scholar study guide sqa advanced higher chemistry unit · pdf filescholar study guide sqa...

SCHOLAR Study Guide

SQA Advanced Higher ChemistryUnit 2Principles of Chemical Reactions

Peter JohnsonHeriot-Watt University

Brian T McKercharBalerno High School

Arthur A SandisonSt 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 foreducational 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 systemor 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 theinformation 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.

AcknowledgementsThanks are due to the members of Heriot-Watt University’s SCHOLAR team who planned andcreated these materials, and to the many colleagues who reviewed the content.

We would like to acknowledge the assistance of the education authorities, colleges, teachersand students who contributed to the SCHOLAR programme and who evaluated these materials.

Grateful acknowledgement is made for permission to use the following material in theSCHOLAR programme:

The Scottish Qualifications Authority for permission to use Past Papers assessments.

The Scottish Government for financial support.

All brand names, product names, logos and related devices are used for identification purposesonly and are trademarks, registered trademarks or service marks of their respective holders.

i

Contents

1 Stoichiometry 11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Using moles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 Volumetric analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.4 Gravimetric analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171.6 Suggestions for further reading . . . . . . . . . . . . . . . . . . . . . . . 181.7 Tutorial questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.8 End of Topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2 Chemical Equilibrium 212.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232.2 The nature of chemical equilibrium . . . . . . . . . . . . . . . . . . . . . 242.3 The equilibrium constant, Kc . . . . . . . . . . . . . . . . . . . . . . . . 262.4 The equilibrium constant, Kp, involving gases . . . . . . . . . . . . . . . 292.5 Homogeneous and heterogeneous equilibria . . . . . . . . . . . . . . . 302.6 Calculations using equilibrium constants . . . . . . . . . . . . . . . . . . 312.7 Factors that alter the composition of an equilibrium mixture . . . . . . . 352.8 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392.9 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 402.10 End of Topic Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

3 Phase Equilibria 413.1 Partition coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 423.2 Solvent extraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 443.3 Chromatography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 533.5 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 543.6 End of Topic Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

4 Acid/base Equilibria 574.1 What are acids and bases? . . . . . . . . . . . . . . . . . . . . . . . . . 584.2 The ionic product of water and the pH scale . . . . . . . . . . . . . . . . 624.3 Dissociation of acids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 684.4 Dissociation of bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . 714.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 734.6 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 744.7 End of Topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

ii CONTENTS

5 Indicators and buffers 775.1 Indicators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 785.2 Buffer solutions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 855.3 Calculating pH and buffer composition. . . . . . . . . . . . . . . . . . . . 905.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 935.5 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 945.6 End of Topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

6 Thermochemistry 956.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 966.2 Bond energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 976.3 Hess’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1006.4 Standard enthalpy changes . . . . . . . . . . . . . . . . . . . . . . . . . 1026.5 Calorimetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1066.6 The Born-Haber cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1086.7 Enthalpy of solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1126.8 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1166.9 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1176.10 End of Topic test. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

7 Reaction Feasibility 1197.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1217.2 The concept of entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1227.3 The second law of thermodynamics . . . . . . . . . . . . . . . . . . . . 1277.4 Gibbs free energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1307.5 Ellingham diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1387.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1437.7 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1447.8 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144

8 Electrochemistry 1458.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1478.2 Electrochemical cells . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1478.3 Standard Electrode Potentials . . . . . . . . . . . . . . . . . . . . . . . . 1548.4 EÆ values and the standard free energy change . . . . . . . . . . . . . . 1628.5 Fuel cells . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1668.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1698.7 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1708.8 End of Topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170

9 Kinetics 1719.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1729.2 Measuring reaction rates . . . . . . . . . . . . . . . . . . . . . . . . . . 1729.3 Rate equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1759.4 Reaction mechanisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1819.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1879.6 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1899.7 End of Topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189

© HERIOT-WATT UNIVERSITY

CONTENTS iii

10 End of Unit 2 Test (NAB) 191

Glossary 193

Further questions 198

Answers to questions and activities 2071 Stoichiometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2072 Chemical Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2133 Phase Equilibria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2174 Acid/base Equilibria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2195 Indicators and buffers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2246 Thermochemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2267 Reaction Feasibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2318 Electrochemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2379 Kinetics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244


1

Topic 1

Stoichiometry

Contents

1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2 Using moles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2.1 Converting moles and mass . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.2.2 Converting moles and volumes for gases . . . . . . . . . . . . . . . . . 6

1.2.3 Determining excess or limiting reactants . . . . . . . . . . . . . . . . . . 71.2.4 Calculating concentrations . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.2.5 Diluting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.3 Volumetric analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.3.1 Titrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.3.2 Redox titrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.3.3 Complexometric volumetric analysis . . . . . . . . . . . . . . . . . . . . 15

1.4 Gravimetric analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171.6 Suggestions for further reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

1.7 Tutorial questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

1.8 End of Topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

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, redox andcomplexometric);

2 TOPIC 1. STOICHIOMETRY

• calculate the concentration of solutions from titrimetric data;

• manipulate results from gravimetric determinations.


1.1. INTRODUCTION 3

1.1 Introduction�

Æ

�

�

Learning Objective

To revise the basic skills, relating to balanced equations and the calculation of formulamasses that are required in stoichiometry

Stoichiometry involves the understanding of the numerical relationships betweenreacting substances. It forms the basis for analytical chemistry. (How much iron isthere in this sample of iron ore? How pure is this sample of anti-viral drug?) It is alsocritical for industrial chemistry. (How many tonnes of sulphuric acid and phosphate rockare required to manufacture this batch of fertiliser? What proportion of monomers arerequired to prepare this plastic?)

These numerical relationships are obtained by consideration of a quantitative reactionin which substances react completely according to the mole ratio given by thebalanced (stoichiometric) equation.

Here are some revision questions for you to try before you proceed with this Topic. If youuse 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: P4O10

Use the relative atomic masses in the data booklet to calculate the relativemolecular masses of the following substances. Give your answers to 1 decimalplace:

Q9: CaO

Q10: Al2(SO4)3

Q11: Chromium(III) oxide

Q12: Butane



Use the numbers from the list below to complete these balanced equations:

[ ] Zn + O2� [ ] ZnO

Li2SO4 + BaCl2� [ ] LiCl + BaSO4

[ ] Ca(NO3)2 + [ ] Na3 PO4�Ca3(PO4)2 + [ ] NaNO3

[ ] C2H4O + [ ] O2 �[ ] CO2 + [ ] H2O

[2], [2], [2], [2], [2], [3], [4], [4], [5], [6].

Q13:

When calcium carbonate reacts with hydrochloric acid, the reaction is described by theequation:

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 hydrochloricacid 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 produceone mole each of calcium chloride, water and carbon dioxide.

a) statement 1 onlyb) statements 1 and 2c) statements 1, 2 and 3.

Q14: When real quantities need to be weighed out in the laboratory or industry, whichstatement is most helpful?

a) statement 1b) statement 2c) statement 3

Balancing equations

An online exercise is provided to help you if you require additional assistance with thismaterial, 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 theamount of a chemical, for example, by weighing a solid, measuring the volume of a gas,determining excess reactants or when diluting solutions.


1.2. USING MOLES 5

1.2.1 Converting moles and mass

To convert from moles to grams or vice versa, use the relationship that one mole isequal 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 areburned? Give your answer to 1 decimal place. The equation is:

CH4 + 2O2 � CO2 + 2H2O.

Q16: How many kilograms (kg) of limestone (CaCO3) are required to produce 168.0 kgof quicklime (CaO) according to the equation below? Give your answer to 1 decimalplace.

CaCO3 � CaO + CO2

Q17: 8.10 g of magnesium was added to excess copper(II) sulphate solution. Howmany grams (to 2 decimal places) of copper metal was produced?

Q18: How many kilograms (kg) of anhydrous sodium carbonate (Na2CO3) are requiredto neutralise a spill of 6 kg of ethanoic acid (C2H4O2)? Give your answer to 1 decimal



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 1molar 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)2Cr2O7(s) � Cr2O3(s) + 4H2O(l) + N2(g)

What volume of gas would you obtain from 5.15 g of ammonium dichromate at 25 ÆC ?The molar volume of nitrogen at 25 ÆC is 24.45 � mol-1.

N.B. If the question had stated the temperature as 125 ÆC, the water would be a gas,so that 1 mole of ammonium dichromate (252 g) would produce 5 moles of gaseousproduct (1 mole of N2 + 4 moles of steam). The molar volume at the higher temperaturefor both of these is 32.7 � mol-1. Under these conditions, the 5.15 g of ammoniumdichromate would produce 3.34� of gas.

Q19: Hydrogen peroxide decomposes according to the equation:

2H2O2(aq)� 2H2O(l) + O2(g)


1.2. USING MOLES 7

Assuming that the molar volume of oxygen is 24.0 � mol-1, how many litres (to 1 decimalplace) could be obtained from 17.0 g of hydrogen peroxide?

Q20: Nitrous oxide (laughing gas, N2O) can be produced by heating ammonium nitrateaccording to the equation:

NH4NO3 �2H2O + N2O

How many litres (to 1 decimal place) of N2O could be obtained from 10.0 g of ammoniumnitrate? The molar volume of N2O is 24.0 � mol-1

.

Q21:

Phosphorus reacts with aqueous sodium hydroxide according to the equation:

P4 + 3NaOH +3H2O�3NaH2PO2 + PH3

How many litres (to 1 decimal place) of phosphine (PH3) could be obtained from 10.3 gof 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 ÆC is 30.0 � mol-1, how many litrescould be obtained by heating 24.5 g of potassium chlorate? Give your answer to 1decimal place.

1.2.3 Determining excess or limiting reactants

In many situations, exact balanced (stoichiometric) amounts of reactants and productsmay not be present. In a situation with two reactants, one or other will be in excess; theother 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 bariumions. How much of the ammonium sulphate remained in the solution at the end ofthe precipitation? Give your answer in grams to 2 decimal places.



Q23: Cadmium ions form a precipitate of cadmium sulphide (CdS) with sodium sulphidesolution.

Cd2+ + Na2S � 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 insolution?

Q24: The residue from photographic processing equipment is being analysed for itssilver ion content by precipitation of silver chloride. The anticipated mass of silver is0.54 g. How many grams (to 2 decimal places) of potassium chloride should be addedto ensure that twicethe required amount is present at the beginning of the reaction?

Q25: 30 g each of ethanoic acid (CH3CO2H) and ethanol (C2H5OH) are reacted toprepare the ester, ethyl ethanoate (CH3CO2C2H5). Which material is in excess?

Q26: Assuming all the limiting reactant in the question above could be used, whatweight 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 solutionwould produce a solution of concentration 0.2 mol � -1.

��

� �� (1.1)


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 potassiumiodate (KIO3) is dissolved in 250 ml of solution.

Q27: Calculate the molar concentration, in mol �-1, of a solution made by dissolving 4.84g of hydrated barium chloride (BaCl2.2H2O) in 100 ml of solution. Give your answer to2 decimal places.

Q28: What is the molar concentration (in mol �-1 to 2 decimal places) of the solutionobtained by dissolving 5.10 g of potassium hydrogenphthalate (C8H5O4K) in 250 ml ofsolution?

Q29: How many grams of hydrated iron(III) chloride (FeCl3.6H2O) are required toprepare 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 bypressing Display Answer.

Q30: You have been asked to prepare 2 litres of 0.025 mol � -1 sodium thiosulphatesolution. How many grams (to 2 decimal places) of solid Na2S2O3.5H2O should youweigh out?

1.2.5 Diluting

When diluting a concentrated solution, it is useful to remember that the number ofmoles of the solute remains the same in the original and diluted solutions; only theamount of solvent will change.

Using this equality of moles before and after dilution, we can use:

M1V1 = M2V2

where M1 and V1 are the molar concentration and volume, respectively, of theconcentrated solution, and M2 and V2 apply to the diluted solution.

Example How many ml of concentrated hydrochloric acid (11 mol � -1) would be required



to prepare 2.5 litres of 2 mol � -1 hydrochloric acid?

Using M1V1 = M2V2

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 � -1concentrated solution. What volume(in ml to 1 decimal place) of this would be required to make 2 litres of 2 mol � -1dilutesolution?

Q32: How many litres (to 1 decimal place) of 0.125 mol �-1 sodium thiosulphate couldbe prepared from 100 ml of 2 mol �-1 solution?

Q33: Potassium ion for intravenous drips should be administered at 0.040 mol � -1. Howmany ml (to 1 decimal place) of 1.0 mol �-1 concentrated solution should be made up to0.5 � to obtain the correct concentration?

Q34: 0.1 ml of an electro-plating solution was diluted to 100 ml. It was then found tocontain 0.005 mol �-1 chromic acid (H2CrO4). What was the concentration of chromiumin 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 quantitativereaction to determine the concentration of another reactant. The reactions used involumetric analysis usually involve acid/base neutralisations, metal complex formationor redox reactions. The procedure, in which the volumes of reacting solutions aredetermined, is called a titration.

A solution of accurately known concentration is called a standard solution. A standardsolution is usually prepared from a primary standard, which is a substance with thefollowing 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 indefinitelywithout 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 molarquantities are reduced.

These characteristics are required to ensure that what is weighed out is anuncontaminated, accurate amount of the material. The most important property is thestability. Many common substances are not stable in air, e.g. sodium hydroxide absorbsboth water and carbon dioxide from air and is therefore unsuitable as a primary standard.

Common primary standards are:


1.3. VOLUMETRIC ANALYSIS 11

• 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) Trueb) False

Q36: It must be stable in air and in solution.

a) Trueb) False

Q37: It must be soluble (probably in water).

a) Trueb) False

Q38: It must be acidic.

a) Trueb) False


1.3.1 Titrations

Once a standard solution of accurately known concentration has been prepared, it canbe used in a titration experiment to determine the concentration of another solution ofa 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 pointis taken to indicate the titration’s end point. In acid/base titrations, there is always achange in pH at the equivalence point, this is often signalled by a change in the colourof 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 wererequired to neutralise 20 ml of barium hydroxide solution completely. So what is theconcentration of barium hydroxide?



