Download SQA Advanced Higher Chemistry Unit 2 Principles of Chemical

SCHOLAR Study Guide
SQA Advanced Higher Chemistry
Unit 2
Principles of Chemical Reactions
Peter Johnson
Heriot-Watt University
Brian T McKerchar
Balerno High School
Arthur A Sandison
St Thomas of Aquin’s High School
Heriot-Watt University
Edinburgh EH14 4AS, United Kingdom.
First published 2001 by Heriot-Watt University.
This edition published in 2009 by Heriot-Watt University SCHOLAR.
Copyright © 2009 Heriot-Watt University.
Members of the SCHOLAR Forum may reproduce this publication in whole or in part for
educational purposes within their establishment providing that no profit accrues at any stage,
Any other use of the materials is governed by the general copyright statement that follows.
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system
or transmitted in any form or by any means, without written permission from the publisher.
Heriot-Watt University accepts no responsibility or liability whatsoever with regard to the
information contained in this study guide.
Distributed by Heriot-Watt University.
SCHOLAR Study Guide Unit 2: Advanced Higher Chemistry
1. Advanced Higher Chemistry
ISBN 978-1-906686-01-7
Printed and bound in Great Britain by Graphic and Printing Services, Heriot-Watt University,
Edinburgh.
Acknowledgements
Thanks are due to the members of Heriot-Watt University’s SCHOLAR team who planned and
created these materials, and to the many colleagues who reviewed the content.
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and students who contributed to the SCHOLAR programme and who evaluated these materials.
Grateful acknowledgement is made for permission to use the following material in the
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i
Contents
1 Stoichiometry
1.1 Introduction . . . . . . . . . . .
1.2 Using moles . . . . . . . . . . .
1.3 Volumetric analysis . . . . . . .
1.4 Gravimetric analysis . . . . . .
1.5 Summary . . . . . . . . . . . .
1.6 Suggestions for further reading
1.7 Tutorial questions . . . . . . . .
1.8 End of Topic test . . . . . . . .
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1
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2 Chemical Equilibrium
2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 The nature of chemical equilibrium . . . . . . . . . . . . . .
2.3 The equilibrium constant, K c . . . . . . . . . . . . . . . . .
2.4 The equilibrium constant, K p , involving gases . . . . . . . .
2.5 Homogeneous and heterogeneous equilibria . . . . . . . .
2.6 Calculations using equilibrium constants . . . . . . . . . . .
2.7 Factors that alter the composition of an equilibrium mixture
2.8 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.9 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.10 End of Topic Test . . . . . . . . . . . . . . . . . . . . . . . .
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3 Phase Equilibria
3.1 Partition coefficients
3.2 Solvent extraction .
3.3 Chromatography . .
3.4 Summary . . . . . .
3.5 Resources . . . . .
3.6 End of Topic Test . .
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4 Acid/base Equilibria
4.1 What are acids and bases? . . . . . . . . .
4.2 The ionic product of water and the pH scale
4.3 Dissociation of acids . . . . . . . . . . . . .
4.4 Dissociation of bases . . . . . . . . . . . .
4.5 Summary . . . . . . . . . . . . . . . . . . .
4.6 Resources . . . . . . . . . . . . . . . . . .
4.7 End of Topic test . . . . . . . . . . . . . . .
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ii
CONTENTS
5 Indicators and buffers
5.1 Indicators . . . . . . . . . . . . . . . .
5.2 Buffer solutions. . . . . . . . . . . . .
5.3 Calculating pH and buffer composition.
5.4 Summary . . . . . . . . . . . . . . . .
5.5 Resources . . . . . . . . . . . . . . .
5.6 End of Topic test . . . . . . . . . . . .
6 Thermochemistry
6.1 Introduction . . . . . . . . .
6.2 Bond energy . . . . . . . .
6.3 Hess’s Law . . . . . . . . .
6.4 Standard enthalpy changes
6.5 Calorimetry . . . . . . . . .
6.6 The Born-Haber cycle . . .
6.7 Enthalpy of solution . . . .
6.8 Summary . . . . . . . . . .
6.9 Resources . . . . . . . . .
6.10 End of Topic test. . . . . . .
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7 Reaction Feasibility
7.1 Introduction . . . . . . . . . . . . . .
7.2 The concept of entropy . . . . . . . .
7.3 The second law of thermodynamics
7.4 Gibbs free energy . . . . . . . . . .
7.5 Ellingham diagrams . . . . . . . . .
7.6 Summary . . . . . . . . . . . . . . .
7.7 Resources . . . . . . . . . . . . . .
7.8 End of topic test . . . . . . . . . . .
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144
8 Electrochemistry
8.1 Introduction . . . . . . . . . . . . . . . . . . . .
8.2 Electrochemical cells . . . . . . . . . . . . . . .
8.3 Standard Electrode Potentials . . . . . . . . . .
8.4 EÆ values and the standard free energy change
8.5 Fuel cells . . . . . . . . . . . . . . . . . . . . .
8.6 Summary . . . . . . . . . . . . . . . . . . . . .
8.7 Resources . . . . . . . . . . . . . . . . . . . .
8.8 End of Topic test . . . . . . . . . . . . . . . . .
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9 Kinetics
9.1 Introduction . . . . . . . .
9.2 Measuring reaction rates
9.3 Rate equations . . . . . .
9.4 Reaction mechanisms . .
9.5 Summary . . . . . . . . .
9.6 Resources . . . . . . . .
9.7 End of Topic test . . . . .
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© H ERIOT-WATT U NIVERSITY
CONTENTS
iii
10 End of Unit 2 Test (NAB)
191
Glossary
193
Further questions
198
Answers to questions and activities
1
Stoichiometry . . . . . . . .
2
Chemical Equilibrium . . . .
3
Phase Equilibria . . . . . . .
4
Acid/base Equilibria . . . . .
5
Indicators and buffers . . . .
6
Thermochemistry . . . . . .
7
Reaction Feasibility . . . . .
8
Electrochemistry . . . . . . .
9
Kinetics . . . . . . . . . . . .
207
207
213
217
219
224
226
231
237
244
© H ERIOT-WATT U NIVERSITY
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1
Topic 1
Stoichiometry
Contents
1.1 Introduction . . . . . . . . . . . . . . . . . . . . .
1.2 Using moles . . . . . . . . . . . . . . . . . . . . .
1.2.1 Converting moles and mass . . . . . . . .
1.2.2 Converting moles and volumes for gases
1.2.3 Determining excess or limiting reactants .
1.2.4 Calculating concentrations . . . . . . . .
1.2.5 Diluting . . . . . . . . . . . . . . . . . . .
1.3 Volumetric analysis . . . . . . . . . . . . . . . . .
1.3.1 Titrations . . . . . . . . . . . . . . . . . .
1.3.2 Redox titrations . . . . . . . . . . . . . . .
1.3.3 Complexometric volumetric analysis . . .
1.4 Gravimetric analysis . . . . . . . . . . . . . . . .
1.5 Summary . . . . . . . . . . . . . . . . . . . . . .
1.6 Suggestions for further reading . . . . . . . . . .
1.7 Tutorial questions . . . . . . . . . . . . . . . . . .
1.8 End of Topic test . . . . . . . . . . . . . . . . . .
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Prerequisite knowledge
Before you begin this Topic, you should be able to:
• calculate the molecular masses of compounds;
• balance chemical equations;
• understand the concept of the mole.
Learning Objectives
After studying this Topic, you should be able to:
• outline the characteristics and preparation of a primary standard;
• calculate the molar concentration of standard solutions;
• understand the principles of volumetric analysis (acid/base,
complexometric);
redox and
2
TOPIC 1. STOICHIOMETRY
• calculate the concentration of solutions from titrimetric data;
• manipulate results from gravimetric determinations.
© H ERIOT-WATT U NIVERSITY
1.1. INTRODUCTION
1.1
Introduction
Learning Objective
Æ
To revise the basic skills, relating to balanced equations and the calculation of formula
masses that are required in stoichiometry
Stoichiometry involves the understanding of the numerical relationships between
reacting substances. It forms the basis for analytical chemistry. (How much iron is
there in this sample of iron ore? How pure is this sample of anti-viral drug?) It is also
critical for industrial chemistry. (How many tonnes of sulphuric acid and phosphate rock
are required to manufacture this batch of fertiliser? What proportion of monomers are
required to prepare this plastic?)
These numerical relationships are obtained by consideration of a quantitative reaction
in which substances react completely according to the mole ratio given by the
balanced (stoichiometric) equation.
Here are some revision questions for you to try before you proceed with this Topic. If you
use the on-line version of these materials, the questions will be marked by the computer.
Write down your answer before you press Display Answer, then compare the two.
Write down the chemical formula for:
Q1: Calcium fluoride
Q2: Iron(III) nitrate
Q3: Barium phosphate
Q4: Potassium dichromate
Write the names of the following substances:
Q5: CaSO4
Q6: CuCl2
Q7: KMnO4
Q8: P4 O10
Use the relative atomic masses in the data booklet to calculate the relative
molecular masses of the following substances. Give your answers to 1 decimal
place:
Q9: CaO
Q10: Al2 (SO4 )3
Q11: Chromium(III) oxide
Q12: Butane
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4
TOPIC 1. STOICHIOMETRY
Use the numbers from the list below to complete these balanced equations:
[ ] Zn + O2
[ ] ZnO
Li2 SO4 + BaCl2
[ ] LiCl + BaSO4
[ ] Ca(NO3 )2 + [ ] Na3 PO4
[ ] C2 H4 O + [ ] O2
Ca3 (PO4 )2 + [ ] NaNO3
[ ] CO2 + [ ] H2 O
[2], [2], [2], [2], [2], [3], [4], [4], [5], [6].
Q13:
When calcium carbonate reacts with hydrochloric acid, the reaction is described by the
equation:
Which of the following statements applies to this equation?
1. Calcium carbonate reacts with hydrochloric acid to produce calcium chloride solution,
water and carbon dioxide.
2. One formula unit of calcium carbonate reacts with two formula units of hydrochloric
acid to produce one formula unit each of calcium chloride, water and carbon dioxide.
3. One mole of calcium carbonate reacts with two moles of hydrochloric acid to produce
one mole each of calcium chloride, water and carbon dioxide.
a) statement 1 only
b) statements 1 and 2
c) statements 1, 2 and 3.
Q14: When real quantities need to be weighed out in the laboratory or industry, which
statement is most helpful?
a) statement 1
b) statement 2
c) statement 3
Balancing equations
An online exercise is provided to help you if you require additional assistance with this
material, or would like to revise this subject.
1.2
Using moles
In the next section, you will see how useful moles are when you need to determine the
amount of a chemical, for example, by weighing a solid, measuring the volume of a gas,
determining excess reactants or when diluting solutions.
© H ERIOT-WATT U NIVERSITY
1.2. USING MOLES
1.2.1
5
Converting moles and mass
To convert from moles to grams or vice versa, use the relationship that one mole is
equal to one gram formula mass.
Example : Using gram formula mass
How many grams of sodium hydroxide are required to precipitate chromium(III)
hydroxide completely from 50.0 g of chromium(III) sulphate?
Q15: How many grams of carbon dioxide are produced when 4.0 g of methane are
burned? Give your answer to 1 decimal place. The equation is:
CH4 + 2O2
CO2 + 2H2 O.
Q16: How many kilograms (kg) of limestone (CaCO 3 ) are required to produce 168.0 kg
of quicklime (CaO) according to the equation below? Give your answer to 1 decimal
place.
CaCO3
CaO + CO2
Q17: 8.10 g of magnesium was added to excess copper(II) sulphate solution. How
many grams (to 2 decimal places) of copper metal was produced?
Q18: How many kilograms (kg) of anhydrous sodium carbonate (Na 2 CO3 ) are required
to neutralise a spill of 6 kg of ethanoic acid (C 2 H4 O2 )? Give your answer to 1 decimal
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TOPIC 1. STOICHIOMETRY
place.
1.2.2
Converting moles and volumes for gases
When working with volumes of gases, use the fact that 1 mole of the gas occupies 1
molar volume. In the following questions an appropriate molar volume will be given.
Example : Using molar volume
When heated ammonium dichromate decomposes according to the equation:
(NH4 )2 Cr2 O7 (s)
Cr2 O3 (s) + 4H2 O(l) + N2 (g)
What volume of gas would you obtain from 5.15 g of ammonium dichromate at 25 Æ C ?
The molar volume of nitrogen at 25 Æ C is 24.45 mol-1 .
N.B. If the question had stated the temperature as 125 Æ C, the water would be a gas,
so that 1 mole of ammonium dichromate (252 g) would produce 5 moles of gaseous
product (1 mole of N2 + 4 moles of steam). The molar volume at the higher temperature
for both of these is 32.7 mol -1 . Under these conditions, the 5.15 g of ammonium
dichromate would produce 3.34 of gas.
Q19: Hydrogen peroxide decomposes according to the equation:
2H2 O2 (aq)
2H2 O(l) + O2 (g)
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1.2. USING MOLES
7
Assuming that the molar volume of oxygen is 24.0 mol -1 , how many litres (to 1 decimal
place) could be obtained from 17.0 g of hydrogen peroxide?
Q20: Nitrous oxide (laughing gas, N 2 O) can be produced by heating ammonium nitrate
according to the equation:
NH4 NO3
2H2 O + N2 O
How many litres (to 1 decimal place) of N 2 O could be obtained from 10.0 g of ammonium
nitrate? The molar volume of N 2 O is 24.0 mol-1
.
Q21:
Phosphorus reacts with aqueous sodium hydroxide according to the equation:
P4 + 3NaOH +3H2 O
3NaH2PO2 + PH3
How many litres (to 1 decimal place) of phosphine (PH 3 ) could be obtained from 10.3 g
of phosphorus, assuming the molar volume is 24.5 mol -1 ?
Q22: Potassium chlorate decomposes when heated according to the equation:
2KClO3
2KCl + 3O2
Assuming that the molar volume of oxygen at 100 Æ C is 30.0 mol-1 , how many litres
could be obtained by heating 24.5 g of potassium chlorate? Give your answer to 1
decimal place.
1.2.3
Determining excess or limiting reactants
In many situations, exact balanced (stoichiometric) amounts of reactants and products
may not be present. In a situation with two reactants, one or other will be in excess; the
other chemical is said to be the limitingreactant.
Example : With excess reactants
In solution, barium ions react with ammonium sulphate to precipitate barium sulphate.
5.00 g of ammonium sulphate was added to a sample containing 3.43 g of barium
ions. How much of the ammonium sulphate remained in the solution at the end of
the precipitation? Give your answer in grams to 2 decimal places.
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TOPIC 1. STOICHIOMETRY
Q23: Cadmium ions form a precipitate of cadmium sulphide (CdS) with sodium sulphide
solution.
Cd2+ + Na2 S
CdS + 2Na+
2.00 g of sodium sulphide was added to a solution containing 1.41 g of cadmium ion
(Cd2+ ). What weight (in grams to 2 decimal places) of sodium sulphide remains in
solution?
Q24: The residue from photographic processing equipment is being analysed for its
silver ion content by precipitation of silver chloride. The anticipated mass of silver is
0.54 g. How many grams (to 2 decimal places) of potassium chloride should be added
to ensure that twicethe required amount is present at the beginning of the reaction?
Q25: 30 g each of ethanoic acid (CH 3 CO2 H) and ethanol (C 2 H5 OH) are reacted to
prepare the ester, ethyl ethanoate (CH 3 CO2 C2 H5 ). Which material is in excess?
Q26: Assuming all the limiting reactant in the question above could be used, what
weight of ester (in grams, to 1 decimal place) could be made?
See further questions on page 198.
1.2.4
Calculating concentrations
A molar solution contains 1 mole of solute in 1 litre of solution.
For example, dissolving 10.6 g of sodium carbonate (0.1 mol) in 0.5 litre of solution
would produce a solution of concentration 0.2 mol -1 .
(1.1)
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1.2. USING MOLES
9
Molar concentrations have the units of mol
-1
or M.
Example : Calculating molar concentrations
Calculate the molar concentration of the solution obtained when 5.35 g of potassium
iodate (KIO3 ) is dissolved in 250 ml of solution.
Q27: Calculate the molar concentration, in mol -1 , of a solution made by dissolving 4.84
g of hydrated barium chloride (BaCl 2 .2H2 O) in 100 ml of solution. Give your answer to
2 decimal places.
Q28: What is the molar concentration (in mol -1 to 2 decimal places) of the solution
obtained by dissolving 5.10 g of potassium hydrogenphthalate (C 8 H5 O4 K) in 250 ml of
solution?
Q29: How many grams of hydrated iron(III) chloride (FeCl 3 .6H2 O) are required to
prepare 500 ml of a 0.05 mol -1 solution? Give your answer to 2 decimal places.
If you use the on-line version of these materials, you can see a worked answer by
pressing Display Answer.
Q30: You have been asked to prepare 2 litres of 0.025 mol -1 sodium thiosulphate
solution. How many grams (to 2 decimal places) of solid Na 2 S2 O3 .5H2 O should you
weigh out?
1.2.5
Diluting
When diluting a concentrated solution, it is useful to remember that the number of
moles of the solute remains the same in the original and diluted solutions; only the
amount of solvent will change.
Using this equality of moles before and after dilution, we can use:
M 1 V 1 = M2 V 2
where M1 and V1 are the molar concentration and volume, respectively, of the
concentrated solution, and M 2 and V2 apply to the diluted solution.
Example How many ml of concentrated hydrochloric acid (11 mol
© H ERIOT-WATT U NIVERSITY
-1 )
would be required
10
TOPIC 1. STOICHIOMETRY
to prepare 2.5 litres of 2 mol
Using
-1
hydrochloric acid?
M 1 V 1 = M2 V 2
11 x V1 = 2 x 2500
So the volume of concentrated acid required is 5000/11 = 454.5 ml.
Q31: Ammonia is often supplied as a 14.8 mol -1 concentrated solution. What volume
(in ml to 1 decimal place) of this would be required to make 2 litres of 2 mol -1 dilute
solution?
Q32: How many litres (to 1 decimal place) of 0.125 mol
be prepared from 100 ml of 2 mol -1 solution?
-1
sodium thiosulphate could
Q33: Potassium ion for intravenous drips should be administered at 0.040 mol -1 . How
many ml (to 1 decimal place) of 1.0 mol -1 concentrated solution should be made up to
0.5 to obtain the correct concentration?
Q34: 0.1 ml of an electro-plating solution was diluted to 100 ml. It was then found to
contain 0.005 mol -1 chromic acid (H2 CrO4 ). What was the concentration of chromium
in the original plating solution in g -1 ? Give your answer to 3 significant figures.
1.3
Volumetric analysis
This technique uses an accurately known concentration of one reagent in a quantitative
reaction to determine the concentration of another reactant. The reactions used in
volumetric analysis usually involve acid/base neutralisations, metal complex formation
or redox reactions. The procedure, in which the volumes of reacting solutions are
determined, is called a titration.
A solution of accurately known concentration is called a standard solution. A standard
solution is usually prepared from a primary standard, which is a substance with the
following characteristics:
• It must be readily available in a high state of purity ( 99.9 %).
• It should be stable in air at normal temperatures, so that it can be stored indefinitely
without change in composition.
• It should be readily soluble (usually in water), and the solution should be stable.
• It should have a reasonably high formula mass, so that errors in weighing molar
quantities are reduced.
These characteristics are required to ensure that what is weighed out is an
uncontaminated, accurate amount of the material. The most important property is the
stability. Many common substances are not stable in air, e.g. sodium hydroxide absorbs
both water and carbon dioxide from air and is therefore unsuitable as a primary standard.
Common primary standards are:
© H ERIOT-WATT U NIVERSITY
1.3. VOLUMETRIC ANALYSIS
• Oxalic acid or anhydrous sodium carbonate for acid/base titrations.
• Ethylenediaminetetraacetic acid (E.D.T.A.) for complexometric titrations.
• Potassium iodate or oxalic acid for redox titrations.
Which of the following statements about primary standards are correct?
Q35: It must be of high purity.
a) True
b) False
Q36: It must be stable in air and in solution.
a) True
b) False
Q37: It must be soluble (probably in water).
a) True
b) False
Q38: It must be acidic.
a) True
b) False
See further questions on page 198.
1.3.1
Titrations
Once a standard solution of accurately known concentration has been prepared, it can
be used in a titration experiment to determine the concentration of another solution of
a substance with which it is known to react in a quantitative reaction
When the reaction is just complete, it is said to have reached the equivalence point.
This is not always readily observable, so that some change associated with this point
is taken to indicate the titration’s end point. In acid/base titrations, there is always a
change in pH at the equivalence point, this is often signalled by a change in the colour
of an indicator to determine the end point.
Example : Titration calculation
In a typical titration you might find that 50 ml of 0.1 mol -1 hydrochloric acid were
required to neutralise 20 ml of barium hydroxide solution completely. So what is the
concentration of barium hydroxide?
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TOPIC 1. STOICHIOMETRY
Alternative methods of performing these calculations are described in the Higher Still
support document ’Unit 2: Principles of Chemical Reactions’.
Q39: 50.0 ml of 0.250 mol -1 potassium hydroxide solution are completely neutralised
by 20.0 ml of sulphuric acid. What is the concentration of the acid in mol -1 ? Give your
answer to 4 decimal places.
Q40: The effluent from a nuclear plant contains nitric acid. If 25.0 ml of the effluent
requires 42.5 ml of 0.010 mol -1 sodium hydroxide for neutralisation, what is the
concentration of the acid in mol -1 ? Give your answer to 3 decimal places.
Q41: What is the concentration of citric acid (a triprotic acid) in lemon juice if 10 ml
of the juice require 12.0 ml of 0.050 mol -1 sodium hydroxide solution for complete
neutralisation? Give your answer to 3 decimal places.
If you use the on-line version of these materials, you can see a worked answer by
pressing Display Answer.
Q42: A flask contains 52.5 ml of 0.15 mol -1 calcium hydroxide (Ca(OH)2 ). How many
millilitres of 0.35 mol -1 sodium carbonate (Na 2 CO3 ) are required to react completely
with the calcium hydroxide in the following reaction? Give your answer to 1 decimal
place.
Na2 CO3 + Ca(OH)2
CaCO3 + 2 NaOH
See further questions on page 198.
1.3.2
Redox titrations
Redox titrations are based on oxidation-reduction reactions. Two common systems use
potassium permanganate and iodine as the oxidising agents.
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1.3. VOLUMETRIC ANALYSIS
13
1.3.2.1 Redox titrations using potassium permanganate
Potassium manganate(VII) (potassium permanganate) is widely used in redox titrations
as it acts as its own indicator. It is decolourised in a redox reaction and therefore the
end point is indicated when a very pale pink colour (slight excess of permanganate) is
observed.
Example : Permanganate titration
’Iron tablets’ for treating anaemia should each contain 25 mg of iron(II). Ten tablets were
dissolved in dilute sulphuric acid and titrated with 0.05 mol -1 potassium permanganate
(KMnO4 ). If 18.0 ml of the permanganate was required, do the tablets contain the correct
quantity of iron?
.
Q43: 25 ml of a solution of sodium nitrite was acidified then titrated with 0.1 mol -1
potassium permanaganate. If 12.5 ml were required for reaction and the half-equation
for nitrite oxidation is:
NO2 - + H2 O
NO3 - + 2H+ + 2e-
what is the concentration of sodium nitrite, in mol
-1
to 3 decimal places?
Q44: Hydrogen peroxide, used as a bleach, is often supplied as a 6% solution. Its
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TOPIC 1. STOICHIOMETRY
concentration may be obtained by reaction with potassium permanganate according to
the reaction:
2MnO4- + 6H+ + 5H2 O2
2Mn2+ + 5O2 + 8H2 O
A solution of hydrogen peroxide was diluted 40 times and 25 ml of this solution required
21.32 ml of 0.02 mol -1 KMnO4 for complete reaction. Calculate the percentage
hydrogen peroxide in the original solution to 1 decimal place.
If you use the on-line version of these materials, you can see a worked answer by
pressing Display Answer.
1.3.2.2 Redox titrations involving iodine
Another common redox system uses iodine. This is useful, since traces of iodine give a
dark blue colour with starch, which can be used as an indicator.
In the next question, an oxidising agent (ozone) quantitatively converts iodide ion to
iodine. The weight of this is then determined by titration with thiosulphate from which
the amount of ozone can be calculated.
Q45: Ozone (O3 ) reacts with iodide ions according to the equation:
O3 + 2I- + H2 O
O2 + I2 + 2OH-
A 50 litre sample of air containing ozone was passed through a solution of potassium
iodide producing iodine. The iodine released reacted completely with 20.0 ml of 0.01
mol -1 sodium thiosulphate. What weight of iodine was produced? Give your answer in
milligrams, to 2 decimal places.
If you use the on-line version of these materials, you can see a worked answer by
pressing Display Answer.
Q46: What weight of ozone was in the air sample?
If you use the on-line version of these materials, you can see a worked answer by
pressing Display Answer.
Q47: Propanone reacts with iodine in alkaline solution according to the equation:
(CH3 )2 CO + 3I2 + 4NaOH
CHI3 + 3NaI + CH3 CO2 Na + 3H2 O
cm3
of a solution of propanone required 18.75 cm 3 of 0.20 mol -1
Titration of 25.0
iodine for complete reaction. What is the molar concentration of propanone, to 2 decimal
places?
1.3.2.3 Other redox titrations
Other oxidising agents often used are cerium(IV), Ce 4+ , and dichromate, Cr2 O7 2- . Here
are two questions involving these.
Q48: 2.25 g of impure cerium(IV) oxide (CeO 2 ) was dissolved in sulphuric acid and
made up to 500 ml. 25 ml of this solution was titrated with 30.0 ml of 0.02 mol -1
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1.3. VOLUMETRIC ANALYSIS
15
ammonium iron(II) sulphate. What is the molar concentration of cerium(IV) ions in the
solution, to 3 decimal places? The ionic equation for the reaction is:
Ce4+ + Fe2+
Ce3+ + Fe3+
Q49: What is the percentage purity of the original cerium(IV) oxide? Express your
answer to the nearest whole number and do not type the percentage sign.%
Q50: A sample of a metallic alloy containing iron was dissolved in sulphuric acid and
titrated with standard potassium dichromate (K 2 Cr2 O7 ) to determine its iron content.
The equation is:
6Fe2+ + Cr2 O7 2- + 14H+
6Fe3+ + 2Cr3+ + 7H2 O
If the original alloy weighed 1.25 g and 29.87 ml of 0.10 mol -1 potassium dichromate
was required for oxidation, calculate the percentage of iron in the sample to the nearest
unit. (Do not type the percentage sign.)%
1.3.3
Complexometric volumetric analysis
This type of analysis is particularly useful for estimating metal ions in solution by using
their ability to form complex ions with certain organic ligands.
Many titrations employ EDTA (ethylenediaminetetraacetic acid) which has multiple
complex-forming sites within one molecule, enabling it to form stable one-to-one
complexes with many metals.
The end point of these titrations is indicated by the colour change in an indicator such
as murexide or eriochrome black.
Example : Complexometric titration
A 50.00 ml aqueous sample containing iron(III) required 21.675 ml of 0.240 mol -1 EDTA
for complete reaction. What is the concentration of iron in the sample in mol -1 ?
Most ions react with EDTA in a 1-to-1 ratio, so use the equation:
The concentration of iron(III) is 0.104 mol
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-1 .
16
TOPIC 1. STOICHIOMETRY
Q51: How many moles of EDTA would be required to form a complex with the strontium
ions in 25 ml of 0.1 mol -1 strontium chloride solution?
Q52: ’Hardness’ in water can be caused by the presence of soluble calcium salts. If a
25.0 ml sample of water required 4.00 ml of 0.010 mol -1 EDTA for complete reaction
with the calcium ion present, what is its concentration in mol -1 , to 4 decimal places?
The complexometric determination of Nickel using EDTA (Unit 2 PPA 1)
30 min
A prescribed practical activity for assessment of Outcome 3 may be carried out (Refer
to SCCC document).
Discuss with your tutor whether the PPA on the determination of the concentration of
Ni2+ by titration with EDTA is to be completed at this stage.
1.4
Gravimetric analysis
Gravimetric analysis is a type of quantitative analysis in which the amount of a specific
chemical in a material is determined by converting it to a product which can be isolated
completely and weighed.
For example, lead sulphate is insoluble in water and when excess sulphate ions are
added to a solution containing lead ions, lead sulphate is precipitated. This can be
separated from other materials by filtration, then dried and weighed to determine the
lead content of the original sample
Example : Gravimetric analysis
A sample of water is known to contain lead ions. When an excess of sodium sulphate
is added to 2.00 litre of the sample, a precipitate of lead sulphate (PbSO 4 ) is produced.
On filtering and drying, it was found to weigh 231.5 mg. What is the concentration of
lead in the water in mg -1 ?
Work from the balanced equation:
Pb2+ (aq) + Na2 SO4 (aq)
PbSO4 (s) + 2Na+ (aq)
• 1 mol of Pb2+ ions produce 1 mol of PbSO 4 precipitate.
• 207.2 mg of Pb 2+ produces 303.3 mg PbSO 4
• (207.2 /303.3) x 231.5 mg Pb 2+ produces 231.5 mg PbSO 4
• 158.15mg of Pb 2+ ion is present in the 2.00 litres of water, the concentration is
79.075 mg -1 .
Q53: Excess potassium chromate was added to a solution of silver nitrate to precipitate
silver chromate (Ag2 CrO4 ). If 11.06 g of silver chromate was produced what weight
of silver nitrate was present in the original solution? Give your answer in grams to 2
decimal places.
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1.5. SUMMARY
17
Q54: Hydrogen sulphide was passed into 100 ml of an acidified solution containing
copper(II) ions to precipitate copper(II) sulphide (CuS). If 4.780 g of copper(II) sulphide
was produced, what is the concentration of copper(II) ions in mol -1 to 2 decimal places?
Q55: In a gravimetric determination of aluminium, an aqueous solution of aluminium
sulphate (Al2 (SO4 )3 ) was treated with an excess of ammonium hydroxide (NH 4 OH) to
precipitate aluminium hydroxide. This was heated and decomposed into aluminium
oxide, which had been weighed. The equations are:
Al2 (SO4 )3 + 6OH2Al(OH)3
2Al(OH)3 + 3SO4 2Al2 O3 + 3H2 O
The solution of aluminium sulphate yield 1.054 g of aluminium oxide. How many grams
of aluminium sulphate were in the original sample?
If you use the on-line version of these materials, you can see a worked answer by
pressing Display Answer.
Q56: Excess sodium carbonate was added to 25.0 ml of a solution containing copper(II)
ion. The precipitate of copper(II) carbonate was filtered off, washed and heated strongly
to convert it to copper(II) oxide. The equations are:
Cu2+ + CO3 2CuCO3
CuCO3
CuO + CO2
If the copper(II) oxide weighed 0.159 g, what is the molar concentration of Cu 2+ in the
original solution? Give your answer to 2 decimal places.
See further questions on page 198.
The gravimetric determination of water in hydrated barium chloride (Unit
2 PPA 2)
A prescribed practical activity for assessment of Outcome 3 may be carried out. (Refer
to SCCC document)
Discuss with your tutor whether the PPA on the determination of the moles of water of
hydration in hydrated barium chloride is to be completed at this stage.
1.5
Summary
The key points of this Topic are:
• Stoichiometry is dependent on the ability to write balanced chemical equations
and to calculate gram formula masses.
• Reacting quantities in mass units use the fact that one mole is one gram formula
mass; reacting quantities in volumes for gases use the fact that one mole of a gas
occupies one molar volume.
• Standard solutions are of accurately known concentrations.
• Primary standards, which can be weighed to produce standard solutions, require
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TOPIC 1. STOICHIOMETRY
certain properties, mainly stability in air and in solution.
• A molar solution contains one mole of solute in one litre of solution.
• Unknown concentrations of substances can be determined by titration, which can
use acid-base, redox or complexometric reactions, in a volumetric analysis.
• Gravimetric analysis estimates the amounts of sustances by isolating a chemical
and weighing it.
1.6
Suggestions for further reading
• Chemistry in Action Michael Freemantle Macmillan Press 2nd ed. 1995 Ch.4
Stoichiometry.
• General Chemistry Darrel D. Ebbing Houghton Mifflin Co. 1996 Ch.4 Calculations
with Chemical Formulae and Equations.
• Chemistry Ken Gadd and Steve Gurr University of Bath 1994 Ch.7 Quantitative
Analysis.
• SCCC Booklet CSYS CHEMISTRY No.10 Chemistry: Certificate of Sixth Year
Studies. Sec. 1 Stoichiometry.
• Chemistry in Context Third Edition Graham Hill and John Holman Nelson 1989
Ch.2 Reacting Quantities and Equations.
• A-level Chemistry E.N. Ramsden Stanley Thornes (Publishers) Ltd. 1985 Ch.3
1.7
Tutorial questions
Q57: How many milligrams (mg) of lead(II) sulphate would be precipitated from a
solution containing 165.6 mg of lead(II) nitrate by the addition of excess sodium
sulphate? Answer to 1 decimal place.
Q58: The catalyst aluminium chloride (AlCl 3 ) is prepared by passing hydrogen chloride
gas over aluminium metal shavings.
2Al(s) + 6HCl(g)
2AlCl3 (s) + 3H2 (g)
Assuming the molar volumes of hydrogen chloride and hydrogen to be 30 mol -1 , how
many litres (to 1 decimal place) of hydrogen chloride would be required to prepare 26.7
g of aluminium chloride?
Q59: How many litres of hydrogen would be produced by the reaction in the previous
question?
Q60: One way to restore the brightness of lead pigments in old masters that have been
dulled by the formation of lead sulphide is to treat with hydrogen peroxide. This converts
the brown lead sulphide into white lead sulphate.
© H ERIOT-WATT U NIVERSITY
1.8. END OF TOPIC TEST
PbS + 4H2 O2
19
PbSO4 + 4H2 O
What mass in grams of hydrogen peroxide should be used to restore a picture containing
5.18 g of lead if a 10% excess is required to ensure effectiveness? Give your answer to
3 significant figures.
See further questions on page 199.
1.8
End of Topic test
An online assessment is provided to help you review this topic.
© H ERIOT-WATT U NIVERSITY
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TOPIC 1. STOICHIOMETRY
© H ERIOT-WATT U NIVERSITY
21
Topic 2
Chemical Equilibrium
Contents
2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 The nature of chemical equilibrium . . . . . . . . . . . . . .
2.3 The equilibrium constant, Kc . . . . . . . . . . . . . . . . .
2.3.1 The units of equilibrium constants . . . . . . . . . .
2.4 The equilibrium constant, Kp , involving gases . . . . . . . .
2.5 Homogeneous and heterogeneous equilibria . . . . . . . .
2.6 Calculations using equilibrium constants . . . . . . . . . . .
2.7 Factors that alter the composition of an equilibrium mixture
2.7.1 Addition of reactant(s) or product(s) . . . . . . . . .
2.7.2 Alteration of pressure . . . . . . . . . . . . . . . . .
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23
24
26
28
29
30
31
35
35
36
2.7.3 Change in temperature .
2.7.4 The influence of catalysts
2.8 Summary . . . . . . . . . . . . .
2.9 Resources . . . . . . . . . . . .
2.10 End of Topic Test . . . . . . . . .
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37
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Prerequisite knowledge
Before you begin this Topic, you should be able to :
• understand stoichiometry and chemical equations (Topic 2.1);
• define the term dynamic chemical equilibrium (Higher);
• describe how concentration, temperature and pressure affect the position of an
equilibrium (Higher).
Learning Objectives
After studying this Topic, you should be able to:
• define the term ’equilibrium constant’,
composition, from the equilibrium equation;
which characterises
• carry out calculations involving Kc and Kp values;
equilibrium
22
TOPIC 2. CHEMICAL EQUILIBRIUM
• predict qualitative changes in the equilibrium state as a consequence of changing
conditions, based on le Chatelier’s principle;
• state the reasons for the equilibrium position being unaffected by the presence of
a catalyst.
© H ERIOT-WATT U NIVERSITY
2.1. INTRODUCTION
2.1
Introduction
’To be in equilibrium is to be in a state of delicate balance, poised between opposing
forces so as to appear to be at rest. Two arms wrestling on a table top, neither able
to budge the other, are in a state of equilibrium. So too is a picture hanging on a wall,
where the downward pull of gravity is balanced exactly by the upward pull of the wire.
In all the natural world there are really only two options: to be in equilibrium, or else to
be approaching it. If you are in equilibrium, there you will stay unless disturbed by some
outside influence. If you are out of equilibrium, as indeed are all living organisms, you
will eventually get there. Everything does; it is only a matter of time.’ (Michael Munowitz,
2000)
This Topic considers the equilibrium state that is achieved by chemical reactions.
First, consider a simple case of physical equilibrium. A boulder at the top of a hill will
remain there until disturbed - it is in a state of equilibrium. When pushed, however, it will
readily move into the valley where it will remain even when displaced slightly.
There is an animation illustrating this point available in the on-line version of this Topic.
Figure 2.1: The position of unstable and stable equilibrium
The boulder’s unstable equilibrium position, shown in Figure 2.1, has moved to a
stable equilibrium. This state usually represents a minimum energy state (in this case,
the lowest gravitational potential energy).
© H ERIOT-WATT U NIVERSITY
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24
TOPIC 2. CHEMICAL EQUILIBRIUM
2.2
The nature of chemical equilibrium
A chemical system in equilibrium shows no changes in macroscopic properties, such as
overall pressure, total volume and concentration of reactants and products. It appears
to be in a completely unchanging state as far as an outside observer is concerned.
Consider a bottle of soda water (carbon dioxide dissolved in water, with free carbon
dioxide above). So long as the system remains closed, the macroscopic properties (e.g.
the pressure of CO2 in the gas and the concentration of the various dissolved materials)
will remain constant - the system is in equilibrium. However, on the microscopic scale
there is change. Carbon dioxide molecules in the gas will bombard the liquid surface
and dissolve; some carbon dioxide molecules in the solution will have sufficient energy
to leave the solution and enter the gas phase.
An animation showing this process is available in the on-line version of this Topic.
At equilibrium these two processes will balance and the number of molecules in the gas
and liquid will always be the same, although the individual molecules will not remain
static. This state is achieved by a dynamic equilibrium between molecules entering
and leaving the liquid, and between carbon dioxide, water and carbonic acid. In other
words, the rate at which carbonic acid is formed from CO 2 and water will be balanced
by carbonic acid dissociating to form CO2 and water.
