydroxide. What do
you observe? This is an example of a spontaneous reaction because the reaction takes place
without any energy being added.
Definition: Spontaneous reaction
A spontaneous reaction is a physical or chemical change that occurs without the addition
of energy.
14.6
Activation energy and the activated complex
From the demonstrations of spontaneous and non-spontaneous reactions, it should be clear that
most reactions will not take place until the system has some minimum amount of energy added
261
14.6
CHAPTER 14. ENERGY CHANGES IN CHEMICAL REACTIONS - GRADE 11
to it. This energy is called the activation energy. Activation energy is the ’threshold energy’
or the energy that must be overcome in order for a chemical reaction to occur.
Definition: Activation energy
Activation energy or ’threshold energy’ is the energy that must be overcome in order for a
chemical reaction to occur.
It is possible to draw an energy diagram to show the energy changes that take place during a
particular reaction. Let’s consider an example:
H2 (g) + F2 (g) → 2HF (g)
[H2 F2 ] (activated complex)
Potential energy
activation
energy
H2 + F2
reactants
∆H = −268
k.J.mol−1
2HF
products
Time
Figure 14.1: The energy changes that take place during an exothermic reaction
The reaction between H2 (g) and F2 (g) (figure 14.1) needs energy in order to proceed, and this is
the activation energy. Once the reaction has started, an in-between, temporary state is reached
where the two reactants combine to give H2 F2 . This state is sometimes called a transition
state and the energy that is needed to reach this state is equal to the activation energy for the
reaction. The compound that is formed in this transition state is called the activated complex.
The transition state lasts for only a very short time, after which either the original bonds reform,
or the bonds are broken and a new product forms. In this example, the final product is HF and
it has a lower energy than the reactants. The reaction is exothermic and ∆H is negative.
Definition: Activated complex
The activated complex is a transitional structure in a chemical reaction that results from the
effective collisions between reactant molecules, and which remains while old bonds break
and new bonds form.
In endothermic reactions, the final products have a higher energy than the reactants. An energy
diagram is shown below (figure 14.2) for the endothermic reaction XY + Z → X + Y Z. In this
example, the activated complex has the formula XYZ. Notice that the activation energy for the
endothermic reaction is much greater than for the exothermic reaction.
teresting The reaction between H and F was considered by NASA (National Aeronautics
Interesting
Fact
Fact
and Space Administration) as a fuel system for rocket boosters because of the
energy that is released during this exothermic reaction.
262
CHAPTER 14. ENERGY CHANGES IN CHEMICAL REACTIONS - GRADE 11
14.6
Potential energy
[XYZ]
activation
energy
X + YZ
products
∆H > 0
XY + Z
reactants
Time
Figure 14.2: The energy changes that take place during an endothermic reaction
Important: Enzymes and activation energy
An enzyme is a catalyst that helps to speed up the rate of a reaction by lowering the
activation energy of a reaction. There are many enzymes in the human body, without which
lots of important reactions would never take place. Cellular respiration is one example of a
reaction that is catalysed by enzymes. You will learn more about catalysts in chapter ??.
Exercise: Energy and reactions
1. Carbon reacts with water according to the following equation:
C + H2 O ⇔ CO + H2 ∆H > 0
(a) Is this reaction endothermic or exothermic?
(b) Give a reason for your answer.
2. Refer to the graph below and then answer the questions that follow:
Potential energy (kJ)
25
0
-15
Time
263
14.7
CHAPTER 14. ENERGY CHANGES IN CHEMICAL REACTIONS - GRADE 11
(a)
(b)
(c)
(d)
14.7
What is the energy of the reactants?
What is the energy of the products?
Calculate ∆H.
What is the activation energy for this reaction?
Summary
• When a reaction occurs, some bonds break and new bonds form. These changes involve
energy.
• When bonds break, energy is absorbed and when new bonds form, energy is released.
• The bond energy is the amount of energy that is needed to break the chemical bond
between two atoms.
• If the energy that is needed to break the bonds is greater than the energy that is released
when new bonds form, then the reaction is endothermic. The energy of the product is
greater than the energy of the reactants.
• If the energy that is needed to break the bonds is less than the energy that is released
when new bonds form, then the reaction is exothermic. The energy of the product is less
than the energy of the reactants.
• An endothermic reaction is one that absorbs energy in the form of heat, while an exothermic reaction is one that releases energy in the form of heat and light.
• The difference in energy between the reactants and the product is called the heat of
reaction and has the symbol ∆H.
• In an endothermic reaction, ∆H is a positive number, and in an exothermic reaction, ∆H
will be negative.
• Photosynthesis, evaporation and the thermal decomposition of limestone, are all examples
of endothermic reactions.
• Combustion reactions and respiration are both examples of exothermic reactions.
• A reaction which proceeds without additional energy being added, is called a spontaneous
reaction.
• Reactions where energy must be supplied for the activation energy to be overcome, are
called non-spontaneous reactions.
• In any reaction, some minimum energy must be overcome before the reaction will proceed.
This is called the activation energy of the reaction.
• The activated complex is the transitional product that is formed during a chemical
reaction while old bonds break and new bonds form.
Exercise: Summary Exercise
1. For each of the following, say whether the statement is true or false. If it is
false, give a reason for your answer.
(a) Energy is released in all chemical reactions.
(b) The condensation of water vapour is an example of an endothermic reaction.
264
CHAPTER 14. ENERGY CHANGES IN CHEMICAL REACTIONS - GRADE 11
14.7
(c)
(d)
2. For
(a)
In an exothermic reaction ∆H is less than zero.
All non-spontaneous reactions are endothermic.
each of the following, choose the one correct answer.
For the following reaction:
A + B ⇔ AB ∆H = -129 kJ.mol−1
i. The energy of the reactants is less than the energy of the product.
ii. The energy of the product is less than the energy of the reactants.
iii. The reaction is non-spontaneous.
iv. The overall energy of the system increases during the reaction.
(b) Consider the following chemical equilibrium:
2NO2 ⇔ N2 O4
Which one of the following graphs best represents the changes in potential
energy that take place during the production of N2 O4 ?
(i)
(ii)
(iii)
3. The cellular respiration reaction is catalysed by enzymes. The equation for the
reaction is:
C6 H12 O6 + 6O2 → 6CO2 + 6H2 O
The change in potential energy during this reaction is shown below:
Potential energy
activation
energy
C6 H12 O6 + 6O2
∆H
6CO2 + 6H2 O
Time
(a) Will the value of ∆H be positive or negative? Give a reason for your
answer.
(b) Explain what is meant by ’activation energy’.
(c) What role do enzymes play in this reaction?
(d) Glucose is one of the reactants in cellular respiration. What important
chemical reaction produces glucose?
(e) Is the reaction in your answer above an endothermic or an exothermic one?
Explain your answer.
(f) Explain why proper nutrition and regular exercise are important in maintaining a healthy body.
265
(iv)
14.7
CHAPTER 14. ENERGY CHANGES IN CHEMICAL REACTIONS - GRADE 11
266
Chapter 15
Types of Reactions - Grade 11
There are many different types of chemical reactions that can take place. In this chapter, we will
be looking at a few of the more common reaction types: acid-base and acid-carbonate reactions,
redox reactions and addition, elimination and substitution reactions.
15.1
Acid-base reactions
15.1.1
What are acids and bases?
In our daily lives, we encounter many examples of acids and bases. In the home, vinegar (acetic
acid), lemon juice (citric acid) and tartaric acid (the main acid found in wine) are common, while
hydrochloric acid, sulfuric acid and nitric acid are examples of acids that are more likely to be
found in laboratories and industry. Hydrochloric acid is also found in the gastric juices in the
stomach. Even fizzy drinks contain acid (carbonic acid), as do tea and wine (tannic acid)! Bases
that you may have heard of include sodium hydroxide (caustic soda), ammonium hydroxide and
ammonia. Some of these are found in household cleaning products. Acids and bases are also
important commercial products in the fertiliser, plastics and petroleum refining industries. Some
common acids and bases, and their chemical formulae, are shown in table 15.1.
Table 15.1: Some common acids and
Acid
Formula
Hydrochoric acid
HCl
Sulfuric acid
H2 SO4
Nitric acid
HNO3
Acetic (ethanoic) acid CH3 COOH
Carbonic acid
H2 CO3
Sulfurous acid
H2 SO3
Phosphoric acid
H3 PO4
bases and their chemical
Base
Sodium hydroxide
Potassium hydroxide
Sodium carbonate
Calcium hydroxide
Magnesium hydroxide
Ammonia
Sodium bicarbonate
formulae
Formula
NaOH
KOH
Na2 CO3
Ca(OH)2
Mg(OH)2
NH3
NaHCO3
Most acids share certain characteristics, and most bases also share similar characteristics. It
is important to be able to have a definition for acids and bases so that they can be correctly
identified in reactions.
15.1.2
Defining acids and bases
A number of definitions for acids and bases have developed over the years. One of the earliest
was the Arrhenius definition. Arrhenius (1887) noticed that water dissociates (splits up) into
hydronium (H3 O+ ) and hydroxide (OH− ) ions according to the following equation:
H2 O ⇔ H3 O+ + OH−
267
15.1
CHAPTER 15. TYPES OF REACTIONS - GRADE 11
Arrhenius described an acid as a compound that increases the concentration of H3 O+ ions
in solution, and a base as a compound that increases the concentration of OH− ions in a
solution. Look at the following examples showing the dissociation of hydrochloric acid and
sodium hydroxide (a base) respectively:
1. HCl + H2 O → H3 O+ + Cl−
Hydrochloric acid in water increases the concentration of H3 O+ ions and is therefore an
acid.
2. NaOH + H2 O → Na+ + OH−
Sodium hydroxide in water increases the concentration of OH− ions and is therefore a
base.
However, this definition could only be used for acids and bases in water. It was important to
come up with a much broader definition for acids and bases.
It was Lowry and Bronsted (1923) who took the work of Arrhenius further to develop a broader
definition for acids and bases. The Bronsted-Lowry model defines acids and bases in terms of
their ability to donate or accept protons.
Definition: Acids and bases
According to the Bronsted-Lowry theory of acids and bases, an acid is a substance that
gives away protons (H+ ), and is therefore called a proton donor. A base is a substance
that takes up protons, and is therefore called a proton acceptor.
Below are some examples:
1. HCl(g) + NH3 (g) → NH4+ + Cl−
In order to decide which substance is a proton donor and which is a proton acceptor, we
need to look at what happens to each reactant. The reaction can be broken down as
follows:
HCl → Cl− + H+ and
NH3 + H+ → NH+
4
From these reactions, it is clear that HCl is a proton donor and is therefore an acid, and
that NH3 is a proton acceptor and is therefore a base.
2. CH3 COOH + H2 O → H3 O+ + CH3 COO−
The reaction can be broken down as follows:
CH3 COOH → CH3 COO− + H+ and
H2 O + H+ → H3 O+
In this reaction, CH3 COOH (acetic acid) is a proton donor and is therefore the acid. In
this case, water acts as a base because it accepts a proton to form H3 O+ .
−
3. NH3 + H2 O → NH+
4 + OH
The reaction can be broken down as follows:
268
For
more
information
on dissociation, refer to
chapter 20.
CHAPTER 15. TYPES OF REACTIONS - GRADE 11
15.1
H2 O → OH− + H+ and
NH3 + H+ → NH+
4
In this reaction, water donates a proton and is therefore an acid in this reaction. Ammonia
accepts the proton and is therefore the base. Notice that in the previous equation, water
acted as a base and that in this equation it acts as an acid. Water can act as both an
acid and a base depending on the reaction. This is also true of other substances. These
substances are called ampholytes and are said to be amphoteric.
Definition: Amphoteric
An amphoteric substance is one that can react as either an acid or base. Examples of
amphoteric substances include water, zinc oxide and beryllium hydroxide.
15.1.3
Conjugate acid-base pairs
Look at the reaction between hydrochloric acid and ammonia to form ammonium and chloride
ions:
−
HCl + NH3 ⇔ NH+
4 + Cl
Looking firstly at the forward reaction (i.e. the reaction that proceeds from left to right), the
changes that take place can be shown as follows:
HCl → Cl− + H+ and
NH3 + H+ → NH+
4
Looking at the reverse reaction (i.e. the reaction that proceeds from right to left), the changes
that take place are as follows:
+
NH+
4 → NH3 + H and
Cl− + H+ → HCl
In the forward reaction, HCl is a proton donor (acid) and NH3 is a proton acceptor (base).
In the reverse reaction, the chloride ion is the proton acceptor (base) and NH+
4 is the proton
donor (acid). A conjugate acid-base pair is two compounds in a reaction that change into
each other through the loss or gain of a proton. The conjugate acid-base pairs for the above
reaction are shown below.
conjugate pair
HCl + NH3
acid1 base2
−
NH+
4 + Cl
acid2 base1
conjugate pair
The reaction between ammonia and water can also be used as an example:
269
15.1
CHAPTER 15. TYPES OF REACTIONS - GRADE 11
conjugate pair
−
NH+
4 + OH
acid2 base1
H2 O + NH3
acid1 base2
conjugate pair
Definition: Conjugate acid-base pair
The term refers to two compounds that transform into each other by the gain or loss of a
proton.
Exercise: Acids and bases
1. In the following reactions, identify (1) the acid and the base in the reactants
and (2) the salt in the product.
(a)
(b)
(c)
(d)
H2 SO4 + Ca(OH)2 → CaSO4 + 2H2 O
CuO + H2 SO4 → CuSO4 + H2 O
H2 O + C6 H5 OH → H3 O+ + C6 H5 O−
HBr + C5 H5 N → (C5 H5 NH+ )Br−
2. In each of the following reactions, label the conjugate acid-base pairs.
(a)
(b)
(c)
(d)
15.1.4
H2 SO4 + H2 O ⇔ H3 O+ + HSO−
4
−
NH+
+
F
⇔
HF
+
NH
3
4
H2 O + CH3 COO− ⇔ CH3 COOH + OH−
H2 SO4 + Cl− ⇔ HCl + HSO−
4
Acid-base reactions
When an acid and a base react, they neutralise each other to form a salt. If the base contains
hydroxide (OH− ) ions, then water will also be formed. The word salt is a general term which
applies to the products of all acid-base reactions. A salt is a product that is made up of the
cation from a base and the anion from an acid. When an acid reacts with a base, they neutralise
each other. In other words, the acid becomes less acidic and the base becomes less basic. Look
at the following examples:
1. Hydrochloric acid reacts with sodium hydroxide to form sodium chloride (the salt) and
water. Sodium chloride is made up of Na+ cations from the base (NaOH) and Cl− anions
from the acid (HCl).
HCl + NaOH → H2 O + NaCl
2. Hydrogen bromide reacts with potassium hydroxide to form potassium bromide (the salt)
and water. Potassium bromide is made up of K+ cations from the base (KOH) and Br−
anions from the acid (HBr).
HBr + KOH → H2 O + KBr
270
CHAPTER 15. TYPES OF REACTIONS - GRADE 11
15.1
3. Hydrochloric acid reacts with sodium hydrocarbonate to form sodium chloride (the salt)
and hydrogen carbonate. Sodium chloride is made up of Na+ cations from the base
(NaHCO3 ) and Cl− anions from the acid (HCl).
HCl + NaHCO3 → H2 CO3 + NaCl
You should notice that in the first two examples, the base contained OH− ions, and therefore the
products were a salt and water. NaCl (table salt) and KBr are both salts. In the third example,
NaHCO3 also acts as a base, despite not having OH− ions. A salt is still formed as one of the
products, but no water is produced.
It is important to realise how important these neutralisation reactions are. Below are some
examples:
• Domestic uses
Calcium oxide (CaO) is put on soil that is too acid. Powdered limestone (CaCO3 ) can
also be used but its action is much slower and less effective. These substances can also
be used on a larger scale in farming and also in rivers.
• Biological uses
Acids in the stomach (e.g. hydrochloric acid) play an important role in helping to digest
food. However, when a person has a stomach ulcer, or when there is too much acid in the
stomach, these acids can cause a lot of pain. Antacids are taken to neutralise the acids so
that they don’t burn as much. Antacids are bases which neutralise the acid. Examples of
antacids are aluminium hydroxide, magnesium hydroxide (’milk of magnesia’) and sodium
bicarbonate (’bicarbonate of soda’). Antacids can also be used to relieve heartburn.
• Industrial uses
Alkaline calcium hydroxide (limewater) can be used to absorb harmful SO2 gas that is
released from power stations and from the burning of fossil fuels.
teresting Bee stings are acidic and have a pH between 5 and 5.5. They can be soothed
Interesting
Fact
Fact
by using substances such as calomine lotion, which is a mild alkali based on zinc
oxide. Bicarbonate of soda can also be used. Both alkalis help to neutralise the
acidic bee sting and relieve some of the itchiness!
271
15.1
CHAPTER 15. TYPES OF REACTIONS - GRADE 11
Important: Acid-base titrations
The neutralisation reaction between an acid and a base can be very useful. If an acidic
solution of known concentration (a standard solution) is added to an alkaline solution until
the solution is exactly neutralised (i.e. it has neither acidic nor basic properties), it is
possible to calculate the exact concentration of the unknown solution. It is possible to do
this because, at the exact point where the solution is neutralised, chemically equivalent
amounts of acid and base have reacted with each other. This type of calculation is called
volumetric analysis. The process where an acid solution and a basic solution are added
to each other for this purpose, is called a titration, and the point of neutralisation is called
the end point of the reaction. So how exactly can a titration be carried out to determine
an unknown concentration? Look at the following steps to help you to understand the
process.
Step 1:
A measured volume of the solution with unknown concentration is put into a flask.
Step 2:
A suitable indicator is added to this solution (bromothymol blue and phenolpthalein are
common indicators).
Step 3:
A volume of the standard solution is put into a burette and is slowly added to the solution
in the flask, drop by drop.
Step 4:
At some point, adding one more drop will change the colour of the unknown solution.
For example, if the solution is basic and bromothymol blue is being used as the indicator in the titration, the bromothymol blue would originally have coloured the solution
blue. At the end point of the reaction, adding one more drop of acid will change the
colour of the basic solution from blue to yellow. Yellow shows that the solution is now acidic.
Step 5:
Record the volume of standard solution that has been added up to this point.
Step 6:
Use the information you have gathered to calculate the exact concentration of the unknown
solution. A worked example is shown below.
Important: Titration calculations
When you are busy with these calculations, you will need to remember the following:
1dm3 = 1 litre = 1000ml = 1000cm3, therefore dividing cm3 by 1000 will give you an answer in
dm3 .
Some other terms and equations which will be useful to remember are shown below:
• Molarity is a term used to describe the concentration of a solution, and is measured in
mol.dm−3 . The symbol for molarity is M. Refer to chapter 13 for more information on
molarity.
• Moles = molarity (mol.dm−3 ) x volume (dm3 )
• Molarity (mol.dm−3 ) = mol / volume
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CHAPTER 15. TYPES OF REACTIONS - GRADE 11
Worked Example 74: Titration calculation
Question: Given the equation:
NaOH + HCl → NaCl + H2 O
25cm3 of a sodium hydroxide solution was pipetted into a conical flask and titrated
with 0.2M hydrochloric acid. Using a suitable indicator, it was found that 15cm3
of acid was needed to neutralise the alkali. Calculate the molarity of the sodium
hydroxide.
Answer
Step 1 : Write down all the information you know about the reaction, and
make sure that the equation is balanced.
NaOH: V = 25 cm3
HCl: V = 15 cm3 ; C = 0.2 M
The equation is already balanced.
Step 2 : Calculate the number of moles of HCl that react according to this
equation.
n
V
Therefore, n(HCl) = M × V (make sure that all the units are correct!)
M=
M = 0.2mol.dm−3
V = 15cm3 = 0.015dm3
Therefore
n(HCl) = 0.2 × 0.015 = 0.003
There are 0.003 moles of HCl that react
Step 3 : Calculate the number of moles of sodium hydroxide in the reaction
Look at the equation for the reaction. For every mole of HCl there is one mole
of NaOH that is involved in the reaction. Therefore, if 0.003 moles of HCl react,
we can conclude that the same quantity of NaOH is needed for the reaction. The
number of moles of NaOH in the reaction is 0.003.
Step 4 : Calculate the molarity of the sodium hydroxide
First convert the volume into dm3 . V = 0.025 dm3 . Then continue with the
calculation.
M=
0.003
n
=
= 0.12
V
0.025
The molarity of the NaOH solution is 0.12 mol.dm3 or 0.12 M
Worked Example 75: Titration calculation
Question: 4.9 g of sulfuric acid is dissolved in water and the final solution has a
volume of 220 cm3 . Using titration, it was found that 20 cm3 of this solution was
273
15.1
15.1
CHAPTER 15. TYPES OF REACTIONS - GRADE 11
able to completely neutralise 10 cm3 of a sodium hydroxide solution. Calculate the
concentration of the sodium hydroxide in mol.dm−3 .
Answer
Step 1 : Write a balanced equation for the titration reaction.
H2 SO4 + 2NaOH → Na2 SO4 + 2H2 O
Step 2 : Calculate the molarity of the sulfuric acid solution.
M = n/V
V = 220 cm3 = 0.22 dm3
n=
4.9g
m
= 0.05mols
=
M
98g.mol−1
Therefore,
M=
0.05
= 0.23mol.dm−3
0.22
Step 3 : Calculate the moles of sulfuric acid that were used in the neutralisation reaction.
Remember that only 20 cm3 of the sulfuric acid solution is used.
M = n/V, therefore n = M × V
n = 0.23 × 0.02 = 0.0046mol
Step 4 : Calculate the number of moles of sodium hydroxide that were
neutralised.
According to the balanced chemical equation, the mole ratio of H2 SO4 to NaOH is
1:2. Therefore, the number of moles of NaOH that are neutralised is 0.0046 × 2 =
0.0092 mols.
Step 5 : Calculate the concentration of the sodium hydroxide solution.
M=
15.1.5
0.0092
n
=
= 0.92M
V
0.01
Acid-carbonate reactions
Activity :: Demonstration : The reaction of acids with carbonates
Apparatus and materials:
Small amounts of sodium carbonate and calcium carbonate (both in powder
form); hydrochloric acid and sulfuric acid; retort stand; two test tubes; two rubber
stoppers for the test tubes; a delivery tube; lime water. The demonstration should
be set up as shown below.
274
CHAPTER 15. TYPES OF REACTIONS - GRADE 11
15.1
delivery tube
rubber stopper
rubber stopper
[glassType=tube,bouchon=true,niveauLiquide1=30]
[glassType=tube,bouchon=true,niveauLiquide1=60]
sodium carbonate &
hydrochloric acid
limewater
Method:
1. Pour limewater into one of the test tubes and seal with a rubber stopper.
2. Pour a small amount of hydrochloric acid into the remaining test tube.
3. Add a small amount of sodium carbonate to the acid and seal the test tube
with the rubber stopper.
4. Connect the two test tubes with a delivery tube.
5. Observe what happens to the colour of the limewater.
6. Repeat the above steps, this time using sulfuric acid and calcium carbonate.
Observations:
The clear lime water turns milky meaning that carbon dioxide has been produced.
When an acid reacts with a carbonate a salt, carbon dioxide and water are formed. Look at the
following examples:
• Nitric acid reacts with sodium carbonate to form sodium nitrate, carbon dioxide and water.
2HNO3 + Na2 CO3 → 2NaNO3 + CO2 + H2 O
• Sulfuric acid reacts with calcium carbonate to form calcium sulphate, carbon dioxide and
water.
H2 SO4 + CaCO3 → CaSO4 + CO2 + H2 O
• Hydrochloric acid reacts with calcium carbonate to form calcium chloride, carbon dioxide
and water.
2HCl + CaCO3 → CaCl2 + CO2 + H2 O
Exercise: Acids and bases
1. The compound NaHCO3 is commonly known as baking soda. A recipe requires
1.6 g of baking soda, mixed with other ingredients, to bake a cake.
275
15.2
CHAPTER 15. TYPES OF REACTIONS - GRADE 11
(a) Calculate the number of moles of NaHCO3 used to bake the cake.
(b) How many atoms of oxygen are there in the 1.6 g of baking soda?
During the baking process, baking soda reacts with an acid to produce
carbon dioxide and water, as shown by the reaction equation below:
+
HCO−
3 (aq) + H (aq) → CO2 (g) + H2 O(l)
(c) Identify the reactant which acts as the Bronsted-Lowry base in this reaction. Give a reason for your answer.
(d) Use the above equation to explain why the cake rises during this baking
process.
(DoE Grade 11 Paper 2, 2007)
2. Label the acid-base conjugate pairs in the following equation:
2−
+
HCO−
3 + H2 O ⇔ CO3 + H3 O
3. A certain antacid tablet contains 22.0 g of baking soda (NaHCO3 ). It is used to
neutralise the excess hydrochloric acid in the stomach. The balanced equation
for the reaction is:
NaHCO3 + HCl → NaCl + H2 O + CO2
The hydrochloric acid in the stomach has a concentration of 1.0 mol.dm−3 .
Calculate the volume of the hydrochloric acid that can be neutralised by the
antacid tablet.
(DoE Grade 11 Paper 2, 2007)
4. A learner is asked to prepare a standard solution of the weak acid, oxalic acid
(COOH)2 2H2 O for use in a titration. The volume of the solution must be 500
cm3 and the concentration 0.2 mol.dm−3 .
(a) Calculate the mass of oxalic acid which the learner has to dissolve to make
up the required standard solution. The leaner titrates this 0.2 mol.dm−3
oxalic acid solution against a solution of sodium hydroxide. He finds that
40 cm3 of the oxalic acid solution exactlt neutralises 35 cm3 of the sodium
hydroxide solution.
(b) Calculate the concentration of the sodium hydroxide solution.
5. A learner finds some sulfuric acid solution in a bottle labelled ’dilute sulfuric
acid’. He wants to determine the concentration of the sulphuric acid solution.
To do this, he decides to titrate the sulphuric acid against a standard potassium
hydroxide (KOH) solution.
(a) What is a standard solution?
(b) Calculate the mass of KOH which he must use to make 300 cm3 of a 0.2
mol.dm−3 KOH solution.
(c) Calculate the pH of the 0.2 mol.dm−3 KOH solution (assume standard
temperature).
(d) Write a balanced chemical equation for the reaction between H2 SO4 and
KOH.
(e) During the titration he finds that 15 cm3 of the KOH solution neutralises
20 cm3 of the H2 SO4 solution. Calculate the concentration of the H2 SO4
solution.
(IEB Paper 2, 2003)
15.2
Redox reactions
A second type of reaction is the redox reaction, in which both oxidation and reduction take
place.
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CHAPTER 15. TYPES OF REACTIONS - GRADE 11
15.2.1
15.2
Oxidation and reduction
If you look back to chapter 4, you will remember that we discussed how, during a chemical
reaction, an exchange of electrons takes place between the elements that are involved. Using
oxidation numbers is one way of tracking what is happening to these electrons in a reaction.
Refer back to section 4.11 if you can’t remember the rules that are used to give an oxidation
number to an element. Below are some examples to refresh your memory before we carry on
with this section!
Examples:
1. CO2
Each oxygen atom has an oxidation number of -2. This means that the charge on two
oxygen atoms is -4. We know that the molecule of CO2 is neutral, therefore the carbon
atom must have an oxidation number of +4.
2. KMnO4
Overall, this molecule has a neutral charge, meaning that the sum of the oxidation numbers of the elements in the molecule must equal zero. Potassium (K) has an oxidation
number of +1, while oxygen (O) has an oxidation number of -2. If we exclude the atom
of manganese (Mn), then the sum of the oxidation numbers equals +1+(-2x4)= -7. The
atom of manganese must therefore have an oxidation number of +7 in order to make the
molecule neutral.
By looking at how the oxidation number of an element changes during a reaction, we can easily
see whether that element is being oxidised or reduced.
Definition: Oxidation and reduction
Oxidation is the loses of an electron by a molecule, atom or ion. Reduction is the gain of
an electron by a molecule, atom or ion.
Example:
Mg + Cl2 → MgCl2
As a reactant, magnesium has an oxidation number of zero, but as part of the product magnesium
chloride, the element has an oxidation number of +2. Magnesium has lost two electrons and
has therefore been oxidised. This can be written as a half-reaction. The half-reaction for this
change is:
Mg → Mg2+ + 2e−
As a reactant, chlorine has an oxidation number of zero, but as part of the product magnesium
chloride, the element has an oxidation number of -1. Each chlorine atom has gained an electron
and the element has therefore been reduced. The half-reaction for this change is:
Cl2 + 2e− → 2Cl−
Definition: Half-reaction
A half reaction is either the oxidation or reduction reaction part of a redox reaction. A
half reaction is obtained by considering the change in oxidation states of the individual
substances that are involved in the redox reaction.
277
15.2
CHAPTER 15. TYPES OF REACTIONS - GRADE 11
Important: Oxidation and reduction made easy!
An easy way to think about oxidation and reduction is to remember:
’OILRIG’ - Oxidation Is Loss of electrons, Reduction Is Gain of electrons.
An element that is oxidised is called a reducing agent, while an element that is reduced is
called an oxidising agent.
15.2.2
Redox reactions
Definition: Redox reaction
A redox reaction is one involving oxidation and reduction, where there is always a change
in the oxidation numbers of the elements involved.
Activity :: Demonstration : Redox reactions
Materials:
A few granules of zinc; 15 ml copper (II) sulphate solution (blue colour), glass
beaker.
zinc granules
copper sulphate
solution
Method:
Add the zinc granules to the copper sulphate solution and observe what happens.
What happens to the zinc granules? What happens to the colour of the solution?
Results:
• Zinc becomes covered in a layer that looks like copper.
• The blue copper sulphate solution becomes clearer.
Cu2+ ions from the CuSO4 solution are reduced to form copper metal. This is
what you saw on the zinc crystals. The reduction of the copper ions (in other words,
their removal from the copper sulphate solution), also explains the change in colour
of the solution. The equation for this reaction is:
Cu2+ + 2e− → Cu
Zinc is oxidised to form Zn2+ ions which are clear in the solution. The equation
for this reaction is:
Zn → Zn2+ + 2e−
The overall reaction is:
Cu2+ (aq) + Zn(s) → Cu(s) + Zn2+ (aq)
Conclusion:
A redox reaction has taken place. Cu2+ ions are reduced and the zinc is oxidised.
278
CHAPTER 15. TYPES OF REACTIONS - GRADE 11
15.2
Below are some further examples of redox reactions:
• H2 + F2 → 2HF can be re-written as two half-reactions:
H2 → 2H+ + 2e− (oxidation) and
F2 + 2e− → 2F− (reduction)
• Cl2 + 2KI → 2KCl + I2 or Cl2 + 2I− → 2Cl− + I2 , can be written as two half-reactions:
Cl2 + 2e− → 2Cl− (reduction) and
2I− → I2 + 2e− (oxidation)
In Grade 12, you will go on to look at electrochemical reactions, and the role that electron
transfer plays in this type of reaction.
Exercise: Redox Reactions
1. Look at the following reaction:
2H2 O2 (l) → 2H2 O(l) + O2 (g)
(a) What is the oxidation number of the oxygen atom in each of the following
compounds?
i. H2 O2
ii. H2 O
iii. O2
(b) Does the hydrogen peroxide (H2 O2 ) act as an oxidising agent or a reducing
agent or both, in the above reaction? Give a reason for your answer.
2. Consider the following chemical equations:
1. Fe(s) → Fe2+ (aq) + 2e−
2. 4H+ (aq) + O2 (g) + 4e− → 2H2 O(l)
Which one of the following statements is correct?
(a)
(b)
(c)
(d)
Fe
Fe
Fe
Fe
is
is
is
is
oxidised and H+ is reduced
reduced and O2 is oxidised
oxidised and O2 is reduced
reduced and H+ is oxidised
(DoE Grade 11 Paper 2, 2007)
3. Which one of the following reactions is a redox reaction?
(a)
(b)
(c)
(d)
HCl + NaOH → NaCl + H2 O
AgNO3 + NaI → AgI + NaNO3
2FeCl3 + 2H2 O + SO2 → H2 SO4 + 2HCl + 2FeCl2
BaCl2 + MgSO4 → MgCl2 + BaSO4
279
15.3
CHAPTER 15. TYPES OF REACTIONS - GRADE 11
15.3
Addition, substitution and elimination reactions
15.3.1
Addition reactions
An addition reaction occurs when two or more reactants combine to form a final product. This
product will contain all the atoms that were present in the reactants. The following is a general
equation for this type of reaction:
A+B → C
Notice that C is the final product with no A or B remaining as a residue.
The following are some examples.
1. The reaction between ethene and bromine to form 1,2-dibromoethane (figure 15.1).
C2 H4 + Br2 → C2 H4 Br2
H
H
C
H
+
C
Br
Br
H
H
H
H
C
C
Br
Br
H
Figure 15.1: The reaction between ethene and bromine is an example of an addition reaction
2. Polymerisation reactions
In industry, making polymers is very important. A polymer is made up of lots of smaller
units called monomers. When these monomers are added together, they form a polymer. Examples of polymers are polyvinylchloride (PVC) and polystyrene. PVC is often
used to make piping, while polystyrene is an important packaging and insulating material.
Polystyrene is made up of lots of styrene monomers which are joined through addition
reactions (figure 15.2). ’Polymerisation’ refers to the addition reactions that eventually
help to form the polystyrene polymer.
CH2
CH
CH2
CH
CH2
CH
CH2
CH
polymerisation
etc
Figure 15.2: The polymerisation of a styrene monomer to form a polystyrene polymer
280
CHAPTER 15. TYPES OF REACTIONS - GRADE 11
15.3
3. The hydrogenation of vegetable oils to form margarine is another example of an addition
reaction. Hydrogenation involves adding hydrogen (H2 ) to an alkene. An alkene is an
organic compound composed of carbon and hydrogen. It contains a double bond between
two of the carbon atoms. If this bond is broken, it means that more hydrogen atoms can
attach themselves to the carbon atoms. During hydrogenation, this double bond is broken,
and more hydrogen atoms are added to the molecule. The reaction that takes place is
shown below. Note that the ’R’ represents any side-chain. A side-chain is simply any
combination of atoms that are attached to the central part of the molecule.
RCHCH2 + H2 → RCH2 CH3
4. The production of the alcohol ethanol from ethene. Ethanol (CH3 CH2 OH) can be made
from alkenes such as ethene (C2 H4 ), through a hydration reaction like the one below. A
hydration reaction is one where water is added to the reactants.
C2 H4 + H2 O → CH3 CH2 OH
A catalyst is needed for this reaction to take place. The catalyst that is most commonly
used is phosphoric acid.
15.3.2
Elimination reactions
An elimination reaction occurs when a reactant is broken up into two products. The general
form of the equation is as follows:
A→B+C
The examples below will help to explain this:
1. The dehydration of an alcohol is one example. Two hydrogen atoms and one oxygen
atom are eliminated and a molecule of water is formed as a second product in the reaction,
along with an alkene.
CH3 CH2 OH → CH2 CH2 + H2 O
H
H
H
C
C
H
OH
H
H
H
C
+
C
H
H
H
H
2. The elimination of potassium bromide from a bromoalkane.
CH3 CH2 Br + KOH → CH2 CH2 + KBr + H2 O
H
H
Br
C
C
H
H
H + KOH
281
H
H
C
C + KBr
H
H
O
+ H2 O
H
15.3
CHAPTER 15. TYPES OF REACTIONS - GRADE 11
3. Ethane cracking is an important industrial process used by SASOL and other petrochemical
industries. Hydrogen is eliminated from ethane (C2 H6 ) to produce an alkene called ethene
(C2 H4 ). Ethene is then used to produce other products such as polyethylene. You will
learn more about these compounds in chapter 23. The equation for the cracking of ethane
looks like this:
C2 H6 → C2 H4 + H2
15.3.3
Substitution reactions
A substitution reaction occurs when an exchange of reactants takes place. The initial reactants
are transformed or ’swopped around’ to give a final product. A simple example of a reaction like
this is shown below:
AB + CD → AC + BD
Some simple examples of substitution reactions are shown below:
CH4 + Cl2 → CH3 Cl + HCl
In this example, a chorine atom and a hydrogen atom are exchanged to create a new product.
2−
−
Cu(H2 O)2+
4 + 4Cl ⇔ Cu(Cl)4 + 4H2 O
Exercise: Addition, substitution and elimination reactions
1. Refer to the diagram below and then answer the questions that follow:
H
Cl
H
C
C
(i)
H
H
H + KOH
H
H
C
C + KCl
H
H
+ H2 O
(a) Is this reaction an example of substitution, elimination or addition?
(b) Give a reason for your answer above.
(c) What type of compound is the reactant marked (i)?
2. The following diagram shows the reactants in an addition reaction.
H
H
C
C + HCl
H
H
(a) Draw the final product in this reaction.
(b) What is the chemical formula of the product?
3. The following reaction takes place:
282
CHAPTER 15. TYPES OF REACTIONS - GRADE 11
H
H
OH
C
C
H
H
H
H2 SO4
H
H
C
C + H2 O
H
H
15.4
Is this reaction a substitution, addition or dehydration reaction? Give a reason
for your answer.
4. Consider the following reaction:
Ca(OH)2 (s) + 2NH4 Cl(s) → CaCl2 (s) + 2NH3 (g) + 2H2 O(g)
Which one of the following best describes the type of reaction which takes
place?
(a) Redox reaction
(b) Acid-base reaction
(c) Dehydration reaction
15.4
Summary
• There are many different types of reactions that can take place. These include acid-base,
acid-carbonate, redox, addition, substitution and elimination reactions.
• The Arrhenius definition of acids and bases defines an acid as a substance that increases
the concentration of hydrogen ions (H+ or H3 O+ ) in a solution. A base is a substance that
increases the concentration of hydroxide ions (OH− ) in a solution. However this definition
only applies to substances that are in water.
• The Bronsted-Lowry definition is a much broader one. An acid is a substance that
donates protons and a base is a substance that accepts protons.
• In different reactions, certain substances can act as both an acid and a base. These
substances are called ampholytes and are said to be amphoteric. Water is an example
of an amphoteric substance.
• A conjugate acid-base pair refers to two compounds in a reaction that change into other
through the loss or gain of a proton.
• When an acid and a base react, they form a salt and water. The salt is made up of a cation
from the base and an anion from the acid. An example of a salt is sodium chloride (NaCl),
which is the product of the reaction between sodium hydroxide (NaOH) and hydrochloric
acid (HCl).
• The reaction between an acid and a base is a neutralisation reaction.
• Titrations are reactions between an acid and a base that are used to calculate the concentration of one of the reacting substances. The concentration of the other reacting
substance must be known.
• In an acid-carbonate reaction, an acid and a carbonate react to form a salt, carbon
dioxide and water.
• A redox reaction is one where there is always a change in the oxidation numbers of the
elements that are involved in the reaction.
• Oxidation is the loss of electrons and reduction is the gain of electrons.
283
15.4
CHAPTER 15. TYPES OF REACTIONS - GRADE 11
• When two or more reactants combine to form a product that contains all the atoms
that were in the reactants, then this is an addition reaction. Examples of addition
reactions include the reaction between ethene and bromine, polymerisation reactions and
hydrogenation reactions.
• A reaction where the reactant is broken down into one or more product, is called an elimination reaction. Alcohol dehydration and ethane cracking are examples of elimination
reactions.
• A substitution reaction is one where the reactants are transformed or swopped around
to form the final product.
Exercise: Summary Exercise
1. Give one word/term for each of the following descriptions:
(a) A chemical reaction during which electrons are transferred
(b) The addition of hydrogen across a double bond
(c) The removal of hydrogen and halogen atoms from an alkane to form an
elkene
2. For each of the following, say whether the statement is true or false. If the
statement is false, re-write the statement correctly.
(a) The conjugate base of NH+
4 is NH3 .
(b) The reactions C + O2 → CO2 and 2KClO3 → 2KCl + 3O2 are examples
of redox reactions.
3. For each of the following questions, choose the one correct statement from the
list provided.
A The following chemical equation represents the formation of the hydronium
ion:
H+ (aq) + H2 O(l) → H3 O+ (aq)
In this reaction, water acts as a Lewis base because it...
i. accepts protons
ii. donates protons
iii. accepts electrons
iv. donates electrons
B What is the concentration (in mol.dm−3 ) of H3 O+ ions in a NaOH solution
which has a pH of 12 at 250 C?
i. 1 × 1012
ii. 1 × 102
iii. 1 × 10−2
iv. 1 × 10−12
(IEB Paper 2, 2005)
C When chlorine water is added to a solution of potassium bromide, bromine
is produced. Which one of the following statements concerning this reaction is correct?
i. Br− is oxidised
ii. Cl2 is oxidised
iii. Br− is the oxidising agent
iv. Cl− is the oxidising agent
(IEB Paper 2, 2005)
4. The stomach secretes gastric juice, which contains hydrochloric acid. The
gastric juice helps with digestion. Sometimes there is an overproduction of
acid, leading to heartburn or indigestion. Antacids, such as milk of magnesia,
can be taken to neutralise the excess acid. Milk of magnesia is only slightly
soluble in water and has the chemical formula Mg(OH)2 .
284
CHAPTER 15. TYPES OF REACTIONS - GRADE 11
a Write a balanced chemical equation to show how the antacid reacts with
the acid.
b The directions on the bottle recommend that children under the age of 12
years take one teaspoon of milk of magnesia, whereas adults can take two
teaspoons of the antacid. Briefly explain why the dosages are different.
c Why is it not advisable to take an overdose of the antacid in the stomach?
Refer to the hydrochloric acid concentration in the stomach in your answer.
In an experiment, 25.0 cm3 of a standard solution of sodium carbonate of
concentration 0.1 mol.dm−3 was used to neutralise 35.0 cm3 of a solution
of hydrochloric acid.
d Write a balanced chemical equation for the reaction.
e Calculate the concentration of the acid.
(DoE Grade 11 Exemplar, 2007)
285
15.4
15.4
CHAPTER 15. TYPES OF REACTIONS - GRADE 11
286
Chapter 16
Reaction Rates - Grade 12
16.1
Introduction
Before we begin this section, it might be useful to think about some different types of reactions
and how quickly or slowly they occur.
Exercise: Thinking about reaction rates
Think about each of the following reactions:
• rusting of metals
• photosynthesis
• weathering of rocks (e.g. limestone rocks being weathered by water)
• combustion
1. For each of the reactions above, write a chemical equation for the reaction that
takes place.
2. How fast is each of these reactions? Rank them in order from the fastest to
the slowest.
3. How did you decide which reaction was the fastest and which was the slowest?
4. Try to think of some other examples of chemical reactions. How fast or slow
is each of these reactions, compared with those listed earlier?
In a chemical reaction, the substances that are undergoing the reaction are called the reactants,
while the substances that form as a result of the reaction are called the products. The reaction
rate describes how quickly or slowly the reaction takes place. So how do we know whether a
reaction is slow or fast? One way of knowing is to look either at how quickly the reactants are
used up during the reaction or at how quickly the product forms. For example, iron and sulfur
react according to the following equation:
F e + S → F eS
In this reaction, we can see the speed of the reaction by observing how long it takes before there
is no iron or sulfur left in the reaction vessel. In other words, the reactants have been used up.
Alternatively, one could see how quickly the iron sulfide product forms. Since iron sulfide looks
very different from either of its reactants, this is easy to do.
In another example:
287
16.1
CHAPTER 16. REACTION RATES - GRADE 12
2M g(s) + O2 → 2M gO(s)
In this case, the reaction rate depends on the speed at which the reactants (solid magnesium
and oxygen gas) are used up, or the speed at which the product (magnesium oxide) is formed.
Definition: Reaction rate
The rate of a reaction describes how quickly reactants are used up or how quickly products
are formed during a chemical reaction. The units used are: moles per second (mols/second
or mol.s−1 ).
The average rate of a reaction is expressed as the number of moles of reactant used up, divided
by the total reaction time, or as the number of moles of product formed, divided by the reaction
time. Using the magnesium reaction shown earlier:
Average reaction rate =
moles M g used
reaction time (s)
or
Average reaction rate =
moles O2 used
reaction time (s)
or
Average reaction rate =
moles M gO produced
reaction time (s)
Worked Example 76: Reaction rates
Question: The following reaction takes place:
4Li + O2 → 2Li2 O
After two minutes , 4 g of Lithium has been used up. Calculate the rate of the
reaction.
Answer
Step 1 : Calculate the number of moles of Lithium that are used up in the
reaction.
n=
4
m
=
= 0.58mols
M
6.94
Step 2 : Calculate the time (in seconds) for the reaction.
t = 2 × 60 = 120s
Step 3 : Calculate the rate of the reaction.
Rate of reaction =
0.58
moles of Lithium used
=
= 0.005
time
120
The rate of the reaction is 0.005 mol.s−1
288
CHAPTER 16. REACTION RATES - GRADE 12
16.2
Exercise: Reaction rates
1. A number of different reactions take place. The table below shows the number
of moles of reactant that are used up in a particular time for each reaction.
Reaction
1
2
3
4
5
Mols used up
2
5
1
3.2
5.9
Time
30 secs
2 mins
1.5 mins
1.5 mins
30 secs
Reaction rate
(a) Complete the table by calculating the rate of each reaction.
(b) Which is the fastest reaction?
(c) Which is the slowest reaction?
2. Two reactions occur simultaneously in separate reaction vessels. The reactions
are as follows:
M g + Cl2 → M gCl2
2N a + Cl2 → 2N aCl
After 1 minute, 2 g of MgCl2 have been produced in the first reaction.
(a) How many moles of MgCl2 are produced after 1 minute?
(b) Calculate the rate of the reaction, using the amount of product that is
produced.
(c) Assuming that the second reaction also proceeds at the same rate, calculate...
i. the number of moles of NaCl produced after 1 minute.
ii. the mass (in g) of sodium that is needed for this reaction to take place.
16.2
Factors affecting reaction rates
Several factors affect the rate of a reaction. It is important to know these factors so that reaction
rates can be controlled. This is particularly important when it comes to industrial reactions, so
that productivity can be maximised. The following are some of the factors that affect the rate
of a reaction.
1. Nature of reactants
Substances have different chemical properties and therefore react differently and at different
rates.
2. Concentration (or pressure in the case of gases)
As the concentration of the reactants increases, so does the reaction rate.
3. Temperature
If the temperature of the reaction increases, so does the rate of the reaction.
4. Catalyst
Adding a catalyst increases the reaction rate.
289
16.2
CHAPTER 16. REACTION RATES - GRADE 12
5. Surface area of solid reactants
Increasing the surface area of the reactants (e.g. if a solid reactant is finely broken up)
will increase the reaction rate.
Activity :: Experiment : The nature of reactants.
Aim:
To determine the effect of the nature of reactants on the rate of a reaction.
Apparatus:
Oxalic acid ((COOH)2 ), iron(II) sulphate (FeSO4 ), potassium permanganate
(KMnO4 ), concentrated sulfuric acid (H2 SO4 ), spatula, test tubes, medicine dropper, glass beaker and glass rod.
H2 SO4
KMnO4
H2 SO4
KMnO4
Test tube 1
Iron (II) sulphate solution
Test tube 2
Oxalic acid solution
Method:
1. In the first test tube, prepare an iron (II) sulphate solution by dissolving about
two spatula points of iron (II) sulphate in 10 cm3 of water.
2. In the second test tube, prepare a solution of oxalic acid in the same way.
3. Prepare a solution of sulfuric acid by adding 1 cm3 of the concentrated acid
to about 4 cm3 of water. Remember always to add the acid to the water, and
never the other way around.
4. Add 2 cm3 of the sulfuric acid solution to the iron(II) and oxalic acid solutions
respectively.
5. Using the medicine dropper, add a few drops of potassium permanganate to
the two test tubes. Once you have done this, observe how quickly each solution
discolours the potassium permanganate solution.
Results:
• You should have seen that the oxalic acid solution discolours the potassium
permanganate much more slowly than the iron(II) sulphate.
2+
• It is the oxalate ions (C2 O2−
ions that cause the discolouration.
4 ) and the Fe
It is clear that the Fe2+ ions act much more quickly than the C2 O2−
4 ions. The
reason for this is that there are no covalent bonds to be broken in the ions
before the reaction can take place. In the case of the oxalate ions, covalent
bonds between carbon and oxygen atoms must be broken first.
Conclusions:
The nature of the reactants can affect the rate of a reaction.
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CHAPTER 16. REACTION RATES - GRADE 12
16.2
teresting Oxalic acids are abundant in many plants. The leaves of the tea plant (Camellia
Interesting
Fact
Fact
sinensis) contain very high concentrations of oxalic acid relative to other plants.
Oxalic acid also occurs in small amounts in foods such as parsley, chocolate, nuts
and berries. Oxalic acid irritates the lining of the gut when it is eaten, and can
be fatal in very large doses.
Activity :: Experiment : Surface area and reaction rates.
Marble (CaCO3 ) reacts with hydrochloric acid (HCl) to form calcium chloride,
water and carbon dioxide gas according to the following equation:
CaCO3 + 2HCl → CaCl2 + H2 O + CO2
Aim:
To determine the effect of the surface area of reactants on the rate of the reaction.
Apparatus:
2 g marble chips, 2 g powdered marble, hydrochloric acid, beaker, two test tubes.
beaker containing dilute
hydrochloric acid
Test tube 1
marble chips
Test tube 2
powdered marble
Method:
1. Prepare a solution of hydrochloric acid in the beaker by adding 2 cm3 of the
concentrated solution to 20 cm3 of water.
2. Place the marble chips and powdered marble into separate test tubes.
3. Add 10 cm3 of the dilute hydrochloric acid to each of the test tubes and observe
the rate at which carbon dioxide gas is produced.
Results:
• Which reaction proceeds the fastest?
• Can you explain this?
Conclusions:
The reaction with powdered marble is the fastest. The smaller the pieces of
marble are, the greater the surface area for the reaction to take place. The greater
the surface area of the reactants, the faster the reaction rate will be.
