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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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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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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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
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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.
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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:
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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!
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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.
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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
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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
172
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
185
10.2
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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CHAPTER 10. ORGANIC MACROMOLECULES - GRADE 12
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
188
OH
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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10.3
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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CHAPTER 10. ORGANIC MACROMOLECULES - GRADE 12
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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10.5
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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10.6
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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CHAPTER 10. ORGANIC MACROMOLECULES - GRADE 12
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 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:
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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
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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.
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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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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.
308
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
312
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?
3SO2 + Cr2 O72− + 2H + → 3SO42− + 2Cr3+ + H2 O
(a)
(b)
(c)
(d)
H+
Cr3+
SO2
Cr2 O2−
7
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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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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+
5. Look at the following half-reactions.
•
•
•
•
Ca2+ + 2e− → Ca
Cl2 + 2e− → 2Cl
F e3+ + 3e− → F e
I2 + 2e− → 2I −
(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?
2−
+
3+
3SO2 + Cr2 O2−
+ H2 O
7 + 2H → 3SO4 + 2Cr
(a)
(b)
(c)
(d)
H+
Cr3+
SO2
Cr2 O2−
7
(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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CHAPTER 17. ELECTROCHEMICAL REACTIONS - GRADE 12
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
345
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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17.6
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 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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23.2
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.
436
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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23.2
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
438
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?
439
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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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.
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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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CHAPTER 23. THE CHEMICAL INDUSTRY - GRADE 12
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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23.5
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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23.5
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).
461
23.6
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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23.6
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.
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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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changing it is not allowed.
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not being considered responsible for modifications made by others.
This License is a kind of “copyleft”, which means that derivative works of the document must
themselves be free in the same sense. It complements the GNU General Public License, which
is a copyleft license designed for free software.
We have designed this License in order to use it for manuals for free software, because free
software needs free documentation: a free program should come with manuals providing the
same freedoms that the software does. But this License is not limited to software manuals; it
can be used for any textual work, regardless of subject matter or whether it is published as a
printed book. We recommend this License principally for works whose purpose is instruction or
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APPLICABILITY AND DEFINITIONS
This License applies to any manual or other work, in any medium, that contains a notice placed
by the copyright holder saying it can be distributed under the terms of this License. Such a
notice grants a world-wide, royalty-free license, unlimited in duration, to use that work under
the conditions stated herein. The “Document”, below, refers to any such manual or work. Any
member of the public is a licensee, and is addressed as “you”. You accept the license if you
copy, modify or distribute the work in a way requiring permission under copyright law.
A “Modified Version” of the Document means any work containing the Document or a portion
of it, either copied verbatim, or with modifications and/or translated into another language.
A “Secondary Section” is a named appendix or a front-matter section of the Document that deals
exclusively with the relationship of the publishers or authors of the Document to the Document’s
overall subject (or to related matters) and contains nothing that could fall directly within that
overall subject. (Thus, if the Document is in part a textbook of mathematics, a Secondary
Section may not explain any mathematics.) The relationship could be a matter of historical
connection with the subject or with related matters, or of legal, commercial, philosophical,
ethical or political position regarding them.
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APPENDIX A. GNU FREE DOCUMENTATION LICENSE
The “Invariant Sections” are certain Secondary Sections whose titles are designated, as being
those of Invariant Sections, in the notice that says that the Document is released under this
License. If a section does not fit the above definition of Secondary then it is not allowed to be
designated as Invariant. The Document may contain zero Invariant Sections. If the Document
does not identify any Invariant Sections then there are none.
The “Cover Texts” are certain short passages of text that are listed, as Front-Cover Texts or
Back-Cover Texts, in the notice that says that the Document is released under this License. A
Front-Cover Text may be at most 5 words, and a Back-Cover Text may be at most 25 words.
A “Transparent” copy of the Document means a machine-readable copy, represented in a format
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formatters. A copy made in an otherwise Transparent file format whose markup, or absence of
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Transparent. An image format is not Transparent if used for any substantial amount of text. A
copy that is not “Transparent” is called “Opaque”.