CO2 + H2 O
« H CO
2
3
In the on-line version of this Topic there is an animation showing the reaction of hydrogen
and iodine reacting to form hydrogen iodide, and an equilibrium being established. You
should view this before attempting the following question.
Q1: At t = 0 there are six molecules of H2 , six of I2 and none of HI. Count the number
of molecules of H2 , I2 , and HI after time, t = 15, t = 30 and t = 70. What do you notice
about them?
This simple animation has reached equilibrium; each time two molecules of hydrogen
and iodine react to form hydrogen iodide. In this case, two molecules of hydrogen iodide
react to form hydrogen and iodine.
Another animation is available in the on-line version of this Topic. This time showing
the decomposition of hydrogen iodide to form hydrogen and iodine and an equilibrium
situation being established. You should view this before attempting the questions that
follow.
Q2: What do you notice about the number of HI, H 2 and I2 molecules after time, t = 35
and t = 70?
Q3: But how do these numbers at equilibrium compare with the previous reaction
starting with hydrogen and iodine?
© H ERIOT-WATT U NIVERSITY
2.2. THE NATURE OF CHEMICAL EQUILIBRIUM
At equilibrium, the rate of production of HI from H 2 and I2 equals the rate of production
of H2 and I2 from HI, therefore the overall composition will not change. This process is
generally shown by the use of reversible arrows.
H2 + I2
« 2HI
An animation is available online showing the progress of the reaction when the reactants
are H2 and I2 . There is also an animation showing the progress of the reaction when HI
is the starting material. Look at the graphs of both these scenarios, shown as Figure 2.2
and Figure 2.3, plotted as concentration versus time.
Figure 2.2: The approach to equilibrium starting with H 2 and I2
Figure 2.3: The approach to equilibrium starting with HI
You should notice that in both situations, the equilibrium concentrations reached are
identical, whether you start with hydrogen and iodine or with hydrogen iodide.
© H ERIOT-WATT U NIVERSITY
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26
TOPIC 2. CHEMICAL EQUILIBRIUM
LEARNING POINT
A chemical reaction is at equilibrium when the composition of the reactants and
products remains constant indefinitely.
This state occurs when the rates of the forward and reverse reactions are equal.
The same equilibrium mixture is obtained whether you start with reactants or
products.
2.3
The equilibrium constant, Kc
Learning Objective
Æ
To be able to define the term ’equilibrium constant’ from the equilibrium equation.
An understanding of the mathematical basis of chemical equilibrium was begun by
Guldberg and Waage in the 1860s with their ’Law of Mass Action’. This stated that
for a generalised chemical reaction:
«
aA + bB
Reactants
cC + dD
Products
at equilibrium, the following expression applied (note that [ ] means concentration in mol
-1 ):
Equilibrium Constant
[C]c [D]d
Kc =
In Terms of
Concentration
[A]a [B]b
Indices (from
balanced equation)
Concentrations
of Products
Concentrations
of Reactants
For example, for the reaction between iron(III) ions and cyanide:
Fe3+(aq) + 6CN-(aq)
Kc =
[Fe(CN)6]3-(aq)
[[Fe(CN)6]3-]
[Fe3+] [CN-]6
Notice that the concentrations of products form the numerator, the concentrations of
reactants are in the denominator and each concentration term is raised to a power equal
to the number of moles in the balanced equation.
© H ERIOT-WATT U NIVERSITY
2.3. THE EQUILIBRIUM CONSTANT, K C
27
Write an equilibrium expression for the following reactions.
Q4: 2Fe3+ (aq) + 3I- (aq)
Q5: H3 PO4 (aq)
« 2Fe
2+ (aq)
« 2H (aq) + HPO
+
4
+ I3 - (aq)
2- (aq)
Returning to the reaction:
H2 + I2
« 2HI
the equilibrium constant is defined as:
[HI]2
[H2 ][I2 ]
Kc =
and at 453 Æ C, it has a value of 50.
Q6:
a)
b)
c)
d)
At 453Æ C which compound is present in greatest concentration?
hydrogen
iodine
hydrogen iodide
all the same concentration
Since the equilibrium constant is the ratio of concentration of products divided by
the concentration of reactants, its actual value gives guidance to the extent of a
reaction once it has reached equilibrium. The greater the value of K c the greater the
concentration of products compared to reactants; in other words, the further the reaction
has gone to completion.
The explosive reaction between hydrogen and fluorine:
H2 + F2
« 2HF
has an equilibrium constant of 1 x 10 47 . At equilibrium, negligible amounts of the
reactants will remain; almost all will have been converted to hydrogen fluoride.
In contrast, the dissociation of chlorine molecules to atoms:
Cl2
« 2Cl
has a Kc value of 1 x 10-38 at normal temperatures, indicating a reaction which hardly
occurs at all under these circumstances.
In time, all reactions can be considered to reach equilibrium. To simplify matters, the
following general assumption is made:
Value of Kc
10-3
Extent of reaction
Effectively no reaction
10-3 to 103
Significant quantities of reactants and
products at equilibrium
103
Reaction is effectively complete
© H ERIOT-WATT U NIVERSITY
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TOPIC 2. CHEMICAL EQUILIBRIUM
A note of caution:
The equilibrium constant gives no indication of the rate at which equilibrium is
achieved. It indicates only the ratios of products to reactants once this state is reached.
You will study reaction rates and feasibility in Topics 9 and 7.
The Kc value for the reaction
Q7:
PCl5
« PCl
3
+ Cl2
is 0.021 at 160Æ C. Which compound is present in greatest concentration at equilibrium?
a)
b)
c)
d)
phosphorus(V) chloride
phosphorus(III) chloride
chlorine
all are the same
The following equilibrium constants apply at room temperature (25 Æ C).
Q8:
« Cu(s) + Zn
(aq) « Cu(s) + Mg
(aq) « Cu(s) + Fe
Zn(s) + Cu2+ (aq)
Mg(s) + Cu2+
2+ (aq)
K = 2 x 1037
2+ (aq)
K = 6 x 1090
2+ (aq)
Fe(s) + Cu2+
K = 3 x 1026
Of the metals Zn, Mg, and Fe, which removes Cu(II) ions from solution most completely?
a) Zn
b) Mg
c) Fe
Q9: In which of the following reactions will the equilibrium lie furthest towards
products?
a)
b)
c)
d)
«
«
2NO2 (g)
N2 O4 (g)
2SO3 (g)
2SO2 (g) + O2 (g)
2NO2 (g)
N2 O4 (g)
2SO3 (g)
2SO2 (g) + O2 (g)
«
«
Kc
Kc
Kc
Kc
at 0Æ C = 159
at 856Æ C = 21.1
at 25Æ C = 14.4
at 636Æ C = 3343
Q10: From the data in the previous question, what do you notice about the value of K c
for the oxidation of SO2 at 856Æ C compared with 636 Æ C?
See further questions on page 200.
From the data in the last question, you can see that the equilibrium constant has different
values as the temperature changes.
LEARNING POINT
At a given temperature, the equilibrium constant for a reaction is a constant value.
Equilibrium constants can change markedly with temperature. The values can
increase or decrease with a change of temperature.
2.3.1
The units of equilibrium constants
Equilibrium constants are now taken to have no units (although some older textbooks
might give them).
© H ERIOT-WATT U NIVERSITY
2.4. THE EQUILIBRIUM CONSTANT, K P , INVOLVING GASES
29
One explanation for this lack of units is that equilibrium constants should be strictly
expressed as the ratio of a property called the activity (not part of the advanced higher
syllabus) of products and reactants. In dilute solutions, the activity is almost the same
as concentration, but since activities are without units, equilibrium constants will have
none.
2.4
The equilibrium constant, Kp , involving gases
Learning Objective
Æ
To define the term ’equilibrium constant’ for gas reactions from the equilibrium
equation.
The equilibrium constant, K c , is expressed in terms of the concentration of reactants and
products. This is the most convenient form to use when solutions are being considered;
but for reactions involving gaseous reactants and products, the equilibrium constant can
be expressed in terms of partial pressures. (Gases inside a closed container each
exert a pressure proportional to the number of moles of the particular gas present.
For example, if two gases are mixed in equimolar amounts and the total pressure is
1 atmosphere, then the partial pressure of each gas is 0.5 atmosphere.)
For the general reaction:
PNH32
Kp
PN2 PH23
Write down an appropriate expression for the equilibrium constant for the following
reactions:
« 2NO(g) + Cl (g)
Q12: 2SO (g) + O (g) « 2SO (g)
Q11: 2NOCl(g)
2
2
2
© H ERIOT-WATT U NIVERSITY
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TOPIC 2. CHEMICAL EQUILIBRIUM
2.5
Homogeneous and heterogeneous equilibria
In the examples above, all the reactants and products have been in the same phase (i.e.
all in liquid/solution or all gaseous). These reactions are homogeneous reactions.
However, considering the reaction:
CaCO3 (s)
« CaO(s) + CO (g)
2
Two of the participating species are solid and one product is a gas.
heterogeneous reaction.
This is a
In the cases where a solid or liquid is present in a reaction, its ’concentration’ is
effectively constant and is given the value 1 in the equilibrium expression (the actual
value is incorporated into the value of K c or Kp ).
In the example above, the solids CaCO 3 and CaO are given the value 1, so that
Kp = PCO2
Consider another reaction, the hydrolysis of methyl ethanoate
CH3 COOCH3(aq) + H2 O(l)
« CH CO H(aq) + CH OH(aq)
3
2
3
In this case, the hydrolysis is carried out in aqueous solution with the concentration of
water remaining effectively constant and so the equilibrium expression is:
Kc
[CH3 CO2 H][CH3 OH]
[CH3 COOCH3 ]
as the value for [H2 O] is taken to be 1.
Q13: Which expression below is correct for the reaction
Zn(s) + Cu2+ (aq)
« Zn
2+ (aq)
+ Cu(s)
a)
Kc =
[Zn2 + ] [Cu]
[Zn] [Cu2 + ]
b)
Kc =
[Zn] [Cu2 + ]
[Zn2 + ] [Cu]
c)
Kc =
[Zn2 + ]
Cu d)
Kc =
[Cu2 + ]
[Zn2 + ]
Q14: Which expression is correct for the reaction
NH4 NO3 (s)
« 2H O(g) + N O(g)
2
2
© H ERIOT-WATT U NIVERSITY
2.6. CALCULATIONS USING EQUILIBRIUM CONSTANTS
31
a)
Kp =
p2H2 O pN2 O
pNH4 NO3
b)
p N2 O
Kp =
pNH4 NO3
c)
Kp = p2H2 O pN2 O
d)
K p = p N2 O
2.6
Calculations using equilibrium constants
Learning Objective
Æ
To be able to carry out calculations involving K c and Kp values
If all the equilibrium concentrations for the reactants and products are known, calculation
of the equilibrium constant is simply a matter of substitution into an appropriate
equilibrium expression.
Example : Calculation of an equilibrium constant
An equilibrium mixture of gaseous O 2 , NO and NO2 at 500 K contains 1.0 x 10 -3 mol
-1 O , 1.9 x 10-3 mol -1 NO and 5.0 x 10 -2 mol -1 NO . Calculate the value of K at
c
2
2
500 K.
Write a balanced equation:
O2 + 2NO
« 2NO
2
Then write the equilibrium expression:
Kc
[NO2 ]2
[NO]2 [O2 ]
and substitute appropriate values:
Kc
The value of Kc at 500 K is 6.9 x 10 5
Alternatively, the equilibrium constant may be given and the concentration of a reactant
or product be required.
© H ERIOT-WATT U NIVERSITY
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TOPIC 2. CHEMICAL EQUILIBRIUM
Example : Calculation of an unknown concentration
For the reaction:
H2 + I2
« 2HI
The equilibrium constant at 700 K is 57.0. If, at equilibrium, the concentration of H 2
equals the concentration of I 2 at 0.08 mol -1 , what is the concentration of HI?
From the chemical equation, the equilibrium expression is:
[HI]2
[H2 ][I2 ]
Kc
We are asked for [HI] so rearranging gives:
[HI] =
and substituting:
Kc [H2 ][I2 ]
57.0 0.08 0.08
[HI]
mol The concentration of HI is 0.60 mol
-1 .
Here are some questions for you to try yourself.
Q15: Calculate the equilibrium concentration of PCl 5 at 300 K when the equilibrium
concentrations of PCl3 and Cl2 are 0.20 mol -1 . The equilibrium constant for the
reaction:
PCl5 (g)
« PCl (g) + Cl
3
at 300 K is 11.5. Calculate your answer in mol
figures.
2
-1
and give your answer to 2 significant
Q16: Calculate the value of K p for the decomposition of nitrosyl bromide at 350 K:
2NOBr
« 2NO + Br
2
given that the equilibrium mixture contains 0.50 atm NOBr, 0.25 atm NO and 0.20 atm
Br2 . Give your answer to 2 significant figures.
Using initial conditions
In other situations, initial conditions are given together with a suitable equilibrium
constant and you are asked to determine equilibrium concentrations. In these cases,
follow the 5 steps below.
© H ERIOT-WATT U NIVERSITY
2.6. CALCULATIONS USING EQUILIBRIUM CONSTANTS
1.
Write a balanced equation for the reaction.
2.
Under the balanced equation, make a list for each substance involved in the
reaction:
1) The initial concentration.
2) The change in concentration on going to equilibrium.
3) The equilibrium concentration.
In constructing the table, define x as the concentration (mol -1 ) or partial pressure
(atm) of one of the substances that reacts on going to equilibrium.
3.
4.
Substitute the equilibrium concentrations into the equilibrium expression for the
reaction and solve for x.
Calculate the equilibrium concentrations from the calculated value of x.
5.
Check your results by substituting them into the equilibrium expression.
Example : Calculating equilibrium concentrations
The equilibrium constant (K c ) for the reaction:
H2 + I2
« 2HI
is 57.0 at 700 K
If 0.10 mol of H2 and 0.10 mol of I 2 react in a 1.0 litre vessel at 700 K, what are the
concentrations of H2 , I2 and HI at equilibrium?
So taking the positive square root of both sides and simplifying:
© H ERIOT-WATT U NIVERSITY
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34
TOPIC 2. CHEMICAL EQUILIBRIUM
7.55(0.10 - )
= 2
0.755 = (2 + 7.55)
=
=
At equilibrium the concentrations are:
[H2 ] = [I2 ] = 0.10 - 0.0791
and [HI] = 2 0.0791
-1
= 0.021 mol
= 0.158 mol
-1
N.B. It is often useful to check that these values can be substituted back into the
equilibrium expression to get the correct value of K c :
(0.158)2
(0.021)2
When taking the square roots (or the solutions to quadratic equations), two values (using
+ve and -ve roots) are found. Only one of these normally gives sensible values in the
physical sense. In this latter example, we chose the +ve root of 57 to give a sensible
answer; taking -7.55 as another mathematically valid solution to 57 would result in x =
0.136. Since there was an initial concentration of H 2 of 0.1, a value of 0.136 would use
more than was available and so should be rejected.
Q17: By following the layout of the example above, calculate the equilibrium
concentrations for the same reaction when the initial amount of iodine is 0.2 mol instead
of 0.1 mol.
A partially worked answer is shown for this in the answer section.
Q18:
The equilibrium constant for the reaction:
N2 O4 (g)
« 2NO (g)
2
is 0.01 at a particular temperature.
Calculate the equilibrium concentrations of NO 2 and N2 O4 at this temperature in a flask
containing initially only N2 O4 at 0.050 mol -1 , giving your answer to 2 decimal places.
Q19: Calculate the equilibrium concentrations in the previous question if the starting
conditions were [N2 O4 ] = 0.0200 mol -1 and [NO2 ] = 0.0300 mol -1 . Answer to 4
decimal places.
Q20: The equilibrium constant (K p ) for the reaction:
O2 + N2
« 2NO
is 4.65 x 10-31 at 25 Æ C. Assuming that the partial pressure of oxygen and nitrogen in
air are 0.21 atm and 0.78 atm respectively, calculate the partial pressure of NO in the
atmosphere at this temperature. Give your answer (in atm), to 3 significant figures.
See further questions on page 200.
© H ERIOT-WATT U NIVERSITY
2.7. FACTORS THAT ALTER THE COMPOSITION OF AN EQUILIBRIUM MIXTURE
2.7
Factors that alter the composition of an equilibrium
mixture
Learning Objective
Æ
To be able to predict changes in the equilibrium position as a consequence of
changing conditions
2.7.1
Addition of reactant(s) or product(s)
At a given temperature the equilibrium constant does not change, so addition of a
reactant, for example, will mean that the concentrations of reactants and products are
no longer satisfying the equilibrium expression and the forward and reverse reactions
continue until equilibrium is re-established.
For example, in the water gas shift reaction:
CO(g) + H2 O(g)
« CO (g) + H (g)
2
2
The equilibrium constant, K c = 4.24 at 800 K.
Starting with 0.2 mol
mixture will contain:
-1
each of CO and H 2 O, you can calculate that the equilibrium
[CO] = [H2 O] = 0.065 mol
and [CO2 ] = [H2 ] = 0.135 mol
If an additional 0.2 mol
since:
-1
-1
-1
of steam is added, the mixture will no longer be at equilibrium,
[CO2 ][H2 ]
= 1.06
[CO][H2 O]
Kc
In order to restore the equilibrium ratios, some of the added steam will react with CO to
form more products (CO2 and H2 ) until the ratios again are correct for the equilibrium
situation.
Using the method for calculating equilibrium concentrations described before, this will
occur when:
[H2 O] = 0.233
[CO] = 0.028
[CO2 ] = [H2 ] = 0.167
Check yourself that these values for equilibrium concentrations fit the equilibrium
expression.
You will notice that adding water has increased the equilibrium concentration of CO 2 and
H2 from 0.135 mol -1 to 0.167 mol -1 .
Q21: A gaseous mixture contains 0.30 mol CO, 0.10 mol H 2 , and 0.02 mol H 2 O, plus
some CH4 , per litre.
© H ERIOT-WATT U NIVERSITY
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36
TOPIC 2. CHEMICAL EQUILIBRIUM
CO(g) + 3H2 (g)
« CH (g) + H O(g)
4
2
The value of Kc at 1200 K is 3.92. What is the concentration of CH 4 in the mixture? Give
your answer to 3 decimal places.
Q22: More H2 was added to the system above to bring its equilibrium concentration to
0.20 mol -1 . What is the concentration of CH4 now?
2.7.2
Alteration of pressure
Le Chatelier’s principle gives a qualitative description of the changes that will occur when
conditions affecting an equilibrium are altered. It states that when a change is applied
to a reaction mixture at equilibrium, reaction will occur in a direction that will reduce the
change.
Applying this to the equilibrium between NO 2 and N2 O4 :
2NO2
«N O
2
4
If the pressure is increased by decreasing the volume of a closed system, the equilibrium
will move to reduce this increased pressure. This can be achieved in this system by two
NO2 molecules combining to form one N 2 O4 molecule, which will occupy less volume
and so reduce the pressure of the system.
An animation of this principle is available in the on-line version of this Topic.
The use of Le Chatelier’s principle in this way applies only to systems containing gases.
If there are equal numbers of moles of gases in the reactants and products, then change
in the equilibrium position will have no effect on the pressure. So, conversely, alteration
of the pressure will not affect the equilibrium position.
© H ERIOT-WATT U NIVERSITY
2.7. FACTORS THAT ALTER THE COMPOSITION OF AN EQUILIBRIUM MIXTURE
For the reactions below, suggest whether an increase in pressure will result in the
reaction going from:
• left to right
• right to left
• or there will be no change
Q23: CaCO3 (s)
« CaO(s) + CO (g)
a) left to right
b) right to left
c) no change
Q24: C(s) + O2 (g)
a) left to right
b) right to left
c) no change
2
« CO (g)
2
« 2NH (g)
Q25: N2 (g) + 3H2 (g)
3
a) left to right
b) right to left
c) no change
Q26: PCl5 (g)
« PCl (g) + Cl (g)
3
2
a) left to right
b) right to left
c) no change
See further questions on page 200.
2.7.3
Change in temperature
A change in temperature will change the value of the equilibrium constant, but even if
the value at the new temperature is not known (which would enable you to calculate the
new equilibrium concentrations) you can determine the direction of the change by using
Le Chatelier’s principle.
Consider an exothermic reaction, such as the industrial manufacture of ethanol from
ethene and steam:
C2 H4 (g) + H2 O(g)
« C H OH(g)
2
5
When the reaction moves to the right, heat is given out. Conversely, when it moves to
the left, heat is absorbed.
Which way will the reaction go if the temperature is raised?
Le Chatelier’s principle states that when the temperature is raised, the system will move
in a direction to reduce the temperature. For an exothermic reaction to absorb heat it
© H ERIOT-WATT U NIVERSITY
37
38
TOPIC 2. CHEMICAL EQUILIBRIUM
must go in reverse, i.e. move from right to left. The equilibrium position for an exothermic
reaction will move in the direction of increased reactants when the temperature is raised.
For the reaction above, increasing temperature will result in more ethene and steam in
the equilibrium reaction mixture. The value of K p will decrease.
To see how temperature affects the equilibrium position of an endothermic reaction
consider:
N2 (g) + O2 (g)
« 2NO(g)
H = +180 kJ mol-1
Q27: When the temperature rises, will the reaction move to absorb or produce heat,
according to Le Chatelier’s principle?
Q28: Will this endothermic reaction move to the right or left?
Q29: Write down an expression for K c for this reaction.
Q30: Will the value of Kc increase or decrease?
This reaction between nitrogen and oxygen can occur in air, but the equilibrium
concentration of NO is very small at normal temperatures, increasing to significant levels
when lightning passes through air.
LEARNING POINT
For an exothermic reaction, the equilibrium constant decreases with increase in
temperature; for an endothermic reaction, the equilibrium constant increases with
temperature.
Q31: Would you expect the dissociation of iodine molecules
I2 (g)
« 2I(g)
to be exothermic or endothermic?
Q32: Would you expect the equilibrium constant for this reaction to increase or
decrease as the temperature increases?
a) increase
b) decrease
Now try this summary question.
Q33: Suggest four ways in which the equilibrium concentration of SO 3 can be
increased in a closed vessel if the only reaction is:
2SO2 (g) + O2 (g)
2.7.4
« 2SO (g)
3
for which Ho = -99 kJ mol-1
The influence of catalysts
Learning Objective
Æ
To be able to state the reasons for the equilibrium position being unaffected by the
presence of a catalyst.
© H ERIOT-WATT U NIVERSITY
2.8. SUMMARY
39
Catalysts affect the rate of a reaction, but since they affect the rate of the forward and
reverse reactions to the same extent, they will not affect the equilibrium position.
The equation for the synthesis of ammonia is:
N2 + 3H2
« 2NH
3
Industrially, in the Haber process, an iron catalyst is used.
You will have studied the rates at which chemical reactions occur and met similar
diagrams to Figure 2.4 during your Higher course. Figure 2.4 shows the energy changes
during the reaction between nitrogen and hydrogen to make ammonia.
700
600
500
400
300
200
100
0
-100
-200
Figure 2.4: Potential Energy Diagram
In simple terms, the rate at which a reaction occurs is related to the height of the
energy barrier (the activation energy). A catalyst reduces the height of this barrier. The
equilibrium position is determined by other factors that are related to the energies of
the products and reactants.
Many industrial processes use catalysts to increase the rate at which products are
formed to assist economic production, but the ratio of concentrations of products to
reactants will always be that determined by the equilibrium constant appropriate to the
conditions.
2.8
Summary
The key points of this Topic are:
• Chemical equilibria are stable dynamic equilibria where the equilibrium
concentrations of reactants and products are defined by an equilibrium constant
© H ERIOT-WATT U NIVERSITY
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TOPIC 2. CHEMICAL EQUILIBRIUM
called Kc given by:
• For gaseous reactions, another constant (K p ) can be defined in terms of the partial
pressures of reactants and products (in atmospheres).
• Both Kc and Kp have values which are constant at a given temperature.
• In homogeneous reactions, the concentrations of all species are in the same
phase; in heterogeneous reactions, the species are in more than one phase. The
concentrations of pure solids or pure liquids are given the value 1 in the equilibrium
equation.
• In response to changes in conditions (addition/removal of reacting species, altered
pressure or temperature) the equilibrium composition will move in a direction to
reduce the effect of the change (Le Chatelier’s principle).
• Catalysts do not affect the equilibrium composition, but will speed its achievement.
2.9
Resources
• Chemistry: McMurry and Fay, Prentice Hall, ISBN 0-13-737776-2
• General Chemistry: Ebbing, Houghton Mifflin, ISBN 0-395-74415-6
• Chemistry in Context: Hill and Holman, Nelson, ISBN 0-17-438401-7
• Chemistry: Fullick and Fullick, Heinemann, ISBN 0-435-57080-3
• Higher Still Support: Advanced Higher Chemistry - Unit 2: Principles of Chemical
Reactions, Learning and Teaching Scotland, ISBN 1 85955 874 7.
• A-level Chemistry: E.N. Ramsden, Stanley Thorne Publishers, ISBN 0-85950154-X
2.10
End of Topic Test
An online assessment is provided to help you review this topic.
© H ERIOT-WATT U NIVERSITY
41
Topic 3
Phase Equilibria
Contents
3.1 Partition coefficients . . . . . . . .
3.2 Solvent extraction . . . . . . . . .
3.2.1 Extraction calculations . . .
3.2.2 Decaffeinated coffee . . . .
3.3 Chromatography . . . . . . . . . .
3.3.1 Paper chromatography . . .
3.3.2 Gas-liquid chromatography
3.4 Summary . . . . . . . . . . . . . .
3.5 Resources . . . . . . . . . . . . .
3.6 End of Topic Test . . . . . . . . . .
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42
44
45
46
48
49
51
53
54
55
Prerequisite knowledge
Before you begin this Topic, you should be able to:
• define equilibrium constants and carry out calculations involving concentrations
and equilibrium constants (Topic 2.2);
• explain the solubility of substances in terms of their bonding and the type of solvent
involved (Higher Unit 1).
Learning Objectives
After studying this Topic, you should be able to:
• explain that a solute distributes itself between two immiscible liquids in a ratio
called the partition coefficient, dependent on the type of solute, type of solvent
and the temperature;
• explain that solvent extraction is an application of the partition of a solute between
two liquids which can be used to purify solutes;
• describe chromatographic techniques as the partition of solutes beween stationary
and mobile phases and explain how data gained by both paper and gas-liquid
chromatography can be used to separate and identify the individual components
of solute mixtures.
42
TOPIC 3. PHASE EQUILIBRIA
3.1
Partition coefficients
Learning Objective
To be able to explain that a solute distributes itself between two immiscible liquids in
a ratio called the partition coefficient, dependent on the type of solute, type of solvent
and the temperature.
Æ
Liquids which do not mix with each other are said to be immiscible. Dissolving a solute
in two immiscible liquids allows the solute to distribute itself between the two liquids.
After a time, equilibrium is established and an equilibrium constant can be written as
shown.
This ratio of the final concentrations of the solute in the two liquids is called the
partition coefficient. Figure 3.1 shows a solute X dissolved in two immiscible solvents:
solvent A (top layer) and solvent B (lower layer).
Figure 3.1: The partition coefficient
Q1:
In which solvent is solute X most soluble?
Q2: Solute X is covalent. Which is the most reasonable conclusion about the nature
of A.
a)
b)
c)
d)
Q3:
It is a polar solvent
It is a non-polar solvent
It is an acidic solvent
It is an ionic solvent
What could be said about solvent B compared to solvent A?
a) B is more polar than A
b) B is less polar than A
c) B is of equal polarity to A
Q4:
Which of these events occurs when equilibrium is reached?
a) exchange between the layers stops
b) movement of the solute stops
© H ERIOT-WATT U NIVERSITY
3.1. PARTITION COEFFICIENTS
43
c) the solvent becomes saturated
d) an equal exchange rate is reached
Q5: Calculate a partition coefficient for this equilibrium.
a)
b)
c)
d)
12
4
3
1
Factors affecting the partition coefficient.
Learning Objective
Æ
To show how the partition coefficient is dependent on the type of solute, the immiscible
liquids involved and the temperature of the equilibrium system.
An extended question analysing the results from a set of experimental observations
In each of experiments 1 to 4, 12 moles of a solute are used (each mole is represented
by a sphere). In experiment 5, 24 moles are used. Equal volumes of three different
immiscible solvents have been used and the experiments allowed to achieve equilibrium.
Figure 3.2
Q6: Compare experiments 1 and 2. Which variable has been altered?
Q7: Has the partition coefficient been affected?
Q8: Which experiment can be compared to 3 to show how different solvents might
affect the partition coefficient?
© H ERIOT-WATT U NIVERSITY
10 min
44
TOPIC 3. PHASE EQUILIBRIA
Q9:
Has the partition coefficient been affected?
See further questions on page 200.
The value of the partition coefficient is dependent on the type of solute, types of solvent
and the temperature of the equilibrium system.
3.2
Solvent extraction
Learning Objective
Æ
To show that solvent extraction is an application of the partition of a solute between
two immiscible liquids.
Solvent extraction is an application of the partition of a solute between two liquids. The
solvents can be used to ’selectively’ separate out the components of a mixture. Suppose
substance A has a partition coefficient of 3.0 in the solvent mixture shown (a ratio of 3:1)
(Figure 3.3). The impurities have a low solubility in the top layer (partition coefficient
approximately equal to zero).
Figure 3.3: Solvent extraction
A mixture of the substances is shaken with the two solvents and the layers allowed to
separate. Situation (a) results. The top layer is then removed and the lower is subjected
to a further extraction using a fresh portion of the top solvent resulting in situation (b).
Q10: How many of the spheres representing A are left?
Q11: How many of the spheres representing A would be left after a third extraction?
If all the top layers are now combined and the solvent evaporated, virtually pure A would
be isolated.
Separation of plutonium and uranium compounds in spent fuel products of the nuclear
industry is mostly done this way.
© H ERIOT-WATT U NIVERSITY
3.2. SOLVENT EXTRACTION
45
The highly radioactive spent fuel is chopped up and dissolved in hot nitric acid. The
first stage in the separation of the uranium (U) and plutonium (Pu) compounds from the
other fission products is achieved by a solvent extraction called a counter current solvent
extraction process. The process is called the PUREX process from (Pu-U-RecoveryEXtraction). More information can be found on the websites listed in the Resources
section at the end of this Topic.
3.2.1
Extraction calculations
Learning Objective
Æ
To be able to carry out simple calculations involving successive extractions of a solute,
given the partition coefficient of the system
Other applications of solvent extraction include the purification of water-soluble organic
acids using a suitable organic solvent such as ether. This will dissolve the acid and leave
the impurities in the water layer. Adding the ether in repeated small quantities is more
efficient than one large volume. The following example illustrates this.
Example : Ether extraction.
An organic acid X is present, along with impurities, in 100 cm 3 of water. 100 cm3 of ether
is available and the partition coefficent of the system is given by the equation shown in :
Calculate the percentage extraction if all the ether is used at once and compare this with
the result if two 50 cm3 portions are used.
Part A. Using all 100 cm3 of ether at once:
So four times as much organic acid is in the ether layer than is in the water layer. 80%
extracted
Part B. Using 50 cm3 ether and repeated extraction.
For the first 50 cm3 of ether:
© H ERIOT-WATT U NIVERSITY
46
TOPIC 3. PHASE EQUILIBRIA
!
For the second 50 cm 3 of ether a further 66.6% of the remaining solute (33.3%) would
be extracted:
In total, the two 50 cm3 portions give:
This is a larger extraction than with a single portion of solvent.
Q12: An aqueous solution of a diprotic organic acid was shaken with ether at 25 o C until
equilibrium was achieved. 25 cm 3 of the ether layer required 30.0 cm 3 of 0.05 mol -1
sodium hydroxide to neutralise and 25 cm 3 of the aqueous layer required 10 cm 3 of 0.05
mol -1 to neutralise. Calculate the partition coefficient (ether / water) at 25 o C.
Q13: Calculate the mass of an organic acid X which can be extracted from 200 cm 3 of
aqueous solution containing 5.0 g of X by shaking it with:
(i) 200 cm3 of a solvent in one portion.
(ii) two 100 cm3 portions of the solvent.
The partition coefficient between the solvent and water = 3.0
3.2.2
Decaffeinated coffee
Learning Objective
Æ
To show that solvent extraction of caffeine from coffee is an application of the partition
of a solute between two liquids
O
H3 C
N
N
O
N
CH3
N
CH3
Figure 3.4: Structure of caffeine
© H ERIOT-WATT U NIVERSITY
3.2. SOLVENT EXTRACTION
Many people wish to have the flavours, sugars and peptides that give coffee its taste,
without having the effects of the caffeine within it (these can include insomnia, irritability,
headaches). Decaffeination of coffee is done by solvent extraction.
Decaffeination of coffee is done by solvent extraction. Dichloromethane CH 2 Cl2 was
used as a solvent in the early processes but its toxic nature led to its replacement
in the nineties with a more environmentally friendly solvent called supercritical carbon
dioxide. Supercritical carbon dioxide exists in closed containers above 73 atmospheres
and 31o C. It behaves both like a gas and a liquid. Passing it through green coffee beans
allows the gas to penetrate the beans and the liquid properties allow it to dissolve out
about 98% of the caffeine. The caffeine is recycled into soft drinks and medicines and
the decaffeinated beans sold.
Figure 3.5: Some decaffeinated foodstuffs
© H ERIOT-WATT U NIVERSITY
47
48
TOPIC 3. PHASE EQUILIBRIA
Figure 3.6: Phase diagram of CO 2
Supercritical carbon dioxide is also used for solvent extraction of flavours in brewing and
extracting aromas and flavours from herbs and spices for use in perfumes.
Q14: Will supercritical CO2 be polar or non-polar?
Q15: Is caffeine therefore polar or non-polar?
Q16: Which should dissolve better in water, caffeinated or decaffeinated coffee? Why
not construct an experiment at home and see if your prediction is correct!
3.3
Chromatography
Learning Objective
Æ
To be able to describe chromatographic techniques as partition of solutes beween
stationary and mobile phases
In the first part of this Topic, you have considered processes that are dependent on the
equilibrium between two immiscible liquids. This section looks at methods of separation
that depend on the equilibrium between different phases. Methods that are widely
used to separate and identify the components of mixtures come under the heading of
chromatography.
There are several different types of chromatography, but the name derives from the early
methods used by the Russian, Michael Tswett, to separate mixtures of plant pigments
© H ERIOT-WATT U NIVERSITY
3.3. CHROMATOGRAPHY
49
into a pattern of coloured components. The Greek words chroma (colour) and graphein
(to write) were chosen to name the method.
All chromatographic methods involve a mobile phase moving over a stationary phase.
Separation occurs because the substances in the mixture have different partition
coefficients between the stationary and mobile phases.
Substances present in the initial mixture which partition more strongly into the stationary
phase will move more slowly than materials which partition more strongly into the mobile
phase.
A simulation of bees and hornets crossing a bed of flowers to mimic chromatography is
on the on-line version of this Topic.
3.3.1
Paper chromatography
Learning Objective
Æ
To be able to explain how data gained from paper chromatography can be used to
identify the components in solute mixtures
In this type of chromatography the mixture of components to be separated is placed as
a small spot close to the bottom of a rectangular piece of absorbent paper (like filter
paper). The bottom of the paper is placed in a shallow pool of solvent in a tank. An
example of the solvent would be an alcohol. Owing to capillary attraction, the solvent is
drawn up the paper, becoming the mobile phase. The solvent front is clearly visible as
the chromatography progresses.
When the paper is removed from the solvent, the various components in the initial spot
have moved different distances up the paper.
A simple simulation of this process, showing chromatography of blue and black inks, is
available on the on-line version of this Topic. The start and final positions are shown in
Figure 3.7 and Figure 3.8 respectively.
Figure 3.7: Start of chromatographic analysis
© H ERIOT-WATT U NIVERSITY
50
TOPIC 3. PHASE EQUILIBRIA
Figure 3.8: End of chromatographic analysis
This separation depends on the different partition coefficients of the various
components. The components are partitioned between the solvent and the water
trapped in the paper. Substances which partition mainly into the solvent mobile phase
will move further up the paper than substances which partition more strongly onto the
stationary phase.
Q17: By observing the chromatography simulation above, which ink (blue or black) has
the most components?
a) blue
b) black
c) both the same.
Q18: Which material in the black ink has stayed longest on the stationary phase, and
has the lowest solvent/water partition coefficient?
a)
b)
c)
d)
e)
red
yellow
dark blue
blue
green
Q19: The red coloured spot has moved furthest, this would indicate that the red
material:
a) has the highest solvent/water partition coefficient
b) has the lowest solvent/water partition coefficient
c) has the lowest molecular mass.
Q20: The movement of materials on paper chromatography is often described by an
Rf value which is the distance travelled by the spot divided by the distance travelled
by the solvent front. As long as the conditions of chromatography remain the same, a
compound will have a constant R f value.
Which colour in the black ink could have an R f value of 0.4?
a) red
© H ERIOT-WATT U NIVERSITY
3.3. CHROMATOGRAPHY
b)
c)
d)
e)
51
yellow
dark blue
blue
green
See further questions on page 201.
Although paper chromatography is still used today, it has been largely replaced by thin
layer chromatography (TLC). In this method, a support of glass or aluminium is coated,
usually with a thin layer of silica or cellulose. The processing is identical to that described
above, but TLC allows a more rapid separation (which prevents the spots spreading too
far) and makes detection of the spots easier. Most materials are not coloured, but can
still be chromatographed.
The invisible spots on paper or thin layer chromatography are revealed by use of a
locating reagent. These react with the compounds in the spots to produce a coloured
derivative. For example, ninhydrin solution can be sprayed onto chromatograms to
reveal amino-acids.
In another TLC detection system, the silica stationary phase is mixed with a fluorescent
dye, so that at the end of the process, viewing the plate under an ultraviolet lamp will
cause the background to glow (often an eerie green) except where there are spots,
which appear black.
3.3.2
Gas-liquid chromatography
Learning Objective
Æ
To be able to explain how data gained from gas-liquid chromatography can be used
to identify the components in solute mixtures
In gas-liquid chromatography (GLC) the stationary phase is a high boiling point liquid
held on an inert, finely-powdered support material, and the mobile phase is a gas (often
called the carrier gas).