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16.2
CHAPTER 16. REACTION RATES - GRADE 12
Activity :: Experiment : Reactant concentration and reaction rate.
Aim:
To determine the effect of reactant concentration on reaction rate.
Apparatus:
Concentrated hydrochloric acid (HCl), magnesium ribbon, two beakers, two test
tubes, measuring cylinder.
Method:
1. Prepare a solution of dilute hydrochloric acid in one of the beakers by diluting
1 part concentrated acid with 10 parts water. For example, if you measure 1
cm3 of concentrated acid in a measuring cylinder and pour it into a beaker, you
will need to add 10 cm3 of water to the beaker as well. In the same way, if you
pour 2 cm3 of concentrated acid into a beaker, you will need to add 20 cm3 of
water. Both of these are 1:10 solutions. Pour 10 cm3 of the 1:10 solution into
a test tube and mark it ’A’. Remember to add the acid to the water, and not
the other way around.
2. Prepare a second solution of dilute hydrochloric acid by diluting 1 part concentrated acid with 20 parts water. Pour 10cm3 of this 1:20 solution into a second
test tube and mark it ’B’.
3. Take two pieces of magnesium ribbon of the same length. At the same time,
put one piece of magnesium ribbon into test tube A and the other into test
tube B, and observe closely what happens.
Mg ribbon
Mg ribbon
Test tube A
1:10 HCl solution
Test tube B
1:20 HCl solution
The equation for the reaction is:
2HCl + M g → M gCl2 + H2
Results:
• Which of the two solutions is more concentrated, the 1:10 or 1:20 hydrochloric
acid solution?
• In which of the test tubes is the reaction the fastest? Suggest a reason for this.
• How can you measure the rate of this reaction?
• What is the gas that is given off?
• Why was it important that the same length of magnesium ribbon was used for
each reaction?
Conclusions:
The 1:10 solution is more concentrated and this reaction therefore proceeds
faster. The greater the concentration of the reactants, the faster the rate of the
reaction. The rate of the reaction can be measured by the rate at which hydrogen
gas is produced.
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CHAPTER 16. REACTION RATES - GRADE 12
16.3
Activity :: Group work : The effect of temperature on reaction rate
1. In groups of 4-6, design an experiment that will help you to see the effect of
temperature on the reaction time of 2 cm of magnesium ribbon and 20 ml of
vinegar. During your group discussion, you should think about the following:
• What equipment will you need?
• How will you conduct the experiment to make sure that you are able to
compare the results for different temperatures?
• How will you record your results?
• What safety precautions will you need to take when you carry out this
experiment?
2. Present your experiment ideas to the rest of the class, and give them a chance
to comment on what you have done.
3. Once you have received feedback, carry out the experiment and record your
results.
4. What can you conclude from your experiment?
16.3
Reaction rates and collision theory
It should be clear now that the rate of a reaction varies depending on a number of factors. But
how can we explain why reactions take place at different speeds under different conditions? Why,
for example, does an increase in the surface area of the reactants also increase the rate of the
reaction? One way to explain this is to use collision theory.
For a reaction to occur, the particles that are reacting must collide with one another. Only a
fraction of all the collisions that take place actually cause a chemical change. These are called
’successful’ collisions. When there is an increase in the concentration of reactants, the chance
that reactant particles will collide with each other also increases because there are more particles
in that space. In other words, the collision frequency of the reactants increases. The number of
successful collisions will therefore also increase, and so will the rate of the reaction. In the same
way, if the surface area of the reactants increases, there is also a greater chance that successful
collisions will occur.
Definition: Collision theory
Collision theory is a theory that explains how chemical reactions occur and why reaction
rates differ for different reactions. The theory assumes that for a reaction to occur the
reactant particles must collide, but that only a certain fraction of the total collisions, the
effective collisions, actually cause the reactant molecules to change into products. This is
because only a small number of the molecules have enough energy and the right orientation
at the moment of impact to break the existing bonds and form new bonds.
When the temperature of the reaction increases, the average kinetic energy of the reactant
particles increases and they will move around much more actively. They are therefore more likely
to collide with one another (Figure 16.1). Increasing the temperature also increases the number
of particles whose energy will be greater than the activation energy for the reaction (refer section
16.5).
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CHAPTER 16. REACTION RATES - GRADE 12
A
A
B
B
A
B
B
A
B
B
16.3
B
A
A
B
A
A
B
B
B
A
A
A
B
Low Temperature
A
High Temperature
Figure 16.1: An increase in the temperature of a reaction increases the chances that the reactant
particles (A and B) will collide because the particles have more energy and move around more.
Exercise: Rates of reaction
Hydrochloric acid and calcium carbonate react according to the following equation:
CaCO3 + 2HCl → CaCl2 + H2 O + CO2
The volume of carbon dioxide that is produced during the reaction is measured
at different times. The results are shown in the table below.
Time (mins)
1
2
3
4
5
6
7
8
9
10
Volume of CO2 produced (cm3 )
14
26
36
44
50
58
65
70
74
77
Note: On a graph of production against time, it is the gradient of the graph that
shows the rate of the reaction.
Questions:
1. Use the data in the table to draw a graph showing the volume of gas that is
produced in the reaction, over a period of 10 minutes.
2. At which of the following times is the reaction fastest? Time = 1 minute; time
= 6 minutes or time = 8 minutes.
3. Suggest a reason why the reaction slows down over time.
4. Use the graph to estimate the volume of gas that will have been produced after
11 minutes.
5. After what time do you think the reaction will stop?
6. If the experiment was repeated using a more concentrated hydrochloric acid
solution...
(a) would the rate of the reaction increase or decrease from the one shown in
the graph?
(b) draw a rough line on the graph to show how you would expect the reaction
to proceed with a more concentrated HCl solution.
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CHAPTER 16. REACTION RATES - GRADE 12
16.4
16.4
Measuring Rates of Reaction
How the rate of a reaction is measured will depend on what the reaction is, and what product
forms. Look back to the reactions that have been discussed so far. In each case, how was the
rate of the reaction measured? The following examples will give you some ideas about other
ways to measure the rate of a reaction:
• Reactions that produce hydrogen gas:
When a metal dissolves in an acid, hydrogen gas is produced. A lit splint can be used
to test for hydrogen. The ’pop’ sound shows that hydrogen is present. For example,
magnesium reacts with sulfuric acid to produce magnesium sulphate and hydrogen.
M g(s) + H2 SO4 → M gSO4 + H2
• Reactions that produce carbon dioxide:
When a carbonate dissolves in an acid, carbon dioxide gas is produced. When carbon
dioxide is passes through limewater, it turns the limewater milky. This is the test for the
presence of carbon dioxide. For example, calcium carbonate reacts with hydrochloric acid
to produce calcium chloride, water and carbon dioxide.
CaCO3 (s) + 2HCl(aq) → CaCl2 (aq) + H2 O(l) + CO2 (g)
• Reactions that produce gases such as oxygen or carbon dioxide:
Hydrogen peroxide decomposes to produce oxygen. The volume of oxygen produced can
be measured using the gas syringe method (figure 16.2). The gas collects in the syringe,
pushing out against the plunger. The volume of gas that has been produced can be read
from the markings on the syringe. For example, hydrogen peroxide decomposes in the
presence of a manganese(IV) oxide catalyst to produce oxygen and water.
2H2 O2 (aq) → 2H2 O(l) + O2 (g)
Gas Syringe System
Gas
[glassType=erlen,niveauLiquide1=40,tubeCoude]
Reactants
Figure 16.2: Gas Syringe Method
• Precipitate reactions:
In reactions where a precipitate is formed, the amount of precipitate formed in a period of
time can be used as a measure of the reaction rate. For example, when sodium thiosulphate
reacts with an acid, a yellow precipitate of sulfur is formed. The reaction is as follows:
N a2 S2 O3 (aq) + 2HCl(aq) → 2N aCl(aq) + SO2 (aq) + H2 O(l) + S(s)
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16.4
CHAPTER 16. REACTION RATES - GRADE 12
One way to estimate the rate of this reaction is to carry out the investigation in a conical
flask and to place a piece of paper with a black cross underneath the bottom of the flask.
At the beginning of the reaction, the cross will be clearly visible when you look into the
flask (figure 16.3). However, as the reaction progresses and more precipitate is formed,
the cross will gradually become less clear and will eventually disappear altogether. Noting
the time that it takes for this to happen will give an idea of the reaction rate. Note that
it is not possible to collect the SO2 gas that is produced in the reaction, because it is very
soluble in water.
[glassType=erlen,niveauLiquide1=40]
Figure 16.3: At the beginning of the reaction beteen sodium thiosulphate and hydrochloric acid,
when no precipitate has been formed, the cross at the bottom of the conical flask can be clearly
seen.
• Changes in mass:
The rate of a reaction that produces a gas can also be measured by calculating the mass
loss as the gas is formed and escapes from the reaction flask. This method can be used for
reactions that produce carbon dioxide or oxygen, but are not very accurate for reactions
that give off hydrogen because the mass is too low for accuracy. Measuring changes in
mass may also be suitable for other types of reactions.
Activity :: Experiment : Measuring reaction rates
Aim:
To measure the effect of concentration on the rate of a reaction.
Apparatus:
• 300 cm3 of sodium thiosulphate (Na2 S2 O3 ) solution. Prepare a solution of
sodium thiosulphate by adding 12 g of Na2 S2 O3 to 300 cm3 of water. This is
solution ’A’.
• 300 cm3 of water
• 100 cm3 of 1:10 dilute hydrochloric acid. This is solution ’B’.
• Six 100 cm3 glass beakers
• Measuring cylinders
• Paper and marking pen
• Stopwatch or timer
Method:
One way to measure the rate of this reaction is to place a piece of paper with a
cross underneath the reaction beaker to see how quickly the cross is made invisible
by the formation of the sulfur precipitate.
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CHAPTER 16. REACTION RATES - GRADE 12
16.5
1. Set up six beakers on a flat surface and mark them from 1 to 6. Under each
beaker you will need to place a piece of paper with a large black cross.
2. Pour 60 cm3 solution A into the first beaker and add 20 cm3 of water
3. Use the measuring cylinder to measure 10 cm3 HCl. Now add this HCl to the
solution that is already in the first beaker (NB: Make sure that you always clean
out the measuring cylinder you have used before using it for another chemical).
4. Using a stopwatch with seconds, record the time it takes for the precipitate
that forms to block out the cross.
5. Now measure 50 cm3 of solution A into the second beaker and add 30 cm3 of
water. To this second beaker, add 10 cm3 HCl, time the reaction and record
the results as you did before.
6. Continue the experiment by diluting solution A as shown below.
Beaker
1
2
3
4
5
6
Solution
(cm3 )
60
50
40
30
20
10
A
Water (cm3 )
20
30
40
50
60
70
Solution
(cm3 )
10
10
10
10
10
10
B
Time
(s)
The equation for the reaction between sodium thiosulphate and hydrochloric acid
is:
N a2 S2 O3 (aq) + 2HCl(aq) → 2N aCl(aq) + SO2 (aq) + H2 O(l) + S(s)
Results:
• Calculate the reaction rate in each beaker. This can be done using the following
equation:
Rate of reaction =
1
time
• Represent your results on a graph. Concentration will be on the x-axis and
reaction rate on the y-axis. Note that the original volume of Na2 S2 O3 can be
used as a measure of concentration.
• Why was it important to keep the volume of HCl constant?
• Describe the relationship between concentration and reaction rate.
Conclusions:
The rate of the reaction is fastest when the concentration of the reactants was
the highest.
16.5
Mechanism of reaction and catalysis
Earlier it was mentioned that it is the collision of particles that causes reactions to occur and
that only some of these collisions are ’successful’. This is because the reactant particles have a
wide range of kinetic energy, and only a small fraction of the particles will have enough energy
to actually break bonds so that a chemical reaction can take place. The minimum energy that
is needed for a reaction to take place is called the activation energy. For more information on
the energy of reactions, refer to chapter 14.
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16.5
CHAPTER 16. REACTION RATES - GRADE 12
Definition: Activation energy
The energy that is needed to break the bonds in reactant molecules so that a chemical
reaction can proceed.
Probability of particle with
that KE
Even at a fixed temperature, the energy of the particles varies, meaning that only some of them
will have enough energy to be part of the chemical reaction, depending on the activation energy
for that reaction. This is shown in figure 16.4. Increasing the reaction temperature has the effect
of increasing the number of particles with enough energy to take part in the reaction, and so the
reaction rate increases.
The distribution of particle kinetic
energies at a fixed temperature.
Average KE
Kinetic Energy of Particle (KE)
Figure 16.4: The distribution of particle kinetic energies at a fixed temperature
A catalyst functions slightly differently. The function of a catalyst is to lower the activation
energy so that more particles now have enough energy to react. The catalyst itself is not changed
during the reaction, but simply provides an alternative pathway for the reaction, so that it needs
less energy. Some metals e.g. platinum, copper and iron can act as catalysts in certain reactions.
In our own human bodies, enzymes are catalysts that help to speed up biological reactions. Catalysts generally react with one or more of the reactants to form a chemical intermediate which
then reacts to form the final product. The chemical intermediate is sometimes called the activated complex.
The following is an example of how a reaction that involves a catalyst might proceed. C represents the catalyst, A and B are reactants and D is the product of the reaction of A and B.
Step 1: A + C → AC
Step 2: B + AC → ABC
Step 3: ABC → CD
Step 4: CD → C + D
In the above, ABC represents the intermediate chemical. Although the catalyst (C) is consumed
by reaction 1, it is later produced again by reaction 4, so that the overall reaction is as follows:
A+B+C→D+C
You can see from this that the catalyst is released at the end of the reaction, completely unchanged.
Definition: Catalyst
A catalyst speeds up a chemical reaction, without being altered in any way. It increases the
reaction rate by lowering the activation energy for a reaction.
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CHAPTER 16. REACTION RATES - GRADE 12
16.5
Energy diagrams are useful to illustrate the effect of a catalyst on reaction rates. Catalysts
decrease the activation energy required for a reaction to proceed (shown by the smaller ’hump’
on the energy diagram in figure 16.5), and therefore increase the reaction rate.
activated complex
Potential energy
activation
energy
products
activation energy
with a catalyst
reactants
Time
Figure 16.5: The effect of a catalyst on the activation energy of a reaction
Activity :: Experiment : Catalysts and reaction rates
Aim:
To determine the effect of a catalyst on the rate of a reaction
Apparatus:
Zinc granules, 0.1 M hydrochloric acid, copper pieces, one test tube and a glass
beaker.
Method:
1. Place a few of the zinc granules in the test tube.
2. Measure the mass of a few pieces of copper and keep them separate from the
rest of the copper.
3. Add about 20 cm3 of HCl to the test tube. You will see that a gas is released.
Take note of how quickly or slowly this gas is released. Write a balanced
equation for the chemical reaction that takes place.
4. Now add the copper pieces to the same test tube. What happens to the rate
at which the gas is produced?
5. Carefully remove the copper pieces from the test tube (do not get HCl on your
hands), rinse them in water and alcohol and then weigh them again. Has the
mass of the copper changed since the start of the experiment?
Results:
During the reaction, the gas that is released is hydrogen. The rate at which the
hydrogen is produced increases when the copper pieces (the catalyst) are added.
The mass of the copper does not change during the reaction.
Conclusions:
The copper acts as a catalyst during the reaction. It speeds up the rate of the
reaction, but is not changed in any way itself.
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16.6
CHAPTER 16. REACTION RATES - GRADE 12
Exercise: Reaction rates
1. For each of the following, say whether the statement is true or false. If it is
false, re-write the statement correctly.
(a) A catalyst increases the energy of reactant molecules so that a chemical
reaction can take place.
(b) Increasing the temperature of a reaction has the effect of increasing the
number of reactant particles that have more energy that the activation
energy.
(c) A catalyst does not become part of the final product in a chemical reaction.
2. 5 g of zinc granules are added to 400 cm3 of 0.5 mol.dm−3 hydrochloric acid.
To investigate the rate of the reaction, the change in the mass of the flask
containing the zinc and the acid was measured by placing the flask on a direct
reading balance. The reading on the balance shows that there is a decrease
in mass during the reaction. The reaction which takes place is given by the
following equation:
Zn(s) + 2HCl(aq) → ZnCl2 (aq) + H2 (g)
(a) Why is there a decrease in mass during the reaction?
(b) The experiment is repeated, this time using 5 g of powdered zinc instead
of granulated zinc. How will this influence the rate of the reaction?
(c) The experiment is repeated once more, this time using 5 g of granulated
zinc and 600 cm3 of 0.5 mol.dm−3 hydrochloric acid. How does the rate
of this reaction compare to the original reaction rate?
(d) What effect would a catalyst have on the rate of this reaction?
(IEB Paper 2 2003)
3. Enzymes are catalysts. Conduct your own research to find the names of common enzymes in the human body and which chemical reactions they play a role
in.
4. 5 g of calcium carbonate powder reacts with 20 cm3 of a 0.1 mol.dm−3 solution
of hydrochloric acid. The gas that is produced at a temperature of 250 C is
collected in a gas syringe.
(a) Write a balanced chemical equation for this reaction.
(b) The rate of the reaction is determined by measuring the volume of gas
thas is produced in the first minute of the reaction. How would the rate
of the reaction be affected if:
i. a lump of calcium carbonate of the same mass is used
ii. 40 cm3 of 0.1 mol.dm−3 hydrochloric acid is used
16.6
Chemical equilibrium
Having looked at factors that affect the rate of a reaction, we now need to ask some important
questions. Does a reaction always proceed in the same direction or can it be reversible? In other
words, is it always true that a reaction proceeds from reactants to products, or is it possible that
sometimes, the reaction will reverse and the products will be changed back into the reactants?
And does a reaction always run its full course so that all the reactants are used up, or can a
reaction reach a point where reactants are still present, but there does not seem to be any further
change taking place in the reaction? The following demonstration might help to explain this.
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CHAPTER 16. REACTION RATES - GRADE 12
16.6
Activity :: Demonstration : Liquid-vapour phase equilibrium
Apparatus and materials:
2 beakers; water; bell jar
Method:
1. Half fill two beakers with water and mark the level of the water in each case.
2. Cover one of the beakers with a bell jar.
3. Leave the beakers and, over the course of a day or two, observe how the water
level in the two beakers changes. What do you notice? Note: You could speed
up this demonstration by placing the two beakers over a bunsen burner to heat
the water. In this case, it may be easier to cover the second beaker with a glass
cover.
Observations:
You should notice that in the beaker that is uncovered, the water level drops
quickly because of evaporation. In the beaker that is covered, there is an initial drop
in the water level, but after a while evaporation appears to stop and the water level
in this beaker is higher than that in the one that is open. Note that the diagram
below shows the situation ate time=0.
bell jar
= condensation
= evaporation
Discussion:
In the first beaker, liquid water becomes water vapour as a result of evaporation
and the water level drops. In the second beaker, evaporation also takes place.
However, in this case, the vapour comes into contact with the surface of the bell
jar and it cools and condenses to form liquid water again. This water is returned to
the beaker. Once condensation has begun, the rate at which water is lost from the
beaker will start to decrease. At some point, the rate of evaporation will be equal to
the rate of condensation above the beaker, and there will be no change in the water
level in the beaker. This can be represented as follows:
liquid ⇔ vapour
In this example, the reaction (in this case, a change in the phase of water) can
proceed in either direction. In one direction there is a change in phase from liquid to
vapour. But the reverse can also take place, when vapour condenses to form water
again.
In a closed system it is possible for reactions to be reversible, such as in the demonstration
above. In a closed system, it is also possible for a chemical reaction to reach equilibrium. We
will discuss these concepts in more detail.
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16.6
CHAPTER 16. REACTION RATES - GRADE 12
16.6.1
Open and closed systems
An open system is one in which matter or energy can flow into or out of the system. In the
liquid-vapour demonstration we used, the first beaker was an example of an open system because
the beaker could be heated or cooled (a change in energy ), and water vapour (the matter ) could
evaporate from the beaker.
A closed system is one in which energy can enter or leave, but matter cannot. The second
beaker covered by the bell jar is an example of a closed system. The beaker can still be heated or
cooled, but water vapour cannot leave the system because the bell jar is a barrier. Condensation
changes the vapour to liquid and returns it to the beaker. In other words, there is no loss of
matter from the system.
Definition: Open and closed systems
An open system is one whose borders allow the movement of energy and matter into and
out of the system. A closed system is one in which only energy can be exchanged, but not
matter.
16.6.2
Reversible reactions
Some reactions can take place in two directions. In one direction the reactants combine to form
the products. This is called the forward reaction. In the other, the products react to form
reactants again. This is called the reverse reaction. A special double-headed arrow is used to
show this type of reversible reaction:
XY + Z ⇔ X + Y Z
So, in the following reversible reaction:
H2 (g) + I2 (g) ⇔ 2HI(g)
The forward reaction is H2 (g) + I2 (g) → 2HI(g). The reverse reaction is 2HI(g) → H2 (g) + I2 (g).
Definition: A reversible reaction
A reversible reaction is a chemical reaction that can proceed in both the forward and reverse
directions. In other words, the reactant and product of one reaction may reverse roles.
Activity :: Demonstration : The reversibility of chemical reactions
Apparatus and materials:
Lime water (Ca(OH)2 ); calcium carbonate (CaCO3 ); hydrochloric acid; 2 test
tubes with rubber stoppers; delivery tube; retort stand and clamp; bunsen burner.
Method and observations:
1. Half-fill a test tube with clear lime water (Ca(OH)2 ).
2. In another test tube, place a few pieces of calcium carbonate (CaCO3 ) and
cover the pieces with dilute hydrochloric acid. Seal the test tube with a rubber
stopper and delivery tube.
3. Place the other end of the delivery tube into the test tube containing the lime
water so that the carbon dioxide that is produced from the reaction between calcium carbonate and hydrochloric acid passes through the lime water. Observe
what happens to the appearance of the lime water.
The equation for the reaction that takes place is:
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CHAPTER 16. REACTION RATES - GRADE 12
16.6
Ca(OH)2 + CO2 → CaCO3 + H2 O
CaCO3 is insoluble and it turns the limewater milky.
4. Allow the reaction to proceed for a while so that carbon dioxide continues to
pass through the limewater. What do you notice? The equation for the reaction
that takes place is:
CaCO3 (s) + H2 O + CO2 → Ca(HCO3 )2
In this reaction, calcium carbonate becomes one of the reactants to produce
hydrogen carbonate (Ca(HCO3 )2 ) and so the solution becomes clear again.
5. Heat the solution in the test tube over a bunsen burner. What do you observe?
You should see bubbles of carbon dioxide appear and the limewater turns milky
again. The reaction that has taken place is:
Ca(HCO3 )2 → CaCO3 (s) + H2 O + CO2
delivery tube
rubber stopper
rubber stopper
[glassType=tube,bouchon=true,niveauLiquide1=30]
[glassType=tube,bouchon=true,niveauLiquide1=60]
calcium carbonate &
hydrochloric acid
limewater
Discussion:
• If you look at the last two equations you will see that the one is the reverse of
the other. In other words, this is a reversible reaction and can be written as
follows:
CaCO3 (s) + H2 O + CO2 ⇔ Ca(HCO3 )2
• Is the forward reaction endothermic or exothermic? Is the reverse reaction
endothermic or exothermic? You should have noticed that the reverse reaction only took place when the solution was heated. Sometimes, changing the
temperature of a reaction can change its direction.
16.6.3
Chemical equilibrium
Using the same reversible reaction that we used in an earlier example:
H2 (g) + I2 (g) ⇔ 2HI(g)
The forward reaction is:
H2 + I2 → 2HI
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16.7
CHAPTER 16. REACTION RATES - GRADE 12
The reverse reaction is:
2HI → H2 + I2
When the rate of the forward reaction and the reverse reaction are equal, the system is said to
be in equilbrium. Figure 16.6 shows this. Initially (time = 0), the rate of the forward reaction
is high and the rate of the reverse reaction is low. As the reaction proceeds, the rate of the
forward reaction decreases and the rate of the reverse reaction increases, until both occur at the
same rate. This is called equilibrium.
Rate of Reaction
Although it is not always possible to observe any macroscopic changes, this does not mean
that the reaction has stopped. The forward and reverse reactions continue to take place and
so microscopic changes still occur in the system. This state is called dynamic equilibrium. In
the liquid-vapour phase equilibrium demonstration, dynamic equilibrium was reached when there
was no observable change in the level of the water in the second beaker even though evaporation
and condensation continued to take place.
H2 +I2 →2HI
equilibrium
2HI→H2 +I2
Time
Figure 16.6: The change in rate of forward and reverse reactions in a closed system
There are, however, a number of factors that can change the chemical equilibrium of a reaction. Changing the concentration, the temperature or the pressure of a reaction can affect
equilibrium. These factors will be discussed in more detail later in this chapter.
Definition: Chemical equilibrium
Chemical equilibrium is the state of a chemical reaction, where the concentrations of the
reactants and products have no net change over time. Usually, this state results when the
forward chemical reactions proceed at the same rate as their reverse reactions.
16.7
The equilibrium constant
Definition: Equilibrium constant
The equilibrium constant (Kc ), relates to a chemical reaction at equilibrium. It can be
calculated if the equilibrium concentration of each reactant and product in a reaction at
equilibrium is known.
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CHAPTER 16. REACTION RATES - GRADE 12
16.7.1
16.7
Calculating the equilibrium constant
Consider the following generalised reaction which takes place in a closed container at a constant
temperature:
A+B ⇔C +D
We know from section 16.2 that the rate of the forward reaction is directly proportional to the
concentration of the reactants. In other words, as the concentration of the reactants increases,
so does the rate of the forward reaction. This can be shown using the following equation:
Rate of forward reaction ∝ [A][B]
or
Rate of forward reaction = k1 [A][B]
Similarly, the rate of the reverse reaction is directly proportional to the concentration of the
products. This can be shown using the following equation:
Rate of reverse reaction ∝ [C][D]
or
Rate of reverse reaction = k2 [C][D]
At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction. This
can be shown using the following equation:
k1 [A][B] = k2 [C][D]
or
k1
[C][D]
=
k2
[A][B]
or, if the constants k1 and k2 are simplified to a single constant, the equation becomes:
kc =
[C][D]
[A][B]
A more general form of the equation for a reaction at chemical equilibrium is:
aA + bB ⇔ cC + dD
where A and B are reactants, C and D are products and a, b, c, and d are the coefficients of
the respective reactants and products. A more general formula for calculating the equilibrium
constant is therefore:
kc =
[C]c [D]d
[A]a [B]b
It is important to note that if a reactant or a product in a chemical reaction is in either the
liquid or solid phase, the concentration stays constant during the reaction. Therefore, these
values can be left out of the equation to calculate kc . For example, in the following reaction:
C(s) + H2 O(g) ⇔ CO(g) + H2 (g)
305
16.7
CHAPTER 16. REACTION RATES - GRADE 12
kc =
[CO][H2 ]
[H2 O]
Important:
1. The constant kc is affected by temperature and so, if the values of k c are being
compared for different reactions, it is important that all the reactions have taken
place at the same temperature.
2. kc values do not have units. If you look at the equation, the units all cancel each
other out.
16.7.2
The meaning of kc values
The formula for kc has the concentration of the products in the numerator and the concentration
of reactants in the denominator. So a high kc value means that the concentration of products
is high and the reaction has a high yield. We can also say that the equilibrium lies far to the
right. The opposite is true for a low kc value. A low kc value means that, at equilibrium, there
are more reactants than products and therefore the yield is low. The equilibrium for the reaction
lies far to the left.
Important: Calculations made easy
When you are busy with calculations that involve the equilibrium constant, the following tips
may help:
1. Make sure that you always read the question carefully to be sure of what you are being asked
to calculate. If the equilibrium constant is involved, make sure that the concentrations you
use are the concentrations at equilibrium, and not the concentrations or quantities that
are present at some other time in the reaction.
2. When you are doing more complicated calculations, it sometimes helps to draw up a table
like the one below and fill in the mole values that you know or those you can calculate.
This will give you a clear picture of what is happening in the reaction and will make sure
that you use the right values in your calculations.
Reactant 1
Reactant 2
Start of reaction
Used up
Produced
Equilibrium
Worked Example 77: Calculating kc
Question: For the reaction:
SO2 (g) + N O2 (g) → N O(g) + SO3 (g)
306
Product 1
CHAPTER 16. REACTION RATES - GRADE 12
the concentration of the reagents is as follows:
[SO3 ] = 0.2 mol.dm−3
[NO2 ] = 0.1 mol.dm−3
[NO] = 0.4 mol.dm−3
[SO2 ] = 0.2 mol.dm−3
Calculate the value of kc .
Answer
Step 1 : Write the equation for kc
kc =
[N O][SO3 ]
[SO2 ][N O2 ]
Step 2 : Fill in the values you know for this equation and calculate kc
kc =
(0.4 × 0.2)
=4
(0.2 × 0.1)
Worked Example 78: Calculating reagent concentration
Question: For the reaction:
S(s) + O2 (g) ⇔ SO2 (g)
1. Write an equation for the equilibrium constant.
2. Calculate the equilibrium concentration of O2 if Kc=6 and [SO2 ] = 3mol.dm−3
at equilibrium.
Answer
Step 1 : Write the equation for kc
kc =
[SO2 ]
[O2 ]
(Sulfur is left out of the equation because it is a solid and its concentration stays
constant during the reaction)
Step 2 : Re-arrange the equation so that oxygen is on its own on one side
of the equation
[O2 ] =
[SO2 ]
kc
Step 3 : Fill in the values you know and calculate [O2 ]
[O2 ] =
3mol.dm−3
= 0.5mol.dm−3
6
Worked Example 79: Equilibrium calculations
Question: Initially 1.4 moles of NH3 (g) is introduced into a sealed 2.0 dm−3 reaction
vessel. The ammonia decomposes when the temperature is increased to 600K and
reaches equilibrium as follows:
307
16.7
16.7
CHAPTER 16. REACTION RATES - GRADE 12
2N H3 (g) ⇔ N2 (g) + 3H2 (g)
When the equilibrium mixture is analysed, the concentration of NH3 (g) is 0.3 mol.dm−3
1. Calculate the concentration of N2 (g) and H2 (g) in the equilibrium mixture.
2. Calculate the equilibrium constant for the reaction at 900 K.
Answer
Step 1 : Calculate the number of moles of NH3 at equilibrium.
c=
n
V
Therefore,
n = c × V = 0.3 × 2 = 0.6mol
Step 2 : Calculate the number of moles of ammonia that react (are ’used
up’) in the reaction.
Moles used up = 1.4 - 0.6 = 0.8 moles
Step 3 : Calculate the number of moles of product that are formed.
Remember to use the mole ratio of reactants to products to do this. In this case,
the ratio of NH3 :N2 :H2 = 2:1:3. Therefore, if 0.8 moles of ammonia are used up in
the reaction, then 0.4 moles of nitrogen are produced and 1.2 moles of hydrogen are
produced.
Step 4 : Complete the following table
Start of reaction
Used up
Produced
Equilibrium
NH3
1.4
0.8
0
0.6
N2
0
0
0.4
0.4
H2
0
0
1.2
1.2
Step 5 : Using the values in the table, calculate [N2 ] and [H2 ]
[N2 ] =
0.4
n
=
= 0.2 mol.dm−3
V
2
[H2 ] =
n
1.2
=
= 0.6 mol.dm−3
V
2
Step 6 : Calculate kc
kc =
(0.6)3 (0.2)
[H2 ]3 [N2 ]
=
= 0.48
[N H3 ]2
(0.3)2
Worked Example 80: Calculating kc
Question: Hydrogen and iodine gas react according to the following equation:
H2 (g) + I2 (g) ⇔ 2HI(g)
When 0.496 mol H2 and 0.181 mol I2 are heated at 450oC in a 1 dm3 container, the
equilibrium mixture is found to contain 0.00749 mol I2 . Calculate the equilibrium
constant for the reaction at 450o C.
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CHAPTER 16. REACTION RATES - GRADE 12
16.7
Answer
Step 1 : Calculate the number of moles of iodine used in the reaction.
Moles of iodine used = 0.181 - 0.00749 = 0.1735 mol
Step 2 : Calculate the number of moles of hydrogen that are used up in the
reaction.
The mole ratio of hydrogen:iodine = 1:1, therefore 0.1735 moles of hydrogen must
also be used up in the reaction.
Step 3 : Calculate the number of moles of hydrogen iodide that are produced.
The mole ratio of H2 :I2 :HI = 1:1:2, therefore the number of moles of HI produced
is 0.1735 × 2 = 0.347 mol.
So far, the table can be filled in as follows:
Start of reaction
Used up
Produced
Equilibrium
H2 (g)
0.496
0.1735
0
0.3225
I2
0.181
0.1735
0
0.0075
2HI
0
0
0.347
0.347
Step 4 : Calculate the concentration of each of the reactants and products
at equilibrium.
n
V
Therefore the equilibrium concentrations are as follows:
[H2 ] = 0.3225 mol.dm−3
[I2 ] = 0.0075 mol.dm−3
[HI] = 0.347 mol.dm−3
c=
Step 5 : Calculate kc
kc =
[HI]
0.347
=
= 143.47
[H2 ][I2 ]
0.3225 × 0.0075
Exercise: The equilibrium constant
1. Write the equilibrium constant expression, Kc for the following reactions:
(a) 2NO(g) + Cl2 (g) ⇔ 2NOCl
(b) H2 (g) + I2 (g) ⇔ 2HI(g)
2. The following reaction takes place:
Fe3+ (aq) + 4Cl− ⇔ FeCl−
4 (aq)
Kc for the reaction is 7.5 × 10−2 mol.dm−3 . At equilibrium, the concentration
−4
of FeCl−
mol.dm−3 and the concentration of free iron (Fe3+ )
4 is 0.95 × 10
−3
is 0.2 mol.dm . Calculate the concentration of chloride ions at equilibrium.
3. Ethanoic acid (CH3 COOH) reacts with ethanol (CH3 CH2 OH) to produce ethyl
ethanoate and water. The reaction is:
CH3 COOH + CH3 CH2 OH → CH3 COOCH2 CH3 + H2 O
At the beginning of the reaction, there are 0.5 mols of ethanoic acid and 0.5
mols of ethanol. At equilibrium, 0.3 mols of ethanoic acid was left unreacted.
The volume of the reaction container is 2 dm3 . Calculate the value of Kc .
309
16.8
16.8
CHAPTER 16. REACTION RATES - GRADE 12
Le Chatelier’s principle
A number of factors can influence the equilibrium of a reaction. These are:
1. concentration
2. temperature
3. pressure
Le Chatelier’s Principle helps to predict what a change in temperature, concentration or
pressure will have on the position of the equilibrium in a chemical reaction. This is very important,
particularly in industrial applications, where yields must be accurately predicted and maximised.
Definition: Le Chatelier’s Principle
If a chemical system at equilibrium experiences a change in concentration, temperature or
total pressure the equilibrium will shift in order to minimise that change.
16.8.1
The effect of concentration on equilibrium
If the concentration of a substance is increased, the equilibrium will shift so that this concentration decreases. So for example, if the concentration of a reactant was increased, the equilibrium
would shift in the direction of the reaction that uses up the reactants, so that the reactant concentration decreases and equilibrium is restored. In the reaction between nitrogen and hydrogen
to produce ammonia:
N2 (g) + 3H2 (g) ⇔ 2N H3 (g)
• If the nitrogen or hydrogen concentration was increased, Le Chatelier’s principle predicts
that equilibrium will shift to favour the forward reaction so that the excess nitrogen and
hydrogen are used up to produce ammonia. Equilibrium shifts to the right.
• If the nitrogen or hydrogen concentration was decreased, the reverse reaction would be
favoured so that some of the ammonia would change back to nitrogen and hydrogen to
restore equilibrium.
• The same would be true if the concentration of the product (NH3 ) was changed. If [NH3 ]
decreases, the forward reaction is favoured and if [NH3 ] increases, the reverse reaction is
favoured.
16.8.2
The effect of temperature on equilibrium
If the temperature of a reaction mixture is increased, the equilibrium will shift to decrease the
temperature. So it will favour the reaction which will use up heat energy, in other words the
endothermic reaction. The opposite is true if the temperature is decreased. In this case, the
reaction that produces heat energy will be favoured, in other words, the exothermic reaction.
The reaction shown below is exothermic (shown by the negative value for ∆ H). This means
that the forward reaction, where nitrogen and hydrogen react to form ammonia, gives off heat.
In the reverse reaction, where ammonia is broken down into hydrogen and nitrogen gas, heat is
used up and so this reaction is endothermic.
e.g. N2 (g) + 3H2 (g) ⇔ 2N H3 (g) and ∆H = −92kJ
An increase in temperature favours the reaction that is endothermic (the reverse reaction) because it uses up energy. If the temperature is increased, then the yield of ammonia (NH3 )
310
CHAPTER 16. REACTION RATES - GRADE 12
16.8
decreases.
A decrease in temperature favours the reaction that is exothermic (the forward reaction) because
it produces energy. Therefore, if the temperature is decreased, then the yield of NH3 increases.
Activity :: Experiment : Le Chatelier’s Principle
Aim:
To determine the effect of a change in concentration and temperature on chemical
equilibrium
Apparatus:
0.2 M CoCl2 solution, concentrated HCl, water, test tube, bunsen burner
Method:
1. Put 4-5 drops of 0.2M CoCl2 solution into a test tube.
2. Add 20-25 drops of concentrated HCl.
3. Add 10-12 drops of water.
4. Heat the solution for 1-2 minutes.
5. Cool the solution for 1 minute under a tap.
6. Observe and record the colour changes that take place during the reaction.
The equation for the reaction that takes place is:
−
e.g. CoCl42− + 6H2 O ⇔ Co(H2 O)2+
6 + 4Cl
{z
}
|
{z
}
|
blue
pink
Results:
Complete your observations in the table below, showing the colour changes that
take place, and also indicating whether the concentration of each of the ions in
solution increases or decreases.
Initial
colour
Final
colour
[Co2+ ]
[Cl− ]
[CoCl2−
4 ]
Add Cl−
Add H2 O
Increase
temp.
Decrease
temp.
Conclusions:
Use your knowledge of equilibrium principles to explain the changes that you
recorded in the table above. Draw a conclusion about the effect of a change in
concentration of either the reactants or products on the equilibrium position. Also
draw a conclusion about the effect of a change in temperature on the equilibrium
position.
311
16.8
CHAPTER 16. REACTION RATES - GRADE 12
16.8.3
The effect of pressure on equilibrium
In the case of gases, we refer to pressure instead of concentration. Similar principles apply as
those that were described before for concentration. When the pressure of a system increases,
there are more particles in a particular space. The equilibrium will shift in a direction that reduces
the number of gas particles so that the pressure is also reduced. To predict what will happen in a
reaction, we need to look at the number of moles of gas that are in the reactants and products.
Look at the example below:
e.g. 2SO2 (g) + O2 (g) ⇔ 2SO3 (g)
In this reaction, two moles of product are formed for every three moles of reactants. If we
increase the pressure on the closed system, the equilibrium will shift to the right because the
forward reaction reduces the number of moles of gas that are present. This means that the
yield of SO3 will increase. The opposite will apply if the pressure on the system decreases. the
equilibrium will shift to the left, and the concentration of SO2 and O2 will increase.
Important: The following rules will help in predicting the changes that take place in
equilibrium reactions:
1. If the forward reaction that forms the product is endothermic, then an increase in
temperature will favour this reaction and the yield of product will increase. Lowering
the temperature will decrease the product yield.
2. If the forward reaction that forms the product is exothermic, then a decrease in
temperature will favour this reaction and the product yield will increase. Increasing
the temperature will decrease the product yield.
3. Increasing the pressure favours the side of the equilibrium with the least number of
gas molecules. This is shown in the balanced symbol equation. This rule applies in
reactions with one or more gaseous reactants or products.
4. Decreasing the pressure favours the side of the equilibrium with the most number of
gas molecules. This rule applies in reactions with one or more gaseous reactants or
products.
5. If the concentration of a reactant (on the left) is increased, then some of it must
change to the products (on the right) for equilibrium to be maintained. The equilibrium position will shift to the right.
6. If the concentration of a reactant (on the left) is decreased, then some of the products
(on the right) must change back to reactants for equilibrium to be maintained. The
equilibrium position will shift to the left.
7. A catalyst does not affect the equilibrium position of a reaction. It only influences
the rate of the reaction, in other words, how quickly equilibrium is reached.
Worked Example 81: Reaction Rates 1
Question: 2N O2 (g) ⇔ 2N O(g) + O2 (g) and ∆H > 0 How will the rate of the
reverse reaction be affected by:
1. a decrease in temperature?
2. the addition of a catalyst?
3. the addition of more NO gas?
Answer
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CHAPTER 16. REACTION RATES - GRADE 12
16.8
1. The rate of the forward reaction will increase since it is the forward reaction that
is exothermix and therefore produces energy to balance the loss of energy from
the decrease in temperature. The rate of the reverse reaction will decrease.
2. The rate of the reverse and the forward reaction will increase.
3. The rate of the reverse reaction will increase so that the extra NO gas is
converted into NO2 gas.
Worked Example 82: Reaction Rates 2
Question:
1. Write a balanced equation for the exothermic reaction between Zn(s) and HCl.
2. Name 3 ways to increase the reaction rate between hydrochloric acid and zinc
metal.
Answer
1. Zn(s) + 2HCl(aq) ⇔ ZnCl2 (aq) + H2 (g)
2. A catalyst could be added, the zinc solid could be ground into a fine powder
to increase its surface area, the HCl concentration could be increased or the
reaction temperature could be increased.
Exercise: Reaction rates and equilibrium
1. The following reaction reaches equilibrium in a closed container:
CaCO3 (s) ⇔ CaO(s) + CO2 (g)
The pressure of the system is increased by decreasing the volume of the container. How will the number of moles and the concentration of the CO2 (g)
have changed when a new equilibrium is reached at the same temperature?
A
B
C
D
moles of CO2
decreased
increased
decreased
decreased
[CO2 ]
decreased
increased
stays the same
increased
(IEB Paper 2, 2003)
2. The following reaction has reached equilibrium in a closed container:
C(s) + H2 O(g) ⇔ CO(g) + H2 (g) ∆H ¿ 0
The pressure of the system is then decreased by increasing the volume of the
container. How will the concentration of the H2 (g) and the value of Kc be
affected when the new equilibrium is established? Assume that the temperature
of the system remains unchanged.
A
B
C
D
[H2 ]
increases
increases
unchanged
decreases
313
Kc
increases
unchanged
unchanged
unchanged
16.8
CHAPTER 16. REACTION RATES - GRADE 12
(IEB Paper 2, 2004)
3. During a classroom experiment copper metal reacts with concentrated nitric
acid to produce NO2 gas, which is collected in a gas syringe. When enough gas
has collected in the syringe, the delivery tube is clamped so that no gas can
escape. The brown NO2 gas collected reaches an equilibrium with colourless
N2 O4 gas as represented by the following equation:
2N O2 (g) ⇔ N2 O4 (g)
Once this equilibrium has been established, there are 0.01 moles of NO2 gas
and 0.03 moles of N2 O4 gas present in the syringe.
(a) A learner, noticing that the colour of the gas mixture in the syringe is no
longer changing, comments that all chemical reactions in the syringe must
have stopped. Is this assumption correct? Explain.
(b) The gas in the syringe is cooled. The volume of the gas is kept constant
during the cooling process. Will the gas be lighter or darker at the lower
temperature? Explain your answer.
(c) The volume of the syringe is now reduced to 75 cm3 by pushing the plunger
in and holding it in the new position. There are 0.032 moles of N2 O4
gas present once the equilibrium has been re-established at the reduced
volume (75 cm3 ). Calculate the value of the equilibrium constant for this
equilibrium.
(IEB Paper 2, 2004)
4. Consider the following reaction, which takes place in a closed container:
A(s) + B(g) → AB(g) ∆H < 0
If you wanted to increase the rate of the reaction, which of the following would
you do?
(a) decrease the concentration of B
(b) decrease the temperature of A
(c) grind A into a fine powder
(d) decrease the pressure
(IEB Paper 2, 2002)
5. Gases X and Y are pumped into a 2 dm3 container. When the container is
sealed, 4 moles of gas X and 4 moles of gas Y are present. The following
equilibrium is established:
2X(g) + 3Y(g) ⇔ X2 Y3
The graph below shows the number of moles of gas X and gas X2 Y3 that are
present from the time the container is sealed.
4
number
of
moles
0,5
30
70
100
time (s)
314
CHAPTER 16. REACTION RATES - GRADE 12
16.9
(a) How many moles of gas X2 Y3 are formed by the time the reaction reaches
equilibrium at 30 seconds?
(b) Calculate the value of the equilibrium constant at t = 50 s.
(c) At 70 s the temperature is increased. Is the forward reaction endothermic
or exothermic? Explain in terms of Le Chatelier’s Principle.
(d) How will this increase in temperature affect the value of the equilibrium
constant?
16.9
Industrial applications
The Haber process is a good example of an industrial process which uses the equilibrium
principles that have been discussed. The equation for the process is as follows:
N2 (g) + 3H2 (g) ⇔ 2N H3 (g) + energy
Since the reaction is exothermic, the forward reaction is favoured at low temperatures, and
the reverse reaction at high temperatures. If the purpose of the Haber process is to produce
ammonia, then the temperature must be maintained at a level that is low enough to ensure that
the reaction continues in the forward direction.
The forward reaction is also favoured by high pressures because there are four moles of reactant
for every two moles of product formed.
The k value for this reaction will be calculated as follows:
k=
[N H3 ]2
[N2 ][H2 ]3
Exercise: Applying equilibrium principles
Look at the values of k calculated for the Haber process reaction at different
temperatures, and then answer the questions that follow:
T oC
25
200
300
400
500
k
6.4
4.4
4.3
1.6
1.5
x
x
x
x
x
102
10−1
10−3
10−4
10−5
1. What happens to the value of k as the temperature increases?
2. Which reaction is being favoured when the temperature is 300 degrees celsius?
3. According to this table, which temperature would be best if you wanted to
produce as much ammonia as possible? Explain.
315
16.10
CHAPTER 16. REACTION RATES - GRADE 12
16.10
Summary
• The rate of a reaction describes how quickly reactants are used up, or how quickly
products form. The units used are moles per second.
• A number of factors can affect the rate of a reaction. These include the nature of the
reactants, the concentration of reactants, temperature of the reaction, the presence or
absence of a catalyst and the surface area of the reactants.
• Collision theory provides one way of explaining why each of these factors can affect the
rate of a reaction. For example, higher temperatures mean increased reaction rates because
the reactant particles have more energy and are more likely to collide successfully with each
other.
• Different methods can be used to measure the rate of a reaction. The method used
will depend on the nature of the product. Reactions that produce gases can be measured
by collecting the gas in a syringe. Reactions that produce a precipitate are also easy to
measure because the precipitate is easily visible.
• For any reaction to occur, a minimum amount of energy is needed so that bonds in the
reactants can break, and new bonds can form in the products. The minimum energy that
is required is called the activation energy of a reaction.
• In reactions where the particles do not have enough energy to overcome this activation
energy, one of two methods can be used to facilitate a reaction to take place: increase the
temperature of the reaction or add a catalyst.
• Increasing the temperature of a reaction means that the average energy of the reactant particles increases and they are more likely to have enough energy to overcome the
activation energy.
• A catalyst is used to lower the activation energy so that the reaction is more likely to
take place. A catalyst does this by providing an alternative, lower energy pathway, for the
reaction.
• A catalyst therefore speeds up a reaction but does not become part of the reaction in
any way.
• Chemical equilibrium is the state of a reaction, where the concentrations of the reactants
and the products have no net change over time. Usually this occurs when the rate of the
forward reaction is the same as the rate of the reverse reaction.
• The equilibrium constant relates to reactions at equilibrium, and can be calculated using
the following equation:
kc =
[C]c [D]d
[A]a [B]b
where A and B are reactants, C and D are products and a, b, c, and d are the coefficients
of the respective reactants and products.
• A high kc value means that the concentration of products at equilibrium is high and the
reaction has a high yield. A low kc value means that the concentration of products at
equilibrium is low and the reaction has a low yield.
• Le Chatelier’s Principle states that if a chemical system at equilibrium experiences a
change in concentration, temperature or total pressure the equilibrium will shift in order
to minimise that change. For example, if the pressure of a gaseous system at eqilibrium
was increased, the equilibrium would shift to favour the reaction that produces the lowest
quantity of the gas. If the temperature of the same system was to increase, the equilibrium
would shift to favour the endothermic reaction. Similar principles apply for changes in
concentration of the reactants or products in a reaction.
• The principles of equilibrium are very important in industrial applications such as the
Haber process, so that productivity can be maximised.
316
CHAPTER 16. REACTION RATES - GRADE 12
Exercise: Summary Exercise
1. For each of the following questions, choose the one correct answer from the
list provided.
(a) Consider the following reaction that has reached equilibrium after some
time in a sealed 1 dm3 flask:
P Cl5 (g) ⇔ P Cl3 (g) + Cl2 (g); ∆H is positive
Which one of the following reaction conditions applied to the system would
decrease the rate of the reverse reaction?
i.
ii.
iii.
iv.
increase the pressure
increase the reaction temperature
continually remove Cl2 (g) from the flask
addition of a suitable catalyst
(IEB Paper 2, 2001)
(b) The following equilibrium constant expression is given for a particular reaction:
Kc = [H2 O]4 [CO2 ]3 /[C3 H8 ][O2 ]5
For which one of the following reactions is the above expression of Kc is
correct?
i.
ii.
iii.
iv.
C3 H8 (g) + 5O2 (g) ⇔ 4H2 O(g) + 3CO2 (g)
4H2 O(g) + 3CO2 (g) ⇔ C3 H8 (g) + 5O2 (g)
2C3 H8 (g) + 7O2 (g) ⇔ 6CO(g) + 8H2 O(g)
C3 H8 (g) + 5O2 (g) ⇔ 4H2 O(l) + 3CO2 (g)
(IEB Paper 2, 2001)
2. 10 g of magnesium ribbon reacts with a 0.15 mol.dm−3 solution of hydrochloric
acid at a temperature of 250 C.