Examples of suitable formats for Transparent copies include plain ASCII without markup, Texinfo
input format, LATEX input format, SGML or XML using a publicly available DTD and standardconforming simple HTML, PostScript or PDF designed for human modification. Examples of
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formats that can be read and edited only by proprietary word processors, SGML or XML for
which the DTD and/or processing tools are not generally available, and the machine-generated
HTML, PostScript or PDF produced by some word processors for output purposes only.
The “Title Page” means, for a printed book, the title page itself, plus such following pages as
are needed to hold, legibly, the material this License requires to appear in the title page. For
works in formats which do not have any title page as such, “Title Page” means the text near the
most prominent appearance of the work’s title, preceding the beginning of the body of the text.
A section “Entitled XYZ” means a named subunit of the Document whose title either is precisely
XYZ or contains XYZ in parentheses following text that translates XYZ in another language.
(Here XYZ stands for a specific section name mentioned below, such as “Acknowledgements”,
“Dedications”, “Endorsements”, or “History”.) To “Preserve the Title” of such a section when
you modify the Document means that it remains a section “Entitled XYZ” according to this
definition.
The Document may include Warranty Disclaimers next to the notice which states that this
License applies to the Document. These Warranty Disclaimers are considered to be included by
reference in this License, but only as regards disclaiming warranties: any other implication that
these Warranty Disclaimers may have is void and has no effect on the meaning of this License.
VERBATIM COPYING
You may copy and distribute the Document in any medium, either commercially or non-commercially,
provided that this License, the copyright notices, and the license notice saying this License applies
to the Document are reproduced in all copies, and that you add no other conditions whatsoever
to those of this License. You may not use technical measures to obstruct or control the reading
or further copying of the copies you make or distribute. However, you may accept compensation
in exchange for copies. If you distribute a large enough number of copies you must also follow
the conditions in section A.
You may also lend copies, under the same conditions stated above, and you may publicly display
copies.
COPYING IN QUANTITY
If you publish printed copies (or copies in media that commonly have printed covers) of the
Document, numbering more than 100, and the Document’s license notice requires Cover Texts,
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APPENDIX A. GNU FREE DOCUMENTATION LICENSE
you must enclose the copies in covers that carry, clearly and legibly, all these Cover Texts: FrontCover Texts on the front cover, and Back-Cover Texts on the back cover. Both covers must also
clearly and legibly identify you as the publisher of these copies. The front cover must present the
full title with all words of the title equally prominent and visible. You may add other material on
the covers in addition. Copying with changes limited to the covers, as long as they preserve the
title of the Document and satisfy these conditions, can be treated as verbatim copying in other
respects.
If the required texts for either cover are too voluminous to fit legibly, you should put the first
ones listed (as many as fit reasonably) on the actual cover, and continue the rest onto adjacent
pages.
If you publish or distribute Opaque copies of the Document numbering more than 100, you must
either include a machine-readable Transparent copy along with each Opaque copy, or state in or
with each Opaque copy a computer-network location from which the general network-using public
has access to download using public-standard network protocols a complete Transparent copy of
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prudent steps, when you begin distribution of Opaque copies in quantity, to ensure that this
Transparent copy will remain thus accessible at the stated location until at least one year after
the last time you distribute an Opaque copy (directly or through your agents or retailers) of that
edition to the public.
It is requested, but not required, that you contact the authors of the Document well before
redistributing any large number of copies, to give them a chance to provide you with an updated
version of the Document.
MODIFICATIONS
You may copy and distribute a Modified Version of the Document under the conditions of
sections A and A above, provided that you release the Modified Version under precisely this
License, with the Modified Version filling the role of the Document, thus licensing distribution
and modification of the Modified Version to whoever possesses a copy of it. In addition, you
must do these things in the Modified Version:
1. Use in the Title Page (and on the covers, if any) a title distinct from that of the Document,
and from those of previous versions (which should, if there were any, be listed in the History
section of the Document). You may use the same title as a previous version if the original
publisher of that version gives permission.