Alternative methods of performing these calculations are described in the Higher Stillsupport document ’Unit 2: Principles of Chemical Reactions’.

Q39: 50.0 ml of 0.250 mol �-1 potassium hydroxide solution are completely neutralisedby 20.0 ml of sulphuric acid. What is the concentration of the acid in mol �-1? Give youranswer to 4 decimal places.

Q40: The effluent from a nuclear plant contains nitric acid. If 25.0 ml of the effluentrequires 42.5 ml of 0.010 mol �-1 sodium hydroxide for neutralisation, what is theconcentration 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 mlof the juice require 12.0 ml of 0.050 mol �-1 sodium hydroxide solution for completeneutralisation? Give your answer to 3 decimal places.


Q42: A flask contains 52.5 ml of 0.15 mol � -1calcium hydroxide (Ca(OH)2). How manymillilitres of 0.35 mol �-1 sodium carbonate (Na2CO3) are required to react completelywith the calcium hydroxide in the following reaction? Give your answer to 1 decimalplace.

Na2CO3 + Ca(OH)2 � CaCO3 + 2 NaOH


1.3.2 Redox titrations

Redox titrations are based on oxidation-reduction reactions. Two common systems usepotassium permanganate and iodine as the oxidising agents.



1.3.2.1 Redox titrations using potassium permanganate

Potassium manganate(VII) (potassium permanganate) is widely used in redox titrationsas it acts as its own indicator. It is decolourised in a redox reaction and therefore theend point is indicated when a very pale pink colour (slight excess of permanganate) isobserved.

Example : Permanganate titration

’Iron tablets’ for treating anaemia should each contain 25 mg of iron(II). Ten tablets weredissolved 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 correctquantity 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-equationfor nitrite oxidation is:

NO2- + H2O � 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



concentration may be obtained by reaction with potassium permanganate according tothe reaction:

2MnO4- + 6H+ + 5H2O2 � 2Mn2+ + 5O2 + 8H2O

A solution of hydrogen peroxide was diluted 40 times and 25 ml of this solution required21.32 ml of 0.02 mol �-1 KMnO4 for complete reaction. Calculate the percentagehydrogen peroxide in the original solution to 1 decimal place.


1.3.2.2 Redox titrations involving iodine

Another common redox system uses iodine. This is useful, since traces of iodine give adark blue colour with starch, which can be used as an indicator.

In the next question, an oxidising agent (ozone) quantitatively converts iodide ion toiodine. The weight of this is then determined by titration with thiosulphate from whichthe amount of ozone can be calculated.

Q45: Ozone (O3) reacts with iodide ions according to the equation:

O3 + 2I- + H2O �O2 + I2 + 2OH-

A 50 litre sample of air containing ozone was passed through a solution of potassiumiodide producing iodine. The iodine released reacted completely with 20.0 ml of 0.01mol �-1 sodium thiosulphate. What weight of iodine was produced? Give your answer inmilligrams, to 2 decimal places.


Q46: What weight of ozone was in the air sample?


Q47: Propanone reacts with iodine in alkaline solution according to the equation:

(CH3)2CO + 3I2 + 4NaOH� CHI3 + 3NaI + CH3CO2Na + 3H2O

Titration of 25.0 cm3 of a solution of propanone required 18.75 cm3 of 0.20 mol �-1

iodine for complete reaction. What is the molar concentration of propanone, to 2 decimalplaces?

1.3.2.3 Other redox titrations

Other oxidising agents often used are cerium(IV), Ce4+, and dichromate, Cr2O72-. Here

are two questions involving these.

Q48: 2.25 g of impure cerium(IV) oxide (CeO2) was dissolved in sulphuric acid andmade up to 500 ml. 25 ml of this solution was titrated with 30.0 ml of 0.02 mol � -1



ammonium iron(II) sulphate. What is the molar concentration of cerium(IV) ions in thesolution, 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 youranswer 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 andtitrated with standard potassium dichromate (K2Cr2O7) to determine its iron content.The equation is:

6Fe2+ + Cr2O72- + 14H+ � 6Fe3+ + 2Cr3+ + 7H2O

If the original alloy weighed 1.25 g and 29.87 ml of 0.10 mol � -1 potassium dichromatewas required for oxidation, calculate the percentage of iron in the sample to the nearestunit. (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 usingtheir ability to form complex ions with certain organic ligands.

Many titrations employ EDTA (ethylenediaminetetraacetic acid) which has multiplecomplex-forming sites within one molecule, enabling it to form stable one-to-onecomplexes with many metals.

The end point of these titrations is indicated by the colour change in an indicator suchas 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 EDTAfor 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 �-1.



Q51: How many moles of EDTA would be required to form a complex with the strontiumions 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 a25.0 ml sample of water required 4.00 ml of 0.010 mol � -1 EDTA for complete reactionwith 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 (Referto SCCC document).

Discuss with your tutor whether the PPA on the determination of the concentration ofNi2+ 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 specificchemical in a material is determined by converting it to a product which can be isolatedcompletely and weighed.

For example, lead sulphate is insoluble in water and when excess sulphate ions areadded to a solution containing lead ions, lead sulphate is precipitated. This can beseparated from other materials by filtration, then dried and weighed to determine thelead content of the original sample

Example : Gravimetric analysis

A sample of water is known to contain lead ions. When an excess of sodium sulphateis added to 2.00 litre of the sample, a precipitate of lead sulphate (PbSO4) is produced.On filtering and drying, it was found to weigh 231.5 mg. What is the concentration oflead in the water in mg �-1?

Work from the balanced equation:

Pb2+(aq) + Na2SO4(aq) � PbSO4(s) + 2Na+(aq)

• 1 mol of Pb2+ ions produce 1 mol of PbSO4 precipitate.

• 207.2 mg of Pb2+produces 303.3 mg PbSO4

• (207.2 /303.3) x 231.5 mg Pb2+ produces 231.5 mg PbSO4

• 158.15mg of Pb2+ ion is present in the 2.00 litres of water, the concentration is79.075 mg �-1.

Q53: Excess potassium chromate was added to a solution of silver nitrate to precipitatesilver chromate (Ag2CrO4). If 11.06 g of silver chromate was produced what weightof silver nitrate was present in the original solution? Give your answer in grams to 2decimal places.


1.5. SUMMARY 17

Q54: Hydrogen sulphide was passed into 100 ml of an acidified solution containingcopper(II) ions to precipitate copper(II) sulphide (CuS). If 4.780 g of copper(II) sulphidewas 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 aluminiumsulphate (Al2(SO4)3) was treated with an excess of ammonium hydroxide (NH4OH) toprecipitate aluminium hydroxide. This was heated and decomposed into aluminiumoxide, which had been weighed. The equations are:

Al2(SO4)3 + 6OH- � 2Al(OH)3 + 3SO42-

2Al(OH)3 � Al2O3 + 3H2O

The solution of aluminium sulphate yield 1.054 g of aluminium oxide. How many gramsof aluminium sulphate were in the original sample?


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 stronglyto convert it to copper(II) oxide. The equations are:

Cu2+ + CO32- �CuCO3

CuCO3 � CuO + CO2

If the copper(II) oxide weighed 0.159 g, what is the molar concentration of Cu2+ in theoriginal solution? Give your answer to 2 decimal places.


The gravimetric determination of water in hydrated barium chloride (Unit2 PPA 2)

A prescribed practical activity for assessment of Outcome 3 may be carried out. (Referto SCCC document)

Discuss with your tutor whether the PPA on the determination of the moles of water ofhydration 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 equationsand to calculate gram formula masses.

• Reacting quantities in mass units use the fact that one mole is one gram formulamass; reacting quantities in volumes for gases use the fact that one mole of a gasoccupies one molar volume.

• Standard solutions are of accurately known concentrations.

• Primary standards, which can be weighed to produce standard solutions, require



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 canuse acid-base, redox or complexometric reactions, in a volumetric analysis.

• Gravimetric analysis estimates the amounts of sustances by isolating a chemicaland 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 Calculationswith Chemical Formulae and Equations.

• Chemistry Ken Gadd and Steve Gurr University of Bath 1994 Ch.7 QuantitativeAnalysis.

• SCCC Booklet CSYS CHEMISTRY No.10 Chemistry: Certificate of Sixth YearStudies. Sec. 1 Stoichiometry.

• Chemistry in Context Third Edition Graham Hill and John Holman Nelson 1989Ch.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 asolution containing 165.6 mg of lead(II) nitrate by the addition of excess sodiumsulphate? Answer to 1 decimal place.

Q58: The catalyst aluminium chloride (AlCl3) is prepared by passing hydrogen chloridegas 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, howmany litres (to 1 decimal place) of hydrogen chloride would be required to prepare 26.7g of aluminium chloride?

Q59: How many litres of hydrogen would be produced by the reaction in the previousquestion?

Q60: One way to restore the brightness of lead pigments in old masters that have beendulled by the formation of lead sulphide is to treat with hydrogen peroxide. This convertsthe brown lead sulphide into white lead sulphate.


1.8. END OF TOPIC TEST 19

PbS + 4H2O2 � PbSO4 + 4H2O

What mass in grams of hydrogen peroxide should be used to restore a picture containing5.18 g of lead if a 10% excess is required to ensure effectiveness? Give your answer to3 significant figures.


1.8 End of Topic test

An online assessment is provided to help you review this topic.


21

Topic 2

Chemical Equilibrium

Contents

2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.2 The nature of chemical equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . 24

2.3 The equilibrium constant, Kc . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2.3.1 The units of equilibrium constants . . . . . . . . . . . . . . . . . . . . . 28

2.4 The equilibrium constant, Kp, involving gases . . . . . . . . . . . . . . . . . . . 29

2.5 Homogeneous and heterogeneous equilibria . . . . . . . . . . . . . . . . . . . 30

2.6 Calculations using equilibrium constants . . . . . . . . . . . . . . . . . . . . . . 31

2.7 Factors that alter the composition of an equilibrium mixture . . . . . . . . . . . 35

2.7.1 Addition of reactant(s) or product(s) . . . . . . . . . . . . . . . . . . . . 35

2.7.2 Alteration of pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

2.7.3 Change in temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

2.7.4 The influence of catalysts . . . . . . . . . . . . . . . . . . . . . . . . . . 38

2.8 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

2.9 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

2.10 End of Topic Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40


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 anequilibrium (Higher).

Learning Objectives


• define the term ’equilibrium constant’, which characterises equilibriumcomposition, from the equilibrium equation;

• carry out calculations involving Kc and Kp values;

22 TOPIC 2. CHEMICAL EQUILIBRIUM

• predict qualitative changes in the equilibrium state as a consequence of changingconditions, based on le Chatelier’s principle;

• state the reasons for the equilibrium position being unaffected by the presence ofa catalyst.


2.1. INTRODUCTION 23

2.1 Introduction

’To be in equilibrium is to be in a state of delicate balance, poised between opposingforces so as to appear to be at rest. Two arms wrestling on a table top, neither ableto 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 tobe approaching it. If you are in equilibrium, there you will stay unless disturbed by someoutside influence. If you are out of equilibrium, as indeed are all living organisms, youwill 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 willremain there until disturbed - it is in a state of equilibrium. When pushed, however, it willreadily 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 astable equilibrium. This state usually represents a minimum energy state (in this case,the lowest gravitational potential energy).



2.2 The nature of chemical equilibrium

A chemical system in equilibrium shows no changes in macroscopic properties, such asoverall pressure, total volume and concentration of reactants and products. It appearsto 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 carbondioxide 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 scalethere is change. Carbon dioxide molecules in the gas will bombard the liquid surfaceand dissolve; some carbon dioxide molecules in the solution will have sufficient energyto 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 gasand liquid will always be the same, although the individual molecules will not remainstatic. This state is achieved by a dynamic equilibrium between molecules enteringand leaving the liquid, and between carbon dioxide, water and carbonic acid. In otherwords, the rate at which carbonic acid is formed from CO2 and water will be balancedby carbonic acid dissociating to form CO2 and water.

CO2 + H2O� H2CO3

In the on-line version of this Topic there is an animation showing the reaction of hydrogenand iodine reacting to form hydrogen iodide, and an equilibrium being established. Youshould 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 numberof molecules of H2, I2, and HI after time, t = 15, t = 30 and t = 70. What do you noticeabout them?

This simple animation has reached equilibrium; each time two molecules of hydrogenand iodine react to form hydrogen iodide. In this case, two molecules of hydrogen iodidereact to form hydrogen and iodine.

Another animation is available in the on-line version of this Topic. This time showingthe decomposition of hydrogen iodide to form hydrogen and iodine and an equilibriumsituation being established. You should view this before attempting the questions thatfollow.

Q2: What do you notice about the number of HI, H2 and I2 molecules after time, t = 35and t = 70?

Q3: But how do these numbers at equilibrium compare with the previous reactionstarting with hydrogen and iodine?


2.2. THE NATURE OF CHEMICAL EQUILIBRIUM 25

At equilibrium, the rate of production of HI from H2 and I2 equals the rate of productionof H2 and I2 from HI, therefore the overall composition will not change. This process isgenerally shown by the use of reversible arrows.

H2 + I2 � 2HI

An animation is available online showing the progress of the reaction when the reactantsare H2 and I2. There is also an animation showing the progress of the reaction when HIis the starting material. Look at the graphs of both these scenarios, shown as Figure 2.2and Figure 2.3, plotted as concentration versus time.

Figure 2.2: The approach to equilibrium starting with H2 and I2

Figure 2.3: The approach to equilibrium starting with HI

You should notice that in both situations, the equilibrium concentrations reached areidentical, whether you start with hydrogen and iodine or with hydrogen iodide.



LEARNING POINT

A chemical reaction is at equilibrium when the composition of the reactants andproducts 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 orproducts.

2.3 The equilibrium constant, Kc�

Æ

�

�

Learning Objective

To be able to define the term ’equilibrium constant’ from the equilibrium equation.