The stationary phase is packed into a tubular column (Figure 3.9) usually of glass or
metal, with a length of 1 to 3 metres and internal diameter about 2 mm. One end of
the column is connected to a gas supply (often nitrogen or helium) via a device which
enables a small volume of liquid sample (containing the mixture to be analysed) to be
injected into the gas stream. The other end of the column is connected to a device which
can detect the presence of compounds in the gas stream.
The column is housed in an oven (Figure 3.9) to enable the temperature to be controlled
throughout the chromatographic analysis. One reason for this is that the materials to be
analysed must be gases during the analysis, so that gas-liquid chromatography is often
carried out at elevated temperatures.
© H ERIOT-WATT U NIVERSITY
52
TOPIC 3. PHASE EQUILIBRIA
Figure 3.9: GLC apparatus
A mixture of the material to be analysed is injected into the gas stream at zero time.
The individual components travel through the packed column at rates which depend on
their partition coefficients between the liquid stationary phase and the gaseous mobile
phase.
The detector is set to measure some change in the carrier gas that signals the presence
of material coming from the end of the column. Some detectors measure the thermal
conductivity of the gas, others burn the material from the column in a hydrogen-air flame
and measure the presence of ions in the flame.
The signal from the detector is recorded and plotted against time to give a series of
peaks each with an individual retention time.
A typical trace (chromatogram) for a complex mixture (premium grade petrol) is shown
in Figure 3.10.
Figure 3.10: Chromatogram of premium grade petrol
As long as the conditions remain constant (same column and stationary phase, same
pressure of gas, and same temperature) the retention time is the same for a given
compound. By chromatographing known materials (standards) the identity of many
peaks in a GLC trace can be established.
© H ERIOT-WATT U NIVERSITY
3.4. SUMMARY
53
The other important feature of many chromatograms is that the area under the peak is
directly proportional to the amount of material present.
Q21: The retention times, under the same conditions as Figure 3.10, for four
compounds are shown in Table 3.1:
Table 3.1:
butane
pentane
benzene
toluene
"
"
"
"
(3.1)
Which of these hydrocarbons is present in the greatest amount in petrol?
a)
b)
c)
d)
butane
pentane
benzene
toluene
Q22: How are retention time (r.t.)
hydrocarbons?
and molecular mass (m) related for these four
a) r.t. increases as m increases
b) r.t. decreases as m increases
c) they are not related
Q23: Using the answer to the previous question, suggest a possible hydrocarbon for
the peak with retention time 20.8 min.
a)
b)
c)
d)
hexane (C6 H14 )
heptane(C 7H16 )
xylene(C8 H10 )
propane(C 3H8 )
3.4
Summary
• A solute mixed with two immiscible liquids will distribute itself in a ratio called the
partition coefficient, dependent upon the temperature, the solvents used and the
type of solute.
• The partition coefficient for a solute X in solvent A (top layer) and solvent B (bottom
layer) is given by:
• Solvent extraction processes are used extensively in industry to separate out
components in a mixture, examples being found in nuclear industry, chemical
industry and food processing.
© H ERIOT-WATT U NIVERSITY
54
TOPIC 3. PHASE EQUILIBRIA
• Chromatography is an analytical technique used to separate and identify the
components in a mixture.
• Chromatographic separations depend on the partition equilibrium between two
phases, one stationary and the other mobile.
• There are several types of chromatography, including paper chromatography and
gas-liquid chromatography.
• In paper chromatography, the stationary phase is the water held on the paper and
the mobile phase is another solvent.
• In gas-liquid chromatography, the stationary phase is a liquid held on a solid
support and the mobile phase is a gas.
3.5
Resources
• Chemistry in Context: Hill and Holman , Nelson ISBN 0-17-438401-7
• Chemistry: E.N.Ramsden, Stanley Thornes, ISBN 0-85950-154-X
• General Chemistry: Ebbing, Houghton Mifflin, ISBN 0-395-74415-6
• Higher Still Support: Chemistry Unit 2: Principles of Chemical Reactions.
(Advanced Higher), Learning and Teaching Scotland, ISBN 1-85955-874-7
• Chemical Storylines: Salters Advanced Chemistry, Heinemann ISBN 0-43563106-3
• Chemical Ideas: Salters Advanced Chemistry, Heinemann, ISBN 0-435-63105-5
• Websites for Purex process:
http://www.uic.com.au/uicchem.htm
http://www-nuen.tamu.edu/COURSES/NU415/reprocessing/purex.htm
http://www.chee.iit.edu/~m3/Introduction.html
• Websites for coffee decaffeination process:
http://www.exicom.org/cew/oct97/awasthi.htm
http://wwwchem.uwimona.edu.jm:1104/lectures/coffee.html
http://www.sciam.com/0697issue/0697working.html
• CD ROM issued by the Royal Society of Chemistry: Practical Chemistry for
Schools and Colleges.
• Video produced by Royal Society of Chemistry (with Glaxo Welcome): Modern
Chemical Techniques
• Booklet produced by Royal Society of Chemistry (with Unilever): Modern Chemical
Techniques
© H ERIOT-WATT U NIVERSITY
3.6. END OF TOPIC TEST
3.6
End of Topic Test
An online assessment is provided to help you review this topic.
© H ERIOT-WATT U NIVERSITY
55
56
TOPIC 3. PHASE EQUILIBRIA
© H ERIOT-WATT U NIVERSITY
57
Topic 4
Acid/base Equilibria
Contents
4.1 What are acids and bases? . . . . . . . . .
4.1.1 The nature of the hydrogen ion . . .
4.2 The ionic product of water and the pH scale
4.2.1 pH scale . . . . . . . . . . . . . . . .
4.3 Dissociation of acids . . . . . . . . . . . . .
4.4 Dissociation of bases . . . . . . . . . . . .
4.5 Summary . . . . . . . . . . . . . . . . . . .
4.6 Resources . . . . . . . . . . . . . . . . . .
4.7 End of Topic test . . . . . . . . . . . . . . .
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58
61
62
65
68
71
73
74
75
Prerequisite knowledge
Before you begin this Topic, you should be able to:
• distinguish between strong and weak acids (and bases) (Higher, Unit 3);
• describe the simple relationship between pH and [H + ] (Higher, Unit 3);
• explain why some salts do not have a pH of 7 (Higher, Unit 3);
• define the equilibrium constant and list the factors that affect the position of
equilibrium (Le Chatelier’s principle) (Topic 2.2);
• explain what is meant by an amphoteric oxide (Topic 1.5).
Learning Objectives
After studying this Topic, you should be able to:
• define the terms - acid, base, conjugate acid, conjugate base- in terms of transfer
of protons;
• explain what is meant by K w , the ionic product of water;
• define the dissociation constant, Ka , for acids and conjugate acids and relate
values for Ka to the strengths of acids and bases;
• calculate the pH of solutions of strong acids and bases from [H + ] and the pH of
solutions of weak acids from Ka values.
58
TOPIC 4. ACID/BASE EQUILIBRIA
4.1
What are acids and bases?
Learning Objective
Æ
To define the terms - acid, base, conjugate acid, conjugate base- in terms of transfer
of protons
The term ’acid’ is familar to us from an early age.
• Acids rot teeth
• Acids dissolve cartoon characters
• Acids eat through metals
There is a popular misconception that all acids are dangerous.
The term ’base’ is much less familiar.
In fact, acids and bases are reacting all around us and within us all the time. Some
examples of common acids and bases are shown in Figure 4.1. About 1.0-1.5 litres per
day of hydrochloric acid is secreted in the human stomach as a digestive juice.
Common Acids
Common Bases
Figure 4.1: Acids and bases in household goods
Acids have been described as substances that dissolve in water to form H + (aq) ions,
whilst bases are substances that react with acids.
However, a better, broader definition was produced independently by Bronsted and
Lowry.
An acid is a substance that is able to donate hydrogen ions (protons) to another
substance, i.e. it is a proton donor.
A base is a substance that accepts hydrogen ions (protons), i.e. it is a proton acceptor.
These definitions can be widely applied, even to reactions which do not occur in aqueous
solution.
© H ERIOT-WATT U NIVERSITY
4.1. WHAT ARE ACIDS AND BASES?
A general acid, formula HA, will dissociate according to Figure 4.2:
Figure 4.2: Acid equilibrium
In the forward reaction, HA loses a proton and so behaves as an acid. In the
reverse reaction, A- accepts a proton, i.e. behaves as a base. A - is said to be the
conjugate base of the acid HA. For every acid, there is a conjugate base formed by
loss of a proton (H + ion).
A base of formula B will accept a proton according to Figure 4.3:
Figure 4.3: Base equilibrium
By similar reasoning, for every base, there will be a conjugate acid formed by gain of
a proton. BH+ is the conjugate acid of the base, B.
Figure 4.4 shows an acid / base reaction. Click on the arrows to navigate through the
following animation.
Figure 4.4: Acid-base equilibrium
Overall, the reaction involves transfer of a proton from one species to another. This
simple transfer is literally of vital importance. Gain or loss of a proton can cause a
change in the structure and function of amino acids, proteins, enzymes and nucleic
acids. So, many biological pathways are controlled by the presence or absence of H +
ions.
© H ERIOT-WATT U NIVERSITY
59
60
TOPIC 4. ACID/BASE EQUILIBRIA
Figure 4.5: Formation of ammonium chloride
Figure 4.6
Q1:
In the reaction shown in Figure 4.6, name the base.
Q2:
In the reaction shown in Figure 4.6, name the acid.
Q3:
In the reaction shown in Figure 4.6, name the conjugate acid.
Q4:
In the reaction shown in Figure 4.6, name the conjugate base.
See further questions on page 201.
Conjugate pairs
10 min
A test to provide practice at identifying acids, bases and their conjugates is available in
the on-line version of this Topic.
An online assessment is provided to help you review this topic.
A water molecule is capable of accepting protons from an acid or donating protons to a
base. Water is therefore amphoteric and perfectly suited to enable the transfer of H +
ions (protons) from one substance to another.
© H ERIOT-WATT U NIVERSITY
4.1. WHAT ARE ACIDS AND BASES?
4.1.1
61
The nature of the hydrogen ion
Learning Objective
Æ
To demonstrate that the hydrogen ion can be represented by H + or H3 O+ depending
on the circumstances.
Li +
are hydrated, as shown in the diagram
In aqueous solution, positive ions like
+
(Figure 4.7). Li is the smallest singly-charged positive metal ion.
Figure 4.7
Q5:
Using the SQA data booklet, what is the ionic radius of the Li + ion in picometres?
A hydrogen ion, H + , is simply a proton that has an estimated radius of 7.7 x 10 -18 m (i.e.
7.7 x 10 -6 pm).
If the proton was scaled up to the size of the ball in a ballpoint pen, the lithium ion on
the same scale would have a diameter of about 5 km, a distance equivalent to about 50
football pitches laid end to end.
A proton is too tiny and highly charged to exist on its own in aqueous solution. Instead,
it bonds to a water molecule using one of the lone pairs of electrons on the oxygen atom
to form a hydronium ion, H 3 O+ (Figure 4.8). This in turn will be bonded to other water
molecules as shown in Figure 4.8.
O
O
H
H
H
H
H
O
H
H
© H ERIOT-WATT U NIVERSITY
O
H
O
H
Figure 4.8
H
H
H
62
TOPIC 4. ACID/BASE EQUILIBRIA
In equations and equilibrium expressions, H + (aq) is often used as a simple shorthand
instead of H3 O+ (aq). H3 O+ (aq) is used when it is necessary to show the behaviour of
water as an acid or a base.
4.2
The ionic product of water and the pH scale
Learning Objective
Æ
To be able to explain the term K w , the ionic product of water
The conductivity of water can be measured using the apparatus shown in Figure 4.9.
The current passed through the liquid is measured. The results for salt solution, tap
water, pure water and hexane are shown in Table 4.1.
Figure 4.9: Conductivity measurements
© H ERIOT-WATT U NIVERSITY
4.2. THE IONIC PRODUCT OF WATER AND THE PH SCALE
Table 4.1: Conductivity data
Substance
Current Reading
0.1 mol l-1 NaCl
Tap water
125 mA
5 mA
20 #A
0 #A
Pure water
Hexane
Q6: Explain why salt solution is a conductor of electricity.
Q7: Explain why hexane is a non-conductor of electricity.
Q8: Pure water is a poor conductor of electricity but not a non-conductor. What
conclusion can be drawn from this?
The dissociation of water can be represented by the equation shown in Figure 4.10:
Figure 4.10: Formation of the hydronium ion
Q9: In the equation shown (Figure 4.10), the water molecule on the left acts as
a) an acid
b) a base
Q10: The hydroxide ion is
a)
b)
c)
d)
the conjugate acid of H 2 O
the conjugate base of H 2 O
the conjugate acid of H3 O+
the conjugate base of H 3 O+
Q11: The reaction (Figure 4.10) is reversible and equilibrium is reached. From the
experimental evidence (Table 4.1), which of these statements is the best description
of the equilibrium position?
a)
b)
c)
d)
The equilibrium lies to the left.
The equilibrium lies well to the left.
The equilibrium lies to the right.
The equilibrium lies well to the right.
Q12: Write an expression for the equilibrium constant for this reaction ( Figure 4.10).
© H ERIOT-WATT U NIVERSITY
63
64
TOPIC 4. ACID/BASE EQUILIBRIA
The concentration of water is effectively constant and can be incorporated into the
equilibrium constant to give a new constant, K w .
or " $ $" " $" Kw is known as the ionic product of water. Like all equilibrium constants, it is
dimensionless but you will see it given incorrectly with units of mol2 -2 in some books.
Figure 4.11
The graph in Figure 4.11 shows how Kw varies with temperature.
Q13: What value does K w have at 25Æ C (298 K)?
a)
b)
c)
d)
1.00
0.68 x 10-14
1.00 x 10-14
1.00 x 1014
Q14: The graph shows that as the temperature increases, K w increase. What does this
suggest about the sign of H Æ for the reaction?
a) It is positive.
b) It is zero.
c) It is negative.
Q15: Explain the answer to the previous question. Write an explanation on paper before
revealing the answer.
© H ERIOT-WATT U NIVERSITY
4.2. THE IONIC PRODUCT OF WATER AND THE PH SCALE
65
Learning point
Kw is the ionic product of water. It is temperature dependent and is represented by K w
= H+ OH- 4.2.1
pH scale
Learning Objective
Æ
To be able to use K w to calculate pH values for strong acids and bases.
In pure water, the concentrations of H + (aq) and OH- (aq) ions will be equal.
The relationship between [H+ ] and [OH- ]
Visit the on-line version of this Topic where you will find a simulation. The simulation uses
a bar graph, shown in Figure 4.12, which can be manipulated to show the connection
between [H+ ] and [OH- ]
10-1
Concentration/mol l-1
10-3
[H+]
[OH-]
10-5
10-7
10-9
10-11
10-13
pH1
pH7
pH13
Figure 4.12: The connection between [H + ] and [OH- ]
Q16: When [H+ ] is 10-7 , what is the pH?
Q17: When [H+ ] is 10-1 , what is the pH?
Q18: When [H+ ] is 10-13 , what is the pH?
Q19: What is the connection between the concentration of H + ions and the pH value?
See further questions on page 201.
Check the value of the expression [H + ] [OH- ] at various pH values.
Kw has a value of 1.00 x 10 -14 at 25Æ C.
© H ERIOT-WATT U NIVERSITY
10 min
66
TOPIC 4. ACID/BASE EQUILIBRIA
If the concentration of H+ ions is known, then the concentration of OH - ions can be
calculated using Kw , and vice versa. A single scale to measure acidity and alkalinity can
be based on the concentration of H + ions. The scale has values ranging from around 0
to 14. The relationship connecting pH and [H + ] is given by:
pH = - log10 H + (4.1)
If an acid is strong (completely dissociated) and monoprotic (e.g. hydrochloric acid,
HCl), the molar concentration of the acid will equal [H + ] and so the pH can be calculated.
Similarly, for a strong alkali, the molar concentration can be used to calculate [OH - ].
Using Kw , the [H+ ] can be calculated and hence the pH.
Calculating pH
30 min
Some worked examples and questions to give practice at calculating pH values for
strong acids and alkalis and at calculating [H + ] and [OH- ] from pH values.
Read through the worked examples carefully. The questions that follow them are
randomly generated so that you can try them as many times as you wish until you are
confident.
Examples
1. pH from [H+ ]
Calculate the pH of a 0.2 mol
-1
solution of hydrochloric acid.
" %"
It is always useful to check your answer by estimating values.
0.02 lies between 0.01 and 0.10, i.e. between 10 -2 and 10-1 .
So the pH must lie between 2 and 1.
2. pH from [OH- ]
Calculate the pH of a solution of 0.006 mol
-1
sodium hydroxide.
© H ERIOT-WATT U NIVERSITY
4.2. THE IONIC PRODUCT OF WATER AND THE PH SCALE
$" " %"
67
"
$" $" Check [H+ ] lies between 10-11 and 10-12 , i.e pH lies between 11 and 12.
Q20: Calculate the pH of a solution that has a H + (aq) concentration of 5 x 10-3 mol
-1 .
Q21: Calculate the pH of a solution that has a H + (aq) concentration of 8 x 10-6 mol
-1 .
Q22: Calculate the pH of a solution that has a OH - (aq) concentration of 6.3 x 10 -2 mol
-1 .
Q23: Calculate the pH of a solution that has a OH - (aq) concentration of 2.9 x 10 -4 mol
-1 .
Using the same relationship (Equation 4.1), the concentration of H+ ions and OH- ions
can be calculated from the pH of the solution.
Example : Concentrations from pH
Calculate the concentration of H + ions and OH- ions in a solution of pH 3.6. Give your
answer to three significant figures.
Step 1
" "
" "
antilog %"
Note: calculators vary slightly in the way in which they antilog numbers. One way is as
follows:
then "
If log " © H ERIOT-WATT U NIVERSITY
68
TOPIC 4. ACID/BASE EQUILIBRIA
Press the 10x key, then type the value (-3.6), followed by ’=’.
If this does not work with your calculator, see your tutor.
Step 2
$" " $" " x x Q24: Calculate the concentration of H + (aq) ions and OH- (aq) ions in a solution of pH
2.3
Q25: Calculate the concentration of H + (aq) ions and OH- (aq) ions in a solution of pH
5.6
Q26: Calculate the concentration of H + (aq) ions and OH- (aq) ions in a solution of pH
11.4
Q27: Calculate the concentration of H + (aq) ions and OH- (aq) ions in a solution of pH
1.9
The relationship between pH and the hydrogen ion concentration is given by
pH = - log10 H + This relationship can be used to calculate the pH for strong acids and alkalis given the
molar concentrations of either H+ or OH- ions.
4.3
Dissociation of acids
Learning Objective
Æ
To be able to define the dissociation constant, K a , for acids. To relate values for Ka to
the strengths of acids. To calculate pH values for weak acids
Water molecules are amphoteric. They are capable of accepting protons from acids
and donating protons to bases.
In the presence of a more powerful acid, HA, water molecules accept protons and the
acid dissociates according to the equation shown in Figure 4.13:
HA(aq) +
H2O(l)
H3O+(aq)
+
A-(aq)
Figure 4.13: Protonation of water
© H ERIOT-WATT U NIVERSITY
4.3. DISSOCIATION OF ACIDS
69
The reaction is reversible and equilibrium is established.
Q28: Write an expression for the equilibrium constant for this reaction as written in the
equation shown in Figure 4.13.
In dilute solutions, the concentration of H 2 O is constant and can be incorporated into
the equilibrium constant to give a new constant:
"$ "
Ka is known as the acid dissociation constant.
If the acid is strong, the equilibrium lies so far to the right that effectively no molecules
of HA remain. Ka will be very large. The acid is 100% dissociated into ions. The reverse
reaction does not take place because the conjugate base is so weak that it cannot
remove protons from H3 O+ ions.
Table 4.2 shows some acids with their Ka values. H+ (aq) has been used instead of H 3 O+
for simplicity.
Table 4.2:
HF(aq)
HIO3 (aq)
H3 PO4 (aq)
H2 SO3 (aq)
«
«
«
«
Ka
H+ (aq)
F- (aq)
+
+ IO3 - (aq)
H+ (aq) + H2 PO3 - (aq)
H+ (aq)
H+ (aq) + HSO3 - (aq)
3.4 x 10-4
1.7 x 10-1
7.9 x 10-3
1.5 x 10-2
Q29: Which is the strongest acid?
a)
b)
c)
d)
HF
HIO3
H3 PO4
H2 SO3
Q30: Which is the weakest acid?
a)
b)
c)
d)
HF
HIO3
H3 PO4
H2 SO3
Learning point
In general, as the strength of the acid increases, the strength of the conjugate base
decreases.
Q31: What is the weakest conjugate base?
a) Fb) IO3 © H ERIOT-WATT U NIVERSITY
70
TOPIC 4. ACID/BASE EQUILIBRIA
c) H2 PO4 d) HSO3 Q32: What is the strongest conjugate base?
a)
b)
c)
d)
FIO3 H2 PO4 HSO3 -
Q33: Write a general statement connecting the value of K a with the strength of an acid.
The data booklet (page 12) shows dissociation constants for various acidic species.
Note that the dissociation constant can also be expressed as pK a where:
pKa = -log10 Ka (compare this with the definition of pH - Equation 4.1)
Both Ka and pKa can be used as a measure of the strength of an acid.
Q34: Which of the following statements would describe a strong acid?
a)
b)
c)
d)
high Ka , high pKa
low Ka , low pKa
high Ka , low pKa
low Ka , low pKa
Q35: On page 12 of the data booklet, which is the strongest acid?
Q36: Which of the following statements would describe a weak acid?
a)
b)
c)
d)
high Ka , high pKa
low Ka , low pKa
high Ka , low pKa
low Ka , high pKa
Q37: On page 12 of the data booklet, which is the weakest acid?
The pH of a weak acid cannot be calculated directly from the molar concentration since
only a small proportion of the molecules are dissociated into ions. However, the pH can
be calculated, provided the K a value for the acid is known.
The required equation is:
%"
%
(4.2)
You must be able to use Equation 4.2 but you do not need to know how to derive it.
However, if you are interested in seeing the derivation, it is available as an example in
the on-line version of this Topic.
The following questions will give you practice in using Equation 4.2. The first two
questions have worked answers that can be displayed. Try them for yourself first. Then
do as many of the others as necessary until you feel comfortable. Page 12 of the SQA
data booklet will be useful. Give all your answers to 2 significant figures.
© H ERIOT-WATT U NIVERSITY
4.4. DISSOCIATION OF BASES
71
Q38: What is the pH of a solution of ethanoic acid of concentration 0.01 mol
Q39: A 0.1 mol
the acid?
-1
-1 ?
solution of a weak acid has a pH of 3.2. What is the pK a value for
Q40: A solution of butanoic acid has a pH of 3.2. What is the molar concentration?
Q41: A solution of hydrofluoric acid has a pH of 2.5. What is the molar concentration?
Q42: What is the pH of a solution of methanoic acid of molar concentration
0.1 mol -1 ?
Q43: What is the pH of a solution of propanoic acid of molar concentration
0.02 mol -1 ?
See further questions on page 201.
4.4
Dissociation of bases
Learning Objective
Æ
To define the dissociation constant, K a , for conjugate acids of bases and relate values
for Ka to the strengths of bases
In the presence of a more powerful base, B, water molecules behave as an acid by
donating protons to the base.
Bases
Two drag and drop exercises are available in the on-line version of this Topic to reinforce
the relationships between acids, bases and their conjugates.
Figure 4.14: Protonation of bases
For a strong base, the equilibrium position lies far to the right. For a weak base, the
equilibrium lies far to the left.
Complete the table Table 4.3 using words from the Word Bank which appears beneath
the table.
Table 4.3:
Strong acid
Strong conjugate acid
100% dissociated
© H ERIOT-WATT U NIVERSITY
10 min
72
TOPIC 4. ACID/BASE EQUILIBRIA
Word Bank
strong base
weak conjugate
base
weak base
weak conjugate acid
weak acid
strong conjugate
base
poorly dissociated 100% dissociated poorly dissociated
The dissociation in aqueous solution of a base of general formula B can be represented
by
The stronger the base B, the weaker is the conjugate acid BH + .
The dissociation of the conjugate acid of base B can be represented by the equation
shown in Figure 4.15:
Figure 4.15: Dissociation of a conjugate base
The conjugate acid dissociation constant, K a , will therefore be as shown in Equation 4.3:
"$
" (4.3)
Note that again the [H 2 O] is assumed to be constant and so is incorporated into the
equilibrium constant.
Ka (and pKa ) values can be used as a measure of the strength of the conjugate acid and
also therefore as a measure of the strength of the base, since the stronger the base the
weaker the conjugate acid.
Ammonia is a familiar example of a base. Use the data booklet (page 12) to answer the
following questions.
Q44: Give the name of the conjugate acid of ammonia.
Q45: What is the pKa value for this conjugate acid?
Q46: Write the expression for the dissociation constant K a for this conjugate acid.
Q47: Is ammonia a weak base or a strong base?
a) weak
b) strong
© H ERIOT-WATT U NIVERSITY
4.5. SUMMARY
73
Many weak bases are organic compounds called amines which are derived from
ammonia by replacing one or more of the hydrogen atoms by other groups. Table 4.4,
shows some examples. Piperidine is a base naturally found in black pepper and codeine
is a naturally occurring amine used as a painkiller.
Table 4.4: Weak bases
Conjugate acid
Base
Formula
Ka
pKa
NH 3
5.6 x 10-10
9.3
CH3 NH2
2.7 x 10-11
10.6
dimethylamine
(CH3 )2 NH
1.9 x 10-11
10.7
piperidine
C5 H10 NH
7.7 x 10-12
11.1
C18 H21 NO3
6.3 x 10-9
?
ammonia
methylamine
codeine
Q48: Calculate pK a for the conjugate acid of codeine. Give your answer to one decimal
place.
Q49: Name the strongest base in Table 4.4.
Q50: Name the weakest base in Table 4.4.
Q51: What effect does replacing the hydrogen atoms in ammonia with methyl groups
have on the strength of the base?
4.5
Summary
• According to the Bronsted-Lowry defintions, an acid is a proton donor (a giver of
H+ ions) and a base is a proton acceptor (a taker of H + ions).
• For every acid there is a conjugate base (formed by loss of a proton) and for every
base there is a conjugate acid (formed by gain of a proton).
• Water molecules can act either as an acid or a base, making water amphoteric.
The ionisation of water can be represented by:
H2 O(l) + H2 O(l)
« H O (aq) + OH (aq)
3
+
-
• The dissociation constant for this ionisation is Kw , the ionic product of water, which
is represented by:
Kw = [H3 O+ ][OH- ] or, more simply, [H+ ][OH- ].
• Kw is temperature dependent and has a value of approximately 1.0 x 10 -14 at 25Æ C.
• The pH scale is based on the [H + ] which are related by pH = -log [H + ]. This
relationship can be used to calculate the concentration of H + ions and OH- ions
since changes in the concentration of one directly change the concentration of the
other to maintain K w .
© H ERIOT-WATT U NIVERSITY
74
TOPIC 4. ACID/BASE EQUILIBRIA
• Acids of general formula HA dissociate in aqueous solution according to the
equation:
HA(aq) + H2 O(l)
« H O (aq) + A (aq)
3
+
-
• The acid dissociation constant, Ka , is given by:
"$ "
•
This can also be represented by pK a where pKa = -log Ka
Both Ka and pKa provide a measure of the strength of an acid. When K a is low (or
pKa is high), the acid is weak. When Ka is high (or pKa is low), the acid is strong.
•
For weak monoprotic acids, pH and pK a are related by:
%"
%
where c is the molar concentration of the weak acid. This relationship allows [H + ]
and [OH- ] to be calculated for weak acids.
• Similarly, bases of general formula B dissociate in aqueous solution according to
the equation:
B(aq) + H2 O(l)
« BH (aq) + OH (aq)
+
-
• The dissociation of the conjugate acid of the base B can be represented by:
BH+ (aq) + H2 O(l)
« B(aq) + H O (aq)
3
+
Values for Ka (and pKa ) for the dissociation of the conjugate acid, BH + , can be
used as a measure of the strength of the base B, since the weaker the conjugate
acid the stronger the base.
4.6
Resources
• Chemistry: McMurry and Fay, Prentice Hall, ISBN 0-13-737776-2
• General Chemistry: Ebbing, Houghton Mifflin, ISBN 0-395-74415-6
• Chemistry in Context: Hill and Holman, Nelson, ISBN 0-17-438401-7
• Higher Still Support: Advanced Higher Chemistry - Unit 2:
Principles of Chemical Reactions, Learning + Teaching Scotland,
ISBN 1 85955 874 7.
• A-level Chemistry: E.N. Ramsden, Stanley Thorne Publishers,
ISBN 0-85950-154-X
© H ERIOT-WATT U NIVERSITY
4.7. END OF TOPIC TEST
Web sites
• http://chemistry.semo.edu/crawford/ch186/lectures/ch15/index.html
• http://www.science.ubc.ca/~chem/tutorials/pH/index.html
4.7
End of Topic test
An online assessment is provided to help you review this topic.
© H ERIOT-WATT U NIVERSITY
75
76
TOPIC 4. ACID/BASE EQUILIBRIA
© H ERIOT-WATT U NIVERSITY
77
Topic 5
Indicators and buffers
Contents
5.1 Indicators . . . . . . . . . . . . . . . .
5.1.1 pH titrations . . . . . . . . . . .
5.2 Buffer solutions. . . . . . . . . . . . .
5.2.1 Acid Buffers . . . . . . . . . . .
5.2.2 Basic buffers. . . . . . . . . . .
5.3 Calculating pH and buffer composition.
5.4 Summary . . . . . . . . . . . . . . . .
5.5 Resources . . . . . . . . . . . . . . .
5.6 End of Topic test . . . . . . . . . . . .
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78
81
85
86
89
90
93
94
94
Prerequisite knowledge
Before you begin this Topic, you should be able to:
• write an expression for the dissociation constant, Ka of a weak acid. ( Topic 2.4 );
• write an expression for the relationship between pH, pK a and molar concentration
( Topic 2.4 );
• use this relationship to calculate the pH of a solution of a weak acid given the
concentration.
Learning Objectives
After studying this Topic, you should be able to:
• explain why the colour of an acid/base indicator is dependent on the pH of the
solution;
• calculate the pH range over which the indicator changes colour from the acid
dissociation constant, KIn , of the indicator;
• choose suitable indicators for different acid/base titrations;
• describe the composition of both acid and basic buffer solutions and explain how
they function;
• calculate the pH of an acid buffer when given the acid dissociation constant and
the composition, and calculate composition when given the other variables.
78
TOPIC 5. INDICATORS AND BUFFERS
5.1
Indicators
Acid/base indicators (or simply indicators) are weak acids which change colour
depending on the pH of the solution.
HIn can be used as a general formula for an indicator and its dissociation can be
represented by this equation:
Figure 5.1: Indicator equilibrium equation
For a good indicator, the undissociated acid, HIn, will have a distinctly different colour
from its conjugate base, In- .
For the indicator litmus, HIn is red and In - is blue. Use this information and Figure 5.1
to answer these questions.
Q1:
In which direction will the equilibrium position move if the[H + ] is increased?
a) left
b) right
Q2:
What colour will the solution be?
Q3: If the [H+ ] is decreased in Figure 5.1 , in which direction will the equilibrium
position move?
a) left
b) right
Q4:
What colour will the solution be now?
See further questions on page 201.
The acid dissociation constant for an indicator HIn is given the symbol K In and is
represented by:
" &
"&
Taking the negative log of both sides gives:
© H ERIOT-WATT U NIVERSITY
5.1. INDICATORS
79
& " "&
%
! "
%"
&
%
%" "&
%"
%
& "&
(5.1)
Equation 5.1 shows that the pH of the solution is determined by the pK In of the indicator
and the ratio of [In- ] to [HIn]. Since these are different colours, the ratio of [In- ] to [HIn]
determines the overall colour of the solution. For a given indicator, the overall colour of
the solution is dependent on the pH of the solution.
Bromothymol blue has a pK In value of 7.0.
Q5:
What value will KIn have? Give your answer in decimal form.
Equation 5.1 can be rearranged to enable the ratio of [In - ] to [HIn] to be calculated. For
bromothymol blue, the HIn form is yellow and the In - form is blue.
Example
Calculate the ratio of [In- ] to [HIn] for bromothymol blue at pH 3.0
%"
& "&
&
"&
%
& "&
%
%"
In other words, for one In- ion, there are 10,000 HIn molecules. The colour of the
solution will be that of the HIn molecules, i.e. it will be yellow.
Q6: Calculate the ratio
&
"&
for bromothymol blue at pH 9.
Q7: What colour will the solution be?
Q8: Calculate the ratio
© H ERIOT-WATT U NIVERSITY
&
"&
80
TOPIC 5. INDICATORS AND BUFFERS
for bromothymol blue at pH 7.
Q9:
What colour will the solution be?
Table 5.1 shows the results of a similar calculation at a number of different pH values.
Table 5.1: Bromothymol blue at various pH values
pH
relative [In- ]
relative [HIn]
colour
2
1
100000
yellow
4
1
1000
yellow
6
1
10
7
8
10
12
1
10
1000
100000
1
1
1
1
yellow
green
blue
blue
blue
This information can be summarised in Figure 5.2
Figure 5.2: Bromothymol blue indicator
It can be seen on the Topic website that at most values of pH the indicator exists in one
of the extreme colours (yellow or blue).
When pH = pK In , the concentrations of In- and HIn are equal and the colour is
intermediate.
Figure 5.3 shows a similar diagram for methyl orange ( pK In = 3.7).
Figure 5.3: Methyl orange indicator
Figure 5.4 shows the diagram for phenolphthalein which is colourless in acid and pink
in alkali.
© H ERIOT-WATT U NIVERSITY
5.1. INDICATORS
81
Figure 5.4: Phenolphthalein indicator
In each of the indicators above, the colour change from the original colour is assumed
to be distinguishable when the [In - ] and [HIn] differ by a factor of about 10.
A change of [H+ ] by a factor of 10 corresponds to a change in pH of 1 unit.
Consequently, the colour of the indicator changes over a pH range given by:
pH = pKIn 1
This is generally true although the range may vary slightly.
The theoretical point at which the colour changes is when:
pH = pKIn (or [H+ ] = KIn ).
5.1.1
pH titrations
When an acid is gradually neutralised by a base, the change in pH can be monitored
using a pH meter. The results can be used to produce a pH titration curve from which
the equivalence point can be identified.
pH titration
A simulation of the production of a pH titration curve with questions to test
understanding.
In the simulation below, 50cm 3 hydrochloric acid of concentration 0.1 mol
neutralised by sodium hydroxide solution.
Use the graph shown as Figure 5.5, to answer the questions.
© H ERIOT-WATT U NIVERSITY
-1
is being
25 min
82
TOPIC 5. INDICATORS AND BUFFERS
Figure 5.5: Titration curve
Q10: What is the pH at the equivalence point?
Q11: What is meant by the equivalence point?
Q12: Use the information in the graph to calculate the concentration (in mol
sodium hydroxide solution.
-1 )
of the
Q13: Between 49.9 cm 3 and 50.0 cm3 , only 0.1 cm3 of alkali was added. What was the
change in pH for this addition?
See further questions on page 202.
During a titration involving a strong acid and base, there is a very rapid change in pH
around the equivalence point.
Table 5.2 shows pH titration curves produced by the different combinations of strong and
weak acids and alkalis. Look closely at the equivalence points in each graph.
© H ERIOT-WATT U NIVERSITY
5.1. INDICATORS
83
Table 5.2: Titration curves
Q14: Which combination has an equivalence point at pH 7?
a)
b)
c)
d)
strong acid/strong alkali
strong acid/weak alkali
weak acid/strong alkali
weak acid/weak alkali
Q15: What is the pH of the salt formed from a strong acid and a strong alkali?
a)
b)
c)
d)
1
5
7
9
Q16: Which of these is the most likely pH of the salt formed from a strong acid and a
weak alkali?
a) 1
b) 5
c) 7
© H ERIOT-WATT U NIVERSITY
84
TOPIC 5. INDICATORS AND BUFFERS
d) 9
Q17: Which of these is the most likely pH of the salt formed from a weak acid and a
strong alkali?
a)
b)
c)
d)
1
5
7
9
See further questions on page 202.
Since the equivalence points occur at different pH for different combinations, different
indicators will be required in each case.
Choosing indicators
Learning Objective
30 min
Æ
To be able to choose an appropriate indicator for a titration experiment.
In the simulation, the pH range for each indicator can be superimposed on the pH
titration curve to enable the best indicator to be chosen for each combination of acid
and alkali.
Figure 5.6 shows pH titration curves onto which information about some indicators has
been added.
Figure 5.6: Two pH titration curves
Table 5.3: pH values of various indicators
Area
indicator
pH range
1
phenolphthalein
8.2-10.0
2
bromothymol blue
6.0-7.6
3
methyl orange
3.0-4.4
© H ERIOT-WATT U NIVERSITY
5.2. BUFFER SOLUTIONS.
85
Q18: Which indicator is most suitable for the titration of a weak acid with a strong alkali?
Q19: Which indicator is most suitable for the titration of a strong acid with a weak alkali?
A suitable indicator for a titration will be one whose colour changes when the pH of
the solution changes most rapidly. The best indicator will be one whose colour change
occurs around the equivalence point. No indicators are suitable for a weak acid/weak
alkali titration since the pH changes only gradually around the equivalence point.
5.2
Buffer solutions.
Learning Objective
To be able to describe the composition and functioning of both acid and basic buffer
solutions. To calculate the pH of an acid buffer when given the acid dissociation
constant and the composition. To calculate composition when given the other
variables
Æ
Small changes in pH can have a surprisingly large effect on a system. For example,
adding a small volume of lemon juice or vinegar to milk changes the protein structure
and curdling occurs. Many processes, particularly in living systems, have to take place
within a precise pH range. Should the pH of blood move 0.5 units outside the range
shown in Table 5.4, the person would become unconscious and die. Evolution has
devised buffer solutions to prevent such changes in pH in the body. A buffer solution
is one in which the pH remains approximately constant when small amounts of acid or
base are added.
Table 5.4
Fluid
pH range
blood
saliva
tears
urine
stomach juices
7.35-7.45
6.4-6.8
7.4
4.8-7.5
1.6-1.8
Biological systems work within precise ranges which buffers keep fairly constant. Why
do you think urine can have such a wide range?