2. List on the Title Page, as authors, one or more persons or entities responsible for authorship
of the modifications in the Modified Version, together with at least five of the principal
authors of the Document (all of its principal authors, if it has fewer than five), unless they
release you from this requirement.
3. State on the Title page the name of the publisher of the Modified Version, as the publisher.
4. Preserve all the copyright notices of the Document.
5. Add an appropriate copyright notice for your modifications adjacent to the other copyright
notices.
6. Include, immediately after the copyright notices, a license notice giving the public permission to use the Modified Version under the terms of this License, in the form shown in the
Addendum below.
7. Preserve in that license notice the full lists of Invariant Sections and required Cover Texts
given in the Document’s license notice.
8. Include an unaltered copy of this License.
9. Preserve the section Entitled “History”, Preserve its Title, and add to it an item stating
at least the title, year, new authors, and publisher of the Modified Version as given on the
Title Page. If there is no section Entitled “History” in the Document, create one stating
the title, year, authors, and publisher of the Document as given on its Title Page, then
add an item describing the Modified Version as stated in the previous sentence.
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APPENDIX A. GNU FREE DOCUMENTATION LICENSE
10. Preserve the network location, if any, given in the Document for public access to a Transparent copy of the Document, and likewise the network locations given in the Document
for previous versions it was based on. These may be placed in the “History” section. You
may omit a network location for a work that was published at least four years before the
Document itself, or if the original publisher of the version it refers to gives permission.
11. For any section Entitled “Acknowledgements” or “Dedications”, Preserve the Title of the
section, and preserve in the section all the substance and tone of each of the contributor
acknowledgements and/or dedications given therein.
12. Preserve all the Invariant Sections of the Document, unaltered in their text and in their
titles. Section numbers or the equivalent are not considered part of the section titles.
13. Delete any section Entitled “Endorsements”. Such a section may not be included in the
Modified Version.
14. Do not re-title any existing section to be Entitled “Endorsements” or to conflict in title
with any Invariant Section.
15. Preserve any Warranty Disclaimers.
If the Modified Version includes new front-matter sections or appendices that qualify as Secondary
Sections and contain no material copied from the Document, you may at your option designate
some or all of these sections as invariant. To do this, add their titles to the list of Invariant
Sections in the Modified Version’s license notice. These titles must be distinct from any other
section titles.
You may add a section Entitled “Endorsements”, provided it contains nothing but endorsements
of your Modified Version by various parties–for example, statements of peer review or that the
text has been approved by an organisation as the authoritative definition of a standard.
You may add a passage of up to five words as a Front-Cover Text, and a passage of up to 25
words as a Back-Cover Text, to the end of the list of Cover Texts in the Modified Version. Only
one passage of Front-Cover Text and one of Back-Cover Text may be added by (or through
arrangements made by) any one entity. If the Document already includes a cover text for the
same cover, previously added by you or by arrangement made by the same entity you are acting
on behalf of, you may not add another; but you may replace the old one, on explicit permission
from the previous publisher that added the old one.
The author(s) and publisher(s) of the Document do not by this License give permission to use
their names for publicity for or to assert or imply endorsement of any Modified Version.
COMBINING DOCUMENTS
You may combine the Document with other documents released under this License, under the
terms defined in section A above for modified versions, provided that you include in the combination all of the Invariant Sections of all of the original documents, unmodified, and list them
all as Invariant Sections of your combined work in its license notice, and that you preserve all
their Warranty Disclaimers.
The combined work need only contain one copy of this License, and multiple identical Invariant
Sections may be replaced with a single copy. If there are multiple Invariant Sections with the
same name but different contents, make the title of each such section unique by adding at the
end of it, in parentheses, the name of the original author or publisher of that section if known,
or else a unique number. Make the same adjustment to the section titles in the list of Invariant
Sections in the license notice of the combined work.
In the combination, you must combine any sections Entitled “History” in the various original
documents, forming one section Entitled “History”; likewise combine any sections Entitled “Acknowledgements”, and any sections Entitled “Dedications”. You must delete all sections Entitled
“Endorsements”.