An understanding of the mathematical basis of chemical equilibrium was begun byGuldberg and Waage in the 1860s with their ’Law of Mass Action’. This stated thatfor a generalised chemical reaction:

aA + bB � cC + dD

Reactants Products

at equilibrium, the following expression applied (note that [ ] means concentration in mol� -1):

For example, for the reaction between iron(III) ions and cyanide:

Fe3+(aq) + 6CN-(aq) [Fe(CN)6]3-(aq)

[Fe3+] [CN-]6K =c

Concentrations of Reactants

Concentrations of Products

Indices (frombalanced equation)

Equilibrium Constant

In Terms ofConcentration

K =[C]c [D]d

[A]a [B]bc

[[Fe(CN)6]3-]

Notice that the concentrations of products form the numerator, the concentrations ofreactants are in the denominator and each concentration term is raised to a power equalto the number of moles in the balanced equation.


2.3. THE EQUILIBRIUM CONSTANT, KC 27

Write an equilibrium expression for the following reactions.

Q4: 2Fe3+(aq) + 3I-(aq)� 2Fe2+(aq) + I3-(aq)

Q5: H3PO4(aq)� 2H+(aq) + HPO42-(aq)

Returning to the reaction:

H2 + I2 � 2HI

the equilibrium constant is defined as:

Kc =[HI]2

[H2][I2]

and at 453 ÆC, it has a value of 50.

Q6: At 453ÆC which compound is present in greatest concentration?

a) hydrogenb) iodinec) hydrogen iodided) all the same concentration

Since the equilibrium constant is the ratio of concentration of products divided bythe concentration of reactants, its actual value gives guidance to the extent of areaction once it has reached equilibrium. The greater the value of Kc the greater theconcentration of products compared to reactants; in other words, the further the reactionhas gone to completion.

The explosive reaction between hydrogen and fluorine:

H2 + F2 � 2HF

has an equilibrium constant of 1 x 1047. At equilibrium, negligible amounts of thereactants 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 hardlyoccurs at all under these circumstances.

In time, all reactions can be considered to reach equilibrium. To simplify matters, thefollowing general assumption is made:

Value of Kc Extent of reaction

� 10-3 Effectively no reaction

10-3 to 103Significant quantities of reactants and

products at equilibrium

� 103 Reaction is effectively complete



A note of caution:

The equilibrium constant gives no indication of the rate at which equilibrium isachieved. 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.

Q7: The Kc value for the reaction

PCl5 � PCl3 + Cl2is 0.021 at 160ÆC. Which compound is present in greatest concentration at equilibrium?

a) phosphorus(V) chlorideb) phosphorus(III) chloridec) chlorined) all are the same

Q8: The following equilibrium constants apply at room temperature (25ÆC).

Zn(s) + Cu2+(aq)� Cu(s) + Zn2+(aq) K = 2 x 1037

Mg(s) + Cu2+(aq)� Cu(s) + Mg2+(aq) K = 6 x 1090

Fe(s) + Cu2+(aq)� Cu(s) + Fe2+(aq) K = 3 x 1026

Of the metals Zn, Mg, and Fe, which removes Cu(II) ions from solution most completely?

a) Znb) Mgc) Fe

Q9: In which of the following reactions will the equilibrium lie furthest towardsproducts?

a) N2O4(g) � 2NO2(g) Kc at 0ÆC = 159b) 2SO2(g) + O2(g) � 2SO3(g) Kc at 856ÆC = 21.1c) N2O4(g) � 2NO2(g) Kc at 25ÆC = 14.4d) 2SO2(g) + O2(g) � 2SO3(g) Kc at 636ÆC = 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 856ÆC compared with 636ÆC?


From the data in the last question, you can see that the equilibrium constant has differentvalues 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 canincrease 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 textbooksmight give them).


2.4. THE EQUILIBRIUM CONSTANT, KP, INVOLVING GASES 29

One explanation for this lack of units is that equilibrium constants should be strictlyexpressed as the ratio of a property called the activity (not part of the advanced highersyllabus) of products and reactants. In dilute solutions, the activity is almost the sameas concentration, but since activities are without units, equilibrium constants will havenone.

2.4 The equilibrium constant, Kp, involving gases�

Æ

�

�

Learning Objective

To define the term ’equilibrium constant’ for gas reactions from the equilibriumequation.

The equilibrium constant, Kc, is expressed in terms of the concentration of reactants andproducts. This is the most convenient form to use when solutions are being considered;but for reactions involving gaseous reactants and products, the equilibrium constant canbe expressed in terms of partial pressures. (Gases inside a closed container eachexert 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 is1 atmosphere, then the partial pressure of each gas is 0.5 atmosphere.)

For the general reaction:

Kp PNH32

PN2 PH2

3

Write down an appropriate expression for the equilibrium constant for the followingreactions:

Q11: 2NOCl(g)� 2NO(g) + Cl2(g)

Q12: 2SO2(g) + O2(g)� 2SO3(g)



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) + CO2(g)

Two of the participating species are solid and one product is a gas. This is aheterogeneous reaction.

In the cases where a solid or liquid is present in a reaction, its ’concentration’ iseffectively constant and is given the value 1 in the equilibrium expression (the actualvalue is incorporated into the value of Kc or Kp).

In the example above, the solids CaCO3 and CaO are given the value 1, so that

Kp = PCO2

Consider another reaction, the hydrolysis of methyl ethanoate

CH3COOCH3(aq) + H2O(l) � CH3CO2H(aq) + CH3OH(aq)

In this case, the hydrolysis is carried out in aqueous solution with the concentration ofwater remaining effectively constant and so the equilibrium expression is:

Kc �[CH3CO2H][CH3OH]

[CH3COOCH3]

as the value for [H2O] is taken to be 1.

Q13: Which expression below is correct for the reaction

Zn(s) + Cu2+(aq)� Zn2+(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

NH4NO3(s) � 2H2O(g) + N2O(g)


2.6. CALCULATIONS USING EQUILIBRIUM CONSTANTS 31

a)

Kp =p2

H2OpN2O

pNH4NO3

b)

Kp =pN2O

pNH4NO3

c)Kp = p2

H2OpN2O

d)Kp = pN2O

2.6 Calculations using equilibrium constants�

Æ

�

�

Learning Objective

To be able to carry out calculations involving Kc and Kp values

If all the equilibrium concentrations for the reactants and products are known, calculationof the equilibrium constant is simply a matter of substitution into an appropriateequilibrium expression.

Example : Calculation of an equilibrium constant

An equilibrium mixture of gaseous O2, NO and NO2 at 500 K contains 1.0 x 10-3 mol �-1 O2, 1.9 x 10-3 mol � -1 NO and 5.0 x 10-2 mol � -1 NO2. Calculate the value of Kc at500 K.

Write a balanced equation:

O2 + 2NO� 2NO2

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 105

Alternatively, the equilibrium constant may be given and the concentration of a reactantor product be required.



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 H2

equals the concentration of I2 at 0.08 mol � -1, what is the concentration of HI?

From the chemical equation, the equilibrium expression is:

Kc �[HI]2

[H2][I2]

We are asked for [HI] so rearranging gives:

[HI] =�

Kc[H2][I2]

and substituting:

[HI] ��

57.0� 0.08� 0.08

��

� �� 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 PCl5 at 300 K when the equilibriumconcentrations of PCl3 and Cl2 are 0.20 mol � -1. The equilibrium constant for thereaction:

PCl5(g) � PCl3(g) + Cl2at 300 K is 11.5. Calculate your answer in mol � -1 and give your answer to 2 significantfigures.

Q16: Calculate the value of Kp for the decomposition of nitrosyl bromide at 350 K:

2NOBr� 2NO + Br2

given that the equilibrium mixture contains 0.50 atm NOBr, 0.25 atm NO and 0.20 atmBr2. Give your answer to 2 significant figures.

Using initial conditions

In other situations, initial conditions are given together with a suitable equilibriumconstant and you are asked to determine equilibrium concentrations. In these cases,follow the 5 steps below.


2.6. CALCULATIONS USING EQUILIBRIUM CONSTANTS 33

1. Write a balanced equation for the reaction.

2. Under the balanced equation, make a list for each substance involved in thereaction:

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. Substitute the equilibrium concentrations into the equilibrium expression for thereaction and solve for x.

4. 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 (Kc) for the reaction:

H2 + I2 � 2HI

is 57.0 at 700 K

If 0.10 mol of H2 and 0.10 mol of I2 react in a 1.0 litre vessel at 700 K, what are theconcentrations of H2, I2 and HI at equilibrium?

��

��

�

��

��

So taking the positive square root of both sides and simplifying:



�7.55(0.10 - �) = 2�

0.755 = �(2 + 7.55)

� =��

��= ��

At equilibrium the concentrations are:

[H2] = [I2] = 0.10 - 0.0791 = 0.021 mol �-1

and [HI] = 2 � 0.0791 = 0.158 mol �-1

N.B. It is often useful to check that these values can be substituted back into theequilibrium expression to get the correct value of Kc:

(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 thephysical sense. In this latter example, we chose the +ve root of 57 to give a sensibleanswer; taking -7.55 as another mathematically valid solution to

�57 would result in x =

0.136. Since there was an initial concentration of H2 of 0.1, a value of 0.136 would usemore than was available and so should be rejected.

Q17: By following the layout of the example above, calculate the equilibriumconcentrations for the same reaction when the initial amount of iodine is 0.2 mol insteadof 0.1 mol.

A partially worked answer is shown for this in the answer section.

Q18:

The equilibrium constant for the reaction:

N2O4(g)� 2NO2(g)

is 0.01 at a particular temperature.

Calculate the equilibrium concentrations of NO2 and N2O4 at this temperature in a flaskcontaining initially only N2O4 at 0.050 mol � -1, giving your answer to 2 decimal places.

Q19: Calculate the equilibrium concentrations in the previous question if the startingconditions were [N2O4] = 0.0200 mol � -1 and [NO2] = 0.0300 mol � -1. Answer to 4decimal places.

Q20: The equilibrium constant (Kp) for the reaction:

O2 + N2 � 2NO

is 4.65 x 10-31 at 25 ÆC. Assuming that the partial pressure of oxygen and nitrogen inair are 0.21 atm and 0.78 atm respectively, calculate the partial pressure of NO in theatmosphere at this temperature. Give your answer (in atm), to 3 significant figures.



2.7. FACTORS THAT ALTER THE COMPOSITION OF AN EQUILIBRIUM MIXTURE 35

2.7 Factors that alter the composition of an equilibriummixture

�

Æ

�

�

Learning Objective

To be able to predict changes in the equilibrium position as a consequence ofchanging conditions

2.7.1 Addition of reactant(s) or product(s)

At a given temperature the equilibrium constant does not change, so addition of areactant, for example, will mean that the concentrations of reactants and products areno longer satisfying the equilibrium expression and the forward and reverse reactionscontinue until equilibrium is re-established.

For example, in the water gas shift reaction:

CO(g) + H2O(g)� CO2(g) + H2(g)

The equilibrium constant, Kc = 4.24 at 800 K.

Starting with 0.2 mol � -1 each of CO and H2O, you can calculate that the equilibriummixture will contain:

[CO] = [H2O] = 0.065 mol � -1

and [CO2] = [H2] = 0.135 mol � -1

If an additional 0.2 mol � -1 of steam is added, the mixture will no longer be at equilibrium,since:

[CO2][H2][CO][H2O]

= 1.06 �� Kc

In order to restore the equilibrium ratios, some of the added steam will react with CO toform more products (CO2 and H2) until the ratios again are correct for the equilibriumsituation.

Using the method for calculating equilibrium concentrations described before, this willoccur when:

[H2O] = 0.233

[CO] = 0.028

[CO2] = [H2] = 0.167

Check yourself that these values for equilibrium concentrations fit the equilibriumexpression.

You will notice that adding water has increased the equilibrium concentration of CO2 andH2 from 0.135 mol � -1 to 0.167 mol � -1.

Q21: A gaseous mixture contains 0.30 mol CO, 0.10 mol H2, and 0.02 mol H2O, plussome CH4, per litre.



CO(g) + 3H2(g)� CH4(g) + H2O(g)

The value of Kc at 1200 K is 3.92. What is the concentration of CH4 in the mixture? Giveyour answer to 3 decimal places.

Q22: More H2 was added to the system above to bring its equilibrium concentration to0.20 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 whenconditions affecting an equilibrium are altered. It states that when a change is appliedto a reaction mixture at equilibrium, reaction will occur in a direction that will reduce thechange.

Applying this to the equilibrium between NO2 and N2O4:

2NO2 � N2O4

If the pressure is increased by decreasing the volume of a closed system, the equilibriumwill move to reduce this increased pressure. This can be achieved in this system by twoNO2 molecules combining to form one N2O4 molecule, which will occupy less volumeand 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 changein the equilibrium position will have no effect on the pressure. So, conversely, alterationof the pressure will not affect the equilibrium position.


2.7. FACTORS THAT ALTER THE COMPOSITION OF AN EQUILIBRIUM MIXTURE 37

For the reactions below, suggest whether an increase in pressure will result in thereaction going from:

• left to right

• right to left

• or there will be no change

Q23: CaCO3(s) � CaO(s) + CO2(g)

a) left to rightb) right to leftc) no change

Q24: C(s) + O2(g) � CO2(g)


Q25: N2(g) + 3H2(g)� 2NH3(g)


Q26: PCl5(g)� PCl3(g) + Cl2(g)



2.7.3 Change in temperature

A change in temperature will change the value of the equilibrium constant, but even ifthe value at the new temperature is not known (which would enable you to calculate thenew equilibrium concentrations) you can determine the direction of the change by usingLe Chatelier’s principle.

Consider an exothermic reaction, such as the industrial manufacture of ethanol fromethene and steam:

C2H4(g) + H2O(g)� C2H5OH(g)

When the reaction moves to the right, heat is given out. Conversely, when it moves tothe 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 movein a direction to reduce the temperature. For an exothermic reaction to absorb heat it



must go in reverse, i.e. move from right to left. The equilibrium position for an exothermicreaction 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 inthe equilibrium reaction mixture. The value of Kp will decrease.

To see how temperature affects the equilibrium position of an endothermic reactionconsider:

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 Kc for this reaction.

Q30: Will the value of Kc increase or decrease?

This reaction between nitrogen and oxygen can occur in air, but the equilibriumconcentration of NO is very small at normal temperatures, increasing to significant levelswhen lightning passes through air.

LEARNING POINT

For an exothermic reaction, the equilibrium constant decreases with increase intemperature; for an endothermic reaction, the equilibrium constant increases withtemperature.