Manufacturing systems also require precise control of pH and buffers are used in
electroplating, photographic work and dye manufacture. Some examples are shown
in Table 5.5
© H ERIOT-WATT U NIVERSITY
86
TOPIC 5. INDICATORS AND BUFFERS
Table 5.5: Applications of pH control
Many pharmacy products try to
match their pH to the pH of the
body tissue.
Electroplating industries need pH
control over their plating solutions.
Buffer solutions are of two types:
• an acid buffer consists of a solution of a weak acid and one of its salts
• a basic buffer consists of a solution of a weak base and one of its salts
If a buffer is to stabilise pH, it must be able to absorb extra acid or alkali if these are
encountered.
5.2.1
Acid Buffers
Learning Objective
Æ
To be able to describe the composition and functioning of an acid buffer solution.
An acid buffer consists of a weak acid represented as HA . It will be slightly dissociated.
Large reserves of HA molecules are present in the buffer.
HA
H
+
A
at equilibrum
Figure 5.7: Dissociation of a weak acid
The weak acid salt MA also present will be completely dissociated. Large reserves of
the A- ion are present in the buffer. This is the conjugate base.
© H ERIOT-WATT U NIVERSITY
5.2. BUFFER SOLUTIONS.
MA
87
M
+
A
complete dissociation
Figure 5.8: Dissociation of a salt of a weak acid
How does the buffer work? If acid is added to the mixture the large reserve of A - ions
will trap the extra hydrogen ions and convert them to the weak acid. This stabilises the
pH. If alkali is added the large reserve of HA molecules will convert the extra OH - to
water. This again stabilises the pH.
A typical example of an acid buffer solution would be ethanoic acid and sodium
ethanoate. The ethanoic acid is only partly dissociated. The sodium ethanoate salt
completely dissociates and provides the conjugate base.
CH3COOH
CH3COO
+
H
(HA reserve)
CH3COONa
CH3COO + Na
(A- reserve)
Figure 5.9: Equilibria in acid buffer solutions
© H ERIOT-WATT U NIVERSITY
88
TOPIC 5. INDICATORS AND BUFFERS
The stable pH of the buffer is due to:
• The weak acid which provides H + to trap added OH - .
• The salt of this acid which provides A- to trap added H + .
Action of a buffer solution.
Learning Objective
10 min
Æ
After completing this activity you should understand how an acid buffer solution
functions by absorbing extra acid or basic ions.
A simulation showing how an acidic buffer works is available in the on-line version of this
Topic.
There are two parts to the on-line simulation:
1. Drag and drop an extra hydrogen ion into the buffer solution, observe the response
of the buffer and read the description.
2. Drag and drop an extra hydroxide ion into the buffer solution, observe the response
and read the description.
Addition of acid to the buffer.
Extra hydrogen ions in the buffer upset the equilibrium situation in the weak acid.
The position of equilibrium shifts (Le Chatelier’s principle) and the large reserves of A ions from the salt allow the H+ ions to be removed. The A - ions provide the conjugate
base.
Addition of hydroxide to the buffer.
Extra hydroxide ions in the buffer react with some H + ions and upset the equilibrium
situation in the weak acid.
The position of equilibrium shifts (Le Chatelier’s principle) and the large reserves of HA
molecules from the weak acid allow the H + ions to be restored.
In an acid buffer, the weak acid supplies hydrogen ions when these are removed by the
addition of a small amount of base. The salt of the weak acid provides the conjugate
base, which can absorb hydrogen ions from addition of small amounts af acid.
© H ERIOT-WATT U NIVERSITY
5.2. BUFFER SOLUTIONS.
5.2.2
Basic buffers.
Learning Objective
Æ
To be able to describe the composition and functioning of a basic buffer solution
A basic buffer consists of a solution of a weak base and one of its salts, e.g. ammonia
solution and ammonium chloride. The ammonia solution is partly ionised and the
ammonium chloride is completely ionised. If hydrogen ions are added, they combine
with ammonia and if hydroxide ions are added, they combine with the ammonium ions
(conjugate acid) provided by the salt (NH 4 Cl).
Figure 5.10: Equilibria in base buffer solutions
The stable pH of the buffer is due to:
• The weak base which provides NH 3 to trap added H + .
• The salt of this base which provides NH 4 + to trap added OH - .
© H ERIOT-WATT U NIVERSITY
89
90
TOPIC 5. INDICATORS AND BUFFERS
Summary of buffer systems
Learning Objective
10 min
Æ
After completing this activity you should be able to write equations representing an
acid buffer system and a basic buffer system.
Two drag and drop exercises are available in the on-line version of this Topic to practise
the writing of equations for buffer systems.
In a basic buffer solution, the weak base removes excess hydrogen ions. The salt of
the weak base provides the conjugate acid, which can supply hydrogen ions if these are
removed by the addition of small amounts of base.
5.3
Calculating pH and buffer composition.
Learning Objective
To be able to calculate the pH of an acid buffer when given the acid dissociation
constant and the composition. To calculate composition when given the other
variables
Æ
A glance at Table 5.4 shows that biological buffer solutions have to operate around
specific pH values. This pH value depends upon two factors. The acid dissociation
constant and the relative proportions of salt and acid. The dissociation constant for a
weak acid HA is given by this expression :
" " "
"
Two assumptions can be made that simplify this expression even further.
1. In a weak acid like HA, which is only very slightly dissociated, the concentration of
HA at equilibrium is approximately the same as the molar concentration put into
the solution.
2. The salt MA completely dissociates.
concentration supplied by the salt.
Therefore [A - ] will effectively be the
© H ERIOT-WATT U NIVERSITY
5.3. CALCULATING PH AND BUFFER COMPOSITION.
91
The expression becomes:
" acid
salt
taking the negative log of each side:
%"
%
log acid
salt
Two important points can be seen from this equation:
• Since
acid
salt
is a ratio, adding water to a buffer will not affect the ratio (it will dilute each equally)
and therefore will not affect the [H + ] which determines the pH.
• If the [acid] = [salt] when the buffer is made up, the pH is the same as the pK a (or
H + = Ka )
This equation allows calculation of pH of an acid buffer from its composition and acid
dissociation constant, or calculation of composition from the other two values. Values
for Ka and pKa are available in the data booklet, page 12.
The next two problems show examples of the two most common type of calculation.
Example : Calculating pH from composition and K a .
Calculate the pH of a buffer solution made with 0.1 mol
=1.7x10-5 mol -1 ) and sodium ethanoate if the salt is added:
a) at 0.1 mol
-1
b) at 0.2 mol
-1
a) With the salt at 0.1 mol
-1
" " " %"
%"
%"
b) With the salt at 0.2 mol
-1
© H ERIOT-WATT U NIVERSITY
acid
salt
log " log -1
-1
ethanoic acid (Ka
92
TOPIC 5. INDICATORS AND BUFFERS
" " " %"
%"
%"
acid
salt
0.5 log " log -1
Notice that doubling the salt concentration has only raised the pH by 0.3 (from 4.77 to
5.07) and in general, the pH of the buffer is tied closely to the pK a value for the weak
acid, in this case ethanoic acid pKa = 4.77 (data booklet rounds this to two significant
figures i.e. 4.8). The ratio of acid to salt effectively provides a ’fine tuning’ of the pH.
Example : Calculating composition from pH and pK a
Calculate the concentrations of ethanoic acid and sodium ethanoate required to make a
buffer solution with a pH of 5.3 (pK a in data booklet).
%"
acid
log
salt
salt
log
acid
salt
acid
log acid
salt
log acid
salt
%
So the ratio of 3.16 to 1 is required and 3.16 moles of sodium ethanoate mixed with one
litre of 1.0 mol -1 ethanoic acid could be used.
© H ERIOT-WATT U NIVERSITY
5.4. SUMMARY
93
Buffer calculations
Learning Objective
To be able to calculate the pH of an acid buffer when given the acid dissociation
constant and the composition, and calculate composition when given the other
variables.
Æ
There are six questions in this set. Try as many as you need to feel confident.
Q20: Calculate the pH of a buffer solution containing 0.10 mol -1 ethanoic acid and
0.50 mol -1 sodium ethanoate (K a is in the data booklet). Give your answer to two
decimal places.
Q21: Calculate the composition of methanoic acid and sodium methanoate required to
make a buffer solution with a pH of 4.0. Quote your answer as a ratio of salt to 1 (so
6.31 to 1 would quote as 6.31).
Q22: A 0.10 mol -1 solution of a weak acid has 0.40 mol -1 of its sodium salt dissolved
in it. The resulting buffer has a pH 5.35. Find the dissociation constant of the acid.
Q23: What pH (to one decimal place) would be expected if 7.20g of sodium benzoate
-1 benzoic acid (sodium benzoate is
was dissolved in one litre of 0.02 mol
C6 H5 COONa, Ka and pKa in data booklet).
See further questions on page 202.
5.4
Summary
• Indicators are weak acids where the acid (HIn) and the conjugate base (In - ) have
different colours.
• The colour of the indicator depends on the relative proportions of HIn and In - ,
which in turn depends on the pH.
• The pH range over which an indicator changes colour depends on K In , the acid
dissociation constant for the indicator.
pH = pKIn 1
• Different titrations require different indicators.
• Buffer solutions are able to keep the pH of a system approximately constant when
small amounts of acid or base are added.
• There are two types of buffer solution, acidic and basic. In each case the solution’s
composition consists of a weak acid (or base) accompanied by a salt of the weak
acid (or base).
• Using these formulae, it is possible to calculate any one of the quantities involved,
© H ERIOT-WATT U NIVERSITY
40 min
94
TOPIC 5. INDICATORS AND BUFFERS
if the other entities are known.
" %"
5.5
%
acid
salt
log acid
salt
Resources
• Chemistry in Context: Hill and Holman , Nelson ISBN 0-17-438401-7
• Chemistry: A. and P. Fullick, Heinemann, ISBN 0-435-57080-3
• General Chemistry: Ebbing, Houghton Mifflin, ISBN 0-395-74415-6
• Chemistry - Advanced Higher : Unit 2: Principles of Chemical Reactions. Higher
Still Support. ISBN 1-85955-874-7
• Chemical Storylines: Salters Advanced Chemistry, Heinemann
ISBN 0-435-63106-3
• Chemical Ideas: Salters Advanced Chemistry, Heinemann, ISBN 0-435-63105-5
• http://chemistry.semo.edu/crawford/ch186/lectures/ch15/index.html
• http://www.science.ubc.ca/~chem/tutorials/pH/launch.html
Some questions in this topic have their origin in SQA publications. Copyright and
permission is gratefully acknowledged.
Electroplating picture courtesy of Sawyer and Smith Corp.
5.6
End of Topic test
An online assessment is provided to help you review this topic.
© H ERIOT-WATT U NIVERSITY
95
Topic 6
Thermochemistry
Contents
6.1
6.2
6.3
6.4
6.5
6.6
6.7
6.8
6.9
6.10
Introduction . . . . . . . . .
Bond energy . . . . . . . .
Hess’s Law . . . . . . . . .
Standard enthalpy changes
Calorimetry . . . . . . . . .
The Born-Haber cycle . . .
Enthalpy of solution . . . .
Summary . . . . . . . . . .
Resources . . . . . . . . .
End of Topic test. . . . . . .
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96
97
100
102
106
108
112
116
117
117
Prerequisite knowledge
Before you begin this Topic, you should be able to:
• describe how exothermic and endothermic enthalpy changes can be illustrated in
potential energy diagrams (Higher Unit 1);
• describe some examples of enthalpy changes and calculate values of enthalpy
change from both experimental evidence and application of Hess‘s Law (Higher
Unit 3);
• describe the formation of ionic and covalent bonds (Unit 1, topic 3 and 4).
Learning Objectives
After studying this Topic, you should be able to:
• define examples of standard enthalpy changes and explain how calorimetry can
be used to measure enthalpy changes;
• apply Hess’s Law to calculations of enthalpy changes and bond enthalpy values;
• use the Born-Haber cycle to calculate enthalpy changes in ionic crystal formation;
• describe the relationships involved in solution processes.
96
TOPIC 6. THERMOCHEMISTRY
6.1
Introduction
Many reactions release energy to their surroundings in the form of heat. These reactions
are described as exothermic. A reaction absorbing heat cools its surroundings and is
described as endothermic. Thermochemistry concerns the study of changes in energy
that occur during chemical reactions. In this Topic, your knowledge of thermochemistry
will be extended to include calculations of enthalpy changes involved in reactions with
both covalent compounds and ionic lattices.
Exothermic:
Figure 6.1: Potential energy diagram of an exothermic reaction
Endothermic:
Figure 6.2: Potential energy diagram of an endothermic reaction
Q1:
What is the value of H in Figure 6.1 expressed in kJ ?
Q2:
What does the symbol E a stand for?
Q3:
What is the value of H in Figure 6.2 expressed in kJ ?
Q4: Why has the reaction described by Figure 6.2 a lower temperature in the reaction
vessel?
© H ERIOT-WATT U NIVERSITY
6.2. BOND ENERGY
6.2
97
Bond energy
Learning Objective
Æ
To be able to use bond enthalpies and mean bond enthalpies to calculate enthalpy
changes in reactions
Covalent molecules are formed when the electrons shared between atoms result in a
lower energy system than the atoms alone before bonding. The process is exothermic
and has a negative enthalpy change, i.e. H is negative.
Figure 6.3: Bond formation in H 2
If the diatomic hydrogen molecule shown in Figure 6.3 was broken apart to form two
hydrogen atoms, the same amount of energy would have to be supplied. The energy
required to break one mole of these bonds is defined as the molar bond enthalpy.
For hydrogen, the molar bond enthalpy is a constant, + 432 kJ mol -1 , quoted in the data
booklet on page 9.
Figure 6.4: Bond breaking in H 2
H2(g)
2H(g)
ΔH = +432 kJ mol-1
Bond enthalpies always refer to breaking bonds under gaseous conditions. This allows
comparisons to be made on an equivalent basis between different bonds.
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98
TOPIC 6. THERMOCHEMISTRY
Bond enthalpy values
Learning Objective
5 min
Æ
To become familiar with databook bond enthalpies.
Use the SQA data booklet to help you answer these questions.
Q5:
How much energy in kJ is needed to break 4g of hydrogen gas into atoms?
Q6:
Enter the name of the halogen molecule which has the strongest bond.
Q7: Which halogen to hydrogen bond is easiest to break? Type the name of the
halogen only.
Q8:
Suggest why nitrogen has a higher bond enthalpy than oxygen.
A molecule like methane, CH 4 , has four carbon to hydrogen bonds within it (see
Figure 6.5) and this bond occurs in different environments in other compounds.
Figure 6.5: Different C-H environments
The exact value of the carbon to hydrogen bond enthalpy depends on which compound
is broken up.
The mean molar bond enthalpy is an average value that is quoted for a bond that can
occur in different molecular environments.
CH4 (g)
C(g) + 4H(g)
H = +1656 kJ mol-1
The four bonds present contribute to this total value, therefore:
'! %(
)* Bond enthalpy values can be used to estimate the enthalpy of a reaction by use of a
thermochemical cycle showing bond dissociation and formation.
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6.2. BOND ENERGY
99
Enthalpy calculations from bond enthalpies
Learning Objective
Æ
After this activity you should be able to calculate enthalpy changes from bond enthalpy
data.
An exercise to calculate enthalpy changes from bond enthalpy data.
Use the values given in Table 6.1 to complete the calculation. The questions will lead
you through.
An animation is available on the website at this point.
Figure 6.6: Reaction energy changes
The next activity uses the information shown to calculate the enthalpy changes in this
reaction.
Table 6.1: Calculation of enthalpy changes
BOND BREAKING (ALWAYS
POSITIVE)
1 mole of H-H bonds
1 mole of Cl-Cl bonds
ENERGY IN
BOND MAKING (ALWAYS NEGATIVE)
=
=
2 moles of H-Cl bonds
=
=
ENERGY OUT
=
Copy down the outline of the calculation and fill in the values you get as responses to
these questions.
Q9: How much energy in kJ is needed to break one mole of H-H bonds?
Q10: How much energy in kJ is needed to break one mole of Cl-Cl bonds?
Q11: What then is the total energy in kJ used up in bond breaking?
Q12: What is the total energy in kJ given out in the making of TWO moles of H-Cl from
isolated atoms?
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10 min
100
TOPIC 6. THERMOCHEMISTRY
See further questions on page 202.
Enthalpies of reaction estimated from bond enthalpies may differ from experimentally
determined values since the mean bond enthalpy is an average from different
environments, i.e. the O-H bond value is a mean of O-H bonds in different molecules,
not just water.
6.3
Hess’s Law
Learning Objective
Æ
To be able to use Hess’s law diagrams of bond making and breaking to calculate
enthalpy changes, and vice-versa
The cycle used to calculate the values of enthalpy changes in the last section is often
called Hess’s Law. This states that the overall reaction enthalpy is the sum of the
reaction enthalpies of each step of the reaction. This is an application of the First Law
of Thermodynamics, which states that energy is conserved. In a chemical reaction, the
energy change in converting reactants into products is the same, regardless of the route
by which the chemical change occurs.
One application of Hess’s Law allows us to calculate an unknown bond enthalpy from
other data. We can calculate the bond enthalpy of the carbon to carbon bond in ethane
from bond enthalpies and the H information below.
2C(g) + 3H2 (g)
C2 H6 (g)
H = -1514 kJ
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6.3. HESS’S LAW
101
Figure 6.7: Bond enthalpy of a C-C bond
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102
TOPIC 6. THERMOCHEMISTRY
Finding a bond enthalpy from enthalpy changes
Learning Objective
10 min
Æ
After this activity you should be able to calculate bond enthalpies from enthalpy of
reaction data.
A question set leading to solving a calculation.
Answer the following questions and complete the calculation. The layout to use is shown
in Figure 6.7. Copy the layout onto paper and fill it in as you answer the questions.
Q13: At the bond breaking stage, how many moles of hydrogen molecules are being
broken?
Q14: Use a databook and calculate the total bond breaking energy requirement in kJ
(remember the sign).
Only one carbon to carbon bond is being made. This is our unknown enthalpy. Write
this in as -x.
Q15: How many moles of carbon to hydrogen bonds are being made?
Q16: Calculate the total bond making energy.
Q17: Complete the arithmetic to solve for x. Try this yourself before displaying the
answer on-line.
6.4
Standard enthalpy changes
The term standard enthalpy change refers to an enthalpy change for a reaction in
which reactants and products are considered to be in their standard states at a specified
temperature.
The standard state of a substance is the most stable state of the substance under
standard conditions and the standard conditions refer to a pressure of one atmosphere
and a specific temperature, usually 298 K (25 Æ C).
Many standard enthalpy changes are given names describing the type of reaction
concerned. The units are always kilojoules per mole (kJ mol -1 ).
The standard molar enthalpy of formation (HÆ f ) is the enthalpy change that occurs
when one mole of a substance is prepared from its elements in their standard states.
The standard molar enthalpy of formation of elements is defined as zero, giving a
’base line’ which allows other enthalpy changes to be measured.
The standard molar enthalpy of combustion (HÆ c ) is the enthalpy change when
one mole of a substance is completely burned in oxygen under standard conditions.
HÆc is always negative since combustion is exothermic. The term ’completely burned’
is important since many elements form more than one oxide, e.g. carbon can form CO
and CO2 .
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6.4. STANDARD ENTHALPY CHANGES
103
Other
enthalpy
changes
like
the
enthalpy of solution
and
the
enthalpy of neutralisation should already be familiar to you from past work and
more are covered in the activity ’Born-Haber cycle’ later in this topic.
Hess’s Law allows calculation of enthalpy changes which are impossible to measure by
experiment.
For example: It is difficult to burn carbon and form carbon monoxide only. So a value
for HÆ f for carbon monoxide would be difficult to determine by experiment.Look at this
Hess’s Law enthalpy cycle, Figure 6.8 :
Figure 6.8: Thermochemical cycle
The overall enthalpy change is independant of the route taken and values for
H3 are measurable.
H 1and
So:
H1 = H2 + H3
H1 =
H3 =
= -394kJ mol-1
standard enthalpy of formation of CO 2
= -271kJ mol-1
-394 =H2 + (-271)
standard enthalpy of combustion of CO
µ
µH
2
= -394 + 271
Standard enthalpy of formation of CO = -123 kJ mol-1
Thus the standard enthalpy of formation of a substance can be calculated from standard
enthalpy changes that are experimentally determined.
Many Hess’s Law problems are best solved by an ’algebraic’ method using the equations
rather than an enthalpy cycle.
As an example: Find a value for the enthalpy change when carbon reacts with carbon
dioxide to give carbon monoxide according to Equation 6.1. The standard molar
enthalpy of combustion of carbon is available in the data booklet and that of carbon
monoxide is -271 kJ mol-1
C(s) + CO2 (g)
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2CO(g)
(6.1)
104
TOPIC 6. THERMOCHEMISTRY
STEP 1
Write the ’target’ equation.
C(s) + CO2 (g)
2CO(g)
STEP 2
Write equations for all the
information given to you.
equation (a)
C(s) + O2 (g)
CO2 (g)
H = ?
H = -394 kJ mol-1
(from databook)
H = -271 kJ mol-1
equation (b)
CO(g) + 1 /2 O2 (g)
STEP 3
Change these equations around by reversing,
multiplying or dividing them, before combining them
to obtain the target equation. Hess’s Law allows you
to treat the H values the same way. The website
animation shows the stages.
CO2 (g)
STEP 3 continued
Reverse equation
(b)
CO2 (g)
CO(g) + 1 /2 O2 (g)
H = +271 kJ mol-1
Multiply this by 2
2CO2 (g)
2CO(g) + O2 (g)
H = +542 kJ mol-1
add this to (a)
C(s) + O2 (g)
+ 2CO2 (g)
CO2 (g)
2CO(g) + O2 (g)
= C(s) + CO (g)
2
2CO(g)
H = -394 kJ mol-1
H = +542 kJ mol-1
H = +148 kJ mol-1
Notice how the number of moles of carbon dioxide reduces to just one mole and the
oxygen molecules cancel out.
Hess’s Law enthalpy cycle.
Learning Objective
10 min
Æ
After this activity you should be able to solve Hess‘s Law problems using an enthalpy
cycle.
A drag and drop exercise practising the enthalpy cycle method is available in the on-line
version of this Topic
Given the information that the standard molar enthalpy of formation of sulphur dioxide
and sulphur trioxide is -297 kJ mol-1 and -395 kJ mol-1 respectively, drag and drop the
expressions in the on-line activity into place to complete the cycle and thus calculate a
value for the standard molar enthalpy of combustion of sulphur dioxide.
These last two examples illustrate an important general law. The standard enthalpy of a
reaction can be calculated from tabulated standard molar enthalpies of formation using
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6.4. STANDARD ENTHALPY CHANGES
105
the relationship:
HÆ =HÆf (products) -HÆf(reactants)
So in the example given previously in Equation 6.1:
C(s) + CO2 (g)
Equation
CO2 (g)
2CO
0
-394
-123
0
-394
-246
C(s)
Given HÆ f
Multiply by number
moles in the equation
of
H = ?
2CO(g)
+
HÆ = HÆf (products) - HÆf(reactants)
HÆ = -246 - (-394)
HÆ = +148 kJ (same result as previously)
Hess’s Law questions
Learning Objective
Æ
After this set of questions you should be able to solve Hess’s Law problems using
algebraic methods.
Do as many of these questions as your tutor suggests. There are 5 questions in this set.
Q18:
Given these two values for enthalpy changes:
(1)
1
/2 N2 (g) + 1 /2 O2 (g)
(2)
1/
2
N2 (g) + O2 (g)
NO(g)
NO2 (g)
H = +90.2 kJ mol-1
H = +33.2 kJ mol-1
Use the algebraic method to find a value for this enthalpy change:
(3)
NO(g) + 1 /2 O2 (g)
NO2 (g)
H = ?
Q19: The molar enthalpy of formation of iron (III) oxide and aluminium oxide are -827
and -1676 kJ mol-1 respectively. Calculate the enthalpy change which takes place in the
thermite reaction (give your answer in kJ mol-1 ).
Fe2 O3 (s) + 2Al(s)
2Fe(s) + Al2 O3 (s)
Q20: The standard molar enthalpy of formation of methane is impossible to measure in
practice.
C(s) + 2H2 (g)
CH4 (g)
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H = ?
30 min
106
TOPIC 6. THERMOCHEMISTRY
Given that the standard molar enthalpy of combustion of methane is H c Æ = - 891 kJ
mol-1 , use data booklet values for the standard molar enthalpy of combustion of carbon
and of hydrogen to calculate a value for the standard molar enthalpy of formation of
methane.
Q21: a) Use the standard molar enthalpies of formation in the table below to calculate
the standard molar enthalpy of combustion of the gas, diborane B 2 H6 .
HÆf / kJ mol-1
B2 H6 (g)
+41.0
H2 O(l)
-286.0
B2 O3 (s)
-1225.0
b) Use the data booklet to help you explain whether ethane or diborane is the better
fuel.
See further questions on page 203.
6.5
Calorimetry
Learning Objective
Æ
To be able to understand the functioning of the bomb calorimeter and use results from
its operation to measure enthalpy changes
Enthalpy changes in combustion reactions can be obtained by experiment using a
calorimeter . You may have tried this last year. The energy released in combustion
heats up water in a container and calculation assumes all the energy is transferred
to the water as heat. The apparatus is not very efficient as some heat is lost to the
surroundings.
A bomb calorimeter can measure the energy release accurately by knowing:
• the mass of substance burned;
• the temperature rise of the calorimeter;
• the heat capacity of the bomb calorimeter (this is the heat energy in kilojoules
required to raise the temperature of the whole apparatus by 1K).
Bomb calorimetry
Learning Objective
15 min
Æ
After completing this activity you should understand how a bomb calorimeter works
and be able to calculate an enthalpy of combustion, H c from sample results.
An activity is available in the on-line version of this Topic in which you can view the parts
of a bomb calorimeter and solve a thermochemical calculation to measure the enthalpy
of combustion of an alcohol.
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6.5. CALORIMETRY
107
Figure 6.9: A bomb calorimeter
1. Crucible
Container holding the sample to be burned. In this case there is 2.0g of ethanol.
2. ’Bomb’
A closed container in which the reaction takes place.
3. Stirrer
Ensures that all the energy released in the reaction is transferred to the
calorimeter.
4. Thermometer
Measures the temperature rise accurately after the ignition. In this reaction the
temperature rose from 16.50 Æ C to 22.24Æ C.
5. Ignition Wires
The sample is ignited by a brief electrical heating.
6. Oxygen Inlet
Excess oxygen is pumped in at high pressure and made available to ensure
complete combustion.
7. Calorimeter
Contains a known mass of water and a calibration experiment shows that a one
degree rise in temperature is caused by 10.30 kJ of energy absorbed (the heat
capacity is thus 10.30 kJ K-1 ).
Calculate the enthalpy of combustion from the data provided. These questions will take
you through the steps needed. Quote all answers to two decimal places.
Q22: How much energy in kJ has to be supplied to raise the temperature of the
calorimeter by one degree?
Q23: What temperature rise (in Æ C or K) actually occurs?
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108
TOPIC 6. THERMOCHEMISTRY
Q24: How much energy in kJ has the ethanol released?
Q25: Calculate the energy change in kJ when one mole of ethanol is burned (remember
the sign).
See further questions on page 203.
6.6
The Born-Haber cycle
Learning Objective
To appreciate that the Born-Haber cycle is an application of Hess’s Law to the
formation of an ionic crystal and can be used to calculate unknown enthalpy changes
within it when the other values are known
Æ
The strength of covalent bonds is measured by the bond enthalpy. In ionic
compounds the comparable enthalpy value is the lattice enthalpy.
The
standard molar enthalpy change of lattice formation (HÆ LATT ) is the enthalpy
change that occurs when one mole of an ionic crystal is formed from the ions in their
gaseous state under standard conditions. Lattice enthalpies cannot be determined
directly, but an application of Hess’s law to the formation of an ionic crystal called the
’Born-Haber cycle’ can be used to calculate the value. This cycle is a closed path that
includes as steps the different enthalpy changes involved in the enthalpy of formation.
For sodium chloride in Figure 6.10 ROUTE 1 is equivalent to the five steps shown in
ROUTE 2.
Figure 6.10: Alternative routes to NaCl solid crystals
When numerical values are known for all the other steps, the lattice enthalpy can be
calculated from Hess’s Law since:
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6.6. THE BORN-HABER CYCLE
109
H by ROUTE 1 = H by ROUTE 2
The Born-Haber cycle
Learning Objective
Æ
After completing this activity you should be able to define the terms involved in the
Born-Haber cycle and be able to solve associated calculations.
An activity in which you can view the definitions of the steps involved in a Born-Haber
cycle and solve an associated calculation to find the enthalpy of lattice formation of a
crystalline solid.
In the on-line version of this Topic you can click on each enthalpy change in turn to see
the definition and take note of the information given here or in the data booklet (write
down each value). The calculation of a value for the standard molar enthalpy change
of lattice formation for sodium chloride will be done after the the information has been
gathered.
Na + (g) +e - +C l (g)
4
Na +
5
- 1
(g) +e + 2 C l 2(g)
Na + (g) +C l - (g)
3
Na (g) + 1 2 Cl 2(g)
2
Na (s) + 1 2 Cl 2(g)
1
Na + C l - (s)
Figure 6.11: Energy steps in the Born-Haber cycle
The steps in the Born-Haber cycle are these:
© H ERIOT-WATT U NIVERSITY
6
30 min
110
TOPIC 6. THERMOCHEMISTRY
1. The standard molar enthalpy of formation (HÆ f ) refers to the enthalpy change
that occurs when one mole of a substance is prepared from its elements in their
standard states. For sodium chloride this is:
ROUTE 1
Na(s) + 1 /2 Cl2 (g)
HÆf = -411 kJ mol-1
Na+ Cl- (s)
2. The standard molar enthalpy of atomisation (H Æ AT ) of an element is the enthalpy
required to produce one mole of isolated gas atoms from the element in its
standard state. HÆ AT is always positive. Use the data book to find the numerical
value of this change and write it down.
Na(s)
HÆAT
Na(g)
3. The first ionisation energy (HÆ I.E. ) of an element is the energy required to remove
one electron from each atom in a mole of gaseous atoms. One mole of gaseous
ions is formed. HÆ IE is always endothermic. Use the data book to find the
numerical value of this change and write it down.
Na(g)
Na+ (g) + e-
HÆI.E.
Subsequent ionisation energies remove successive electrons, i.e. the second
ionisation energy would be:
Na+ (g)
Na2+ (g) + e-
Na(g)
Na2+ (g) +2 e-
Note: For this change:
The value of this enthalpy change would be the sum of the first and second
ionisation energies. (+502 +4560)kJ mol -1
4. The molar bond enthalpy (HÆ BOND ) of a diatomic molecule is the enthalpy
required to break one mole of the bonds. In this case, only half a mole is involved.
Notice that for chlorine, the molar bond enthalpy is identical to the molar enthalpy
of atomisation since the element is a gas.
Use the data book to find the numerical value of this change and write it down.
Cl (g)
2
HÆ
BOND
Cl(g)
5. The electron affinity (HÆ E.A. ) is usually defined as the enthalpy change for the
process of adding one electron to each atom in one mole of isolated atoms in the
gaseous state. Use the data book to find this value.
Cl(g) + e-
Cl- (g)
HÆE.A.
As with ionisation enthalpies, second and subsequent electron affinities can be
written. The enthalpy change involved in producing an ion X 2- would be the sum
of the first and second electron affinities.
6. The standard molar enthalpy change of lattice formation (HÆ LATT ) is the
enthalpy change which occurs when one mole of an ionic crystal is formed from
the ions in their gaseous state under standard conditions.
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6.6. THE BORN-HABER CYCLE
111
The standard molar enthalpy change of lattice formation is the quantity the Born-Haber
cycle will help us to calculate since:
H by ROUTE 1
=
H by ROUTE 2
=
2+3+4+5+6
HÆf
=
HÆAT + HÆI.E. + 1/2 HÆBOND +
HÆE.A. + HÆLATT
HÆLATT
=
HÆf - (HÆAT + HÆI.E. +
1 / H Æ
Æ
2
BOND + H E.A. )
Step1
So
Q26: Substitute in the numerical values you took note of in the activity and find a
value in kJ mol-1 for the standard molar enthalpy of lattice formation of sodium chloride
(remember the sign).
A Born-Haber cycle for lithium fluoride
Learning Objective
Æ
After completing this activity you should be familiar with the steps involved in making
up a Born-Haber cycle for the formation of an ionic crystal.
A drag and drop exercise is available in the on-line version of this Topic to become
familiar with the enthalpy changes involved in a Born-Haber cycle.
10 min
Drag and drop the expressions in the on-line version into place to complete the cycle
for lithium fluoride. Use the data booklet to source values for all the enthalpy changes
(pages 9,10 and 17) and calculate a value for the standard molar enthalpy of formation
of lithium fluoride.
Use paper and pencil to draw a Born-Haber Cycle diagram showing all the steps involved
in the formation of lithium fluoride. Use the data booklet to source values for all the
enthalpy changes (pages 9,10 and 17) and calculate a value for the standard molar
enthalpy of formation of lithium fluoride.
Q27: Substitute in the numerical values you noted from the data booklet to solve for
the standard molar enthalpy of formation of lithium fluoride. Do this exercise on paper
before revealing the solution.
Further questions to practise Born-Haber cycles
Learning Objective
Æ
To be able to solve Born-Haber cycle questions for ionic crystal formation.
A question set leading to a standard molar enthalpy of formation of copper (II) chloride.
Answer the questions.
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10 min
112
TOPIC 6. THERMOCHEMISTRY
Q28: If a Born-Haber cycle was drawn for copper (II) chloride (CuCl 2 ), what value in kJ
mol-1 (remember the sign) would be used to represent the change:
Cu(g)
Cu2+ (g) + 2e-
Hint: This is a 2+ ion!
Q29: What value in kJ would be used to represent the change:
Cl2 (g)
2Cl(g)
Q30: What value in kJ would be used to represent the change:
2Cl(g)
2Cl- (g)
Q31: Calculate a value for the standard molar enthalpy of formation of copper (II)
chloride in kJ mol-1 given the data booklet values for the other changes.
6.7
Enthalpy of solution
Learning Objective
To be able to link lattice enthalpy, enthalpy of solution and hydration enthalpy in a
thermochemical cycle.
Æ
To define each term and perform calculations
The lattice enthalpy of an ionic compound gives an indication of the strength of the
bonding between the ions. When an ionic compound dissolves, it is useful to think of
the process in two stages:
1)
Energy has to be supplied to break down the lattice. This
stage is endothermic.
2)
The free ions become surrounded by water molecules.
The formation of new bonds releases energy.
Together these enthalpy changes give an overall enthalpy change for the solution
process.
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6.7. ENTHALPY OF SOLUTION
113
Figure 6.12: An ionic solid dissolving in water
These changes can be represented by a Born-Haber cycle.
The enthalpy of solution (HÆ SOLN ) is defined as the enthalpy change when one mole
of a substance is dissolved completely in water.
The enthalpy of hydration (HÆ HYD ) is the enthalpy change when one mole of
individual gaseous ions is completely hydrated, i.e:
En+ (g)
En+ (aq)
and
En- (g)
© H ERIOT-WATT U NIVERSITY
En- (aq)
114
TOPIC 6. THERMOCHEMISTRY
Figure 6.13: Enthalpy of solution of NaCl
The standard molar enthalpy of solution of an ionic compound thus involves lattice and
hydration enthalpy steps and is given by:
"soln
"hyd
"
latt
Notice that the lattice enthalpy is defined as the enthalpy associated with lattice
formation. This would be exothermic with a value of -769 kJ mol -1 (see the dotted
line on the diagram). It is the lattice breaking enthalpy (+769 kJ mol -1 ) which is involved
here. If using the formula, be careful with the signs!
Q32: Why does Figure 6.13 show two separate values for the enthalpy of hydration?
Q33: Calculate a value for H SOLN in kJ mol-1 from Figure 6.13. Remember the sign.
Q34: In terms of heat, is this solution process
a) exothermic?
b) endothermic
Q35: Would the temperature reading on a thermometer in the solution as the salt
dissolved
a) go down
b) go up?
c) stay the same?
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6.7. ENTHALPY OF SOLUTION
115
This hand-warmer
contains a large
amount of sodium
ethanoate salt.
When a small
metal disc inside
is bent,
crystallisation of
the salt is initiated
and the pad
warms up. The
salt can be made
to dissolve again
by placing the bag
in boiling water for
a few minutes.
Figure 6.14: A hand-warmer
Q36: Is the crystallisation of sodium ethanoate salt exothermic or endothermic?
(Crystallisation is the opposite of solution).
Q37: For sodium ethanoate, which has the higher numerical value; the enthalpy due to
hydration or lattice?
Enthalpy of solution
Learning Objective
Æ
After completing this activity you should know how enthalpy of solution relates to
lattice enthalpy and hydration enthalpy.
A drag and drop exercise is available in the on-line version of this Topic relating solution,
lattice and hydration enthalpies.
In the on-line exercise drag and drop the expressions into place to complete the cycle
showing the enthalpy of solution of potassium bromide. Answer the questions which
follow.
Q38: Use a data booklet to find values enabling you to calculate a value for
KBr in kJ mol-1 .
H soln of
Q39: In terms of heat, this solution process is therefore
a) exothermic
b) endothermic
Q40: The temperature reading on a thermometer in the solution as the salt dissolved
would therefore
a) go down
b) go up
c) stay the same
© H ERIOT-WATT U NIVERSITY
10 min
116
TOPIC 6. THERMOCHEMISTRY
Enthalpy Signs and Calculations
An online exercise is provided to help you if you require additional assistance with this
material, or would like to revise this subject.
6.8
Summary
• The First Law of Thermodynamics states that energy can be changed from one
form to another but it cannot be created or destroyed. This allows thermochemical
cycles called Hess’s Law cycles to be used to calculate unknown enthalpy values;
if a reaction can take place by more than route, the overall enthalpy change is the
same, whichever route is taken.
So in a reaction where A is converted into B by two routes as shown:
Hess’s Law calculations can also be done using an algebraic method.