(a) Write a balanced chemical equation for the reaction.
(b) State two ways of increasing the rate of production of H2 (g).
(c) A table of the results is given below:
Time elapsed (min) Vol of H2 (g) (cm3 )
0
0
0.5
17
1.0
25
1.5
30
2.0
33
2.5
35
3.0
35
i. Plot a graph of volume versus time for these results.
ii. Explain the shape of the graph during the following two time intervals:
t = 0 to t = 2.0 min and then t = 2.5 and t = 3.0 min by referring
to the volume of H2 (g) produced.
(IEB Paper 2, 2001)
3. Cobalt chloride crystals are dissolved in a beaker containing ethanol and then
a few drops of water are added. After a period of time, the reaction reaches
equilibrium as follows:
−
CoCl42− (blue) +6H2 O ⇔ Co(H2 O)2+
6 (pink) +4Cl
The solution, which is now just blue, is poured into three test tubes. State,
in each case, what colour changes will be observed (if any) if the following are
added in turn to each test tube:
(a) 1 cm3 of distilled water
(b) A few crystals of sodium chloride
317
16.10
16.10
CHAPTER 16. REACTION RATES - GRADE 12
(c) The addition of dilute hydrochloric acid to the third test tube causes the
solution to turn pink. Explain why this occurs.
(IEB Paper 2, 2001)
318
Chapter 17
Electrochemical Reactions - Grade
12
17.1
Introduction
Chapter 15 in Grade 11 discussed oxidation, reduction and redox reactions. Oxidation involves
a loss of electrons and reduction involves a gain of electrons. A redox reaction is a reaction
where both oxidation and reduction take place. What is common to all of these processes is that
they involve a transfer of electrons and a change in the oxidation state of the elements that are
involved.
Exercise: Oxidation and reduction
1. Define the terms oxidation and reduction.
2. In each of the following reactions say whether the iron in the reactants is
oxidised or reduced.
(a)
(b)
(c)
(d)
(e)
F e → F e2+ + 2e−
F e3+ + e− → F e2+
F e2 O3 → F e
F e2+ → F e3+ + e−
F e2 O3 + 2Al → Al2
3. In each of the following equations, say which elements in the reactants are
oxidised and which are reduced.
(a)
(b)
(c)
(d)
CuO(s) + H2 (g) → Cu(s) + H2 O(g)
2N O(g) + 2CO(g) → N2 (g) + 2CO2 (g)
M g(s) + F eSO4 (aq) → M gSO4 (aq) + F e(s)
Zn(s) + 2AgN O3 (aq) → 2Ag + Zn(N O3 )2 (aq)
4. Which one of the substances listed below acts as the oxidising agent in the
following reaction?
(a)
(b)
(c)
(d)
H+
Cr3+
SO2
Cr2 O2−
7
3SO2 + Cr2 O72− + 2H + → 3SO42− + 2Cr3+ + H2 O
319
17.2
CHAPTER 17. ELECTROCHEMICAL REACTIONS - GRADE 12
In Grade 11, an experiment was carried out to see what happened when zinc granules are added
to a solution of copper(II) sulphate. In the experiment, the Cu2+ ions from the copper(II)
sulphate solution were reduced to copper metal, which was then deposited in a layer on the zinc
granules. The zinc atoms were oxidised to form Zn2+ ions in the solution. The half reactions
are as follows:
Cu2+ (aq) + 2e− → Cu(s) (reduction half reaction)
Zn(s) → Zn2+ (aq) + 2e− (oxidation half reaction)
The overall redox reaction is:
Cu2+ (aq) + Zn → Cu(s) + Zn2+ (aq)
There was an increase in the temperature of the reaction when you carried out this experiment.
Is it possible that this heat energy could be converted into electrical energy? In other words, can
we use a chemical reaction where there is an exchange of electrons, to produce electricity? And
if this is possible, what would happen if an electrical current was supplied to cause some type
of chemical reaction to take place?
An electrochemical reaction is a chemical reaction that produces a voltage, and therefore a
flow of electrical current. An electrochemical reaction can also be the reverse of this process, in
other words if an electrical current causes a chemical reaction to take place.
Definition: Electrochemical reaction
If a chemical reaction is caused by an external voltage, or if a voltage is caused by a chemical
reaction, it is an electrochemical reaction.
Electrochemistry is the branch of chemistry that studies these electrochemical reactions. In
this chapter, we will be looking more closely at different types of electrochemical reactions, and
how these can be used in different ways.
17.2
The Galvanic Cell
Activity :: Experiment : Electrochemical reactions
Aim:
To investigate the reactions that take place in a zinc-copper cell
Apparatus:
zinc plate, copper plate, measuring balance, zinc sulphate (ZnSO4 ) solution (1
mol.dm−3 ), copper sulphate (CuSO4 ) solution (1 mol.dm−3 ), two 250 ml beakers,
U-tube, Na2 SO4 solution, cotton wool, ammeter, connecting wire.
Method:
1. Measure the mass of the copper and zinc plates and record your findings.
2. Pour about 200 ml of the zinc sulphate solution into a beaker and put the zinc
plate into it.
3. Pour about 200 ml of the copper sulphate solution into the second beaker and
place the copper plate into it.
4. Fill the U-tube with the Na2 SO4 solution and seal the ends of the tubes with
the cotton wool. This will stop the solution from flowing out when the U-tube
is turned upside down.
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17.2
5. Connect the zinc and copper plates to the ammeter and observe whether the
ammeter records a reading.
6. Place the U-tube so that one end is in the copper sulphate solution and the
other end is in the zinc sulphate solution. Is there a reading on the ammeter?
In which direction is the current flowing?
7. Take the ammeter away and connect the copper and zinc plates to each other
directly using copper wire. Leave to stand for about one day.
8. After a day, remove the two plates and rinse them first with distilled water,
then with alcohol and finally with ether. Dry the plates using a hair dryer.
9. Weigh the zinc and copper plates and record their mass. Has the mass of the
plates changed from the original measurements?
Note: A voltmeter can also be used in place of the ammeter. A voltmeter will
measure the potential difference across the cell.
electron flow
A
+
+
Cu
-
salt
bridge
CuSO4(aq)
Zn
ZnSO4(aq)
Results:
During the experiment, you should have noticed the following:
• When the U-tube containing the Na2 SO4 solution was absent, there was no
reading on the ammeter.
• When the U-tube was connected, a reading was recorded on the ammeter.
• After the plates had been connected directly to each other and left for a day,
there was a change in their mass. The mass of the zinc plate decreased, while
the mass of the copper plate increased.
• The direction of electron flow is from the zinc plate towards the copper plate.
Conclusions:
When a zinc sulphate solution containing a zinc plate is connected by a U-tube
to a copper sulphate solution containing a copper plate, reactions occur in both
solutions. The decrease in mass of the zinc plate suggests that the zinc metal has
been oxidised. The increase in mass of the copper plate suggests that reduction has
occurred here to produce more copper metal. This will be explained in detail below.
17.2.1
Half-cell reactions in the Zn-Cu cell
The experiment above demonstrated a zinc-copper cell. This was made up of a zinc half cell
and a copper half cell.
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Definition: Half cell
A half cell is a structure that consists of a conductive electrode surrounded by a conductive
electrolyte. For example, a zinc half cell could consist of a zinc metal plate (the electrode)
in a zinc sulphate solution (the electrolyte).
How do we explain what has just been observed in the zinc-copper cell?
• Copper plate
At the copper plate, there was an increase in mass. This means that Cu2+ ions from the
copper sulphate solution were deposited onto the plate as atoms of copper metal. The
half-reaction that takes place at the copper plate is:
Cu2+ + 2e− → Cu (Reduction half reaction)
Another shortened way to represent this copper half-cell is Cu2+ /Cu.
• Zinc plate
At the zinc plate, there was a decrease in mass. This means that some of the zinc goes
into solution as Z2+ ions. The electrons remain on the zinc plate, giving it a negative
charge. The half-reaction that takes place at the zinc plate is:
Zn → Zn2+ + 2e− (Oxidation half reaction)
The shortened way to represent the zinc half-cell is Zn/Zn2+ .
The overall reaction is:
Zn + Cu2+ + 2e− → Zn2+ + Cu + 2e− or, if we cancel the electrons:
Zn + Cu2+ → Zn2+ + Cu
For this electrochemical cell, the standard notation is:
Zn|Zn2+ ||Cu2+ |Cu
where
| =
|| =
a phase boundary (solid/aqueous)
the salt bridge
In the notation used above, the oxidation half-reaction at the anode is written on the left, and
the reduction half-reaction at the cathode is written on the right. In the Zn-Cu electrochemical
cell, the direction of current flow in the external circuit is from the zinc electrode (where there
has been a build up of electrons) to the copper electrode.
17.2.2
Components of the Zn-Cu cell
In the zinc-copper cell, the copper and zinc plates are called the electrodes. The electrode
where oxidation occurs is called the anode, and the electrode where reduction takes place is
called the cathode. In the zinc-copper cell, the zinc plate is the anode and the copper plate is
the cathode.
Definition: Electrode
An electrode is an electrical conductor that is used to make contact with a metallic part
of a circuit. The anode is the electrode where oxidation takes place. The cathode is the
electrode where reduction takes place.
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17.2
The zinc sulphate and copper sulphate solutions are called the electrolyte solutions.
Definition: Electrolyte
An electrolyte is a substance that contains free ions and which therefore behaves as an
electrical conductor.
The U-tube also plays a very important role in the cell. In the Zn/Zn2+ half-cell, there is a build
up of positive charge because of the release of electrons through oxidation. In the Cu2+ /Cu halfcell, there is a decrease in the positive charge because electrons are gained through reduction.
This causes a movement of SO2−
4 ions into the beaker where there are too many positive ions,
in order to neutralise the solution. Without this, the flow of electrons in the outer circuit stops
completely. The U-tube is called the salt bridge. The salt bridge acts as a transfer medium
that allows ions to flow through without allowing the different solutions to mix and react.
Definition: Salt bridge
A salt bridge, in electrochemistry, is a laboratory device that is used to connect the oxidation
and reduction half-cells of a galvanic cell.
17.2.3
The Galvanic cell
In the zinc-copper cell the important thing to notice is that the chemical reactions that take place
at the two electrodes cause an electric current to flow through the outer circuit. In this type
of cell, chemical energy is converted to electrical energy. These are called galvanic cells.
The zinc-copper cell is one example of a galvanic cell. A galvanic cell (which is also sometimes
referred to as a voltaic or electrochemical cell) consists of two metals that are connected by
a salt bridge between the individual half-cells. A galvanic cell generates electricity using the
reactions that take place at these two metals, each of which has a different reaction potential.
So what is meant by the ’reaction potential’ of a substance? Every metal has a different half
reaction and different dissolving rates. When two metals with different reaction potentials are
used in a galvanic cell, a potential difference is set up between the two electrodes, and the result
is a flow of current through the wire that connects the electrodes. In the zinc-copper cell, zinc
has a higher reaction potential than copper and therefore dissolves more readily into solution.
The metal ’dissolves’ when it loses electrons to form positive metal ions. These electrons are
then transferred through the connecting wire in the outer circuit.
Definition: Galvanic cell
A galvanic (voltaic) cell is an electrochemical cell that uses a chemical reaction between
two dissimilar electrodes dipped in an electrolyte, to generate an electric current.
teresting It was the Italian physician and anatomist Luigi Galvani who marked the birth
Interesting
Fact
Fact
of electrochemistry by making a link between chemical reactions and electricity.
In 1780, Galvani discovered that when two different metals (copper and zinc for
example) were connected together and then both touched to different parts of a
nerve of a frog leg at the same time, they made the leg contract. He called this
”animal electricity”. While many scientists accepted his ideas, another scientist,
Alessandro Volta, did not. In 1800, because of his professional disagreement over
the galvanic response that had been suggested by Luigi Galvani, Volta developed
the voltaic pile, which was very similar to the galvanic cell. It was the work of
these two men that paved the way for all electrical batteries.
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CHAPTER 17. ELECTROCHEMICAL REACTIONS - GRADE 12
Worked Example 83: Understanding galvanic cells
Question: For the following cell:
Zn|Zn2+ ||Ag + |Ag
1. Give the anode and cathode half-reactions.
2. Write the overall equation for the chemical reaction.
3. Give the direction of the current in the external circuit.
Answer
Step 1 : Identify the oxidation and reduction reactions
In the standard notation format, the oxidation reaction is written on the left and the
reduction reaction on the right. So, in this cell, zinc is oxidised and silver ions are
reduced.
Step 2 : Write the two half reactions
Oxidation half-reaction:
Zn → Zn2+ + 2e−
Reduction half-reaction:
Ag + + e− → Ag
Step 3 : Combine the half-reactions to get the overall equation.
When you combine the two half-reactions, all the reactants must go on the left side
of the equation and the products must go on the right side of the equation. The
overall equation therefore becomes:
Zn + Ag+ + e− → Zn2+ + 2e− + Ag
Note that this equation is not balanced. This will be discussed later in the chapter.
Step 4 : Determine the direction of current flow
A build up of electrons occurs where oxidation takes place. This is at the zinc
electrode. Current will therefore flow from the zinc electrode to the silver electrode.
17.2.4
Uses and applications of the galvanic cell
The principles of the galvanic cell are used to make electrical batteries. In science and technology, a battery is a device that stores chemical energy and makes it available in an electrical
form. Batteries are made of electrochemical devices such as one or more galvanic cells, fuel
cells or flow cells. Batteries have many uses including in torches, electrical appliances (long-life
alkaline batteries), digital cameras (lithium battery), hearing aids (silver-oxide battery), digital
watches (mercury battery) and military applications (thermal battery). Refer to chapter 23 for
more information on batteries.
The galvanic cell can also be used for electroplating. Electroplating occurs when an electrically
conductive object is coated with a layer of metal using electrical current. Sometimes, electroplating is used to give a metal particular properties such as corrosion protection or wear resistance.
At other times, it can be for aesthetic reasons for example in the production of jewellery. This
will be discussed in more detail later in this chapter.
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CHAPTER 17. ELECTROCHEMICAL REACTIONS - GRADE 12
17.3
Exercise: Galvanic cells
1. The following half-reactions take place in an electrochemical cell:
Fe → Fe3+ + 3e−
Fe2+ + 2e− → Fe
(a)
(b)
(c)
(d)
(e)
Which is the oxidation half-reaction?
Which is the reduction half-reaction?
Name one oxidising agent.
Name one reducing agent.
Use standard notation to represent this electrochemical cell.
2. For the following cell:
M g|M g 2+||M n2+ |M n
(a)
(b)
(c)
(d)
(e)
(f)
(g)
Give the cathode half-reaction.
Give the anode half-reaction.
Give the overall equation for the electrochemical cell.
What metals could be used for the electrodes in this electrochemical cell.
Suggest two electrolytes for this electrochemical cell.
In which direction will the current flow?
Draw a simple sketch of the complete cell.
3. For the following cell:
Sn|Sn2+ ||Ag + |Ag
(a)
(b)
(c)
(d)
17.3
Give the cathode half-reaction.
Give the anode half-reaction.
Give the overall equation for the electrochemical cell.
Draw a simple sketch of the complete cell.
The Electrolytic cell
In section 17.2, we saw that a chemical reaction that involves a transfer of electrons, can be used
to produce an electric current. In this section, we are going to see whether the ’reverse’ process
applies. In other words, is it possible to use an electric current to force a particular chemical
reaction to occur, which would otherwise not take place? The answer is ’yes’, and the type of
cell that is used to do this, is called an electrolytic cell.
Definition: Electrolytic cell
An electrolytic cell is a type of cell that uses electricity to drive a non-spontaneous reaction.
An electrolytic cell is activated by applying an electrical potential across the anode and cathode
to force an internal chemical reaction between the ions that are in the electrolyte solution. This
process is called electrolysis.
Definition: Electrolysis
In chemistry and manufacturing, electrolysis is a method of separating bonded elements and
compounds by passing an electric current through them.
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CHAPTER 17. ELECTROCHEMICAL REACTIONS - GRADE 12
Activity :: Demonstration : The movement of coloured ions
A piece of filter paper is soaked in an ammonia-ammonium chloride solution and
placed on a microscope slide. The filter paper is then connected to a supply of
electric current using crocodile clips and connecting wire as shown in the diagram
below. A line of copper chromate solution is placed in the centre of the filter paper.
The colour of this solution is initially green-brown.
negative ions
copper chromate (green brown)
+
positive ions
+
-
-
+ -
+ start of reaction
after 20 minutes
The current is then switched on and allowed to run for about 20 minutes. After
this time, the central coloured band disappears and is replaced by two bands, one
yellow and the other blue, which seem to have separated out from the first band of
copper chromate.
Explanation:
• The cell that is used to supply an electric current sets up a potential difference
across the circuit, so that one of the electrodes is positive and the other is
negative.
• The chromate (CrO2−
4 ) ions in the copper chromate solution are attracted
to the positive electrode, while the Cu2+ ions are attracted to the negative
electrode.
Conclusion:
The movement of ions occurs because the electric current in the outer circuit
sets up a potential difference between the two electrodes.
Similar principles apply in the electrolytic cell, where substances that are made of ions can be
broken down into simpler substances through electrolysis.
17.3.1
The electrolysis of copper sulphate
There are a number of examples of electrolysis. The electrolysis of copper sulphate is just one.
Activity :: Demonstration : The electrolysis of copper sulphate
Two copper electrodes are placed in a solution of blue copper sulphate and are
connected to a source of electrical current as shown in the diagram below. The
current is turned on and the reaction is left for a period of time.
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CHAPTER 17. ELECTROCHEMICAL REACTIONS - GRADE 12
+
+
–
–
positive anode
negative cathode
copper electrode
copper electrode
SO2−
4
17.3
Cu2+
CuSO4 solution
Observations:
• The initial blue colour of the solution remains unchanged.
• It appears that copper has been deposited on one of the electrodes but dissolved
from the other.
Explanation:
• At the negative cathode, positively charged Cu2+ ions are attracted to the
negatively charged electrode. These ions gain electrons and are reduced to
form copper metal, which is deposited on the electrode. The half-reaction that
takes place is as follows:
Cu2+ (aq) + 2e− → Cu(s) (reduction half reaction)
• At the positive anode, copper metal is oxidised to form Cu2+ ions. This is
why it appears that some of the copper has dissolved from the electrode. The
half-reaction that takes place is as follows:
Cu(s) → Cu2+ (aq) + 2e− (oxidation half reaction)
• The amount of copper that is deposited at one electrode is approximately the
same as the amount of copper that is dissolved from the other. The number
of Cu2+ ions in the solution therefore remains almost the same and the blue
colour of the solution is unchanged.
Conclusion:
In this demonstration, an electric current was used to split CuSO4 into its component ions, Cu2+ and SO2−
4 . This process is called electrolysis.
17.3.2
The electrolysis of water
Water can also undergo electrolysis to form hydrogen gas and oxygen gas according to the
following reaction:
2H2 O(l) → 2H2 (g) + O2 (g)
This reaction is very important because hydrogen gas has the potential to be used as an energy source. The electrolytic cell for this reaction consists of two electrodes (normally platinum
metal), submerged in an electrolyte and connected to a source of electric current.
The reduction half-reaction that takes place at the cathode is as follows:
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CHAPTER 17. ELECTROCHEMICAL REACTIONS - GRADE 12
2H2 O(l) + 2e− → H2 (g) + 2OH − (aq)
The oxidation half-reaction that takes place at the anode is as follows:
2H2 O(l) → O2 (g) + 4H + (aq) + 4e−
17.3.3
A comparison of galvanic and electrolytic cells
It should be much clearer now that there are a number of differences between a galvanic and an
electrolytic cell. Some of these differences have been summarised in table 17.1.
Item
Metals used for electrode
Charge of the anode
Charge of the cathode
The electrolyte solution/s
Energy changes
Applications
Galvanic cell
Two metals with different
reaction potentials are used
as electrodes
negative
positive
The electrolyte solutions
are kept separate from one
another, and are connected
only by a salt bridge
Chemical potential energy
from chemical reactions is
converted to electrical energy
Run batteries, electroplating
Electrolytic cell
The same metal can be
used for both the cathode
and the anode
positive
negative
The cathode and anode are
in the same electrolyte
An external supply of electrical energy causes a chemical reaction to occur
Electrolysis e.g. of water,
NaCl
Table 17.1: A comparison of galvanic and electrolytic cells
Exercise: Electrolyis
1. An electrolytic cell consists of two electrodes in a silver chloride (AgCl) solution,
connected to a source of current. A current is passed through the solution and
Ag+ ions are reduced to a silver metal deposit on one of the electrodes.
(a) Give the equation for the reduction half-reaction.
(b) Give the equation for the oxidation half-reacion.
2. Electrolysis takes place in a solution of molten lead bromide (PbBr) to produce
lead atoms.
(a) Draw a simple diagram of the electrolytic cell.
(b) Give equations for the half-reactions that take place at the anode and
cathode, and include these in the diagram.
(c) On your diagram, show the direction in which current flows.
17.4
Standard Electrode Potentials
If a voltmeter is connected in the circuit of an electrochemical cell, a reading is obtained. In
other words, there is a potential difference between the two half cells. In this section, we are
going to look at this in more detail to try to understand more about the electrode potentials
of each of the electrodes in the cell. We are going to break this section down so that you build
up your understanding gradually. Make sure that you understand each subsection fully before
moving on, otherwise it might get confusing!
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CHAPTER 17. ELECTROCHEMICAL REACTIONS - GRADE 12
17.4.1
17.4
The different reactivities of metals
All metals have different reactivities. When metals react, they give away electrons and form
positive ions. But some metals do this more easily than others. Look at the following two half
reactions:
Zn → Zn2+ + 2e−
Cu → Cu2+ + 2e−
Of these two metals, zinc is more reactive and is more likely to give away electrons to form Zn2+
ions in solution, than is copper.
17.4.2
Equilibrium reactions in half cells
Let’s think back to the Zn-Cu electrochemical cell. This cell is made up of two half cells and
the reactions that take place at each of the electrodes are as follows:
Zn → Zn2+ + 2e−
Cu2+ + 2e− → Cu
At the zinc electrode, the zinc metal loses electrons and forms Zn2+ ions. The electrons are
concentrated on the zinc metal while the Zn2+ ions are in solution. But some of the ions will be
attracted back to the negatively charged metal, will gain their electrons again and will form zinc
metal. A dynamic equilibrium is set up between the zinc metal and the Zn2+ ions in solution
when the rate at which ions are leaving the metal is equal to the rate at which they are joining
it again. The situation looks something like the diagram in figure 17.1.
--2+
-
- --
2+
2+
zinc metal
concentration of electrons on metal surface
2+
Zn2+ ions in solution
2+
Figure 17.1: Zinc loses electrons to form positive ions in solution. The electrons accumulate on
the metal surface.
The equilibrium reaction is represented like this:
Zn2+ (aq) + 2e− ⇔ Zn(s)
(NOTE: By convention, the ions are written on the left hand side of the equation)
In the zinc half cell, the equilibrium lies far to the left because the zinc loses electrons easily
to form Zn2+ ions. We can also say that the zinc is oxidised and that it is a strong reducing agent.
At the copper electrode, a similar process takes place. The difference though is that copper is
not as reactive as zinc and so it does not form ions as easily. This means that the build up of
electrons on the copper electrode is less (figure 17.2).
The equilibrium reaction is shown like this:
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17.4
CHAPTER 17. ELECTROCHEMICAL REACTIONS - GRADE 12
-
-
-2+
copper metal
concentration of electrons on metal surface
Cu2+ ions in solution
2+
Figure 17.2: Zinc loses electrons to form positive ions in solution. The electrons accumulate on
the metal surface.
Cu2+ (aq) + 2e− ⇔ Cu(s)
The equation lies far to the right because most of the copper is present as copper metal rather
than as Cu2+ ions. In this half reaction, the Cu2+ ions are reduced.
17.4.3
Measuring electrode potential
If we put the two half cells together, a potential difference is set up in two places in the Zn-Cu
cell:
1. There is a potential difference between the metal and the solution surrounding it because
one is more negative than the other.
2. There is a potential difference between the Zn and Cu electrodes because one is more
negative than the other.
It is the potential difference (recorded as a voltage) between the two electrodes that causes
electrons, and therefore current, to flow from the more negative electrode to the less negative
electrode.
The problem though is that we cannot measure the potential difference (voltage) between a
metal and its surrounding solution in the cell. To do this, we would need to connect a voltmeter
to both the metal and the solution, which is not possible. This means we cannot measure the
exact electrode potential (Eo V) of a particular metal. The electrode potential describes the
ability of a metal to give up electrons. And if the exact electrode potential of each of the
electrodes involved can’t be measured, then it is difficult to calculate the potential difference
between them. But what we can do is to try to describe the electrode potential of a metal
relative to another substance. We need to use a standard reference electrode for this.
17.4.4
The standard hydrogen electrode
Before we look at the standard hydrogen electrode, it may be useful to have some more understanding of the ideas behind a ’reference electrode’. Refer to the Tip box on ’Understanding the
ideas behind a reference electrode’ before you read further.
Important: Understanding the ideas behind a reference electrode
Adapted from www.chemguide.co.uk
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CHAPTER 17. ELECTROCHEMICAL REACTIONS - GRADE 12
17.4
Let’s say that you have a device that you can use to measure heights from some distance away.
You want to use this to find out how tall a particular person is. Unfortunately, you can’t see
their feet because they are standing in long grass. Although you can’t measure their absolute
height, what you can do is to measure their height relative to the post next to them. Let’s say
that person A for example is 15 cm shorter than the height of the post. You could repeat this
for a number of other people (B and C). Person B is 30 cm shorter than the post and person C
is 10 cm taller than the post.
A
B
C
You could summarise your findings as follows:
Person
A
B
C
Height relative to post (cm)
-15
-30
+10
Although you don’t know any of their absolute heights, you can rank them in order, and do some
very simple sums to work out exactly how much taller one is than another. For example, person
C is 25 cm taller than A and 40 cm taller than B.
As mentioned earlier, it is difficult to measure the absolute electrode potential of a particular
substance, but we can use a reference electrode (similar to the ’post’ in the Tip box example)
that we use to calculate relative electrode potentials for these substances. The reference elctrode
that is used is the standard hydrogen electrode (figure 17.3).
Definition: Standard hydrogen electrode
The standard hydrogen electrode is a redox electrode which forms the basis of the scale of
oxidation-reduction potentials. The actual electrode potential of the hydrogen electrode is
estimated to be 4.44 0.02 V at 250 C, but its standard electrode potential is said to be zero
at all temperatures so that it can be used as for comparison with other electrodes. The
hydrogen electrode is based on the following redox half cell:
2H+ (aq) + 2e− → H2 (g)
A standard hydrogen electrode consists of a platinum electrode in a solution containing H+ ions.
The solution (e.g. H2 SO4 ) that contains the H+ ions has a concentration of 1 mol.dm−3 . As
the hydrogen gas bubbles over the platinum electrode, an equilibrium is set up between hydrogen
molecules and hydrogen ions in solution. The reaction is as follows:
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17.4
CHAPTER 17. ELECTROCHEMICAL REACTIONS - GRADE 12
b
H2 gas
(1 x atmosphereric pressure)
c
b
Pt
cb
b
c
c
b
[H3 O+ ]
(1 mol.dm−3 )
25 ◦ C
c
b
Figure 17.3: The standard hydrogen electrode
2H + (aq) + 2e− ⇔ H2 (g)
The position of this equilibrium can change if you change some of the conditions (e.g. concentration, temperature). It is therefore important that the conditions for the standard hydrogen
electrode are standardised as follows: pressure = 100 kPa (1atm); temperature = 298 K (250 C)
and concentration = 1 mol.dm−3 .
In order to use the hydrogen electrode, it needs to be attached to the electrode system that
you are investigating. For example, if you are trying to determine the electrode potential of
copper, you will need to connect the copper half cell to the hydrogen electrode; if you are trying
to determine the electrode potential of zinc, you will need to connect the zinc half cell to the
hydrogen electrode and so on. Let’s look at the examples of zinc and copper in more detail.
1. Zinc
Zinc has a greater tendency than hydrogen to form ions, so if the standard hydrogen
electrode is connected to the zinc half cell, the zinc will be relatively more negative because
the electrons that are released when zinc is oxidised will accumulate on the metal. The
equilibria on the two electrodes are as follows:
Zn2+ (aq) + 2e− ⇔ Zn(s)
2H + (aq) + 2e− ⇔ H2 (g)
In the zinc half-reaction, the equilibrium lies far to the left and in the hydrogen halfreaction, the equilibrium lies far to the right. A simplified representation of the cell is
shown in figure 17.4.
The voltmeter measures the potential difference between the charge on these electrodes. In
this case, the voltmeter would read 0.76 and would show that Zn is the negative electrode
(i.e. it has a relatively higher number of electrons).
2. Copper
Copper has a lower tendency than hydrogen to form ions, so if the standard hydrogen
electrode is connected to the copper half cell, the hydrogen will be relatively more negative.
The equilibria on the two electrodes are as follows:
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CHAPTER 17. ELECTROCHEMICAL REACTIONS - GRADE 12
17.4
V
-
H electrode
(less negative) -
-----
-
Zn electrode with electrons
Figure 17.4: When zinc is connected to the standard hydrogen electrode, relatively few electrons
build up on the platinum (hydrogen) electrode. There are lots of electrons on the zinc electrode.
Cu2+ (aq) + 2e− ⇔ Cu(s)
2H + (aq) + 2e− ⇔ H2 (g)
In the copper half-reaction, the equilibrium lies far to the right and in the hydrogen halfreaction, the equilibrium lies far to the left. A simplified representation of the cell is shown
in figure 17.5.
V
H electrode
-----
-
-
Cu electrode
-
Figure 17.5: When copper is connected to the standard hydrogen electrode, relatively few electrons build up on the copper electrode. There are lots of electrons on the hydrogen electrode.
The voltmeter measures the potential difference between the charge on these electrodes. In
this case, the voltmeter would read 0.34 and would show that Cu is the positive electrode
(i.e. it has a relatively lower number of electrons).
17.4.5
Standard electrode potentials
The voltages recorded earlier when zinc and copper were connected to a standard hydrogen
electrode are in fact the standard electrode potentials for these two metals. It is important
to remember that these are not absolute values, but are potentials that have been measured
relative to the potential of hydrogen if the standard hydrogen electrode is taken to be zero.
Important: Conventions and voltage sign
By convention, the hydrogen electrode is written on the left hand side of the cell. The sign of
the voltage tells you the sign of the metal electrode.
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CHAPTER 17. ELECTROCHEMICAL REACTIONS - GRADE 12
In the examples we used earlier, zinc’s electrode potential is actually -0.76 and copper is +0.34.
So, if a metal has a negative standard electrode potential, it means it forms ions easily. The
more negative the value, the easier it is for that metal to form ions. If a metal has a positive
standard electrode potential, it means it does not form ions easily. This will be explained in more
detail below.
Luckily for us, we do not have to calculate the standard electrode potential for every metal. This
has been done already and the results are recorded in a table of standard electrode potentials
(table 17.2).
A few examples from the table are shown in table 17.3. These will be used to explain some of
the trends in the table of electrode potentials.
Refer to table 17.3 and notice the following trends:
• Metals at the top of series (e.g. Li) have more negative values. This means they ionise
easily, in other words, they release electrons easily. These metals are easily oxidised and
are therefore good reducing agents.
• Metal ions at the bottom of the table are good at picking up electrons. They are easily
reduced and are therefore good oxidising agents.
• The reducing ability (i.e. the ability to act as a reducing agent) of the metals in the table
increases as you move up in the table.
• The oxidising ability of metals increases as you move down in the table.
Worked Example 84: Using the table of Standard Electrode Potentials
Question:
The following half-reactions take place in an electrochemical cell:
Cu2+ + 2e− ⇔ Cu
Ag− + e− ⇔ Ag
1. Which of these reactions will be the oxidation half-reaction in the cell?
2. Which of these reactions will be the reduction half-reaction in the cell?
Answer
Step 5 : Determine the electrode potential for each metal
From the table of standard electrode potentials, the electrode potential for the copper half-reaction is +0.34 V. The electrode potential for the silver half-reaction is
+0.80 V.
Step 6 : Use the electrode potential values to determine which metal is
oxidised and which is reduced
Both values are positive, but silver has a higher positive electrode potential than
copper. This means that silver does not form ions easily, in other words, silver is
more likely to be reduced. Copper is more likely to be oxidised and to form ions more
easily than silver. Copper is the oxidation half-reaction and silver is the reduction
half-reaction.
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CHAPTER 17. ELECTROCHEMICAL REACTIONS - GRADE 12
Half-Reaction
Li+ + e− ⇋ Li
K + + e− ⇋ K
Ba2+ + 2e− ⇋ Ba
Ca2+ + 2e− ⇋ Ca
N a+ + e − ⇋ N a
M g 2+ + 2e− ⇋ M g
M n2+ + 2e− ⇋ M n
2H2O + 2e− ⇋ H2 (g) + 2OH −
Zn2+ + 2e− ⇋ Zn
Cr2+ + 2e− ⇋ Cr
F e2+ + 2e− ⇋ F e
Cr3+ + 3e− ⇋ Cr
Cd2+ + 2e− ⇋ Cd
Co2+ + 2e− ⇋ Co
N i2+ + 2e− ⇋ N i
Sn2+ + 2e− ⇋ Sn
P b2+ + 2e− ⇋ P b
F e3+ + 3e− ⇋ F e
2H + + 2e− ⇋ H2 (g)
S + 2H + + 2e− ⇋ H2 S(g)
Sn4+ + 2e− ⇋ Sn2+
Cu2+ + e− ⇋ Cu+
SO42+ + 4H + + 2e− ⇋ SO2 (g) + 2H2 O
Cu2+ + 2e− ⇋ Cu
2H2 O + O2 + 4e− ⇋ 4OH −
Cu+ + e− ⇋ Cu
I2 + 2e− ⇋ 2I −
O2 (g) + 2H + + 2e− ⇋ H2 O2
F e3+ + e− ⇋ F e2+
N O3− + 2H + + e− ⇋ N O2 (g) + H2 O
Hg 2+ + 2e− ⇋ Hg(l)
Ag + + e− ⇋ Ag
N O3− + 4H + + 3e− ⇋ N O(g) + 2H2 O
Br2 + 2e− ⇋ 2Br−
O2 (g) + 4H + + 4e− ⇋ 2H2 O
M nO2 + 4H + + 2e− ⇋ M n2+ + 2H2 O
Cr2 O72− + 14H + + 6e− ⇋ 2Cr3+ + 7H2 O
Cl2 + 2e− ⇋ 2Cl−
Au3+ + 3e− ⇋ Au
M nO4− + 8H + + 5e− ⇋ M n2+ + 4H2 O
Co3+ + e− ⇋ Co2+
F2 + 2e− ⇋ 2F −
17.4
E0V
-3.04
-2.92
-2.90
-2.87
-2.71
-2.37
-1.18
-0.83
-0.76
-0.74
-0.44
-0.41
-0.40
-0.28
-0.25
-0.14
-0.13
-0.04
0.00
0.14
0.15
0.16
0.17
0.34
0.40
0.52
0.54
0.68
0.77
0.78
0.78
0.80
0.96
1.06
1.23
1.28
1.33
1.36
1.50
1.52
1.82
2.87
Table 17.2: Standard Electrode Potentials
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CHAPTER 17. ELECTROCHEMICAL REACTIONS - GRADE 12
Half-Reaction
Li+ + e− ⇋ Li
Zn2+ + 2e− ⇋ Zn
F e3+ + 3e− ⇋ F e
2H + + 2e− ⇋ H2 (g)
Cu2+ + 2e− ⇋ Cu
Hg 2+ + 2e− ⇋ Hg(l)
Ag + + e− ⇋ Ag
E0V
-3.04
-0.76
-0.04
0.00
0.34
0.78
0.80
Table 17.3: A few examples from the table of standard electrode potentials
Important: Learning to understand the question in a problem.
Before you tackle this problem, make sure you understand exactly what the question is
asking. If magnesium is able to displace silver from a solution of silver nitrate, this means
that magnesium metal will form magnesium ions and the silver ions will become silver metal.
In other words, there will now be silver metal and a solution of magnesium nitrate. This
will only happen if magnesium has a greater tendency than silver to form ions. In other
words, what the question is actually asking is whether magnesium or silver forms ions more
easily.
Worked Example 85: Using the table of Standard Electrode Potentials
Question: Is magnesium able to displace silver from a solution of silver nitrate?
Answer
Step 1 : Determine the half-reactions that would take place if magnesium
were to displace silver nitrate.
The half-reactions are as follows:
M g 2+ + 2e− ⇔ M g
Ag + + e− ⇔ Ag
Step 2 : Use the table of electrode potentials to see which metal forms
ions more easily.
Looking at the electrode potentials for the magnesium and silver reactions:
For the magnesium half-reaction: Eo V = -2.37
For the silver half-reaction: Eo V = 0.80
This means that magnesium is more easily oxidised than silver and the equilibrium in this half-reaction lies to the left. The oxidation reaction will occur
spontaneously in magnesium. Silver is more easily reduced and the equilibrium
lies to the right in this half-reaction. It can be concluded that magnesium will
displace silver from a silver nitrate solution so that there is silver metal and
magnesium ions in the solution.
Exercise: Table of Standard Electrode Potentials
1. In your own words, explain what is meant by the ’electrode potential’ of a
metal.
2. Give the standard electrode potential for each of the following metals:
(a) magnesium
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17.4
(b) lead
(c) nickel
3. Refer to the electrode potentials in table 17.3.
(a)
(b)
(c)
(d)
Which of the metals is most likely to be oxidised?
Which metal is most likely to be reduced?
Which metal is the strongest reducing agent?
In the copper half-reaction, does the equilibrium position for the reaction
lie to the left or to the right? Explain your answer.
(e) In the mercury half-reaction, does the equilibrium position for the reaction
lie to the left or to the right? Explain your answer.
(f) If silver was added to a solution of copper sulphate, would it displace the
copper from the copper sulphate solution? Explain your answer.
4. Use the table of standard electrode potentials to put the following in order from
the strongest oxidising agent to the weakest oxidising agent.
•
•
•
•
Cu2+
MnO−
4
Br2
Zn2+
•
•
•
•
Ca2+ + 2e− → Ca
Cl2 + 2e− → 2Cl
F e3+ + 3e− → F e
I2 + 2e− → 2I −
5. Look at the following half-reactions.
(a) Which substance is the strongest oxidising agent?
(b) Which substance is the strongest reducing agent?
6. Which one of the substances listed below acts as the oxidising agent in the
following reaction?
(a)
(b)
(c)
(d)
H+
Cr3+
SO2
Cr2 O2−
7
2−
+
3+
3SO2 + Cr2 O2−
+ H2 O
7 + 2H → 3SO4 + 2Cr
(IEB Paper 2, 2004)
7. If zinc is added to a solution of magnesium sulphate, will the zinc displace the
magnesium from the solution? Give a detailed explanation for your answer.
17.4.6
Combining half cells
Let’s stay with the example of the zinc and copper half cells. If we combine these cells as we
did earlier in the chapter (section 17.2), the following two equilibria exist:
Zn2+ + 2e− ⇔ Zn(E 0 = −0.76V )
Cu2+ + 2e− ⇔ Cu(E 0 = +0.34V )
We know from demonstrations, and also by looking at the sign of the electrode potential, that
when these two half cells are combined, zinc will be the oxidation half-reaction and copper will be
the reduction half-reaction. A voltmeter connected to this cell will show that the zinc electrode
is more negative than the copper electrode. The reading on the meter will show the potential
difference between the two half cells. This is known as the electromotive force (emf) of the
cell.
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Definition: Electromotive Force (emf)
The emf of a cell is defined as the maximum potential difference between two electrodes or
half cells in a voltaic cell. emf is the electrical driving force of the cell reaction. In other
words, the higher the emf, the stronger the reaction.
Definition: Standard emf (E0cell )
Standard emf is the emf of a voltaic cell operating under standard conditions (i.e. 100 kPa,
concentration = 1 mol.dm−3 and temperature = 298 K). The symbol 0 denotes standard
conditions.
When we want to represent this cell, it is shown as follows:
Zn|Zn2+ (1mol.dm−3 )||Cu2+ (1mol.dm−3 )|Cu
The anode half cell (where oxidation takes place) is always written on the left. The cathode
half cell (where reduction takes place) is always written on the right.
It is important to note that the potential difference across a cell is related to the extent to which
the spontaneous cell reaction has reached equilibrium. In other words, as the reaction proceeds
and the concentration of reactants decreases and the concentration of products increases, the
reaction approaches equilibrium. When equilibrium is reached, the emf of the cell is zero and
the cell is said to be ’flat’. There is no longer a potential difference between the two half cells,
and therefore no more current will flow.
17.4.7
Uses of standard electrode potential
Standard electrode potentials have a number of different uses.
Calculating the emf of an electrochemical cell
To calculate the emf of a cell, you can use any one of the following equations:
E0(cell) = E0 (right) - E0 (left) (’right’ refers to the electrode that is written on the right in
standard cell notation. ’Left’ refers to the half-reaction written on the left in this notation)
E0(cell) = E0 (reduction half reaction) - E0 (oxidation half reaction)
E0(cell) = E0 (oxidising agent) - E0 (reducing agent)
E0(cell) = E0 (cathode) - E0 (anode)
So, for the Zn-Cu cell,
E0(cell) = 0.34 - (-0.76)
= 0.34 + 0.76
= 1.1 V
Worked Example 86: Calculating the emf of a cell
Question: The following reaction takes place:
Cu(s) + Ag + (aq) → Cu2+ (aq) + Ag(s)
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CHAPTER 17. ELECTROCHEMICAL REACTIONS - GRADE 12
1. Represent the cell using standard notation.
2. Calculate the cell potential (emf) of the electrochemical cell.
Answer
Step 1 : Write equations for the two half reactions involved
Cu2+ + 2e− ⇔ Cu (Eo V = 0.16V)
Ag + + e− ⇔ Ag (Eo V = 0.80V)
Step 2 : Determine which reaction takes place at the cathode and
which is the anode reaction
Both half-reactions have positive electrode potentials, but the silver half-reaction
has a higher positive value. In other words, silver does not form ions easily, and
this must be the reduction half-reaction. Copper is the oxidation half-reaction.
Copper is oxidised, therefore this is the anode reaction. Silver is reduced and so
this is the cathode reaction.
Step 3 : Represent the cell using standard notation
Cu|Cu2+ (1mol.dm−3 )||Ag + (1mol.dm−3 )|Ag
Step 4 : Calculate the cell potential
E0(cell) = E0 (cathode) - E0 (anode)
= +0.80 - (+0.34)
= +0.46 V
Worked Example 87: Calculating the emf of a cell
Question: Calculate the cell potential of the electrochemical cell in which the following reaction takes place, and represent the cell using standard notation.
M g(s) + 2H + (aq) → M g2+(aq) + H2 (g)
Answer
Step 1 : Write equations for the two half reactions involved
M g 2+ + 2e− ⇔ M g (Eo V = -2.37)
2H + + 2e− ⇔ H2 (Eo V = 0.00)
Step 2 : Determine which reaction takes place at the cathode and
which is the anode reaction
From the overall equation, it is clear that magnesium is oxidised and hydrogen
ions are reduced in this reaction. Magnesium is therefore the anode reaction and
hydrogen is the cathode reaction.
Step 3 : Represent the cell using standard notation
M g|M g 2+ ||H + |H2
Step 4 : Calculate the cell potential
E0(cell) = E0 (cathode) - E0 (anode)
= 0.00 - (-2.37)
= +2.37 V
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17.4
17.4
CHAPTER 17. ELECTROCHEMICAL REACTIONS - GRADE 12
Predicting whether a reaction will take place spontaneously
Look at the following example to help you to understand how to predict whether a reaction will
take place spontaneously or not.
In the reaction,
P b2+ (aq) + 2Br− (aq) → Br2 (l) + P b(s)
the two half reactions are as follows:
P b2+ + 2e− ⇔ P b (-0.13 V)
Br2 + 2e− ⇔ 2Br− (+1.06 V)
Important: Half cell reactions
You will see that the half reactions are written as they appear in the table of standard electrode
potentials. It may be useful to highlight the reacting substance in each half reaction. In this
case, the reactants are Pb2+ and Br− ions.
Look at the electrode potential for the first half reaction. The negative value shows that lead
loses electrons easily, in other words it is easily oxidised. The reaction would normally proceed
from right to left (i.e. the equilibrium lies to the left), but in the original equation, the opposite
is happening. It is the Pb2+ ions that are being reduced to lead. This part of the reaction is
therefore not spontaneous. The positive electrode potential value for the bromine half-reaction
shows that bromine is more easily reduced, in other words the equilibrium lies to the right. The
spontaneous reaction proceeds from left to right. This is not what is happening in the original
equation and therefore this is also not spontaneous. Overall it is clear then that the reaction will
not proceed spontaneously.
Worked Example 88: Predicting whether a reaction is spontaneous
Question: Will copper react with dilute sulfuric acid (H2 SO4 )? You are given the
following half reactions:
Cu2+ (aq) + 2e− ⇔ Cu(s) (E0 = +0.34 V)
2H + (aq) + 2e− ⇔ H2 (g) (E0 = 0 V)
Answer
Step 5 : For each reaction, look at the electrode potentials and decide in
which direction the equilibrium lies
In the first half reaction, the positive electrode potential means that copper does
not lose electrons easily, in other words it is more easily reduced and the equilibrium
position lies to the right. Another way of saying this is that the spontaneous reaction
is the one that proceeds from left to right, when copper ions are reduced to copper
metal.
In the second half reaction, the spontaneous reaction is from right to left.
Step 6 : Compare the equilibrium positions to the original reaction
What you should notice is that in the original reaction, the reactants are copper
(Cu) and sulfuric acid (2H+ ). During the reaction, the copper is oxidised and the
hydrogen ions are reduced. But from an earlier step, we know that neither of these
half reactions will proceed spontaneously in the direction indicated by the original
reaction. The reaction is therefore not spontaneous.
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CHAPTER 17. ELECTROCHEMICAL REACTIONS - GRADE 12
17.4
Important:
A second method for predicting whether a reaction is spontaneous
Another way of predicting whether a reaction occurs spontaneously, is to look at the sign of the
emf value for the cell. If the emf is positive then the reaction is spontaneous. If the emf is
negative, then the reaction is not spontaneous.
Balancing redox reactions
We will look at this in more detail in the next section.
Exercise: Predicting whether a reaction will take place spontaneously
1. Predict whether the following reaction will take place spontaneously or not.
Show all your working.
2Ag(s) + Cu2+ (aq) → Cu(s) + 2Ag + (aq)
2. Zinc metal reacts with an acid, H+ (aq) to produce hydrogen gas.
(a) Write an equation for the reaction, using the table of electrode potentials.
(b) Predict whether the reaction will take place spontaneously. Show your
working.
3. Four beakers are set up, each of which contains one of the following solutions:
(a)
(b)
(c)
(d)
Mg(NO3 )2
Ba(NO3 )2
Cu(NO3 )2
Al(NO3 )2
Iron is added to each of the beakers. In which beaker will a spontaneous
reaction take place?
4. Which one of the following solutions can be stored in an aluminium container?
(a)
(b)
(c)
(d)
Cu(SO)4
Zn(SO)4
NaCl
Pb(NO3 )2
Exercise: Electrochemical cells and standard electrode potentials
1. An electrochemical cell is made up of a copper electrode in contact with a
copper nitrate solution and an electrode made of an unknown metal M in
contact with a solution of MNO3 . A salt bridge containing a KNO3 solution
joins the two half cells. A voltmeter is connected across the electrodes. Under
standard conditions the reading on the voltmeter is 0.46V.
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17.5
CHAPTER 17. ELECTROCHEMICAL REACTIONS - GRADE 12
V
Cu
Salt bridge (KNO3 )
Cu(NO3 )2 (aq)
M
MNO3 (aq)
The reaction in the copper half cell is given by:
Cu → Cu2+ + 2e−
(a) Write down the standard conditions which apply to this electrochemical
cell.
(b) Identify the metal M. Show calculations.
(c) Use the standard electrode potentials to write down equations for the:
i. cathode half-reaction
ii. anode half-reaction
iii. overall cell reaction
(d) What is the purpose of the salt bridge?
(e) Explain why a KCl solution would not be suitable for use in the salt bridge
in this cell.
(IEB Paper 2, 2004)
2. Calculate the emf for each of the following standard electrochemical cells:
(a)
M g|M g 2+ ||H + |H2
(b)
F e|F e3+ ||F e2+ |F e
(c)
Cr|Cr2+||Cu2+ |Cu
(d)
P b|P b2+ ||Hg 2+ |Hg
3. Given the following two half-reactions:
• F e3+ (aq) + e− ⇔ F e2+ (aq)
• M nO4− (aq) + 8H + (aq) + 5e− ⇔ M n2+ (aq) + 4H2 O(l)
(a) Give the standard electrode potential for each half-reaction.
(b) Which reaction takes place at the cathode and which reaction takes place
at the anode?
(c) Represent the electrochemical cell using standard notation.
(d) Calculate the emf of the cell
17.5
Balancing redox reactions
Half reactions can be used to balance redox reactions. We are going to use some worked examples
to help explain the method.
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CHAPTER 17. ELECTROCHEMICAL REACTIONS - GRADE 12
Worked Example 89: Balancing redox reactions
Question: Magnesium reduces copper (II) oxide to copper. In the process, magnesium is oxidised to magnesium ions. Write a balanced equation for this reaction.
Answer
Step 1 : Write down the unbalanced oxidation half reaction.
M g → M g 2+
Step 2 : Balance the number of atoms on both sides of the equation.