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 ordecrease as the temperature increases?

a) increaseb) decrease

Now try this summary question.

Q33: Suggest four ways in which the equilibrium concentration of SO3 can beincreased in a closed vessel if the only reaction is:

2SO2(g) + O2(g)� 2SO3(g) for which �Ho = -99 kJ mol-1

2.7.4 The influence of catalysts�

Æ

�

�

Learning Objective

To be able to state the reasons for the equilibrium position being unaffected by thepresence of a catalyst.


2.8. SUMMARY 39

Catalysts affect the rate of a reaction, but since they affect the rate of the forward andreverse reactions to the same extent, they will not affect the equilibrium position.

The equation for the synthesis of ammonia is:

N2 + 3H2 � 2NH3

Industrially, in the Haber process, an iron catalyst is used.

You will have studied the rates at which chemical reactions occur and met similardiagrams to Figure 2.4 during your Higher course. Figure 2.4 shows the energy changesduring the reaction between nitrogen and hydrogen to make ammonia.

0

100

200

300

400

500

600

700

-100

-200

Figure 2.4: Potential Energy Diagram

In simple terms, the rate at which a reaction occurs is related to the height of theenergy barrier (the activation energy). A catalyst reduces the height of this barrier. Theequilibrium position is determined by other factors that are related to the energies ofthe products and reactants.

Many industrial processes use catalysts to increase the rate at which products areformed to assist economic production, but the ratio of concentrations of products toreactants will always be that determined by the equilibrium constant appropriate to theconditions.

2.8 Summary

The key points of this Topic are:

• Chemical equilibria are stable dynamic equilibria where the equilibriumconcentrations of reactants and products are defined by an equilibrium constant



called Kc given by:

��

��

• For gaseous reactions, another constant (Kp) can be defined in terms of the partialpressures 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 samephase; in heterogeneous reactions, the species are in more than one phase. Theconcentrations of pure solids or pure liquids are given the value 1 in the equilibriumequation.

• In response to changes in conditions (addition/removal of reacting species, alteredpressure or temperature) the equilibrium composition will move in a direction toreduce 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 ChemicalReactions, Learning and Teaching Scotland, ISBN 1 85955 874 7.

• A-level Chemistry: E.N. Ramsden, Stanley Thorne Publishers, ISBN 0-85950-154-X

2.10 End of Topic Test



41

Topic 3

Phase Equilibria

Contents

3.1 Partition coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

3.2 Solvent extraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

3.2.1 Extraction calculations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

3.2.2 Decaffeinated coffee . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

3.3 Chromatography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

3.3.1 Paper chromatography . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

3.3.2 Gas-liquid chromatography . . . . . . . . . . . . . . . . . . . . . . . . . 51

3.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

3.5 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

3.6 End of Topic Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55



• define equilibrium constants and carry out calculations involving concentrationsand equilibrium constants (Topic 2.2);

• explain the solubility of substances in terms of their bonding and the type of solventinvolved (Higher Unit 1).

Learning Objectives


• explain that a solute distributes itself between two immiscible liquids in a ratiocalled the partition coefficient, dependent on the type of solute, type of solventand the temperature;

• explain that solvent extraction is an application of the partition of a solute betweentwo liquids which can be used to purify solutes;

• describe chromatographic techniques as the partition of solutes beween stationaryand mobile phases and explain how data gained by both paper and gas-liquidchromatography can be used to separate and identify the individual componentsof solute mixtures.

42 TOPIC 3. PHASE EQUILIBRIA

3.1 Partition coefficients�

Æ

�

�

Learning Objective

To be able to explain that a solute distributes itself between two immiscible liquids ina ratio called the partition coefficient, dependent on the type of solute, type of solventand the temperature.

Liquids which do not mix with each other are said to be immiscible. Dissolving a solutein 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 asshown.

� ��

This ratio of the final concentrations of the solute in the two liquids is called thepartition 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 natureof A.

a) It is a polar solventb) It is a non-polar solventc) It is an acidic solventd) It is an ionic solvent

Q3: What could be said about solvent B compared to solvent A?

a) B is more polar than Ab) B is less polar than Ac) B is of equal polarity to A

Q4: Which of these events occurs when equilibrium is reached?

a) exchange between the layers stopsb) movement of the solute stops


3.1. PARTITION COEFFICIENTS 43

c) the solvent becomes saturatedd) an equal exchange rate is reached

Q5: Calculate a partition coefficient for this equilibrium.

a) 12b) 4c) 3d) 1

Factors affecting the partition coefficient.

10 min

�

Æ

�

�

Learning Objective

To show how the partition coefficient is dependent on the type of solute, the immiscibleliquids 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 representedby a sphere). In experiment 5, 24 moles are used. Equal volumes of three differentimmiscible 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 mightaffect the partition coefficient?



Q9: Has the partition coefficient been affected?


The value of the partition coefficient is dependent on the type of solute, types of solventand the temperature of the equilibrium system.

3.2 Solvent extraction�

Æ

�

�

Learning Objective

To show that solvent extraction is an application of the partition of a solute betweentwo immiscible liquids.

Solvent extraction is an application of the partition of a solute between two liquids. Thesolvents can be used to ’selectively’ separate out the components of a mixture. Supposesubstance 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 coefficientapproximately equal to zero).

Figure 3.3: Solvent extraction

A mixture of the substances is shaken with the two solvents and the layers allowed toseparate. Situation (a) results. The top layer is then removed and the lower is subjectedto 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 wouldbe isolated.

Separation of plutonium and uranium compounds in spent fuel products of the nuclearindustry is mostly done this way.


3.2. SOLVENT EXTRACTION 45

The highly radioactive spent fuel is chopped up and dissolved in hot nitric acid. Thefirst stage in the separation of the uranium (U) and plutonium (Pu) compounds from theother fission products is achieved by a solvent extraction called a counter current solventextraction process. The process is called the PUREX process from (Pu-U-Recovery-EXtraction). More information can be found on the websites listed in the Resourcessection at the end of this Topic.

3.2.1 Extraction calculations�

Æ

�

�

Learning Objective

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 organicacids using a suitable organic solvent such as ether. This will dissolve the acid and leavethe impurities in the water layer. Adding the ether in repeated small quantities is moreefficient than one large volume. The following example illustrates this.

Example : Ether extraction.

An organic acid X is present, along with impurities, in 100 cm3 of water. 100 cm3 of etheris 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 withthe 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:



�

��

��

��

��

��

�

�

� �� !

For the second 50 cm3 of ether a further 66.6% of the remaining solute (33.3%) wouldbe 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 25oC untilequilibrium was achieved. 25 cm3 of the ether layer required 30.0 cm3 of 0.05 mol � -1

sodium hydroxide to neutralise and 25 cm3 of the aqueous layer required 10 cm3 of 0.05mol � -1 to neutralise. Calculate the partition coefficient (ether / water) at 25oC.

Q13: Calculate the mass of an organic acid X which can be extracted from 200 cm3 ofaqueous 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�

Æ

�

�

Learning Objective

To show that solvent extraction of caffeine from coffee is an application of the partitionof a solute between two liquids

N

N N

N

O

O

CH3

H3C

CH3

Figure 3.4: Structure of caffeine


3.2. SOLVENT EXTRACTION 47

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 CH2Cl2 wasused as a solvent in the early processes but its toxic nature led to its replacementin the nineties with a more environmentally friendly solvent called supercritical carbondioxide. Supercritical carbon dioxide exists in closed containers above 73 atmospheresand 31oC. It behaves both like a gas and a liquid. Passing it through green coffee beansallows the gas to penetrate the beans and the liquid properties allow it to dissolve outabout 98% of the caffeine. The caffeine is recycled into soft drinks and medicines andthe decaffeinated beans sold.

Figure 3.5: Some decaffeinated foodstuffs



Figure 3.6: Phase diagram of CO2

Supercritical carbon dioxide is also used for solvent extraction of flavours in brewing andextracting 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? Whynot construct an experiment at home and see if your prediction is correct!

3.3 Chromatography�

Æ

�

�

Learning Objective

To be able to describe chromatographic techniques as partition of solutes beweenstationary and mobile phases

In the first part of this Topic, you have considered processes that are dependent on theequilibrium between two immiscible liquids. This section looks at methods of separationthat depend on the equilibrium between different phases. Methods that are widelyused to separate and identify the components of mixtures come under the heading ofchromatography.

There are several different types of chromatography, but the name derives from the earlymethods used by the Russian, Michael Tswett, to separate mixtures of plant pigments


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 partitioncoefficients between the stationary and mobile phases.

Substances present in the initial mixture which partition more strongly into the stationaryphase will move more slowly than materials which partition more strongly into the mobilephase.

A simulation of bees and hornets crossing a bed of flowers to mimic chromatography ison the on-line version of this Topic.

3.3.1 Paper chromatography�

Æ

�

�

Learning Objective

To be able to explain how data gained from paper chromatography can be used toidentify the components in solute mixtures

In this type of chromatography the mixture of components to be separated is placed asa small spot close to the bottom of a rectangular piece of absorbent paper (like filterpaper). The bottom of the paper is placed in a shallow pool of solvent in a tank. Anexample of the solvent would be an alcohol. Owing to capillary attraction, the solvent isdrawn up the paper, becoming the mobile phase. The solvent front is clearly visible asthe chromatography progresses.

When the paper is removed from the solvent, the various components in the initial spothave moved different distances up the paper.

A simple simulation of this process, showing chromatography of blue and black inks, isavailable on the on-line version of this Topic. The start and final positions are shown inFigure 3.7 and Figure 3.8 respectively.

Figure 3.7: Start of chromatographic analysis



Figure 3.8: End of chromatographic analysis

This separation depends on the different partition coefficients of the variouscomponents. The components are partitioned between the solvent and the watertrapped in the paper. Substances which partition mainly into the solvent mobile phasewill move further up the paper than substances which partition more strongly onto thestationary phase.

Q17: By observing the chromatography simulation above, which ink (blue or black) hasthe most components?

a) blueb) blackc) both the same.

Q18: Which material in the black ink has stayed longest on the stationary phase, andhas the lowest solvent/water partition coefficient?

a) redb) yellowc) dark blued) bluee) green

Q19: The red coloured spot has moved furthest, this would indicate that the redmaterial:

a) has the highest solvent/water partition coefficientb) has the lowest solvent/water partition coefficientc) has the lowest molecular mass.

Q20: The movement of materials on paper chromatography is often described by anRf value which is the distance travelled by the spot divided by the distance travelledby the solvent front. As long as the conditions of chromatography remain the same, acompound will have a constant Rf value.

Which colour in the black ink could have an Rf value of 0.4?

a) red


3.3. CHROMATOGRAPHY 51

b) yellowc) dark blued) bluee) green


Although paper chromatography is still used today, it has been largely replaced by thinlayer 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 describedabove, but TLC allows a more rapid separation (which prevents the spots spreading toofar) and makes detection of the spots easier. Most materials are not coloured, but canstill be chromatographed.

The invisible spots on paper or thin layer chromatography are revealed by use of alocating reagent. These react with the compounds in the spots to produce a colouredderivative. For example, ninhydrin solution can be sprayed onto chromatograms toreveal amino-acids.

In another TLC detection system, the silica stationary phase is mixed with a fluorescentdye, so that at the end of the process, viewing the plate under an ultraviolet lamp willcause the background to glow (often an eerie green) except where there are spots,which appear black.

3.3.2 Gas-liquid chromatography�

Æ

�

�

Learning Objective

To be able to explain how data gained from gas-liquid chromatography can be usedto identify the components in solute mixtures

In gas-liquid chromatography (GLC) the stationary phase is a high boiling point liquidheld on an inert, finely-powdered support material, and the mobile phase is a gas (oftencalled the carrier gas).

The stationary phase is packed into a tubular column (Figure 3.9) usually of glass ormetal, with a length of 1 to 3 metres and internal diameter about 2 mm. One end ofthe column is connected to a gas supply (often nitrogen or helium) via a device whichenables a small volume of liquid sample (containing the mixture to be analysed) to beinjected into the gas stream. The other end of the column is connected to a device whichcan 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 controlledthroughout the chromatographic analysis. One reason for this is that the materials to beanalysed must be gases during the analysis, so that gas-liquid chromatography is oftencarried out at elevated temperatures.



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 ontheir partition coefficients between the liquid stationary phase and the gaseous mobilephase.

The detector is set to measure some change in the carrier gas that signals the presenceof material coming from the end of the column. Some detectors measure the thermalconductivity of the gas, others burn the material from the column in a hydrogen-air flameand measure the presence of ions in the flame.

The signal from the detector is recorded and plotted against time to give a series ofpeaks each with an individual retention time.

A typical trace (chromatogram) for a complex mixture (premium grade petrol) is shownin Figure 3.10.

Figure 3.10: Chromatogram of premium grade petrol

As long as the conditions remain constant (same column and stationary phase, samepressure of gas, and same temperature) the retention time is the same for a givencompound. By chromatographing known materials (standards) the identity of manypeaks in a GLC trace can be established.


3.4. SUMMARY 53

The other important feature of many chromatograms is that the area under the peak isdirectly proportional to the amount of material present.

Q21: The retention times, under the same conditions as Figure 3.10, for fourcompounds 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) butaneb) pentanec) benzened) toluene

Q22: How are retention time (r.t.) and molecular mass (m) related for these fourhydrocarbons?

a) r.t. increases as m increasesb) r.t. decreases as m increasesc) they are not related

Q23: Using the answer to the previous question, suggest a possible hydrocarbon forthe peak with retention time 20.8 min.

a) hexane (C6H14)b) heptane(C7H16)c) xylene(C8H10)d) propane(C3H8)

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 thetype of solute.

• The partition coefficient for a solute X in solvent A (top layer) and solvent B (bottomlayer) is given by:

� ��

• Solvent extraction processes are used extensively in industry to separate outcomponents in a mixture, examples being found in nuclear industry, chemicalindustry and food processing.



• Chromatography is an analytical technique used to separate and identify thecomponents in a mixture.

• Chromatographic separations depend on the partition equilibrium between twophases, one stationary and the other mobile.

• There are several types of chromatography, including paper chromatography andgas-liquid chromatography.