• Standard molar enthalpy changes (H Æ ) refer to enthalpy changes where
reactants and products are in their standard (most stable) state at 298 K and 1
atmosphere pressure. There are a variety of defined standard enthalpy changes
referring to different reaction types, e.g. HÆ c is a standard molar enthalpy of
combustion. Calorimeters can be used to measure the energy change in a reaction
like combustion. Calculations of the enthalpy change on complete combustion of
one mole of the substance yields H Æ c .
• The standard enthalpy of a reaction can be calculated from tables of standard
enthalpies of formation.
HÆf (products) - HÆf (reactants)
• Molar bond enthalpy is the energy required to break one mole of bonds.
X - Y(g)
X(g) + Y(g)
Mean molar bond enthalpies are average values for a bond (like C-H) which exists
in different molecular environments. Values are found in the data booklet or can
be calculated from enthalpy change information.
Enthalpy changes in reactions can be calculated from the enthalpies involved in
the bond breaking and bond making steps.
© H ERIOT-WATT U NIVERSITY
6.9. RESOURCES
117
• Born-Haber cycles are enthalpy diagrams applied to the formation of ionic crystals
and can be used to determine enthalpies of lattice formation that cannot be
determined by experiment. Standard molar enthalpy of lattice formation is the
enthalpy change on formation of one mole of a crystal from gaseous ions under
standard conditions. Other steps in the Born-Haber cycle include enthalpy of
atomisation, ionisation enthalpy, bond enthalpy, electron affinity and enthalpy of
formation.
• The standard molar enthalpy of solution of an ionic compound involves lattice and
hydration enthalpy steps and is given by:
"soln
6.9
"hyd
"
latt
Resources
• Chemistry in Context: Hill and Holman, Nelson, ISBN 0-17-438401-7
• Chemistry: A. and P. Fullick, Heinemann, ISBN 0-435-57080-3
• General Chemistry: Ebbing, Houghton Mifflin, ISBN 0-395-74415-6
• Chemistry - Advanced Higher: Unit 2: Principles of Chemical Reactions
Higher Still Support, ISBN 1-85955-874-7
• Chemical Storylines: Salters Advanced Chemistry, Heinemann
ISBN 0-435-63106-3
• Chemical Ideas: Salters Advanced Chemistry, Heinemann, ISBN 0-435-63105-5
6.10
End of Topic test.
An online assessment is provided to help you review this topic.
© H ERIOT-WATT U NIVERSITY
118
TOPIC 6. THERMOCHEMISTRY
© H ERIOT-WATT U NIVERSITY
119
Topic 7
Reaction Feasibility
Contents
7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.2 The concept of entropy . . . . . . . . . . . . . . . . . . . . .
7.2.1 Entropy and spontaneity . . . . . . . . . . . . . . . .
7.2.2 Calculating entropy values . . . . . . . . . . . . . .
7.3 The second law of thermodynamics . . . . . . . . . . . . .
7.4 Gibbs free energy . . . . . . . . . . . . . . . . . . . . . . .
7.4.1 Calculating GÆ values . . . . . . . . . . . . . . . .
7.4.2 Free energy change under non-standard conditions
7.5 Ellingham diagrams . . . . . . . . . . . . . . . . . . . . . .
7.5.1 Extraction of metals . . . . . . . . . . . . . . . . . .
7.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.7 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.8 End of topic test . . . . . . . . . . . . . . . . . . . . . . . .
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121
122
124
125
127
130
133
136
138
140
143
144
144
Prerequisite knowledge
After studying this Topic, you should be able to:
• describe standard enthalpy changes involved in covalent and ionic bonding (Topic
2.6);
• use Hess’s law in calculations involving enthalpy changes including bond
enthalpies and Born-Haber cycles. (Topic 2.6).
Learning Objectives
Before you begin this Topic, you should be able to:
• define examples of standard entropy changes and describe how entropy changes
in reaction systems are temperature dependant;
• state the second and third laws of thermodynamics in terms of entropy and
describe how the total entropy change can be expressed in terms of Gibbs free
energy;
• solve problems and carry out calculations involving Gibbs free energy, enthalpy
and entropy;
120
TOPIC 7. REACTION FEASIBILITY
• relate feasibility and equilibrium position to Gibbs free energy values;
• use Ellingham diagrams to predict reaction conditions.
© H ERIOT-WATT U NIVERSITY
7.1. INTRODUCTION
7.1
121
Introduction
Exothermic reactions have a negative value for the enthalpy change and release heat
to the surroundings. In contrast, endothermic reactions have a positive enthalpy change
and heat is absorbed from the surroundings as the reaction takes place.
A hundred years ago, notable chemists like P. Berthelot in Paris and J. Thomsen in
Copenhagen felt that a general criterion for predicting reaction spontaneity could be
assumed from the sign of the enthalpy change. If H had a negative value, the reaction
would occur on its own accord, without help from us, and if H had a positive value, the
reaction could not occur by itself.
Indeed, almost all exothermic reactions are spontaneous at standard conditions for
example, iron rusting:
2Fe(s) + 3 /2 O2 (g) + 3H2 O(l)
H = - 791 kJ
2Fe(OH)3 (s)
The opposite reaction, rusting reversing to give pure iron, with a positive enthalpy
change, never occurs spontaneously at standard conditions.
Unfortunately, this simple rule is not universal. At 1 atmosphere and 25 Æ C, the change
shown here when ice melts is spontaneous, even though the enthalpy change is
endothermic.
H2 O(s)
H2 O(l)
H = + 6.01 kJ mol-1
Similarly, liquid water evaporates under standard conditions even though the enthalpy
change is endothermic.
H2 O(l)
H2 O(g)
H = + 44.1 kJ mol-1
Figure 7.1: Formation of water vapour from ice
There are also many instances where reactions that are not spontaneous at room
temperature become spontaneous at higher temperatures.
An example is the
decomposition of limestone:
CaCO3(s)
CaO(s) + CO2 (g)
H = + 178.0 kJ mol-1
Limestone deposits are stable over centuries. If the temperature is raised above 1100 K
however, the limestone decomposes to give off carbon dioxide.
In other words, there is more to the question of spontaneity than simply a consideration
of the enthalpy change and it would seem that temperature is an important factor. The
other important factor is entropy.
© H ERIOT-WATT U NIVERSITY
122
TOPIC 7. REACTION FEASIBILITY
7.2
The concept of entropy
Learning Objective
To understand the concept of entropy;
To define standard entropy changes;
To describe how entropy is temperature dependent;
To relate entropy and spontaneity;
Æ
To carry out relevant calculations.
The entropy of a system is the degree of disorder of the system. The greater the
disorder, the greater the entropy. Low entropy is associated with strongly ordered
substances.
Table 7.1: Entropy changes
Increasing Entropy
Decreasing Entropy
A puddle dries up on a warm day as the A builder uses a pile of loose bricks to
liquid becomes water vapour. The disorder construct a wall. The order of the system
(entropy) increases.
increases. Entropy falls.
Heating ammonium nitrate forms one mole
of dinitrogen oxide and two moles of
steam. Three moles of gas formed. The
disorder (entropy) increases.
When the individual ions in a crystal come
together they take up a set position. The
disorder of the system falls.
Entropy
decreases.
NH4 NO3 (s)
Na+ (aq) + Cl- (aq)
N2 O(g) + 2H2 O(g)
NaCl(s)
Entropy is the degree of disorder of a system, and the examples shown here illustrate
that the term can be applied to areas other than chemistry!
A typical teenagers bedroom has a high
degree of disorder. A high entropy.
A jigsaw is a highly ordered system. When
made up, it has low entropy.
Figure 7.2: Examples of low and high entropy
© H ERIOT-WATT U NIVERSITY
7.2. THE CONCEPT OF ENTROPY
Entropy and temperature
Learning Objective
Æ
After completion of this activity you should understand the role of temperature and
changes of state on entropy values.
An animation relating entropy changes to temperature and changes of state.
Use the slide control to vary the temperature of the water molecules. Note the changes
in entropy and answer the question set which follows.
Q1: What is the entropy value in J K -1 mol-1 of a perfect water crystal at zero Kelvin?
Q2: What name is given to the change occuring at 273 K?
Q3: Is the change from 270 K to 275 K accompanied by a positive or negative change
in entropy?
Q4: Is this accompanied by an increase or decrease in the disorder of the molecules
in the system?
See further questions on page 203.
Entropy increases as temperature increases. Changes of state involve increases in
entropy. The entropy of a perfect crystal at zero Kelvin is zero.
The melting of ice and the evaporation of water are both examples of endothermic
reactions which proceed spontaneously.
© H ERIOT-WATT U NIVERSITY
123
10 min
124
TOPIC 7. REACTION FEASIBILITY
H2 O(s)
H2 O(l)
H = + 6.01 kJ mol-1
H2 O(l)
H2 O(g)
H = + 44.1 kJ mol-1
In each of these reactions, although H is positive, it is the increase in the entropy of
the system that is the driving force behind the reactions. This change in entropy, given
the symbol S, more than compensates for the positive enthalpy change.
7.2.1
Entropy and spontaneity
Learning Objective
Æ
To show how a reaction that has no change in enthalpy can proceed spontaneously
by having an increase in entropy.
How can an increase in entropy provide the drive necessary to allow a reaction to
proceed spontaneously? An example of a change where there is no change in the
enthalpy value will help illustrate this.
Before mixing
After mixing
Figure 7.3: Bromine and air mixing
In Figure 7.3 a coloured gas is placed in the left-hand side and air is present in the righthand side. Both are at 1 atmosphere pressure and 25 Æ C. If the partition is removed, the
two types of gas mix spontaneously. There is no temperature change; the only driving
force is the desire to increase the disorder. (Think for example of how a really bad smell
released in a corner of a room soon spreads to fill it!)
No matter how long the containers are left, the gases will not spontaneously unmix. How
many rooms have you been in where the oxygen suddenly all goes to one side and the
nitrogen to the other? (How likely is it that the ’bad smell’ will all move and concentrate
in its original corner!)
Figure 7.4 shows the reverse of Figure 7.3 to illustrate how improbable this would be.
Before ’unmixing’!
After ’unmixing’!
Figure 7.4: Bromine and air mixture separating
The laws of chance and probability dictate that the increased disorder, and therefore the
higher entropy, will prevail.
© H ERIOT-WATT U NIVERSITY
7.2. THE CONCEPT OF ENTROPY
7.2.2
Calculating entropy values
Learning Objective
To be able to estimate the change in entropy in a system by application of some
general principles.
Æ
To calculate precise entropy changes in a system from standard entropy values.
Since the entropy of a substance depends on the order of the system, when a solid
crystal is cooled to absolute zero (zero kelvin), all the translational motion of the particles
is eliminated and each particle has a well defined location, i.e. it is 100% ordered. The
entropy is therefore zero. This is one version of the Third Law of Thermodynamics
125
As temperature is increased, entropy increases. As with enthalpy values, it is normal
to quote standard entropy values for substances as the entropy value for the standard
state of the substance.
Notice that the unit of entropy values is joules per kelvin per mole (J K -1 mol-1 ). Be
careful with this, since enthalpy values are normally in kilojoules per mole and in
problems involving these quantities the units must be harmonised.
In a chemical reaction system, the standard entropy change (S Æ ) can be calculated
from the standard entropy values of the reactants and products.
SÆ = SÆPRODUCTS - SÆREACTANTS
It is normally possible, however, to get an idea of whether entropy increases or
decreases in a system by inspection of the equation. Actual values of standard entropies
are in the SQA data booklet on page 16.
Estimating and calculating spontaneity
Learning Objective
Æ
After this activity you should be able to estimate the sign of the entropy change in a
reaction and calculate the change in standard entropy using values from data tables.
A data handling exercise involving entropy values.
Look at the information outlining the general principles behind estimating entropy
changes and the data table of standard entropy values (Table 7.2). You will also need
access to the SQA data booklet. Answer the questions which follow.
GENERAL PRINCIPLES:
1. Gases have a higher entropy than liquids, which in turn have a higher entropy
value than solids.
2. The more complex the molecule, the higher the entropy.
3. Breaking a large molecule into smaller molecules increases entropy.
4. Dissolving increases entropy.
© H ERIOT-WATT U NIVERSITY
30 min
126
TOPIC 7. REACTION FEASIBILITY
Table 7.2: Data table of standard entropy values
SUBSTANCE
Standard Entropy S Æ / J K-1 mol-1
Diamond, C(s)
2.4
Sodium chloride, NaCl (s)
72.4
Water (ice), H2 O(s)
48
Water (liquid), H2 O(l)
70
Water (gas), H2 O(g)
188.7
Methane (g)
186.2
Ethane (g)
229.5
Propane (g)
269.9
Butane (g)
310.1
Pentane (g)
261.1
Give one-word answers to the following questions.
Q5:
Name the substance with the most organised structure on the list.
Q6:
Does it have the highest or lowest standard entropy value?
Q7:
Which substance shows a sequence which agrees with general principle 1?
Q8:
Name the family which agrees with general principle 2.
See further questions on page 203.
For each of the next six examples, make a judgement based on the general principles,
saying whether you would expect the entropy change of the chemicals to be:
A positive
Q9:
B negative
C stay the same
A salt dissolving in water.
Q10: Salad dressing separating into oil and vinegar.
Q11: Perfume spreading through a room.
Q12: 2Mg(s) + O2 (g)
2MgO(s)
Q13: The data given in Table 7.2 allows us to calculate a value for the following change
when steam condenses to water at standard conditions.
H2 O(g)
H2 O(l)
S Æ = SÆPRODUCTS - SÆREACTANTS
S Æ = 70.0 - 188.7
S Æ = - 118.7 J K-1 mol-1
© H ERIOT-WATT U NIVERSITY
7.3. THE SECOND LAW OF THERMODYNAMICS
127
Use a similar method and additional data from the SQA data booklet to calculate the
standard entropy change for this reaction. Do this on paper (now!) before displaying the
answer.
2H2 (g) + O2 (g)
2H2 O(g)
Q14: The entropy change for this reaction during the formation of ammonia in the Haber
process is given by:
N2 (g) + 3H2 (g)
2NH3 (g)
SÆ = - 99.5 J K-1 mol-1
Use values in the data booklet to calculate a standard entropy for ammonia.
Q15: Calculate the entropy change involved in the formation of one mole of aluminium
oxide from the elements under standard conditions.
Q16: Calculate the entropy change in J K -1 mol-1 involved in the formation of calcium
chloride from the elements. (Remember the sign.)
See further questions on page 203.
The direction of entropy change in a reaction can be estimated by application of some
general principles. The change in standard entropies in a reaction system can be
calculated from:
SÆ = SÆPRODUCTS - SÆREACTANTS
7.3
The second law of thermodynamics
Learning Objective
To understand that the total entropy of a reaction system and its surroundings always
increases for a spontaneous process.
Æ
To solve associated problems.
An increase in entropy provides a driving force towards a reaction proceeding
spontaneously. There are, however, processes that proceed spontaneously that seem
to involve an entropy decrease.
For example, steam condenses to water at room temperature:
H2 O(g)
H2 O(l)
SÆ = - 118.7 J K-1 mol-1
HÆ = - 44.1 kJ mol-1
In this case, the enthalpy change is also negative. This outpouring of energy from the
system (- HSYSTEM ) is transferred to the surroundings .
© H ERIOT-WATT U NIVERSITY
128
TOPIC 7. REACTION FEASIBILITY
Figure 7.5:
The heat is transferred to the cold surface of the window and to the air around that area.
This increases the disorder or entropy of the surroundings. (Just think of the scalding
effect that would occur if your hand were placed in the steam - the disorder of the skin
would increase!)
In general terms, heat energy released by a reaction system into the surroundings
increases the entropy of the surroundings. If heat is absorbed by a reaction from the
surroundings, this will decrease the entropy of the surroundings.
In the case of steam condensing, the entropy gain of the surroundings is equal to the
energy lost (- H) of the chemical system divided by the temperature:
"
+
The entropy change in the condensation situation therefore requires consideration of two
entropy changes. The change in the system itself and the change in the surroundings
must be added together, and for a spontaneous change to occur this total entropy
change must be positive.
SÆTOTAL = SÆSYSTEM + SÆSURROUNDINGS
In the case of steam condensing, the heat given out can be used to calculate a value for
SÆSURROUNDINGS:
"
+
* And a calculation of S Æ SYSTEM from the data booklet gives:
* * All this means that although the entropy of a system itself may drop, the process itself will
© H ERIOT-WATT U NIVERSITY
7.3. THE SECOND LAW OF THERMODYNAMICS
129
still be a natural, spontaneous change if the drop is compensated by a larger increase
in entropy of the surroundings.
Expressed another way, this is the Second Law of Thermodynamics The total entropy
of a reaction system and its surroundings always increases for a spontaneous change.
A word of caution: spontaneous does not mean ’fast’. It means ’able to occur without
needing work to bring it about’. Thermodynamics is concerned with the direction of
change and not the rate of change. The rate of change of a reaction is studied in more
detail in Topic 2.9 Kinetics.
Melting ice
Learning Objective
After this activity you should be able to use data booklet information to calculate
the total entropy change in a system and its surroundings and therefore predict
sponteneity.
Æ
A step-by-step data handling exercise to calculate the total entropy of a system and
surroundings.
Look at the animation of ice melting at 25 Æ C. Use the hints in each part of the problem
to carry out the calculation yourself before displaying the answer.
Figure 7.6: Melting of ice
A block of ice melts at 25Æ C with an enthalpy change of + 6.01 kJ mol -1 . Use this value
and values of entropy in to illustrate the Second Law of Thermodynamics, i.e. show that
STOTAL is positive.
Q17: Step 1 involves calculating a value for the entropy change in the surroundings,
using the value for the enthalpy change given. Try this on paper yourself before revealing
the answer.
Q18: Step 2 involves calculating a value for the entropy change in the system. Try this
on paper yourself before revealing the answer.
Q19: Step 3 involves calculating a value for the total entropy change from the first two
steps. Try this yourself.
Here are two further examples to try.
Q20: Calcium carbonate, present in limestone, is stable under normal atmospheric
conditions. When it is in a volcanic area and it gets very hot, it can thermally decompose.
Given the following information, use the Second Law of Thermodynamics to show why
limestone is stable at 25Æ C but not at 1500 Æ C.
© H ERIOT-WATT U NIVERSITY
20 min
130
TOPIC 7. REACTION FEASIBILITY
CaCO3 (s)
HÆ = +178 kJ mol-1
SÆ = +161 J K-1 mol-1
CaO(s) + CO2 (g)
Q21: Graphite has been converted into diamond by the use of extreme pressure and
temperature. Given the following information and values of entropy in the data booklet,
show why diamond can not be made from graphite at 1 atmosphere pressure, either at
room temperature or 5000 Æ C.
C(s)graphite
H = +2.0 kJ mol-1
C(s)diamond
The total entropy of a reaction system and its surroundings always increases for a
spontaneous process and can be calculated from the temperature and the enthalpy
change for the system.
7.4
Gibbs free energy
Learning Objective
To understand how the total entropy is normally expressed in terms of Gibbs free
energy.
To understand that the direction of spontaneous change is in the direction of
decreasing free energy and that the standard free energy change for a reaction can
be calculated in a variety of ways.
Æ
Having to consider what happens to the surroundings in a chemical system is a bit of
a nuisance, since chemists really just wish to look at the reaction itself and be able to
predict the feasibility of its occurence. This can be done by rearranging the equation
which best represents the Second Law of Thermodynamics.
'
" "
+
+
Multiplying both sides by -T gives the following expression:
-TSÆ TOTAL = HÆ SYSTEM - TSÆ SYSTEM
The term -TSÆ TOTAL is replaced by a new symbol
standard Gibbs free energy change. The expression becomes:
G Æ
called
the
GÆ = HÆ - TSÆ
Since GÆ = -TSÆ TOTAL and the second law states that spontaneous changes have
SÆTOTAL positive, it follows that the direction of spontaneous change is in the direction
© H ERIOT-WATT U NIVERSITY
7.4. GIBBS FREE ENERGY
131
of decreasing free energy. Under standard conditions, G Æ will be negative. Under
non-standard conditions, all spontaneous changes have G negative.
Predicting spontaneity using free energy changes
Learning Objective
After this activity you should be able to predict the sign of G and say whether a
reacton will be spontaneous or not from a consideration of temperature, enthalpy and
entropy.
Æ
An on-line activity to complete a table showing a summary of variables involved in
determining the sign of G.
Read through the four possible combinations of variables involved in calculating a sign
for G and then complete the summary table by dragging and dropping the correct
entries into the frame.
The sign of the Gibbs free energy change in a reaction depends upon the positive or
negative nature of H and S and sometimes depends upon temperature.
© H ERIOT-WATT U NIVERSITY
15 min
132
TOPIC 7. REACTION FEASIBILITY
Figure 7.7:
Now work out which statement goes into each box in turn before dropping it into place.
Don’t do this by trial and error, think it through first!
The feasibility of a chemical reaction can be predicted from a consideration of the signs
of HÆ and SÆ .
© H ERIOT-WATT U NIVERSITY
7.4. GIBBS FREE ENERGY
7.4.1
Calculating
133
GÆ values
Learning Objective
Æ
To be able to calculate standard free energy changes from a variety of data sources
The standard free energy change for a reaction can be calculated from the standard
enthalpy and standard entropy changes.
GÆ = HÆ - TSÆ
Example Calculate a value for G Æ and thus predict whether or not the combustion of
graphite is feasible. Enthalpy and entropy values can be found in the data booklet.
C(s)graphite + O2 (g)
CO2 (g)
HÆ = standard enthalpy of combustion of graphite = -394 kJ mol -1 (data booklet)
,
-
-
! " "
!
')
* )* The reaction has a negative G Æ and is thermodynamically feasible.
In the same way that standard enthalpies of formation can be used to calculate standard
enthalpy changes for a reaction (Topic 2.6 - Standard enthalpy changes), the standard
free energy change of a reaction can be calculated from the standard free energies of
formation of reactants and products.
Example Calculate a value for G Æ for this equilibrium reaction given that the standard
free energies of formation of nitrogen dioxide and dinitrogen tetroxide are +52 kJ mol -1
and +98 kJ mol-1 respectively.
2NO2
-
-
-
«N O
2
4
-# $
)* # $
This reaction is therefore feasible in the direction that the equation shows (since G Æ is
negative). This means that the equilibrium composition would favour the products over
the reactants. If the value of GÆ had worked out at +6 kJ mol -1 , the thermodynamics
would be predicting that the reaction was non-feasible. The equilibrium composition
would favour the reactants over the products.
© H ERIOT-WATT U NIVERSITY
134
TOPIC 7. REACTION FEASIBILITY
It follows then that when G Æ = 0, the products and reactants will be equally favoured.
This fact can be useful in calculating the temperature at which a reaction becomes
feasible. This will be the temperature around which G Æ changes sign. The reaction will
proceed spontaneously in a favoured direction until the composition is reached when
GÆ = 0. If values for HÆ and SÆ are known, or can be calculated, then T (the
temperature) is the only unknown in the equation.
GÆ = HÆ - TSÆ
Example Calculate the temperature at which limestone (calcium carbonate) becomes
thermally unstable.
HÆ = +178 kJ mol-1
CaCO3(s)
CaO(s) + CO2 (g)
SÆ = +161 J K-1 mol-1
GÆ = HÆ - TSÆ
+
+
* + * Temperature when feasible = 1105.6 K (832.6 Æ C)
Above this temperature, GÆ is negative and decomposition occurs.
© H ERIOT-WATT U NIVERSITY
7.4. GIBBS FREE ENERGY
135
Calculations involving free energy changes
Learning Objective
After this activity you should be able to calculate values for GÆ from standard free
energy of formation data. You should also be able to calculate the temperature at
which a reaction becomes feasible.
Æ
A set of tutorial examples involving calculations of free energy changes.
Try at least three of the tutorial examples.
Q22: Use the table of standard free energies of formation to calculate values of G Æ for
these two reactions and thus predict whether or not the reaction is spontaneous .
a) 2Mg(s) + CO2 (g)
2MgO(s) + C(s)
2Cu(s) + CO2 (g)
b) 2CuO(s) + C(s)
Table 7.3:
SUBSTANCE
GÆFORMATION / kJ mol-1
CO2
-394
MgO
-569
ZnO
-318
CuO
-130
All elements
0
Q23: Calculate the standard free energy change at both 400 K and 1000 K for the
reaction:
HÆf /kJ mol-1
SÆ /J K-1 mol-1
MgCO3 (s)
MgO(s) +
CO2 (g)
-1113
-602
-394
66
27
214
Q24: Use the data given, along with data booklet values to calculate the temperature at
which the Haber process becomes feasible.
HÆf /kJ mol-1
«
N2 (g) +
3H2 (g)
0
0
-46.4
?
?
193.2
SÆ /J K-1 mol-1
Q25: Given these reaction values for oxides of nitrogen:
4NO(g)
2N2 O(g) + O2 (g)
2NO2 (g)
2NO(g) + O2 (g)
Æ
a) Calculate G for this reaction:
2N2 O(g) + 3O2 (g)
GÆ = -139.56 kJ
GÆ = -69.70 kJ
« 4NO (g)
2
b) Say whether the equilibrium position favours reactants or products.
© H ERIOT-WATT U NIVERSITY
2NH3 (g)
60 min
136
TOPIC 7. REACTION FEASIBILITY
See further questions on page 203.
Know about standard free energy changes for a reaction.
Understand the feasibility of a reaction and how the temperature at which a reaction
becomes feasible can be calculated from thermochemical data.
7.4.2
Free energy change under non-standard conditions
Learning Objective
Æ
To know that a reaction will proceed spontaneously in the forward direction until the
composition is reached where G = zero
The values of GÆ with which we have been working are based on the free energy
values of reactants and products under standard conditions. However, the free energy
change, G, during the course of a reaction when a mixture of reactants and products
is present, goes through a minimum.
Figure 7.8: Variation of free energy with composition
As the reaction proceeds, the percentage of reactants falls and the percentage of
products rises. At the minimum point, the value of G = 0 and this dictates the
equilibrium percentage of reactants and products.
G is different from GÆ because, as soon as the reaction starts:
• the standard conditions no longer apply;
• the newly produced product mixes with the now depleted reactants, increasing
entropy and decreasing free energy.
Figure 7.9 contains three graphs which show the situation for different values of the G Æ
change. In each case, the reaction proceeds spontaneously in the forwards direction
until the composition is reached where G = 0.
© H ERIOT-WATT U NIVERSITY
7.4. GIBBS FREE ENERGY
137
Figure 7.9: Free energy graphs for the general reaction, reactants (A) going to products
(B)
In graph 1, (top left), GÆ is negative, therefore there is more B than A at equilibrium.
In graph 2, (top right), GÆ is positive, therefore there is more A than B at equilibrium.
In graph 3, (bottom middle),
equilibrium.
GÆ
is 0. There will be equal amounts of A and B at
A word of warning: It must always be remembered that although a negative value of
GÆ predicts a feasible reaction, it says nothing about the rate of reaction. It may well
be that the reaction has a high activation energy barrier that prevents it taking place at
a measureable speed. Graph 1 (top left) could be slow to reach equilibrium and graph
2 (top right) could be fast. The website animation illustrates this point.
The combustion of the chemicals
in a match is an exothermic
reaction with an increase in
entropy. The free energy change is
therefore negative but the high
activation energy prevents burning
before striking.
Figure 7.10: Thermodynamics of burning a match
This is a suitable point to try the Prescribed Practical Activity on ’Verification of a
thermodynamic prediction.’
© H ERIOT-WATT U NIVERSITY
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TOPIC 7. REACTION FEASIBILITY
PPA - Verification of a thermodynamic prediction
Learning Objective
90 min
Æ
To be able to calculate the theoretical decomposition temperature for sodium
hydrogencarbonate and verify this experimentally.
A calculation from given values of enthalpy and entropy is carried out and an experiment
undertaken to verify the results.
Consult with your tutor whether this PPA for the assessment of outcome 3 may be carried
out (Refer to SCCC Advanced Higher Chemistry Unit 2 PPA 4).
7.5
Ellingham diagrams
Learning Objective
To be able to plot variations of free energy change with temperature and use
the subsequent graphs to predict the conditions under which reactions can occur,
particularly the extraction of a metal from its ores.
Æ
The numerical value of G alters with temperature. This is a consequence of the
entropy term being temperature dependent and it is possible to calculate values of G
for differing temperatures.
GÆ = HÆ - TSÆ
is rearranged to give
GÆ = -TSÆ + HÆ
This has the same format as the equation of a straight line, y = mx + c, with a gradient of SÆ and an intercept of HÆ on the y-axis. Graphing values of GÆ against temperature
like this yields graphs called Ellingham diagrams.
Q26: Look at the Equation 7.1 for a reaction forming a mixture called ’water gas’.
Calculate a value for the temperature at which this reaction becomes feasible.
C(s) + H2 O(g)
CO(g) + H2 (g)
(7.1)
HÆ = +131.3 kJ mol-1
SÆ = +133.8 J mol-1
Plotting an Ellingham diagram
Learning Objective
20 min
Æ
After this activity you should understand how Ellingham diagrams can be used to
predict the conditions under which a reaction can occur.
A data handling activity constructing Ellingham diagrams and using these to predict the
conditions under which a reaction can occur.
© H ERIOT-WATT U NIVERSITY
7.5. ELLINGHAM DIAGRAMS
139
Read through the text and follow the instruction carefully.
This table shows values of GÆ over a temperature range for the formation of water gas
(Equation 7.1).
Temperature / K
200
400
600
800
1000
1200
GÆ / kJ mol-1
104
78
51
24
-3
-29
Plot these points on the on-line graph or on graph paper.
The line you have drawn refers to:
C(s) + H2 O(g)
CO(g) + H2 (g)
This reaction can be considered as a combination of these two oxide formations:
C(s) + 1 /2 O2 (g)
H2 (g) + 1 /2 O2 (g)
CO(g)
(7.2)
H2 O(g)
(7.3)
If the second equation (Equation 7.3) was reversed and added to the first (Equation 7.2)
then the formation of water gas (Equation 7.1) results.
Both Equation 7.2 and Equation 7.3 can have their values for
temperatures calculated and plotted onto the same graph.
G Æ
at various
Now return to your graph and plot the values for carbon monoxide and water formation
on the same axes.
FORMATION OF CO(g)
Temperature / K
200
400
600
800
1000
1200
GÆ / kJ mol-1
-128
-146
-164
-182
-200
-218
Temperature / K
200
400
600
800
1000
1200
GÆ / kJ mol-1
-223
-224
-215
-206
-197
-188
FORMATION OF H 2 O(g)
INTERPRETATION
The lines relating G to temperature for the two equations Equation 7.2 and
Equation 7.3 intersect at 981.3 K when G = 0. The carbon and hydrogen are both
capable of reacting with oxygen but at any temperature above 981.3 K, the carbon is
capable of winning oxygen from the water molecule and forcing the second equation
(Equation 7.3) to reverse. CONSIDER 1000 K
C(s) + 1 /2 O2 (g)
H2 (g) + 1 /2 O2 (g)
CO(g)
H2 O(g)
GÆ = -200 kJ mol-1
GÆ = -197 kJ mol-1
By reversing the second equation and adding to the first, the result is:
© H ERIOT-WATT U NIVERSITY
140
TOPIC 7. REACTION FEASIBILITY
C(s) + H2 O(g)
CO(g) + H2 (g)
GÆ = -3 kJ mol-1
At 1000 K the formation of water gas is feasible and spontaneous.
CONSIDER 800 K
C(s) + 1 /2 O2 (g)
H2 (g) + 1 /2 O2 (g)
CO(g)
H2 O(g)
GÆ = -182 kJ mol-1
GÆ = -206 kJ mol-1
By reversing the second equation and adding to the first, the result is:
C(s) + H2 O(g)
CO(g) + H2 (g)
GÆ = +24 kJ mol-1
At 800 K the formation of water gas is not thermodynamically feasible or spontaneous.
The Ellingham diagram provides a simple clear picture of the relationship between the
different reactions and allows prediction of the conditions under which combinations of
individual reactions become feasible.
In any Ellingham diagram, the reaction with the most negative G value at a chosen
temperature will operate as written. Any reaction with an equation above this value can
be reversed. The point of intersection of any two lines shows the temperature at which
G is zero and the overall reaction becomes feasible. In general, the lower of two lines
drawn goes ’as written’.
7.5.1
Extraction of metals
Learning Objective
Æ
To show that Ellingham diagrams can be used to predict the conditions required to
extract a metal from its oxide
Ellingham diagrams plot values of G Æ against temperature. If the lines are drawn for
metal oxide formation reactions, these can be used to predict the conditions required to
extract a metal from its oxide. This requires the formation of the metal oxide process
to be reversed. Any chemical used to aid the reversing of this process must provide
enough free energy to supply this reversal. It is normal to write all reactions that are on
the graph to involve one mole of oxygen (so that oxygen is removed when two equations
are combined).
Interpreting Ellingham diagrams
Learning Objective
20 min
Æ
After this activity you should be able to use Ellingham diagrams to predict the
conditions required to extract a metal from its ore.
An interpretation exercise using an Ellingham diagram and answering questions that
lead to an understanding of metal ore extraction.
Answer these questions on paper before displaying the explanation. In each case refer
to the Ellingham diagram in Figure 7.11
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7.5. ELLINGHAM DIAGRAMS
141
Figure 7.11: Ellingham diagram
Q27: Which oxide (in Figure 7.11) could be broken down by heat alone at 1000 K?
(Hint: at 1000 K the G value of the reversed reaction needs to be negative.)
Q28: Above which temperature would the breakdown of zinc oxide become feasible by
heat alone?
Q29: Use the graph Figure 7.11 to calculate the
carbon reduces zinc oxide at:
GÆ value for the reaction in which
a) 1000 K
b) 1500 K
2C(s) + 2ZnO(s)
2Zn(s) + 2CO(g)
Try each calculation for a) and b) on paper by following this route.
• (i) Write down the target equation.
• (ii) Write the equations for carbon combustion and zinc combustion, along with their
GÆ values from the graph at 1000 K in Figure 7.11.
• (iii) Write the reversed equation for the zinc combustion remembering to reverse
GÆ.
• (iv) Add this new equation to the carbon equation and note the sign on G Æ . Is the
reaction feasible or not?
Q30: At what temperature does the reduction of zinc oxide by carbon become feasible?
See further questions on page 204.
Try the next two questions yourself, on paper.
Q31: This Ellingham diagram, Figure 7.12, shows the reactions involved in the blast
furnace reduction of iron(II) oxide with carbon.
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TOPIC 7. REACTION FEASIBILITY
-200
O2
® 2C
O2
+
2CO
2FeO
O2®
2Fe +
G0
-1
kJ mol
-400
2C
+O
2®
2CO
-600
500
1000
1500
2000
Temperature / K
Figure 7.12: Ellingham diagram for the reduction of iron(II) oxide
a) Write the combined equation showing the reduction of iron(II) oxide by carbon at 1500
K.
b) Calculate the standard free energy change at this temperature.
c) At what temperature does the reduction of iron(II) oxide by carbon become feasible?
d) At what temperatures will it be thermodynamically feasible for carbon monoxide to
reduce iron(II) oxide?
e) Can you suggest a reason (apart from the temperature) why carbon monoxide might
be more efficient than carbon at reducing iron(II) oxide?
Q32: Although magnesium ores are very abundant in the Earth’s crust, the very high
reactivity of magnesium makes it difficult to extract the metal. During the Second World
War, magnesium was manufactured by reduction of its oxide by carbon.
2MgO + 2C
2CO + 2Mg
Examine the Ellingham diagram, Figure 7.13, and answer the questions which follow.
Figure 7.13: Ellingham diagram
a) In what temperature range is the above process thermodynamically feasible?
© H ERIOT-WATT U NIVERSITY
7.6. SUMMARY
143
b) Describe two problems that the operation of the process at this temperature would
present.
c) Use the Ellingham diagram to calculate G Æ for the production of magnesium in the
following:
2MgO + Si
« SiO
+ 2Mg
at 1500 K
In industry, the extraction of magnesium from magnesium oxide using silicon involves
two modifications.
2
(i) A mixture of calcium oxide and magnesium oxide is used, and the calcium oxide
reacts with the silicon oxide produced.
CaO + SiO2
CaSiO3
(GÆ = -92 kJ mol-1 at 1500 K)
(ii) The gaseous magnesium formed is continuously removed from the reaction mixture.
Use this information to answer the questions below.
d) Calculate GÆ for the reaction:
CaSiO 3 + 2Mg
CaO + 2MgO + Si
at 1500 K
e) Explain why the removal of magnesium from the reaction mixture helps the process.
7.6
Summary
• The entropy of a system is the degree of disorder of the system. This increases
with temperature, with the sharpest increases at the changes of state. Standard
entropies (SÆ ) are quoted at 298 K and 1 atmosphere pressure. A perfect crystal
has zero entropy at zero Kelvin.
• Calculations of the standard entropy change for a reaction can be made from the
standard entropy values for reactants and products:
So
S
SoPRODUCTS
o
REACTANTS
• Spontaneous changes involve an increase in the total entropy for a process that
is the sum of the entropy changes for the system and its surroundings, given by:
STOTAL
SSYSTEM SSURROUNDINGS
• The entropy change in the surroundings can be calculated from:
SSURROUNDINGS
TH
• The free energy change (Gibbs free energy change)
accounting for the surroundings as well as the system.
Go
Ho
TS
G
provides a way of
o
Calculations of the standard free energy change for a reaction can be made by
using standard enthalpy and standard entropy values in this equation.
© H ERIOT-WATT U NIVERSITY
144
TOPIC 7. REACTION FEASIBILITY
• The standard free energy change in a reaction can also be calculated from a
knowledge of the standard free energy of formation of the reactants and products
in this equation:
Go GoPRODUCTS GoREACTANTS
A negative value for G indicates the thermodynamic feasibility of a reaction.
It does not, however, give any information about the rate of reaction.
Sometimes reactions which have a positive G value can be made feasible by
changing the reaction conditions, particularly temperature. The temperature at
which a reaction becomes feasible can be calculated from values of H Æ and
SÆ.
• Reactions proceed spontaneously in a forward direction until the composition
where G = 0 is reached, and at this point the system is in equilibrium.
• The variation of G with temperature can be plotted onto graphs called Ellingham
diagrams. These graphs can be used to predict the conditions under which
reactions can occur, and in particular, to predict conditions necessary to extract
metals from their ores.