You are allowed to add hydrogen ions (H+ ) and water molecules if the reaction takes
place in an acid medium. If the reaction takes place in a basic medium, you can add
either hydroxide ions (OH− ) or water molecules. In this case, there is one magnesium atom on the left and one on the right, so no additional atoms need to be added.
Step 3 : Once the atoms are balanced, check that the charges balance.
Charges can be balanced by adding electrons to either side. The charge on the left
of the equation is 0, but the charge on the right is +2. Therefore, two electrons
must be added to the right hand side so that the charges balance. The half reaction
is now:
M g → M g 2+ + 2e−
Step 4 : Repeat the above steps, but this time using the reduction half
reaction.
The reduction half reaction is:
Cu2+ → Cu
The atoms balance but the charges don’t. Two electrons must be added to the right
hand side.
Cu2+ + 2e− → Cu
Step 5 : Multiply each half reaction by a suitable number so that the number
of electrons released in the oxidation half reaction is made equal to the
number of electrons that are accepted in the reduction half reaction.
No multiplication is needed because there are two electrons on either side.
Step 6 : Combine the two half reactions to get a final equation for the overall
reaction.
M g + Cu2+ + 2e− → M g 2+ + Cu + 2e− (The electrons on either side cancel
and you get...)
M g + Cu2+ → M g 2+ + Cu
Step 7 : Do a final check to make sure that the equation is balanced
In this case, it is.
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17.5
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CHAPTER 17. ELECTROCHEMICAL REACTIONS - GRADE 12
Worked Example 90: Balancing redox reactions
Question: Chlorine gas oxidises Fe(II) ions to Fe(III) ions. In the process, chlorine
is reduced to chloride ions. Write a balanced equation for this reaction.
Answer
Step 1 : Write down the oxidation half reaction.
F e2+ → F e3+
Step 2 : Balance the number of atoms on both sides of the equation.
There is one iron atom on the left and one on the right, so no additional atoms need
to be added.
Step 3 : Once the atoms are balanced, check that the charges balance.
The charge on the left of the equation is +2, but the charge on the right is +3.
Therefore, one electron must be added to the right hand side so that the charges
balance. The half reaction is now:
F e2+ → F e3+ + e−
Step 4 : Repeat the above steps, but this time using the reduction half
reaction.
The reduction half reaction is:
Cl2 → Cl−
The atoms don’t balance, so we need to multiply the right hand side by two to fix
this. Two electrons must be added to the left hand side to balance the charges.
Cl2 + 2e− → 2Cl−
Step 5 : Multiply each half reaction by a suitable number so that the number
of electrons released in the oxidation half reaction is made equal to the
number of electrons that are accepted in the reduction half reaction.
We need to multiply the oxidation half reaction by two so that the number of electrons
on either side are balanced. This gives:
2F e2+ → 2F e3+ + 2e−
Step 6 : Combine the two half reactions to get a final equation for the overall
reaction.
2F e2+ + Cl2 → 2F e3+ + 2Cl−
Step 7 : Do a final check to make sure that the equation is balanced
The equation is balanced.
Worked Example 91: Balancing redox reactions in an acid medium
Question: The following reaction takes place in an acid medium:
Cr2 O72− + H2 S → Cr3+ + S
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CHAPTER 17. ELECTROCHEMICAL REACTIONS - GRADE 12
Write a balanced equation for this reaction.
Answer
Step 1 : Write down the oxidation half reaction.
Cr2 O72− → Cr3+
Step 2 : Balance the number of atoms on both sides of the equation.
We need to multiply the right side by two so that the number of Cr atoms will
balance. To balance the oxygen atoms, we will need to add water molecules to the
right hand side.
Cr2 O72− → 2Cr3+ + 7H2 O
Now the oxygen atoms balance but the hydrogens don’t. Because the reaction takes
place in an acid medium, we can add hydrogen ions to the left side.
Cr2 O72− + 14H + → 2Cr3+ + 7H2 O
Step 3 : Once the atoms are balanced, check that the charges balance.
The charge on the left of the equation is (-2+14) = +12, but the charge on the
right is +6. Therefore, six electrons must be added to the left hand side so that the
charges balance. The half reaction is now:
Cr2 O72− + 14H + + 6e− → 2Cr3+ + 7H2 O
Step 4 : Repeat the above steps, but this time using the reduction half
reaction.
The reduction half reaction after the charges have been balanced is:
S 2− → S + 2e−
Step 5 : Multiply each half reaction by a suitable number so that the number
of electrons released in the oxidation half reaction is made equal to the
number of electrons that are accepted in the reduction half reaction.
We need to multiply the reduction half reaction by three so that the number of
electrons on either side are balanced. This gives:
3S 2− → 3S + 6e−
Step 6 : Combine the two half reactions to get a final equation for the overall
reaction.
Cr2 O72− + 14H + + 3S 2− → 3S + 2Cr3+ + 7H2 O
Step 7 : Do a final check to make sure that the equation is balanced
Worked Example 92: Balancing redox reactions in an alkaline medium
Question: If ammonia solution is added to a solution that contains cobalt(II) ions, a
complex ion is formed, called the hexaaminecobalt(II) ion (Co(NH3 )2+
6 ). In a chemical reaction with hydrogen peroxide solution, hexaaminecobalt ions are oxidised by
hydrogen peroxide solution to the hexaaminecobalt(III) ion Co(NH3 )3+
6 . Write a
balanced equation for this reaction.
Answer
Step 1 : Write down the oxidation half reaction
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17.5
17.5
CHAPTER 17. ELECTROCHEMICAL REACTIONS - GRADE 12
3+
Co(N H3 )2+
6 → Co(N H3 )6
Step 2 : Balance the number of atoms on both sides of the equation.
The number of atoms are the same on both sides.
Step 3 : Once the atoms are balanced, check that the charges balance.
The charge on the left of the equation is +2, but the charge on the right is +3.
One elctron must be added to the right hand side to balance the charges in the
equation.The half reaction is now:
3+
−
Co(N H3 )2+
6 → Co(N H3 )6 + e
Step 4 : Repeat the above steps, but this time using the reduction half
reaction.
Although you don’t actually know what product is formed when hydrogen peroxide
is reduced, the most logical product is OH− . The reduction half reaction is:
H2 O2 → OH −
After the atoms and charges have been balanced, the final equation for the reduction
half reaction is:
H2 O2 + 2e− → 2OH −
Step 5 : Multiply each half reaction by a suitable number so that the number
of electrons released in the oxidation half reaction is made equal to the
number of electrons that are accepted in the reduction half reaction.
We need to multiply the oxidation half reaction by two so that the number of electrons
on both sides are balanced. This gives:
3+
−
2Co(N H3 )2+
6 → 2Co(N H3 )6 + 2e
Step 6 : Combine the two half reactions to get a final equation for the overall
reaction.
3+
−
2Co(N H3 )2+
6 + H2 O2 → 2Co(N H3 )6 + 2OH
Step 7 : Do a final check to make sure that the equation is balanced
Exercise: Balancing redox reactions
1. Balance the following equations.
(a) HN O3 + P bS → P bSO4 + N O + H2 O
(b) N aI + F e2 (SO4 )3 → I2 + F eSO4 + N a2 SO4
2. Manganate(VII) ions (MnO−
4 ) oxidise hydrogen peroxide (H2 O2 ) to oxygen
gas. The reaction is done in an acid medium. During the reaction, the manganate(VII) ions are reduced to manganese(II) ions (Mn2+ ). Write a balanced
equation for the reaction.
3. Chlorine gas is prepared in the laboratory by adding concentrated hydrochloric
acid to manganese dioxide powder. The mixture is carefully heated.
(a) Write down a balanced equation for the reaction which takes place.
(b) Using standard electrode potentials, show by calculations why this mixture
needs to be heated.
(c) Besides chlorine gas which is formed during the reaction, hydrogen chloride
gas is given off when the conentrated hydrochloric acid is heated. Explain
why the hydrogen chloride gas is removed from the gas mixture when the
gas is bubbled through water.
(IEB Paper 2, 2004)
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CHAPTER 17. ELECTROCHEMICAL REACTIONS - GRADE 12
17.6
4. The following equation can be deduced from the table of standard electrode
potentials:
+
3+
2Cr2 O2−
(aq) + 3O2 (g) + 8H2 O(l) (E0 =
7 (aq) + 16H (aq) → 4Cr
+0.10V)
This equation implies that an acidified solution of aqueous potassium dichromate (orange) should react to form Cr3+ (green). Yet aqueous laboratory
solutions of potassium dichromate remain orange for years. Which ONE of the
following best explains this?
(a)
(b)
(c)
(d)
Laboratory solutions of aqueous potassium dichromate are not acidified
The E0 value for this reaction is only +0.10V
The activation energy is too low
The reaction is non-spontaneous
(IEB Paper 2, 2002)
5. Sulfur dioxide gas can be prepared in the laboratory by heating a mixture of
copper turnings and concentrated sulfuric acid in a suitable flask.
(a) Derive a balanced ionic equation for this reaction using the half-reactions
that take place.
(b) Give the E0 value for the overall reaction.
(c) Explain why it is necessary to heat the reaction mixture.
(d) The sulfur dioxide gas is now bubbled through an aqueous solution of
potassium dichromate. Describe and explain what changes occur during
this process.
(IEB Paper 2, 2002)
17.6
Applications of electrochemistry
Electrochemistry has a number of different uses, particularly in industry. We are going to look
at a few examples.
17.6.1
Electroplating
Electroplating is the process of using electrical current to coat an electrically conductive object
with a thin layer of metal. Mostly, this application is used to deposit a layer of metal that has
some desired property (e.g. abrasion and wear resistance, corrosion protection, improvement of
aesthetic qualities etc.) onto a surface that doesn’t have that property. Electro-refining (also
sometimes called electrowinning is electroplating on a large scale. Electrochemical reactions are
used to deposit pure metals from their ores. One example is the electrorefining of copper.
Copper plays a major role in the electrical reticulation industry as it is very conductive and is
used in electric cables. One of the problems though is that copper must be pure if it is to be
an effective current carrier. One of the methods used to purify copper, is electro-winning. The
copper electro-winning process is as follows:
1. Bars of crude (impure) copper containing other metallic impurities is placed on the anodes.
2. The cathodes are made up of pure copper with few impurities.
3. The electrolyte is a solution of aqueous CuSO4 and H2 SO4 .
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CHAPTER 17. ELECTROCHEMICAL REACTIONS - GRADE 12
4. When current passes through the cell, electrolysis takes place. The impure copper anode
dissolves to form Cu2+ ions in solution. These positive ions are attracted to the negative
cathode, where reduction takes place to produce pure copper metal. The reactions that
take place are as follows:
At the anode:
Cu(s) → Cu2+ (aq) + 2e−
At the cathode:
Cu+2 (aq) + 2e− → Cu(s)
(> 99%purity)
5. The other metal impurities (Zn, Au, Ag, Fe and Pb) do not dissolve and form a solid
sludge at the bottom of the tank or remain in solution in the electrolyte.
+
–
+
–
negative cathode
positive anode
impure copper electrode
pure copper electrode
Cu
2+
Figure 17.6: A simplified diagram to illustrate what happens during the electrowinning of copper
17.6.2
The production of chlorine
Electrolysis can also be used to produce chlorine gas from brine/seawater (NaCl). This is sometimes referred to as the ’Chlor-alkali’ process. The reactions that take place are as follows:
At the anode the reaction is:
2Cl− → Cl2 (g) + 2e−
whereas at the cathode, the following happens:
2N a+ + 2H2 O + 2e− → 2N a+ + 2OH − + H2
The overall reaction is:
2N a+ + 2H2 O + 2Cl− → 2N a+ + 2OH − + H2 + Cl2
Chlorine is a very important chemical. It is used as a bleaching agent, a disinfectant, in solvents,
pharmaceuticals, dyes and even plastics such as polyvinlychloride (PVC).
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CHAPTER 17. ELECTROCHEMICAL REACTIONS - GRADE 12
+
17.7
–
+
–
positive anode
negative cathode
electrode
electrode
Na+
Cl−
NaCl solution
Figure 17.7: The electrolysis of sodium chloride
17.6.3
Extraction of aluminium
Aluminum metal is a commonly used metal in industry where its properties of being both light
and strong can be utilized. It is also used in the manufacture of products such as aeroplanes
and motor cars. The metal is present in deposits of bauxite which is a mixture of silicas, iron
oxides and hydrated alumina (Al2 O3 x H2 O).
Electrolysis can be used to extract aluminum from bauxite. The process described below produces
99% pure aluminum:
1. Aluminum is melted along with cryolite (N a3 AlF6 ) which acts as the electrolyte. Cryolite
helps to lower the melting point and dissolve the ore.
2. The anode carbon rods provide sites for the oxidation of O2− and F − ions. Oxygen and
flourine gas are given off at the anodes and also lead to anode consumption.
3. At the cathode cell lining, the Al3+ ions are reduced and metal aluminum deposits on the
lining.
4. The AlF63− electrolyte is stable and remains in its molten state.
The basic electrolytic reactions involved are as follows: At the cathode:
Al+3 + 3e−
→ Al(s)
(99%purity)
At the anode:
2O2−
→ O2 (g) + 4e−
The overall reaction is as follows:
2Al2 O3
→ 4Al + 3O2
The only problem with this process is that the reaction is endothermic and large amounts of
electricity are needed to drive the reaction. The process is therefore very expensive.
17.7
Summary
• An electrochemical reaction is one where either a chemical reaction produces an external
voltage, or where an external voltage causes a chemical reaction to take place.
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17.7
CHAPTER 17. ELECTROCHEMICAL REACTIONS - GRADE 12
• In a galvanic cell a chemical reaction produces a current in the external circuit. An
example is the zinc-copper cell.
• A galvanic cell has a number of components. It consists of two electrodes, each of
which is placed in a separate beaker in an electrolyte solution. The two electrolytes are
connected by a salt bridge. The electrodes are connected two each other by an external
circuit wire.
• One of the electrodes is the anode, where oxidation takes place. The cathode is the
electrode where reduction takes place.
• In a galvanic cell, the build up of electrons at the anode sets up a potential difference
between the two electrodes, and this causes a current to flow in the external circuit.
• A galvanic cell is therefore an electrochemical cell that uses a chemical reaction between
two dissimilar electrodes dipped in an electrolyte to generate an electric current.
• The standard notation for a galvanic cell such as the zinc-copper cell is as follows:
Zn|Zn2+ ||Cu2+ |Cu
where
| =
|| =
a phase boundary (solid/aqueous)
the salt bridge
• The galvanic cell is used in batteries and in electroplating.
• An electrolytic cell is an electrochemical cell that uses electricity to drive a non-spontaneous
reaction. In an electrolytic cell, electrolysis occurs, which is a process of separating elements and compounds using an electric current.
• One example of an electrolytic cell is the electrolysis of copper sulphate to produce copper
and sulphate ions.
• Different metals have different reaction potentials. The reaction potential of metals (in
other words, their ability to ionise), is recorded in a standard table of electrode potential.
The more negative the value, the greater the tendency of the metal to be oxidised. The
more positive the value, the greater the tendency of the metal to be reduced.
• The values on the standard table of electrode potentials are measured relative to the
standard hydrogen electrode.
• The emf of a cell can be calculated using one of the following equations:
E0(cell) = E0 (right) - E0 (left)
E0(cell) = E0 (reduction half reaction) - E0 (oxidation half reaction)
E0(cell) = E0 (oxidising agent) - E0 (reducing agent)
E0(cell) = E0 (cathode) - E0 (anode)
• It is possible to predict whether a reaction is spontaneous or not, either by looking at the
sign of the cell’s emf or by comparing the electrode potentials of the two half cells.
• It is possible to balance redox equations using the half-reactions that take place.
• There are a number of important applications of electrochemistry. These include electroplating, the production of chlorine and the extraction of aluminium.
Exercise: Summary exercise
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CHAPTER 17. ELECTROCHEMICAL REACTIONS - GRADE 12
1. For each of the following, say whether the statement is true or false. If it is
false, re-write the statement correctly.
(a) The anode in an electrolytic cell has a negative charge.
(b) The reaction 2KClO3 → 2KCl + 3O2 is an example of a redox reaction.
(c) Lead is a stronger oxidising agent than nickel.
2. For each of the following questions, choose the one correct answer.
(a) Which one of the following reactions is a redox reaction?
i. HCl + N aOH → N aCl + H2 O
ii. AgN O3 + N aI → AgI + N aN O3
iii. 2F eCl3 + 2H2 O + SO2 → H2 SO4 + 2HCl + 2F eCl2
iv. BaCl2 + M gSO4 → M gCl2 + BaSO4
(IEB Paper 2, 2003)
(b) Consider the reaction represented by the following equation:
−
−
Br2(l) + 2Iaq
→ 2Braq
+ I2(s)
Which one of the following statements about this reaction is correct?
i. bromine is oxidised
ii. bromine acts as a reducing agent
iii. the iodide ions are oxidised
iv. iodine acts as a reducing agent
(IEB Paper 2, 2002)
(c) The following equations represent two hypothetical half-reactions:
X2 + 2e− ⇔ 2X − (+1.09 V) and
Y + + e− ⇔ Y (-2.80 V)
Which one of the following substances from these half-reactions has the
greatest tendency to donate electrons?
i. X−
ii. X2
iii. Y
iv. Y+
(d) Which one of the following redox reactions will not occur spontaneously
at room temperature?
i. M n + Cu2+ → M n2+ + Cu
ii. Zn + SO42− + 4H + → Zn2+ + SO2 + 2H2 O
iii. F e3+ + 3N O2 + 3H2 O → F e + 3N O3− + 6H +
iv. 5H2 S + 2M nO4− + 6H + → 5S + 2M n2+ + 8H2 O
(e) Which statement is CORRECT for a Zn-Cu galvanic cell that operates
under standard conditions?
i. The concentration of the Zn2+ ions in the zinc half-cell gradually decreases.
ii. The concentration of the Cu2+ ions in the copper half-cell gradually
increases.
iii. Negative ions migrate from the zinc half-cell to the copper half-cell.
iv. The intensity of the colour of the electrolyte in the copper half-cell
gradually decreases.
(DoE Exemplar Paper 2, 2008)
3. In order to investigate the rate at which a reaction proceeds, a learner places a
beaker containing concentrated nitric acid on a sensitive balance. A few pieces
of copper metal are dropped into the nitric acid.
(a) Use the relevant half-reactions from the table of Standard Reduction Potentials to derive the balanced nett ionic equation for the reaction that
takes place in the beaker.
(b) What chemical property of nitric acid is illustrated by this reaction?
(c) List three observations that this learner would make during the investigation.
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17.7
17.7
CHAPTER 17. ELECTROCHEMICAL REACTIONS - GRADE 12
(IEB Paper 2, 2005)
4. The following reaction takes place in an electrochemical cell:
Cu(s) + 2AgN O3 (aq) → Cu(N O3 )2 (aq) + 2Ag(s)
(a) Give an equation for the oxidation half reaction.
(b) Which metal is used as the anode?
(c) Determine the emf of the cell under standard conditions.
(IEB Paper 2, 2003)
5. The nickel-cadmium (NiCad) battery is small and light and is made in a sealed
unit. It is used in portable appliances such as calculators and electric razors.
The following two half reactions occur when electrical energy is produced by
the cell.
Half reaction 1: Cd(s) + 2OH− (aq) → Cd(OH)2 (s) + 2e−
Half reaction 2: NiO(OH)(s) + H2 O(l) + e− → Ni(OH)2 (s) + OH− (aq)
(a) Which half reaction (1 or 2) occurs at the anode? Give a reason for your
answer.
(b) Which substance is oxidised?
(c) Derive a balanced ionic equation for the overall cell reaction for the discharging process.
(d) Use your result above to state in which direction the cell reaction will
proceed (forward or reverse) when the cell is being charged.
(IEB Paper 2, 2001)
6. An electrochemical cell is constructed by placing a lead rod in a porous pot
containing a solution of lead nitrate (see sketch). The porous pot is then placed
in a large aluminium container filled with a solution of aluminium sulphate. The
lead rod is then connected to the aluminium container by a copper wire and
voltmeter as shown.
V
copper wire
lead rod
porous pot
Al2 (SO4 )3 (aq)
Pb(NO3 )2
(aq)
aluminium container
(a) Define the term reduction.
(b) In which direction do electrons flow in the copper wire? (Al to Pb or Pb
to Al)
(c) Write balanced equations for the reactions that take place at...
i. the anode
ii. the cathode
(d) Write a balanced nett ionic equation for the reaction which takes place in
this cell.
(e) What are the two functions of the porous pot?
(f) Calculate the emf of this cell under standard conditions.
(IEB Paper 2, 2005)
352
Part IV
Chemical Systems
353
Chapter 18
The Water Cycle - Grade 10
18.1
Introduction
You may have heard the word ’cycle’ many times before. Think for example of the word ’bicycle’
or the regular ’cycle tests’ that you may have at school. A cycle is a series of events that repeats
itself. In the case of a bicycle, the wheel turns through a full circle before beginning the motion
again, while cycle tests happen regularly, normally every week or every two weeks. Because a
cycle repeats itself, it doesn’t have a beginning or an end.
Our Earth is a closed system. This means that it can exchange energy with its surroundings
(i.e. the rest of the solar system), but no new matter is brought into the system. For this reason,
it is important that all the elements and molecules on Earth are recycled so that they are never
completely used up. In the next two sections, we are going to take a closer look at two cycles
that are very important for life on Earth. They are the water cycle and the nitrogen cycle.
18.2
The importance of water
For many people, it is so easy to take water for granted, and yet life on Earth would not exist
were it not for this extraordinary compound. Not only is it believed that the first forms of life
actually started in water, but most of the cells in living organisms contain between 70% and
95% water. Here in the cells, water acts as a solvent and helps to transport vital materials such
as food and oxygen to where they are needed, and also removes waste products such as carbon
dioxide and ammonia from the body. For many animals and plants, water is their home. Think
for example of fish and amphibians that live either all or part of the time in rivers, dams and the
oceans. In other words, if water did not exist, no life would be possible.
Apart from allowing life to exist, water also has a number of other functions. Water shapes the
landscape around us by wearing away at rocks and also transports and deposits sediments on
floodplains and along coastal regions. Water also plays a very important role in helping to regulate Earth’s climate. We will discuss this again later in the chapter. As humans we use water in
our homes, in industry, in mining, irrigation and even as a source of electricitiy in hydro-electric
schemes. In fact, if we were able to view Earth from space, we would see that almost three
quarters of our planet’s surface is covered in water. It is because of this that Earth is sometimes
called the ’Blue Planet’. Most of this water is stored in the oceans, with the rest found in ice
(e.g. glaciers), groundwater (e.g. boreholes), surface water (e.g. rivers, lakes, estuaries, dams)
and in the atmosphere as clouds and water vapour.
teresting In the search for life on other planets, one of the first things that scientists look
Interesting
Fact
Fact
for is water. However, most planets are either too close to the sun (and therefore
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18.3
CHAPTER 18. THE WATER CYCLE - GRADE 10
too hot) for water to exist in liquid form, or they are too far away and therefore
too cold. So, even if water were to be found, the conditions are unlikely to allow
it to exist in a form that can support the diversity of life that we see on Earth.
18.3
The movement of water through the water cycle
The water cycle is the continuous movement of water over, above, and beneath the Earth’s
surface. As water moves, it changes phase between liquid (water), solid (ice) and gas (water
vapour). It is powered by solar energy and, because it is a cycle, it has no beginning or end.
Definition: The Water Cycle
The water cycle is the continuous circulation of water across the Earth. The water cycle is
driven by solar radiation and it includes the atmosphere, land, surface water and groundwater. As water moves through the cycle, it changes state between liquid, solid, and gas
phases. The actual movement of water from one part of the cycle to another (e.g. from
river to ocean) is the result of processes such as evaporation, precipitation, infiltration and
runoff.
The movement of water through the water cycle is shown in figure 18.1. In the figure, each
process within this cycle is numbered. Each process will be described below.
1. The source of energy
The water cycle is driven by the sun, which provides the heat energy that is needed for
many of the other processes to take place.
2. Evaporation
When water on the earth’s surface is heated by the sun, the average energy of the water
molecules increases and some of the molecules are able to leave the liquid phase and
become water vapour. This is called evaporation. Evaporation is the change of water from
a liquid to a gas as it moves from the ground, or from bodies of water like the ocean,
rivers and dams, into the atmosphere.
3. Transpiration
Transpiration is the evaporation of water from the aerial parts of plants, especially the
leaves but also from the stems, flowers and fruits. This is another way that liquid water
can enter the atmosphere as a gas.
4. Condensation
When evaporation takes place, water vapour rises in the atmosphere and cools as the
altitude (height above the ground) increases. As the temperature drops, the energy of the
water vapour molecules also decreases, until the molecules don’t have enough energy to
stay in the gas phase. At this point, condensation occurs. Condensation is the change of
water from water vapour (gas) into liquid water droplets in the air. Clouds, fog and mist
are all examples of condensation. A cloud is actually a collection of lots and lots of tiny
water droplets. This mostly takes place in the upper atmosphere but can also take place
close to the ground if there is a significant temperature change.
teresting Have you ever tried breathing out on a very cold day? It looks as though
Interesting
Fact
Fact
you are breathing out smoke! The moist air that you breathe out is much
warmer than the air outside your body. As this warm, moist air comes into
356
CHAPTER 18. THE WATER CYCLE - GRADE 10
18.3
1
SUN
Condensation forms
clouds
4
Rain falls onto the
soil, flows into the
rivers or seeps into
the soil
5
Rain falls
directly into
rivers, dams
or the
oceans
7
2
Surface water
Evaporation
3
Some rain seeps into
the soil and becomes
part of the ground water
supply
6
Ground water may feed into
rivers or will eventually lead
into the sea
Figure 18.1: The water cycle
contact with the colder air outside, its temperature drops very quickly and
the water vapour in the air you breathe out condenses. The ’smoke’ that
you see is actually formed in much the same way as clouds form in the upper
atmosphere.
5. Precipitation
Precipitation occurs when water falls back to the earth’s surface in the form of rain or
snow. Rain will fall as soon as a cloud becomes too saturated with water droplets. Snow is
similar to rain, except that it is frozen. Snow only falls if temperatures in the atmosphere
are around freezing. The freeing point of water is 00 C).
6. Infiltration
If precipitation occurs, some of this water will filter into the soil and collect underground.
This is called infiltration. This water may evaporate again from the soil at a later stage,
or the underground water may seep into another water body.
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18.3
CHAPTER 18. THE WATER CYCLE - GRADE 10
7. Surface runoff
This refers to the many ways that water moves across the land. This includes surface runoff
such as when water flows along a road and into a drain, or when water flows straight across
the sand. It also includes channel runoff when water flows in rivers and streams. As it
flows, the water may infiltrate into the ground, evaporate into the air, become stored in
lakes or reservoirs, or be extracted for agricultural or other human uses.
Important: It is important to realise that the water cycle is all about energy exchanges.
The sun is the original energy source. Energy from the sun heats the water and causes
evaporation. This energy is stored in water vapour as latent heat. When the water vapour
condenses again, the latent heat is released, and helps to drive circulation in the atmosphere.
The liquid water falls to earth, and will evaporate again at a later stage. The atmospheric
circulation patterns that occur because of these exchanges of heat are very important in
influencing climate patterns.
Activity :: Experiment : The Water Cycle
Materials:
Tile or piece of plastic (e.g. lid of ice-cream container) to make a hill slope; glass
fish tank with a lid; beaker with ice cubes; lamp; water
Set up a model of the water cycle as follows:
lamp
ice cubes
glass tank
slope
water
1. Lean the plastic against one side so that it creates a ’hill slope’ as shown in the
diagram.
2. Pour water into the bottom of the tank until about a quarter of the hill slope
is covered.
3. Close the fish tank lid.
4. Place the beaker with ice on the lid directly above the hill slope.
5. Turn the lamp on and position it so that it shines over the water.
6. Leave the model like this for 20-30 minutes and then observe what happens.
Make sure that you don’t touch the lamp as it will be very hot!
Observation questions:
1. Which parts of the water cycle can you see taking place in the model?
2. Which parts of the water cycle are not represented in the model?
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CHAPTER 18. THE WATER CYCLE - GRADE 10
18.4
3. Can you think of how those parts that are not shown could be represented?
4. What is the energy source in the model? What would the energy source be in
reality?
5. What do you think the function of the ice is in the beaker?
18.4
The microscopic structure of water
In many ways, water behaves very differently from other liquids. These properties are directly
related to the microscopic structure of water, and more specifically to the shape of the molecule
and its polar nature, and to the bonds that hold water molecules together.
18.4.1
The polar nature of water
Every water molecule is made up of one oxygen atom that is bonded to two hydrogen atoms.
When atoms bond, the nucleus of each atom has an attractive force on the electrons of the other
atoms. This ’pull’ is stronger in some atoms than in others and is called the electronegativity of
the atom. In a water molecule, the oxygen atom has a higher electronegativty than the hydrogen
atoms and therefore attracts the electrons more strongly. The result is that the oxygen atom
has a slightly negative charge and the two hydrogen atoms each have a slightly positive charge.
The water molecule is said to be polar because the electrical charge is not evenly distributed
in the molecule. One part of the molecule has a different charge to other parts. You will learn
more about this in chapter 4.
Oxygen
(slightly negative charge)
Hydrogen
O
(slightly positive charge)
H
H
Hydrogen
(slightly positive charge)
Figure 18.2: Diagrams showing the structure of a water molecule. Each molecule is made up of
two hydrogen atoms that are attached to one oxygen atom.
18.4.2
Hydrogen bonding in water molecules
In every water molecule, the forces that hold the individual atoms together are called intramolecular forces. But there are also forces between different water molecules. These are
called intermolecular forces (figure 18.3). You will learn more about these at a later stage, but
for now it is enough to know that in water, molecules are held together by hydrogen bonds.
Hydrogen bonds are a much stronger type of intermolecular force than those found in many
other substances, and this affects the properties of water.
Important: Intramolecular and intermolecular forces
359
18.5
CHAPTER 18. THE WATER CYCLE - GRADE 10
If you find these terms confusing, remember that ’intra’ means within (i.e. the forces within
a molecule). An introvert is someone who doesn’t express emotions and feelings outwardly.
They tend to be quieter and keep to themselves. ’Inter’ means between (i.e. the forces between
molecules). An international cricket match is a match between two different countries.
intermolecular forces
H
intramolecular forces
O
O
H
O
O
H
O
O
Figure 18.3: Intermolecular and intramolecular forces in water. Note that the diagram on the
left only shows intermolecular forces. The intramolecular forces are between the atoms of each
water molecule.
18.5
The unique properties of water
Because of its polar nature and the strong hydrogen bonds between its molecules, water has
some special properties that are quite different to those of other substances.
1. Absorption of infra-red radiation
The polar nature of the water molecule means that it is able to absorb infra-red radiation
(heat) from the sun. As a result of this, the oceans and other water bodies act as heat
reservoirs, and are able to help moderate the Earth’s climate.
2. Specific heat
Definition: Specific heat
Specific heat is the amount of heat energy that is needed to increase the temperature of a
substance by one degree.
Water has a high specific heat, meaning that a lot of energy must be absorbed by water
before its temperature changes.
Activity :: Demonstration : The high specific heat of water
(a) Pour about 100 ml of water into a glass beaker.
(b) Place the beaker on a stand and heat it over a bunsen burner for about 2
minutes.
(c) After this time, carefully touch the side of the beaker (Make sure you touch
the glass very lightly because it will be very hot and may burn you!). Then
use the end of a finger to test the temperature of the water.
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CHAPTER 18. THE WATER CYCLE - GRADE 10
18.5
What do you notice? Which of the two (glass or water) is the hottest?
You have probably observed this phenomenon if you have boiled water in a pot on the
stove. The metal of the pot heats up very quickly, and can burn your fingers if you touch
it, while the water may take several minutes before its temperature increases even slightly.
How can we explain this in terms of hydrogen bonding? Remember that increasing the
temperature of a substance means that its particles will move more quickly. However,
before they can move faster, the bonds between them must be broken. In the case of
water, these bonds are strong hydrogen bonds, and so a lot of energy is needed just to
break these, before the particles can start moving faster.
It is the high specific heat of water and its ability to absorb infra-red radiation that allows
it to regulate climate. Have you noticed how places that are closer to the sea have less
extreme daily temperatures than those that are inland? During the day, the oceans heat
up slowly, and so the air moving from the oceans across land is cool. Land temperatures
are cooler than they would be if they were further from the sea. At night, the oceans lose
the heat that they have absorbed very slowly, and so sea breezes blowing across the land
are relatively warm. This means that at night, coastal regions are generally slightly warmer
than areas that are further from the sea.
By contrast, places further from the sea experience higher maximum temperatures, and
lower minimum temperatures. In other words, their temperature range is higher than that
for coastal regions. The same principle also applies on a global scale. The large amount of
water across Earth’s surface helps to regulate temperatures by storing infra-red radiation
(heat) from the sun, and then releasing it very slowly so that it never becomes too hot or
too cold, and life is able to exist comfortably. In a similar way, water also helps to keep
the temperature of the internal environment of living organisms relatively constant. This
is very important. In humans, for example, a change in body temperature of only a few
degrees can be deadly.
3. Melting point and boiling point
The melting point of water is 00 C and its boiling point is 1000C. This large difference
between the melting and boiling point is very important because it means that water can
exist as a liquid over a large range of temperatures. The three phases of water are shown
in figure 18.4.
4. High heat of vaporisation
Definition: Heat of vaporisation
Heat of vaporisation is the energy that is needed to change a given quantity of a substance
into a gas.
The strength of the hydrogen bonds between water molecules also means that it has a
high heat of vaporisation. ’Heat of vaporisation’ is the heat energy that is needed to
change water from the liquid to the gas phase. Because the bonds between molecules are
strong, water has to be heated to 1000 C before it changes phase. At this temperature,
the molecules have enough energy to break the bonds that hold the molecules together.
The heat of vaporisation for water is 40.65 kJ/mol. It is very lucky for life on earth that
water does have a high heat of vaporisation. Can you imagine what a problem it would
be if water’s heat of vaporisation was much lower? All the water that makes up the cells
in our bodies would evaporate and most of the water on earth would no longer be able to
exist as a liquid!
5. Less dense solid phase
Another unusual property of water is that its solid phase (ice) is less dense than its liquid
phase. You can observe this if you put ice into a glass of water. The ice doesn’t sink to
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18.5
CHAPTER 18. THE WATER CYCLE - GRADE 10
Gas (water vapour)
co n
eva dens
p o a ti o
rat n
io n
on
a ti
lim o n
s u b a ti
re- blim
su
Liquid
Solid (ice)
freezing
melting
Figure 18.4: Changes in phase of water
the bottom of the glass, but floats on top of the liquid. This phenomenon is also related to
the hydrogen bonds between water molecules. While other materials contract when they
solidify, water expands. The ability of ice to float as it solidifies is a very important factor
in the environment. If ice sank, then eventually all ponds, lakes, and even the oceans would
freeze solid as soon as temperatures dropped below freezing, making life as we know it
impossible on Earth. During summer, only the upper few inches of the ocean would thaw.
Instead, when a deep body of water cools, the floating ice insulates the liquid water below,
preventing it from freezing and allowing life to exist under the frozen surface.
Figure 18.5: Ice cubes floating in water
teresting Antarctica, the ’frozen continent’, has one of the world’s largest and deepest
Interesting
Fact
Fact
freshwater lakes. And this lake is hidden beneath 4 kilometres of ice! Lake
Vostok is 200 km long and 50 km wide. The thick, glacial blanket of ice acts
as an insulator, preventing the water from freezing.
6. Water as a solvent
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CHAPTER 18. THE WATER CYCLE - GRADE 10
18.6
Water is also a very good solvent, meaning that it is easy for other substances to dissolve
in it. It is very seldom, in fact, that we find pure water. Most of the time, the water that
we drink and use has all kinds of substances dissolved in it. It is these that make water
taste different in different areas. So why, then, is it important that water is such a good
solvent? We will look at just a few examples.
• Firstly, think about the animals and plants that live in aquatic environments such
as rivers, dams or in the sea. All of these living organisms either need oxygen for
respiration or carbon dioxide for photosynthesis, or both. How do they get these
gases from the water in which they live? Oxygen and carbon dioxide are just two
of the substances that dissolve easily in water, and this is how plants and animals
obtain the gases that they need to survive. Instead of being available as gases in the
atmosphere, they are present in solution in the surrounding water.
• Secondly, consider the fact that all plants need nitrogen to grow, and that they absorb
this nitrogen from compounds such as nitrates and nitrates that are present in the
soil. The question remains, however, as to how these nitrates and nitrites are able to
be present in the soil at all, when most of the Earth’s nitrogen is in a gaseous form
in the atmosphere. Part of the answer lies in the fact that nitrogen oxides, which
are formed during flashes of lightning, can be dissolved in rainwater and transported
into the soil in this way, to be absorbed by plants. The other part of the answer lies
in the activities of nitrogen-fixing bacteria in the soil, but this is a topic that we will
return to in a later section.
It should be clear now, that water is an amazing compound, and that without its unique properties, life on Earth would definitely not be possible.
Exercise: The properties of water
1. A learner returns home from school on a hot afternoon. In order to get cold
water to drink, she adds ice cubes to a glass of water. She makes the following
observations:
• The ice cubes float in the water.
• After a while the water becomes cold and the ice cubes melt.
(a) What property of ice cubes allows them to float in the water?
(b) Briefly explain why the water gets cold when the ice cubes melt.
(c) Briefly describe how the property you mentioned earlier affects the survival
of aquatic life during winter.
2. Which properties of water allow it to remain in its liquid phase over a large
temperature range? Explain why this is important for life on earth.
18.6
Water conservation
Water is a very precious substance and yet far too often, earth’s water resources are abused and
taken for granted. How many times have you walked past polluted rivers and streams, or seen
the flow of water in a river reduced to almost nothing because of its extraction for industrial and
other uses? And if you were able to test the quality of the water you see, you would probably
be shocked. Often our water resources are contaminated with chemicals such as pesticides and
fertilisers. If water is to continue playing all the important functions that were discussed earlier,
it is vital that we reduce the impact of humans on these resources.
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CHAPTER 18. THE WATER CYCLE - GRADE 10
Activity :: Group work : Human impacts on the water cycle
Read the following extract from an article, entitled ’The Effects of Urbanisation
on the Water Cycle’ by Susan Donaldson, and then answer the questions that follow.
As our communities grow, we notice many visible changes including housing developments, road networks, expansion of services and more. These
changes have an impact on our precious water resources, with pollution
of water being one of many such impacts. To understand these impacts
you will need to have a good knowledge of the water cycle!
It is interesting to note that the oceans contain most of earth’s water
(about 97%). Of the freshwater supplies on earth, 78% is tied up in polar
ice caps and snow, leaving only a very small fraction available for use by
humans. Of the available fresh water, 98% is present as groundwater,
while the remaining 2% is in the form of surface water. Because our
usable water supply is so limited, it is vitally important to protect water
quality. Within the water cycle, there is no ’new’ water ever produced on
the earth. The water we use today has been in existence for billions of
years. The water cycle continually renews and refreshes this finite water
supply.
So how exactly does urbanisation affect the water cycle? The increase
in hard surfaces (e.g. roads, roofs, parking lots) decreases the amount
of water that can soak into the ground. This increases the amount of
surface runoff. The runoff water will collect many of the pollutants that
have accumulated on these surfaces (e.g. oil from cars) and carry them
into other water bodies such as rivers or the ocean. Because there is less
infiltration, peak flows of stormwater runoff are larger and arrive earlier,
increasing the size of urban floods. If groundwater supplies are reduced
enough, this may affect stream flows during dry weather periods because
it is the groundwater that seeps to the surface at these times.
Atmospheric pollution can also have an impact because condensing water
vapour will pick up these pollutants (e.g. SO2 , CO2 and NO2 ) and return
them to earth into other water bodies. However, while the effects of
urbanisation on water quality can be major, these impacts can be reduced
if wise decisions are made during the process of development.
Questions
1. In groups, try to explain...
(a) what is meant by ’urbanisation’
(b) how urbanisation can affect water quality
2. Explain why it is so important to preserve the quality of our water supplies.
3. The article gives some examples of human impacts on water quality. In what
other ways do human activities affect water quality?
4. What do you think some of the consequences of these impacts might be for
humans and other forms of life?
5. Imagine that you are the city manager in your own city or the city closest to
you. What changes would you introduce to try to protect the quality of water
resources in your urban area?
6. What measures could be introduced in rural areas to protect water quality?
Apart from the pollution of water resources, the overuse of water is also a problem. In looking
at the water cycle, it is easy sometimes to think that water is a never-ending resource. In a sense
this is true because water cannot be destroyed. However, the availability of water may vary from
place to place. In South Africa for example, many regions are extremely dry and receive very
little rainfall. The same is true for many other parts of the world, where the scarcity of water
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CHAPTER 18. THE WATER CYCLE - GRADE 10
18.6
is a life and death issue. The present threat of global warming is also likely to affect water
resources. Some climate models suggest that rising temperatures could increase the variability
of climate and decrease rainfall in South Africa. With this in mind, and remembering that South
Africa is already a dry country, it is vitally important that we manage our water use carefully. In
addition to this, the less water there is available, the more likely it is that water quality will also
decrease. A decrease in water quality limits how water can be used and developed.
At present, the demands being placed on South Africa’s water resources are large. Table 18.1
shows the water requirements that were predicted for the year 2000. The figures in the table
were taken from South Africa’s National Water Resource Strategy, produced by the Department
of Water Affairs and Forestry in 2004. In the table, ’rural’ means water for domestic use and
stock watering in rural areas, while ’urban’ means water for domestic, industrial and commercial
use in the urban area. ’Afforestation’ is included because many plantations reduce stream flow
because of the large amounts of water they need to survive.
Table 18.1: The predicted water requirements for various water management areas in South
Africa for 2000 (million m3 /annum)
Water management
area
Irrigation
Urban
Rural
Limpopo
Thukela
Upper Vaal
Upper Orange
Breede
Country total
238
204
114
780
577
7920
34
52
635
126
39
2897
28
31
43
60
11
574
Mining
and bulk
industrial
14
46
173
2
0
755
Power
generation
7
1
80
0
0
297
Afforestation Total
1
0
0
0
6
428
Activity :: Case Study : South Africa’s water requirements
Refer to table 18.1 and then answer the following questions:
1. Which water management area in South Africa has the highest need for water...
(a)
(b)
(c)
(d)
in the mining and industry sector?
for power generation?
in the irrigation sector?
Suggest reasons for each of your answers above.
2. For South Africa as a whole...
(a) Which activity uses the most water?
(b) Which activity uses the least water?
3. Complete the following table, by calculating the percentage (%) that each
activity contributes to the total water requirements in South Africa for the year
2000.
Water use activity
Irrigation
Urban
Rural
Mining and bulk industry
Power generation
Afforestation
% of SA’s total water requirements
365
322
334
1045
968
633
12871
18.7
CHAPTER 18. THE WATER CYCLE - GRADE 10
Table 18.2: The available water yield in South Africa in 2000 for various water management
areas (million m3 /annum)
Water management Surface
Ground Irrigation Urban
Mining Total loarea
water
and
cal yield
bulk
industrial
Limpopo
160
98
8
15
0
281
Thukela
666
15
23
24
9
737
Upper Vaal
598
32
11
343
146
1130
Upper Orange
4311
65
34
37
0
4447
Breede
687
109
54
16
0
866
Country total
10240
1088
675
970
254
13227
Now look at table 18.2, which shows the amount of water available in South Africa during 2000.
In the table, ’usable return flow’ means the amount of water that can be reused after it has been
used for irrigation, urban or mining.
Activity :: Case Study : Water conservation
Refer to table 18.2 and then answer the following questions:
1. Explain what is meant by...
(a) surface water
(b) ground water
2. Which water management area has the...
(a)
(b)
(c)
(d)
lowest surface water yield?
highest surface water yield?
lowest total yield?
highest total yield?
3. Look at the country’s total water requirements for 2000 and the total available
yield.
(a) Calculate what percentage of the country’s water yield is already being
used up.
(b) Do you think that the country’s total water requirements will increase or
decrease in the coming years? Give a reason for your answer.
4. South Africa is already placing a huge strain on existing water resources. In
groups of 3-4, discuss ways that the country’s demand for water could be
reduced. Present your ideas to the rest of the class for discussion.
18.7
Summary
• Water is critical for the survival of life on Earth. It is an important part of the cells of
living organisms and is used by humans in homes, industry, mining and agriculture.
• Water moves between the land and sky in the water cycle. The water cycle describes
the changes in phase that take place in water as it circulates across the Earth. The water
cycle is driven by solar radiation.
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CHAPTER 18. THE WATER CYCLE - GRADE 10
18.7
• Some of the important processes that form part of the water cycle are evaporation, transpiration, condensation, precipitation, infiltration and surface runoff. Together these processes
ensure that water is cycled between the land and sky.
• It is the microscopic structure of water that determines its unique properties.
• Water molecules are polar and are held together by hydrogen bonds. These characteristics
affect the properties of water.
• Some of the unique properties of water include its ability to absorb infra-red radiation, its
high specific heat, high heat of vaporisation and the fact that the solid phase of water is
less dense that its liquid phase.
• These properties of water help it to sustain life on Earth by moderating climate, regulating
the internal environment of living organisms and allowing liquid water to exist below ice,
even if temperatures are below zero.
• Water is also a good solvent. This property means that it is a good transport medium
in the cells of living organisms, and that it can dissolve gases and other compounds that
may be needed by aquatic plants and animals.
• Human activities threaten the quality of water resources through pollution and altered
runoff patterns.
• As human populations grow, there is a greater demand for water. In many areas, this
demand exceeds the amount of water available for use. Managing water wisely is important
in ensuring that there will always be water available both for human use, and to maintain
natural ecosystems.
Exercise: Summary Exercise
1. Give a word or term for each of the following phrases:
(a)
(b)
(c)
(d)
The
The
The
The
continuous circulation of water across the earth.
change in phase of water from gas to liquid.
movement of water across a land surface.
temperature at which water changes from liquid to gas.
2. In each of the following multiple choice questions, choose the one correct answer
from the list provided.
(a) Many of the unique properties of water (e.g. its high specific heat and
high boiling point) are due to:
i. strong covalent bonds between the hydrogen and oxygen atoms in each
water molecule
ii. the equal distribution of charge in a water molecule
iii. strong hydrogen bonds between water molecules
iv. the linear arrangement of atoms in a water molecule
(b) Which of the following statements is false?
i. Most of the water on earth is in the oceans.
ii. The hardening of surfaces in urban areas results in increased surface
runoff.
iii. Water conservation is important because water cannot be recycled.
iv. Irrigation is one of the largest water users in South Africa.
3. The sketch below shows a process that leads to rainfall in town X. The town
has been relying only on rainfall for its water supply because it has no access
to rivers or tap water. A group of people told the community that they will
never run out of rainwater because it will never stop raining.
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18.7
CHAPTER 18. THE WATER CYCLE - GRADE 10
Cloud
P2
P1
Town X
Sea
(a) List the processes labelled P1 and P2 that lead to rainfall in town X.
(b) Is this group of people correct in saying that town X will never run out of
rainwater? Justify your answer using the sketch.
Recently, the amount of rainwater has decreased significantly. Various
reasons have been given to explain the drought. Some of the community
members are blaming this group who told them that it will never stop
raining.
(c) What scientific arguments can you use to convince the community members that this group of people should not be blamed for the drought?
(d) What possible strategies can the community leaders adopt to ensure that
they have a regular supply of water.
368
Chapter 19
Global Cycles: The Nitrogen Cycle
- Grade 10
19.1
Introduction
The earth’s atmosphere is made up of about 78% nitrogen, making it the largest pool of this
gas. Nitrogen is essential for many biological processes. It is in all amino acids, proteins and
nucleic acids. As you will see in a later chapter, these compounds are needed to build tissues,
transport substances around the body, and control what happens in living organisms. In plants,
much of the nitrogen is used in chlorophyll molecules which are needed for photosynthesis and
growth.
So, if nitrogen is so essential for life, how does it go from being a gas in the atmosphere to being
part of living organisms such as plants and animals? The problem with nitrogen is that it is an
’inert’ gas, which means that it is unavailable to living organisms in its gaseous form. This is
because of the strong triple bond between its atoms that makes it difficult to break. Something
needs to happen to the nitrogen gas to change it into a form that it can be used. And at some
later stage, these new compounds must be converted back into nitrogen gas so that the amount
of nitrogen in the atmosphere stays the same. This process of changing nitrogen into different
forms is called the nitrogen cycle (figure 19.1).
Definition: The nitrogen cycle
The nitrogen cycle is a biogeochemical cycle that describes how nitrogen and nitrogencontaining compounds are changed in nature.
Very broadly, the nitrogen cycle is made up of the following processes:
• Nitrogen fixation - The process of converting inert nitrogen gas into more useable nitrogen
compounds such as ammonia.
• Nitrification - The conversion of ammonia into nitrites and then into nitrates, which can
be absorbed and used by plants.
• Denitrification - The conversion of nitrates back into nitrogen gas in the atmosphere.
We are going to look at each of these processes in more detail.
19.2
Nitrogen fixation
Nitrogen fixation is needed to change gaseous nitrogen into forms such as ammonia that are more
useful to living organisms. Some fixation occurs in lightning strikes and in industrial processes,
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19.2
CHAPTER 19. GLOBAL CYCLES: THE NITROGEN CYCLE - GRADE 10
Nitrogen in the Atmosphere
Industrial
fixation
Nitrogen fixation
by bacteria
Animals obtain
nitrates from plants
Decomposers
e.g. bacteria
Ammonia
(NH3-)
Nitrification by
nitrifying bacteria
Plant consumption
Atmosphere
Soil
Denitrification returns nitrogen to the atmosphere
Lightning
fixation
Nitrites
(NO-2)
Nitrification
Nitrates
(NO-3)
Figure 19.1: A simplified diagram of the nitrogen cycle
but most fixation is done by different types of bacteria living either in the soil or in parts of the
plants.