• In paper chromatography, the stationary phase is the water held on the paper andthe mobile phase is another solvent.

• In gas-liquid chromatography, the stationary phase is a liquid held on a solidsupport 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


• 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 0-435-63106-3

• 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 forSchools and Colleges.

• Video produced by Royal Society of Chemistry (with Glaxo Welcome): ModernChemical Techniques

• Booklet produced by Royal Society of Chemistry (with Unilever): Modern ChemicalTechniques


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

http://www.exicom.org/cew/oct97/awasthi.htm

http://wwwchem.uwimona.edu.jm:1104/lectures/coffee.html

http://www.sciam.com/0697issue/0697working.html


3.6 End of Topic Test



57

Topic 4

Acid/base Equilibria

Contents

4.1 What are acids and bases? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 584.1.1 The nature of the hydrogen ion . . . . . . . . . . . . . . . . . . . . . . . 61

4.2 The ionic product of water and the pH scale . . . . . . . . . . . . . . . . . . . . 624.2.1 pH scale . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

4.3 Dissociation of acids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 684.4 Dissociation of bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 714.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 734.6 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 744.7 End of Topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75



• 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 ofequilibrium (Le Chatelier’s principle) (Topic 2.2);

• explain what is meant by an amphoteric oxide (Topic 1.5).

Learning Objectives


• define the terms - acid, base, conjugate acid, conjugate base- in terms of transferof protons;

• explain what is meant by Kw, the ionic product of water;

• define the dissociation constant, Ka, for acids and conjugate acids and relatevalues for Ka to the strengths of acids and bases;

• calculate the pH of solutions of strong acids and bases from [H+] and the pH ofsolutions of weak acids from Ka values.

58 TOPIC 4. ACID/BASE EQUILIBRIA

4.1 What are acids and bases?�

Æ

�

�

Learning Objective

To define the terms - acid, base, conjugate acid, conjugate base- in terms of transferof 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. Someexamples of common acids and bases are shown in Figure 4.1. About 1.0-1.5 litres perday 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 andLowry.

An acid is a substance that is able to donate hydrogen ions (protons) to anothersubstance, 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 aqueoussolution.


4.1. WHAT ARE ACIDS AND BASES? 59

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 thereverse reaction, A- accepts a proton, i.e. behaves as a base. A- is said to be theconjugate base of the acid HA. For every acid, there is a conjugate base formed byloss 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 ofa 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 thefollowing animation.

Figure 4.4: Acid-base equilibrium

Overall, the reaction involves transfer of a proton from one species to another. Thissimple transfer is literally of vital importance. Gain or loss of a proton can cause achange in the structure and function of amino acids, proteins, enzymes and nucleicacids. So, many biological pathways are controlled by the presence or absence of H+

ions.



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.


Conjugate pairs

10 min

A test to provide practice at identifying acids, bases and their conjugates is available inthe on-line version of this Topic.


A water molecule is capable of accepting protons from an acid or donating protons to abase. Water is therefore amphoteric and perfectly suited to enable the transfer of H+

ions (protons) from one substance to another.


4.1. WHAT ARE ACIDS AND BASES? 61

4.1.1 The nature of the hydrogen ion�

Æ

�

�

Learning Objective

To demonstrate that the hydrogen ion can be represented by H+ or H3O+ dependingon the circumstances.

In aqueous solution, positive ions like Li+ are hydrated, as shown in the diagram(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 onthe same scale would have a diameter of about 5 km, a distance equivalent to about 50football 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 atomto form a hydronium ion, H3O+ (Figure 4.8). This in turn will be bonded to other watermolecules as shown in Figure 4.8.

HO

HH

HO

HH

OH

H O

O

HH

H

H

Figure 4.8



In equations and equilibrium expressions, H+(aq) is often used as a simple shorthandinstead of H3O+(aq). H3O+(aq) is used when it is necessary to show the behaviour ofwater as an acid or a base.

4.2 The ionic product of water and the pH scale�

Æ

�

�

Learning Objective

To be able to explain the term Kw, 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, tapwater, pure water and hexane are shown in Table 4.1.

Figure 4.9: Conductivity measurements


4.2. THE IONIC PRODUCT OF WATER AND THE PH SCALE 63

Table 4.1: Conductivity data

Substance Current Reading

0.1 mol l-1 NaCl 125 mATap water 5 mA

Pure water 20 #A

Hexane 0 #A

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. Whatconclusion 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 acidb) a base

Q10: The hydroxide ion is

a) the conjugate acid of H2Ob) the conjugate base of H2Oc) the conjugate acid of H3O+

d) the conjugate base of H3O+

Q11: The reaction (Figure 4.10) is reversible and equilibrium is reached. From theexperimental evidence (Table 4.1), which of these statements is the best descriptionof the equilibrium position?

a) The equilibrium lies to the left.b) The equilibrium lies well to the left.c) The equilibrium lies to the right.d) The equilibrium lies well to the right.

Q12: Write an expression for the equilibrium constant for this reaction ( Figure 4.10).



The concentration of water is effectively constant and can be incorporated into theequilibrium constant to give a new constant, Kw.

�� "�$

�� $"�

�

or �� "�

� �$"�

�

Kw is known as the ionic product of water. Like all equilibrium constants, it isdimensionless 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 Kw have at 25ÆC (298 K)?

a) 1.00b) 0.68 x 10-14

c) 1.00 x 10-14

d) 1.00 x 1014

Q14: The graph shows that as the temperature increases, Kw increase. What does thissuggest about the sign of �HÆ 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 beforerevealing the answer.



Learning point

Kw is the ionic product of water. It is temperature dependent and is represented by Kw

= �H+��OH-�

4.2.1 pH scale�

Æ

�

�

Learning Objective

To be able to use Kw 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-]

10 min

Visit the on-line version of this Topic where you will find a simulation. The simulation usesa bar graph, shown in Figure 4.12, which can be manipulated to show the connectionbetween [H+] and [OH-]

10-13

10-11

10-9

10-7

10-5

10-3

10-1

pH1 pH7 pH13

[H+]

[OH-]

Co

nce

ntr

atio

n/m

ol

l-1

Figure 4.12: The connection between [H+] and [OH-]

Q16: When [H+] is 10-7, what is the pH?



Q19: What is the connection between the concentration of H+ ions and the pH value?


Check the value of the expression [H+] [OH-] at various pH values.

Kw has a value of 1.00 x 10-14 at 25ÆC.



If the concentration of H+ ions is known, then the concentration of OH- ions can becalculated using Kw, and vice versa. A single scale to measure acidity and alkalinity canbe based on the concentration of H+ ions. The scale has values ranging from around 0to 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 forstrong acids and alkalis and at calculating [H+] and [OH-] from pH values.

Read through the worked examples carefully. The questions that follow them arerandomly generated so that you can try them as many times as you wish until you areconfident.

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.



�$"�

��

� � ��

�� "�

� �$"�

��"�

��

��

�$"��

��

� ��

� ��

%" � � ��

��

� ��

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- ionscan 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 youranswer to three significant figures.

Step 1

%" � � ��"�

��

�"�

��

�"�

��

"�� antilog ��

� ��

� ��

Note: calculators vary slightly in the way in which they antilog numbers. One way is asfollows:

If log�"�

��

then�"�

��



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 pH2.3




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 themolar concentrations of either H+ or OH- ions.

4.3 Dissociation of acids�

Æ

�

�

Learning Objective

To be able to define the dissociation constant, Ka, for acids. To relate values for Ka tothe strengths of acids. To calculate pH values for weak acids

Water molecules are amphoteric. They are capable of accepting protons from acidsand donating protons to bases.

In the presence of a more powerful acid, HA, water molecules accept protons and theacid dissociates according to the equation shown in Figure 4.13:

HA(aq) + H2O(l) H3O+(aq) + A-(aq)

Figure 4.13: Protonation of water


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 theequation shown in Figure 4.13.

In dilute solutions, the concentration of H2O is constant and can be incorporated intothe 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 moleculesof HA remain. Ka will be very large. The acid is 100% dissociated into ions. The reversereaction does not take place because the conjugate base is so weak that it cannotremove protons from H3O+ ions.

Table 4.2 shows some acids with their Ka values. H+(aq) has been used instead of H3O+

for simplicity.

Table 4.2:

Ka

HF(aq) � H+(aq) + F-(aq) 3.4 x 10-4

HIO3(aq) � H+(aq) + IO3-(aq) 1.7 x 10-1

H3PO4(aq) � H+(aq) + H2PO3-(aq) 7.9 x 10-3

H2SO3(aq) � H+(aq) + HSO3-(aq) 1.5 x 10-2

Q29: Which is the strongest acid?

a) HFb) HIO3

c) H3PO4

d) H2SO3

Q30: Which is the weakest acid?

a) HFb) HIO3

c) H3PO4

d) H2SO3

Learning point

In general, as the strength of the acid increases, the strength of the conjugate basedecreases.

Q31: What is the weakest conjugate base?

a) F-

b) IO3-



c) H2PO4-

d) HSO3-

Q32: What is the strongest conjugate base?

a) F-

b) IO3-

c) H2PO4-

d) HSO3-

Q33: Write a general statement connecting the value of Ka 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 pKa 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) high Ka, high pKa

b) low Ka, low pKa

c) high Ka, low pKa

d) 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) high Ka, high pKa

b) low Ka, low pKa

c) high Ka, low pKa

d) 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 sinceonly a small proportion of the molecules are dissociated into ions. However, the pH canbe calculated, provided the Ka 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 inthe on-line version of this Topic.

The following questions will give you practice in using Equation 4.2. The first twoquestions have worked answers that can be displayed. Try them for yourself first. Thendo as many of the others as necessary until you feel comfortable. Page 12 of the SQAdata booklet will be useful. Give all your answers to 2 significant figures.


4.4. DISSOCIATION OF BASES 71

Q38: What is the pH of a solution of ethanoic acid of concentration 0.01 mol � -1?

Q39: A 0.1 mol � -1 solution of a weak acid has a pH of 3.2. What is the pKa value forthe acid?

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 concentration0.1 mol � -1?

Q43: What is the pH of a solution of propanoic acid of molar concentration0.02 mol � -1?


4.4 Dissociation of bases�

Æ

�

�

Learning Objective

To define the dissociation constant, Ka, for conjugate acids of bases and relate valuesfor Ka to the strengths of bases

In the presence of a more powerful base, B, water molecules behave as an acid bydonating protons to the base.

Bases

10 minTwo drag and drop exercises are available in the on-line version of this Topic to reinforcethe 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, theequilibrium lies far to the left.

Complete the table Table 4.3 using words from the Word Bank which appears beneaththe table.

Table 4.3:

Strong acid

Strong conjugate acid

100% dissociated



Word Bank

� strong base � weak base � weak acid

� weak conjugatebase

� weak conjugate acid � strong conjugatebase

� poorly dissociated � 100% dissociated � poorly dissociated

The dissociation in aqueous solution of a base of general formula B can be representedby

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 equationshown in Figure 4.15:

Figure 4.15: Dissociation of a conjugate base

The conjugate acid dissociation constant, Ka, will therefore be as shown in Equation 4.3:

�� "�$

��

��"��(4.3)

Note that again the [H2O] is assumed to be constant and so is incorporated into theequilibrium constant.

Ka (and pKa) values can be used as a measure of the strength of the conjugate acid andalso therefore as a measure of the strength of the base, since the stronger the base theweaker the conjugate acid.

Ammonia is a familiar example of a base. Use the data booklet (page 12) to answer thefollowing 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 Ka for this conjugate acid.

Q47: Is ammonia a weak base or a strong base?

a) weakb) strong


4.5. SUMMARY 73

Many weak bases are organic compounds called amines which are derived fromammonia 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 codeineis a naturally occurring amine used as a painkiller.

Table 4.4: Weak bases

Conjugate acid

Base Formula Ka pKa

ammonia NH3 5.6 x 10-10 9.3

methylamine CH3NH2 2.7 x 10-11 10.6

dimethylamine (CH3)2NH 1.9 x 10-11 10.7

piperidine C5H10NH 7.7 x 10-12 11.1

codeine C18H21NO3 6.3 x 10-9 ?

Q48: Calculate pKa for the conjugate acid of codeine. Give your answer to one decimalplace.

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 groupshave 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 everybase 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:

H2O(l) + H2O(l) � H3O+(aq) + OH-(aq)

• The dissociation constant for this ionisation is Kw, the ionic product of water, whichis represented by:

Kw = [H3O+][OH-] or, more simply, [H+][OH-].

• Kw is temperature dependent and has a value of approximately 1.0 x 10-14 at 25ÆC.

• The pH scale is based on the [H+] which are related by pH = -log [H+]. Thisrelationship can be used to calculate the concentration of H+ ions and OH- ionssince changes in the concentration of one directly change the concentration of theother to maintain Kw.



• Acids of general formula HA dissociate in aqueous solution according to theequation:

HA(aq) + H2O(l)� H3O+(aq) + A-(aq)

• The acid dissociation constant, Ka, is given by:

�� "�$

��

�"��

•

This can also be represented by pKa where pKa = -log Ka

Both Ka and pKa provide a measure of the strength of an acid. When Ka is low (orpKa is high), the acid is weak. When Ka is high (or pKa is low), the acid is strong.

•

For weak monoprotic acids, pH and pKa 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 tothe equation:

B(aq) + H2O(l)� BH+(aq) + OH-(aq)

• The dissociation of the conjugate acid of the base B can be represented by:

BH+(aq) + H2O(l) � B(aq) + H3O+(aq)

Values for Ka (and pKa) for the dissociation of the conjugate acid, BH+, can beused as a measure of the strength of the base B, since the weaker the conjugateacid the stronger the base.




• 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



Web sites

• http://chemistry.semo.edu/crawford/ch186/lectures/ch15/index.html

• http://www.science.ubc.ca/~chem/tutorials/pH/index.html




http://chemistry.semo.edu/crawford/ch186/lectures/ch15/index.html

http://www.science.ubc.ca/~chem/tutorials/pH/index.html

77

Topic 5

Indicators and buffers

Contents

5.1 Indicators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 785.1.1 pH titrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

5.2 Buffer solutions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 855.2.1 Acid Buffers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 865.2.2 Basic buffers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

5.3 Calculating pH and buffer composition. . . . . . . . . . . . . . . . . . . . . . . . 905.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 935.5 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 945.6 End of Topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94



• write an expression for the dissociation constant, Ka of a weak acid. ( Topic 2.4 );

• write an expression for the relationship between pH, pKa and molar concentration( Topic 2.4 );

• use this relationship to calculate the pH of a solution of a weak acid given theconcentration.