470
APPENDIX A. GNU FREE DOCUMENTATION LICENSE
COLLECTIONS OF DOCUMENTS
You may make a collection consisting of the Document and other documents released under
this License, and replace the individual copies of this License in the various documents with a
single copy that is included in the collection, provided that you follow the rules of this License
for verbatim copying of each of the documents in all other respects.
You may extract a single document from such a collection, and distribute it individually under
this License, provided you insert a copy of this License into the extracted document, and follow
this License in all other respects regarding verbatim copying of that document.
AGGREGATION WITH INDEPENDENT WORKS
A compilation of the Document or its derivatives with other separate and independent documents
or works, in or on a volume of a storage or distribution medium, is called an “aggregate” if the
copyright resulting from the compilation is not used to limit the legal rights of the compilation’s
users beyond what the individual works permit. When the Document is included an aggregate,
this License does not apply to the other works in the aggregate which are not themselves derivative
works of the Document.
If the Cover Text requirement of section A is applicable to these copies of the Document, then if
the Document is less than one half of the entire aggregate, the Document’s Cover Texts may be
placed on covers that bracket the Document within the aggregate, or the electronic equivalent
of covers if the Document is in electronic form. Otherwise they must appear on printed covers
that bracket the whole aggregate.
TRANSLATION
Translation is considered a kind of modification, so you may distribute translations of the Document under the terms of section A. Replacing Invariant Sections with translations requires
special permission from their copyright holders, but you may include translations of some or
all Invariant Sections in addition to the original versions of these Invariant Sections. You may
include a translation of this License, and all the license notices in the Document, and any Warranty Disclaimers, provided that you also include the original English version of this License and
the original versions of those notices and disclaimers. In case of a disagreement between the
translation and the original version of this License or a notice or disclaimer, the original version
will prevail.
If a section in the Document is Entitled “Acknowledgements”, “Dedications”, or “History”, the
requirement (section A) to Preserve its Title (section A) will typically require changing the actual
title.
TERMINATION
You may not copy, modify, sub-license, or distribute the Document except as expressly provided
for under this License. Any other attempt to copy, modify, sub-license or distribute the Document
is void, and will automatically terminate your rights under this License. However, parties who
have received copies, or rights, from you under this License will not have their licenses terminated
so long as such parties remain in full compliance.
FUTURE REVISIONS OF THIS LICENSE
The Free Software Foundation may publish new, revised versions of the GNU Free Documentation
License from time to time. Such new versions will be similar in spirit to the present version, but
may differ in detail to address new problems or concerns. See http://www.gnu.org/copyleft/.
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APPENDIX A. GNU FREE DOCUMENTATION LICENSE
Each version of the License is given a distinguishing version number. If the Document specifies
that a particular numbered version of this License “or any later version” applies to it, you have the
option of following the terms and conditions either of that specified version or of any later version
that has been published (not as a draft) by the Free Software Foundation. If the Document does
not specify a version number of this License, you may choose any version ever published (not as
a draft) by the Free Software Foundation.
ADDENDUM: How to use this License for your documents
To use this License in a document you have written, include a copy of the License in the document
and put the following copyright and license notices just after the title page:
c YEAR YOUR NAME. Permission is granted to copy, distribute and/or
Copyright modify this document under the terms of the GNU Free Documentation License,
Version 1.2 or any later version published by the Free Software Foundation; with no
Invariant Sections, no Front-Cover Texts, and no Back-Cover Texts. A copy of the
license is included in the section entitled “GNU Free Documentation License”.
If you have Invariant Sections, Front-Cover Texts and Back-Cover Texts, replace the “with...Texts.”
line with this:
with the Invariant Sections being LIST THEIR TITLES, with the Front-Cover Texts being LIST,
and with the Back-Cover Texts being LIST.
If you have Invariant Sections without Cover Texts, or some other combination of the three,
merge those two alternatives to suit the situation.
If your document contains nontrivial examples of program code, we recommend releasing these
examples in parallel under your choice of free software license, such as the GNU General Public
License, to permit their use in free software.
472
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