1. Biological fixation
Some bacteria are able to fix nitrogen. They use an enzyme called nitrogenase to combine
gaseous nitrogen with hydrogen to form ammonia. The bacteria then use some of this
ammonia to produce their own organic compounds, while what is left of the ammonia
becomes available in the soil.
Some of these bacteria are free-living, in other words they live in the soil. Others live in
the root nodules of legumes (e.g. soy, peas and beans). Here they form a mutualistic
relationship with the plant. The bacteria get carbohydrates (food) from the plant and,
in exchange, produce ammonia which can be converted into nitrogen compounds that are
essential for the survival of the plant. In nutrient-poor soils, planting lots of legumes can
help to enrich the soil with nitrogen compounds.
A simplified equation for biological nitrogen fixation is:
N2 + 8H + + 8e− → 2N H3 + H2
Energy is used in the process, but this is not shown in the above equation.
Another important source of ammonia in the soil is decomposition. When animals and
plants die, the nitrogen compounds that were present in them are broken down and converted into ammonia. This process is carried out by decomposition bacteria and fungi in
the soil.
2. Industrial nitrogen fixation
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CHAPTER 19. GLOBAL CYCLES: THE NITROGEN CYCLE - GRADE 10
19.3
In the Haber-Bosch process, nitrogen (N2 ) is converted together with hydrogen gas (H2 )
into ammonia (NH3 ) fertiliser. This is an artificial process.
3. Lightning
In the atmosphere, lightning and photons are important in the reaction between nitrogen
(N2 ) and oxygen (O2 ) to form nitric oxide (NO) and then nitrates.
teresting It is interesting to note that by cultivating legumes, using the Haber-Bosch
Interesting
Fact
Fact
process to manufacture chemical fertilisers and increasing pollution from vehicles
and industry, humans have more than doubled the amount of nitrogen that would
normally be changed from nitrogen gas into a biologically useful form. This has
serious environmental consequences.
19.3
Nitrification
Nitrification involves two biological oxidation reactions: firstly, the oxidation of ammonia with
oxygen to form nitrite (NO−
2 ) and secondly the oxidation of these nitrites into nitrates.
1. N H3 + O2 → N O2− + 3H + + 2e− (production of nitrites)
2. N O2− + H2 O → N O3− + 2H + + 2e− (production of nitrates)
Nitrification is an important step in the nitrogen cycle in soil because it converts the ammonia
(from the nitrogen fixing part of the cycle) into nitrates, which are easily absorbed by the roots
of plants. This absorption of nitrates by plants is called assimilation. Once the nitrates have
been assimilated by the plants, they become part of the plants’ proteins. These plant proteins
are then available to be eaten by animals. In other words, animals (including humans) obtain
their own nitrogen by feeding on plants. Nitrification is performed by bacteria in the soil, called
nitrifying bacteria.
Activity :: Case Study : Nitrates in drinking water
Read the information below and then carry out your own research to help you
answer the questions that follow.
The negatively charged nitrate ion is not held onto soil particles and so can
be easily washed out of the soil. This is called leaching. In this way, valuable
nitrogen can be lost from the soil, reducing the soil’s fertility. The nitrates can
then accumulate in groundwater, and eventually in drinking water. There are strict
regulations that control how much nitrate can be present in drinking water, because
nitrates can be reduced to highly reactive nitrites by microorganisms in the gut.
Nitrites are absorbed from the gut and bind to haemoglobin (the pigment in blood
that helps to transport oxygen around the body). This reduces the ability of the
haemoglobin to carry oxygen. In young babies this can lead to respiratory distress,
a condition known as ”blue baby syndrome”.
1. How is nitrate concentration in water measured?
2. What concentration of nitrates in drinking water is considered acceptable? You
can use drinking water standards for any part of the world, if you can’t find any
for South Africa.
3. What is ’blue baby syndrome’ and what are the symptoms of the disease?
371
19.4
19.4
CHAPTER 19. GLOBAL CYCLES: THE NITROGEN CYCLE - GRADE 10
Denitrification
Denitrification is the process of reducing nitrate and nitrite into gaseous nitrogen. The process
is carried out by denitrification bacteria. The nitrogen that is produced is returned to the atmosphere to complete the nitrogen cycle.
The equation for the reaction is:
2N O3− + 10e− + 12H + → N2 + 6H2 O
19.5
Human Influences on the Nitrogen Cycle
Humans have contributed significantly to the nitrogen cycle in a number of ways.
• Both artificial fertilisation and the planting of nitrogen fixing crops, increase the amount
of nitrogen in the soil. In some ways this has positive effects because it increases the fertility
of the soil, and means that agricultural productivity is high. On the other hand, however, if
there is too much nitrogen in the soil, it can run off into nearby water courses such as rivers,
or can become part of the groundwater supply as we mentioned earlier. Increased nitrogen
in rivers and dams can lead to a problem called eutrophication. Eutrophication is a process
where water bodies such as rivers, estuaries, dams and slow-moving streams receive excess
nutrients (e.g. nitrogen and phosphorus compounds) that stimulate excessive plant growth.
Sometimes this can cause certain plant species to be favoured over the others and one
species may ’take over’ the ecosystem, resulting in a decrease in plant diversity. This
is called a ’bloom’. Eutrophication also affects water quality. When the plants die and
decompose, large amounts of oxygen are used up and this can cause other animals in the
water to die.
Activity :: Case Study : Fertiliser use in South Africa
Refer to the data table below, which shows the average fertiliser use (in
kilograms per hectare or kg/ha) over a number of years for South Africa and
the world. Then answer the questions that follow:
SA
World
1965
27.9
34.0
1970
42.2
48.9
1975
57.7
63.9
1980
80.3
80.6
1985
66.6
86.7
1990
54.9
90.9
1995
48.5
84.9
2000
47.1
88.2
2002
61.4
91.9
1. On the same set of axes, draw two line graphs to show how fertiliser use
has changed in SA and the world between 1965 and 2002.
2. Describe the trend you see for...
(a) the world
(b) South Africa
3. Suggest a reason why the world’s fertiliser use has changed in this way over
time.
4. Do you see the same pattern for South Africa?
5. Try to suggest a reason for the differences you see in the fertiliser use data
for South Africa.
6. One of the problems with increased fertiliser use is that there is a greater
chance of nutrient runoff into rivers and dams, and therefore a greater
danger of eutrophication. In groups of 5-6, discuss the following questions:
(a) What could farmers do to try to reduce the risk of nutrient runoff from
fields into water systems? Try to think of at least 3 different strategies
that they could use.
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CHAPTER 19. GLOBAL CYCLES: THE NITROGEN CYCLE - GRADE 10
19.6
(b) Imagine you are going to give a presentation on eutrophication to a
group of farmers who know nothing about it. How will you educate
them about the dangers? How will you convince them that it is in their
interests to change their farming practices? Present your ideas to the
class.
• Atmospheric pollution is another problem. The main culprits are nitrous oxide (N2 O),
nitric oxide (NO) and nitrogen dioxide (NO2 ). Most of these gases result either from emissions from agricultural soils (and particularly artificial fertilisers), or from the combustion
of fossil fuels in industry or motor vehicles. The combustion (burning) of nitrogen-bearing
fuels such as coal and oil releases this nitrogen as N2 or NO gases. Both NO2 and NO can
combine with water droplets in the atmosphere to form acid rain. Furthermore, both NO
and NO2 contribute to the depletion of the ozone layer and some are greenhouse gases.
In high concentrations these gases can contribute towards global warming.
19.6
The industrial fixation of nitrogen
A number of industrial processes are able to fix nitrogen into different compounds and then
convert these compounds into fertilisers. In the descriptions below, you will see how atmospheric
nitrogen is fixed to produce ammonia, how ammonia is then reacted with oxygen to form nitric
acid and how nitric acid and ammonia are then used to produce the fertiliser, ammonium nitrate.
• Preparation of ammonia (NH3 )
The industrial preparation of ammonia is known as the Haber-Bosch process. At a high
pressure and a temperature of approximately 5000 C, and in the presence of a suitable
catalyst (usually iron), nitrogen and hydrogen react according to the following equation:
N2 + 3H2 → 2N H3
Ammonia is used in the preparation of artficial fertilisers such as (NH4 )2 SO4 and is also
used in cleaning agents and cooling installations.
teresting Fritz Haber and Carl Bosch were the two men responsible for developing
Interesting
Fact
Fact
the Haber-Bosch process. In 1918, Haber was awarded the Nobel Prize in
Chemistry for his work. The Haber-Bosch process was a milestone in industrial chemistry because it meant that nitrogenous fertilisers were cheaper
and much more easily available. At the time, this was very important in
providing food for the growing human population.
Haber also played a major role in the development of chemical warfare in
World War I. Part of this work included the development of gas masks with
absorbent filters. He also led the teams that developed chlorine gas and
other deadly gases for use in trench warfare. His wife, Clara Immerwahr,
also a chemist, opposed his work on poison gas and committed suicide with
his service weapon in their garden. During the 1920s, scientists working at
his institute also developed the cyanide gas formulation Zyklon B, which
was used as an insecticide and also later, after he left the programme, in the
Nazi extermination camps.
Haber was Jewish by birth, but converted from Judaism in order to be more
accepted in Germany. Despite this, he was forced to leave the country in
1933 because he was Jewish ’by definition’ (his mother was Jewish). He died
in 1934 at the age of 65. Many members of his extended family died in the
Nazi concentration camps, possibly gassed by Zyklon B.
373
19.7
CHAPTER 19. GLOBAL CYCLES: THE NITROGEN CYCLE - GRADE 10
• Preparation of nitric acid (HNO3 )
Nitric acid is used to prepare fertilisers and explosives. The industrial preparation of nitric
acid is known as the Ostwald process. The Ostwald process involves the conversion of
ammonia into nitric acid in various stages:
Firstly, ammonia is heated with oxygen in the presence of a platinum catalyst to form nitric
oxide and water.
4N H3 (g) + 5O2 (g) → 4N O(g) + 6H2 O(g)
Secondly, nitric oxide reacts with oxygen to form nitrogen dioxide. This gas is then readily
absorbed by the water to produce nitric acid. A portion of nitrogen dioxide is reduced back
to nitric oxide.
2N O(g) + O2 (g) → 2N O2 (g)
3N O2 (g) + H2 O(l) → 2HN O3 (aq) + N O(g)
The NO is recycled, and the acid is concentrated to the required strength by a process
called distillation.
• Preparation of ammonium nitrate
Ammonium nitrate is used as a fertiliser, as an explosive and also in the preparation of
’laughing gas’ which is used as an anaesthetic. Ammonium nitrate is prepared by reacting
ammonia with nitric acid:
N H3 + HN O3 → N H4 N O3
Activity :: Debate : Fertiliser use
Divide the class into two groups to debate the following topic:
Increasing the use of artificial fertilisers is the best solution to meet the growing
food needs of the world’s human population.
One group should take the position of agreeing with the statement, and the other
should disagree. In your groups, discuss reasons why you have the opinion that you
do, and record some notes of your discussion. Your teacher will then explain to you
how to proceed with the debate.
19.7
Summary
• Nitrogen is essential for life on earth, since it forms part of amino acids, proteins and
nucleic acids.
• The atmosphere is composed mostly of nitrogen gas, but the gas is inert, meaning that
it is not available to living organisms in its gaseous form.
• The nitrogen cycle describes how nitrogen and nitrogen-containing compounds are changed
into different forms in nature.
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CHAPTER 19. GLOBAL CYCLES: THE NITROGEN CYCLE - GRADE 10
19.7
• The nitrogen cycle consists of three major processes: nitrogen fixation, nitrification and
denitrification.
• Nitrogen fixation is the conversion of atmospheric nitrogen into compounds such as
ammonia, that are more easily used.
• Nitrogen can be fixed biologically through the actions of bacteria, industrially through
the Haber-Bosch process or by lightning.
• Nitrification converts ammonia into nitrites and nitrates, which can be easily assimilated
by plants.
• Denitrification converts nitrites and nitrates back into gaseous nitrogen to complete the
nitrogen cycle.
• Humans have had a number of impacts on the nitrogen cycle. The production of artificial
fertilisers for example, means that there is a greater chance of runoff into water systems.
In some cases, eutrophication may occur.
• Eutrophication is the enrichment of water systems with excess nutrients, which may
stimulate excessive plant growth at the expense of other parts of the ecosystem.
• Many nitrogen gases such as NO, N2 O and NO2 are released by agricultural soils and
artificial fertilisers. These gases may combine with water vapour in the atmosphere and
result in acid rain. Some of these gases are also greenhouse gases and may contribute
towards global warming.
• A number of industrial processes are used to produce articifical fertilisers.
• The Haber-Bosch process converts atmsopheric nitrogen into ammonia.
• The Ostwald process reacts ammonia with oxygen to produce nitric acid, which is used
in the preparation of fertilisers and explosives.
• If ammonia and nitric acid react, the product is ammonium nitrate, which is used as a
fertiliser and as an explosive.
Exercise: Summary Exercise
1. Look at the diagram and the descriptions of the nitrogen cycle earlier in the
chapter:
(a) Would you describe the changes that take place in the nitrogen cycle as
chemical or physical changes? Explain your answer.
(b) Are the changes that take place in the water cycle physical or chemical
changes? Explain your answer.
2. Explain what is meant by each of the following terms:
(a) nitrogen fixing
(b) fertiliser
(c) eutrophication
3. Explain why the fixing of atmospheric nitrogen is so important for the survival
of life on earth.
4. Refer to the diagram below and then answer the questions that follow:
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19.7
CHAPTER 19. GLOBAL CYCLES: THE NITROGEN CYCLE - GRADE 10
N2
(1)
(2)
(3)
(4)
(5)
(a) Explain the role of decomposers in the nitrogen cycle.
(b) If the process taking place at (3) is nitrification, then label the processes
at (1) and (5).
(c) Identify the nitrogen products at (2) and (4).
(d) On the diagram, indicate the type of bacteria that are involved in each
stage of the nitrogen cycle.
(e) In industry, what process is used to produce the compound at 2?
(f) Does the diagram above show a ’cycle’ ? Explain your answer.
5. NO and NO2 are both nitrogen compounds:
(a) Explain how each of these compounds is formed?
(b) What effect does each of these compounds have in the environment?
6. There are a number of arguments both ’for’ and ’against’ the use of artificial
fertilisers. Draw a table to summarise the advantages and disadvantages of
their use.
376
Chapter 20
The Hydrosphere - Grade 10
20.1
Introduction
As far as we know, the Earth we live on is the only planet that is able to support life. Among
other things, Earth is just the right distance from the sun to have temperatures that are suitable
for life to exist. Also, the Earth’s atmosphere has exactly the right type of gases in the right
amounts for life to survive. Our planet also has water on its surface, which is something very
unique. In fact, Earth is often called the ’Blue Planet’ because most of it is covered in water.
This water is made up of freshwater in rivers and lakes, the saltwater of the oceans and estuaries,
groundwater and water vapour. Together, all these water bodies are called the hydrosphere.
20.2
Interactions of the hydrosphere
It is important to realise that the hydrosphere interacts with other global systems, including the
atmosphere, lithosphere and biosphere.
• Atmosphere
When water is heated (e.g. by energy from the sun), it evaporates and forms water vapour.
When water vapour cools again, it condenses to form liquid water which eventually returns
to the surface by precipitation e.g. rain or snow. This cycle of water moving through the
atmosphere, and the energy changes that accompany it, is what drives weather patterns
on earth.
• Lithosphere
In the lithosphere (the ocean and continental crust at the Earth’s surface), water is an
important weathering agent, which means that it helps to break rock down into rock
fragments and then soil. These fragments may then be transported by water to another
place, where they are deposited. This is called erosion. These two process i.e. weathering
and erosion, help to shape the earth’s surface. You can see this for example in rivers.
In the upper streams, rocks are eroded and sediments are transported down the river and
deposited on the wide flood plains lower down. On a bigger scale, river valleys in mountains
have been carved out by the action of water, and cliffs and caves on rocky beach coastlines,
are also the result of weathering and erosion by water.
• Biosphere
In the biosphere, land plants absorb water through their roots and then transport this
through their vascular (transport) system to stems and leaves. This water is needed in
photosynthesis, the food production process in plants. Transpiration (evaporation of water
from the leaf surface) then returns water back to the atmosphere.
377
20.3
20.3
CHAPTER 20. THE HYDROSPHERE - GRADE 10
Exploring the Hydrosphere
The large amount of water on our planet is something quite unique. In fact, about 71% of
the earth is covered by water. Of this, almost 97% is found in the oceans as saltwater, about
2.2% occurs as a solid in ice sheets, while the remaining amount (less than 1%) is available as
freshwater. So from a human perspective, despite the vast amount of water on the planet, only a
very small amount is actually available for human consumption (e.g. drinking water). Before we
go on to look more closely at the chemistry of the hydrosphere, we are going to spend some time
exploring a part of the hydrosphere, in order to start appreciating what a complex and beautiful
part of the world it is.
Activity :: Investigation : Investigating the hydrosphere
1. Choosing a study site:
For this exercise, you can choose any part of the hydrosphere that you would
like to explore. This may be a rock pool, a lake, river, wetland or even just a
small pond. The guidelines below will apply best to a river investigation, but
you can ask similar questions and gather similar data in other areas. When
choosing your study site, consider how accessible it is (how easy is it to get
to?) and the problems you may experience (e.g. tides, rain).
2. Collecting data:
Your teacher will provide you with the equipment you need to collect the following data. You should have at least one study site where you will collect
data, but you might decide to have more if you want to compare your results
in different areas. This works best in a river, where you can choose sites down
its length.
(a) Chemical data
Measure and record data such as temperature, pH, conductivity and dissolved oxygen at each of your sites. You may not know exactly what
these measurements mean right now, but it will become clearer later in the
chapter.
(b) Hydrological data
Measure the water velocity of the river and observe how the volume of water
in the river changes as you move down its length. You can also collect a
water sample in a clear bottle, hold it to the light and see whether the
water is clear or whether it has particles in it.
(c) Biological data
What types of animals and plants are found in or near this part of the
hydrosphere? Are they specially adapted to their environment?
Record your data in a table like the one shown below:
Site 1
Site 2
Site 3
Temperature
pH
Conductivity
Dissolved oxygen
Animals and plants
3. Interpreting the data:
Once you have collected and recorded your data, think about the following
questions:
• How does the data you have collected vary at different sites?
• Can you explain these differences?
• What effect do you think temperature, dissolved oxygen and pH have on
animals and plants that are living in the hydrosphere?
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CHAPTER 20. THE HYDROSPHERE - GRADE 10
20.4
• Water is seldom ’pure’. It usually has lots of things dissolved (e.g. Mg,
Ca and NO−
3 ions) or suspended (e.g. soil particles, debris) in it. Where
do these substances come from?
• Are there any human activities near this part of the hydrosphere? What
effect could these activities have on the hydrosphere?
20.4
The Importance of the Hydrosphere
It is so easy sometimes to take our hydrosphere for granted, and we seldom take the time to
really think about the role that this part of the planet plays in keeping us alive. Below are just
some of the very important functions of water in the hydrosphere:
• Water is a part of living cells
Each cell in a living organism is made up of almost 75% water, and this allows the cell
to function normally. In fact, most of the chemical reactions that occur in life, involve
substances that are dissolved in water. Without water, cells would not be able to carry
out their normal functions, and life could not exist.
• Water provides a habitat
The hydrosphere provides an important place for many animals and plants to live. Many
−
+
gases (e.g. CO2 , O2 ), nutrients e.g. nitrate (NO−
3 ), nitrite (NO2 ) and ammonium (NH4 )
2+
2+
ions, as well as other ions (e.g. Ca and Mg ) are dissolved in water. The presence of
these substances is critical for life to exist in water.
• Regulating climate
You may remember from chapter ?? that one of water’s unique characteristics is its high
specific heat. This means that water takes a long time to heat up, and also a long time
to cool down. This is important in helping to regulate temperatures on earth so that they
stay within a range that is acceptable for life to exist. Ocean currents also help to disperse
heat.
• Human needs
Humans use water in a number of ways. Drinking water is obviously very important, but
water is also used domestically (e.g. washing and cleaning) and in industry. Water can
also be used to generate electricity through hydropower.
These are just a few of the very important functions that water plays on our planet. Many of the
functions of water relate to its chemistry and to the way in which it is able to dissolve substances
in it.
20.5
Ions in aqueous solution
As we mentioned earlier, water is seldom pure. Because of the structure of the water molecule,
it is able to dissolve substances in it. This is very important because if water wasn’t able to do
this, life would not be able to survive. In rivers and the oceans for example, dissolved oxygen
means that organisms are still able to respire (breathe). For plants, dissolved nutrients are also
available. In the human body, water is able to carry dissolved substances from one part of the
body to another.
Many of the substances that dissolve are ionic, and when they dissolve they form ions in solution.
We are going to look at how water is able to dissolve ionic compounds, and how these ions
maintain a balance in the human body, how they affect water hardness, and how specific ions
determine the pH of solutions.
379
20.5
20.5.1
CHAPTER 20. THE HYDROSPHERE - GRADE 10
Dissociation in water
You may remember from chapter 5 that water is a polar molecule (figure 20.1). This means
that one part of the molecule has a slightly positive charge and the other part has a slightly
negative charge.
δ+
H2 O
δ−
Figure 20.1: Water is a polar molecule
It is the polar nature of water that allows ionic compounds to dissolve in it. In the case of
sodium chloride (NaCl) for example, the positive sodium ions (Na+ ) will be attracted to the
negative pole of the water molecule, while the negative chloride ions (Cl− ) will be attracted to
the positive pole of the water molecule. In the process, the ionic bonds between the sodium
and chloride ions are weakened and the water molecules are able to work their way between the
individual ions, surrounding them and slowly dissolving the compound. This process is called
dissociation. A simplified representation of this is shown in figure 20.2.
Definition: Dissociation
Dissociation in chemistry and biochemistry is a general process in which ionic compounds
separate or split into smaller molecules or ions, usually in a reversible manner.
Cl−
δ+
H2 O
δ−
Cl− δ + H2 O δ − N a+ δ − H2 O δ + Cl−
δ−
H2 O
δ+
Cl−
Figure 20.2: Sodium chloride dissolves in water
The dissolution of sodium chloride can be represented by the following equation:
N aCl(s) → N a+ (aq) + Cl− (aq)
The symbols s (solid), l (liquid), g (gas) and aq (material is dissolved in water) are written after
the chemical formula to show the state or phase of the material. The dissolution of potassium
sulphate into potassium and sulphate ions is shown below as another example:
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CHAPTER 20. THE HYDROSPHERE - GRADE 10
20.5
K2 SO4 (s) → 2K + (aq) + SO42− (aq)
Remember that molecular substances (e.g. covalent compounds) may also dissolve, but most
will not form ions. One example is sugar.
C6 H12 O6 (s) ⇔ C6 H12 O6 (aq)
There are exceptions to this and some molecular substances will form ions when they dissolve.
Hydrogen chloride for example can ionise to form hydrogen and chloride ions.
HCl(g) → H + (aq) + Cl− (aq)
teresting The ability of ionic compounds to dissolve in water is extremely important in
Interesting
Fact
Fact
the human body! The body is made up of cells, each of which is surrounded
by a membrane. Dissolved ions are found inside and outside of body cells, in
different concentrations. Some of these ions are positive (e.g. Mg2+ ) and some
are negative (e.g. Cl− ). If there is a difference in the charge that is inside and
outside the cell, then there is a potential difference across the cell membrane.
This is called the membrane potential of the cell. The membrane potential
acts like a battery and affects the movement of all charged substances across the
membrane. Membrane potentials play a role in muscle functioning, digestion,
excretion and in maintaining blood pH, to name just a few. The movement of
ions across the membrane can also be converted into an electric signal that can
be transferred along neurons (nerve cells), which control body processes. If ionic
substances were not able to dissociate in water, then none of these processes
would be possible! It is also important to realise that our bodies can lose ions
such as Na+ , K+ , Ca2+ , Mg2+ , and Cl− , for example when we sweat during
exercise. Sports drinks such as Lucozade and Powerade are designed to replace
these lost ions so that the body’s normal functioning is not affected.
Exercise: Ions in solution
1. For each of the following, say whether the substance is ionic or molecular.
(a)
(b)
(c)
(d)
potassium nitrate (KNO3 )
ethanol (C2 H5 OH)
sucrose sugar (C12 H22 O11
sodium bromide (NaBr)
2. Write a balanced equation to show how each of the following ionic compounds
dissociate in water.
(a)
(b)
(c)
(d)
sodium sulphate (Na2 SO4 )
potassium bromide (KBr)
potassium permanganate (KMNO4 )
sodium phosphate (Na3 PO4 )
381
20.5
20.5.2
CHAPTER 20. THE HYDROSPHERE - GRADE 10
Ions and water hardness
Definition: Water hardness
Water hardness is a measure of the mineral content of water. Minerals are substances such
as calcite, quartz and mica that occur naturally as a result of geological processes.
Hard water is water that has a high mineral content. Water that has a low mineral content
is known as soft water. If water has a high mineral content, it usually contains high levels of
metal ions, mainly calcium (Ca) and magnesium (Mg). The calcium enters the water from either
CaCO3 (limestone or chalk) or from mineral deposits of CaSO4 . The main source of magnesium
is a sedimentary rock called dolomite, CaMg(CO3 )2 . Hard water may also contain other metals
as well as bicarbonates and sulphates.
teresting The simplest way to check whether water is hard or soft is to use the lather/froth
Interesting
Fact
Fact
test. If the water is very soft, soap will lather more easily when it is rubbed
against the skin. With hard water this won’t happen. Toothpaste will also not
froth well in hard water.
A water softener works on the principle of ion exchange. Hard water passes through a media
bed, usually made of resin beads that are supersaturated with sodium. As the water passes
through the beads, the hardness minerals (e.g. calcium and magnesium) attach themselves to
the beads. The sodium that was originally on the beads is released into the water. When the
resin becomes saturated with calcium and magnesium, it must be recharged. A salt solution is
passed through the resin. The sodium replaces the calcium and magnesium, and these ions are
released into the waste water and discharged.
20.5.3
The pH scale
The concentration of specific ions in solution, affects whether the solution is acidic or basic. You
will learn about acids and bases in chapter 15. Acids and bases can be described as substances
that either increase or decrease the concentration of hydrogen (H+ or H3 O+) ions in a solution.
An acid increases the hydrogen ion concentration in a solution, while a base decreases the
hydrogen ion concentration. pH is used to measure whether a substance is acidic or basic
(alkaline).
Definition: pH
pH is a measure of the acidity or alkalinity of a solution. The pH scale ranges from 0 to 14.
Solutions with a pH less than seven are acidic, while those with a pH greater than seven
are basic (alkaline). pH 7 is considered neutral.
pH can be calculated using the following equation:
pH = −log[H + ]
or
pH = −log[H3 O+ ]
The brackets in the above equation are used to show concentration in mol.dm−3 .
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CHAPTER 20. THE HYDROSPHERE - GRADE 10
20.5
Worked Example 93: pH calculations
Question: Calculate the pH of a solution where the concentration of hydrogen ions
is 1 × 10−7 mol.dm−3 .
Answer
Step 1 : Determine the concentration of hydrogen ions in mol.dm−3
In this example, the concentration has been given and is 1 × 10−7 mol.dm−3
Step 2 : Substitute this value into the pH equation and calculate the pH
value
pH = -log[H+]
= -log(1 × 10−7 )
=7
Worked Example 94: pH calculations
Question: In a solution of ethanoic acid, the following equilibrium is established:
CH3 COOH(aq) + H2 O ⇔ CH3 COO− (aq) + H3 O+
The concentration of CH3 COO− ions is found to be 0.003 mol.dm−3 . Calculate the
pH of the solution.
Answer
Step 1 : Determine the concentration of hydrogen ions in the solution
According to the balanced equation for this reaction, the mole ratio of CH3 COO−
ions to H3 O+ ions is the same, therefore the concentration of these two ions in the
solution will also be the same. So, [H3 O+ ] = 0.003 dm−3 .
Step 2 : Substitute this value into the pH equation and calculate the pH
value
pH = -log[H3O+ ]
= -log(0.003)
= 2.52
Understanding pH is very important. In living organisms, it is necessary to maintain a constant
pH so that chemical reactions can occur under optimal conditions.
Important: It may also be useful for calculations involving the pH scale, to know that the
following equation can also be used:
[H3 O+ ][OH− ] = 1 × 10−14
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20.5
CHAPTER 20. THE HYDROSPHERE - GRADE 10
teresting A build up of acid in the human body can be very dangerous. Lactic acidosis
Interesting
Fact
Fact
is a condition caused by the buildup of lactic acid in the body. It leads to
acidification of the blood (acidosis) and can make a person very ill. Some of
the symptoms of lactic acidosis are deep and rapid breathing, vomiting, and
abdominal pain. In the fight against HIV, lactic acidosis is a problem. One of
the antiretrovirals (ARV’s) that is used in anti-HIV treatment is Stavudine (also
known as Zerit or d4T). One of the side effects of Stavudine is lactic acidosis,
particularly in overweight women. If it is not treated quickly, it can result in
death.
In agriculture, farmers need to know the pH of their soils so that they are able to plant the
right kinds of crops. The pH of soils can vary depending on a number of factors such as
rainwater, the kinds of rocks and materials from which the soil was formed and also human
influences such as pollution and fertilisers. The pH of rain water can also vary and this too has
an effect on agriculture, buildings, water courses, animals and plants. Rainwater is naturally
acidic because carbon dioxide in the atmosphere combines with water to form carbonic acid.
Unpolluted rainwater has a pH of approximately 5.6. However, human activities can alter the
acidity of rain and this can cause serious problems such as acid rain.
Exercise: Calculating pH
1. Calculate the pH of each of the following solutions:
(a) A 0.2 mol.dm−3 KOH solution
(b) A 0.5 mol.dm−3 HCl solution
2. What is the concentration (in mol.dm−3 ) of H3 O+ ions in a NaOH solution
which has a pH of 12?
3. The concentrations of hydronium and hydroxyl ions in a typical sample of
seawater are 10−8 mol.dm−3 and 10−6 mol.dm−3 respectively.
(a) Is the seawater acidic or basic?
(b) What is the pH of the seawater?
(c) Give a possible explanation for the pH of the seawater.
(IEB Paper 2, 2002)
20.5.4
Acid rain
The acidity of rainwater comes from the natural presence of three substances (CO2 , NO, and
SO2 ) in the lowest layer of the atmosphere. These gases are able to dissolve in water and
therefore make rain more acidic than it would otherwise be. Of these gases, carbon dioxide
(CO2 ) has the highest concentration and therefore contributes the most to the natural acidity
of rainwater. We will look at each of these gases in turn.
Definition: Acid rain
Acid rain refers to the deposition of acidic components in rain, snow and dew. Acid rain
occurs when sulfur dioxide and nitrogen oxides are emitted into the atmosphere, undergo
chemical transformations, and are absorbed by water droplets in clouds. The droplets then
fall to earth as rain, snow, mist, dry dust, hail, or sleet. This increases the acidity of the
soil, and affects the chemical balance of lakes and streams.
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CHAPTER 20. THE HYDROSPHERE - GRADE 10
20.5
1. Carbon dioxide
Carbon dioxide reacts with water in the atmosphere to form carbonic acid (H2 CO3 ).
CO2 + H2 O → H2 CO3
The carbonic acid dissociates to form hydrogen and hydrogen carbonate ions. It is the
presence of hydrogen ions that lowers the pH of the solution, making the rain acidic.
H2 CO3 → H + + HCO3−
2. Nitric oxide
Nitric oxide (NO) also contributes to the natural acidity of rainwater and is formed during
lightning storms when nitrogen and oxygen react. In air, NO is oxidised to form nitrogen
dioxide (NO2 ). It is the nitrogen dioxide which then reacts with water in the atmosphere
to form nitric acid (HNO3 ).
3N O2 (g) + H2 O → 2HN O3 (aq) + N O(g)
The nitric acid dissociates in water to produce hydrogen ions and nitrate ions. This again
lowers the pH of the solution, making it acidic.
HN O3 → H + + N O3−
3. Sulfur dioxide
Sulfur dioxide in the atmosphere first reacts with oxygen to form sulfur trioxide, before
reacting with water to form sulfuric acid.
2SO2 + O2 → 2SO3
SO3 + H2 O → H2 SO4
Sulfuric acid dissociates in a similar way to the previous reactions.
H2 SO4 → HSO4− + H +
Although these reactions do take place naturally, human activities can greatly increase the concentration of these gases in the atmosphere, so that rain becomes far more acidic than it would
otherwise be. The burning of fossil fuels in industries, vehicles etc is one of the biggest culprits.
If the acidity of the rain drops below 5, it is referred to as acid rain.
Acid rain can have a very damaging effect on the environment. In rivers, dams and lakes, increased acidity can mean that some species of animals and plants will not survive. Acid rain can
also degrade soil minerals, producing metal ions that are washed into water systems. Some of
these ions may be toxic e.g. Al3+ . From an economic perspective, altered soil pH can drastically
affect agricultural productivity.
Acid rain can also affect buildings and monuments, many of which are made from marble and
limestone. A chemical reaction takes place between CaCO3 (limestone) and sulfuric acid to
produce aqueous ions which can be easily washed away. The same reaction can occur in the
lithosphere where limestone rocks are present e.g. limestone caves can be eroded by acidic
rainwater.
H2 SO4 + CaCO3 → CaSO4 .H2 O + CO2
385
20.6
CHAPTER 20. THE HYDROSPHERE - GRADE 10
Activity :: Investigation : Acid rain
You are going to test the effect of ’acid rain’ on a number of substances.
Materials needed:
samples of chalk, marble, zinc, iron, lead, dilute sulfuric acid, test tubes, beaker,
glass dropper
Method:
1. Place a small sample of each of the following substances in a separate test
tube: chalk, marble, zinc, iron and lead
2. To each test tube, add a few drops of dilute sulfuric acid.
3. Observe what happens and record your results.
Discussion questions:
• In which of the test tubes did reactions take place? What happened to the
sample substances?
• What do your results tell you about the effect that acid rain could have on each
of the following: buildings, soils, rocks and geology, water ecosystems?
• What precautions could be taken to reduce the potential impact of acid rain?
20.6
Electrolytes, ionisation and conductivity
Conductivity in aqueous solutions, is a measure of the ability of water to conduct an electric
current. The more ions there are in the solution, the higher its conductivity.
Definition: Conductivity
Conductivity is a measure of a solution’s ability to conduct an electric current.
20.6.1
Electrolytes
An electrolyte is a material that increases the conductivity of water when dissolved in it.
Electrolytes can be further divided into strong electrolytes and weak electrolytes.
Definition: Electrolyte
An electrolyte is a substance that contains free ions and behaves as an electrically conductive
medium. Because they generally consist of ions in solution, electrolytes are also known as
ionic solutions.
1. Strong electrolytes
A strong electrolyte is a material that ionises completely when it is dissolved in water:
AB(s,l,g) → A+ (aq) + B − (aq)
This is a chemical change because the original compound has been split into its component ions and bonds have been broken. In a strong electrolyte, we say that the extent
of ionisation is high. In other words, the original material dissociates completely so that
there is a high concentration of ions in the solution. An example is a solution of potassium
nitrate:
KN O3 (s) → K + (aq) + N O3− (aq)
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CHAPTER 20. THE HYDROSPHERE - GRADE 10
20.6
2. Weak electrolytes
A weak electrolyte is a material that goes into solution and will be surrounded by water
molecules when it is added to water. However, not all of the molecules will dissociate into
ions. The extent of ionisation of a weak electrolyte is low and therefore the concentration
of ions in the solution is also low.
AB(s,l,g) → AB(aq) ⇔ A+ (aq) + B − (aq)
The following example shows that, in the final solution of a weak electrolyte, some of the
original compound plus some dissolved ions are present.
C2 H3 O2 H(l) → C2 H3 O2 H ⇔ C2 H3 O2− (aq) + H + (aq)
20.6.2
Non-electrolytes
A non-electrolyte is a material that does not increase the conductivity of water when dissolved
in it. The substance goes into solution and becomes surrounded by water molecules, so that the
molecules of the chemical become separated from each other. However, although the substance
does dissolve, it is not changed in any way and no chemical bonds are broken. The change is a
physical change. In the oxygen example below, the reaction is shown to be reversible because
oxygen is only partially soluble in water and comes out of solution very easily.
C2 H5 OH(l) → C2 H5 OH(aq)
O2 (g) ⇔ O2 (aq)
20.6.3
Factors that affect the conductivity of water
The conductivity of water is therefore affected by the following factors:
• The type of substance that dissolves in water
Whether a material is a strong electrolyte (e.g. potassium nitrate, KNO3 ), a weak electrolyte (e.g. acetate, C2 H3 O2 H) or a non-electrolyte (e.g. sugar, alcohol, oil) will affect
the conductivity of water because the concentration of ions in solution will be different in
each case.
• The concentration of ions in solution
The higher the concentration of ions in solution, the higher its conductivity will be.
• Temperature
The warmer the solution the higher the solubility of the material being dissolved, and
therefore the higher the conductivity as well.
Activity :: Experiment : Electrical conductivity
Aim:
To investigate the electrical conductivities of different substances and solutions.
Apparatus:
solid salt (NaCl) crystals; different liquids such as distilled water, tap water,
seawater, benzene and alcohol; solutions of salts e.g. NaCl, KBr; a solution of
an acid (e.g. HCl) and a solution of a base (e.g. NaOH); torch cells; ammeter;
conducting wire, crocodile clips and 2 carbon rods.
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20.6
CHAPTER 20. THE HYDROSPHERE - GRADE 10
Method:
Set up the experiment by connecting the circuit as shown in the diagram below.
In the diagram, ’X’ represents the substance or solution that you will be testing.
When you are using the solid crystals, the crocodile clips can be attached directly to
each end of the crystal. When you are using solutions, two carbon rods are placed
into the liquid, and the clips are attached to each of the rods. In each case, complete
the circuit and allow the current to flow for about 30 seconds. Observe whether the
ammeter shows a reading.
battery
Ammeter
A
test substance
X
crocodile clip
Results:
Record your observations in a table similar to the one below:
Test substance
Ammeter reading
What do you notice? Can you explain these observations?
Remember that for electricity to flow, there needs to be a movement of charged
particles e.g. ions. With the solid NaCl crystals, there was no flow of electricity
recorded on the ammeter. Although the solid is made up of ions, they are held
together very tightly within the crystal lattice, and therefore no current will flow.
Distilled water, benzene and alcohol also don’t conduct a current because they are
covalent compounds and therefore do not contain ions.
The ammeter should have recorded a current when the salt solutions and the acid
and base solutions were connected in the circuit. In solution, salts dissociate into
their ions, so that these are free to move in the solution. Acids and bases behave
in a similar way, and dissociate to form hydronium and oxonium ions. Look at the
following examples:
KBr → K+ + Br−
NaCl → Na+ + Cl−
HCl + H2 O → H3 O+ + Cl−
NaOH → Na+ + OH−
Conclusions:
Solutions that contain free-moving ions are able to conduct electricity because of
the movement of charged particles. Solutions that do not contain free-moving ions
do not conduct electricity.
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CHAPTER 20. THE HYDROSPHERE - GRADE 10
20.7
teresting Conductivity in streams and rivers is affected by the geology of the area where
Interesting
Fact
Fact
the water is flowing through. Streams that run through areas with granite
bedrock tend to have lower conductivity because granite is made of materials
that do not ionise when washed into the water. On the other hand, streams
that run through areas with clay soils tend to have higher conductivity because
the materials ionise when they are washed into the water. Pollution can also affect conductivity. A failing sewage system or an inflow of fertiliser runoff would
raise the conductivity because of the presence of chloride, phosphate, and nitrate
(ions) while an oil spill (non-ionic) would lower the conductivity. It is very important that conductivity is kept within a certain acceptable range so that the
organisms living in these water systems are able to survive.
20.7
Precipitation reactions
Sometimes, ions in solution may react with each other to form a new substance that is insoluble.
This is called a precipitate.
Definition: Precipitate
A precipitate is the solid that forms in a solution during a chemical reaction.
Activity :: Demonstration : The reaction of ions in solution
Apparatus and materials:
4 test tubes; copper(II) chloride solution; sodium carbonate solution; sodium
sulphate solution
CuCl2
CuCl2
Na2 CO3
Na2 SO4
Method:
1. Prepare 2 test tubes with approximately 5 ml of dilute Cu(II)chloride solution
in each
2. Prepare 1 test tube with 5 ml sodium carbonate solution
3. Prepare 1 test tube with 5 ml sodium sulphate solution
4. Carefully pour the sodium carbonate solution into one of the test tubes containing copper(II) chloride and observe what happens
5. Carefully pour the sodium sulphate solution into the second test tube containing
copper(II) chloride and observe what happens
Results:
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20.7
CHAPTER 20. THE HYDROSPHERE - GRADE 10
1. A light blue precipitate forms when sodium carbonate reacts with copper(II)
chloride
2. No precipitate forms when sodium sulphate reacts with copper(II) chloride
It is important to understand what happened in the previous demonstration. We will look at
what happens in each reaction, step by step.
1. Reaction 1: Sodium carbonate reacts with copper(II) chloride
When these compounds react, a number of ions are present in solution: Cu2+ , Cl− , N a+
and CO32− .
Because there are lots of ions in solution, they will collide with each other and may recombine in different ways. The product that forms may be insoluble, in which case a precipitate
will form, or the product will be soluble, in which case the ions will go back into solution.
Let’s see how the ions in this example could have combined with each other:
Cu2+ + CO2−
3 → CuCO3
Cu2+ + 2Cl− → CuCl2
Na+ + Cl− → NaCl
Na+ + CO2−
3 → Na2 CO3
You can automatically exclude the reactions where sodium carbonate and copper(II) chloride are the products because these were the initial reactants. You also know that sodium
chloride (NaCl) is soluble in water, so the remaining product (copper carbonate) must be
the one that is insoluble. It is also possible to look up which salts are soluble and which
are insoluble. If you do this, you will find that most carbonates are insoluble, therefore the
precipitate that forms in this reaction must be CuCO3 . The reaction that has taken place
between the ions in solution is as follows:
2N a+ + CO32− + Cu2+ + 2Cl− → CuCO3 + 2N a+ + 2Cl−
2. Reaction 2: Sodium sulphate reacts with copper(II) chloride
The ions that are present in solution are Cu2+ , Cl− , N a+ and SO42− .
The ions collide with each other and may recombine in different ways. The possible combinations of the ions are as follows:
Cu2+ + SO2−
4 → CuSO4
Cu2+ + 2Cl− → CuCl2
Na+ + Cl− → NaCl
Na+ + SO2−
4 → Na2 SO4
If we look up which of these salts are soluble and which are insoluble, we see that most
chlorides and most sulphates are soluble. This is why no precipitate forms in this second
reaction. Even when the ions recombine, they immediately separate and go back into
solution. The reaction that has taken place between the ions in solution is as follows:
2N a+ + SO42− + Cu2+ + 2Cl− → 2N a+ + SO42− + Cu2+ + 2Cl−
Table 20.1 shows some of the general rules about the solubility of different salts based on a
number of investigations:
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CHAPTER 20. THE HYDROSPHERE - GRADE 10
20.8
Table 20.1: General rules for the solubility of salts
Salt
Nitrates
Potassium, sodium and ammonium salts
Chlorides
Sulphates
Carbonates
20.8
Solubility
All are soluble
All are soluble
All are soluble except silver chloride, lead(II)chloride and mercury(II)chloride
All
are
soluble
except
lead(II)sulphate,
barium sulphate and calcium sulphate
All are insoluble except those of
potassium, sodium and ammonium
Testing for common anions in solution
It is also possible to carry out tests to determine which ions are present in a solution.
20.8.1
Test for a chloride
Prepare a solution of the unknown salt using distilled water and add a small amount of silver
nitrate solution. If a white precipitate forms, the salt is either a chloride or a carbonate.
−
Cl− + Ag+ + NO−
3 → AgCl + NO3 (AgCl is white precipitate)
−
+
CO2−
3 + 2Ag + 2NO3 → Ag2 CO3 + 2NO2 (Ag2 CO3 is white precipitate)
The next step is to treat the precipitate with a small amount of concentrated nitric acid. If
the precipitate remains unchanged, then the salt is a chloride. If carbon dioxide is formed, and
the precipitate disappears, the salt is a carbonate.
AgCl + HNO3 → (no reaction; precipitate is unchanged)
Ag2 CO3 + 2HNO3 → 2AgNO3 + H2 O + CO2 (precipitate disappears)
20.8.2
Test for a sulphate
Add a small amount of barium chloride solution to a solution of the test salt. If a white precipitate
forms, the salt is either a sulphate or a carbonate.
2+
SO2−
+ Cl− → BaSO4 + Cl− (BaSO4 is a white precipitate)
4 + Ba
2+
CO2−
+ Cl− → BaCO3 + Cl− (BaCO3 is a white precipitate)
3 + Ba
If the precipitate is treated with nitric acid, it is possible to distinguish whether the salt is a
sulphate or a carbonate (as in the test for a chloride).
BaSO4 + HNO3 → (no reaction; precipitate is unchanged)
BaCO3 + 2HNO3 → Ba(NO3 )2 + H2 O + CO2 (precipitate disappears)
391
20.8
20.8.3
CHAPTER 20. THE HYDROSPHERE - GRADE 10
Test for a carbonate
If a sample of the dry salt is treated with a small amount of acid, the production of carbon
dioxide is a positive test for a carbonate.
Acid + CO32− → CO2
If the gas is passed through limewater and the solution becomes milky, the gas is carbon dioxide.
Ca(OH)2 + CO2 → CaCO3 + H2 O (It is the insoluble CaCO3 precipitate that makes the
limewater go milky)
20.8.4
Test for bromides and iodides
As was the case with the chlorides, the bromides and iodides also form precipitates when they
are reacted with silver nitrate. Silver chloride is a white precipitate, but the silver bromide and
silver iodide precipitates are both pale yellow. To determine whether the precipitate is a bromide
or an iodide, we use chlorine water and carbon tetrachloride (CCl4 ).
Chlorine water frees bromine gas from the bromide, and colours the carbon tetrachloride a reddish brown.
Chlorine water frees iodine gas from an iodide, and colours the carbon tetrachloride is coloured
purple.
Exercise: Precipitation reactions and ions in solution
1. Silver nitrate (AgNO3 ) reacts with potassium chloride (KCl) and a white precipitate is formed.
(a) Write a balanced equation for the reaction that takes place.
(b) What is the name of the insoluble salt that forms?
(c) Which of the salts in this reaction are soluble?
2. Barium chloride reacts with sulfuric acid to produce barium sulphate and hydrochloric acid.
(a) Write a balanced equation for the reaction that takes place.
(b) Does a precipitate form during the reaction?
(c) Describe a test that could be used to test for the presence of barium
sulphate in the products.
3. A test tube contains a clear, colourless salt solution. A few drops of silver
nitrate solution are added to the solution and a pale yellow precipitate forms.
Which one of the following salts was dissolved in the original solution?
(a)
(b)
(c)
(d)
NaI
KCl
K2 CO3
Na2 SO4
(IEB Paper 2, 2005)
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CHAPTER 20. THE HYDROSPHERE - GRADE 10
20.9
20.9
Threats to the Hydrosphere
It should be clear by now that the hydrosphere plays an extremely important role in the survival of
life on Earth, and that the unique properties of water allow various important chemical processes
to take place which would otherwise not be possible. Unfortunately for us however, there are
a number of factors that threaten our hydrosphere, and most of these threats are because of
human activities. We are going to focus on two of these issues: overuse and pollution and look
at ways in which these problems can possibly be overcome.
1. Overuse of water
We mentioned earlier that only a very small percentage of the hydrosphere’s water is
available as freshwater. However, despite this, humans continue to use more and more
water to the point where water consumption is fast approaching the amount of water
that is available. The situation is a serious one, particularly in countries such as South
Africa which are naturally dry, and where water resources are limited. It is estimated that
between 2020 and 2040, water supplies in South Africa will no longer be able to meet the
growing demand for water in this country. This is partly due to population growth, but
also because of the increasing needs of industries as they expand and develop. For each
of us, this should be a very scary thought. Try to imagine a day without water...difficult
isn’t it? Water is so much a part of our lives, that we are hardly aware of the huge part
that it plays in our daily lives.
Activity :: Discussion : Creative water conservation
As populations grow, so do the demands that are placed on dwindling water
resources. While many people argue that building dams helps to solve this watershortage problem, the reality is that dams are only a temporary solution, and
that they often end up doing far more ecological damage than good. The only
sustainable solution is to reduce the demand for water, so that water supplies
are sufficient to meet this. The more important question then is how to do this.
Discussion:
Divide the class into groups, so that there are about five people in each.
Each group is going to represent a different sector within society. Your teacher
will tell you which sector you belong to from the following: Farming, industry,
city management or civil society (i.e. you will represent the ordinary ’man on
the street’). In your groups, discuss the following questions as they relate to the
group of people you represent: (Remember to take notes during your discussions,
and nominate a spokesperson to give feedback to the rest of the class on behalf
of your group)
• What steps could be taken by your group to conserve water?
• Why do you think these steps are not being taken?
• What incentives do you think could be introduced to encourage this group
to conserve water more efficiently?
2. Pollution
Pollution of the hydrosphere is also a major problem. When we think of pollution, we
sometimes only think of things like plastic, bottles, oil and so on. But any chemical that
is present in the hydrosphere in an amount that is not what it should be is a pollutant.
Animals and plants that live in the hydrosphere are specially adapted to surviving within a
certain range of conditions. If these conditions are changed (e.g. through pollution), these
organisms may not be able to survive. Pollution then, can affect entire aquatic ecosystems.