Learning Objectives


• explain why the colour of an acid/base indicator is dependent on the pH of thesolution;

• calculate the pH range over which the indicator changes colour from the aciddissociation 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 howthey function;

• calculate the pH of an acid buffer when given the acid dissociation constant andthe 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 colourdepending on the pH of the solution.

HIn can be used as a general formula for an indicator and its dissociation can berepresented by this equation:

Figure 5.1: Indicator equilibrium equation

For a good indicator, the undissociated acid, HIn, will have a distinctly different colourfrom its conjugate base, In-.

For the indicator litmus, HIn is red and In- is blue. Use this information and Figure 5.1to answer these questions.

Q1: In which direction will the equilibrium position move if the[H+] is increased?

a) leftb) right

Q2: What colour will the solution be?

Q3: If the [H+] is decreased in Figure 5.1 , in which direction will the equilibriumposition move?

a) leftb) right

Q4: What colour will the solution be now?


The acid dissociation constant for an indicator HIn is given the symbol KIn and isrepresented by:

�� "�� &��

�"&��

Taking the negative log of both sides gives:


5.1. INDICATORS 79

� �� "�

�� &��

�"&��

�� %��

��! � ��"�

�� %"

�� %�� %" � ��&��

�"&��

�� %" � %�� &��

�"&��(5.1)

Equation 5.1 shows that the pH of the solution is determined by the pKIn of the indicatorand 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 ofthe solution is dependent on the pH of the solution.

Bromothymol blue has a pKIn 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. Forbromothymol 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 thesolution will be that of the HIn molecules, i.e. it will be yellow.

Q6: Calculate the ratio�&��

�"&��

for bromothymol blue at pH 9.


Q8: Calculate the ratio�&��

�"&��



for bromothymol blue at pH 7.


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 yellow

7 1 1 green

8 10 1 blue10 1000 1 blue12 100000 1 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 oneof the extreme colours (yellow or blue).

When pH = pKIn, the concentrations of In- and HIn are equal and the colour isintermediate.

Figure 5.3 shows a similar diagram for methyl orange ( pKIn = 3.7).

Figure 5.3: Methyl orange indicator

Figure 5.4 shows the diagram for phenolphthalein which is colourless in acid and pinkin alkali.


5.1. INDICATORS 81

Figure 5.4: Phenolphthalein indicator

In each of the indicators above, the colour change from the original colour is assumedto 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 monitoredusing a pH meter. The results can be used to produce a pH titration curve from whichthe equivalence point can be identified.

pH titration

25 min

A simulation of the production of a pH titration curve with questions to testunderstanding.

In the simulation below, 50cm3 hydrochloric acid of concentration 0.1 mol �-1 is beingneutralised by sodium hydroxide solution.

Use the graph shown as Figure 5.5, to answer the questions.



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 � -1) of thesodium hydroxide solution.

Q13: Between 49.9 cm3 and 50.0 cm3, only 0.1 cm3 of alkali was added. What was thechange in pH for this addition?


During a titration involving a strong acid and base, there is a very rapid change in pHaround the equivalence point.

Table 5.2 shows pH titration curves produced by the different combinations of strong andweak acids and alkalis. Look closely at the equivalence points in each graph.


5.1. INDICATORS 83

Table 5.2: Titration curves

Q14: Which combination has an equivalence point at pH 7?

a) strong acid/strong alkalib) strong acid/weak alkalic) weak acid/strong alkalid) weak acid/weak alkali

Q15: What is the pH of the salt formed from a strong acid and a strong alkali?

a) 1b) 5c) 7d) 9

Q16: Which of these is the most likely pH of the salt formed from a strong acid and aweak alkali?

a) 1b) 5c) 7



d) 9

Q17: Which of these is the most likely pH of the salt formed from a weak acid and astrong alkali?

a) 1b) 5c) 7d) 9


Since the equivalence points occur at different pH for different combinations, differentindicators will be required in each case.

Choosing indicators

30 min

�

Æ

�

�

Learning Objective

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 pHtitration curve to enable the best indicator to be chosen for each combination of acidand alkali.

Figure 5.6 shows pH titration curves onto which information about some indicators hasbeen 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


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 ofthe solution changes most rapidly. The best indicator will be one whose colour changeoccurs around the equivalence point. No indicators are suitable for a weak acid/weakalkali titration since the pH changes only gradually around the equivalence point.

5.2 Buffer solutions.�

Æ

�

�

Learning Objective

To be able to describe the composition and functioning of both acid and basic buffersolutions. To calculate the pH of an acid buffer when given the acid dissociationconstant and the composition. To calculate composition when given the othervariables

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 structureand curdling occurs. Many processes, particularly in living systems, have to take placewithin a precise pH range. Should the pH of blood move 0.5 units outside the rangeshown in Table 5.4, the person would become unconscious and die. Evolution hasdevised buffer solutions to prevent such changes in pH in the body. A buffer solutionis one in which the pH remains approximately constant when small amounts of acid orbase are added.

Table 5.4

Fluid pH range

blood 7.35-7.45saliva 6.4-6.8tears 7.4urine 4.8-7.5

stomach juices 1.6-1.8

Biological systems work within precise ranges which buffers keep fairly constant. Whydo you think urine can have such a wide range?

Manufacturing systems also require precise control of pH and buffers are used inelectroplating, photographic work and dye manufacture. Some examples are shownin Table 5.5



Table 5.5: Applications of pH control

Many pharmacy products try tomatch their pH to the pH of thebody tissue.

Electroplating industries need pHcontrol 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 areencountered.

5.2.1 Acid Buffers�

Æ

�

�

Learning Objective

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 ofthe A- ion are present in the buffer. This is the conjugate base.



MA +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- ionswill trap the extra hydrogen ions and convert them to the weak acid. This stabilises thepH. If alkali is added the large reserve of HA molecules will convert the extra OH- towater. This again stabilises the pH.

A typical example of an acid buffer solution would be ethanoic acid and sodiumethanoate. The ethanoic acid is only partly dissociated. The sodium ethanoate saltcompletely dissociates and provides the conjugate base.

CH3COOH

CH3COO +

CH3COO + H

CH3COONa Na

(HA reserve)

(A- reserve)

Figure 5.9: Equilibria in acid buffer solutions



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.

10 min

�

Æ

�

�

Learning Objective

After completing this activity you should understand how an acid buffer solutionfunctions by absorbing extra acid or basic ions.

A simulation showing how an acidic buffer works is available in the on-line version of thisTopic.

There are two parts to the on-line simulation:

1. Drag and drop an extra hydrogen ion into the buffer solution, observe the responseof the buffer and read the description.

2. Drag and drop an extra hydroxide ion into the buffer solution, observe the responseand 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 conjugatebase.

Addition of hydroxide to the buffer.

Extra hydroxide ions in the buffer react with some H+ ions and upset the equilibriumsituation in the weak acid.

The position of equilibrium shifts (Le Chatelier’s principle) and the large reserves of HAmolecules 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 theaddition of a small amount of base. The salt of the weak acid provides the conjugatebase, which can absorb hydrogen ions from addition of small amounts af acid.



5.2.2 Basic buffers.�

Æ

�

�

Learning Objective

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. ammoniasolution and ammonium chloride. The ammonia solution is partly ionised and theammonium chloride is completely ionised. If hydrogen ions are added, they combinewith ammonia and if hydroxide ions are added, they combine with the ammonium ions(conjugate acid) provided by the salt (NH4Cl).

Figure 5.10: Equilibria in base buffer solutions

The stable pH of the buffer is due to:

• The weak base which provides NH3 to trap added H+.

• The salt of this base which provides NH4+ to trap added OH-.



Summary of buffer systems

10 min

�

Æ

�

�

Learning Objective

After completing this activity you should be able to write equations representing anacid buffer system and a basic buffer system.

Two drag and drop exercises are available in the on-line version of this Topic to practisethe writing of equations for buffer systems.

In a basic buffer solution, the weak base removes excess hydrogen ions. The salt ofthe weak base provides the conjugate acid, which can supply hydrogen ions if these areremoved by the addition of small amounts of base.

5.3 Calculating pH and buffer composition.�

Æ

�

�

Learning Objective

To be able to calculate the pH of an acid buffer when given the acid dissociationconstant and the composition. To calculate composition when given the othervariables

A glance at Table 5.4 shows that biological buffer solutions have to operate aroundspecific pH values. This pH value depends upon two factors. The acid dissociationconstant and the relative proportions of salt and acid. The dissociation constant for aweak 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 ofHA at equilibrium is approximately the same as the molar concentration put intothe solution.

2. The salt MA completely dissociates. Therefore [A-] will effectively be theconcentration supplied by the salt.


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 pKa (orH+ = Ka)

This equation allows calculation of pH of an acid buffer from its composition and aciddissociation constant, or calculation of composition from the other two values. Valuesfor 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 Ka.

Calculate the pH of a buffer solution made with 0.1 mol � -1 ethanoic acid (Ka

=1.7x10-5mol � -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

�"�

�� acid�

�salt��"�

��

��"�

�� - 1

%" �� log�"�

�� log

��

�%" � � ��

%" � ��

b) With the salt at 0.2 mol � -1



�"�

�� acid�

�salt��"�

��

��"�

�� 0.5 �� - 1

%" � � log�"�

�� log

� ��

�%" � � ��

%" � ��

Notice that doubling the salt concentration has only raised the pH by 0.3 (from 4.77 to5.07) and in general, the pH of the buffer is tied closely to the pKa value for the weakacid, in this case ethanoic acid pKa = 4.77 (data booklet rounds this to two significantfigures i.e. 4.8). The ratio of acid to salt effectively provides a ’fine tuning’ of the pH.

Example : Calculating composition from pH and pKa

Calculate the concentrations of ethanoic acid and sodium ethanoate required to make abuffer solution with a pH of 5.3 (pKa in data booklet).


�� log�acid��salt�

log�acid��salt�

� � ��

log�salt��acid�

� � ��

�salt��acid�

� ��

So the ratio of 3.16 to 1 is required and 3.16 moles of sodium ethanoate mixed with onelitre of 1.0 mol � -1 ethanoic acid could be used.


5.4. SUMMARY 93

Buffer calculations

40 min

�

Æ

�

�

Learning Objective

To be able to calculate the pH of an acid buffer when given the acid dissociationconstant and the composition, and calculate composition when given the othervariables.

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 and0.50 mol � -1 sodium ethanoate (Ka is in the data booklet). Give your answer to twodecimal places.

Q21: Calculate the composition of methanoic acid and sodium methanoate required tomake a buffer solution with a pH of 4.0. Quote your answer as a ratio of salt to 1 (so6.31 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 dissolvedin 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 benzoatewas dissolved in one litre of 0.02 mol � -1 benzoic acid (sodium benzoate isC6H5COONa, Ka and pKa in data booklet).


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 aciddissociation 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 whensmall amounts of acid or base are added.

• There are two types of buffer solution, acidic and basic. In each case the solution’scomposition consists of a weak acid (or base) accompanied by a salt of the weakacid (or base).

• Using these formulae, it is possible to calculate any one of the quantities involved,



if the other entities are known.

�"�

�� acid�

�salt�


5.5 Resources• Chemistry in Context: Hill and Holman , Nelson ISBN 0-17-438401-7

• Chemistry: A. and P. Fullick, Heinemann, ISBN 0-435-57080-3


• Chemistry - Advanced Higher : Unit 2: Principles of Chemical Reactions. HigherStill Support. ISBN 1-85955-874-7

• Chemical Storylines: Salters Advanced Chemistry, HeinemannISBN 0-435-63106-3


• 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 andpermission is gratefully acknowledged.

Electroplating picture courtesy of Sawyer and Smith Corp.




http://chemistry.semo.edu/crawford/ch186/lectures/ch15/index.html

http://www.science.ubc.ca/~chem/tutorials/pH/launch.html

95

Topic 6

Thermochemistry

Contents

6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

6.2 Bond energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

6.3 Hess’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

6.4 Standard enthalpy changes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102

6.5 Calorimetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106

6.6 The Born-Haber cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

6.7 Enthalpy of solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

6.8 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

6.9 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

6.10 End of Topic test. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117



• describe how exothermic and endothermic enthalpy changes can be illustrated inpotential energy diagrams (Higher Unit 1);

• describe some examples of enthalpy changes and calculate values of enthalpychange from both experimental evidence and application of Hess‘s Law (HigherUnit 3);

• describe the formation of ionic and covalent bonds (Unit 1, topic 3 and 4).

Learning Objectives


• define examples of standard enthalpy changes and explain how calorimetry canbe 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 reactionsare described as exothermic. A reaction absorbing heat cools its surroundings and isdescribed as endothermic. Thermochemistry concerns the study of changes in energythat occur during chemical reactions. In this Topic, your knowledge of thermochemistrywill be extended to include calculations of enthalpy changes involved in reactions withboth 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 Ea 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 reactionvessel?


6.2. BOND ENERGY 97

6.2 Bond energy�

Æ

�

�

Learning Objective

To be able to use bond enthalpies and mean bond enthalpies to calculate enthalpychanges in reactions

Covalent molecules are formed when the electrons shared between atoms result in alower energy system than the atoms alone before bonding. The process is exothermicand has a negative enthalpy change, i.e. �H is negative.

Figure 6.3: Bond formation in H2

If the diatomic hydrogen molecule shown in Figure 6.3 was broken apart to form twohydrogen atoms, the same amount of energy would have to be supplied. The energyrequired 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 databooklet on page 9.

Figure 6.4: Bond breaking in H2

H2(g) 2H(g) ΔH = +432 kJ mol-1

Bond enthalpies always refer to breaking bonds under gaseous conditions. This allowscomparisons to be made on an equivalent basis between different bonds.



Bond enthalpy values

5 min

�

Æ

�

�

Learning Objective

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 thehalogen only.