7.7
Resources
• Chemistry in Context: Hill and Holman, Nelson, ISBN 0-17-438401-7
• Chemistry: A. and P. Fullick, Heinemann, ISBN 0-435-57080-3
• General Chemistry: Ebbing, Houghton Mifflin, ISBN 0-395-74415-6
• Chemistry - Advanced Higher: Unit 2: Principles of Chemical Reactions
Higher Still Support, ISBN 1-85955-874-7
• Chemical Storylines: Salters Advanced Chemistry, Heinemann
ISBN 0-435-63106-3
• Chemical Ideas: Salters Advanced Chemistry, Heinemann, ISBN 0-435-63105-5
7.8
End of topic test
An online assessment is provided to help you review this topic.
© H ERIOT-WATT U NIVERSITY
145
Topic 8
Electrochemistry
Contents
8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.2 Electrochemical cells . . . . . . . . . . . . . . . . . . . . . . . .
8.2.1 Shorthand cell conventions for electrochemical cells . .
8.3 Standard Electrode Potentials . . . . . . . . . . . . . . . . . . .
8.3.1 The Standard Hydrogen Electrode . . . . . . . . . . . .
8.3.2 The Electrochemical Series and associated calculations
8.4 EÆ values and the standard free energy change . . . . . . . . .
8.5 Fuel cells . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.7 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.8 End of Topic test . . . . . . . . . . . . . . . . . . . . . . . . . .
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147
147
151
154
155
158
162
166
169
170
170
Prerequisite knowledge
Before you begin this Topic, you should be able to:
• define and use the terms reduction, oxidation and redox;
• write ion-electron half equations and combine these to produce redox equations;
(Higher, Unit 1)
• derive balanced ion-electron half equations which contain H + ions and H2 O
molecules; (Higher, Unit 1)
• define the terms reducing agent and oxidising agent and identify these in redox
reactions; (Higher, Unit 1)
• understand what is meant by the standard free energy change (G Æ ). (Topic 7)
Learning Objectives
At the end of this Topic, you should be able to:
• describe the make-up of an electrochemical cell using the IUPAC cell convention;
• state what is meant by the cell e.m.f. and describe the factors that affect its
magnitude;
146
TOPIC 8. ELECTROCHEMISTRY
• explain what is meant by the standard electrode potential and explain how such
values are measured;
• calculate the e.m.f. of a cell under standard conditions (E Æ ) and relate this to the
standard free energy change (G Æ );
• explain how fuel cells are different from electrochemical cells and give some
possible benefits of their use.
© H ERIOT-WATT U NIVERSITY
8.1. INTRODUCTION
8.1
147
Introduction
Electrochemistry is concerned with chemical reactions in which electrons are transferred
from one reactant to another (redox reactions) and with the relationships between the
electron transfer and the electron currents generated or used in the process. This Topic
concentrates on reactions which generate electricity. These reactions spontaneously
release energy and so must be accompanied by a decrease in standard free energy
(i.e. GÆ must be negative).
Table 8.1: Uses of batteries
Courtesy of International Fuel Cells.
The applications of these reactions range from the lead-acid batteries in cars to the tiny
silver oxide or lithium batteries that power heart pacemakers, and from the humble torch
battery to the fuel cells which provide electricity for the Space Shuttle (Table 8.1).
Future applications may include the production of hydrogen by a photochemical process
or the power source in environmentally friendly cars.
8.2
Electrochemical cells
Learning Objective
Æ
To explain how an electrode potential arises and how two electrodes can be combined
to produce an electrical cell
© H ERIOT-WATT U NIVERSITY
148
TOPIC 8. ELECTROCHEMISTRY
When zinc metal is placed in contact
with a solution containing Zn 2+ (aq)
ions, a few of the metal atoms ionise.
This leaves electrons on the metal
that can be accepted by other Zn 2+
ions.
Figure 8.1: Zinc metal electrode
An equilibrium is set up: Zn(s)
« Zn
2+ (aq)
+ 2e-
For zinc, this equilibrium lies slightly
towards the ions.
Thus there is
an increase in the positive charge
in the solution and an increase in
the negative charge on the metal.
In other words, the metal becomes
slightly negative with respect to the
solution. There is also a very slight
decrease in mass of the electrode.
Figure 8.2: Zinc metal electrode at equilibrium
With copper metal and a solution of copper(II) ions, the same processes are possible
and an equilibrium is established:
Cu(s)
« Cu
2+ (aq)
+ 2e© H ERIOT-WATT U NIVERSITY
8.2. ELECTROCHEMICAL CELLS
149
In this case however, the equilibrium lies slightly towards the atoms. Electrons are
removed from the metal to reduce some of the copper(II) ions.
Q1: Will the copper metal be positive or negative with respect to the solution?
Q2: There will be a very slight change in mass. Will the copper metal gain or lose
mass?
a) gain
b) lose
Q3: Which of the two metals is more reactive?
a) zinc
b) copper
Q4: Before revealing the answer, write on paper a general statement linking the
reactivity of a metal with the equilibrium between the metal and its ions.
In both cases, there is a slight difference in charge between metal and solution and so
a potential difference is set up. This is known as the electrode potential.
Learning Point
In general, a potential difference is set up whenever a metal is placed in contact with
its ions in solution.
If zinc metal is placed in a solution containing copper(II) ions, a displacement reaction
takes place because zinc atoms have a tendency to lose electrons and copper(II) ions
have a tendency to gain electrons.
Zn(s) + Cu2+ (aq)
Zn2+ (aq) + Cu(s)
The reaction is spontaneous and exothermic.
Q5: Are the copper(II) ions
a) reduced?
b) oxidised?
Q6: Are the copper(II) ions acting as
a) an oxidising agent?
b) a reducing agent?
Q7: Describe the reaction in terms of electrons.
The same overall reaction can take place in an electrochemical cell (Figure 8.3) but the
energy is not released as heat but instead mostly as electricity. This is because the
two reactants are kept separate and electron transfer takes place through an external
connecting wire.
© H ERIOT-WATT U NIVERSITY
150
TOPIC 8. ELECTROCHEMISTRY
Figure 8.3: Zn-Cu cell
Zinc - copper cell.
15 min
A simulation is available in the on-line version of this Topic showing how the zinc-copper
cell produces electricity and some questions to check understanding.
Use the diagram in Figure 8.3 to answer the following questions.
Q8:
Write the ion-electron equation for the process taking place at the zinc electrode.
Q9: Write the ion-electron equation for the process taking place at the copper
electrode.
Q10: What is moving as the electric current in the wire?
Q11: What is the purpose of the salt bridge?
See further questions on page 204.
An electrochemical cell is composed of two half-cells between which electrical contact
is made by an electrolyte, often in the form of a salt bridge (ion bridge). Such a cell is
able to produce an electric current because there is a potential difference between the
two electrodes.
The electrode at which oxidation occurs is called the anode. In the above cell, the zinc
electrode is the anode because the reaction that occurs is:
Zn(s)
Zn2+ (aq) + 2e© H ERIOT-WATT U NIVERSITY
8.2. ELECTROCHEMICAL CELLS
151
The electrode at which reduction occurs is called the cathode. In the above cell, the
copper electrode is the cathode because copper(II) ions are reduced to copper atoms.
Cu2+ (aq) + 2e-
Cu(s)
When no current is drawn, the electric potential difference between the two electrodes
is known as the electromotive force or e.m.f. of the cell. This is given the symbol, E,
and is measured in volts. The e.m.f. (or voltage) of a cell is a measure of the cell’s ability
to push electrons around the external circuit.
8.2.1
Shorthand cell conventions for electrochemical cells
Learning Objective
Æ
To use the cell convention. To work out the overall cell reaction and to describe a cell
using the cell notation given the cell reaction
There is a standard shorthand notation that is used to represent an electrochemical cell.
The zinc/copper cell would be represented by the following expression:
Zn(s) Zn2+ (aq)
Cu
2+ (aq)
Cu(s)
• The anode is always written to the left and the cathode to the right.
• The single vertical line represents a phase boundary (a boundary between two
different states, i.e. solid and solution).
• The double vertical line represents the junction between the two solutions (i.e. the
salt bridge).
When written in this way, the electrons flow from the electrode on the left to the electrode
on the right. This is the spontaneous direction for the cell and the e.m.f. will always be
positive for this direction.
electron flow
phase boundary
Zn(s)
Zn2+(aq)
Cu2+(aq)
Anode half-cell
Cu(s)
Cathode half-cell
Salt bridge
Figure 8.4: An electrochemical cell
Sometimes one of the reactants in the redox reaction is a gas or an ion in solution.
In such cases, an inert electrode such as carbon or platinum is used to make contact
between the external circuit and the solution, e.g.
© H ERIOT-WATT U NIVERSITY
152
TOPIC 8. ELECTROCHEMISTRY
Figure 8.5 represents a half-cell in which hydrogen
gas is oxidised to hydrogen ions on a carbon rod.
Figure 8.5
Figure 8.6
Figure 8.6 represents a half-cell in which chlorine
gas is reduced to chloride ions on a platinum
electrode.
Figure 8.7
Figure 8.7 represents a half-cell in which iron(II)
ions are oxidised to iron(III) ions at a platinum
electrode.
In Figure 8.5 and Figure 8.6, note the extra phase boundary between the solution and
gas. In Figure 8.7, note the comma between Fe 2+ (aq) and Fe3+ (aq) since these are both
in the same phase (both in solution).
A word of warning: Sometimes a cell diagram may be drawn with the anode to the right
of the diagram. However, no matter which way round the diagram is, the cell notation
must have the anode on the left. A useful memory aid is that the anode and cathode are
in alphabetical order, i.e anode (to the left) before cathode (to the right).
Example : Information from cell notation
The following questions refer to this electrochemical cell:
Mg(s) Mg2+ (aq)
Ag (aq) Ag(s)
+
Write a balanced equation for the overall cell reaction. First, you will need to work out
ion-electron half equations for each electrode process.
Oxidation half equation
Oxidation occurs at the anode. Anode is always to the left hand side.
Mg2+ (aq)
i.e. Mg(s)
Oxidation - electrons as products.
so, Mg(s)
Mg2+ (aq) + 2e- .
Reduction half equation
Reduction occurs at the cathode. Cathode is always to the right hand side.
i.e. Ag+ (aq)
Ag(s)
Reduction - electrons as reactants.
so, Ag+ (aq) + e-
Ag(s)
Combining half equations
Add the half equations together so that the electrons cancel out.
Double the silver equation:
© H ERIOT-WATT U NIVERSITY
8.2. ELECTROCHEMICAL CELLS
153
2Ag+(aq) + 2e-
2Ag(s)
Mg2+ (aq) + 2e-
Mg(s)
Add the two equations:
Mg(s) + 2Ag+ (aq)
Mg2+ (aq) + 2Ag(s)
For each of the following questions, write your answer on paper before revealing the
correct answer.
Write balanced equations for the following electrochemical cells:
Zn (aq) Zn(s)
Q13: Cr(s) Cr (aq) Cu (aq) Cu(s)
Q14: Pt(s) H (g) H (aq) Cl (g) Cl (aq) Pt(s)
Q15: Ni(s) Ni (aq) H (g) H (aq) Pt(s)
Q12: Mg Mg2+ (aq)
2+
3+
2
2+
+
+
-
2
+
2
See further questions on page 204.
Example : Writing cell notations
Write the cell notation for a cell containing copper in a solution of copper (II) ions
connected to silver in a solution of silver ions.
Copper is more reactive than silver and electrons will flow from copper to silver.
1. Write the oxidation half cell to the left.
Copper is oxidised:
(aq) Cu(s) Cu2+ (aq)
2. Add the salt bridge: Cu(s) Cu 2+
3. Write the reduction half cell to the right.
Silver ions are reduced to silver:
Ag+ (aq) + e-
Ag(s)
i.e. Ag+ (aq) Ag(s)
4. Add this to the right hand side to complete the cell:
Cu(s) Cu2+ (aq) Ag+ (aq) Ag(s)
In each of the following questions, write the cell notation for the cell that is described.
Write your answers on paper before revealing the answers.
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TOPIC 8. ELECTROCHEMISTRY
Q16: Write the cell notation for a cell in which zinc metal in zinc(II) sulphate solution is
connected to magnesium metal in magnesium sulphate solution.
Q17: Write the cell notation for a cell in which the following reactions occur:
Ni2+ (aq) + 2e-
Ni(s)
Pb2+ (aq) + 2e-
Pb(s)
Q18: Write the cell notation for a cell in which copper is oxidised to Cu 2+ ions by chlorine
gas on a platinum electrode in a solution of chloride ions.
Q19: Write the cell notation for a cell in which the following reactions occur:
H2 (g)
2H+ (aq) + 2e-
Hg2+ (aq) + 2e-
Hg(l)
See further questions on page 204.
8.3
Standard Electrode Potentials
Learning Objective
Æ
To outline the use of the standard hydrogen electrode in determining absolute values
of electrode potentials.
The cell e.m.f. (E) is the electrical potential difference between the two electrodes when
no current is drawn and so depends on the electrode potential of each electrode.
Clearly the size of the cell e.m.f. depends on the nature of the half-reactions taking
place.
The individual electrode potentials depend on the position of the equilibrium between
the metal and its ions. i.e.
« Zn
Cu(s) « Cu
Zn(s)
2+ (aq)
+ 2e-
2+ (aq)
+ 2e-
The position of these equilibria, as with all equilibria, will depend on the temperature
and the concentrations of the species involved.
Learning Point
The cell e.m.f. depends on the concentration, the temperature and the type of cell
(i.e. the reaction taking place in the cell).
In order to compare the e.m.f. of different cells, the measurements must be made
under the same conditions. A set of standard conditions have been defined for the
measurement of a cell e.m.f.
Pressure of any gas -
1 atmosphere
Concentration of solutions Temperature -
1 mol -1
298 K (25Æ C)
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155
These are the same standard conditions used for the measurements of thermodynamic
quantities in Topics 6 and 7 Unit 2.
The absolute value of the electrode potential of a half cell cannot be determined
experimentally since by its definition redox reactions require two half reactions. It is
impossible to have an oxidation without a reduction. It is only possible to measure the
overall cell voltage.
This problem is overcome by using a reference standard electrode.
8.3.1
The Standard Hydrogen Electrode
If one half-cell reaction is chosen as a standard and arbitrarily assigned a value of
zero, then a number of cells can be set up using this standard electrode. The voltage
measured for any such cell can then be assigned to the other half-cell reaction.
The standard chosen is the standard hydrogen electrode. (Figure 8.8).
Figure 8.8: Standard hydrogen electrode
A
Hydrogen gas at 298 K
and 1 atmosphere
pressure.
B
A glass tube with holes
to allow hydrogen gas to
escape.
C
Platinum foil coated with D
platinum black (very fine
powdered platinum).
A solution containing
ions
of
H+ (aq)
concentration 1 mol
-1
The half-cell reaction is:
2H+ (aq) + 2e-
« H (g)
2
and the standard hydrogen electrode is given an arbitrary value of 0.00 V.
If a cell is set up under standard conditions using a hydrogen electrode and copper
electrode, a voltage of +0.34 V is recorded (Figure 8.9). This is measured using a high
resistance voltmeter to ensure that only a tiny current is drawn.
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TOPIC 8. ELECTROCHEMISTRY
Figure 8.9: Copper-hydrogen cell
This voltage can be assigned as the standard electrode potential for the reaction
occurring at the copper electrode.
But what reaction is taking place - oxidation of copper atoms or reduction of copper (II)
ions? In this case, from the meter connections, we can deduce that electrons flow from
the hydrogen, through the meter, to the copper.
Q20: Which electrode is the anode?
a) Copper electrode
b) Hydrogen electrode
Q21: Write the ion-electron equation for the anode half-reaction.
Q22: Write the ion-electron equation for the cathode half-reaction.
Q23: Write the cell notation for this cell.
The reaction occurring at the copper electrode is a reduction. So under standard
conditions, we can say that the standard reduction potential of copper is +0.34 V.
By convention, electrode potentials are quoted as standard reduction potentials and
are given the symbol, EÆ .
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8.3. STANDARD ELECTRODE POTENTIALS
So,
Cu2+ (aq) + 2e-
157
EÆ = +0.34 V
Cu(s)
If a similar cell is set up using a zinc electrode (Figure 8.10), the electron flow is from
zinc to hydrogen and the measured voltage under standard conditions is +0.76 V.
Figure 8.10: Zinc-hydrogen cell
Q24: Which electrode is the anode?
a) Zinc electrode
b) Hydrogen electrode
Q25: Write the ion-electron equation for the anode reaction.
Q26: Write the ion-electron equation for the cathode reaction.
Q27: Write a balanced equation for the overall cell reaction.
See further questions on page 204.
Summarising:
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« Zn(s)
2H (aq) + 2e « H (g)
Cu (aq) + 2e « Cu(s)
Zn2+ (aq) + 2e+
-
2+
-
2
EÆ = -0.76 V
EÆ = 0.00 V
EÆ = +0.34 V
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TOPIC 8. ELECTROCHEMISTRY
8.3.2
The Electrochemical Series and associated calculations
Learning Objective
To use the Electrochemical Series to estimate the relative strengths of oxidising
agents and reducing agents. To use standard reduction potentials to calculate cell
e.m.f. and predict the direction of a redox reaction
Æ
The standard reduction potentials of a very large number of half-cells have been
measured against the standard hydrogen electrode. These values have been organised
into a list called the Electrochemical Series, part of which is shown on page 11 of the
Data Booklet.
Usually, the reduction potentials are arranged in numerical order, with the lowest (most
negative) values at the top and the highest (most positive) values at the bottom.
A high positive value means that the half-reaction is likely to go as written. A high
negative value means that the half-reaction is likely to go in the opposite direction.
Note that all half-reactions are reversible.
In an electrochemical cell, one half-reaction will go as written (the reduction half) and the
other will go in the opposite direction (the oxidation half). Since an electrode potential
can be assigned to both half-reactions, the e.m.f. of the cell can be calculated.
Examples
1. Calculation of e.m.f.
Calculate the e.m.f. under standard conditions of the zinc-copper cell.
Zn(s) Zn2+ (aq)
Cu
2+ (aq)
Cu(s)
Anode - oxidation:
Zn(s)
Zn2+ (aq) + 2e-
Reduction Potential (E Æ ) = -0.76 V
Reversing the sign gives +0.76 V
µ the zinc contribution is +0.76 V.
Cathode - reduction:
Cu2+ (aq) + 2e-
Cu(s)
Reduction Potential (E Æ ) = +0.34 V
Reaction goes as written
µ the copper contribution is +0.34 V.
The cell e.m.f. is the sum of the two electrode potentials
i.e. +0.76 + 0.34 = 1.10 V
The calculated cell e.m.f. is 1.10 V
2. A quicker way
Calculate the e.m.f. under standard conditions of a magnesium-copper cell.
Mg(s) Mg2+ (aq) Cu2+ (aq) Cu(s)
EÆ (cell) = EÆ (right) - EÆ (left)
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159
where EÆ (right) and EÆ (left) are the standard reduction potentials.
In this cell, EÆ (right) is the reduction potential of copper, i.e. +0.34 V.
EÆ (left) is the reduction potential of magnesium, i.e. -2.37 V.
So, EÆ (cell) = +0.34 V - (-2.37 V) = +2.71 V
So, the calculated e.m.f. = +2.71 V
You can now gain practice at these calculations. For each of the following cells, calculate
the e.m.f. in volts, assuming standard conditions (include the sign but not the units).
Q28: Mg(s) Mg2+ (aq)
Zn
2+ (aq)
Zn(s)
Q30: Pt(s) H (g) H (aq) Cl (g) Cl (aq) Pt(s)
Q31: Ni(s) Ni (aq) Br (aq), Br (aq) Pt(s)
Q29: Cr(s) Cr3+ (aq) Cu2+ (aq) Cu(s)
2
2+
+
-
2
2
-
See further questions on page 204.
The table of standard reduction potentials (E Æ values) on page 11 of the Data Booklet
can also be used to estimate the relative strengths of reducing agents and oxidising
agents. It must be emphasised that E Æ values refer to standard conditions. If the
conditions are significantly different, then any predictions based on E Æ values may be
wrong.
Oxidising agents and reducing agents
A series of questions on oxidising and reducing agents designed to familiarise students
with the Electrochemical Series and enable them to make generalisations about the
relative strengths of oxidising agents and reducing agents.
Look at the Electrochemical Series on page 11 of the Data Book and use the information
to answer the following questions. Write your answers on paper before displaying the
answer.
Remember that the more positive the value for a standard reduction potential, the more
likely the reaction is to go as a reduction.
Q32: Which of the following reductions is the easiest?
a)
b)
c)
d)
Na(s)
Na+ (aq) + e2F- (aq)
F2 (g) + 2e
2Br- (aq)
Br2 (g) + 2e
2+
Cu(s)
Cu (aq) + 2e
Q33: In the Electrochemical Series shown, where will the easiest reduction be found?
a) at the top of the table
b) at the bottom of the table
Q34: What is meant by an oxidising agent?
Q35:
In the reduction,
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TOPIC 8. ELECTROCHEMISTRY
F2 (g) + 2e-
2F-
is F2 an oxidising agent or a reducing agent?
See further questions on page 204.
Remember that the more negative the value for a standard reduction potential, the more
likely the reaction is to go as an oxidation.
Q36: Where in the Electrochemical Series in the data booklet will you find the reaction
which is most likely to go in the opposite direction?
a) top of the table
b) bottom of the table
Q37: What is meant by a reducing agent?
Q38: On which side of an equation will you find the reducing agent?
Q39: Where in the table will the best reducing agents be found.
a)
b)
c)
d)
Top left
Top right
Bottom left
Bottom right
See further questions on page 204.
All the equations in the Electrochemical Series are written in the form:
Oxidising Agent + electrons
« Reducing Agent.
It follows that the best oxidising agents are found at the bottom of the series on the left
hand side, and the best reducing agents are found at the top and on the right hand side.
The diagram in Figure 8.11 is a useful summary.
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8.3. STANDARD ELECTRODE POTENTIALS
Figure 8.11: Redox summary
It can be used in the following set of questions which will give you more practice in
identifying and comparing oxidising agents and reducing agents.
If you feel comfortable with this concept, you may only need to do the first two or three
questions. If not, do them all.
Q40: From the grid Figure 8.12, enter the letter of the box containing the best oxidising
agent.
Figure 8.12: Answer grid
Q41: From the grid Figure 8.12, enter the letter of the box containing the best reducing
agent.
Q42: From the grid Figure 8.12, enter the letter of the substance that could be used as
an oxidising agent or a reducing agent.
Q43: From the grid Figure 8.13, enter the letter of the box containing the best oxidising
agent.
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TOPIC 8. ELECTROCHEMISTRY
Figure 8.13: Answer grid
Q44: From the grid Figure 8.13, enter the letter of the box containing the best reducing
agent.
See further questions on page 205.
It also follows that a half-reaction will be able to force any half-reaction above it in the
series to go in reverse (i.e. as an oxidation). This can be described as the ’Anticlockwise
Rule’.
Figure 8.14: The ’anticlockwise’ rule
Any pair of half-reactions combined in this way will produce a positive value for the cell
e.m.f. Under standard conditions, the reaction will be spontaneous in this direction. E Æ
values therefore provide a means for predicting the direction of a redox reaction.
A positive EÆ value is obtained if the reaction takes place in the direction written. The
standard free change (G Æ ) has to be negative for a reaction to do this. These two
quantities are connected.
8.4
EÆ values and the standard free energy change
Learning Objective
Æ
To define the G Æ in terms of the cell e.m.f. and use cell e.m.f to calculate GÆ
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8.4. EÆ VALUES AND THE STANDARD FREE ENERGY CHANGE
Consider a zinc-copper cell under standard conditions. The cell e.m.f. (E Æ ) when no
current is drawn is +1.10 V. The positive sign means that the reaction is spontaneous in
the direction:
Zn(s) + Cu2+ (aq)
« Zn
2+ (aq)
+ Cu(s)
i.e. the equilibrium lies in favour of the products.
A negative value for GÆ indicates the same. In any cell, the amount of energy available
is going to depend upon the E Æ value and the relationship is likely to have the form:
GÆ = -(constant) x EÆ
The constant in any reaction system will depend on the number of electrons involved in
the cell redox exchange and on the charge that each electron carries, i.e.
GÆ = -nFEÆ
In the zinc-copper cell, using one mole of each (under standard conditions), 2 moles of
electrons are exchanged. Each mole of electrons carries 96500 coulombs of charge
(the Faraday constant).
For the zinc-copper cell, the potential difference is +1.10 V.
In the cell reaction, two moles of electrons are transferred. So,
GÆ = -2 x 1.10 x 96500 J
This would represent the maximum amount of energy available if one mole of zinc
reacted in the cell under thermodynamically reversible conditions. For this reaction,
G Æ =
-2 x 1.10 x 96500 J mol-1
= -212.3 kJ mol-1
For any cell, operating under conditions of thermodynamic reversibility, the standard free
energy for the cell reaction is given by the expression:
Figure 8.15: Relationship between G and E Æ
Example : Calculation of G Æ from cell e.m.f.
Calculate GÆ for the nickel-silver cell.
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TOPIC 8. ELECTROCHEMISTRY
1) Calculate the cell e.m.f.
Write the two half-equations with their E Æ values.
« Ni(s)
Ag (aq) + e « Ag(s)
EÆ = -0.23 V
Ni2+ (aq) + 2e+
EÆ = +0.80 V
-
Using the Anticlockwise rule, the silver half-reaction will be able to force the nickel halfreaction into reverse. The nickel contribution to the e.m.f. will become +0.23 V.
cell e.m.f. = 0.80 + 0.23 V
= +1.03 V
2) Work out the number of moles of electrons transferred.
Ni2+ (aq) + 2e-
Oxidation
Ni(s)
Reduction
Ag+ (aq) + e-
Ag(s)
Doubling the reduction equation and combining (so that the electrons cancel out) gives:
Ni(s) + 2Ag+ (aq)
Ni2+ (aq) + 2Ag(s)
So each mole of nickel transfers two moles of electrons.
i.e.
3) Calculate GÆ :
-
. /
* * )* Note: Although you need to double the silver half-equation to work out the overall cell
reaction, you must not double the E Æ value for silver when calculating the cell e.m.f.
The cell voltage is not proportional to the amount of material present.
Calculation of standard free energy change.
A set of questions to give practice at calculating GÆ from the cell e.m.f.
20 min
Calculate the standard free energy change in each of the following questions. Give your
answer in kJ mol-1 to 4 significant figures (and remember to include the sign).
Q45: A silver oxide-zinc battery used in a hearing aid delivers a voltage of 1.60 V. The
cell reaction is:
Zn(s) + Ag2 O(s)
ZnO(s) + 2Ag(s)
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8.4. EÆ VALUES AND THE STANDARD FREE ENERGY CHANGE
165
Q46: A lead-acid storage cell has a standard cell potential of 1.924 V. The anode
reaction is:
Pb(s)
Pb2+ (aq) + 2e-
Six such cells are connected in series in a car battery.
Q47: Some rechargeable lithium batteries deliver 3.0 V.
Li(s)
Li+ (aq) + e-
They are used in calculators, watches and cameras because of their light weight and
high voltage.
Q48: Ni(s) Ni2+ (aq)
Cu
2+ (aq)
Cu(s)
See further questions on page 205.
Calculating standard entropy changes.
Description of an experiment which enables H Æ to be calculated. Calculation of
allows SÆ for the reaction to be calculated.
GÆ
The diagram below shows an experiment in which solid copper is added to silver (I)
nitrate solution.
Use the information provided in Figure 8.16, to answer the questions which follow.
Figure 8.16: What happens next?
1)
Copper powder:
0.635 g
2)
Excess silver (I) nitrate: 5)
Volume of solution = 25 cm 3
3)
mass = 4)
Initial Temperature = 25 Æ C
6)
Final Temperature = 39 Æ C
Blue
solution
2+
Cu (aq) ions.
containing
Precipitate of silver metal.
Q49: What is the relationship between the enthalpy change and the temperature rise?
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TOPIC 8. ELECTROCHEMISTRY
Q50: Use this relationship to calculate the energy released in this experiment.
Q51: Now use the mass of copper to calculate the standard enthalpy change,
kJ mol-1
H Æ in
Q52: The same reaction can be carried out in an electrochemical cell. Calculate the
cell e.m.f. under standard conditions, in volts. Remember to include the sign.
See further questions on page 205.
8.5
Fuel cells
Learning Objective
Æ
To be able to describe similarities and also the main difference between fuel cells and
conventional electrochemical cells
Like batteries, fuels use redox reactions to generate electricity. In a battery, the chemical
energy is stored in the materials that make up the battery. When one of the reactants is
used up, the battery stops working and must be recharged or replaced. In a fuel cell, the
reactants are continually fed into the cell. The fuel is passed into the anode and oxygen
(or air) into the cathode. As long as this supply continues, the fuel cell keeps producing
electricity.
The simplest fuel cells use hydrogen as the fuel but methane, ethanol and even petrol
or diesel fuel cells are also possible.
Figure 8.17 shows a diagram of one type of fuel cell that uses hydrogen as the fuel.
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8.5. FUEL CELLS
167
Figure 8.17: Fuel cell
At the anode, hydrogen gas is oxidised at the platinum electrode to form protons (H +
ions) and electrons. Electrons move round the external circuit and the protons migrate
through the electrolyte to the cathode where they react with oxygen and electrons on
the platinum electrode. Comparison of the fuel cell (Figure 8.17) with the zinc-copper
cell ( Figure 8.3) shows similarities.
Q53: Which part of the fuel cell performs the same function as the salt bridge?
Q54: Describe the fuel cell using standard cell notation.
Q55: Predict the cell e.m.f. under standard conditions. (Type the value without the sign
or units)
Q56: Write the overall cell equation.
See further questions on page 205.
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TOPIC 8. ELECTROCHEMISTRY
Table 8.2: Fuel cells for transport
Courtesy of Ballard Power Systems and Courtesy of Mazda and Breakthrough
Breakthrough Technologies Institute, Inc. Technologies Institute, Inc.
Fuel cells hold a great deal of promise for the future, although many problems still remain
to be solved. This type of fuel cell would be the ideal solution for cutting down, if not
eliminating, car and bus exhaust emissions (Table 8.2).
If hydrogen is to be used as the fuel, the obvious source would be water. Extracting
hydrogen from water is the first problem. Electrolysis, even using renewable sources of
energy, still remains expensive. Photochemical processes and biological methods are
being investigated. Storage of the hydrogen also remains a safety problem. Liquefaction
of hydrogen is difficult due to the very low boiling point. Absorption onto metals as
metal hydrides is promising but still expensive, as is absorption by carbon nanotubes.
However, there is optimism that these problems will be overcome.
Table 8.3: Methanol fuel cells
The use of other fuels such as
methanol is promising, although CO 2
will still be a product. Indeed a
number of car manufacturers are
developing cars powered by a
methanol fuel cell with performance
approaching that of petrol-engined
cars.
Courtesy of Opel and Breakthrough
Technologies Institute, Inc.
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8.6. SUMMARY
169
Table 8.4: Power generation by fuel cells
Other types of fuel cell, although
unsuitable for vehicles, are in use for
power generation in hospitals, office
buildings and factories.
Courtesy of Ballard Power Systems and
Breakthrough Technologies Institute, Inc.
Fuel cells are no longer of interest only to NASA scientists. With continued research,
their use will spread.
Watch this space!!
8.6
Summary
• Electrochemical cells consist of two half-cells between which electrical contact is
made by an electrolyte, often in the form of a salt-bridge.
• Each half-cell generates a potential difference that causes electrons to move in
the external circuit. The overall cell voltage (e.m.f.) is the sum of the electrode
potentials of each electrode. The size of this voltage depends on the cell reaction,
temperature and concentrations.
•
An electrochemical cell can be described by a standard cell notation, e.g.
Zn(s) Zn2+ (aq)
Cu
2+ (aq)
Anode
Cu(s)
Cathode
• Individual electrode potentials cannot be measured but values are obtained by
comparison with the standard hydrogen electrode.
• Standard Reduction Potentials (E Æ values) can be used to:
a) calculate a cell e.m.f. under standard conditions.
b) estimate relative strengths of oxidising agents and reducing agents.
c) calculate standard free energy changes from the relationship:
G Æ = -nFEÆ.
• Fuel cells operate like electrochemical cells, the only difference being that the fuel
for the reaction is continually provided from an external source.
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8.7
Resources
• Chemistry: McMurry and Fay, Prentice Hall, ISBN 0-13-737776-2
• General Chemistry: Ebbing, Houghton Mifflin, ISBN 0-395-74415-6
• Chemistry in Context: Hill and Holman, Nelson, ISBN 0-17-438401-7
• Chemistry: Fullick and Fullick, Heinemann, ISBN 0-435-57080-3
• Higher Still Support: Advanced Higher Chemistry - Unit 2: Principles of Chemical
Reactions, Learning and Teaching Scotland, ISBN 1 85955 874 7.
• Chemical Ideas: Salters Advanced Chemistry, Heinemann, ISBN 0-435-63105-5
Websites:
http://chemistry.about.com/science/chemistry/msub26.htm
http://www.fuelcells.org/
8.8
End of Topic test
An online assessment is provided to help you review this topic.
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Topic 9
Kinetics
Contents
9.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
172
9.2 Measuring reaction rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.3 Rate equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
172
175
9.3.1 Order of reaction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.4 Reaction mechanisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
177
181
9.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.6 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
187
189
9.7 End of Topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
189
Prerequisite knowledge
Before you begin this Topic, you should be able to:
• list factors affecting the rate of a chemical reaction (i.e. concentration of reactants,
surface areas, temperature, catalysts); (Standard Grade)
• use simple collision theory to explain the effects of these factors on the rate.
(Higher, Unit 1)
Learning Objectives
After studying this Topic, you should be able to:
• explain what is meant by the overall order of a chemical reaction and the order
with respect to a given reactant;
• construct, interpret and use rate equations, in which k is the rate constant;
• explain what is meant by a reaction mechanism and its rate determining step.
172
TOPIC 9. KINETICS
9.1
Introduction
Chemical thermodynamics considers the starting point and finishing point of a chemical
process and asks the questions:
• Is the reaction feasible?
• How far will the reaction go?
At no stage does time appear in any of the thermodynamic equations. A feasible
(spontaneous) reaction may be very fast (e.g. neutralisation of an acid when mixed
with an alkali) or very slow (e.g. the rusting of iron).
Chemical kinetics asks the question:
• How fast does the reaction go?
If thermodynamics is about start and finish, kinetics is about what happens along the
way.
In industry, ’time is money’. It is obviously very important that chemicals are produced
as quickly and as efficiently as possible.
In this Topic, your knowledge of the factors which affect the rate of a reaction will be
put on a more mathematical basis. In particular, the Topic will investigate the effects of
changing the concentrations of reactants on the reaction rate. The information obtained
will be used to gain an insight into reaction mechanisms.
9.2
Measuring reaction rates
Learning Objective
Æ
To be able to choose appropriate methods to monitor the rate of a chemical reaction
During the course of any chemical reaction, the concentrations of reactants will decrease
and the concentrations of products will increase.
The rate of a reaction can therefore be defined either as the rate of consumption of a
reactant or as the rate of formation of a product, i.e.
0
! or
0
%!
or
0
or
0
%!
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173
The method used to measure the change in concentration of reactant or product
depends on the reaction being studied.
The most convenient methods involve monitoring a physical quantity which changes
during the course of the reaction, e.g.
Table 9.1: Physical properties suitable for measuring rate
The mass of the apparatus (when a gas is being released)
The volume of a gaseous product
The pH of a solution (when H + or OH- ions are used or produced)
The conductivity of a solution (when the number or nature of the ions present
changes)
The colour (when products and reactants have different colours)
Q1: Which of the above methods (Table 9.1) could not be used to monitor the rate of
this reaction:
CaCO3(s) + 2HCl(aq)
CaCl2(aq) + H2O(l) + CO2(g)
Q2: Which of the methods shown in (Table 9.1) could be used to monitor the rate of
the reaction between iodine and propanone.
I2(aq) + CH3COCH3(aq)
propanone
CH3COCH2I(aq) + H+(aq) + I-(aq)
iodopropanone
Change in concentration with time
In this activity, experimental data are used to find out what happens to the concentrations
of reactants and products and the reaction rate during the course of a reaction.
The reaction between bromine and methanoic acid is a convenient reaction to study.
Br2 (aq) + HCOOH(aq)
2Br - (aq) + 2H+ (aq) + CO2 (g)
The concentration of bromine can be monitored using a colorimeter and the volume of
carbon dioxide produced can also be measured. The following data were obtained from
an experiment.
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Table 9.2: Concentration and volume against time - sample data
Time / s
0
30
60
90
120
240
360
600
Concentration of Br 2 / mol
12.0 x 10-3
11.0x 10-3
10.1 x 10-3
9.3 x 10-3
8.6 x 10-3
6.4 x 10-3
4.8 x 10-3
3.7 x 10-3
-1
Volume of CO2 / cm3
0
22
42
59
75
123
158
191
Use the data in Table 9.2 to answer the following questions. Where appropriate, answer
on paper before revealing the answer.
Q3: What is the average rate, in cm 3 per second, of CO2 production over the first 30
seconds?
Q4: What is the average rate with respect to bromine over the first 30 seconds?
Remember to include units.
Q5: Use the data in Table 9.2 to calculate the average rate of reaction in cm 3 s-1
between 120 and 240 seconds with respect to carbon dioxide.
Q6: Calculate (to 2 significant figures) the rate with respect to bromine over the same
time period.
See further questions on page 205.
The reaction rate decreases as the reaction proceeds because the concentration of
reactants decreases.
It is difficult to measure precisely the rate of reaction at any given time (the instantaneous
rate), since the rate is constantly changing. It is possible to measure the average rate
over a particular time interval using the relationship:
0
Clearly the shorter the time interval, the closer we will get to the instantaneous rate.
Figure 9.1 shows the changes in concentration of reactant and product for a simple
reaction.
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9.3. RATE EQUATIONS
175
Figure 9.1: Reactant and product concentration with time
Line X shows the tangent to the product curve at time = t 1 .
The instantaneous rate at time = t 1 is given by the gradient of the tangent at that time,
i.e.
! !
0
! %!
!