The most common forms of pollution in the hydrosphere are waste products from humans
and from industries, nutrient pollution e.g. fertiliser runoff which causes eutrophication
(this will be discussed in a later section) and toxic trace elements such as aluminium,
mercury and copper to name a few. Most of these elements come from mines or from
industries.
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20.10
CHAPTER 20. THE HYDROSPHERE - GRADE 10
It is important to realise that our hydrosphere exists in a delicate balance with other systems,
and that disturbing this balance can have serious consequences for life on this planet.
Activity :: Group Project : School Action Project
There is a lot that can be done within a school to save water. As a class,
discuss what actions could be taken by your class to make people more aware of how
important it is to conserve water.
20.10
Summary
• The hydrosphere includes all the water that is on Earth. Sources of water include freshwater (e.g. rivers, lakes), saltwater (e.g. oceans), groundwater (e.g. boreholes) and water
vapour. Ice (e.g. glaciers) is also part of the hydrosphere.
• The hydrosphere interacts with other global systems, including the atmosphere, lithosphere and biosphere.
• The hydrosphere has a number of important functions. Water is a part of all living cells,
it provides a habitat for many living organisms, it helps to regulate climate, and it is used
by humans for domestic, industrial and other use.
• The polar nature of water means that ionic compounds dissociate easily in aqueous
solution into their component ions.
• Ions in solution play a number of roles. In the human body for example, ions help to
regulate the internal environment (e.g. controlling muscle function, regulating blood pH).
Ions in solution also determine water hardness and pH.
• Water hardness is a measure of the mineral content of water. Hard water has a high
mineral concentration and generally also a high concentration of metal ions e.g. calcium
and magnesium. The opposite is true for soft water.
• pH is a measure of the concentration of hydrogen ions in solution. The formula used to
calculate pH is as follows:
pH = -log[H3O+ ] or pH = -log[H+ ]
A solution with a pH less than 7 is considered acidic and more than 7 is considered basic
(or alkaline). A neutral solution has a pH of 7.
• Gases such as CO2 , NO2 and SO2 dissolve in water to form weak acid solutions. Rain is
naturally acidic because of the high concentrations of carbon dioxide in the atmosphere.
Human activities such as burning fossil fuels, increase the concentration of these gases in
the atmosphere, resulting in acid rain.
• Conductivity is a measure of a solution’s ability to conduct an electric current.
• An electrolyte is a substance that contains free ions, and is therefore able to conduct an
electric current. Electrolytes can be divided into strong and weak electrolytes, based on
the extent to which the substance ionises in solution.
• A non-electrolyte cannot conduct an electric current because it dooes not contain free
ions.
• The type of substance, the concentration of ions and the temperature of the solution,
affect its conductivity.
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CHAPTER 20. THE HYDROSPHERE - GRADE 10
20.10
• A precipitate is formed when ions in solution react with each other to form an insoluble
product. Solubility ’rules’ help to identify the precipitate that has been formed.
• A number of tests can be used to identify whether certain anions are present in a solution.
• Despite the importance of the hydrosphere, a number of factors threaten it. These include
overuse of water, and pollution.
Exercise: Summary Exercise
1. Give one word for each of the following descriptions:
(a)
(b)
(c)
(d)
the change in phase of water from a gas to a liquid
a charged atom
a term used to describe the mineral content of water
a gas that forms sulfuric acid when it reacts with water
2. Match the information in column A with the information in column B by writing
only the letter (A to I) next to the question number (1 to 7)
Column A
1. A polar molecule
2. molecular solution
3. Mineral that increases water hardness
4. Substance that increases the hydrogen ion concentration
5. A strong electrolyte
6. A white precipitate
7. A non-conductor of electricity
Column B
A. H2 SO4
B. CaCO3
C. NaOH
D. salt water
E. calcium
F. carbon dioxide
G. potassium nitrate
H. sugar water
I. O2
3. For each of the following questions, choose the one correct answer from the
list provided.
(a) Which one of the following substances does not conduct electricity in the
solid phase but is an electrical conductor when molten?
i. Cu
ii. PbBr2
iii. H2 O
iv. I2
(IEB Paper 2, 2003)
(b) The following substances are dissolved in water. Which one of the solutions
is basic?
i. sodium nitrate
ii. calcium sulphate
iii. ammonium chloride
iv. potassium carbonate
(IEB Paper 2, 2005)
4. The concentration of hydronium and hydroxyl ions in a typical sample of seawater are 10−8 and 10−6 respectively.
(a) Is the seawater acidic or basic?
(b) Calculate the pH of this seawater.
5. Three test tubes (X, Y and Z) each contain a solution of an unknown potassium
salt. The following observations were made during a practical investigation to
identify the solutions in the test tubes:
A: A white precipitate formed when silver nitrate (AgNO3 ) was added to test
tube Z.
395
20.10
CHAPTER 20. THE HYDROSPHERE - GRADE 10
B: A white precipitate formed in test tubes X and Y when barium chloride
(BaCl2 ) was added.
C: The precipitate in test tube X dissolved in hydrochloric acid (HCl) and a gas
was released.
D: The precipitate in test tube Y was insoluble in hydrochloric acid.
(a) Use the above information to identify the solutions in each of the test tubes
X, Y and Z.
(b) Write a chemical equation for the reaction that took place in test tube X
before hydrochloric acid was added.
(DoE Exemplar Paper 2 2007)
396
Chapter 21
The Lithosphere - Grade 11
(NOTE TO SELF: Need map showing major mining areas in SA)
21.1
Introduction
If we were to cut the Earth in half we would see that our planet is made up of a number of layers,
namely the core at the centre, the mantle and the outer crust (figure 21.1). The core is made
up mostly of iron. The mantle, which lies between the core and the crust, consists of molten
rock, called magma which moves continuously because of convection currents. The crust is
the thin, hard outer layer that ’floats’ on the magma of the mantle. It is the upper part of the
mantle and the crust that make up the lithosphere (’lith’ means ’types of stone’ and ’sphere’
refers to the round shape of the earth). Together, the lithosphere, hydrosphere and atmosphere
make up the world as we know it.
Atmosphere
Crust
Upper Mantle
Mantle
Outer Core
Inner Core
Figure 21.1: A cross-section through the earth to show its different layers
Definition: Lithosphere
The lithosphere is the solid outermost shell of our planet. The lithosphere includes the crust
and the upper part of the mantle, and is made up of material from both the continents and
the oceans on the Earth’s surface.
397
21.2
CHAPTER 21. THE LITHOSPHERE - GRADE 11
Earlier chapters have focused on the hydrosphere (chapter 20) and the atmosphere (chapter 22).
The lithosphere is also very important, not only because it is the surface on which we live, but
also because humans gain many valuable resources from this part of the planet.
21.2
The chemistry of the earth’s crust
The crust is made up of about 80 elements, which occur in over 2000 different compounds
and minerals. However, most of the mass of crustal material is made up of only 8 of these
elements. These are oxygen (O), silica (Si), aluminium (Al), iron (Fe), calcium (Ca), sodium
(Na), potassium (K) and magnesium (Mg). These metal elements are seldom found in their
pure form, but are usually part of other more complex minerals. A mineral is a compound that
is formed through geological processes, which give it a particular structure. A mineral could
be a pure element, but more often minerals are made up of many different elements combined.
Quartz is just one example. It is a mineral that is made up of silica and oxygen. Some more
examples are shown in table 21.1.
Definition: Mineral
Minerals are natural compounds formed through geological processes. The term ’mineral’
includes both the material’s chemical composition and its structure. Minerals range in
composition from pure elements to complex compounds.
Table 21.1: Table showing examples of minerals and their chemistry
Chemistry
Comments
SiO2 (silicon dioxide)
Quartz is used for glass, in electrical components, optical lenses and in building
stone
Gold
Au (pure element) or Gold is often found in a group of minerAuTe2 (Calaverite, a gold als called the tellurides. Calaverite is a
mineral)
mineral that belongs to this group, and is
the most common gold-bearing mineral.
Gold has an affinity for tellurium (Te).
Hematite
Fe2 O3 (iron oxide)
Iron usually occurs in iron oxide minerals
or as an alloy of iron and nickel.
Orthoclase KAlSi3 O8 (potassium alu- Orthoclase belongs to the feldspar group
minium silicate)
of minerals.
Copper
Cu (pure element) or copper can be mined as a pure element
Cu2 (CO3 )(OH)2
(mala- or as a mineral such as malachite.
chite or copper carbonate
hydroxide)
Mineral
Quartz
A rock is a combination of one or more minerals. Granite for example, is a rock that is made up
of minerals such as SiO2 , Al2 O3 , CaO, K2 O, Na2 O and others. There are three different types
of rocks, igneous, sedimentary and metamorphic. Igneous rocks (e.g. granite, basalt) are
formed when magma is brought to the earth’s surface as lava, and then solidifies. Sedimentary
rocks (e.g. sandstone, limestone) form when rock fragments, organic matter or other sediment
particles are deposited and then compacted over time until they solidify. Metamorphic rock is
formed when any other rock types are subjected to intense heat and pressure over a period of
time. Examples include slate and marble.
Many of the elements that are of interest to us (e.g. gold, iron, copper), are unevenly distributed
in the lithosphere. In places where these elements are abundant, it is profitable to extract them
(e.g. through mining) for economic purposes. If their concentration is very low, then the cost
of extraction becomes more than the money that would be made if they were sold. Rocks that
contain valuable minerals are called ores. As humans, we are particularly interested in the ores
that contain metal elements, and also in those minerals that can be used to produce energy.
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CHAPTER 21. THE LITHOSPHERE - GRADE 11
21.3
Definition: Ore
An ore is a volume of rock that contains minerals which make it valuable for mining.
teresting A gemstone (also sometimes called a gem or semi-precious stone), is a
Interesting
Fact
Fact
highly attractive and valuable piece of mineral which, when cut and polished,
is used in jewelry and other adornments. Examples of gemstones are amethyst,
diamond, cat’s eye and sapphire.
Exercise: Rocks and minerals
1. Where are most of the earth’s minerals concentrated?
2. Explain the difference between a mineral, a rock and an ore.
3. Carry out your own research to find out which elements are found in the following minerals:
(a) gypsum
(b) calcite
(c) topaz
4. Which minerals are found in the following rocks?
(a) basalt
(b) sandstone
(c) marble
21.3
A brief history of mineral use
Many of the minerals that are important to humans are metals such as gold, aluminium, copper
and iron. Throughout history, metals have played a very important role in making jewelery, tools,
weapons and later machinery and other forms of technology. We have become so used to having
metals around us that we seldom stop to think what life might have been like before metals
were discovered. During the Stone Age for example, stones were used to make tools. Slivers
of stone were cut from a rock and then sharpened. In Africa, some of the stone tools that have
been found date back to 2.5 million years ago!
It was the discovery of metals that led to some huge advances in agriculture, warfare, transport
and even cookery. One of the first metals to be discovered was gold, probably because of its
beautiful shiny appearance. Gold was used mostly to make jewelery because it was too soft to
make harder tools. Later, copper became an important metal because it could be hammered
into shape, and it also lasted a lot longer than the stone that had previously been used in knives,
cooking utensils and weapons. Copper can also be melted and then put into a mould to re-shape
it. This is known as casting.
At about the time that copper was in widespread use, it was discovered that if certain kinds of
stones are heated to high enough temperatures, liquid metals flow from them. These rocks are
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21.4
CHAPTER 21. THE LITHOSPHERE - GRADE 11
ores, and contain the metal minerals inside them. The process of heating mineral ores in this
way is called smelting. It was also found that ores do not only occur at the earth’s surface, but
also deep below it. This discovery led to the beginning of mining.
But humans’ explorations into the world of metals did not end here! In some areas, the ores of
iron and tin were found close together. The cast alloy of these two metals is bronze. Bronze is a
very useful metal because it produces a sharper edge than copper. Another important discovery
was that of iron. Iron is the most abundant metal at the earth’s surface but it is more difficult
to work with than copper or tin. It is very difficult to extract iron from its ore because it has
an extremely high melting point, and only specially designed furnaces are able to produce the
temperatures that are needed. An important discovery was that if iron is heated in a furnace
with charcoal, some of the carbon in the charcoal is transferred to the iron, making the metal
even harder. If this hot metal has its temperature reduced very suddenly, it becomes even harder
and produces steel. Today, steel is a very important part of industry and construction.
teresting Originally it was believed that much of Africa’s knowledge of metals and their
Interesting
Fact
Fact
uses was from the Middle East. But this may not be the case. More recent studies
have shown that iron was used far earlier than it would have been if knowledge
of this metal had started in the Middle East. Many metal technologies may
in fact have developed independently in Africa and in many African countries,
metals have an extremely important place in society. In Nigeria’s Yoruba country
for example, iron has divine status because it is used to make instruments for
survival. ’Ogun’, the God of Iron, is seen as the protector of the kingdom.
21.4
Energy resources and their uses
Apart from minerals and ores, the products of the lithosphere are also important in meeting our
energy needs.
Coal is one of the most important fuels that is used in the production of electricity. Coal is formed
from organic material when plants and animals decompose, leaving behind organic remains that
accumulate and become compacted over millions of years under sedimentary rock. The layers of
compact organic material that can be found between sedimentary layers, are coal. When coal is
burned, a large amount of heat energy is released, which is used to produce electricity. South
Africa is the world’s sixth largest coal producer, with Mpumalanga contributing about 83
Another element that is found in the crust, and which helps to meet our energy needs, is uranium. Uranium produces energy through the process of nuclear fission. Neutrons are aimed
at the nucleii of the uranium atoms in order to split them. When the nucleus of a uranium
atom is split, a large amount of energy is released as heat. This heat is used to produce steam,
which turns turbines to generate electricity. Uranium is produced as a by-product of gold in
some mines in the Witwatersrand, and as a by-product in some copper mines, for example in
Phalaborwa. This type of nuclear power is relatively environmentally friendly since it produces
low gas emissions. However, the process does produce small amounts of radioactive wastes ,
which must be carefully disposed of in order to prevent contamination.
Oil is another product of the lithosphere which is critical in meeting our fuel needs. While most
of South Africa’s oil is imported and then processed at a refinery in either Durban, Cape Town
or Sasolburg, some is extracted from coal. The technology behind this type of extraction has
largely been developed by SASOL (South African Coal, Oil and Gas Corporation). Oil, like coal,
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21.5
is organic in origin and normally forms from organic deposits on the ocean floor. Oil requires
unique geological and geochemical conditions in order to be produced. Part of this process
involves the burial of organic-rich sediments under extremely high temperatures and pressures.
The oil that is produced is then pushed out into nearby sedimentary layers. Oil will then move
upwards until it is trapped by an impermeable rock layer. It accumulates here, and can then be
accessed by oil rigs and other advanced equipment.
21.5
Mining and Mineral Processing: Gold
21.5.1
Introduction
Gold was discovered in South Africa in the late 1800’s and since then has played a very important
role in South Africa’s history and economy. Its discovery brought many foreigners into South
Africa, who were lured by the promises of wealth. They set up small mining villages, which later
grew into larger settlements, towns and cities. One of the first of these settlements was the
beginning of present-day Johannesburg, also known as ’Egoli’ or ’Place of Gold’.
Most of South Africa’s gold is concentrated in the ’Golden Arc’, which stretches from Johannesburg to Welkom. Geologists believe that, millions of years ago, this area was a massive inland
lake. Rivers feeding into this lake brought sand, silt, pebbles and fine particles of gold and
deposited them over a long period of time. Eventually these deposits accumulated and became
compacted to form gold-bearing sedimentary rock or gold reefs. It is because of this complex,
but unique, set of circumstances that South Africa’s gold deposits are so concentrated in that
area. In other countries like Zimbabwe, gold occurs in smaller ’pockets’, which are scattered
over a much greater area.
21.5.2
Mining the Gold
A number of different techniques can be used to mine gold. The three most common methods
in South Africa are panning, open cast and shaft mining.
1. Panning
Panning for gold is a manual technique that is used to sort gold from other sediments.
Wide, shallow pans are filled with sand and gravel (often from river beds) that may contain
gold. Water is added and the pans are shaken so that the gold is sorted from the rock and
other materials. Because gold is much more dense, it settles to the bottom of the pan.
Pilgrim’s Rest in Mpumalanga, was the first site for gold panning in South Africa.
2. Open cast mining
This is a form of surface mining. Surface layers of rock and sediments are removed so that
the deeper gold rich layers can be reached. This type of mining is not suitable if the gold
is buried very deep below the surface.
3. Shaft mining
South Africa’s thin but extensive gold reefs slope at an angle underneath the ground, and
this means that some deposits are very deep and often difficult to reach. Shaft mining is
needed to reach the gold ore. After the initial drilling, blasting and equipping of a mine
shaft, tunnels are built leading outwards from the main shaft so that the gold reef can
be reached. Shaft mining is a dangerous operation, and roof supports are needed so that
the rock does not collapse. There are also problems of the intense heat and high pressure
below the surface which make shaft mining very complex, dangerous and expensive. A
diagram illustrating open cast and shaft mining is shown in figure 21.2.
21.5.3
Processing the gold ore
For every ton of ore that is mined, only a very small amount of gold is extracted. A number
of different methods can be used to separate gold from its ore, but one of the more common
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21.5
CHAPTER 21. THE LITHOSPHERE - GRADE 11
b b bb bbb
b b bb
b bb b b
b bb b b
1
Gold deposits
Gold seam close to surface
Open cast mining at shallow gold seam
2
Sloping gold seams
3
Main underground shaft
4
Tunnel from main shaft to access gold
5
1
b b bb b b b
b
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b bb b b
b
b b b b b
2
b
b bb b b b b b b b
b b bb bb bbb bbbb b b bb bb bb bb b bb bbbb bb b b bb
b b
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b b b b b b b bb
3
b bb b bb bb b bb b b bb b b b b
b b b b bb
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b b b bbb b bbb b b b b b b b bb b b b b b
b b b b b b b bb b b b b bb bb b b b b b b b b b
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bb b b
b b bb b b b
b b bb b
b
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5
4
Figure 21.2: Diagram showing open cast and shaft mining
methods is called gold cyanidation.
In the process of gold cyanidation, the ore is crushed and then cyanide solution is added so
that the gold particles are chemically dissolved from the ore. In this step of the process, gold is
oxidised. Zinc dust is then added to the cyanide solution. The zinc takes the place of the gold,
so that the gold is precipitated out of the solution. This process is shown in figure 21.3.
teresting Another method that is used to process gold is called the ’carbon-in-pulp’ (CIP)
Interesting
Fact
Fact
method. This method makes use of the high affinity that activated carbon has
for gold, and there are three stages to the process. The first stage involves the
absorption of gold in the cyanide solution to carbon. In the elution stage, gold
is removed from the carbon into an alkaline cyanide solution. In the final stage,
electro-winning is used to remove gold from the solution through a process of
electrolysis. Gold that has been removed is deposited on steel wool electrodes.
Both the carbon and the acid are then treated so that they can be re-used.
21.5.4
Characteristics and uses of gold
Gold has a number of uses because of its varied and unique characteristics. Below is a list of
some of these characteristics that have made gold such a valuable metal:
• Shiny
Gold’s beautiful appearance has made it one of the favourite metals for use in jewelery.
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21.5
STEP 1 - The ore is crushed until it is in fine pieces
STEP 2 - A sodium cyanide (NaCN) solution is mixed with the finely ground rock
4Au + 8N aCN + O2 + 2H2 O → 4N aAu(CN )2 + 4N aOH
Gold is oxidised.
STEP 3 - The gold-bearing solution can then be separated from the
remaining solid ore through filtration
STEP 4 - Zinc is added. Zinc replaces the gold in the gold-cyanide solution.
The gold is precipitated from the solution.
This is the reduction part of the reaction.
Figure 21.3: Flow diagram showing how gold is processed
• Durable
Gold does not tarnish or corrode easily, and therefore does not deteriorate in quality. It is
sometimes used in dentistry to make the crowns for teeth.
• Malleable and ductile
Since gold can be bent and twisted into shape, as well as being flattened into very thin
sheets, it is very useful in fine wires and to produce sheets of gold.
• Good conductor
Gold is a good conductor of electricity and is therefore used in transistors, computer circuits
and telephone exchanges.
• Heat ray reflector
Because gold reflects heat very effectively, it is used in space suits and in vehicles. It is
also used in the protective outer coating of artificial satellites. One of the more unusual
applications of gold is its use in firefighting, where a thin layer of gold is placed in the
masks of the firefighters to protect them from the heat.
Activity :: Case Study : Dropping like a gold balloon
Read the article below, which has been adapted from one that appeared in the
Financial Mail on 15th April 2005 and then answer the questions that follow.
As recently as 1980, South Africa accounted for over 70% of world gold
production. In 2004, that figure was a dismal 14%. Chamber of Mines
figures showed that SA’s annual gold production last year slipped to its
lowest level since 1931.
Chamber economist Roger Baxter says the ’precipitous’ fall in production
was caused by the dual impact of the fall in the rand gold price due to the
strong rand, and the continued upward rise in costs. Many of these costs,
laments Baxter, are ’costs we do not have control over’. These include
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water, transport, steel and labour costs, which rose by 13% on average in
2004.
He provides a breakdown of the cost components faced by mines:
• Water prices have risen by 10% per year for the past 3 years
• Steel prices have increased by double-digit rates for each of the past
3 years
• Spoornet’s tariffs rose 35% in 2003 and 16.5% in 2004
• Labour costs, which make up 50% of production costs, rose above
inflation in 2003 and 2004
At these costs, and at current rand gold prices, about 10 mines, employing
90 000 people, are marginal or loss-making, says Baxter.
1. Refer to the table below showing SA’s gold production in tons between 1980
and 2004.
Year
1980
1985
1990
1995
2000
2004
Production
(t)
675
660
600
525
425
340
Draw a line graph to illustrate these statistics.
2. What percentage did South Africa’s gold production contribute towards global
production in:
(a) 1980
(b) 2004
3. Outline two reasons for this drop in gold production.
4. Briefly explain how the increased cost of resources such as water contributes
towards declining profitability in gold mines.
5. Suggest a reason why the cost of steel might affect gold production.
6. Suggest what impact a decrease in gold production is likely to have on...
(a) South Africa’s economy
(b) mine employees
7. Find out what the current price of gold is. Discuss why you think gold is so
expensive.
21.5.5
Environmental impacts of gold mining
However, despite the incredible value of gold and its usefulness in a variety of applications, all
mining has an environmental cost. The following are just a few of the environmental impacts of
gold mining:
• Resource consumption
Gold mining needs large amounts of electricity and water.
• Poisoned water
Acid from gold processing can leach into nearby water systems such as rivers, causing
damage to animals and plants, as well as humans that may rely on that water for drinking.
The disposal of other toxic waste (e.g. cyanide) can also have a devastating effect on
biodiversity.
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21.5
• Solid waste
This applies particularly to open pit mines, where large amounts of soil and rock must be
displaced in order to access the gold reserves. Processing the gold ore also leaves solid
waste behind.
• Air pollution
Dust from open pit mines, as well as harmful gases such as sulfur dioxide and nitrogen
dioxide which are released from the furnaces, contribute to air pollution.
• Threaten natural areas
Mining activities often encroach on protected areas and threaten biodiversity in their operation areas.
Activity :: Discussion : Mine rehabilitation
There is a growing emphasis on the need to rehabilitate old mine sites that are
no longer in use. If it is too difficult to restore the site to what it was before, then
a new type of land use might be decided for that area. Any mine rehabilitation
programme should aim to achieve the following:
• ensure that the site is safe and stable
• remove pollutants that are contaminating the site
• restore the biodiversity that was there before mining started
• restore waterways to what they were before mining
There are different ways to achieve these goals. Plants for example, can be used
to remove metals from polluted soils and water, and can also help to stabilise the
soil so that other vegetation can grow. Land contouring can help to restore drainage
in the area.
Discussion:
In groups of 3-4, discuss the following questions:
1. What are the main benefits of mine rehabilitation?
2. What are some of the difficulties that may be experienced in trying to rehabilitate a mine site?
3. Suggest some creative ideas that could be used to encourage mining companies
to rehabilitate old sites.
4. One rehabilitation project that has received a lot of publicity is the rehabilitation
of dunes that were mined for titanium by Richards Bay Minerals (RBM). As a
group, carry out your own research to try to find out more about this project.
• What actions did RBM take to rehabilitate the dunes?
• Was the project successful?
• What were some of the challenges faced?
Exercise: Gold mining
Mapungubwe in the Limpopo Province is evidence of gold mining in South Africa
as early as 1200. Today, South Africa is a world leader in the technology of gold
mining. The following flow diagram illustrates some of the most important steps in
the recovery of gold from its ore.
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CHAPTER 21. THE LITHOSPHERE - GRADE 11
Gold
bearing ore
A
NaAu(CN)2
B
Gold
precipitate
C
Pure gold
1. Name the process indicated by A.
2. During process A, gold is extracted from the ore. Is gold oxidised or reduced
during this process?
3. Use oxidation numbers to explain your answer to the question above.
4. Name the chemical substance that is used in process B.
5. During smelting (illustrated by C in the diagram), gold is sent into a calcining
furnace. Briefly explain the importance of this process taking place in the
furnace.
6. The recovery of gold can have a negative impact on water in our country, if
not managed properly. State at least one negative influence that the recovery
of gold can have on water resources and how it will impact on humans and the
environment.
21.6
Mining and mineral processing: Iron
Iron is one of the most abundant metals on Earth. Its concentration is highest in the core, and
lower in the crust. It is extracted from iron ore and is almost never found in its elemental form.
Iron ores are usually rich in iron oxide minerals and may vary in colour from dark grey to rusty
red. Iron is usually found in minerals such as magnetite (Fe3 O4 ) and hematite (Fe2 O3 ). Iron
ore also contains other elements, which have to be removed in various ways. These include silica
(Si), phosphorus (P), aluminium (Al) and sulfur (S).
21.6.1
Iron mining and iron ore processing
One of the more common methods of mining for iron ore is open cast mining. Open cast
mining is used when the iron ore is found near the surface. Once the ore has been removed, it
needs to be crushed into fine particles before it can be processed further.
As mentioned earlier, iron is commonly found in the form of iron oxides. To create pure iron,
the ore must be smelted to remove the oxygen.
Definition: Smelting
Smelting is a method used to extract a metal from its ore and then purify it.
Smelting usually involves heating the ore and also adding a reducing agent (e.g. carbon) so that
the metal can be freed from its ore. The bonds between iron and oxygen are very strong, and
therefore it is important to use an element that will form stronger bonds with oxygen that the
iron. This is why carbon is used. In fact, carbon monoxide is the main ingredient that is needed
to strip oxygen from iron. These reactions take place in a blast furnace.
A blast furnace is a huge steel container many metres high and lined with heat-resistant material.
In the furnace the solid raw materials, i.e. iron ore, carbon (in the form of ’coke’, a type of coal)
and a flux (e.g. limestone) are fed into the top of the furnace and a blast of heated air is forced
◦
into the furnace from the bottom. Temperatures in a blast furnace can reach 2000 C. A simple
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21.6
diagram of a blast furnace is shown in figure 21.4. The equations for the reactions that take
place are shown in the flow diagram below.
STEP 1: Production of carbon monoxide
C + O2 → CO2
CO2 + C → 2CO
STEP 2: Reduction of iron oxides takes place in a number of stages to produce iron.
3F e2 O3 + CO → 2F e3 O4 + CO2
F e3 O4 + CO → 3F eO + CO2
F eO + CO → F e + CO2
STEP 3: Fluxing
The flux is used to melt impurities in the ore. A common flux is limestone (CaCO3 ). Common
impurities are silica, phosphorus (makes steel brittle), aluminium and sulfur (produces SO2
gases during smelting and interferes with the smellting process).
CaCO3 → CaO + CO2
CaO + SiO2 → CaSiO3
In step 3, the calcium carbonate breaks down into calcium oxide and carbon dioxide. The calcium
oxide then reacts with silicon dioxide (the impurity) to form a slag. In this case the slag is the
CaSiO3 . The slag melts in the furnace, whereas the silicon dioxide would not have, and floats
on the more dense iron. This can then be separated and removed.
iron
oxide
(haematite)
limestone
coke
waste
gases
Step 2
waste
gases
carbon
monoxide
+
iron oxide
dioxide
+
iron
dioxide
+
carbon
carbon
monoxide
carbon
+
oxygen
carbon
dioxide
carbon
Step 1
carbon
air
blast
molten
slag
iron
forms
carbon
monoxide forms
carbon dioxide
forms
air
blast
molten
iron
Figure 21.4: A blast furnace, showing the reactions that take place to produce iron
21.6.2
Types of iron
Iron is the most used of all the metals. Its combination of low cost and high strength make it
very important in applications such as industry, automobiles, the hulls of large ships and in the
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structural components of buildings. Some of the different forms that iron can take include:
• Pig iron is raw iron and is the direct product when iron ore and coke are smelted. It has
between 4% and 5% carbon and contains varying amounts of contaminants such as sulfur,
silicon and phosphorus. Pig iron is an intermediate step between iron ore, cast iron and
steel.
• Wrought iron is commercially pure iron and contains less than 0.2% carbon. It is tough,
malleable and ductile. Wrought iron does not rust quickly when it is used outdoors. It has
mostly been replaced by mild steel for ’wrought iron’ gates and blacksmithing. Mild steel
does not have the same corrosion resistance as true wrought iron, but is cheaper and more
widely available.
• Steel is an alloy made mostly of iron, but also containing a small amount of carbon.
Elements other than carbon can also be used to make alloy steels. These include manganese
and tungsten. By varying the amounts of the alloy elements in the steel, the following
characteristics can be altered: hardness, elasticity, ductility and tensile strength.
• Corrugated iron is actually sheets of galvanised steel that have been rolled to give them
a corrugated pattern. Corrugated iron is a common building material.
One problem with iron and steel is that pure iron and most of its alloys rust. These products therefore need to be protected from water and oxygen, and this is done through painting,
galvanisation and plastic coating.
teresting Iron is also a very important element in all living organisms. One important
Interesting
Fact
Fact
role that iron plays is that it is a component of the protein haemoglobin which
is the protein in blood. It is the iron in the haemoglobin that helps to attract
and hold oxygen so that this important gas can be transported around the body
in the blood, to where it is needed.
21.6.3
Iron in South Africa
The primary steel industry is an important part of the South African economy and it generates
a great deal of foreign exchange.
• About 40 million tons of iron ore is mined annually in South Africa. Approximately 15
million tons are consumed locally, and the remaining 25 million tons are exported.
• South Africa is ranked about 20th in the world in terms of its crude steel production.
• South Africa is the largest steel producer in Africa.
• South Africa exports crude steel and finished steel products, and a lot is also used locally.
• Some of the products that are manufactured in South Africa include reinforcing bars,
railway track material, wire rod, plates and steel coils and sheets.
Exercise: Iron
Iron is usually extracted from heamatite (iron(III)oxide). Iron ore is mixed with
limestone and coke in a blast furnace to produce the metal. The following incomplete
word equations describe the extraction process:
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21.7
A coke + oxygen → gasX
B gasX + coke → gasY
C iron(III)oxide + gasY → iron + gasX
1. Name the gases X and Y.
2. Write a balanced chemical equation for reaction C.
3. What is the function of gas Y in reaction C?
4. Why is limestone added to the reaction mixture?
5. Briefly describe the impact that the mining of iron has on the economy and
the environment in our country.
(DoE Exemplar Paper, Grade 11, 2007)
21.7
Mining and mineral processing: Phosphates
A phosphate is a salt of phosphoric acid (H3 PO4 ). Phosphates are the naturally occurring
form of the element phosphorus. Phosphorus is seldom found in its pure elemental form, and
phosphate therefore refers to a rock or ore that contains phosphate ions. The chemical formula
for the phosphate ion is PO3−
4 .
21.7.1
Mining phosphates
Phosphate is found in beds in sedimentary rock, and has to be quarried to access the ore. A
quarry is a type of open pit mine that is used to extract ore. In South Africa, the main phosphate
producer is at the Palaborwa alkaline igneous complex, which produces about 3 million tons of
ore per year. The ore is crushed into a powder and is then treated with sulfuric acid to form
a superphosphate (Ca(H2 PO4 )2 ), which is then used as a fertilizer. In the equation below, the
phosphate mineral is calcium phosphate (Ca3 (PO4 )2 .
Ca3 (P O4 )2 + 2H2 SO4 → Ca(H2 P O4 )2 + 2CaSO4
Alternatively, the ore can be treated with concentrated phosphoric acid, in which case the reaction
looks like this:
3Ca3 (P O4 )2 .CaF2 + 4H3 P O4 + 9H2 O → 9Ca(H2 P O4 )2 + CaF2
21.7.2
Uses of phosphates
Phosphates are mostly used in agriculture. Phosphates are one of the three main nutrients
needed by plants, and they are therefore an important component of fertilisers to stimulate
plant growth.
teresting Exploring the lithosphere for minerals is not a random process! Geologists
Interesting
Fact
Fact
help to piece together a picture of what past environments might have been
like, so that predictions can be made about where minerals might have a high
concentration. Geophysicists measure gravity, magnetics and the electrical
properties of rocks to try to make similar predictions, while geochemists sample
the soils at the earth’s surface to get an idea of what lies beneath them. You
can see now what an important role scientists play in mineral exploration!
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CHAPTER 21. THE LITHOSPHERE - GRADE 11
Exercise: Phosphates
Rock phosphate [Ca10 (PO4 )6 F2 ], mined from open pit mines at Phalaborwa, is
an important raw material in the production of fertilisers. The following two reactions
are used to transform rock phosphate into water soluble phosphates:
A: Ca10 (PO4 )6 F2 + 7X + 3H2 O → 3Ca(H2 PO4 )2 H2 O + 2HF + 7CaSO4
B: Ca10 (PO4 )6 F2 + 14Y + 10H2 O → 10Ca(H2 PO4 )2 H2 O + 2HF
1. Identify the acids represented by X and Y.
2. Despite similar molecular formulae, the products Ca(H2 PO4 )2 formed in the
two reactions have different common names. Write down the names for each
of these products in reactions A and B.
3. Refer to the products in reactions A and B and write down TWO advantages
of reaction B over reaction A.
4. Why is rock phosphate unsuitable as fertiliser?
5. State ONE advantage and ONE disadvantage of phosphate mining.
(DoE Exemplar Paper, Grade 11, 2007)
Activity :: Case Study : Controversy on the Wild Coast - Titanium mining
Read the extract below, which has been adapted from an article that appeared in
the Mail and Guardian on 4th May 2007, and then answer the questions that follow.
A potentially violent backlash looms in Pondoland over efforts by an Australian company to persuade villagers to back controversial plans to mine
an environmentally sensitive strip of the Wild Coast. The mining will
take place in the Xolobeni dunes, south of Port Edward. The application
has outraged environmental groups, largely because the proposed mining
areas form part of the Pondoland centre of endemism, which has more
species than the United Kingdom, some of which are endemic and facing
extinction.
Exploratory drilling revealed Xolobeni has the world’s 10th largest titanium
deposit, worth about R11 billion. The amount of money that will be
spent over the mine’s 22 years, including a smelter, is estimated at R1.4
billion. The Australian mining company predicts that 570 direct jobs will
be created.
But at least two communities fiercely oppose the mining plans. Some
opponents are former miners who fear Gauteng’s mine dumps and compounds will be replicated on the Wild Coast. Others are employees of failed
ecotourism ventures, who blame the mining company for their situation.
Many are suspicious of outsiders. The villagers have also complained
that some of the structures within the mining company are controlled
by business leaders with political connections, who are in it for their own
gain. Intimidation of people who oppose the mining has also been alleged.
Headman Mandoda Ndovela was shot dead after his outspoken criticism
of the mining.
Mzamo Dlamini, a youth living in one of the villages that will be affected
by the mining, said 10% of the Amadiba ’who were promised riches by
the mining company’ support mining. ’The rest say if people bring those
machines, we will fight.’
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21.8
1. Explain what the following words means:
(a) endemic
(b) smelter
(c) ecotourism
2. What kinds of ’riches’ do you think the Amadiba people have been promised
by the mining company?
3. In two columns, list the potential advantages and disadvantages of mining
in this area.
4. Imagine that you were one of the villagers in this area. Write down three
questions that you would want the mining company to answer before you made
a decision about whether to oppose the mining or not. Share your ideas with
the rest of the class.
5. Imagine that you are an environmentalist. What would your main concerns be
about the proposed mining project? Again share your answers with the rest of
the class.
21.8
Energy resources and their uses: Coal
The products of the lithosphere are also important in meeting our energy needs. Coal is one
product that is used to produce energy. In South Africa, coal is particularly important because
most of our electricity is generated using coal as a fuel. South Africa is the world’s sixth largest
coal producer, with Mpumalanga contributing about 83% of our total production. Other areas
in which coal is produced, include the Free State, Limpopo and KwaZulu-Natal. One of the
problems with coal however, is that it is a non-renewable resource, meaning that once all resources have been used up, it cannot simply be produced again. Burning coal also produces large
quantities of greenhouse gases, which may play a role in global warming. At present, ESKOM,
the South African government’s electric power producer, is the coal industry’s main customer.
21.8.1
The formation of coal
Coal is what is known as a fossil fuel. A fossil fuel is a hydrocarbon that has been formed
from organic material such as the remains of plants and animals. When plants and animals
decompose, they leave behind organic remains that accumulate and become compacted over
millions of years under sedimentary rock. Over time, the heat and pressure in these parts of the
earth’s crust also increases, and coal is formed. When coal is burned, a large amount of heat
energy is released, which is used to produce electricity. Oil is also a fossil fuel and is formed in
a similar way.
Definition: Fossil Fuel
A fossil fuel is a hydrocarbon that is formed from the fossilised remains of dead plants and
animals that have been under conditions of intense heat and pressure for millions of years.
21.8.2
How coal is removed from the ground
Coal can be removed from the crust in a number of different ways. The most common methods
used are strip mining, open cast mining and underground mining.
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1. Strip mining
Strip mining is a form of surface mining that is used when the coal reserves are very
shallow. The overburden (overlying sediment) is removed so that the coal seams can be
reached. These sediments are replaced once the mining is finished, and in many cases,
attempts are made to rehabilitate the area.
2. Open cast mining
Open cast mining is also a form of surface mining, but here the coal deposits are too deep
to be reached using strip mining. One of the environmental impacts of open cast mining
is that the overburden is dumped somewhere else away from the mine, and this leaves a
huge pit in the ground.
3. Underground mining
Undergound mining is normally used when the coal seams are amuch deeper, usually at a
depth greater than 40 m. As with shaft mining for gold, the problem with underground
mining is that it is very dangerous, and there is a very real chance that the ground could
collapse during the mining if it is not supported. One way to limit the danger is to use
pillar support methods, where some of the ground is left unmined so that it forms pillars to
support the roof. All the other surfaces underground will be mined. Using another method
called longwalling, the roof is allowed to collapse as the mined-out area moves along. In
South Africa, only a small percentage of coal is mined in this way.
21.8.3
The uses of coal
Although in South Africa, the main use of coal is to produce electricity, it can also be used for
other purposes.
1. Electricity
In order to generate electricity, solid coal must be crushed and then burned in a furnace
with a boiler. A lot of steam is produced and this is used to spin turbines which then
generate electricity.
2. Gasification
If coal is broken down and subjected to very high temperatures and pressures, it forms a
synthesis gas, which is a mix of carbon dioxide and hydrogen gases. This is very important
in the chemical industry (refer to chapter 23).
3. Liquid fuels
Coal can also be changed into liquid fuels like petrol and diesel using the Fischer-Tropsch
process. In fact, South Africa is one of the leaders in this technology (refer to chapter 23).
The only problem is that producing liquid fuels from coal, rather than refining petroleum
that has been drilled, releases much greater amounts of carbon dioxide into the atmosphere,
and this contributes further towards global warming.
21.8.4
Coal and the South African economy
In South Africa, the coal industry is second only to the gold industry. More than this, South
Africa is one of the world’s top coal exporters, and also one of the top producers. Of the coal
that is produced, most is used locally to produce electricity and the rest is used in industry and
domestically.
The problem with coal however, is that it is a non-renewable resource which means that once
all the coal deposits have been mined, there will be none left. Coal takes such a long time to
form, and requires such specific environmental conditions, that it would be impossible for coal to
re-form at a rate that would keep up with humankind’s current consumption. It is therefore very
important that South Africa, and other countries that rely on coal, start to look for alternative
energy resources.
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21.8.5
21.8
The environmental impacts of coal mining
There are a number of environmental impacts associated with coal mining.
• Visual impact and landscape scars
Coal mining leaves some very visible scars on the landscape, and destroys biodiversity
(e.g. plants, animals). During strip mining and open cast mining, the visual impact is
particularly bad, although this is partly reduced by rehabilitation in some cases.
• Spontaneous combustion and atmospheric pollution
Coal that is left in mine dumps may spontaneously combust, producing large amounts of
sulfurous smoke which contributes towards atmospheric pollution.
• Acid formation
Waste products from coal mining have a high concentration of sulfur compounds. When
these compounds are exposed to water and oxygen, sulfuric acid is formed. If this acid
washes into nearby water systems, it can cause a lot of damage to the ecosystem. Acid
can also leach into soils and alter its acidity. This in turn affects what will be able to grow
there.
• Global warming
As was discussed earlier, burning coal to generate electricity produces carbon dioxide and
nitrogen oxides which contribute towards global warming (refer to chapter 22). Another
gas that causes problems is methane. All coal contains methane, and deeper coal contains
the most methane. As a greenhouse gas, methane is about twenty times more potent than
carbon dioxide.
teresting It is easy to see how mining, and many other activities including industry and veInteresting
Fact
Fact
hicle transport, contribute towards Global Warming. It was for this reason that
South Africa joined the Carbon Sequestration Leadership Forum (CSLF).
The forum is an international climate change initiative that focuses on developing cost effective technologies to separate and capture carbon dioxide from the
atmosphere so that it can be stored in some way. The CSLF also aims to make
these technologies as widely available as possible.
Exercise: Coal in South Africa
The following advertisement appeared in a local paper:
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21.9
CHAPTER 21. THE LITHOSPHERE - GRADE 11
ENERGY STARTS WITH SOUTH AFRICAN COAL
bbb
b
bbb
bbb
bbb
bbb
bb
bbb
bb
es
itiv ves
addervati uring
o
N p r e s c o lo
0
No tificial re 230
r
o
a
f
e
N o e st b
B
b
b
b
b
Coal is as old as the hills and
just as natural.
It’s like juice from real fruit.
Coal SOUTH AFRICA
1. ”Coal is as old as the hills, and just as natural.” Is this statement TRUE?
Motivate your answer by referring to how coal is formed.
2. Coal is a non-renewable energy source. Quote a statement from the advertisement that gives an indication that coal is non-renewable. Give a reason for
your choice.
3. Is coal actually a healthy source of energy? Motivate your answer by referring to all influences that coal and coal mining have on both humans and the
environment.
4. Why is coal used as a primary energy source in South Africa?
(DoE Exemplar Paper 2, Grade 11, 2007)
21.9
Energy resources and their uses: Oil
Oil is another product of the lithosphere which is very important in meeting our fuel needs.
21.9.1
How oil is formed
Oil is formed in a very similar way to coal, except that the organic material is laid down in
oceans. Organisms such as zooplankton and algae settle to the ocean floor and become buried
under layers of mud. Over time, as these layers of sediment accumulate and the heat and pressure
also increase, the organic material changes to a waxy material called kerogen. Eventually, with
continuing heat and pressure, liquid and gas hydrocarbons are formed. These hydrocarbons
are lighter than rock and therefore move upwards through the rock layers before being trapped
by an impermeable layer. Here the oil will slowly accumulate until there is enough that it can
be accessed by oil rigs and other equipment. Crude oil or petroleum, is actually a mixture of
hydrocarbons (mostly alkanes) of different lengths, ranging from 5 carbons to 18 carbons in the
hydrocarbon chain. If the mixture contains mostly short hydrocarbons, then it is a gas called
natural gas. As the hydrocarbon chains in the mixture become longer, the product becomes
more and more solid. Coal is made up of the longest hydrocarbons. For more information on
hydrocarbons, refer to chapter 9.
21.9.2
Extracting oil
When enough oil has accumulated in a well, it becomes economically viable to try to extract it
either through drilling or pumping. If the pressure in the oil reservoir is high, the oil is forced
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21.10
naturally to the surface. This is known as primary recovery of oil. If the pressure is low, then
pumps must be used to extract it. This is known as secondary recovery. When the oil is very
difficult to extract, steam injection into the reservoir can be used to heat the oil, reduce its
viscosity and make it easier to extract.
While most of South Africa’s oil is imported and then processed at a refinery in either Durban,
Cape Town or Sasolburg, some is extracted from coal, as discussed in section 21.8.
21.9.3
Other oil products
Oil can also be used to make a variety of different products. You will find more information on
this in chapter 23.
• Fractional distillation
Fractional distillation is the separation of a mixture into the parts that make it up. In
oil refineries, crude oil is separated into useful products such as asphalt, diesel, fuel oil,
gasoline, kerosine, liquid petroleum gas (LPG) and tar, to name just a few.
• Cracking
There are two types of cracking, steam cracking and hydrocracking. Cracking is used
to change heavy hydrocarbons such as petroleum into lighter hydrocarbons such as fuels
(LPG and gasoline), plastics (ethylene) and other products that are needed to make fuel
gas (propylene).
21.9.4
The environmental impacts of oil extraction and use
Some of the key environmental impacts associated with the extraction and use of oil are as
follows:
• Pollution
Exploring the oceans for oil, and the actual drilling process, can result in major pollution.
• Ecosystem impacts
Dredging the ocean floors for oil can disrupt seabed ecosystems.
• Global warming
Burning oil as a fuel source produces carbon dioxide, which contributes towards global
warming.
21.10
Alternative energy resources
As the world’s population increases, so does the demand for energy. As we have already mentioned, many of our energy resources are non-renewable and will soon run out. In addition,
many of the fuels that we use produce large amounts of greenhouse gases, which can contribute
towards global warming. If we are to maintain the quality and health of our planet, and also
meet our growing need for energy, we will need to investigate alternative energy resources. In
this next section, we are going to take a closer look at some of these possible alternatives. Many
of these options are very controversial, and may have both pros and cons.
• Nuclear power
Another element that is found in the crust, and which helps to meet our energy needs, is
uranium. Uranium produces energy through the process of nuclear fission (chapter ??).
Neutrons are aimed at the nucleii of the uranium atoms in order to split them. When the
nucleus of a uranium atom is split, a large amount of energy is released as heat. This
heat is used to produce steam, which turns turbines to generate electricity. Uranium is
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CHAPTER 21. THE LITHOSPHERE - GRADE 11
produced as a by-product of gold in some mines in the Witwatersrand, and as a by-product
in some copper mines, for example in Palaborwa. Many people regard this type of nuclear
power as relatively environmentally friendly because it doesn’t produce a lot of greenhouse
gases. However, generating nuclear power does produce radioactive wastes, which must
be carefully disposed of in order to prevent contamination. There are also concerns around
leaking of nuclear materials.
• Natural gas
Natural gas is formed in a similar way to oil and is often located above oil deposits in the
earth’s crust. ’Natural gas’ refers to a hydrocarbon gas, composed mostly of methane. It
is highly combustible and produces low emissions.
In June 2002, construction began on a pipeline that would stretch for 865 km between
Mozambique and South Africa. Mozambique has large sources of under-utilised natural gas
and so an agreement was reached between SASOL and the South African and Mozambican
governments to build the pipeline, which would transport natural gas from Mozambique to
South Africa. The benefits of natural gas include the fact that it is a clean-burning fossil
fuel and few by-products are emitted as pollutants. It is also an economical and efficient
energy source as the gas can easily be piped directly to a customer’s facility.
• Biofuels
In many parts of the world, ethanol is currently being used as a substitute for crude
petroleum. Ethanol can be produced through the fermentation of sugar-containing products such as sugar cane. One of the problems with this however, is the vast areas of land
that are needed to cultivate the necessary crops. Crops such as maize can also be used in
the process. In South Africa, a company called ’Ethanol Africa’ has been set up by commercial farmers to convert their surplus maize into environmentally-friendly biofuel, and
plans are underway to establish ethanol plants in some of the maize-producing provinces.
• Hydropower
Hydropower produces energy from the action of falling water. As water falls from a height,
the energy is used to turn turbines which produce electricity. However, for hydropower to
be effective, a large dam is needed to store water. The building of a dam comes with
its own set of problems such as the expense of construction, as well as the social and
environmental impacts of relocating people (if the area is populated),and disrupting a
natural river course.
• Solar energy
Solar energy is energy from the sun. The sun’s radiation is trapped in solar panels and
is then converted into electricity. While this process is environmentally friendly, and solar
energy is a renewable resource, the supply of radiation is not constant (think for example
of cloudy days, and nights), and the production of electricity is not efficient. Solar energy
can however meet small energy needs such as the direct heating of homes.
• Geothermal energy
This type of energy comes from the natural heat below the Earth’s surface. If hot underground steam can be tapped and brought to the surface, it has the potential to produce
electricity.
Activity :: Discussion : Using energy wisely
The massive power cuts or ’load shedding’ that South Africans began to experience at the beginning of 2008, were a dramatic wake-up call to the growing energy
crisis that the country faces.
There are alternative energy sources available, but they will take years to become
functional, and many of them have their own problems. Another way to look at the
problem, is to put the focus on reducing how much energy is used rather than
focusing only on ways to meet the growing demand.
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21.11
1. In your groups, discuss ways that each of the following groups of people could
save energy.
(a) industries
(b) domestic users
(c) farmers
2. Discuss creative incentives that could be used to encourage each of these groups
to reduce their energy consumption.
21.11
Summary
• The lithosphere is the solid, outermost part of our planet and contains many important
metal elements such as gold and iron, as well as products that are needed to produce
energy.
• These elements are seldom found in their pure form, but rather as minerals in rocks.
• A mineral is formed through geological processes. It can either be a pure element (e.g.
gold) or may consist of a number of different elements e.g. the gold-bearing mineral
calaverite (AuTe2 ).