Q8: Suggest why nitrogen has a higher bond enthalpy than oxygen.

A molecule like methane, CH4, has four carbon to hydrogen bonds within it (seeFigure 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 compoundis broken up.

The mean molar bond enthalpy is an average value that is quoted for a bond that canoccur 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 athermochemical cycle showing bond dissociation and formation.


6.2. BOND ENERGY 99

Enthalpy calculations from bond enthalpies

10 min

�

Æ

�

�

Learning Objective

After this activity you should be able to calculate enthalpy changes from bond enthalpydata.

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 leadyou 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 thisreaction.

Table 6.1: Calculation of enthalpy changes

BOND BREAKING (ALWAYSPOSITIVE)

BOND MAKING (ALWAYS NEGATIVE)

1 mole of H-H bonds = 2 moles of H-Cl bonds =

1 mole of Cl-Cl bonds =

ENERGY IN = ENERGY OUT =

Copy down the outline of the calculation and fill in the values you get as responses tothese 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 fromisolated atoms?




Enthalpies of reaction estimated from bond enthalpies may differ from experimentallydetermined values since the mean bond enthalpy is an average from differentenvironments, 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�

Æ

�

�

Learning Objective

To be able to use Hess’s law diagrams of bond making and breaking to calculateenthalpy changes, and vice-versa

The cycle used to calculate the values of enthalpy changes in the last section is oftencalled Hess’s Law. This states that the overall reaction enthalpy is the sum of thereaction enthalpies of each step of the reaction. This is an application of the First Lawof Thermodynamics, which states that energy is conserved. In a chemical reaction, theenergy change in converting reactants into products is the same, regardless of the routeby which the chemical change occurs.

One application of Hess’s Law allows us to calculate an unknown bond enthalpy fromother data. We can calculate the bond enthalpy of the carbon to carbon bond in ethanefrom bond enthalpies and the �H information below.

2C(g) + 3H2(g) � C2H6(g) �H = -1514 kJ


6.3. HESS’S LAW 101

Figure 6.7: Bond enthalpy of a C-C bond



Finding a bond enthalpy from enthalpy changes

10 min

�

Æ

�

�

Learning Objective

After this activity you should be able to calculate bond enthalpies from enthalpy ofreaction data.

A question set leading to solving a calculation.

Answer the following questions and complete the calculation. The layout to use is shownin 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 beingbroken?

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. Writethis 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 theanswer on-line.

6.4 Standard enthalpy changes

The term standard enthalpy change refers to an enthalpy change for a reaction inwhich reactants and products are considered to be in their standard states at a specifiedtemperature.

The standard state of a substance is the most stable state of the substance understandard conditions and the standard conditions refer to a pressure of one atmosphereand a specific temperature, usually 298 K (25ÆC).

Many standard enthalpy changes are given names describing the type of reactionconcerned. The units are always kilojoules per mole (kJ mol-1).

The standard molar enthalpy of formation (�HÆ

f) is the enthalpy change that occurswhen 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 whenone mole of a substance is completely burned in oxygen under standard conditions.�HÆ

c 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 COand CO2.


6.4. STANDARD ENTHALPY CHANGES 103

Other enthalpy changes like the enthalpy of solution and theenthalpy of neutralisation should already be familiar to you from past work andmore 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 byexperiment.

For example: It is difficult to burn carbon and form carbon monoxide only. So a valuefor �HÆ

f for carbon monoxide would be difficult to determine by experiment.Look at thisHess’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 �H 1and�H3 are measurable.

So:

�H1 = �H2 + �H3

�H1 = standard enthalpy of formation of CO2 = -394kJ mol-1

�H3 = standard enthalpy of combustion of CO = -271kJ mol-1

� -394 =�H2 + (-271)

��H2 = -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 standardenthalpy changes that are experimentally determined.

Many Hess’s Law problems are best solved by an ’algebraic’ method using the equationsrather than an enthalpy cycle.

As an example: Find a value for the enthalpy change when carbon reacts with carbondioxide to give carbon monoxide according to Equation 6.1. The standard molarenthalpy of combustion of carbon is available in the data booklet and that of carbonmonoxide is -271 kJ mol-1

C(s) + CO2(g) � 2CO(g) (6.1)



STEP 1 Write the ’target’ equation.

C(s) + CO2(g) � 2CO(g) �H = ?

STEP 2 Write equations for all theinformation given to you.

equation (a) C(s) + O2(g) � CO2(g) �H = -394 kJ mol-1

(from databook)

equation (b) CO(g) + 1/2 O2(g) � CO2(g) �H = -271 kJ mol-1

STEP 3 Change these equations around by reversing,multiplying or dividing them, before combining themto obtain the target equation. Hess’s Law allows youto treat the �H values the same way. The websiteanimation shows the stages.

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) � CO2(g) �H = -394 kJ mol-1

+ 2CO2(g) � 2CO(g) + O2(g) �H = +542 kJ mol-1��

= C(s) + CO2(g) �2CO(g) �H = +148 kJ mol-1

Notice how the number of moles of carbon dioxide reduces to just one mole and theoxygen molecules cancel out.

Hess’s Law enthalpy cycle.

10 min

�

Æ

�

�

Learning Objective

After this activity you should be able to solve Hess‘s Law problems using an enthalpycycle.

A drag and drop exercise practising the enthalpy cycle method is available in the on-lineversion of this Topic

Given the information that the standard molar enthalpy of formation of sulphur dioxideand sulphur trioxide is -297 kJ mol-1 and -395 kJ mol-1 respectively, drag and drop theexpressions in the on-line activity into place to complete the cycle and thus calculate avalue for the standard molar enthalpy of combustion of sulphur dioxide.

These last two examples illustrate an important general law. The standard enthalpy of areaction can be calculated from tabulated standard molar enthalpies of formation using


6.4. STANDARD ENTHALPY CHANGES 105

the relationship:

�HÆ =��HÆ

f(products) -��HÆ

f(reactants)

So in the example given previously in Equation 6.1:

C(s) + CO2(g) � 2CO(g) �H = ?

Equation C(s) + CO2(g) � 2CO

Given �HÆ

f 0 -394 -123

Multiply by number ofmoles in the equation 0 -394 -246

�HÆ = ��HÆ

f(products) - ��HÆ

f(reactants)

�HÆ = -246 - (-394)

�HÆ = +148 kJ (same result as previously)

Hess’s Law questions

30 min

�

Æ

�

�

Learning Objective

After this set of questions you should be able to solve Hess’s Law problems usingalgebraic 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) � NO(g) �H = +90.2 kJ mol-1

(2) 1/2 N2(g) + O2(g) � NO2(g) �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 -827and -1676 kJ mol-1 respectively. Calculate the enthalpy change which takes place in thethermite reaction (give your answer in kJ mol-1).

Fe2O3(s) + 2Al(s) � 2Fe(s) + Al2O3(s)

Q20: The standard molar enthalpy of formation of methane is impossible to measure inpractice.

C(s) + 2H2(g) � CH4(g) �H = ?



Given that the standard molar enthalpy of combustion of methane is �HcÆ = - 891 kJ

mol-1, use data booklet values for the standard molar enthalpy of combustion of carbonand of hydrogen to calculate a value for the standard molar enthalpy of formation ofmethane.

Q21: a) Use the standard molar enthalpies of formation in the table below to calculatethe standard molar enthalpy of combustion of the gas, diborane B2H6.

�HÆ

f / kJ mol-1

B2H6(g) +41.0H2O(l) -286.0B2O3(s) -1225.0

b) Use the data booklet to help you explain whether ethane or diborane is the betterfuel.


6.5 Calorimetry�

Æ

�

�

Learning Objective

To be able to understand the functioning of the bomb calorimeter and use results fromits operation to measure enthalpy changes

Enthalpy changes in combustion reactions can be obtained by experiment using acalorimeter . You may have tried this last year. The energy released in combustionheats up water in a container and calculation assumes all the energy is transferredto the water as heat. The apparatus is not very efficient as some heat is lost to thesurroundings.

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 kilojoulesrequired to raise the temperature of the whole apparatus by 1K).

Bomb calorimetry

15 min

�

Æ

�

�

Learning Objective

After completing this activity you should understand how a bomb calorimeter worksand be able to calculate an enthalpy of combustion, �Hc from sample results.

An activity is available in the on-line version of this Topic in which you can view the partsof a bomb calorimeter and solve a thermochemical calculation to measure the enthalpyof combustion of an alcohol.


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 thecalorimeter.

4. Thermometer

Measures the temperature rise accurately after the ignition. In this reaction thetemperature rose from 16.50ÆC to 22.24ÆC.

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 ensurecomplete combustion.

7. Calorimeter

Contains a known mass of water and a calibration experiment shows that a onedegree rise in temperature is caused by 10.30 kJ of energy absorbed (the heatcapacity is thus 10.30 kJ K-1).

Calculate the enthalpy of combustion from the data provided. These questions will takeyou 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 thecalorimeter by one degree?

Q23: What temperature rise (in ÆC or K) actually occurs?



Q24: How much energy in kJ has the ethanol released?

Q25: Calculate the energy change in kJ when one mole of ethanol is burned (rememberthe sign).


6.6 The Born-Haber cycle�

Æ

�

�

Learning Objective

To appreciate that the Born-Haber cycle is an application of Hess’s Law to theformation of an ionic crystal and can be used to calculate unknown enthalpy changeswithin it when the other values are known

The strength of covalent bonds is measured by the bond enthalpy. In ioniccompounds the comparable enthalpy value is the lattice enthalpy. Thestandard molar enthalpy change of lattice formation (�HÆ

LATT) is the enthalpychange that occurs when one mole of an ionic crystal is formed from the ions in theirgaseous state under standard conditions. Lattice enthalpies cannot be determineddirectly, 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 thatincludes 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 inROUTE 2.

Figure 6.10: Alternative routes to NaCl solid crystals

When numerical values are known for all the other steps, the lattice enthalpy can becalculated from Hess’s Law since:


6.6. THE BORN-HABER CYCLE 109

��H by ROUTE 1 = ��H by ROUTE 2

The Born-Haber cycle

30 min

�

Æ

�

�

Learning Objective

After completing this activity you should be able to define the terms involved in theBorn-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-Habercycle and solve an associated calculation to find the enthalpy of lattice formation of acrystalline solid.

In the on-line version of this Topic you can click on each enthalpy change in turn to seethe definition and take note of the information given here or in the data booklet (writedown each value). The calculation of a value for the standard molar enthalpy changeof lattice formation for sodium chloride will be done after the the information has beengathered.

�

�

� �

�

�

��

��

��

��

��

��

��

Figure 6.11: Energy steps in the Born-Haber cycle

The steps in the Born-Haber cycle are these:



1. The standard molar enthalpy of formation (�HÆ

f) refers to the enthalpy changethat occurs when one mole of a substance is prepared from its elements in theirstandard states. For sodium chloride this is:

ROUTE 1 Na(s) + 1/2 Cl2(g) � Na+ Cl-(s) �HÆ

f = -411 kJ mol-1

2. The standard molar enthalpy of atomisation (�HÆ

AT) of an element is the enthalpyrequired to produce one mole of isolated gas atoms from the element in itsstandard state. �HÆ

AT is always positive. Use the data book to find the numericalvalue of this change and write it down.

Na(s) � Na(g) �HÆ

AT

3. The first ionisation energy (�HÆ

I.E.) of an element is the energy required to removeone electron from each atom in a mole of gaseous atoms. One mole of gaseousions is formed. �HÆ

IE is always endothermic. Use the data book to find thenumerical value of this change and write it down.

Na(g) � Na+(g) + e- �HÆ

I.E.

Subsequent ionisation energies remove successive electrons, i.e. the secondionisation energy would be:

Na+(g)� Na2+(g) + e-

Note: For this change:

Na(g)� Na2+(g) +2 e-

The value of this enthalpy change would be the sum of the first and secondionisation energies. (+502 +4560)kJ mol-1

4. The molar bond enthalpy (�HÆ

BOND) of a diatomic molecule is the enthalpyrequired 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 enthalpyof atomisation since the element is a gas.

Use the data book to find the numerical value of this change and write it down.

�

�Cl2(g) � Cl(g)

�

�� HÆ

BOND

5. The electron affinity (�HÆ

E.A.) is usually defined as the enthalpy change for theprocess of adding one electron to each atom in one mole of isolated atoms in thegaseous state. Use the data book to find this value.

Cl(g) + e- � Cl-(g) �HÆ

E.A.

As with ionisation enthalpies, second and subsequent electron affinities can bewritten. The enthalpy change involved in producing an ion X 2- would be the sumof the first and second electron affinities.

6. The standard molar enthalpy change of lattice formation (�HÆ

LATT) is theenthalpy change which occurs when one mole of an ionic crystal is formed fromthe ions in their gaseous state under standard conditions.


6.6. THE BORN-HABER CYCLE 111

The standard molar enthalpy change of lattice formation is the quantity the Born-Habercycle will help us to calculate since:

��H by ROUTE 1 = ��H by ROUTE 2

Step1 = 2 + 3 + 4 + 5 + 6

�HÆ

f= �HÆ

AT + �HÆ

I.E. + 1/2 �HÆ

BOND +

�HÆ

E.A. + �HÆ

LATT

So �HÆ

LATT= �HÆ

f - (�HÆ

AT + �HÆ

I.E. +

1/2 �HÆ

BOND + �HÆ

E.A.)

Q26: Substitute in the numerical values you took note of in the activity and find avalue 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

10 min

�

Æ

�

�

Learning Objective

After completing this activity you should be familiar with the steps involved in makingup 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 becomefamiliar with the enthalpy changes involved in a Born-Haber cycle.

Drag and drop the expressions in the on-line version into place to complete the cyclefor 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 formationof lithium fluoride.

Use paper and pencil to draw a Born-Haber Cycle diagram showing all the steps involvedin the formation of lithium fluoride. Use the data booklet to source values for all theenthalpy changes (pages 9,10 and 17) and calculate a value for the standard molarenthalpy of formation of lithium fluoride.

Q27: Substitute in the numerical values you noted from the data booklet to solve forthe standard molar enthalpy of formation of lithium fluoride. Do this exercise on paperbefore revealing the solution.

Further questions to practise Born-Haber cycles

10 min

�

Æ

�

�

Learning Objective

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.