The signs ensure that the rate is always positive. Line Y shows the tangent at time t
= 0. The gradient of this line will give the initial rate of the reaction. Since the initial
concentration of a reactant or product is accurately known, and since the reaction is
fastest at the start, initial rates of reaction are often used in kinetic studies.
9.3
Rate equations
Learning Objective
Æ
To be able to derive a rate equation from experimental data.
Figure 9.2 shows that the concentration of a reactant affects the rate of the reaction.
What is the mathematical relationship between concentration and rate?
To investigate this, it is necessary to carry out a number of experiments in which the
concentration is altered.
The following graphs (Figure 9.2) show the results obtained for the catalytic
decomposition of hydrogen peroxide.
2H2 O2 (aq)
2H2 O( ) + O2 (g)
Only the concentrations of H2 O2 were changed. All other variables were kept constant.
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Q7:
What variables would need to be controlled?
Figure 9.2: Catalytic decomposition of hydrogen peroxide
The initial concentrations of hydrogen peroxide used were:
/%
/%
/%
" $ " $ " $ " $ /%
From the results and the graphs, it is possible to calculate the average rate over the first
10 seconds. However, it is normal to determine initial rates of reaction (the gradient at
time = 0 secs) and use these values in subsequent calculations. (Table 9.3)
Table 9.3: Rates of oxygen evolution
Expt
A
B
C
D
[H2 O2 ] / mol
-1
Average rate over
10 s / cm3 s-1
0.50
0.38
0.20
0.10
0.40
0.32
0.16
0.08
Initial rate / cm3 s-1
0.51
0.40
0.21
0.11
On graph paper, plot a graph of initial rate of reaction against initial concentration.
Q8:
What shape of graph is obtained?
Q9: What is the relationship between rate of reaction and the concentration of
hydrogen peroxide.
If the enzyme catalase is used, the rate of reaction will depend also on the concentration
of catalase, Further experiments can be done to show that the rate of reaction is also
directly proportional to the catalase, i.e.
Rate
[H O ] and Rate [catalase]
2
2
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9.3. RATE EQUATIONS
177
Combining these gives:
Rate
[H O ] [catalase]
2
2
or
0
) " $
(9.1)
This expression (Equation 9.1) is known as the rate equation and k is known as the
rate constant.
Learning Point
It is important to note that the rate equation and the rate constant can only be
determined experimentally. They cannot be obtained from the stoichiometric equation.
9.3.1
Order of reaction
Learning Objective
Æ
To be able work out the order of reaction with respect to individual reactants, the
overall order and the rate constant from experimental data.
For a general reaction,
A+B
products
The rate equation will have the form:
0
) $
(9.2)
m and n usually have values of 0, 1 or 2 and are characteristic of the particular reaction.
They define the order of the reaction.
The order of a reaction with respect to a particular reactant is the power to which the
concentration of that reactant is raised in the rate equation.
In Equation 9.2, the order of reaction with respect to A is m and with respect to B is n.
The overall order of a reaction is the sum of the powers to which the concentrations
of all reactants are raised.
In Equation 9.2, the overall order of the reaction is (m + n).
Q10: For the catalytic decomposition of hydrogen peroxide (Equation 9.1) what is the
order of reaction with respect to hydrogen peroxide?
Q11: What is the order of reaction with respect to catalase?
Q12: What is the overall order of the reaction?
Sometimes it is possible to work out the order of a reaction without plotting the final
graph of initial rate against concentration. In simple cases, an inspection of the initial
rate data can be enough.
For example, look at Table 9.3 (shown again below) which shows the initial rate data for
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the catalytic decomposition of hydrogen peroxide.
Expt
[H2 O2 ] / mol
-1
0.40
0.32
0.16
0.08
A
B
C
D
Initial rate / cm3 s-1
Average rate over
10 s / cm3 s-1
0.50
0.38
0.20
0.10
0.51
0.40
0.21
0.11
Compare experiments B and C. The initial concentration used in experiment B is twice
that used in C. You will notice that the initial rate in experiment B is twice that of C, within
experimental error.
Similarly, the initial concentration in C is twice that in D and again the initial rate in C is
twice that in D.
The rate is directly proportional to [H2 O2 ] and the reaction is first order with respect to
H2 O2.
Example : Rate equations and constants
Consider the reaction between nitrogen monoxide (nitric oxide) and hydrogen.
2NO(g) + 2H2 (g)
N2 (g) + 2H2 O(g)
Use the following data to work out the rate equation and calculate the rate constant. All
concentrations are initial concentrations.
Experiment
[NO] / mol
-1
-1
[H2 ] / mol
Initial rate /
mol
1
1 x 10-3
1 x 10-3
2 x 10-3
2
3
-1
s-1
6.0 x 10-5
1 x 10-3
2 x 10-3
1 x 10-3
1.2 x 10-4
2.4 x 10-4
The rate equation will be of the form:
Rate = k [NO]m [H2 ]n
We can use the data to find values for m and n.
Compare experiments 1 and 2. The [H 2 ] has been doubled. The initial rate has also
been doubled.
Rate
[H ]
i.e. n = 1
2
So the reaction is first order with respect to H 2 .
Compare experiments 1 and 3. The [NO] has been doubled. In this case, the initial rate
has increased four-fold. The reaction is not first order with respect to NO.
If m = 2, rate
[NO]
2
Doubling [NO] will quadruple the rate since 2 2 = 4. The reaction is therefore second
order with respect to NO, and m = 2. The rate equation is therefore:
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9.3. RATE EQUATIONS
179
Rate = k [NO]2 [H2 ]
The overall order of reaction is3 (since m + n = 2 + 1).
Since k is a constant for the reaction, any one of the sets of data can be used to evaluate
k. Using experiment 1 and substituting values:
)
) )
Units of k.
The units of k can be worked out from the units of the other quantities in the rate
equation. To work out the units of k for the previous example (a third order reaction):
)
) )
So k = 6.0 x 10 4 mol-2 l2 s-1
The units of k depend on the overall order of the reaction.
Rate equation
Overall order
Rate = k[A]0
0
Rate = k[A]1
1
Rate = k[A]2
Rate = k[A][B]
2
mol-1
s-1
2
mol-1
s-1
Rate = k[A]2 [B]
3
mol-2
Units of k
-1
mol
s-1
s-1
2
s-1
It is important to point out again that the rate equation can only be derived
experimentally. It cannot be obtained from the balancing numbers in the balanced
equation. In the reaction in the previous example, two moles of NO react with two
moles of H2 . Equal numbers of moles react but the experimentally determined order of
reaction with respect to each reactant is different.
Orders and rate constants
A series of questions to give practice at working out orders of reaction, rate equations
and calculating rate constants.
For any questions which are not marked by the computer, answer first on paper before
revealing the correct answer.
The next two questions refer to the following reaction:
Bromide ions are oxidised by bromate ions (BrO 3 - ) in acidic solution according to the
equation:
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5Br- (aq) + BrO3 - (aq) + 6H+ (aq)
3Br2 (aq) + 3H2 O( )
By experiment, the reaction is found to be first order with respect to both bromide and
bromate but second order with respect to hydrogen ions.
Q13: Write the rate equation for this reaction.
Q14: What is the overall order of the reaction?
The next four questions refer to the hydrolysis of urea in the presence of the enzyme,
urease.
The rate equation for the reaction is found by experiment to be:
Rate = k [urea][urease]
Q15: What is the overall order of reaction?
Q16: What is the order with respect to water?
Q17: What is the order with respect to urea?
Q18: What is the order with respect to urease?
The next four questions refer to the decomposition of dinitrogen pentoxide, N 2 O5 :
2N2 O5 (g)
4NO2 (g) + O2 (g)
Experiments were carried out in which the initial concentration was changed and the
initial rate of reaction was measured. The following data were obtained.
[N2 O5 ] / mol
-1
Initial Rate / mol
0.05
0.10
0.20
-1
s-1
2.2 x 10-5
4.4 x 10-5
8.8 x 10-5
Q19: Write the rate equation for the reaction.
Q20: What will be the units of the rate constant?
Q21: Calculate the rate constant for the reaction. (Enter your answer in normal decimal
form, e.g. 0.001, and do not include units.)
Q22: If the initial concentration was 0.07 mol -1 , calculate the initial rate of reaction in
mol -1 s-1 . (Again, answer in normal decimal form, and do not include units.)
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181
The next three questions refer to the following reaction:
Iodide ions are oxidised in acidic solution to triiodide ions, I 3 - , by hydrogen peroxide.
H2 O2 (aq) + 3I- (aq) + 2H+ (aq)
I3 - (aq) + 2H2 O( )
The following initial rate data were obtained:
Experiment
1
2
3
4
Initial Concentrations / mol
-1
[H2 O2 ]
[I- ]
[H+ ]
0.02
0.04
0.02
0.02
0.02
0.02
0.04
0.02
0.001
0.001
0.001
0.002
Initial Rate /
mol -1 s-1
9.2 x 10-6
1.84 x 10-5
1.84 x 10-5
9.2 x 10-6
Q23: From the above data, write the rate equation.
Q24: What will be the units of k?
Q25: Calculate the value for the rate constant.
9.4
Reaction mechanisms
Learning Objective
Æ
To be able to explain what is meant by reaction mechanism and rate-determining step
(RDS) and relate the rate equation to the RDS
At Higher grade, simple collision theory was used to explain how reaction rates were
affected by changes in concentration, temperature and surface area (if one of the
reactants was solid).
Consider the reaction between two diatomic elements A and B:
A2 (g) + B2 (g)
2AB(g)
Reaction can occur when a molecule of A collides with a molecule of B with sufficient
energy and the correct orientation. A transition state forms which breaks down to give
the products.
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TOPIC 9. KINETICS
Figure 9.3: Formation of a transition state
If there is insufficient energy or the wrong orientation, the molecules simply bounce
apart.
This is a simple, one-step reaction. If the concentration of either reactant is doubled, the
number of successful collisions will be doubled. In other words, the reaction will be first
order with respect to both A2 and B2 , i.e.
Rate = k [A2 ][B2 ]
Experiments confirm this rate equation. Note that only for a simple one-step process
can the stoichiometry be used to work out the rate equation.
Most reactions are not so simple, e.g.
2NO(s) + 2H2 (g)
N2 (g) + 2H2 O(g)
The balanced equation shows two molecules of NO reacting with two moles of H 2 .
If this were to occur by a simple one-step reaction, then two molecules of NO would
have to collide simultaneously with two molecules of H 2 . This simultaneous collision of
four molecules is highly unlikely. Even a three molecule collision is statistically unlikely.
Most chemical reactions which proceed at a measurable rate are believed to occur by
a series of simple steps, each involving one or two molecules or ions. Such a series of
simple reaction steps is known as a reaction mechanism. The sum of all the individual
steps must equal the stoichiometric (balanced) equation. The reaction mechanism must
also fit the experimentally determined rate equation.
This poses a problem. Do we need to know how fast each step is in order to work out
the overall rate?
As an analogy, consider this production line in the bottling plant in a distillery.
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183
Figure 9.4: A whisky production line
There are three parts and the maximum capacity of each is:
2 bottles filled per minute.
Part 2
Filler
Capper
Part 3
Labeller
60 bottles labelled per minute.
Part 1
120 bottles capped per minute.
The production line is switched on. After 30 seconds a bottle has been filled and is
passed to the Capper which caps it in 0.5 seconds and passes it to the Labeller which
takes a further second to label it. So after 31.5 seconds we have completed one bottle.
After one minute, the second bottle is full and is immediately capped by the Capper,
which has been twiddling its thumbs waiting for the bottle to arrive.
Q26: How many seconds will it take to fill, cap and label 100 bottles?
Q27: How many seconds will it take to fill, cap and label 1000 bottles?
Clearly it does not matter how fast the Capper and Labeller are. The overall process is
governed by how fast the bottles are filled. Bottle filling is the rate determining step.
In any chemical reaction mechanism, one step will be significantly slower than
the others and this step will determine the overall reaction rate, i.e. it will be
the rate determining step (RDS for short).
Now consider the reaction between nitrogen dioxide and carbon monoxide.
NO2 (g) + CO(g)
NO(g) + CO2 (g)
Q28: If this was a simple reaction, what would the rate equation be?
In fact, experimental evidence shows that the rate equation is
Rate = k [NO2 ]2
The reaction must occur by two or more steps.
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A possible mechanism is as follows:
2NO2
NO3 + CO
NO3 + NO
NO2 + CO2
slow step
very fast step
The first step very slowly produces the intermediate, NO3 . As soon as it is formed, step
2 occurs. The NO3 reacts with a CO molecule to form the product, CO 2 . The rate of
formation of CO2 does not depend on the rate of reaction between the NO 3 and CO.
Instead it depends on the rate of formation of NO 3 . The first step is like the ’bottle filling’
and so is described as the rate determining step.
In this case, the slower step (RDS) involves two molecules of NO 2 . The reaction is
second order with respect to NO 2 and the rate equation is:
Rate = k [NO2 ]2 .
Learning Point
In general, the overall rate of a reaction depends on the rate of the slowest step in the
mechanism. The rate equation provides information about the rate-determining step.
The reaction is zero order with respect to CO because CO is not involved in the RDS
and only gets involved in a fast step which takes place after the RDS. Changes in the
concentration of CO will not affect the reaction rate.
It must be emphasised that this reaction mechanism is a suggestion and there may be
other possible mechanisms which also fit the rate equation. If further evidence could be
found for the existence of the intermediate, NO 3 , this would strengthen the case for this
mechanism.
We can now explain the apparent discrepancy between the balancing numbers in the
stoichiometric equation and the orders with respect to the individual reactants. The
order of reaction with respect to a particular reactant refers to the number of molecules
(or ions) of that reactant that take part in the RDS.
From the example above, it is clear that a reactant can have zero order if it does not
feature in the RDS.
It is also possible for a substance to appear in the rate equation but not in the
stoichiometric equation, e.g. a catalyst is not a reactant but must feature in the ratedetermining step if it is to speed up the reaction. (Equation 9.1)
The acid catalysed reaction between iodine and propanone is another example of this:
This reaction is found to be first order with respect to propanone (CH 3 COCH3 ) and also
first order with respect to H+ ions which act as a catalyst.
To find out the order with respect to iodine, try the PPA.
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9.4. REACTION MECHANISMS
185
PPA - Kinetics of the acid-catalysed propanone / iodine reaction.
Consult with your tutor whether the practical work for this PPA should be be carried out
at this point. A simulation of the experiment is available on-line.
The reaction between propanone and iodine
is first order with respect to propanone and first order with respect to hydrogen ions
which catalyse the reaction. The order with respect to iodine can be determined
by taking a reaction mixture in which the initial concentration of propanone and
hydrogen ions are very much larger than that of iodine. With such conditions, only
the concentration of iodine will vary significantly during the reaction and this will allow
us to see what effect it has on the reaction rate.
The course of the reaction can be followed by monitoring the concentration of iodine.
This involves removing samples from the reaction mixture from time to time and
analysing them for iodine.
This simulated practical is not like a normal simulation, where you press the
button and the work is done for you, in this simulation you must prepare the
solutions, take the readings and decide how to plot the graphs.
NOTE: To progress through the experiment, click on the "tabs" at the top of the program
to change to different screens
Complete the tasks on each screen and then move on to the next tab to the right.
You can browse the different screens if you want at any time, in fact it’s probably a good
idea to have a look at them all before you start. As you might expect, some things later
in the experiment won’t work unless you’ve actually run the experiment.
If you get stuck, use the
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button at the top right of each screen.
186
TOPIC 9. KINETICS
Questions on reaction mechanisms
20 min
A number of questions to give practice at interpreting rate data and relating these to
reaction mechanisms.
For any questions which are not marked by the computer, answer first on paper before
revealing the correct answer.
Q29: Which of the following reactions is most likely to occur by a simple one-step
process?
a)
b)
c)
d)
4HBr + O2
H2 S + Cl2
2NO + O2
2H2 + O2
2H2 O + 2Br2
S + 2HCl
2NO2
2H2 O
Q30: Which of the following reactions is least likely to occur by a simple one-step
process?
a)
b)
c)
d)
4HBr + O2
H2 S + Cl2
2NO + O2
2H2 + O2
2H2 O + 2Br2
S + 2HCl
2NO2
2H2 O
The next six questions refer to the reaction between propanone and bromine in alkaline
solution.
The balanced equation is:
CH3 COCH3 (aq) + Br2 (aq) + OH- (aq)
CH3 COCH2 Br(aq) + H2 O( ) + Br- (aq)
The experimentally determined rate equation is:
Rate = k [CH3 COCH3 ][OH- ]
Use this information to select True or False for each of the following statements.
Q31: The reaction is first order with respect to bromine.
a) True
b) False
Q32: The reaction involves a simple one-step process.
a) True
b) False
Q33: The reaction is second order overall.
a) True
b) False
Q34: The rate determining step involves one molecule of propanone and one molecule
of bromine.
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9.5. SUMMARY
187
a) True
b) False
See further questions on page 206.
The next five questions refer to a reaction involving hydrogen peroxide and bromide ions
in aqueous solution.
H2 O2 + BrH2 O2 + BrO-
BrO- + H2 O
Step 1
Br- + H2 O + O2
Step 2
Q35: What is the equation for the overall reaction ?
Q36: What is the role played by the Br - ion?
Q37: What role is played by the BrO - ion?
Q38: If step 1 is the rate determining step, which of the following is the rate equation?
a)
b)
c)
d)
Rate = k [H2 O2 ]
Rate = k [H2 O2 ][Br- ]
Rate = k [H2 O2 ]2
Rate = k [H2 O2 ]2 [Br- ]
Most reactions occur by a series of simple steps - a reaction mechanism. The
rate determining step (RDS) is the slowest step in the mechanism. Experimentally
determined rate equations give information about the RDS and can be used to suggest
possible mechanisms.
9.5
Summary
• The rate of a chemical reaction normally depends on the concentrations of the
reactants.
• For a first order reaction, the rate is directly proportional to the concentration of
one reactant and the rate can be expressed as:
Rate = k [A]
where k is the rate constant and [A] is the concentration of reactant A in mol
• In general, for a simple reaction step such as:
nA + mB
products
the rate equation is of the form
Rate = k[A]n [B]m
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TOPIC 9. KINETICS
• The order of the reaction with respect to A is n and the order with respect to B is
m.
• The overall order of reaction is n + m.
• The rate equation, and hence value for k, is always derived experimentally, usually
from data obtained by varying the initial concentrations and measuring initial rates.
• Most reactions occur by a series of simple steps known as a reaction mechanism.
• The overall rate of reaction is determined by the rate of the slowest step - the rate
determining step.
• Rate equations can provide evidence for a proposed reaction mechanism but
cannot provide proof as other possible reaction mechanisms may also give the
same rate equation.
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9.6. RESOURCES
9.6
Resources
• Chemistry: McMurry and Fay, Prentice Hall, ISBN 0-13-737776-2
• General Chemistry: Ebbing, Houghton Mifflin, ISBN 0-395-74415-6
• Chemistry in Context: Hill and Holman, Nelson, ISBN 0-17-438401-7
• Chemistry: Fullick and Fullick, Heinemann, ISBN 0-435-57080-3
• Higher Still Support: Advanced Higher Chemistry - Unit 2: Principles of Chemical
Reactions, Learning and Teaching Scotland, ISBN 1 85955 874 7.
• A-level Chemistry: E.N. Ramsden, Stanley Thorne Publishers, ISBN 0-85950154-X
9.7
End of Topic test
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TOPIC 9. KINETICS
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Topic 10
End of Unit 2 Test (NAB)
Contents
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TOPIC 10. END OF UNIT 2 TEST (NAB)
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GLOSSARY
Glossary
acid
A substance that is able to donate hydrogen ions (protons) to another substance,
i.e. it is a proton donor
amphoteric
A substance that can behave as an acid in some situations, but also as a base in
other situations, is described as amphoteric
anode
The electrode at which oxidation occurs
balanced (stoichiometric) equation.
Balanced (stoichiometric) equations define the ratios of moles of reactants and
products.
base
A substance that accepts hydrogen ions (protons), i.e. it is a proton acceptor
buffer solution
A solution in which the pH remains approximately constant when small amounts
of acid or base are added
cathode
The electrode at which reduction occurs
chromatography
Chromatography is an analytical method where mixtures are separated into
their components by partitioning between a stationary and mobile phase. The
stationary/mobile phases are solid/liquid in paper and thin layer chromatography,
and liquid/gas in gas-liquid chromatography
closed
A closed system has no exchange of matter or energy with its surroundings
conjugate acid
For every base, there will a conjugate acid formed by gain of a proton (H + )
conjugate base
for every acid, there is a conjugate base formed by loss of a proton (H + ion)
dynamic equilibrium
A dynamic equilibrium is achieved when the rates of two opposing processes
become equal, so that no net change results
electrode potential
The difference in electric potential between the electrode (metal) and the ions in
solution
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193
194
GLOSSARY
electromotive force
The electric potential difference between the electrodes in a cell when no current
is drawn, i.e. the maximum potential difference
end point.
The end point of a titration is the point when the reaction is shown to be complete.
This is frequently signalled by the change in colour of an indicator.
enthalpy of hydration
The enthalpy change when one mole of individual gaseous ions is completely
hydrated, i.e:
En+ (g)
En+ (aq)
and
En- (g)
En- (aq)
enthalpy of neutralisation
The enthalpy change when the acid is neutralised to form one mole of water
enthalpy of solution
The enthalpy change when one mole of a substance is dissolved completely in
water
entropy
The entropy of a system is the degree of disorder of the system. The greater the
disorder, the greater the entropy. Low entropy is associated with strongly ordered
substances
equivalence point
the equivalence point in a titration experiment is reached when the reaction
between the titrant (added from the burette) and the titrate (in the flask) is just
complete.
equivalence point.
the equivalence point in a titration experiment is reached when the reaction
between the titrant (added from the burette) and the titrate (in the flask) is just
complete.
excess
Excess reactants are present in greater than stoichiometric amounts and will not
be completely used at the end of the reaction.
hess’s Law
Hess’s Law states that the overall reaction enthalpy is the sum of the reaction
enthalpies of each step of the reaction.
heterogeneous reaction
Heterogeneous reactions have reactants and/or products in more than one phase
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GLOSSARY
195
homogeneous reactions
Homogeneous reactions have all the reactants and products in the same phase
intermediate
A species that is formed in one step of a reaction mechanism and then used up in
a subsequent step
limiting
Limiting reactants are present in smallest stoichiometric amounts and will be
completely used to produce products whose amounts will be determined by the
initial quantity of limiting reactant.
mean molar bond enthalpy
An average value that is quoted for a bond that can occur in different molecular
environments
molar bond enthalpy
The molar bond enthalpy for a diatomic molecule X-Y is the energy required to
break one mole of X-Y bonds, that is for the process:
X - Y(g)
X(g) + Y(g)
molar solution
A molar solution contains 1 mole of solute in 1 litre of solution.
order of a reaction
The power to which the concentration of a particular reactant is raised in the rate
equation
overall order of a reaction
The sum of the powers to which the concentrations of all reactants in the rate
equation are raised in the rate equation
partition coefficient
The ratio of the concentrations of a solute in two immiscible liquids is called the
partition coefficient
primary standard
A primary standard is a stable, soluble solid which can be accurately weighed and
dissolved in a known volume of solution to produce a standard solution.
quantitative reaction
A quantitative reaction is one in which the reactants react completely according to
ratios in the balanced stoichiometric equation.
rate constant
In a rate equation of the form
0
) $
k is the rate constant and has a constant value for a given reaction at a particular
temperature
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196
GLOSSARY
rate determining step
The slowest step in a reaction mechanism that governs the overall rate
rate equation
An equation that tells how the reaction rate depends on the concentration of each
reactant
reaction mechanism
The series of simple steps by which a chemical reaction occurs
retention time
Retention time is the time taken for an individual peak to traverse the gas-liquid
chromatographic column after the injection time
second Law of Thermodynamics
The total entropy of a reaction system and its surroundings always increases for a
spontaneous change
stable equilibrium
A stable equilibrium state is one which is regained after a small disturbance from
this state has occurred
standard conditions
Conditions at a pressure of one atmosphere and a specific temperature (298
K)(25Æ C).
standard enthalpy change
The enthalpy change for a reaction in which reactants and products are considered
to be in their standard states at a specified temperature
standard Gibbs free energy
The standard Gibbs free energy change for a reaction is related to the standard
enthalpy and entropy changes by
GÆ = HÆ - TSÆ
The direction of spontaneous change is in the direction of decreasing free energy.
standard molar enthalpy change of lattice formation
The enthalpy change that occurs when one mole of an ionic crystal is formed from
the ions in their gaseous state under standard conditions
standard molar enthalpy of combustion
The enthalpy change when one mole of a substance is completely burned in
oxygen under standard conditions.
standard molar enthalpy of formation
The enthalpy change that occurs when one mole of a substance is produced from
its elements in their standard states
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GLOSSARY
standard reduction potential
The electrode potential for a reduction half-cell measured under standard
conditions (relative to the standard hydrogen electrode)
standard solution
A standard solution is one with an accurately known concentration. It can be
prepared by weighing a primary standard and dissolving it in a known volume of
solution, or by titrating against another standard solution.
standard state
The most stable state of a substance or element under standard conditions
stoichiometric
Stoichiometric chemical processes involve quantitative reactions that use exact
whole numbers of particles (atoms, ions or molecules) in fixed, characteristic
ratios.
third Law of Thermodynamics
The Third Law of Thermodynamics states that the entropy of a perfect crystal at 0
K is zero
titration
A titration is a procedure which is used to find the concentration of solutions.
transition state
The arrangement of atoms at the maximum in the potential energy profile for a
reaction
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198
FURTHER QUESTIONS
Further questions
Topic 1: Stoichiometry
Questions continued from page 8.
Q61: Ethanoic acid is prepared industrially by reacting liquid methanol with carbon
monoxide gas.
CH3 OH( ) + CO(g)
CH3 CO2 H( )
If 15.0 g of methanol and 10.0 g of carbon monoxide were placed in a reaction vessel,
which is the limiting reactant?
Q62: How many grams (to 2 decimal places) of ethanoic acid could be produced by the
reaction above?
Questions continued from page 11.
Q63: It must be coloured.
a) True
b) False
Q64: It must have a low boiling point.
a) True
b) False
Q65: It must be a liquid.
a) True
b) False
Q66: It must have a relatively high formula mass.
a) True
b) False
Questions continued from page 12.
Q67: What volume (in ml to 2 decimal places) of 2 mol -1 sodium carbonate would be
required to neutralise 1 ml of concentrated ’syrupy’ phosphoric acid (15 mol -1 )?
Questions continued from page 16.
Q68: When sodium carbonate solution is added to aqueous copper(II) ions, the
precipitate in fact consists of a mixture of copper(II) carbonate and hydroxide. Would
this affect the analysis above? Give an explanation for your answer.
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FURTHER QUESTIONS
199
Questions continued from page 18.
Q69: When plaster of Paris [(CaSO 4 )2. H2 O] sets after it has been mixed with water, it
forms gypsum (CaSO4 .2H2 O). How many grams of water (to 1 decimal place) should
be added to 580 g of plaster of Paris to ensure that it is converted completely to gypsum
with 27 g of water remaining?
Q70: There are only 5 g of copper(II) sulphate (CuSO 4.5H2 O) left in a bottle. What
volume of 0.2 mol -1 solution could be made from this weight? Give your answer to the
nearest unit. ml
Q71: The insecticide DDT is known to increase in concentration in living tissues as
it moves along a food chain. Its concentration in the water of Lake Michigan is 2 x
10-5 mg -1 . If it increases in concentration 300,000 times in fish tissues, estimate the
concentration (in mg -1 to 1 decimal place) in the trout in Lake Michigan.
Q72: A further increase to 5.1 g DDT kg -1 tissue is found in ospreys. What is the
increase in concentration in going from fish to these birds?
Q73: An indigestion remedy (antacid) contains an aqueous suspension of magnesium
hydroxide (Mg(OH)2 ). This suspension was diluted 50 times and 10.0 ml of this reacted
completely with 24.0 ml of 0.15 mol -1 hydrochloric acid. What is the concentration of
magnesium hydroxide in the original suspension in mol -1 , to 1 decimal place?
Q74: What is the concentration of the magnesium hydroxide in g
to 1 decimal place.
-1 ?
Give your answer
Q75: Alcohol levels in blood can be determined by a redox titration with potassium
dichromate according to the balanced equation:
C2 H5 OH(aq) + 2Cr2 O7 2- (aq) + 16H+(aq)
2CO2 (g) + 4Cr3+ (aq) + 11H2 O(l)
What is the blood alcohol level in mol -1 (to 3 decimal places) if 8.76 ml of 0.050 mol
K2 Cr2 O7 is required for titration of a 10.026 ml sample of blood?
-1
Q76: It is more usual to express blood alcohol levels in mg alcohol / 100 ml blood. What
is the alcohol level in the question above in these units? Express your answer to 3
significant figures.
Q77: Titration with solutions of potassium bromate (KBrO 3 ) can be used to determine
the concentration of As(III). What is the molar concentration of As(III) in a solution if
22.35 ml of 0.100 mol -1 KBrO3 is needed to titrate 50.00 ml of the As(III) solution? The
balanced equation is:
3H3 AsO3 + BrO3 -
Br- + 3H3 AsO4
Give your answer to 3 decimal places.
Q78: In a titration of Cu2+ ion with ethylene diamine, 25.0 ml of 0.10 mol -1 Cu2+ reacted
completely with 37.5 ml of 0.20 mol -1 ethylene diamine. How many ethylene diamine
molecules complex with each copper(II) ion?
Q79: The zinc ions in a 0.9328 g sample of foot powder was titrated with 18.52 ml of
0.0233 mol -1 EDTA. Calculate the percentage (to 3 decimal places) of zinc in this
sample.
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200
FURTHER QUESTIONS
Q80: 17.870 g of chlorophyll were heated in an oxygen atmosphere to destroy the
organic part and leave a residue of magnesium oxide. If this residue weighed 0.806
g, what is the percentage of magnesium in chlorophyll? Give your answer to 2 decimal
places.
Q81: The formula mass of chlorophyll is 893.5. How many atoms of magnesium are
there in each molecule? Enter your answer as a number.
Q82: Gold has compounds containing gold(I) ion or gold(III) ion. A compound of gold
and chlorine was treated with a solution of silver nitrate, AgNO 3, to convert the chloride
ion in the compound to a precipitate of AgCl. A 162.7 mg sample of the gold compound
gave 100.3 mg AgCl. Calculate the percentage of chlorine in the gold compound, to 2
decimal places.
Q83: Decide whether the formula is AuCl or AuCl 3 .
Topic 2: Chemical Equilibrium
Questions continued from page 28.
Q34: Would the process to manufacture SO 3 be more productive at 636 Æ C or 856Æ C?
Questions continued from page 34.
Q35: Around lightning, the temperature of the air is raised to approximately 2400 Æ C.
At this temperature, Kp for the previous reaction is increased to 2.44 x 10 -3 . What is
the equilibrium partial pressure of NO around a lightning strike? Assume that the initial
partial pressures of nitrogen and oxygen are 0.78 and 0.21 respectively.
Questions continued from page 37.
Q36: 3Fe(s) + 4H2 O(g)
a) left to right
b) right to left
c) no change
Q37: 2NO(g) + Cl2 (g)
« Fe O (s) + 4H (g)
3
4
2
« 2NOCl(g)
a) left to right
b) right to left
c) no change
Topic 3: Phase Equilibria
Factors affecting the partition coefficient. (page 43)
Q24: Which experiment used a different solute? (Figure 3.2)
Q25: Does the partition coefficient depend on the type of solute?
Q26: Experiment 5 used twice as much solute as experiment 3. What effect has this
had on the partition coefficient?
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FURTHER QUESTIONS
201
a) increases the partition coefficient
b) decreases the partition coefficient
c) no effect
Questions continued from page 50.
Q27: Both the blue and dark blue spots from both the original inks have moved similar
distances. What might you conclude from this?
Topic 4: Acid/base Equilibria
Questions continued from page 60.
Q52: What type of bond is formed?
Q53: What type of substance is ammonium chloride?
a) acid
b) base
c) salt
The relationship between [H+] and [OH-] (page 65)
Q54: At pH 1, what is the OH - ?
Q55: At this pH, what value is obtained by multiplying H + by OH- ?
Q56: At pH 7, what value is obtained by multiplying H + by OH- ?
Questions continued from page 70.
Q57: What is the pH of a solution of sulphurous acid of molar concentration
0.05 mol -1 ?
Q58: What is the pH of a solution of carbonic acid of molar concentration
0.001 mol -1 ?
Q59: A solution of benzoic acid has a pH of 3.0. What is the molar concentration?
Q60: A solution of methanoic acid has a pH of 2.4. What is the molar concentration?
Topic 5: Indicators and buffers
Questions continued from page 78.
Q24: Which of the following expressions represents the acid dissociation constant of
HIn in Figure 5.1.
a)
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"& " $
"$ &
202
FURTHER QUESTIONS
b)
"&
"$ &
"$ &
"& " $
"$ &
"&
c)
d)
pH titration (page 81)
Q25: Which of these statements is true?
a)
b)
c)
d)
The alkali is more concentrated than the acid.
The pH rises rapidly at the beginning.
The alkali is less concentrated than the acid.
The pH changes rapidly only around the equivalence point.
Questions continued from page 83.
Q26: Using the graphs in Table 5.2 and your previous answers, write a general
statement about the pH at the equivalence point in an acid/ alkali titration. Then display
the answer.
Buffer calculations (page 93)
Q27: To prepare 1 litre of a buffer solution which would maintain a pH 5.5, 0.6g of
ethanoic acid was used. What mass in grams of sodium ethanoate should the solution
contain? Answer to one decimal place.
Q28: One of the systems which maintains the pH of blood at 7.40 involves the acid
H2 PO4 - and the salt containing HPO 4 2- . Calculate the ratio of salt to acid in blood (K a
and pKa of H2 PO4 - in data booklet).
Topic 6: Thermochemistry
Enthalpy calculations from bond enthalpies (page 99)
Q41: Remembering that the total bond breaking is positive and bond making is
negative, what is the overall enthalpy change in this reaction?
Q42:
Use the same layout on paper and values in the data booklet to work through this on-line
example before displaying the answer to check your working.
Calculate the enthalpy change involved in this reaction:
2H2 (g) + O2 (g)
2H2 O(g)
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FURTHER QUESTIONS
203
Hess’s Law questions (page 105)
Q43: Solid magnesium chloride exists in two forms: anhydrous (MgCl 2 ) and hydrated
(MgCl2 .6H2 O).
Calculate the enthalpy change for converting anhydrous magnesium chloride to
hydrated magnesium chloride, given the following enthalpies of formation:
HÆf (hydrated magnesium chloride)
HÆf (anhydrous magnesium chloride)
HÆf (water)
= -2500 kJ mol-1
= -642 kJ mol-1
= -286 kJ mol-1
Bomb calorimetry (page 106)
Q44: What value in kJ mol-1 is given in the data booklet for the standard molar enthalpy
of combustion of ethanol? (again: remember the sign).
Q45: The difference could be due to heat losses or the non-standard conditions, but
assuming the conditions to be standard, calculate the percentage efficiency of this
experiment.
Topic 7: Reaction Feasibility
Entropy and temperature (page 123)
Q33: What change occurs between 370 K and 375 K?
Q34: Does the disorder of the system increase or decrease?
Q35: Is there a positive or negative change in entropy?
Q36: Explain why the slope of the line above 373 K is far steeper than below 273 K.
Estimating and calculating spontaneity (page 125)
Q37: Explain why there is a drop from butane to pentane.
Q38: N2 (g) + 3H2 (g)
Q39: Zn(s) + Cu2+ (aq)
2NH3 (g)
Zn2+ (aq) + Cu(s)
Calculations involving free energy changes (page 135)
Q40: Chloroform was one of the first anaesthetics used in surgery. At the boiling point
of any liquid, the gas and liquid are in equilibrium. Use this information to calculate a
boiling point for chloroform.
CHCl3 (l)
CHCl3 (g)
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HÆ = 31.4 kJ mol-1
SÆ = 94.2 J K-1 mol-1
204
FURTHER QUESTIONS
Interpreting Ellingham diagrams (page 140)
Q41: The data booklet gives the melting point of zinc as approximately 700 K. What
happens to the entropy and what effect does it have on the gradient of the graph? (point
A)
Q42: What causes the further little ’kink’ in the zinc line at 1180 K?
Topic 8: Electrochemistry
Zinc - copper cell. (page 150)
Q57: What properties should the substance used for the salt bridge have?
examples of suitable substances.
Give
Q58: Why does the current stop flowing?
Questions continued from page 153.
Q59: Al(s) Al3+ (aq)
Pb
2+ (aq)
Pb(s)
Questions continued from page 153.
Q60: Write the cell notation for a cell in which this cell reaction occurs:
Fe2+ (aq) + Ag+ (aq)
Fe3+ (aq) + Ag(s)
Questions continued from page 157.
Q61: What will be the value of the standard reduction potential (E Æ ) of zinc? (Remember
to show the sign and units).
Questions continued from page 159.
Q63: Pt(s) Sn
Q62: Al(s) Al3+ (aq)
Pb
2+ (aq),
Pb(s)
(aq) Fe (aq), Fe
2+ (aq)
Sn4+
3+
2+ (aq)
Pt(s)
Oxidising agents and reducing agents (page 159)
Q64: Where in the table will the best oxidising agents be found?
a)
b)
c)
d)
Top left
Top right
Bottom left
Bottom right
Q65: Draw a diagram to summarise this information.
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FURTHER QUESTIONS
205
Questions continued from page 161.
Q66: From the grid Figure 8.13, enter the letter of the substance that could be used as
an oxidising agent or a reducing agent.
Q67: From the grid Figure 10.1, enter the letter of the box containing the best oxidising
agent.
Figure 10.1: Answer grid
Q68: From the grid Figure 10.1, enter the letter of the box containing the best reducing
agent.
Calculation of standard free energy change. (page 164)
Q69: Al(s) + 3 /2 I2 (s)
Q70: Fe(s) Fe2+ (aq)
Al3+ (aq) + 3 I- (aq)
Ag (aq) Ag(s)
+
Calculating standard entropy changes. (page 165)
Q71: Use this answer to calculate the standard free energy change, G Æ , in kJ mol-1 to
one decimal place.
Q72: What is the relationship between free energy, enthalpy and entropy?