• A rock is an aggregate of a number of different minerals.
• An ore is rock that contains minerals which make it valuable for mining.
• Minerals have been used throughout history. As new metals and minerals were discovered,
important growth took place in industry, agriculture and technology.
• Gold is one of the most important metals in the history of South Africa. It was the discovery
of gold that led to an influx of fortune-seeking foreigners, and a growth in mining villages
and towns.
• Most of South Africa’s gold is concentrated in the ’Golden Arc’ in the area between
Johannesburg and Welkom.
• Three methods are used to obtain gold from the lithosphere: panning, open cast mining
and shaft mining.
• Gold ore must be processed so that the metal can be removed. One way to process the
ore after it has been crushed is a method called gold cyanidation. A cyanide solution is
added to the crushed ore so that a gold-cyanide solution is formed. Zinc is then added to
this solution so that the gold is precipitated out.
• Gold has a number of important characteristics which make it a useful metal for jewelery
and other applications. The metal is shiny, durable, malleable, ductile, is a good conductor
of electricity and is also a good heat reflector.
• Gold mining has a number of environmental impacts, which include resource consumption, air pollution, poisoned water, solid waste and the destruction of biodiversity in natural
areas.
• Mine rehabilitation is one way of restoring old mine sites to what they were like before.
• Iron is another important metal and is used in industry, furniture and building materials.
• Iron is usually found in minerals such as iron oxides. These minerals must be processed
to remove the metal.
• When iron ore is processed, a blast furnace is used. The iron ore, carbon and a flux are
added to the top of the furnace and hot air is blasted into the bottom of the furnace.
A number of reactions occur in the furnace to finally remove the iron from its ore. Iron
oxides are reduced by carbon monoxide to produce iron.
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• Iron can occur in a number of forms, depending on its level of purity and carbon content.
It can also occur in an alloy e.g. steel.
• Phosphates are found in sedimentary rock, which must be quarried to access the ore.
• Phosphates react with phosphoric acid or sulfuric acid to produce a superphosphate
(Ca(H2 PO4 )2 ), which is an important component in fertilisers.
• The products of the lithosphere are also important in meeting energy needs. Coal and
oil can be extracted from the lithosphere for this purpose.
• Coal and oil are both fossil fuels. A fossil fuel is a hydrocarbon that has been formed
from the fossilsed remains of plants and animals that have been under conditions of high
heat and pressure over a long period of time.
• Coal and oil are non-renewable resources, meaning that once they have been used up,
no more can be produced.
• Coal can be removed from the ground using strip mining, open cast mining or underground mining.
• Coal is burned to produce energy, which can be used to generate electricity. Coal can
also be used to produce liquid fuels or a syngas which can be converted into other useful
products for the chemical industry.
• Some of the environmental impacts associated with coal mining include landscape scars,
spontaneous combustion, acid formation and global warming.
• Oil is also a fossil fuel but it forms in the oceans. It can extracted using either pumping
or drilling, depending on the pressure of the oil.
• Fractional distillation of oil can be used to make products such as diesel, gasoline and
liquid petroleum gas.
• Cracking can be used to convert heavy hydrocarbons to light hydrocarbons.
• The environmental impacts of oil extraction and use are similar to those for coal.
• In view of the number of environmental impacts associated with the extraction and use
of coal and oil, other alternative energy sources should be considered. These include
nuclear power, biofuels, hydropower and a number of others. All of these alternatives have
their own advantages and disadvantages.
Exercise: Summary Exercise
1. Give one word to describe each of the following phrases:
(a)
(b)
(c)
(d)
earth’s crust together with the upper layer of the mantle
a mineral containing silica and oxygen
an alloy of iron and tin
a manual technique used to sort gold from other sediments
2. For each of the following questions, choose the one correct answer from the
list provided.
(a) One of the main reasons that South Africa’s gold industry has been so
economically viable is that...
i. gold panning can be used as an additional method to extract gold
ii. open cast mining can be used to extract gold reserves
iii. South Africa’s geological history is such that its gold reserves are concentrated in large reefs
iv. South Africa has large amounts of water to use in mining
(b) The complete list of reactants in an iron blast furnace is...
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CHAPTER 21. THE LITHOSPHERE - GRADE 11
i.
ii.
iii.
iv.
3.
carbon and oxygen
coal, oxygen, iron ore and limestone
carbon, oxygen and iron ore
coal, air, iron ore and slag
More profits, more poisons
In the last three decades, gold miners have made use of cyanidation to
recover gold from the ore. Over 99% of gold from ore can be extracted
in this way. It allows miners to obtain gold flakes - too small for the
eye to see. Gold can also be extracted from the waste of old operations
which sometimes leave as much as a third of the gold behind.
The left-over cyanide can be re-used, but is more often stored in a
pond behind a dam or even dumped directly into a local river. A
teaspoonful of 2% solution of cyanide can kill a human adult.
Mining companies insist that cyanide breaks down when exposed to
sunlight and oxygen which render it harmless. They also point to
scientific studies that show that cyanide swallowed by fish will not
’bio-accumulate’, which means it does not pose a risk to anyone who
eats the fish. In practice, cyanide solution that seeps into the ground
will not break down because of the absence of sunlight. If the cyanide
solution is very acidic, it could turn into cyanide gas, which is toxic
to fish. On the other hand, if the solution is alkaline the cyanide does
not break down.
There are no reported cases of human death from cyanide spills. If
you don’t see corpses, everything is okay.
(a) What is cyanidation?
(b) What type of chemical reaction takes place during this process: precipitation, acid-base or redox?
(c) Is the pH of the solution after cyanidation greater than, less than or equal
to 7?
(d) How is solid gold recovered from this solution?
(e) Refer to cyanidation and discuss the meaning of the heading of this extract:
More profits, more poisons.
(DoE Grade 11 Paper 2, 2007)
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21.11
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CHAPTER 21. THE LITHOSPHERE - GRADE 11
420
Chapter 22
The Atmosphere - Grade 11
Our earth is truly an amazing planet! Not only is it exactly the right distance from the sun to
have temperatures that will support life, but it is also one of the only planets to have liquid water
on its surface. In addition, our earth has an atmosphere that has just the right composition to
allow life to exist. The atmosphere is the layer of gases that surrounds the earth. We may not
always be aware of them, but without these gases, life on earth would definitely not be possible.
The atmosphere provides the gases that animals and plants need for respiration (breathing) and
photosynthesis (the production of food), it helps to keep temperatures on earth constant and
also protects us from the sun’s harmful radiation.
In this chapter, we are going to take a closer look at the chemistry of the earth’s atmosphere
and at some of the human activities that threaten the delicate balance that exists in this part of
our planet.
22.1
The composition of the atmosphere
Earth’s atmosphere is a mixture of gases. Two important gases are nitrogen and oxygen, which
make up about 78.1% and 20.9% of the atmosphere respectively. A third gas, Argon, contributes
about 0.9%, and a number of other gases such as carbon dioxide, methane, water vapour, helium
and ozone make up the remaining 0.1%. In an earlier chapter, we discussed the importance of
nitrogen as a component of proteins, the building blocks of life. Similarly, oxygen is essential for
life because it is the gas we need for respiration. We will discuss the importance of some of the
other gases later in this chapter.
teresting
Interesting
Fact
Fact
The earth’s early atmosphere was very different from what it is today. When the earth
formed around 4.5 billion years ago, there was probably no atmosphere. Some scientists
believe that the earliest atmosphere contained gases such as water vapour, carbon dioxide,
nitrogen and sulfur which were released from inside the planet as a result of volcanic activity.
Many scientists also believe that the first stage in the evolution of life, around 4 billion
years ago, needed an oxygen-free environment. At a later stage, these primitive forms of
plant life began to release small amounts of oxygen into the atmosphere as a product of
photosynthesis. During photosynthesis, plants use carbon dioxide, water and sunlight to
produce simple sugars. Oxygen is also released in the process.
6CO2 + 6H2 O + sunlight → C6 H12 O6 + 6O2
This build-up of oxygen in the atmosphere eventually led to the formation of the ozone layer,
which helped to filter the sun’s harmful UV radiation so that plants were able to flourish
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CHAPTER 22. THE ATMOSPHERE - GRADE 11
in different environments. As plants became more widespread and photosythesis increased,
so did the production of oxygen. The increase in the amount of oxygen in the atmosphere
would have allowed more forms of life to exist.
If you have ever had to climb to a very high altitude (altitude means the ’height’ in the atmosphere), you will have noticed that it becomes very difficult to breathe, and many climbers suffer
from ’altitude sickness’ before they reach their destination. This is because the density of gases
becomes less as you move higher in the atmosphere. It is gravity that holds the atmosphere
close to the earth. As you move higher, this force weakens slightly and so the gas particles
become more spread out. In effect, when you are at a high altitude, the gases in the atmosphere
haven’t changed, but there are fewer oxygen molecules in the same amount of air that you are
able to breathe.
Definition: Earth’s atmosphere
The Earth’s atmosphere is a layer of gases that surround the planet, and which are held there
by the Earth’s gravity. The atmosphere contains roughly 78.1% nitrogen, 20.9% oxygen,
0.9% argon, 0.038% carbon dioxide, trace amounts of other gases, and a variable amount of
water vapour. This mixture of gases is commonly known as air. The atmosphere protects
life on Earth by absorbing ultraviolet solar radiation and reducing temperature extremes
between day and night.
22.2
The structure of the atmosphere
The earth’s atmosphere is divided into different layers, each with its own particular characteristics
(figure 22.1).
22.2.1
The troposphere
The troposphere is the lowest level in the atmosphere, and it is the part in which we live. The
troposphere varies in thickness, and extends from the ground to a height of about 7km at the
poles and about 18km at the equator. An important characteristic of the troposphere is that its
temperature decreases with an increase in altitude. In other words, as you climb higher, it will
get colder. You will have noticed this if you have climbed a mountain, or if you have moved
from a city at a high altitude to one which is lower; the average temperature is often lower where
the altitude is higher. This is because the troposphere is heated from the ’bottom up’. In other
words, places that are closer to the Earth’s surface will be warmer than those at higher altitudes.
The heating of the atmosphere will be discussed in more detail later in this chapter.
The word troposphere comes from the Greek tropos, meaning turning or mixing. The troposphere is the most turbulent part of the atmosphere and is the part where our weather takes
place. Weather is the state of the air at a particular place and time e.g. if it is warm or cold,
wet or dry, and how cloudy or windy it is. Generally, jet aircraft fly just above the troposphere
to avoid all this turbulence.
22.2.2
The stratosphere
Above the troposphere is another layer called the stratosphere, where most long distance aircraft fly. The stratosphere extends from altitudes of 10 to 50km. If you have ever been in an
aeroplane and have looked out the window once you are well into the flight, you will have noticed
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CHAPTER 22. THE ATMOSPHERE - GRADE 11
22.2
120
110
Thermosphere
100
90
80
70
Mesosphere
Height (km)
60
50
40
Stratosphere
30
20
Troposphere
10
0
-100
-90
-80
-70
-60
-50
-40
-30
-20
Temperature (◦ C)
-10
0
10
20
Figure 22.1: A generalised diagram showing the structure of the atmosphere to a height of 110
km
that you are actually flying above the level of the clouds. As we have already mentioned, clouds
and weather occur in the troposphere, whereas the stratosphere has very stable atmospheric
conditions and very little turbulence. It is easy to understand why aircraft choose to fly here!
The stratosphere is different from the troposphere because its temperature increases as altitude
increases. This is because the stratosphere absorbs solar radiation directly, meaning that the
upper layers closer to the sun will be warmer. The upper layers of the stratosphere are also
warmer because of the presence of the ozone layer. Ozone (O3 ) is formed when solar radiation
splits an oxygen molecule (O2 ) into two atoms of oxygen. Each individual atom is then able to
combine with an oxygen molecule to form ozone. The two reactions are shown below:
O2 → O + O
O + O2 → O3
The change from one type of molecule to another produces energy, and this contributes to higher
temperatures in the upper part of the stratosphere. An important function of the ozone layer
is to absorb UV radiation and reduce the amount of harmful radiation that reaches the Earth’s
surface.
Extension: CFCs and the ozone layer
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CHAPTER 22. THE ATMOSPHERE - GRADE 11
You may have heard people talking about ’the hole in the ozone layer’. What do
they mean by this and do we need to be worried about it?
Most of the earth’s ozone is found in the stratosphere and this limits the amount
of UV radiation that reaches the earth. However, human activities have once again
disrupted the chemistry of the atmosphere. Chlorofluorocarbons (CFC’s) are compounds found in aerosol cans, fridges and airconditioners. In aerosol cans, it is the
CFC’s that cause the substance to be sprayed outwards. The bad side of CFC’s is
that, when they are released into the atmosphere, they break down ozone molecules
so that the ozone is no longer able to protect us as much from UV rays. The ’ozone
hole’ is actually a thinning of the ozone layer approximately above Antarctica. Let’s
take a closer look at the chemical reactions that are involved in breaking down ozone:
1. When CFC’s react with UV radiation, a carbon-chlorine bond in the chlorofluorocarbon breaks and a new compound is formed, with a chlorine atom.
CF Cl3 + U V → CF Cl2 + Cl
2. The single chlorine atom reacts with ozone to form a molecule of chlorine
monoxide and oxygen gas. In the process, ozone is destroyed.
Cl + O3 → ClO + O2
3. The chlorine monoxide then reacts with a free oxygen atom (UV radiation
breaks O2 down into single oxygen atoms) to form oxygen gas and a single
chlorine atom.
ClO + O → Cl + O2
4. The chlorine atom is then free to attack more ozone molecules, and the process
continues. A single CFC molecule can destroy 100 000 ozone molecules.
One possible consequence of ozone depletion is an increase in the incidence
of skin cancer because there is more UV radiation reaching earth’s surface. CFC
replacements are now being used to reduce emissions, and scientists are trying to
find ways to restore ozone levels in the atmosphere.
22.2.3
The mesosphere
The mesosphere is located about 50-80/85km above Earth’s surface. Within this layer, temperature decreases with increasing altitude. Temperatures in the upper mesosphere can fall as
low as -100◦C in some areas. Millions of meteors burn up daily in the mesosphere because of
collisions with the gas particles that are present in this layer. This leads to a high concentration
of iron and other metal atoms.
22.2.4
The thermosphere
The thermosphere exists at altitudes above 80 km. In this part of the atmosphere, ultraviolet
(UV) and shorter X-Ray radiation from the sun cause neutral gas atoms to be ionised. At these
radiation frequencies, photons from the solar radiation are able to dislodge electrons from neutral atoms and molecules during a collision. A plasma is formed, which consists of negative free
electrons and positive ions. This part of the atmosphere is called the ionosphere. At the same
time that ionisation takes place however, an opposing process called recombination also begins.
Some of the free electrons are drawn to the positive ions, and combine again with them if they
are in close enough contact. Since the gas density increases at lower altitudes, the recombination process occurs more often here because the gas molecules and ions are closer together. The
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CHAPTER 22. THE ATMOSPHERE - GRADE 11
22.2
ionisation process produces energy which means that the upper parts of the thermosphere, which
are dominated by ionisation, have a higher temperature than the lower layers where recombination takes place. Overall, temperature in the thermosphere increases with an increase in altitude.
Extension: The ionosphere and radio waves
The ionosphere is of practical importance because it allows radio waves to be transmitted. A radio wave is a type of electromagnetic radiation that humans use to
transmit information without wires. When using high-frequency bands, the ionosphere is used to reflect the transmitted radio beam. When a radio wave reaches
the ionosphere, the electric field in the wave forces the electrons in the ionosphere
into oscillation at the same frequency as the radio wave. Some of the radio wave
energy is given up to this mechanical oscillation. The oscillating electron will then
either recombine with a positive ion, or will re-radiate the original wave energy back
downward again. The beam returns to the Earth’s surface, and may then be reflected
back into the ionosphere for a second bounce.
teresting
Interesting
Fact
Fact
The ionosphere is also home to the auroras. Auroras are caused by the collision of charged
particles (e.g. electrons) with atoms in the earth’s upper atmosphere. Charged particles are
energised and so, when they collide with atoms, the atoms also become energised. Shortly
afterwards, the atoms emit the energy they have gained, as light. Often these emissions
are from oxygen atoms, resulting in a greenish glow (wavelength 557.7 nm) and, at lower
energy levels or higher altitudes, a dark red glow (wavelength 630 nm). Many other colours
can also be observed. For example, emissions from atomic nitrogen are blue, and emissions
from molecular nitrogen are purple. Auroras emit visible light (as described above), and
also infra-red, ultraviolet and x-rays, which can be observed from space.
Exercise: The composition of the atmosphere
1. Complete the following summary table by providing the missing information for
each layer in the atmosphere.
Atmospheric
Height (km)
Gas composition General characlayer
teristics
Troposphere
0-18
Turbulent; part
of
atmosphere
where weather
occurs
Ozone reduces
harmful radiation
reaching Earth
Mesosphere
High
concentration of metal
atoms
more than 80
km
2. Use your knowledge of the atmosphere to explain the following statements:
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CHAPTER 22. THE ATMOSPHERE - GRADE 11
(a) Athletes who live in coastal areas need to acclimatise if they are competing
at high altitudes.
(b) Higher incidences of skin cancer have been recorded in areas where the
ozone layer in the atmosphere is thin.
(c) During a flight, turbulence generally decreases above a certain altitude.
22.3
Greenhouse gases and global warming
22.3.1
The heating of the atmosphere
As we mentioned earlier, the distance of the earth from the sun is not the only reason that
temperatures on earth are within a range that is suitable to support life. The composition of the
atmosphere is also critically important.
The earth receives electromagnetic energy from the sun in the visible spectrum. There are also
small amounts of infrared and ultraviolet radiation in this incoming solar energy. Most of the
radiation is shortwave radiation, and it passes easily through the atmosphere towards the earth’s
surface, with some being reflected before reaching the surface. At the surface, some of the energy is absorbed, and this heats up the earth’s surface. But the situation is a little more complex
than this.
A large amount of the sun’s energy is re-radiated from the surface back into the atmosphere as
infrared radiation, which is invisible. As this radiation passes through the atmosphere, some of
it is absorbed by greenhouse gases such as carbon dioxide, water vapour and methane. These
gases are very important because they re-emit the energy back towards the surface. By doing
this, they help to warm the lower layers of the atmosphere even further. It is this ’re-emission’ of
heat by greenhouse gases, combined with surface heating and other processes (e.g. conduction
and advection) that maintain temperatures at exactly the right level to support life. Without
the presence of greenhouse gases, most of the sun’s energy would be lost and the Earth would
be a lot colder than it is! A simplified diagram of the heating of the atmosphere is shown in
figure 22.2.
22.3.2
The greenhouse gases and global warming
Many of the greenhouse gases occur naturally in small quantities in the atmosphere. However,
human activities have greatly increased their concentration, and this has led to a lot of concern
about the impact that this could have in increasing global temperatures. This phenomenon is
known as global warming. Because the natural concentrations of these gases are low, even a
small increase in their concentration as a result of human emissions, could have a big effect on
temperature. But before we go on, let’s look at where some of these human gas emissions come
from.
• Carbon dioxide (CO2 )
Carbon dioxide enters the atmosphere through the burning of fossil fuels (oil, natural gas,
and coal), solid waste, trees and wood products, and also as a result of other chemical
reactions (e.g. the manufacture of cement). Carbon dioxide can also be removed from
the atmosphere when it is absorbed by plants during photosynthesis.
• Methane (CH4 )
Methane is emitted when coal, natural gas and oil are produced and transported. Methane
emissions can also come from livestock and other agricultural practices and from the decay
of organic waste.
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CHAPTER 22. THE ATMOSPHERE - GRADE 11
22.3
sun
Outgoing long-wave
infrared radiation
Incoming
short-wave
solar radiation
atmosphere
infrared radiation
is absorbed and
re-emitted by
greenhouse gases
in the atmosphere
earth’s surface
Figure 22.2: The heating of the Earth’s atmosphere
• Nitrous oxide (N2 O)
Nitrous oxide is emitted by agriculture and industry, and when fossil fuels and solid waste
are burned.
• Fluorinated gases (e.g. hydrofluorocarbons, perfluorocarbons, and sulfur hexafluoride)
These gases are all synthetic, in other words they are man-made. They are emitted from a
variety of industrial processes. Fluorinated gases are sometimes used in the place of other
ozone-depleting substances (e.g. CFC’s). These are very powerful greenhouse gases, and
are sometimes referred to as High Global Warming Potential gases (’High GWP gases’).
Overpopulation is a major problem in reducing greenhouse gas emissions, and in slowing down
global warming. As populations grow, their demands on resources (e.g. energy) increase, and
so does their production of greenhouse gases.
Extension: Ice core drilling - Taking a look at earth’s past climate
Global warming is a very controversial issue. While many people are convinced
that the increase in average global temperatures is directly related to the increase in
atmospheric concentrations of carbon dioxide, others argue that the climatic changes
we are seeing are part of a natural pattern. One way in which scientists are able to
understand what is happening at present, is to understand the earth’s past atmosphere, and the factors that affected its temperature.
So how, you may be asking, do we know what the earth’s past climate was like?
One method that is used is ice core drilling. Antarctica is the coldest continent
on earth, and because of this there is very little melting that takes place. Over
thousands of years, ice has accumulated in layers and has become more and more
compacted as new ice is added. This is partly why Antarctica is also on average one
of the highest continents! On average, the ice sheet that covers Antarctica is 2500
m thick, and at its deepest location, is 4700 m thick.
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As the snow is deposited on top of the ice sheet each year, it traps different
chemicals and impurities which are dissolved in the ice. The ice and impurities hold
information about the Earth’s environment and climate at the time that the ice was
deposited. Drilling an ice core from the surface down, is like taking a journey back
in time. The deeper into the ice you venture, the older the layer of ice. By analysing
the gases and oxygen isotopes that are present (along with many other techniques)
in the ice at various points in the earth’s history, scientists can start to piece together
a picture of what the earth’s climate must have been like.
Top layers are the most
recently deposited
Increasing age
Bottom layers are
the oldest
One of the most well known ice cores was the one drilled at a Russian station
called Vostok in central Antarctica. So far, data has been gathered for dates as far
back as 160 000 years!
Activity :: Case Study : Looking at past climatic trends
Make sure that you have read the ’Information box’ on ice core drilling before
you try this activity.
The values in the table below were extrapolated from data obtained by scientists
studying the Vostok ice core. ’Local temperature change’ means by how much the
temperature at that time was different from what it is today. For example, if the
local temperature change 160 000 years ago was -9◦ C, this means that atmospheric
temperatures at that time were 9◦ C lower than what they are today. ’ppm’ means
’parts per million’ and is a unit of measurement for gas concentrations.
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CHAPTER 22. THE ATMOSPHERE - GRADE 11
Years before present
(x 1000)
Local temperature
change (◦ C)
160
150
140
130
120
110
100
90
80
70
60
50
40
30
20
10
0 (1850)
Present
-9
-10
-10
-3
+1
-4
-8
-5
-6
-8
-9
-7
-8
-7
-9
-2
-0.5
22.3
Carbon
dioxide
(ppm)
190
205
240
280
278
240
225
230
220
250
190
220
180
225
200
260
280
371
Questions
1. On the same set of axes, draw graphs to show how temperature and carbon
dioxide concentrations have changed over the last 160 000 years. Hint: ’Years
before present’ will go on the x-axis, and should be given negative values.
2. Compare the graphs that you have drawn. What do you notice?
3. Is there a relationship between temperature and the atmospheric concentration
of carbon dioxide?
4. Do these graphs prove that temperature changes are determined by the concentration of gases such as carbon dioxide in the atmosphere? Explain your
answer.
5. What other factors might you need to consider when analysing climatic trends?
22.3.3
The consequences of global warming
Activity :: Group Discussion : The impacts of global warming
In groups of 3-4, read the following extracts and then answer the questions that
follow.
By 2050 Warming to Doom Million Species, Study Says
By 2050, rising temperatures exacerbated by human-induced belches of
carbon dioxide and other greenhouse gases could send more than a million
of Earth’s land-dwelling plants and animals down the road to extinction,
according to a recent study. ”Climate change now represents at least
as great a threat to the number of species surviving on Earth as habitatdestruction and modification,” said Chris Thomas, a conservation biologist
at the University of Leeds in the United Kingdom.
The researchers worked independently in six biodiversity-rich regions around
the world, from Australia to South Africa, plugging field data on species
distribution and regional climate into computer models that simulated the
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CHAPTER 22. THE ATMOSPHERE - GRADE 11
ways species’ ranges are expected to move in response to temperature and
climate changes. According to the researchers’ collective results, the predicted range of climate change by 2050 will place 15 to 35 percent of the
1 103 species studied at risk of extinction.
National Geographic News, 12 July 2004
Global Warming May Dry Up Africa’s Rivers, Study Suggests
Many climate scientists already predict that less rain will fall annually in
parts of Africa within 50 years due to global warming. Now new research
suggests that even a small decrease in rainfall on the continent could
cause a drastic reduction in river water, the lifeblood for rural populations
in Africa.
A decrease in water availability could occur across about 25 percent of
the continent, according to the new study. Hardest hit would be areas in
northwestern and southern Africa, with some of the most serious effects
striking large areas of Botswana and South Africa.
To predict future rainfall, the scientists compared 21 of what they consider
to be the best climate change models developed by research teams around
the world. On average, the models forecast a 10 to 20% drop in rainfall
in northwestern and southern Africa by 2070. With a 20% decrease, Cape
Town would be left with just 42% of its river water, and ”Botswana would
completely dry up,” de Wit said. In parts of northern Africa, river water
levels would drop below 50%.
Less river water would have serious implications not just for people but
for the many animal species whose habitats rely on regular water supplies.
National Geographic News, 3 March 2006
Discussion questions
1. What is meant by ’biodiversity’ ?
2. Explain why global warming is likely to cause a loss of biodiversity.
3. Why do you think a loss of biodiversity is of such concern to conservationists?
4. Suggest some plant or animal species in South Africa that you think might be
particularly vulnerable to extinction if temperatures were to rise significantly.
Explain why you chose these species.
5. In what way do people, animals and plants rely on river water?
6. What effect do you think a 50% drop in river water level in some parts of Africa
would have on the people living in these countries?
7. Discuss some of the other likely impacts of global warming that we can expect
(e.g. sea level rise, melting of polar ice caps, changes in ocean currents).
22.3.4
Taking action to combat global warming
Global warming is a major concern at present. A number of organisations, panels and research
bodies have been working to gather accurate and relevant information so that a true picture of
our current situation can be painted. One important orgaisation that you may have heard of
is the Intergovernmental Panel on Climate Change (IPCC). The IPCC was established in
1988 by two United Nations organizations, the World Meteorological Organization (WMO) and
the United Nations Environment Programme (UNEP), to evaluate the risk of climate change
brought on by humans. You may also have heard of the Kyoto Protocol, which will be discussed
a little later.
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CHAPTER 22. THE ATMOSPHERE - GRADE 11
22.4
Activity :: Group Discussion : World carbon dioxide emissions
The data in the table below shows carbon dioxide emissions from the consumption
of fossil fuels (in million metric tons of carbon dioxide).
Region or Country
United States
Brazil
France
UK
Saudi Arabia
Botswana
South Africa
India
World Total
1980
4754
186
487
608
175
1.26
234
299
18333
1985
4585
187
394
588
179
1.45
298
439
19412
1990
5013
222
368
598
207
2.68
295
588
21426
1995
5292
288
372
555
233
3.44
344
867
22033
2000
5815
345
399
551
288
4.16
378
1000
23851
2004
5912
336
405
579
365
3.83
429
1112
27043
Questions
1. Using a coloured pen, highlight those countries that are ’developed’ and those
that are ’developing’.
2. Explain why CO2 emissions are so much higher in developed countries than in
developing countries.
3. How does South Africa compare to the other developing countries, and also to
the developed countries?
Carbon dioxide emissions are a major problem worldwide. The Kyoto Protocol
was signed in Kyoto, Japan in December 1997. Its main objective was to reduce
global greenhouse gas emissions by encouraging countries to become signatories to
the guidelines that had been laid out in the protocol. These guidelines set targets for
the world’s major producers to reduce their emissions within a certain time. However,
some of the worst contributors to greenhouse gas emissions (e.g. USA) were not
prepared to sign the protocol, partly because of the potential effect this would have on
the country’s economy, which relies on industry and other ’high emission’ activities.
Panel discussion
Form groups with 5 people in each. Each person in the group must adopt one
of the following roles during the discussion:
• the owner of a large industry
• an environmental scientist
• an economist
• a politician
• a chairperson for the discussion
In your group, you are going to discuss some of the economic and environmental
implications for a country that decides to sign the Kyoto Protocol. Each person will
have the opportunity to express the view of the character they have adopted. You
may ask questions of the other people, or challenge their ideas, provided that you
ask permission from the chairperson first.
22.4
Summary
• The atmosphere is the layer of gases that surrounds Earth. These gases are important in
sustaining life, regulating temperature and protecting Earth from harmful radiation.
• The gases that make up the atmosphere are nitrogen, oxygen, carbon dioxide and others
e.g. water vapour, methane.
• There are four layer in the atmosphere, each with their own characteristics.
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CHAPTER 22. THE ATMOSPHERE - GRADE 11
• The troposphere is the lowest layer and here, temperature decreases with an increase in
altitude. The troposphere is where weather occurs.
• The next layer is the stratosphere where temperature increases with an increase in altitude
because of the presence of ozone in this layer, and the direct heating from the sun.
• The depletion of the ozone layer is largely because of CFC’s, which break down ozone
through a series of chemical reactions.
• The mesosphere is characterised by very cold temperatures and meteor collisions. The
mesosphere contains high concentrations of metal atoms.
• In the thermosphere, neutral atoms are ionised by UV and X-ray radiation from the sun.
Temperature increases with an increase in altitude because of the energy that is released
during this ionisation process, which occurs mostly in the upper thermosphere.
• The thermosphere is also known as the ionosphere, and is the part of the atmosphere
where radio waves can be transmitted.
• The auroras are bright coloured skies that occur when charged particles collide with atoms
in the upper atmosphere. Depending on the type of atom, energy is released as light at
different wavelengths.
• The Earth is heated by radiation from the sun. Incoming radiation has a short wavelength
and some is absorbed directly by the Earth’s surface. However, a large amount of energy
is re-radiated as longwave infrared radiation.
• Greenhouse gases such as carbon dioxide, water vapour and methane absorb infrared
radiation and re-emit it back towards the Earth’s surface. In this way, the bottom layers
of the atmsophere are kept much warmer than they would be if all the infrared radiation
was lost.
• Human activities such as the burning of fossil fuels, increase the concentration of greenhouse gases in the atmosphere and may contribute towards global warming.
• Some of the impacts of global warming include changing climate patterns, rising sea levels and a loss of biodiversity, to name a few. Interventions are needed to reduce this
phenomenon.
Exercise: Summary Exercise
1. The atmosphere is a relatively thin layer of gases which support life and provide protection to living organisms. The force of gravity holds the atmosphere
against the earth. The diagram below shows the temperatures associated with
the various layers that make up the atmosphere and the altitude (height) from
the earth’s surface.
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CHAPTER 22. THE ATMOSPHERE - GRADE 11
22.4
120
E
110
100
D
90
80
70
C
Height (km) 60
50
40
B
30
20
10
0
-100
-90 -80 -70 -60 -50 -40 -30 -20 -10
Temperature (◦ C)
A
0
10 20
(a) Write down the names of the layers A, B and D of the atmosphere.
(b) In which one of the layers of the atmosphere is ozone found?
(c) Give an explanation for the decrease in temperature as altitude increases
in layer A.
(d) In layer B, there is a steady increase in temperature as the altitude increases. Write down an explanation for this trend.
2.
Planet Earth in Danger
It is now accepted that greenhouse gases are to blame for planet earth
getting warmer. The increase in the number of sudden floods in Asia
and droughts in Africa; the rising sea level and increasing average temperatures are global concerns. Without natural greenhouse gases,like
carbon dioxide and water vapour,life on earth is not possible. However, the increase in levels of carbon dioxide in the atmosphere since
the Industrial Revolution is of great concern. Greater disasters are to
come, which will create millions of climate refugees. It is our duty to
take action for the sake of future generations who will pay dearly for
the wait-and-see attitude of the current generation. Urgent action to
reduce waste is needed. Global warming is a global challenge and calls
for a global response now, not later.
(Adapted from a speech by the French President, Jacques Chirac)
(a) How do greenhouse gases, such as carbon dioxide, heat up the earth’s
surface?
(b) Draw a Lewis structure for the carbon dioxide molecule
(c) The chemical bonds within the carbon dioxide molecule are polar. Support
this statement by performing a calculation using the table of electronegativities.
(d) Classify the carbon dioxide molecule as polar or non-polar. Give a reason
for your answer.
(e) Suggest ONE way in which YOU can help to reduce the emissions of
greenhouse gases.
3. Plants need carbon dioxide (CO2 ) to manufacture food. However, the engines of motor vehicles cause too much carbon dioxide to be released into the
atmosphere.
(a) State the possible consequence of having too much carbon dioxide in the
atmosphere.
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22.4
CHAPTER 22. THE ATMOSPHERE - GRADE 11
(b) Explain two possible effects on humans if the amount of carbon dioxide in
the atmosphere becomes too low.
(DoE Exemplar Paper Grade 11, 2007)
434
Chapter 23
The Chemical Industry - Grade 12
23.1
Introduction
The chemical industry has been around for a very long time, but not always in the way we
think of it today! Dyes, perfumes, medicines and soaps are all examples of products that have
been made from chemicals that are found in either plants or animals. However, it was not until
the time of the Industrial Revolution that the chemical industry as we know it today began to
develop. At the time of the Industrial Revolution, the human population began to grow very
quickly and more and more people moved into the cities to live. With this came an increase
in the need for things like paper, glass, textiles and soaps. On the farms, there was a greater
demand for fertilisers to help produce enough food to feed all the people in cities and rural areas.
Chemists and engineers responded to these growing needs by using their technology to produce
a variety of new chemicals. This was the start of the chemical industry.
In South Africa, the key event that led to the growth of the chemical industry was the discovery
of diamonds and gold in the late 1800’s. Mines needed explosives so that they could reach the
diamonds and gold-bearing rock, and many of the main chemical companies in South Africa
developed to meet this need for explosives. In this chapter, we are going to take a closer look
at one of South Africa’s major chemical companies, Sasol, and will also explore the chloralkali
and fertiliser industries.
23.2
Sasol
Oil and natural gas are important fuel resources. Unfortunately, South Africa has no large
oil reserves and, until recently, had very little natural gas. One thing South Africa does have
however, is large supplies of coal. Much of South Africa’s chemical industry has developed
because of the need to produce oil and gas from coal, and this is where Sasol has played a very
important role.
Sasol was established in 1950, with its main aim being to convert low grade coal into petroleum
(crude oil) products and other chemical feedstocks. A ’feedstock’ is something that is used to
make another product. Sasol began producing oil from coal in 1955.
teresting The first interest in coal chemistry started as early as the 1920’s. In the early
Interesting
Fact
Fact
1930’s a research engineer called Etienne Rousseau was employed to see whether
oil could be made from coal using a new German technology called the FischerTropsch process. After a long time, and after many negotiations, Rousseau
was given the rights to operate a plant using this new process. As a result,
the government-sponsored ’South African Coal, Oil and Gas Corporation Ltd’
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CHAPTER 23. THE CHEMICAL INDUSTRY - GRADE 12
(commonly called ’Sasol’) was formed in 1950 to begin making oil from coal. A
manufacturing plant was established in the Free State and the town of Sasolburg developed around this plant. Production began in 1955. In 1969, the
Natref crude oil refinery was established, and by 1980 and 1982 Sasol Two and
Sasol Three had been built at Secunda.
23.2.1
Sasol today: Technology and production
Today, Sasol is an oil and gas company with diverse chemical interests. Sasol has three main
areas of operation: Firstly, coal to liquid fuels technology, secondly the production of crude
oil and thirdly the conversion of natural gas to liquid fuel.
1. Coal to liquid fuels
Sasol is involved in mining coal and converting it into synthetic fuels, using the FischerTropsch technology. Figure 23.1 is a simplified diagram of the process that is involved.
Coal mining
Coal
gasification
Condensates from the gas are cooled
to produce tars, oils and pitches.
Ammonia, sulfur and phenolics are
also recovered.
Crude synthesis
gas
(Sasol/Lurgi
process)
Gas
purification
SAS
reactor
Figure 23.1: The gasification of coal to produce liquid fuels
Coal gasification is also known as the Sasol/Lurgi gasification process, and involves converting low grade coal to a synthesis gas. Low grade coal has a low percentage carbon, and
contains other impurities. The coal is put under extremely high pressure and temperature
in the presence of steam and oxygen. The gas that is produced has a high concentration
of hydrogen (H2 ) and carbon monoxide (CO). That is why it is called a ’synthesis gas’,
because it is a mixture of more than one gas.
In the Sasol Advanced Synthol (SAS) reactors, the gas undergoes a high temperature
Fischer-Tropsch conversion. Hydrogen and carbon monoxide react under high pressure and
temperature and in the presence of an iron catalyst, to produce a range of hydrocarbon
products. Below is the generalised equation for the process. Don’t worry too much about
the numbers that you see in front of the reactants and products. It is enough just to see
that the reaction of hydrogen and carbon monoxide (the two gases in the synthesis gas)
produces a hydrocarbon and water.
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C1 to C20
hydrocarbons
CHAPTER 23. THE CHEMICAL INDUSTRY - GRADE 12
23.2
(2n + 1)H2 + nCO → Cn H2n+2 + nH2 O
A range of hydrocarbons are produced, including petrol, diesel, jet fuel, propane, butane,
ethylene, polypropylene, alcohols and acetic acids.
Important: Different types of fuels
It is important to understand the difference between types of fuels and the terminology that
is used for them. The table below summarises some of the fuels that will be mentioned in
this chapter.
Compound
Petroleum
(crude oil)
Natural gas
Paraffin wax
Petrol (gasoline)
Diesel
Liquid Petroleum
Gas (LPG)
Paraffin
Jet fuel
Description
A naturally occurring liquid that forms in the earth’s
lithosphere (see section 21.9 in chapter 21). It is a
mixture of hydrocarbons, mostly alkanes, ranging
from C5 H12 to C18 H38 .
Natural gas has the same origin as petroleum, but
is made up of shorter hydrocarbon chains.
This is made up of longer hydrocarbon chains, making it a solid compound.
A liquid fuel that is derived from petroleum, but
which contains extra additives to increase the octane rating of the fuel. Petrol is used as a fuel in
combustion engines.
Diesel is also derived from petroleum, but is used in
diesel engines.
LPG is a mixture of hydrocarbon gases, and is used
as a fuel in heating appliances and vehicles. Some
LPG mixtures contain mostly propane, while others are mostly butane. LPG is manufactured when
crude oil is refined, or is extracted from natural gas
supplies in the ground.
This is a technical name for the alkanes, but refers
specifically to the linear alkanes. Isoparaffin refers
to non-linear alkanes.
A type of aviation fuel designed for use in jet engined
aircraft. It is an oil-based fuel and contains additives
such as antioxidants, corrosion inhibitors and icing
inhibitors.
You will notice in the diagram that Sasol doesn’t only produce liquid fuels, but also a
variety of other chemical products. Sometimes it is the synthetic fuels themselves that are
used as feedstocks to produce these chemical products. This is done through processes
such as hydrocracking and steamcracking. Cracking is when heavy hydrocarbons are
converted to simpler light hydrocarbons (e.g. LPG and petrol) through the breaking of
C-C bonds. A heavy hydrocarbon is one that has a high number of hydrogen and carbon
atoms (more solid), and a light hydrocarbon has fewer hydrogen and carbon atoms and is
either a liquid or a gas.
Definition: Hydrocracking
Hydrocracking is a cracking process that is assisted by the presence of an elevated partial
pressure of hydrogen gas. It produces chemical products such as ethane, LPG, isoparaffins,
jet fuel and diesel.
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CHAPTER 23. THE CHEMICAL INDUSTRY - GRADE 12
Definition: Steam cracking
Steam cracking occurs under very high temperatures. During the process, a liquid or gaseous
hydrocarbon is diluted with steam and then briefly heated in a furnace at a temperature of
about 8500 C. Steam cracking is used to convert ethane to ethylene. Ethylene is a chemical
that is needed to make plastics. Steam cracking is also used to make propylene, which is
an important fuel gas.
2. Production of crude oil
Sasol obtains crude oil off the coast of Gabon (a country in West Africa) and refines this
at the Natref refinery (figure 23.2). Sasol also sells liquid fuels through a number of service
stations.
Oil processed
at Natref
refinery
Imported
crude oil
Linear-chained hydrocarbons
e.g. waxes, paraffins
and diesel.
Figure 23.2: Crude oil is refined at Sasol’s Natref refinery to produce liquid fuels
3. Liquid fuels from natural gas
Sasol produces natural gas in Mozambique and is expanding its ’gas to fuel’ technology.
The gas undergoes a complex process to produce linear-chained hydrocarbons such as
waxes and paraffins (figure 23.3).
Mozambique
natural
gas
Autothermal
reactor
Sasol Slurry
Phase F-T
reactor
Linear-chained hydrocarbons
e.g. waxes and paraffins
Figure 23.3: Conversion of natural gas to liquid fuels
In the autothermal reactor, methane from natural gas reacts with steam and oxygen over
an iron-based catalyst to produce a synthesis gas. This is a similar process to that involved
in coal gasification. The oxygen is produced through the fractional distillation of air.
Definition: Fractional distillation
Fractional distillation is the separation of a mixture into its component parts, or fractions.
Since air is made up of a number of gases (with the major component being nitrogen),
fractional distillation can be used to separate it into these different parts.
The syngas is then passes through a Sasol Slurry Phase Distillate (SSPD) process. In
this process, the gas is reacted at far lower temperatures than in the SAS reactors. Apart
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CHAPTER 23. THE CHEMICAL INDUSTRY - GRADE 12
23.2
from hard wax and candle wax, high quality diesel can also be produced in this process.
Residual gas from the SSPD process is sold as pipeline gas while some of the lighter
hydrocarbons are treated to produce kerosene and paraffin. Ammonia is also produced,
which can be used to make fertilisers.
teresting Sasol is a major player in the emerging Southern African natural gas industry,
Interesting
Fact
Fact
after investing 1.2 billion US dollars to develop onshore gas fields in central
Mozambique. Sasol has been supplying natural gas from Mozambique’s Temane
field to customers in South Africa since 2004.
Exercise: Sasol processes
Refer to the diagrams summarising the three main Sasol processes, and use these
to answer the following questions:
1. Explain what is meant by each of the following terms:
(a)
(b)
(c)
(d)
(e)
crude oil
hydrocarbon
coal gasification
synthetic fuel
chemical feedstock
2. (a) What is diesel?
(b) Describe two ways in which diesel can be produced.
3. Describe one way in which lighter chemical products such as ethylene, can be
produced.
4. Coal and oil play an important role in Sasol’s technology.
(a) In the table below, summarise the similarities and differences between coal,
oil and natural gas in terms of how they are formed (’origin’), their general
chemical formula and whether they are solid, liquid or gas.
Coal
Oil
Natural gas
Origin
General
chemical
formula
Solid, liquid
or gas
(b) In your own words, describe how coal is converted into liquid fuels.
(c) Explain why Sasol’s ’coal to liquid fuels’ technology is so important in
meeting South Africa’s fuel needs.
(d) Low grade coal is used to produce liquid fuels. What is the main use of
higher grade coal in South Africa?
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23.2
CHAPTER 23. THE CHEMICAL INDUSTRY - GRADE 12
Activity :: Case Study : Safety issues and risk assessments
Safety issues are important to consider when dealing with industrial processes.
Read the following extract that appeared in the Business report on 6th February
2006, and then discuss the questions that follow.
Cape Town - Sasol, the petrochemicals group, was likely to face prosecution on 10 charges of culpable homicide after an explosion at its Secunda
plant in 2004 in which 10 people died, a Cape Town labour law specialist said on Friday. The specialist, who did not want to be named, was
speaking after the inquiry into the explosion was concluded last Tuesday.
It was convened by the labour department.
The evidence led at the inquiry showed a failure on the part of the company to conduct a proper risk assessment and that: Sasol failed to identify
hazards associated with a high-pressure gas pipeline running through the
plant, which had been shut for extensive maintenance work, in the presence of hundreds of people and numerous machines, including cranes,
fitters, contractors, and welding and cutting machines. Because there
had never been a risk assessment, the hazard of the high-pressure pipeline
had never been identified.
Because Sasol had failed to identify the risk, it did not take any measures
to warn people about it, mark the line or take precautions. There had also
been inadequacy in planning the shutdown work. In the face of a barrage
of criticism for the series of explosions that year, Sasol embarked on a comprehensive programme to improve safety at its operations and appointed
Du Pont Safety Resources, the US safety consultancy, to benchmark the
petrochemical giant’s occupational health and safety performance against
international best practice.
1. Explain what is meant by a ’risk assessment’.
2. Imagine that you have been asked to conduct a risk assessment of the Sasol/Lurgi
gasification process. What information would you need to know in order to do
this assessment?
3. In groups, discuss the importance of each of the following in ensuring the safety
of workers in the chemical industry:
• employing experienced Safety, Health and Environment personnel
• regular training to identify hazards
• equipment maintenance and routine checks
4. What other precautions would you add to this list to make sure that working
conditions are safe?
23.2.2
Sasol and the environment
From its humble beginnings in 1950, Sasol has grown to become a major contributor towards the
South African economy. Today, the industry produces more than 150 000 barrels of fuels and
petrochemicals per day, and meets more than 40% of South Africa’s liquid fuel requirements. In
total, more than 200 fuel and chemical products are manufactured at Sasolburg and Secunda,
and these products are exported to over 70 countries worldwide. This huge success is largely due
to Sasol’s ability to diversify its product base. The industry has also helped to provide about 170
000 jobs in South Africa, and contributes around R40 billion to the country’s Gross Domestic
Product (GDP).
However, despite these obvious benefits, there are always environmental costs associated with
industry. Apart from the vast quantities of resources that are needed in order for the industry to
operate, the production process itself produces waste products and pollutants.
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CHAPTER 23. THE CHEMICAL INDUSTRY - GRADE 12
23.2
Exercise: Consumption of resources
Any industry will always use up huge amounts of resources in order to function
effectively, and the chemical industry is no exception. In order for an industry to
operate, some of the major resources that are needed are energy to drive many of
the processes, water, either as a coolant or as part of a process and land for mining
or operations.
Refer to the data table below which shows Sasol’s water use between 2002 and
2005 (Sasol Sustainable Development Report 2005 ), and answer the questions that
follow.
Water use (1000m3 )
River water
Potable water
Total
2002
113 722
15 126
157 617
2003
124 179
10 552
178 439
2004
131 309
10 176
173 319
2005
124 301
10 753
163 203
1. Explain what is meant by ’potable’ water.
2. Describe the trend in Sasol’s water use that you see in the above statistics.
3. Suggest possible reasons for this trend.
4. List some of the environmental impacts of using large amounts of river water
for industry.
5. Suggest ways in which these impacts could be reduced
Exercise: Industry and the environment
Large amounts of gases and pollutants are released during production, and when
the fuels themselves are used. Refer to the table below, which shows greenhouse gas
and atmospheric pollution data for Sasol between 2002 and 2005, and then answer
the questions that follow. (Source: Sasol Sustainable Development Report 2005 )
Greenhouse gases and air
pollutants (kilotonnes)
Carbon dioxide (CO2 )
Hydrogen sulfide (H2 S)
Nitrogen oxides (N Ox )
Sulfur dioxide (SO2 )
2002
2003
2004
2005
57 476
118
168
283
62 873
105
173
239
66 838
102
178
261
60 925
89
166
222
1. Draw line graphs to show how the quantity of each pollutant produced has
changed between 2002 and 2005.
2. Describe what you see in the graphs, and suggest a reason for this trend.
3. Explain what is meant by each of the following terms:
(a) greenhouse gas
(b) global warming
4. Describe some of the possible effects of global warming.
5. When sulfur dioxide is present in the atmosphere, it may react with water
vapour to produce weak sulfuric acid. In the same way, nitrogen dioxide and
water vapour react to form nitric acid. These reactions in the atmosphere may
cause acid rain. Outline some of the possible consequences of acid rain.
6. Many industries are major contributors towards environmental problems such as
global warming, environmental pollution, over-use of resources and acid rain.
Industries are in a difficult position: On one hand they must meet the ever
increasing demands of society, and on the other, they must achieve this with
as little environmental impact as possible. This is a huge challenge.
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23.3
CHAPTER 23. THE CHEMICAL INDUSTRY - GRADE 12
• Work in groups of 3-4 to discuss ways in which industries could be encouraged (or in some cases forced) to reduce their environmental impact.
• Elect a spokesperson for each group, who will present your ideas to the
class.
• Are the ideas suggested by each group practical?
• How easy or difficult do you think it would be to implement these ideas in
South Africa?
teresting Sasol is very aware of its responsibility towards creating cleaner fuels. From
Interesting
Fact
Fact
1st January 2006, the South African government enforced a law to prevent lead
from being added to petrol. Sasol has complied with this. One branch of Sasol,
Sasol Technology also has a bio-diesel research and development programme
focused on developing more environmentally friendly forms of diesel. One way
to do this is to use renewable resources such as soybeans to make diesel. Sasol
is busy investigating this new technology.
23.3
The Chloralkali Industry
The chlorine-alkali (chloralkali) industry is an important part of the chemical industry, and produces chlorine and sodium hydroxide through the electrolysis of salt (NaCl). The main raw
material is brine which is a saturated solution of sodium chloride (NaCl) that is obtained from
natural salt deposits.
The products of this industry have a number of important uses. Chlorine is used to purify water,
and is used as a disinfectant. It is also used in the manufacture of many every-day items such
as hypochlorous acid, which is used to kill bacteria in drinking water. Chlorine is also used in
paper production, antiseptics, food, insecticides, paints, petroleum products, plastics (such as
polyvinyl chloride or PVC), medicines, textiles, solvents, and many other consumer products.
Many chemical products such as chloroform and carbon tetrachloride also contain chlorine.
Sodium hydroxide (also known as ’caustic soda’) has a number of uses, which include making
soap and other cleaning agents, purifying bauxite (the ore of aluminium), making paper and
making rayon (artificial silk).
23.3.1
The Industrial Production of Chlorine and Sodium Hydroxide
Chlorine and sodium hydroxide can be produced through a number of different reactions. However, one of the problems is that when chlorine and sodium hydroxide are produced together, the
chlorine combines with the sodium hydroxide to form chlorate (ClO− ) and chloride (Cl− ) ions.
This produces sodium chlorate, NaClO, a component of household bleach. To overcome this
problem the chlorine and sodium hydroxide must be separated from each other so that they don’t
react. There are three industrial processes that have been designed to overcome this problem,
and to produce chlorine and sodium hydroxide. All three methods involve electrolytic cells
(chapter 17).
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CHAPTER 23. THE CHEMICAL INDUSTRY - GRADE 12
23.3
Important: Electrolytic cells
Electrolytic cells are used to split up or loosen ions. They are made up of an electrolyte and
two electrodes, the cathode and the anode. An electrolytic cell is activated by applying
an external electrical current. This creates an electrical potential across the cathode and
anode, and forces a chemical reaction to take place in the electrolyte. Cations flow towards
the cathode and are reduced. Anions flow to the anode and are oxidised. Two new products
are formed, one product at the cathode and one at the anode.
1. The Mercury Cell
In the mercury-cell (figure 23.4), brine passes through a chamber which has a carbon
electrode (the anode) suspended from the top. Mercury flows along the floor of this
chamber and acts as the cathode. When an electric current is applied to the circuit,
chloride ions in the electrolyte are oxidised to form chlorine gas.
−
2Cl−
(aq) → Cl2(g) + 2e
At the cathode, sodium ions are reduced to sodium.
−
2Na+
(aq) + 2e → 2Na(Hg)
The sodium dissolves in the mercury, forming an amalgam of sodium and mercury. The
amalgam is then poured into a separate vessel, where it decomposes into sodium and mercury. The sodium reacts with water in the vessel and produces sodium hydroxide (caustic
soda) and hydrogen gas, while the mercury returns to the electrolytic cell to be used again.
2Na(Hg) + 2H2 O(l) → 2NaOH(aq) + H2(g)
Cl2
NaCl
Main vessel
Carbon anode (+)
NaCl
Secondary vessel
mercury cathode (-)
sodium
mercury
amalgam
H2
NaOH
mercury returned to the
electrolytic cell
H2 O
Figure 23.4: The Mercury Cell
This method, however, only produces a fraction of the chlorine and sodium hydroxide that
is used by industry as it has certain disadvantages: mercury is expensive and toxic, and
although it is returned to the electrolytic cell, some always escapes with the brine that has
been used. The mercury reacts with the brine to form mercury(II) chloride. In the past
this effluent was released into lakes and rivers, causing mercury to accumulate in fish and
other animals feeding on the fish. Today, the brine is treated before it is discharged so
that the environmental impact is lower.
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CHAPTER 23. THE CHEMICAL INDUSTRY - GRADE 12
2. The Diaphragm Cell
In the diaphragm-cell (figure 23.5), a porous diaphragm divides the electrolytic cell, which
contains brine, into an anode compartment and a cathode compartment. The brine is
introduced into the anode compartment and flows through the diaphragm into the cathode
compartment. When an electric current passes through the brine, the salt’s chlorine ions
and sodium ions move to the electrodes. Chlorine gas is produced at the anode. At the
cathode, sodium ions react with water, forming caustic soda and hydrogen gas. Some salt
remains in the solution with the caustic soda and can be removed at a later stage.
+
Cl2
–
H2
NaCl
anode
cathode
porous diaphragm
Figure 23.5: Diaphragm Cell
This method uses less energy than the mercury cell, but the sodium hydroxide is not as
easily concentrated and precipitated into a useful substance.
teresting To separate the chlorine from the sodium hydroxide, the two half-cells were
Interesting
Fact
Fact
traditionally separated by a porous asbestos diaphragm, which needed to
be replaced every two months. This was damaging to the environment, as
large quantities of asbestos had to be disposed. Today, the asbestos is being
replaced by other polymers which do not need to be replaced as often.
3. The Membrane Cell
The membrane cell (figure 23.6) is very similar to the diaphragm cell, and the same reactions occur. The main difference is that the two electrodes are separated by an ion-selective
membrane, rather than by a diaphragm. The structure of the membrane is such that it
allows cations to pass through it between compartments of the cell. It does not allow
anions to pass through. This has nothing to do with the size of the pores, but rather with
the charge on the ions. Brine is pumped into the anode compartment, and only the positively charged sodium ions pass into the cathode compartment, which contains pure water.
At the positively charged anode, Cl− ions from the brine are oxidised to Cl2 gas.
2Cl− → Cl2(g) + 2e−
At the negatively charged cathode, hydrogen ions in the water are reduced to hydrogen
gas.
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CHAPTER 23. THE CHEMICAL INDUSTRY - GRADE 12
+
Cl2
23.3
–
H2
NaCl
NaOH
anode
cathode
H2 O
membrane
NaCl
Figure 23.6: Membrane Cell
−
2H+
(aq) + 2e → H2(g)
The N a+ ions flow through the membrane to the cathode compartment and react with
the remaining hydroxide (OH − ) ions from the water to form sodium hydroxide (NaOH).
The chloride ions cannot pass through, so the chlorine does not come into contact with the
sodium hydroxide in the cathode compartment. The sodium hydroxide is removed from
the cell. The overall equation is as follows:
2NaCl + 2H2 0 → Cl2 + H2 + 2NaOH
The advantage of using this method is that the sodium hydroxide that is produced is
very pure because it is kept separate from the sodium chloride solution. The caustic soda
therefore has very little salt contamination. The process also uses less electricity and is
cheaper to operate.
Exercise: The Chloralkali industry
1. Refer to the flow diagram below which shows the reactions that take place in
the membrane cell, and then answer the questions that follow.
445
23.3
CHAPTER 23. THE CHEMICAL INDUSTRY - GRADE 12
(b)
ANODE
CATHODE
NaCl is added
to this compartment
(a)
Cl− ions
H+ ions are reduced to H2 gas
Na+ ions in solution
OH− ions in solution
(c)
Na+ and OH−
ions react to
form NaOH
(a) What liquid is present in the cathode compartment at (a)?
(b) Identify the gas that is produced at (b).
(c) Explain one feature of this cell that allows the Na+ and OH− ions to react
at (c).
(d) Give a balanced equation for the reaction that takes place at (c).
2. Summarise what you have learnt about the three types of cells in the chloralkali
industry by completing the table below:
Mercury cell
Diaphragm
cell
Membrane
cell
Main raw material
Mechanism of separating Cl2 and
NaOH
Anode reaction
Cathode reaction
Purity of NaOH produced
Energy consumption
Environmental
impact
23.3.2
Soaps and Detergents
Another important part of the chloralkali industry is the production of soaps and detergents.
You will remember from an earlier chapter, that water has the property of surface tension. This
means that it tends to bead up on surfaces and this slows down the wetting process and makes
cleaning difficult. You can observe this property of surface tension when a drop of water falls
onto a table surface. The drop holds its shape and does not spread. When cleaning, this surface
tension must be reduced so that the water can spread. Chemicals that are able to do this
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CHAPTER 23. THE CHEMICAL INDUSTRY - GRADE 12
23.3
are called surfactants. Surfactants also loosen, disperse and hold particles in suspension, all
of which are an important part of the cleaning process. Soap is an example of one of these
surfactants. Detergents contain one or more surfactants. We will go on to look at these in more
detail.
Definition: Surfactant
A surfactant is a wetting agent that lowers the surface tension of a liquid, allowing it to
spread more easily.
1. Soaps
In chapter 10, a number of important biological macromolecules were discussed, including
carbohydrates, proteins and nucleic acids. Fats are also biological macromolecules. A fat
is made up of an alcohol called glycerol, attached to three fatty acids (figure 23.7). Each
fatty acid is made up of a carboxylic acid attached to a long hydrocarbon chain. An oil
has the same structure as a fat, but is a liquid rather than a solid. Oils are found in plants
(e.g. olive oil, sunflower oil) and fats are found in animals.
glycerol
fatty acids
O
H
H
C
O
C
O
(CH2 )14 CH3
H
C
O
C
O
(CH2 )14 CH3
H
C
O
C
(CH2 )14 CH3
H
Figure 23.7: The structure of a fat, composed of an alcohol and three fatty acids
To make soap, sodium hydroxide (NaOH) or potassium hydroxide (KOH) must be added
to a fat or an oil. During this reaction, the glycerol is separated from the hydrocarbon
chain in the fat, and is replaced by either potassium or sodium ions (figure 23.8). Soaps
are the water-soluble sodium or potassium salts of fatty acids.
teresting Soaps can be made from either fats or oils. Beef fat is a common source of
Interesting
Fact
Fact
fat, and vegetable oils such as palm oil are also commonly used.
Fatty acids consist of two parts: a carboxylic acid group and a hydrocarbon chain. The
hydrocarbon chain is hydrophobic, meaning that it is repelled by water. However, it is
attracted to grease, oils and other dirt. The carboxylic acid is hydrophilic, meaning that
it is attracted to water. Let’s imagine that we have added soap to water in order to clean
a dirty rugby jersey. The hydrocarbon chain will attach itself to the soil particles in the
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23.3
CHAPTER 23. THE CHEMICAL INDUSTRY - GRADE 12
H
H
C
O
O
C
O
H
(CH2 )14 CH3
O
H
C
OH
Na+ − O
C
O
(CH2 )14 CH3
+ 3NaOH
H
C
O
C
O
(CH2 )14 CH3
H
C
OH + Na+ − O
C
O
(CH2 )14 CH3
H
C
O
C
(CH2 )14 CH3
H
C
OH
Na+ − O
C
(CH2 )14 CH3
H
H
glycerol
sodium salts of fatty acids
Figure 23.8: Sodium hydroxide reacts with a fat to produce glycerol and sodium salts of the
fatty acids
jersey, while the carboxylic acid will be attracted to the water. In this way, the soil is pulled
free of the jersey and is suspended in the water. In a washing machine or with vigourous
handwashing, this suspension can be rinsed off with clean water.
Definition: Soap
Soap is a surfactant that is used with water for washing and cleaning. Soap is made by
reacting a fat with either sodium hydroxide (NaOH) or potassium hydroxide (KOH).
2. Detergents
Definition: Detergent
Detergents are compounds or mixtures of compounds that are used to assist cleaning. The
term is often used to distinguish between soap and other chemical surfactants for cleaning.
Detergents are also cleaning products, but are composed of one or more surfactants.
Depending on the type of cleaning that is needed, detergents may contain one or more of
the following:
• Abrasives to scour a surface.
• Oxidants for bleaching and disinfection.
• Enzymes to digest proteins, fats or carbohydrates in stains. These are called biological
detergents.
Exercise: The choralkali industry
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CHAPTER 23. THE CHEMICAL INDUSTRY - GRADE 12
1. Raw material into
main reaction vessel
2.Chlorine is produced
5. H2 gas released
23.3
3. Na-Hg amalgam breaks into
Na and Hg in
second reaction
vessel
4. NaOH is produced
1. The diagram above shows the sequence of steps that take place in the mercury
cell.
(a) Name the ’raw material’ in step 1.
(b) Give the chemical equation for the reaction that produces chlorine in step
2.
(c) What other product is formed in step 2.
(d) Name the reactants in step 4.
2. Approximately 30 million tonnes of chlorine are used throughout the world
annually. Chlorine is produced industrially by the electrolysis of brine. The
diagram represents a membrane cell used in the production of Cl2 gas.
Cl2 +
NaCl
– H
2
NaOH
anode
cathode
H2 O
membrane
NaCl
(a) What ions are present in the electrolyte in the left hand compartment of
the cell?
(b) Give the equation for the reaction that takes place at the anode.
(c) Give the equation for the reaction that takes place at the cathode.
(d) What ion passes through the membrane while these reactions are taking
place?
Chlorine is used to purify drinking water and swimming pool water. The
substance responsible for this process is the weak acid, hypochlorous acid
(HOCl).
(e) One way of putting HOCl into a pool is to bubble chlorine gas through
the water. Give an equation showing how bubbling Cl2 (g) through water
produces HOCl.
(f) A common way of treating pool water is by adding ’granular chlorine’.
Granular chlorine consists of the salt calcium hypochlorite, Ca(OCl)2 . Give
an equation showing how this salt dissolves in water. Indicate the phase
of each substance in the equation.
(g) The OCl− ion undergoes hydrolysis , as shown by the following equation:
OCl− + H2 O ⇔ HOCl + OH−
Will the addition of granular chlorine to pure water make the water acidic,
basic or will it remain neutral? Briefly explain your answer.
(IEB Paper 2, 2003)
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23.4
CHAPTER 23. THE CHEMICAL INDUSTRY - GRADE 12
23.4
The Fertiliser Industry
23.4.1
The value of nutrients
Nutrients are very important for life to exist. An essential nutrient is any chemical element
that is needed for a plant to be able to grow from a seed and complete its life cycle. The same
is true for animals. A macronutrient is one that is required in large quantities by the plant or
animal, while a micronutrient is one that only needs to be present in small amounts for a plant
or an animal to function properly.
Definition: Nutrient
A nutrient is a substance that is used in an organism’s metabolism or physiology and which
must be taken in from the environment.
In plants, the macronutrients include carbon (C), hydrogen (H), oxygen (O), nitrogen (N),
phosphorus (P) and potassium (K). The source of each of these nutrients for plants, and their
function, is summarised in table 23.1. Examples of micronutrients in plants include iron, chlorine,
copper and zinc.
Table 23.1: The source and function of the macronutrients in plants
Nutrient
Source
Function
Carbon
Carbon dioxide in the Component
of
organic
air
molecules such as carbohydrates, lipids and proteins
Hydrogen
Water from the soil
Component
of
organic
molecules
Oxygen
Water from the soil
Component
of
organic
molecules
Nitrogen
Nitrogen compounds Part of plant proteins and
in the soil
chlorophyll.
Also boosts
plant growth.
Phosphorus
Phosphates in the soil Needed for photosynthesis,
blooming and root growth
Potassium
Soil
Building proteins, part of
chlorophyll and reduces diseases in plants
Animals need similar nutrients in order to survive. However since animals can’t photosynthesise,
they rely on plants to supply them with the nutrients they need. Think for example of the human
diet. We can’t make our own food and so we either need to eat vegetables, fruits and seeds (all
of which are direct plant products) or the meat of other animals which would have fed on plants
during their life. So most of the nutrients that animals need are obtained either directly or indirectly from plants. Table 23.2 summarises the functions of some of the macronutrients in animals.
Micronutrients also play an important function in animals. Iron for example, is found in haemoglobin,
the blood pigment that is responsible for transporting oxygen to all the cells in the body.
Nutrients then, are essential for the survival of life. Importantly, obtaining nutrients starts with
plants, which are able either to photosynthesise or to absorb the required nutrients from the soil.
It is important therefore that plants are always able to access the nutrients that they need so
that they will grow and provide food for other forms of life.
23.4.2
The Role of fertilisers
Plants are only able to absorb soil nutrients in a particular form. Nitrogen for example, is absorbed as nitrates, while phosphorus is absorbed as phosphates. The nitrogen cycle (chapter
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CHAPTER 23. THE CHEMICAL INDUSTRY - GRADE 12
23.4
Table 23.2: The functions of animal macronutrients
Nutrient
Function
Carbon
Component of organic compounds
Hydrogen
Component of organic compounds
Oxygen
Component of organic compounds
Nitrogen
Component of nucleic acids and
proteins
Phosphorus
Component of nucleic acids and
phospholipids
Potassium
Helps in coordination and regulating the water balance in the body
19) describes the process that is involved in converting atmospheric nitrogen into a form that
can be used by plants.
However, all these natural processes of maintaining soil nutrients take a long time. As populations
grow and the demand for food increases, there is more and more strain on the land to produce
food. Often, cultivation practices don’t give the soil enough time to recover and to replace
the nutrients that have been lost. Today, fertilisers play a very important role in restoring soil
nutrients so that crop yields stay high. Some of these fertilisers are organic (e.g. compost,
manure and fishmeal), which means that they started off as part of something living. Compost
for example is often made up of things like vegetable peels and other organic remains that have
been thrown away. Others are inorganic and can be made industrially. The advantage of these
commercial fertilisers is that the nutrients are in a form that can be absorbed immediately by
the plant.
Definition: Fertiliser
A fertiliser is a compound that is given to a plant to promote growth. Fertilisers usually
provide the three major plant nutrients and most are applied via the soil so that the nutrients
are absorbed by plants through their roots.
When you buy fertilisers from the shop, you will see three numbers on the back of the packet e.g.
18-24-6. These numbers are called the NPK ratio, and they give the percentage of nitrogen,
phosphorus and potassium in that fertiliser. Depending on the types of plants you are growing,
and the way in which you would like them to grow, you may need to use a fertiliser with a
slightly different ratio. If you want to encourage root growth in your plant for example, you
might choose a fertiliser with a greater amount of phosphorus. Look at the table below, which
gives an idea of the amounts of nitrogen, phosphorus and potassium there are in different types
of fertilisers. Fertilisers also provide other nutrients such as calcium, sulfur and magnesium.
Table 23.3: Common grades of some fertiliser materials
Description
Grade (NPK %)
Ammonium nitrate
34-0-0
Urea
46-0-0
Bone Meal
4-21-1
Seaweed
1-1-5
Starter fertilisers
18-24-6
Equal NPK fertilisers
12-12-12
High N, low P and medium K fertilisers
25-5-15
23.4.3
The Industrial Production of Fertilisers
The industrial production of fertilisers may involve several processes.
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CHAPTER 23. THE CHEMICAL INDUSTRY - GRADE 12
1. Nitrogen fertilisers
Making nitrogen fertilisers involves producing ammonia, which is then reacted with oxygen to produce nitric acid. Nitric acid is used to acidify phosphate rock to produce nitrogen
fertilisers. The flow diagram below illustrates the processes that are involved. Each of these
steps will be examined in more detail.
(a) HABER PROCESS
The production of ammonia
from nitrogen and hydrogen
(b) OSTWALD PROCESS
Production of nitric acid
from ammonia and oxygen
(c) NITROPHOSPHATE PROCESS
Acidification of phosphate rock
with nitric acid to produce phosphoric
acid and calcium nitrate
Figure 23.9: Flow diagram showing steps in the production of nitrogen fertilisers
(a) The Haber Process
The Haber process involves the reaction of nitrogen and hydrogen to produce ammonia. Nitrogen is produced through the fractional distillation of air. Fractional
distillation is the separation of a mixture (remember that air is a mixture of different
gases) into its component parts through various methods. Hydrogen can be produced through steam reforming. In this process, a hydrocarbon such as methane
reacts with water to form carbon monoxide and hydrogen according to the following
equation:
CH4 + H2 O → CO + 3H2
Nitrogen and hydrogen are then used in the Haber process. The equation for the
Haber process is:
N2 (g) + 3H2 (g) → 2NH3 (g)
(The reaction takes place in the presence of an iron (Fe) catalyst under conditions of
200 atmospheres (atm) and 450-500 degrees Celsius)
teresting The Haber process developed in the early 20th century, before the start
Interesting
Fact
Fact
of World War 1. Before this, other sources of nitrogen for fertilisers
had included saltpeter (N aN O3 ) from Chile and guano. Guano is the
droppings of seabirds, bats and seals. By the 20th century, a number
of methods had been developed to ’fix’ atmospheric nitrogen. One of
these was the Haber process, and it advanced through the work of two
German men, Fritz Haber and Karl Bosch (The process is sometimes also
referred to as the ’Haber-Bosch process’). They worked out what the
best conditions were in order to get a high yield of ammonia, and found
these to be high temperature and high pressure. They also experimented
with different catalysts to see which worked best in that reaction. During
World War 1, the ammonia that was produced through the Haber process
was used to make explosives. One of the advantages for Germany was
that, having perfected the Haber process, they did not need to rely on
other countries for the chemicals that they needed to make them.
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23.4
(b) The Ostwald Process
The Ostwald process is used to produce nitric acid from ammonia. Nitric acid can
then be used in reactions that produce fertilisers. Ammonia is converted to nitric
acid in two stages. First, it is oxidised by heating with oxygen in the presence of a
platinum catalyst to form nitric oxide and water. This step is strongly exothermic,
making it a useful heat source.
4NH3 (g) + 5O2 (g) → 4NO(g) + 6H2 O(g)
Stage two, which combines two reaction steps, is carried out in the presence of water.
Initially nitric oxide is oxidised again to yield nitrogen dioxide:
2NO(g) + O2 (g) → 2NO2 (g)
This gas is then absorbed by the water to produce nitric acid. Nitric oxide is also a
product of this reaction. The nitric oxide (NO) is recycled, and the acid is concentrated to the required strength.
3NO2 (g) + H2 O(l) → 2HNO3 (aq) + NO(g)
(c) The Nitrophosphate Process
The nitrophosphate process involves acidifying phosphate rock with nitric acid to
produce a mixture of phosphoric acid and calcium nitrate:
Ca3 (PO4 )2 + 6HNO3 + 12H2 O → 2H3 PO4 + 3Ca(NO3 )2 + 12H2 O
When calcium nitrate and phosphoric acid react with ammonia, a compound fertiliser
is produced.
Ca(NO3 )2 + 4H3 PO4 + 8NH3 → CaHPO4 + 2NH4 NO3 + 8(NH4 )2HPO4
If potassium chloride or potassium sulphate is added, the result will be NPK fertiliser.
(d) Other nitrogen fertilisers
• Urea ((NH2 )2 CO) is a nitrogen-containing chemical product which is produced
on a large scale worldwide. Urea has the highest nitrogen content of all solid
nitrogeneous fertilisers in common use (46.4%) and is produced by reacting ammonia with carbon dioxide.
Two reactions are involved in producing urea:
i. 2NH3 + CO2 → H2 N − COONH4
ii. H2 N − COONH4 → (NH2 )2 CO + H2 O
• Other common fertilisers are ammonium nitrate and ammonium sulphate. Ammonium nitrate is formed by reacting ammonia with nitric acid.
NH3 + HNO3 → NH4 NO3
Ammonium sulphate is formed by reacting ammonia with sulphuric acid.
2NH3 + H2 SO4 → (NH4 )2 SO4
2. Phosphate fertilisers
The production of phosphate fertilisers also involves a number of processes. The first is
the production of sulfuric acid through the contact process. Sulfuric acid is then used
in a reaction that produces phosphoric acid. Phosphoric acid can then be reacted with
phosphate rock to produce triple superphosphates.
(a) The production of sulfuric acid
Sulfuric acid is produced from sulfur, oxygen and water through the contact process.
In the first step, sulfur is burned to produce sulfur dioxide.
S(s) + O2 (g) → SO2 (g)
This is then oxidised to sulfur trioxide using oxygen in the presence of a vanadium(V)
oxide catalyst.
2SO2 + O2 (g) → 2SO3 (g)
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CHAPTER 23. THE CHEMICAL INDUSTRY - GRADE 12
Finally the sulfur trioxide is treated with water to produce 98-99% sulfuric acid.
SO3 (g) + H2 O(l) → H2 SO4 (l)
(b) The production of phosphoric acid
The next step in the production of phosphate fertiliser is the reaction of sulfuric
acid with phosphate rock to produce phosphoric acid (H3 PO4 ). In this example, the
phosphate rock is fluoropatite (Ca5 F(PO4 )3 ).
Ca5 F(PO4 )3 + 5H2 SO4 + 10H2 O → 5CaSO4 2H2 O + HF + 3H3 PO4
(c) The production of phosphates and superphosphates
When concentrated phosphoric acid reacts with ground phosphate rock, triple superphosphate is produced.
3Ca3 (PO4 )2 CaF2 + 4H3 PO4 + 9H2 O → 9Ca(H2 PO4 )2 + CaF2
3. Potassium
Potassium is obtained from potash, an impure form of potassium carbonate (K2 CO3 ).
Other potassium salts (e.g. KCl AND K2 O) are also sometimes included in fertilisers.
23.4.4
Fertilisers and the Environment: Eutrophication
Eutrophication is the enrichment of an ecosystem with chemical nutrients, normally by compounds that contain nitrogen or phosphorus. Eutrophication is considered a form of pollution
because it promotes plant growth, favoring certain species over others. In aquatic environments,
the rapid growth of certain types of plants can disrupt the normal functioning of an ecosystem,
causing a variety of problems. Human society is impacted as well because eutrophication can
decrease the resource value of rivers, lakes, and estuaries making recreational activities less enjoyable. Health-related problems can also occur if eutrophic conditions interfere with the treatment
of drinking water.
Definition: Eutrophication
Eutrophication refers to an increase in chemical nutrients in an ecosystem. These chemical
nutrients usually contain nitrogen or phosphorus.
In some cases, eutrophication can be a natural process that occurs very slowly over time. However, it can also be accelerated by certain human activities. Agricultural runoff, when excess
fertilisers are washed off fields and into water, and sewage are two of the major causes of eutrophication. There are a number of impacts of eutrophication.
• A decrease in biodiversity (the number of plant and animal species in an ecosystem)
When a system is enriched with nitrogen, plant growth is rapid. When the number of
plants increases in an aquatic system, they can block light from reaching deeper. Plants
also consume oxygen for respiration, and if the oxygen content of the water decreases too
much, this can cause other organisms such as fish to die.
• Toxicity
Sometimes, the plants that flourish during eutrophication can be toxic and may accumulate
in the food chain.
teresting South Africa’s Department of Water Affairs and Forestry has a ’National EuInteresting
Fact
Fact
trophication Monitoring Programme’ which was set up to monitor eutrophication
in impoundments such as dams, where no monitoring was taking place.
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CHAPTER 23. THE CHEMICAL INDUSTRY - GRADE 12
23.4
Despite the impacts, there are a number of ways of preventing eutrophication from taking place.
Cleanup measures can directly remove the excess nutrients such as nitrogen and phosphorus
from the water. Creating buffer zones near farms, roads and rivers can also help. These act as
filters and cause nutrients and sediments to be deposited there instead of in the aquatic system.
Laws relating to the treatment and discharge of sewage can also help to control eutrophication.
A final possible intervention is nitrogen testing and modeling. By assessing exactly how much
fertiliser is needed by crops and other plants, farmers can make sure that they only apply just
enough fertiliser. This means that there is no excess to run off into neighbouring streams during
rain. There is also a cost benefit for the farmer.
Activity :: Discussion : Dealing with the consequences of eutrophication
In many cases, the damage from eutrophication is already done. In groups, do
the following:
1. List all the possible consequences of eutrophication that you can think of.
2. Suggest ways to solve these problems, that arise because of eutrophication.
Exercise: Chemical industry: Fertilisers
Why we need fertilisers
There is likely to be a gap between food production and demand in several parts of the world by 2020. Demand is influenced by population
growth and urbanisation, as well as income levels and changes in dietary
preferences.
The facts are as follows:
• There is an increasing world population to feed
• Most soils in the world used for large-scale, intensive production of
crops lack the necessary nutrients for the crops
Conclusion: Fertilisers are needed!
The flow diagram below shows the main steps in the industrial preparation of
two important solid fertilisers.
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23.5
CHAPTER 23. THE CHEMICAL INDUSTRY - GRADE 12
Air
Methane
Nitrogen
Hydrogen
Haber process
Fertiliser C
H2 SO4
Process
Y
NH3
NO
Brown
gas
Liquid E
Fertiliser D
1. Write down the balanced chemical equation for the formation of the brown gas.
2. Write down the name of process Y.
3. Write down the chemical formula of liquid E.
4. Write down the chemical formulae of fertilisers C and D respectively.
The following extract comes from an article on fertilisers:
A world without food for its people
A world with an environment poisoned through the actions of man
Are two contributing factors towards a disaster scenario.
5. Write down THREE ways in which the use of fertilisers poisons the environment.
23.5
Electrochemistry and batteries
You will remember from chapter 17 that a galvanic cell (also known as a voltaic cell) is a type of
electrochemical cell where a chemical reaction produces electrical energy. The emf of a galvanic
cell is the difference in voltage between the two half cells that make it up. Galvanic cells have a
number of applications, but one of the most important is their use in batteries. You will know
from your own experience that we use batteries in a number of ways, including cars, torches,
sound systems and cellphones to name just a few.
23.5.1
How batteries work
A battery is a device in which chemical energy is directly converted to electrical energy. It
consists of one or more voltaic cells, each of which is made up of two half cells that are connected
in series by a conductive electrolyte. The voltaic cells are connected in series in a battery. Each
cell has a positive electrode (cathode), and a negative electrode (anode). These do not touch
each other but are immersed in a solid or liquid electrolyte.
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CHAPTER 23. THE CHEMICAL INDUSTRY - GRADE 12
23.5
Each half cell has a net electromotive force (emf) or voltage. The voltage of the battery is the
difference between the voltages of the half-cells. This potential difference between the two half
cells is what causes an electric current to flow.
Batteries are usually divided into two broad classes:
• Primary batteries irreversibly transform chemical energy to electrical energy. Once the
supply of reactants has been used up, the battery can’t be used any more.
• Secondary batteries can be recharged, in other words, their chemical reactions can be
reversed if electrical energy is supplied to the cell. Through this process, the cell returns to
its original state. Secondary batteries can’t be recharged forever because there is a gradual
loss of the active materials and electrolyte. Internal corrosion can also take place.
23.5.2
Battery capacity and energy
The capacity of a battery, in other words its ability to produce an electric charge, depends on
a number of factors. These include:
• Chemical reactions
The chemical reactions that take place in each of a battery’s half cells will affect the voltage
across the cell, and therefore also its capacity. For example, nickel-cadmium (NiCd) cells
measure about 1.2V, and alkaline and carbon-zinc cells both measure about 1.5 volts.
However, in other cells such as Lithium cells, the changes in electrochemical potential
are much higher because of the reactions of lithium compounds, and so lithium cells can
produce as much as 3 volts or more. The concentration of the chemicals that are involved
will also affect a battery’s capacity. The higher the concentration of the chemicals, the
greater the capacity of the battery.
• Quantity of electrolyte and electrode material in cell
The greater the amount of electrolyte in the cell, the greater its capacity. In other words,
even if the chemistry in two cells is the same, a larger cell will have a greater capacity than
a small one. Also, the greater the surface area of the electrodes, the greater will be the
capacity of the cell.
• Discharge conditions
A unit called an Ampere hour (Ah) is used to describe how long a battery will last. An
ampere hour (more commonly known as an amp hour) is the amount of electric charge
that is transferred by a current of one ampere for one hour. Battery manufacturers use a
standard method to rate their batteries. So, for example, a 100 Ah battery will provide a
current of 5 A for a period of 20 hours at room temperature. The capacity of the battery
will depend on the rate at which it is discharged or used. If a 100 Ah battery is discharged
at 50 A (instead of 5 A), the capacity will be lower than expected and the battery will run
out before the expected 2 hours.
The relationship between the current, discharge time and capacity of a battery is expressed
by Peukert’s law:
Cp = I k t
In the equation, ’Cp ’ represents the battery’s capacity (Ah), I is the discharge current (A),
k is the Peukert constant and t is the time of discharge (hours).
23.5.3
Lead-acid batteries
In a lead-acid battery, each cell consists of electrodes of lead (Pb) and lead (IV) oxide (PbO2 )
in an electrolyte of sulfuric acid (H2 SO4 ). When the battery discharges, both electrodes turn
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CHAPTER 23. THE CHEMICAL INDUSTRY - GRADE 12
into lead (II) sulphate (PbSO4 ) and the electrolyte loses sulfuric acid to become mostly water.
The chemical half reactions that take place at the anode and cathode when the battery is discharging are as follows:
−
0
Anode (oxidation): Pb(s) + SO2−
4 (aq) ⇔ PbSO4 (s) + 2e (E = -0.356 V)
+
−
0
Cathode (reduction): PbO2 (s) + SO2−
4 (aq) + 4H + 2e ⇔ PbSO4 (s) + 2H2 O(l) (E = 1.685
V)
The overall reaction is as follows:
PbO2 (s) + 4H+ (aq) + 2SO2−
4 (aq) + Pb(s) → 2PbSO4 (s) + 2H2 O(l)
The emf of the cell is calculated as follows:
EMF = E (cathode)- E (anode)
EMF = +1.685 V - (-0.356 V)
EMF = +2.041 V
Since most batteries consist of six cells, the total voltage of the battery is approximately 12 V.
One of the important things about a lead-acid battery is that it can be recharged. The recharge
reactions are the reverse of those when the battery is discharging.
The lead-acid battery is made up of a number of plates that maximise the surface area on which
chemical reactions can take place. Each plate is a rectangular grid, with a series of holes in it.
The holes are filled with a mixture of lead and sulfuric acid. This paste is pressed into the holes
and the plates are then stacked together, with suitable separators between them. They are then
placed in the battery container, after which acid is added (figure 23.10).
-
+
Lead anode plates
Lead cathode plates
coated with PbO2
H2 SO4
Figure 23.10: A lead-acid battery
Lead-acid batteries have a number of applications. They can supply high surge currents, are
relatively cheap, have a long shelf life and can be recharged. They are ideal for use in cars,
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CHAPTER 23. THE CHEMICAL INDUSTRY - GRADE 12
23.5
where they provide the high current that is needed by the starter motor. They are also used in
forklifts and as standby power sources in telecommunication facilities, generating stations and
computer data centres. One of the disadvantages of this type of battery is that the battery’s
lead must be recycled so that the environment doesn’t become contaminated. Also, sometimes
when the battery is charging, hydrogen gas is generated at the cathode and this can cause a
small explosion if the gas comes into contact with a spark.
23.5.4
The zinc-carbon dry cell
A simplified diagram of a zinc-carbon cell is shown in figure 23.11.
metal cap
carbon rod (cathode)
zinc case
manganese (IV) oxide
paste of NH4 Cl
separator between zinc
and the electrolyte
Figure 23.11: A zinc-carbon dry cell
A zinc-carbon cell is made up of an outer zinc container, which acts as the anode. The cathode
is the central carbon rod, surrounded by a mixture of carbon and manganese (IV) oxide (MnO2 ).
The electrolyte is a paste of ammonium chloride (NH4 Cl). A fibrous fabric separates the two
electrodes, and a brass pin in the centre of the cell conducts electricity to the outside circuit.
The paste of ammonium chloride reacts according to the following half-reaction:
−
2NH+
4 (aq) + 2e → 2NH3 (g) + H2 (g)
The manganese(IV) oxide in the cell removes the hydrogen produced above, according to the
following reaction:
2MnO2 (s) + H2 (g) → Mn2 O3 (s) + H2 O(l)
The combined result of these two reactions can be represented by the following half reaction,
which takes place at the cathode:
−
Cathode: 2NH+
4 (aq) + 2MnO2 (s) + 2e → Mn2 O3 (s) + 2NH3 (g) + H2 O(l)
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CHAPTER 23. THE CHEMICAL INDUSTRY - GRADE 12
The anode half reaction is as follows:
Anode: Zn(s) → Zn2+ + 2e−
The overall equation for the cell is:
2+
0
Zn(s) + 2MnO2 (s) + 2NH+
4 → Mn2 O3 (s) + H2 O + Zn(NH3 )2 (aq) (E = 1.5 V)
Alkaline batteries are almost the same as zinc-carbon batteries, except that the electrolyte
is potassium hydroxide (KOH), rather than ammonium chloride. The two half reactions in an
alkaline battery are as follows:
Anode: Zn(s) + 2OH− (aq) → Zn(OH)2 (s) + 2e−
Cathode: 2MnO2 (s) + H2 O(l) + 2e− → Mn2 O3 (s) + 2OH− (aq)
Zinc-carbon and alkaline batteries are cheap primary batteries and are therefore very useful in
appliances such as remote controls, torches and radios where the power drain is not too high.
The disadvantages are that these batteries can’t be recycled and can leak. They also have a
short shelf life. Alkaline batteries last longer than zinc-carbon batteries.
teresting The idea behind today’s common ’battery’ was created by Georges Leclanche
Interesting
Fact
Fact
in France in the 1860’s. The anode was a zinc and mercury alloyed rod, the
cathode was a porous cup containing crushed MnO2 . A carbon rod was inserted
into this cup. The electrolyte was a liquid solution of ammonium chloride, and
the cell was therefore called a wet cell. This was replaced by the dry cell in the
1880’s. In the dry cell, the zinc can which contains the electrolyte, has become
the anode, and the electrolyte is a paste rather than a liquid.
23.5.5
Environmental considerations
While batteries are very convenient to use, they can cause a lot of damage to the environment.
They use lots of valuable resources as well as some potentially hazardous chemicals such as lead,
mercury and cadmium. Attempts are now being made to recycle the different parts of batteries
so that they are not disposed of in the environment, where they could get into water supplies,
rivers and other ecosystems.
Exercise: Electrochemistry and batteries
A dry cell, as shown in the diagram below, does not contain a liquid electrolyte.
The electrolyte in a typical zinc-carbon cell is a moist paste of ammonium chloride
and zinc chloride.
(NOTE TO SELF: Insert diagram)
The paste of ammonium chloride reacts according to the following half-reaction:
−
2NH+
4 (aq) + 2e → 2NH3 (g) + H2 (g) (a)
Manganese(IV) oxide is included in the cell to remove the hydrogen produced
during half-reaction (a), according to the following reaction:
2MnO2 2(s) + H2 (g) → Mn2 O3 (s) + H2 O(l) (b)
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CHAPTER 23. THE CHEMICAL INDUSTRY - GRADE 12
23.6
The combined result of these two half-reactions can be represented by the following half reaction:
−
2NH+
4 (aq) + 2MnO2 (s) + 2e → Mn2 O3 (s) + 2NH3 (g) + H2 O(l) (c)
1. Explain why it is important that the hydrogen produced in half-reaction (a) is
removed by the manganese(IV) oxide.
In a zinc-carbon cell, such as the one above, half-reaction (c) and the halfreaction that takes place in the Zn/Zn2+ half-cell, produce an emf of 1,5 V
under standard conditions.
2. Write down the half-reaction occurring at the anode.
3. Write down the net ionic equation occurring in the zinc-carbon cell.
4. Calculate the reduction potential for the cathode half-reaction.
5. When in use the zinc casing of the dry cell becomes thinner, because it is
oxidised. When not in use, it still corrodes. Give a reason for the latter observation.
6. Dry cells are generally discarded when ’flat’. Why is the carbon rod the most
useful part of the cell, even when the cell is flat?
(DoE Exemplar Paper 2, 2007)
23.6
Summary
• The growth of South Africa’s chemical industry was largely because of the mines, which
needed explosives for their operations. One of South Africa’s major chemical companies is
Sasol. Other important chemical industries in the country are the chloralkali and fertiliser
industries.
• All countries need energy resources such as oil and natural gas. Since South Africa doesn’t
have either of these resources, Sasol technology has developed to convert coal into liquid
fuels.
• Sasol has three main operation focus areas: Firstly, the conversion of coal to liquid fuel,
secondly the production and refinement of crude oil which has been imported, and thirdly
the production of liquid fuels from natural gas.
• The conversion of coal to liquid fuels involves a Sasol/Lurgi gasification process, followed
by the conversion of this synthesis gas into a range of hydrocarbons, using the FischerTropsch technology in SAS reactors.
• Heavy hydrocarbons can be converted into light hydrcarbons through a process called
cracking. Common forms of cracking are hydrocracking and steam cracking.
• With regard to crude oil, Sasol imports crude oil from Gabon and then refines this at the
Natref refinery.
• Gas from Mozambique can be used to produce liquid fuels, through two processes: First,
the gas must pass through an autothermal reactor to produce a synthesis gas. Secondly,
this synthesis gas is passed through a Sasol Slurry Phase Distillate process to convert
the gas to hydrocarbons.
• All industries have an impact on the environment through the consumption of natural
resources such as water, and through the production of pollution gases such as carbon
dioxide, hydrogen sulfides, nitrogen oxides and others.
• The chloralkali industry produces chlorine and sodium hydroxide. The main raw material is brine (NaCl).
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CHAPTER 23. THE CHEMICAL INDUSTRY - GRADE 12
• In industry, electrolytic cells are used to split the sodium chloride into its component
ions to produce chlorine and sodium hydroxide. One of the challenges in this process is to
keep the products of the electrolytic reaction (i.e. the chlorine and the sodium hydroxide)
separate so that they don’t react with each other. Specially designed electrolytic cells are
needed to do this.
• There are three types of electrolytic cells that are used in this process: mercury cell, the
diaphragm cell and the membrane cell.
• The mercury cell consists of two reaction vessels. The first reaction vessel contains
a mercury cathode and a carbon anode. An electric current passed through the brine
produces Cl− and Na+ ions. The Cl− ions are oxidised to form chlorine gas at the anode.
Na+ ions combine with the mercury cathode to form a sodium-mercury amalgam. The
sodium-mercury amalgam passes into the second reaction vessel containing water, where
the Na+ ions react with hydroxide ions from the water. Sodium hydroxide is the product
of this reaction.
• One of the environmental impacts of using this type of cell, is the use of mercury, which
is highly toxic.
• In the diaphragm cell, a porous diaphragm separates the anode and the cathode compartments. Chloride ions are oxidised to chlorine gas at the anode, while sodium ions produced
at the cathode react with water to produce sodium hydroxide.
• The membrane cell is very similar to the diaphragm cell, except that the anode and
cathode compartments are separated by an ion-selective membrane rather than by a
diaphragm. Brine is only pumped into the anode compartment. Positive sodium ions pass
through the membrane into the cathode compartment, which contains water. As with
the other two cells, chlorine gas is produced at the anode and sodium hydroxide at the
cathode.
• One use of sodium hydroxide is in the production of soaps and detergents, and so this
is another important part of the chloralkali industry.
• To make soap, sodium hydroxide or potassium hydroxide react with a fat or an oil. In the
reaction, the sodium or potassium ions replace the alcohol in the fat or oil. The product,
a sodium or potassium salt of a fatty acid, is what soap is made of.
• The fatty acids in soap have a hydrophilic and a hydrophobic part in each molecule, and
this helps to loosen dirt and clean items.
• Detergents are also cleaning products, but are made up of a mixture of compounds. They
may also have other components added to them to give certain characteristics. Some of
these additives may be abrasives, oxidants or enzymes.
• The fertiliser industry is another important chemical industry.
• All plants need certain macronutrients (e.g. carbon, hydrogen, oxygen, potassium, nitrogen and phosphorus) and micronutrients (e.g. iron, chlorine, copper and zinc) in order
to survive. Fertilisers provide these nutrients.
• In plants, most nutrients are obtained from the atmosphere or from the soil.
• Animals also need similar nutrients, but they obtain most of these directly from plants or
plant products. They may also obtain them from other animals, whcih may have fed on
plants during their life.
• The fertiliser industry is very important in ensuring that plants and crops receive the correct
nutrients in the correct quantities to ensure maximum growth.
• Nitrogen fertilisers can be produced industrially using a number of chemical processes:
The Haber process reacts nitrogen and hydrogen to produce ammonia; the Ostwald
process reacts oxygen and ammonia to produce nitric acid; the nitrophosphate process
reacts nitric acid with phosphate rock to produce compound fertilisers.
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CHAPTER 23. THE CHEMICAL INDUSTRY - GRADE 12
23.6
• Phosphate fertilisers are also produced through a series of reactions. The contact process produces sulfuric acid. Sulfuric acid then reacts with phosphate rock to produce
phosphoric acid, after which phosphoric acid reacts with ground phosphate rock to produce fertilisers such as triple superphosphate.
• Potassium is obtained from potash.
• Fertilisers can have a damaging effect on the environment when they are present in high
quantities in ecosystems. They can lead to eutrophication. A number of preventative
actions can be taken to reduce these impacts.
• Another important part of the chemical industry is the production of batteries.
• A battery is a device that changes chemical energy into electrical energy.
• A battery consists of one or more voltaic cells, each of which is made up of two half
cells that are connected in series by a conductive electrolyte. Each half cell has a net
electromotive force (emf) or voltage. The net voltage of the battery is the difference
between the voltages of the half-cells. This potential difference between the two half cells
is what causes an electric current to flow.
• A primary battery cannot be recharged, but a secondary battery can be recharged.
• The capacity of a battery depends on the chemical reactions in the cells, the quantity
of electrolyte and electrode material in the cell, and the discharge conditions of the
battery.
• The relationship between the current, discharge time and capacity of a battery is expressed
by Peukert’s law:
Cp = I k t
In the equation, ’Cp ’ represents the battery’s capacity (Ah), I is the discharge current (A),
k is the Peukert constant and t is the time of discharge (hours).
• Two common types of batteries are lead-acid batteries and the zinc-carbon dry cell.
• In a lead-acid battery, each cell consists of electrodes of lead (Pb) and lead (IV) oxide
(PbO2 ) in an electrolyte of sulfuric acid (H2 SO4 ). When the battery discharges, both
electrodes turn into lead (II) sulphate (PbSO4 ) and the electrolyte loses sulfuric acid to
become mostly water.
• A zinc-carbon cell is made up of an outer zinc container, which acts as the anode. The
cathode is the central carbon rod, surrounded by a mixture of carbon and manganese (IV)
oxide (MnO2 ). The electrolyte is a paste of ammonium chloride (NH4 Cl). A fibrous fabric
separates the two electrodes, and a brass pin in the centre of the cell conducts electricity
to the outside circuit.
• Despite their many advantages, batteries are made of potentially toxic materials and can
be damaging to the environment.
Exercise: Summary Exercise
1. Give one word or term for each of the following descriptions:
(a)
(b)
(c)
(d)
(e)
A solid organic compound that can be used to produce liquid fuels.
The process used to convert heavy hydrocarbons into light hydrocarbons.
The process of separating nitrogen from liquid air.
The main raw material in the chloralkali industry.
A compound given to a plant to promote growth.
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CHAPTER 23. THE CHEMICAL INDUSTRY - GRADE 12
(f) An electrolyte used in lead-acid batteries.
2. Indicate whether each of the following statements is true or false. If the statement is false, rewrite the statement correctly.
(a) The longer the hydrocarbon chain in an organic compound, the more likely
it is to be a solid at room temperature.
(b) The main elements used in fertilisers are nitrogen, phosphorus and potassium.
(c) A soap molecule is composed of an alcohol molecule and three fatty acids.
(d) During the industrial preparation of chlorine and sodium hydroxide, chemical energy is converted to electrical energy.
3. For each of the following questions, choose the one correct answer from the
list provided.
(a) The sequence of processes that best describes the conversion of coal to
liquid fuel is:
i. coal → gas purification → SAS reactor → liquid hydrocarbon
ii. coal → autothermal reactor → Sasol slurry phase F-T reactor → liquid
hydrocarbon
iii. coal → coal purification → synthesis gas → oil
iv. coal → coal gasification → gas purification → SAS reactor → liquid
hydrocarbons
(b) The half-reaction that takes place at the cathode of a mercury cell in the
chloralkali industry is:
i. 2Cl− → Cl2 + 2e−
ii. 2Na+ + 2e− → 2Na
iii. 2H+ + 2e− → H2
iv. NaCl + H2 O → NaOH + HCl
(c) In a zinc-carbon dry cell...
i. the electrolyte is manganese (IV) oxide
ii. zinc is oxidised to produce electrons
iii. zinc is reduced to produce electrons
iv. manganese (IV) dioxide acts as a reducing agent
4. Chloralkali manufacturing process
The chloralkali (also called ’chlorine-caustic’) industry is one of the largest
electrochemical technologies in the world. Chlorine is produced using three
types of electrolytic cells. The simplified diagram below shows a membrane
cell.
Power supply
Gas A
Gas B
Saturated
NaCl
+
M
Depleted
NaCl
N
NaOH
(a) Give two reasons why the membrane cell is the preferred cell for the preparation of chlorine.
(b) Why do you think it is advisable to use inert electrodes in this process?
(c) Write down the equation for the half-reaction taking place at electrode M.
(d) Which gas is chlorine gas? Write down only Gas A or Gas B.
464
CHAPTER 23. THE CHEMICAL INDUSTRY - GRADE 12
23.6
(e) Briefly explain how sodium hydroxide forms in this cell.
(DoE Exemplar Paper 2,2007)
5. The production of nitric acid is very important in the manufacture of fertilisers. Look at the diagram below, which shows part of the fertiliser production
process, and then answer the questions that follow.
(1)
N2 (g)
+ H2 (3)
(2)
+ O2
NO + H2 O
Ostwald Process
(4)
(a)
(b)
(c)
(d)
(e)
Name
Name
Name
Name
Name
the process at (1).
the gas at (2).
the process at (3) that produces gas (2).
the product at (4).
two fertilisers that can be produced from nitric acid.
6. A lead-acid battery has a number of different components. Match the description in Column A with the correct word or phrase in Column B. All the
descriptions in Column A relate to lead-acid batteries.
Column A
The electrode metal
Electrolyte
A product of the overall cell reaction
An oxidising agent in the cathode half-reaction
Type of cells in a lead-acid battery
465
Column B
Lead sulphate
Mercury
Electrolytic
Lead
Sulfuric acid
Ammonium chloride
Lead oxide
Galvanic
23.6
CHAPTER 23. THE CHEMICAL INDUSTRY - GRADE 12
466
Appendix A
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APPENDIX A. GNU FREE DOCUMENTATION LICENSE
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APPENDIX A. GNU FREE DOCUMENTATION LICENSE
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APPENDIX A. GNU FREE DOCUMENTATION LICENSE
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470
APPENDIX A. GNU FREE DOCUMENTATION LICENSE
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APPENDIX A. GNU FREE DOCUMENTATION LICENSE
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