Q73: Using this relationship and your answers above, calculate the standard entropy
change at 298K (25 Æ C), in J K-1 mol-1 .
Q74: By considering the overall redox equation, comment on the sign of S Æ .
Questions continued from page 167.
Q75: Apart from generating electricity, what advantage would the use of such a cell
have on the space shuttle?
Topic 9: Kinetics
Change in concentration with time (page 173)
Q39: What happens to the rate of reaction as the reaction proceeds?
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206
FURTHER QUESTIONS
Questions on reaction mechanisms (page 186)
Q40: The following mechanism fits the rate equation.
a) True
b) False
Q41: The following mechanism fits the rate equation.
a) True
b) False
Topic 10: End of Unit 2 Test (NAB)
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ANSWERS: TOPIC 1
Answers to questions and activities
1 Stoichiometry
Answers from page 3.
Q1: CaF2
Q2: Fe(NO3 )3
Q3: Ba3 (PO4 )2
Q4: K2 Cr2 O7
Answers from page 3.
Q5: Calcium sulphate
Q6: Copper(II) chloride
Q7: Potassium permanganate or potassium manganate(VII)
Q8: Phosphorus(V) oxide
Answers from page 3.
Q9: 56.0
Q10: 342.3
Q11: 152.0
Q12: 58.0
Answers from page 4.
Q13: c) statements 1, 2 and 3.
Q14: c) statement 3
Answers from page 5.
Q15: 11.0
Q16: 300.0
Q17: 21.17
Q18: 5.3
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207
208
ANSWERS: TOPIC 1
Answers from page 6.
Q19: 6.0
Q20: 3.0
Q21: 2.0
Q22: 9.0
Answers from page 8.
Q23: 1.02
Q24: 0.75
Q25: ethanol
Q26: 44.0
Answers from page 9.
Q27: 0.20
Q28: 0.10
Q29: This time, use Equation 1.1 in the form:
• Weight (moles) = Molar concentration (mol
-1 )
x Volume (litres)
• So moles = 0.05 x 0.5 = 0.025
• Then use the GFM to convert moles to grams.
• For FeCl3 .6H2 O the GFM is 270.3, so 0.025 moles is 6.76 g.
• 6.76 g of hydrated iron(III) chloride are required to prepare 500 ml of 0.05 mol
solution.
-1
Q30: 12.41
Answers from page 10.
Q31: 270.3
Q32: 1.6
Q33: 20.0
Q34: 260
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ANSWERS: TOPIC 1
209
Answers from page 11.
Q35: a) True
Q36: a) True
Q37: a) True
Q38: b) False
Answers from page 12.
Q39: 0.3125
Q40: 0.017
Q41:
Q42: 22.5
Answers from page 13.
Q43: 0.125
Q44: Calculate the molar concentration of the diluted peroxide solution first.
• (M1 x V1 )/n1 = (M2 x V2 )/n2
• (M1 x 25)/5 = (0.02 x 21.32)/2
• M1 = 0.04264
• Then multiply by the dilution factor to obtain the molar concentration of the orginal
peroxide solution.
• Molar concentration is 1.7056 mol
-1
• The molar mass of H2 O2 is 34, so multiplying this by the molar concentration gives
58.0 g -1 . The hydrogen peroxide concentration is therefore 5.8 g/100ml (or 5.8%).
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210
ANSWERS: TOPIC 1
Answers from page 14.
Q45: First, use the thiosulphate titration information to calculate the number of moles of
iodine released.
• The equation for reduction of iodine by thiosulphate is:
• I2 + 2S2 O3 2-
2I- + S4 O6 2-
• From (M1 x V1 )/n1 = (M2 x V2 )/n2
• Number of moles of iodine is M1 x V1 (litres) = 0.0001
• so the weight of iodine produced is 0.0001 x 253.8 (GFM for I 2 ) = 25.38 mg
Q46:
• From the equation for ozone and iodide calculate the weight of ozone given the
weight of iodine produced.
• 1 mol of ozone produces 1 mol of iodine, so 0.0001 mol of ozone will produce
0.0001 mol of iodine.
• 0.0001 mol of ozone is 0.0048 g.
• So 4.8 mg of ozone was present in the 50 litre air sample.
Q47: 0.05
Answers from page 14.
Q48: 0.024
Q49: 92
Q50: 80
Answers from page 16.
Q51: 0.0025
Q52: 0.0016
Answers from page 16.
Q53: 11.33
Q54: 0.50
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ANSWERS: TOPIC 1
211
Al2(SO4)3 + 6OH2Al(OH)3
2Al(OH)3 + 3SO42Al2O3 + 3H2O
The solution of Al2(SO4)3 yielded 1.054 g of Al2O3. What weight of aluminium
sulphate was in the original sample?
You should remember that in any procedure to determine an element, that
element must be conserved throughout the operation. The number of moles of
aluminium atoms is the same in the initial aluminium sulphate solution and the final
aluminium oxide regardless of what happens in between.
Moles of Al2(SO4)3 = Moles of Al2O3 (= 0.5 x moles of Al atoms)
REQUIRED
Q55:
GIVEN
GIVEN
REQUIRED
Al2O3
Al2(SO4)3
3 x 16.0 = 48.0
2 x 27.0 = 54.0
1 mol = 102.0 g
12 x 16.0 = 192.0
3 x 32.1 = 96.3
2 x 27.0 = 54.0
1 mol = 342.3 g
1 mol of Al2O3 is produced from 1 mol of Al2(SO4)3
102.0 g of Al2O3 is produced from 342.3 g Al2(SO4)3
1 g of Al2O3 is produced from 342.3/102 g Al2(SO4)3
1.054 g of Al2O3 is produced from 1.054 x 342.3/102 g Al2(SO4)3
= 3.54 g
The original solution contained 3.54 g of aluminium sulphate.
Q56: 0.08
Answers from page 18.
Q57: 151.7
Q58: 18.0
Q59: 9.0
Q60: 3.74
Further answers
Answers from page 198.
Q61: carbon monoxide
Q62: 21.43
Answers from page 198.
Q63: b) False
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ANSWERS: TOPIC 1
Q64: b) False
Q65: b) False
Q66: a) True
Answers from page 198.
Q67: 11.25
Answers from page 198.
Q68: It would not affect the result because the precipitated hydroxide would also
produce oxide (like the carbonate) when heated.
Answers from page 199.
Q69: 134.9
Q70: 100
Q71: 6.0
Q72: 850
Q73: 9.0
Q74: 524.7
Q75: 0.022
Q76: 101
Q77: 0.134
Q78: 3
Q79: 3.025
Q80: 2.72
Q81: 1
Q82: 15.26
Q83: AuCl. The % chlorine in AuCl is 15.26. AuCl
3
contains 35.09% chlorine.
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ANSWERS: TOPIC 2
213
2 Chemical Equilibrium
Answers from page 24.
Q1: At t = 15 there are 5, 5 and 2, respectively. At t = 30 and t = 70 there is no change
with four of each.
Answers from page 24.
Q2: There are four of each at both times.
Q3: They are the same.
Answers from page 27.
Q4:
Kc =
[Fe2 + ]2 [I3- ]
[Fe3 + ]2 [I - ]3
Q5:
Kc =
[H + ]2 [HPO24 - ]
[H3 PO4 ]
Answers from page 27.
Q6: c) hydrogen iodide
Answers from page 28.
Q7: a) phosphorus(V) chloride
Q8: b) Mg
Q9: d)
2SO2 (g) + O2 (g)
« 2SO (g)
Kc at 636Æ C = 3343
3
Q10: The value increases from 21 to 3343 as the temperature drops.
Answers from page 29.
Q11:
Kp =
p2NO pCl2
p2NOCl
Q12:
Kp =
© H ERIOT-WATT U NIVERSITY
pSO3
pO2 2 p2SO2
214
ANSWERS: TOPIC 2
Answers from page 30.
Q13: c)
Kc =
[Zn2 + ]
Cu Q14: c)
Kp = p2H2 O pN2 O
Answers from page 32.
Q15: 0.0035
Q16: 0.050
Answers from page 34.
Q17:
x = 0.094 or 0.229 but only 0.094 is a sensible solution.
µ [H ] = 0.1 - x = 0.006 mol
2
[I2 ] = 0.2 - x = 0.106 mol
[HI] = 2x = 0.188 mol
-1
-1
-1
Q18:
x = 0.01
µ [N O ] = 0.05 - x = 0.04 mol
2
4
[NO2 ] = 2x = 0.02 mol
-1
-1
Q19:
x = -0.0092
This is a negative number, which means that in reaching equilibrium, the backward
reaction occurs, increasing the concentration of N 2 O4 .
so [N2 O4 ] = 0.02 - x = 0.0292
[NO2 ] = 0.03 + 2x = 0.0116
Q20: 2.76 x 10-16
Answers from page 35.
Q21: 0.059
Q22: 0.470
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ANSWERS: TOPIC 2
Answers from page 37.
Q23: b) right to left
Q24: c) no change
Q25: a) left to right
Q26: b) right to left
Answers from page 38.
Q27: absorb
Q28: right
Q29:
Q30: increase
Answers from page 38.
Q31: Endothermic, because energy always has to be supplied to break bonds.
Q32: a) increase
Answers from page 38.
Q33: Increase the pressure (3 moles of reactant produce 2 moles of product);
decrease the temperature (exothermic reaction will move forward to try and increase the
temperature); addition of SO 2 (adding reactant will push equilibrium towards products);
addition of oxygen.
Further answers
Answers from page 200.
Q34: 636Æ C
Answers from page 200.
Q35:
You should have calculated x to be 1.65 x 10 -4 .
so [O2 ] = 0.21 - x = 0.2098 atm
[N2 ] = 0.78 - x = 0.7798 atm
[NO] = 2x = 0.00033 atm
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216
ANSWERS: TOPIC 2
Answers from page 200.
Q36: c) no change
Q37: a) left to right
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ANSWERS: TOPIC 3
3 Phase Equilibria
Answers from page 42.
Q1: A
Q2: b) It is a non-polar solvent
Q3: a) B is more polar than A
Q4: d) an equal exchange rate is reached
Q5: c) 3
Factors affecting the partition coefficient. (page 43)
Q6: temperature
Q7: yes
Q8: 1
Q9: yes
Answers from page 44.
Q10: Only one, as a further 3:1 ratio has been set up and three have been extracted.
Q11: Probably none. Although for clarity only a few spheres are shown and in reality
millions of particles are involved. Reaching true 100% extraction is very difficult.
Answers from page 46.
Q12: Partition coefficient = 3.0. This comes about since the concentration of acid in the
ether layer is 0.03 mol -1 and in the aqueous layer is 0.01 mol -1 .
Q13: (i) 75% of 5 g = 3.75 g and (ii) In first batch of 100 cm 3 there would be 3 g of solute
and in the second batch 1.2 g giving a total of 4.2 g.
Answers from page 48.
Q14: non-polar
Q15: non-polar
Q16: decaffeinated
Answers from page 50.
Q17: b) black
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218
ANSWERS: TOPIC 3
Q18: e) green
Q19: a) has the highest solvent/water partition coefficient
Q20: c) dark blue
Answers from page 53.
Q21: b) pentane
Q22: a) r.t. increases as m increases
Q23: c) xylene(C8 H10 )
Further answers
Factors affecting the partition coefficient. (page 43)
Q24: 4
Q25: yes
Q26: c) no effect
Answers from page 201.
Q27: They are probably the same dyes in both cases since the R f values would be
the same. If they were different materials, they would probably have moved different
distances.
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ANSWERS: TOPIC 4
219
4 Acid/base Equilibria
Answers from page 60.
Q1: ammonia
Q2: hydrogen chloride
Q3: ammonium ion
Q4: chloride ion
Answers from page 61.
Q5: 68
Answers from page 63.
Q6: Salt solution contains ions and therefore it is an electrolyte.
Q7: Hexane is completely covalent and therefore contains no ions.
Q8: Pure water must contain a small number of ions.
Answers from page 63.
Q9: b) a base
Q10: b) the conjugate base of H 2 O
Q11: b) The equilibrium lies well to the left.
Q12:
"$ $" " $
Answers from page 64.
Q13: c) 1.00 x 10-14
Q14: a) It is positive.
Q15:
As the temperature increases, Kw increases, i.e. the equilibrium moves to the right.
Raising the temperature always favours the endothermic process.
reaction must be endothermic and therefore H Æ is positive.
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220
ANSWERS: TOPIC 4
The relationship between [H+] and [OH-] (page 65)
Q16: 7
Q17: 1
Q18: 13
Q19:
pH is the negative power of the Hydrogen ion concentration.
This is the origin of the term pH.
Calculating pH (page 66)
Q20: The pH is 2.30
Q21: The pH is 5.10
Q22: The pH is 12.80
Q23: The pH is 10.46
Q24:
[H+ ] = 0.005 mol
-1
[OH- ] = 1.995 x 10-12 mol
-1
Q25:
[H+ ] = 2.511 x 10-6 mol
-1
[OH- ] = 3.982 x 10-9 mol
-1
Q26:
[H+ ] = 3.98 x 10-12 mol
-1
[OH- ] = 2.51 x 10-3 mol
-1
Q27:
[H+ ] = 1.26 x 10-2 mol
-1
[OH- ] = 7.94 x 10-13 mol
-1
Answers from page 69.
Q28:
"$ " " $
Answers from page 69.
Q29: b) HIO3
Q30: a) HF
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221
Answers from page 69.
Q31: b) IO3 Q32: a) FQ33:
The higher the value of K a the stronger the acid.
Strong acids will have Ka values very much greater than 1. (These are not normally
quoted.)
Weak acids have Ka values less than 1.
Answers from page 70.
Q34: c) high Ka , low pKa
Q35: sulphurous acid (pK a 1.8)
Q36: d) low Ka , high pKa
Q37: the hydrogen phosphate ion, HPO 4 2- (pKa 12.7)
Answers from page 70.
Q38: From page 12 of the data booklet,
%
%"
and
% Q39: 5.4
Q40: 0.025
Q41: From the data booklet,
%
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ANSWERS: TOPIC 4
%"
doubling rearranging
%
%
% %"
%"
antilog Q42: 2.4
Q43: 3.3
Answers from page 72.
Q44: ammonium ion
Q45: 9.3
Q46:
or
1" "
1 "
1 " "$
1 "
Q47: a) weak
Answers from page 73.
Q48: 8.2
Q49: piperidine
Q50: codeine
Q51:
The more methyl groups the stronger the base becomes.
The Ka value decreases (and pK a increases), meaning that the conjugate acid is getting
weaker. Hence the base is getting stronger.
Further answers
Answers from page 201.
Q52: A dative covalent bond (or coordinate bond) is formed since both electrons in the
bond originate from the same atom.
Q53: c) salt
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ANSWERS: TOPIC 4
The relationship between [H+] and [OH-] (page 65)
Q54: OH- is 10-13 .
Q55: 1 x 10 -14 .
Q56: 1 x 10 -14 .
Answers from page 201.
Q57: 1.6
Q58: 4.7
Q59: 0.016
Q60: 0.10
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224
ANSWERS: TOPIC 5
5 Indicators and buffers
Answers from page 78.
Q1:
a) left
Q2:
red
Q3:
b) right
Q4:
blue
Answers from page 79.
Q5:
0.0000001
Answers from page 79.
Q6:
100
Q7:
blue
Q8:
1
Q9:
green
pH titration (page 81)
Q10: 7
Q11: At the equivalence point, the exact amount of alkali has been added to neutralise
the acid; no more, no less.
Q12: 0.1
Q13: 3
Answers from page 83.
Q14: a) strong acid/strong alkali
Q15: c) 7
Q16: b) 5
Q17: d) 9
Choosing indicators (page 84)
Q18: The pH of the equivalence point falls within the pH range over which
phenolphthalein changes colour. So there will be a sharp endpoint. Both the other
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ANSWERS: TOPIC 5
225
indicators will change colour gradually. For methyl orange, the colour change takes
place long before the equivalence point.
Q19: For both methyl orange and bromothymol blue, the pH of the solution is changing
rapidly over the indicators pH range. So there will be a sharp endpoint even although
the equivalence point falls in neither range.
Buffer calculations (page 93)
Q20: 5.47
Q21: 1.58
Q22: 1.79 x 10-5
Q23: 4.6
Further answers
Answers from page 201.
Q24: d)
"$ &
"&
pH titration (page 81)
Q25: d) The pH changes rapidly only around the equivalence point.
Answers from page 202.
Q26: The pH at the equivalence point is the same as the pH of the salt formed.
combination
pH of salt
strong acid/strong alkali
7
strong acid/weak alkali
7
weak acid/strong alkali
7
weak acid/weak alkali
depends on relative strengths
Buffer calculations (page 93)
Q27: 4.1
Q28: 1.58
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ANSWERS: TOPIC 6
6 Thermochemistry
Answers from page 96.
Q1:
-894
Q2:
Activation Energy
Q3:
137
Q4: The reaction shown in Figure 6.2 has a lower reaction vessel temperature
because endothermic reactions draw heat from the reaction mixture and vessel.
Bond enthalpy values (page 98)
Q5:
864
Q6:
chlorine
Q7:
iodine
Q8: Nitrogen atoms are bonded together by a triple bond which requires more energy
to break than a double bond, as in oxygen.
Enthalpy calculations from bond enthalpies (page 99)
Q9:
432
Q10: 243
Q11: 675
Q12: 856
Finding a bond enthalpy from enthalpy changes (page 102)
Q13: 3
Q14: +1296
Q15: 6
Q16: -2484-x
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227
Q17:
You might like to compare this value with the data booklet value and suggest a reason
for the difference (refer to Figure 6.5 for a hint).
Hess’s Law questions (page 105)
Q18:
H(2)
H(3)
H(3)
H(3)
= H(1) + H(3)
= H(2) - H(1)
= +33.2 - 90.2 kJ mol-1
= -57 kJ mol-1
Q19: -849
Q20:
By using the algebraic method, you should reach this point:-
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ANSWERS: TOPIC 6
HÆf (methane)
HÆf (methane)
= HÆ c (carbon) + 2HÆ c (hydrogen) + ( - HÆ c (methane) )
= - 75 kJ mol-1
Q21: a) Using HÆ = HÆ f (products) - HÆ f (reactants)
The reaction: B2 H6 (g) + 3O2 (g)
B 2 H6
Equation
B2 O3 (s) + 3H2 O(l)
+ 3O2
B 2 O3
+3H2 O
0
-1225.0
-286.0
0
-1225.0
-858.0
given Hf (kJ +41.0
mol-1 )
Multiply by
+41.0
moles present
H Æ
= (-1225.0 - 858.0) - (+41.0 + 0) kJ
= -2083.0 - 41.0
Æ
= - 2124.0 kJ
H
b) The data book quotes ethane as having a standard molar enthalpy of combustion of
-1560 kJ mol-1 . So, mole for mole, the diborane is the one which would release most
energy. (Try working out whether one gram of each would still leave diborane as the
better fuel in terms of heat released per g of fuel.)
Bomb calorimetry (page 106)
Q22: 10.30
Q23: 5.74
Q24: 59.12
Q25: -1359.76
Answers from page 111.
Q26:
HÆLATT
HÆ LATT
HÆ LATT
HÆ LATT
=
=
=
=
HÆf - (HÆ AT + HÆI.E. + 1/2 HÆBOND + HÆE.A.)
-411 - ( +109 + 502 + (1 /2 x 243) + (-348.7))
-411 - (+383.8)
-794.8 kJ mol-1
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ANSWERS: TOPIC 6
A Born-Haber cycle for lithium fluoride (page 111)
Q27:
HÆf
=
HÆf
HÆf
HÆf
=
HÆAT + HÆI.E. + 1/2 HÆBOND + HÆE.A. + HÆLATT
+159 + 526 + (1 /2 x 155) + (-3282) + (-1030)
= +762.5 - 1358.2
=
- 595.7 kJ mol-1
Further questions to practise Born-Haber cycles (page 111)
Q28: +2721
Q29: +243
Q30: The value would be 2 x (-348.7) kJ since there are two moles of atoms having
electrons added.
Q31: -170.4
Answers from page 114.
Q32: There are two values, -418 and -338 kJ because there are two types of ion being
hydrated.
Q33: +13
Q34: b) endothermic
Q35: a) go down
Answers from page 115.
Q36: exothermic
Q37: hydration
Enthalpy of solution (page 115)
Q38: +16
Q39: b) endothermic
Q40: a) go down
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230
ANSWERS: TOPIC 6
Further answers
Enthalpy calculations from bond enthalpies (page 99)
Q41: Since:
= +675 kJ
BOND BREAKING
= -856 kJ
BOND MAKING
OVERALL ENTHALPY CHANGE = -181 kJ
Q42:
BOND BREAKING (ALWAYS POSITIVE) BOND MAKING (ALWAYS NEGATIVE)
2 mole of H-H bonds
=864
1 mole of O=O bonds
ENERGY IN
=497
=+1361
4 moles of H-Cl bonds
=4x458
ENERGY OUT
=-1832
OVERALL ENERGY CHANGE = - 471 kJ
Hess’s Law questions (page 105)
Q43: a) Using HÆ = HÆ f (products) - HÆ f (reactants)
The reaction: MgCl2 (s) + 6H2 O(l)
Equation
given Hf (kJ
mol-1 )
Multiply by
moles present
HÆ
HÆ
H Æ
MgCl2 .6H2 O(s)
MgCl2 (s)
+ 6H2 O(l)
MgCl2 .6H2 O(s)
-642
-286
-2500
-642
-1716
-2500
= HÆ f (products) - HÆ f (reactants)
= (-2500) - (-642 + (-1716)) kJ
= (-2500 + 2358)
= - 142 kJ
Bomb calorimetry (page 106)
Q44: -1367
Q45: 99.47
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231
7 Reaction Feasibility
Entropy and temperature (page 123)
Q1: 0
Q2: melting
Q3: positive
Q4: increase
Estimating and calculating spontaneity (page 125)
Q5: diamond
Q6: lowest
Q7: water
Q8: alkanes
Q9: A - the dissolving increases entropy (general principle 4).
Q10: B - separation into two layers is more organised.
Q11: A - more disordered than in the bottle!
Q12: B - gaseous oxygen becomes ordered in the solid product.
Q13:
S Æ = SÆPRODUCTS - SÆREACTANTS
S Æ = 2 x 188.7 - (2 x 130.7 + 205.2)
S Æ = 377.4 - 466.6
S Æ = - 89.2 J K-1
So, for one mole of water (as a gas) being formed the value would be - 44.6 J K -1 mol-1 .
Notice that the negative sign shows a decrease in entropy. This is expected since 3
moles of gas reactants become two moles of products.
Q14:
S Æ = SÆPRODUCTS - SÆREACTANTS
- 99.5 = 2 x SÆ AMMONIA - (191.6 + 3 x 130.7)
2 x S Æ AMMONIA = - 99.5 + 583.7
S Æ AMMONIA = + 242.1 J K-1 mol-1
Q15:
SÆ = - 313.5 J K-1 mol-1
Q16: -156.3
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ANSWERS: TOPIC 7
Melting ice (page 129)
Q17:
"
+
* (heat is absorbed from the surroundings)
Q18:
! " " * Entropy of system increases as water melts.
Q19:
* The process leads to an increase in the total entropy of the system and surroundings.
The process is spontaneous.
Q20:
At 25Æ C (298 K), STOTAL is negative (-436 J K-1 mol-1 )
At 1500ÆC (1773 K), STOTAL is positive (+60.6 J K-1 mol-1 )
Q21:
At 25Æ C (298 K), STOTAL = negative (approx -10 J K-1 mol-1 )
At 5000ÆC (5273 K), STOTAL = negative (approx -3.7 J K-1 mol-1 )
It is not thermodynamically feasible at either temperature.
Calculations involving free energy changes (page 135)
Q22: a) 2Mg(s) + CO2 (g)
-
2MgO(s) + C(s)
-! " -" )* The reaction where magnesium reduces carbon dioxide is feasible under standard
conditions.
b) 2CuO(s) + C(s)
2Cu(s) + CO2 (g)
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ANSWERS: TOPIC 7
233
-! " -" )* -
The reaction where carbon reduces copper(II) oxide is feasible under standard
conditions.
Q23:
HÆ = +117 kJ mol-1
SÆ = +175 J K-1 mol-1
GÆ = HÆ - TSÆ
GÆ400 = +47 kJ mol-1
GÆ1000 = -58 kJ mol-1
Since
At 400 K
At 1000 K
This reaction is feasible only at higher temperatures.
Q24:
* * " -
" +
+
Q25:
a) GÆ = +0.16 kJ
b) Equilibrium position favours the reactants.
Answers from page 138.
Q26:
-
' %
, , +
+
" + + This reaction becomes feasible at 708.4 Æ C and above.
Interpreting Ellingham diagrams (page 140)
Q27: Silver(I) oxide. At 1000 K, G is +60 kJ mol-1 on the graph. Even with no other
chemical involved, this reverses to breakdown silver(I) oxide with GÆ = -60 kJ mol-1
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ANSWERS: TOPIC 7
Q28: Above 2200 K approximately. This would allow G to be negative for:
2Zn + O 2
2ZnO
Q29:
a)
(i) At 1000 K the target equation is:
(ii)
2C(s) + 2ZnO(s)
2CO(g)
2C(s) + O2 (g)
2Zn(s) + O2 (g)
2ZnO(s)
(iii) Reverse the zinc equation:
2ZnO(s)
2Zn(s) + 2CO(g)
GÆ = -400 kJ mol-1
GÆ = -500 kJ mol-1
GÆ = +500 kJ mol-1
2Zn(s) + O2 (g)
(iv) Adding to the carbon equation gives:
2C(s) + O2 (g) + 2ZnO(s)
2CO(g) + 2Zn(s) + O2 (g)
The oxygen on each side cancels out giving:
2C(s) + 2ZnO(s)
2CO(g) + 2Zn(s)
GÆ = +100 kJ mol-1
So at 1000 K the reaction is not feasible.
b)
(i) At 1500 K the target equation is the same:
(ii)
2C(s) + 2ZnO(s)
2CO(g)
2C(s) + O2 (g)
2Zn(s) + O2 (g)
2ZnO(s)
(iii) Reverse the zinc equation:
2ZnO(s)
2Zn(s) + 2CO(g)
GÆ = -500 kJ mol-1
GÆ = -300 kJ mol-1
GÆ = +300 kJ mol-1
2Zn(s) + O2 (g)
(iv) Adding to the carbon equation gives:
2C(s) + O2 (g) + 2ZnO(s)
2CO(g) + 2Zn(s) + O2 (g)
The oxygen on each side cancels out giving:
2C(s) + 2ZnO(s)
2CO(g) + 2Zn(s)
GÆ = -200 kJ mol-1
So at 1500 K the reaction is feasible.
Q30: Where the two lines cross,
feasible.
G Æ
= 0. Above this temperature, the reaction is
Remember: The lower of the two lines operates as written and the upper line will be
reversed.
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235
Answers from page 141.
Q31:
a) 2FeO(s) + 2C(s)
2Fe(s) + 2CO(g)
b) GÆ = -155 kJ mol-1
c) Above 1010 K
d) Below 980 K
e) It is a gas and can mix better with the solid iron(II) oxide.
Q32:
a) above about 2100 K
b) The high cost of maintaining temperature. The fact that magnesium is a gas at this
temperature.
c) GÆ = +160 kJ mol-1 (there would be some leeway in this figure).
d) GÆ = +252 kJ mol-1 (dependent on your answer to part (c)).
e) Keeps the equilibrium reaction below from going in the reverse direction.
2MgO + Si
« SiO
2
+ 2Mg
Further answers
Entropy and temperature (page 123)
Q33: boiling
Q34: increase
Q35: positive
Q36:
The degree of disorder (entropy) of the water molecules as a gas above 373 K increases
rapidly as the temperature increases. The molecules also vibrate and move much more
freely. Below 273 K, the water molecules in ice vibrate more as they are heated but their
movement is still restricted and thus their entropy only increases slowly.
Estimating and calculating spontaneity (page 125)
Q37: Butane is a gas with a high degree of disorder, and although pentane is a more
complex molecule, it is a liquid under standard conditions and therefore has a more
ordered structure.
Q38: B - there are four moles of gas reactants, only two moles of gas products.
Q39: C - although the zinc conforms to general principle 4, the copper ions go the other
way by the same amount.
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ANSWERS: TOPIC 7
Calculations involving free energy changes (page 135)
Q40: Boiling point = 333 K or 60 Æ C
Interpreting Ellingham diagrams (page 140)
Q41: As the zinc melts, the disorder (entropy) increases. Since the gradient is given by
-S (from the straight line G = -TS + H), the slope of the line changes.
Q42: Zinc vaporises at 1180 K with an increase in entropy and a subsequent change in
the gradient of the line on the Ellingham diagram.
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237
8 Electrochemistry
Answers from page 149.
Q1:
Copper will be positive with
respect to the solution.
Q2: a) gain
Q3: a) zinc
Q4:
The more reactive the metal, the more the equilibrium will move towards the ions.
The more reactive the metal, the more negative it will be with respect to the solution.
Answers from page 149.
Q5: a) reduced?
Q6: a) an oxidising agent?
Q7: Zinc atoms lose electrons and copper(II) ions gain them, i.e.
transferred from zinc to copper (II) ions.
electrons are
Zinc - copper cell. (page 150)
Q8: Zn(s)
Zn2+ (aq) + 2e-
Q9: Cu2+ (aq) + 2e-
Cu(s)
Q10: electrons
Q11: To complete the circuit by allowing the movement of ions between the two
solutions. For this reason, it is often known as an ion bridge.
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ANSWERS: TOPIC 8
Answers from page 153.
Q12:
Mg2+ (aq) + 2e-
Oxidation is Mg(s)
Reduction is Zn2+ (aq) + 2eMg(s) + Zn2+ (aq)
Zn(s)
Mg2+ (aq) + Zn(s)
Q13:
Cr3+ (aq) + 3e-
Oxidation is Cr(s)
Reduction is Cu+ (aq) + e-
Cu(s)
Reduction must be multiplied by 3
Cr(s) + 3Cu+ (aq)
Cr3+ (aq) + 3Cu(s)
2H+ (aq) + 2Cl- (aq)
Q14: H2 (g) + Cl2 (g)
Q15: Ni(s) + 2H+ (aq)
Ni2+ (aq) + H2 (g)
Answers from page 153.
Q16: Magnesium more reactive
Mg2+ (aq)
-
flow from Mg
Zn(s)
Q17: Ni(s)Ni (aq)Pb (aq) Pb(s)
Q18: Cu(s) Cu (aq) Cl (g)Cl (aq) Pt(s)
Mg(s)
µe
2+
2+
2+
Q19:
Zn.
Zn2+ (aq)
-
2
Pt(s) H2 (g) H+ (aq) Hg2+ (aq) Hg(l)
Or
Pt(s) H2 (g) H+ (aq) Hg2+ (aq) Hg(l) Pt(s)
Answers from page 156.
Q20: b) Hydrogen electrode
2H+ (aq) + 2e-
Q21: H2 (g)
Q22: Cu2+ (aq) + 2eQ23:
Pt(s) H2 (g) H+ (aq)
Cu(s)
Cu
2+ (aq)
Cu(s)
Note that in this case, the diagram (Figure 8.9) has the hydrogen electrode on the right
hand side. So we cannot work out the cell notation by looking at the diagram.
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239
Answers from page 157.
Q24: a) Zinc electrode
Q25: Zn(s)
Zn2+ (aq) + 2e-
Q26: 2H+ (aq) + 2e-
H2 (g)
Q27: Zn(s) + 2H+ (aq)
Zn2+ (aq) + H2 (g)
Answers from page 159.
Q28: +1.61
Q29: +1.08
Q30: +1.36
Q31: +1.30
Oxidising agents and reducing agents (page 159)
Q32: b) F2 (g) + 2e-
2F- (aq)
Q33: b) at the bottom of the table
Q34: An oxidising agent is a substance that causes something else to be oxidised. If
the other substance is oxidised, the oxidising agent is reduced.
Q35: oxidising agent
Q36: a) top of the table
Q37: A reducing agent is one that causes something else to be reduced. If the other
substance is reduced, the reducing agent must be oxidised.
Q38: right
Q39: b) Top right
Answers from page 161.
Q40: D
Q41: C
Q42: E
Q43: B
Q44: A
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ANSWERS: TOPIC 8
Calculation of standard free energy change. (page 164)
Q45: -308.8
Q46: -371.3
Q47: -289.5
Q48: -110.0
Calculating standard entropy changes. (page 165)
Q49:
H = -cmT
Where: H
is the enthalpy change in kJ.
c
is the specific heat capacity of water (4.18 kJ kg-1 Æ C-1 ).
m
is the mass of substance being heated in kg.
is the change in temperature in Æ C.
(If the temperature rises, T is positive and hence H is negative - an exothermic
process.)
T
Q50:
*
)*
"
Q51:
%% %% " )* " )* Q52: +0.46
Answers from page 167.
Q53: proton exchange membrane
Q54: Pt(s) H2 (g) H+ (aq)
H (aq) O (g) Pt(s)
+
2
Q55: 1.23
Q56: 2H2 (g) + O2 (g)
2H2 O(l)
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ANSWERS: TOPIC 8
241
Further answers
Zinc - copper cell. (page 150)
Q57: The substance should be:
1. ionic (i.e. an electrolyte)
2. soluble
3. and should not react with the other solutions.
Examples include potassium nitrate, sodium chloride and sodium sulphate.
Q58: The current will stop flowing whenever one of the reactants is completely used up.
In the simulation, copper(II) ions were used up first.
Answers from page 204.
Q59:
Oxidation is Al(s)
Al3+ (aq) + 3e-
Reduction is Pb2+ (aq) + 2e-
Pb(s)
The oxidation must be multiplied by 2 and the reduction must be multiplied by 3.
2Al(s) + 3Pb2+ (aq)
2Al3+ (aq) + 3Pb(s)
Answers from page 204.
Q60: Pt(s) Fe2+ (aq), Fe3+ (aq) Ag+ (aq) Ag(s)
Answers from page 204.
Q61: EÆ = -0.76 V
The voltmeter reading of +0.76 V is assigned to the half-cell reaction occurring at the
zinc electrode, since the value for hydrogen is zero. In the cell, zinc atoms are oxidised
to zinc(II) ions. So the oxidation potential is +0.76 V. The sign must be reversed to get
the reduction potential.
Answers from page 204.
Q62: +1.55
Q63: +0.62
Oxidising agents and reducing agents (page 159)
Q64: c) Bottom left
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ANSWERS: TOPIC 8
Q65: A possible diagram is:
Figure 10.2: Summary of oxidising/reducing agents
Answers from page 205.
Q66: C
Q67: D
Q68: F
Calculation of standard free energy change. (page 164)
Q69: -642.7
Q70: -239.3
Calculating standard entropy changes. (page 165)
Q71: -88.8
Q72:
GÆ = HÆ - TSÆ
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ANSWERS: TOPIC 8
Q73:
243
- " + + " -
" + -
" )* * )* * * ) 2 . )* *
2 -
! Q74: The negative sign means a local decrease in entropy, i.e. the system has become
more ordered.
From the equation, it can be seen that:
1 mole solid
+
2 moles solution
2 moles solid
+
1 mole solution
(ordered)
(disordered)
(ordered)
(disordered)
The products are more ordered than the reactants, i.e. the products have a lower
entropy.
Answers from page 205.
Q75: Production of water for drinking. The cell would also produce heat which could be
useful.
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ANSWERS: TOPIC 9
9 Kinetics
Answers from page 173.
Q1:
Only colour cannot be used since none of the reactants or products is coloured.
A gas is produced and its volume will increase as the reaction proceeds. Also the mass
of the apparatus will decrease as the gas escapes. You will probably have used these
methods at Standard Grade or Higher.
Acid is used up in the reaction. So the pH will increase.
Acids are good conductors of electricity and the total number of ions will decrease as
the reaction proceeds. Consequently the conductivity will decrease.
Q2:
Changes in pH and conductivity can be used since H + (aq) ions are produced.
Iodine is the only coloured substance present so the solution will get paler as the
reaction proceeds. The change in colour can be monitored using a colorimeter.
Change in concentration with time (page 173)
Q3:
0
Q4:
0
Remember that the rate will always have a positive value. Since
and
0
the signs cancel out.
Q5:
0.4
Q6:
1.8 x 10-5 mol
-1
s-1
Answers from page 176.
Q7:
• Concentration of catalyst if homogeneous
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ANSWERS: TOPIC 9
• Mass and particle size of catalyst if heterogeneous
• Temperature
• Pressure
Answers from page 176.
Q8: A straight line graph is obtained.
Q9:
The rate is directly proportional to [H2 O2 ]
i.e. Rate
[H O ]
2
2
or Rate = constant x [H2 O2 ]
Answers from page 177.
Q10: 1
Q11: 1
Q12: 2
Orders and rate constants (page 179)
Q13: Rate = k [Br- ][BrO3 - ][H+ ]2
Q14: 4
Q15: 2
Q16: 0
Q17: 1
Q18: 1
Q19: Rate = k [N2 O5 ]
Q20: s-1
Q21: 0.00044
Q22: 0.0000308
Q23:
Rate = k [H2 O2 ][I- ]
or
Rate = k [H2 O2 ][I- ][H+ ]0
Q24: mol-1
s-1
Q25: 0.023
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246
ANSWERS: TOPIC 9
Answers from page 183.
Q26: 3001.5
Q27: 30001.5 seconds.
The filler takes 30 s to fill one bottle, so the thousandth bottle will be filled after 1000 x
30 s and a further 1.5 s will be needed to cap and label it.
Answers from page 183.
Q28: Rate = k [NO2 ][CO]
Questions on reaction mechanisms (page 186)
Q29: b) H2 S + Cl2
S + 2HCl
Q30: a) 4HBr + O2
2H2 O + 2Br2
Q31: b) False
Q32: b) False
Q33: a) True
Q34: b) False
Q35: 2H2 O2
2H2 O + O2
Q36: catalyst
Q37: intermediate
Q38: b) Rate = k [H2 O2 ][Br- ]
Further answers
Change in concentration with time (page 173)
Q39: The reaction slows down.
The rate depends on the concentration of the reactants. As the reactants are used up,
the concentration decreases and so the rate decreases.
You can see this more clearly if you plot graphs to show the change in concentration of
bromine with time and the change in volume of CO 2 with time.
Questions on reaction mechanisms (page 186)
Q40: b) False
Q41: a) True
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