Q28: If a Born-Haber cycle was drawn for copper (II) chloride (CuCl2), what value in kJmol-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�

Æ

�

�

Learning Objective

To be able to link lattice enthalpy, enthalpy of solution and hydration enthalpy in athermochemical cycle.

To define each term and perform calculations

The lattice enthalpy of an ionic compound gives an indication of the strength of thebonding between the ions. When an ionic compound dissolves, it is useful to think ofthe process in two stages:

1) Energy has to be supplied to break down the lattice. Thisstage 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 solutionprocess.


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 moleof a substance is dissolved completely in water.

The enthalpy of hydration (�HÆ

HYD) is the enthalpy change when one mole ofindividual gaseous ions is completely hydrated, i.e:

En+(g) � En+(aq)

and

En-(g) � En-(aq)



Figure 6.13: Enthalpy of solution of NaCl

The standard molar enthalpy of solution of an ionic compound thus involves lattice andhydration enthalpy steps and is given by:

�"�soln � �"�

hyd � �"�latt

Notice that the lattice enthalpy is defined as the enthalpy associated with latticeformation. This would be exothermic with a value of -769 kJ mol-1 (see the dottedline on the diagram). It is the lattice breaking enthalpy (+769 kJ mol-1) which is involvedhere. 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 �HSOLN 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 saltdissolved

a) go downb) go up?c) stay the same?


6.7. ENTHALPY OF SOLUTION 115

This hand-warmercontains a largeamount of sodiumethanoate salt.When a smallmetal disc insideis bent,crystallisation ofthe salt is initiatedand the padwarms up. Thesalt can be madeto dissolve againby placing the bagin boiling water fora 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 tohydration or lattice?

Enthalpy of solution

10 min

�

Æ

�

�

Learning Objective

After completing this activity you should know how enthalpy of solution relates tolattice 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 cycleshowing the enthalpy of solution of potassium bromide. Answer the questions whichfollow.

Q38: Use a data booklet to find values enabling you to calculate a value for �Hsoln ofKBr in kJ mol-1.

Q39: In terms of heat, this solution process is therefore

a) exothermicb) endothermic

Q40: The temperature reading on a thermometer in the solution as the salt dissolvedwould therefore

a) go downb) go upc) stay the same



Enthalpy Signs and Calculations

An online exercise is provided to help you if you require additional assistance with thismaterial, 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 thermochemicalcycles 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 thesame, 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 (�HÆ) refer to enthalpy changes wherereactants and products are in their standard (most stable) state at 298 K and 1atmosphere pressure. There are a variety of defined standard enthalpy changesreferring to different reaction types, e.g. �HÆ

c is a standard molar enthalpy ofcombustion. Calorimeters can be used to measure the energy change in a reactionlike combustion. Calculations of the enthalpy change on complete combustion ofone mole of the substance yields �HÆ

c.

• The standard enthalpy of a reaction can be calculated from tables of standardenthalpies of formation.

��HÆ


f(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 existsin different molecular environments. Values are found in the data booklet or canbe calculated from enthalpy change information.

Enthalpy changes in reactions can be calculated from the enthalpies involved inthe bond breaking and bond making steps.


6.9. RESOURCES 117

• Born-Haber cycles are enthalpy diagrams applied to the formation of ionic crystalsand can be used to determine enthalpies of lattice formation that cannot bedetermined by experiment. Standard molar enthalpy of lattice formation is theenthalpy change on formation of one mole of a crystal from gaseous ions understandard conditions. Other steps in the Born-Haber cycle include enthalpy ofatomisation, ionisation enthalpy, bond enthalpy, electron affinity and enthalpy offormation.

• The standard molar enthalpy of solution of an ionic compound involves lattice andhydration enthalpy steps and is given by:

�"�soln � �"�

hyd � �"�latt

6.9 Resources• Chemistry in Context: Hill and Holman, Nelson, ISBN 0-17-438401-7



• Chemistry - Advanced Higher: Unit 2: Principles of Chemical ReactionsHigher Still Support, ISBN 1-85955-874-7



6.10 End of Topic test.



119

Topic 7

Reaction Feasibility

Contents

7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1217.2 The concept of entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122

7.2.1 Entropy and spontaneity . . . . . . . . . . . . . . . . . . . . . . . . . . . 1247.2.2 Calculating entropy values . . . . . . . . . . . . . . . . . . . . . . . . . 125

7.3 The second law of thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . 1277.4 Gibbs free energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130

7.4.1 Calculating �GÆ values . . . . . . . . . . . . . . . . . . . . . . . . . . . 1337.4.2 Free energy change under non-standard conditions . . . . . . . . . . . 136

7.5 Ellingham diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1387.5.1 Extraction of metals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140

7.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1437.7 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1447.8 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144



• describe standard enthalpy changes involved in covalent and ionic bonding (Topic2.6);

• use Hess’s law in calculations involving enthalpy changes including bondenthalpies and Born-Haber cycles. (Topic 2.6).

Learning Objectives


• define examples of standard entropy changes and describe how entropy changesin reaction systems are temperature dependant;

• state the second and third laws of thermodynamics in terms of entropy anddescribe how the total entropy change can be expressed in terms of Gibbs freeenergy;

• solve problems and carry out calculations involving Gibbs free energy, enthalpyand entropy;

120 TOPIC 7. REACTION FEASIBILITY

• relate feasibility and equilibrium position to Gibbs free energy values;

• use Ellingham diagrams to predict reaction conditions.



7.1 Introduction

Exothermic reactions have a negative value for the enthalpy change and release heatto the surroundings. In contrast, endothermic reactions have a positive enthalpy changeand 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 inCopenhagen felt that a general criterion for predicting reaction spontaneity could beassumed from the sign of the enthalpy change. If �H had a negative value, the reactionwould occur on its own accord, without help from us, and if �H had a positive value, thereaction could not occur by itself.

Indeed, almost all exothermic reactions are spontaneous at standard conditions forexample, iron rusting:

2Fe(s) + 3/2O2(g) + 3H2O(l) � 2Fe(OH)3(s) �H = - 791 kJ

The opposite reaction, rusting reversing to give pure iron, with a positive enthalpychange, never occurs spontaneously at standard conditions.

Unfortunately, this simple rule is not universal. At 1 atmosphere and 25ÆC, the changeshown here when ice melts is spontaneous, even though the enthalpy change isendothermic.

H2O(s) � H2O(l) �H = + 6.01 kJ mol-1

Similarly, liquid water evaporates under standard conditions even though the enthalpychange is endothermic.

H2O(l) � H2O(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 roomtemperature become spontaneous at higher temperatures. An example is thedecomposition 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 Khowever, the limestone decomposes to give off carbon dioxide.

In other words, there is more to the question of spontaneity than simply a considerationof the enthalpy change and it would seem that temperature is an important factor. Theother important factor is entropy.



7.2 The concept of entropy�

Æ

�

�

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;

To carry out relevant calculations.

The entropy of a system is the degree of disorder of the system. The greater thedisorder, the greater the entropy. Low entropy is associated with strongly orderedsubstances.

Table 7.1: Entropy changes

Increasing Entropy Decreasing Entropy

A puddle dries up on a warm day as theliquid becomes water vapour. The disorder(entropy) increases.

A builder uses a pile of loose bricks toconstruct a wall. The order of the systemincreases. Entropy falls.

Heating ammonium nitrate forms one moleof dinitrogen oxide and two moles ofsteam. Three moles of gas formed. Thedisorder (entropy) increases.

When the individual ions in a crystal cometogether they take up a set position. Thedisorder of the system falls. Entropydecreases.

NH4NO3(s) � N2O(g) + 2H2O(g) Na+(aq) + Cl-(aq) � NaCl(s)

Entropy is the degree of disorder of a system, and the examples shown here illustratethat the term can be applied to areas other than chemistry!

A typical teenagers bedroom has a highdegree of disorder. A high entropy.

A jigsaw is a highly ordered system. Whenmade up, it has low entropy.

Figure 7.2: Examples of low and high entropy


7.2. THE CONCEPT OF ENTROPY 123

Entropy and temperature

10 min

�

Æ

�

�

Learning Objective

After completion of this activity you should understand the role of temperature andchanges 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 changesin 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 changein entropy?

Q4: Is this accompanied by an increase or decrease in the disorder of the moleculesin the system?


Entropy increases as temperature increases. Changes of state involve increases inentropy. The entropy of a perfect crystal at zero Kelvin is zero.

The melting of ice and the evaporation of water are both examples of endothermicreactions which proceed spontaneously.



H2O(s) � H2O(l) �H = + 6.01 kJ mol-1

H2O(l) � H2O(g) �H = + 44.1 kJ mol-1

In each of these reactions, although �H is positive, it is the increase in the entropy ofthe system that is the driving force behind the reactions. This change in entropy, giventhe symbol �S, more than compensates for the positive enthalpy change.

7.2.1 Entropy and spontaneity�

Æ

�

�

Learning Objective

To show how a reaction that has no change in enthalpy can proceed spontaneouslyby having an increase in entropy.

How can an increase in entropy provide the drive necessary to allow a reaction toproceed spontaneously? An example of a change where there is no change in theenthalpy 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 right-hand side. Both are at 1 atmosphere pressure and 25ÆC. If the partition is removed, thetwo types of gas mix spontaneously. There is no temperature change; the only drivingforce is the desire to increase the disorder. (Think for example of how a really bad smellreleased 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. Howmany rooms have you been in where the oxygen suddenly all goes to one side and thenitrogen to the other? (How likely is it that the ’bad smell’ will all move and concentratein 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 thehigher entropy, will prevail.


7.2. THE CONCEPT OF ENTROPY 125

7.2.2 Calculating entropy values�

Æ

�

�

Learning Objective

To be able to estimate the change in entropy in a system by application of somegeneral principles.

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 solidcrystal is cooled to absolute zero (zero kelvin), all the translational motion of the particlesis eliminated and each particle has a well defined location, i.e. it is 100% ordered. Theentropy is therefore zero. This is one version of the Third Law of Thermodynamics

As temperature is increased, entropy increases. As with enthalpy values, it is normalto quote standard entropy values for substances as the entropy value for the standardstate of the substance.

Notice that the unit of entropy values is joules per kelvin per mole (J K-1 mol-1). Becareful with this, since enthalpy values are normally in kilojoules per mole and inproblems involving these quantities the units must be harmonised.

In a chemical reaction system, the standard entropy change (�SÆ) can be calculatedfrom 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 ordecreases in a system by inspection of the equation. Actual values of standard entropiesare in the SQA data booklet on page 16.

Estimating and calculating spontaneity

30 min

�

Æ

�

�

Learning Objective

After this activity you should be able to estimate the sign of the entropy change in areaction 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 entropychanges and the data table of standard entropy values (Table 7.2). You will also needaccess 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 entropyvalue 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.



Table 7.2: Data table of standard entropy values

SUBSTANCE Standard Entropy SÆ/ J K-1 mol-1

Diamond, C(s) 2.4

Sodium chloride, NaCl(s) 72.4

Water (ice), H2O(s) 48

Water (liquid), H2O(l) 70

Water (gas), H2O(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.


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 B negative C stay the same

Q9: 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 changewhen steam condenses to water at standard conditions.

H2O(g) � H2O(l)

�S Æ = �SÆPRODUCTS - �SÆREACTANTS

�S Æ = 70.0 - 188.7�S Æ = - 118.7 J K-1 mol-1


7.3. THE SECOND LAW OF THERMODYNAMICS 127

Use a similar method and additional data from the SQA data booklet to calculate thestandard entropy change for this reaction. Do this on paper (now!) before displaying theanswer.

2H2(g) + O2(g) � 2H2O(g)

Q14: The entropy change for this reaction during the formation of ammonia in the Haberprocess 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 aluminiumoxide from the elements under standard conditions.

Q16: Calculate the entropy change in J K-1mol-1 involved in the formation of calciumchloride from the elements. (Remember the sign.)


The direction of entropy change in a reaction can be estimated by application of somegeneral principles. The change in standard entropies in a reaction system can becalculated from:

�SÆ = �SÆPRODUCTS - �SÆREACTANTS

7.3 The second law of thermodynamics�

Æ

�

�

Learning Objective

To understand that the total entropy of a reaction system and its surroundings alwaysincreases for a spontaneous process.

To solve associated problems.

An increase in entropy provides a driving force towards a reaction proceedingspontaneously. There are, however, processes that proceed spontaneously that seemto involve an entropy decrease.

For example, steam condenses to water at room temperature:

H2O(g) � H2O(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 thesystem (- �HSYSTEM) is transferred to the surroundings .



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 scaldingeffect that would occur if your hand were placed in the steam - the disorder of the skinwould increase!)

In general terms, heat energy released by a reaction system into the surroundingsincreases the entropy of the surroundings. If heat is absorbed by a reaction from thesurroundings, this will decrease the entropy of the surroundings.

In the case of steam condensing, the entropy gain of the surroundings is equal to theenergy lost (- �H) of the chemical system divided by the temperature:

�� "�

��

+

The entropy change in the condensation situation therefore requires consideration of twoentropy changes. The change in the system itself and the change in the surroundingsmust be added together, and for a spontaneous change to occur this total entropychange 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 �SÆ

SYSTEM from the data booklet gives:

�� *��

��

�� *��

All this means that although the entropy of a system itself may drop, the process itself will


7.3. THE SECOND LAW OF THERMODYNAMICS 129

still be a natural, spontaneous change if the drop is compensated by a larger increasein entropy of the surroundings.

Expressed another way, this is the Second Law of Thermodynamics The total entropyof 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 withoutneeding work to bring it about’. Thermodynamics is concerned with the direction ofchange and not the rate of change. The rate of change of a reaction is studied in moredetail in Topic 2.9 Kinetics.

Melting ice

20 min

�

Æ

�

�

Learning Objective

After this activity you should be able to use data booklet information to calculatethe total entropy change in a system and its surroundings and therefore predictsponteneity.

A step-by-step data handling exercise to calculate the total entropy of a system andsurroundings.

Look at the animation of ice melting at 25ÆC. Use the hints in each part of the problemto carry out the calculation yourself before displaying the answer.

Figure 7.6: Melting of ice

A block of ice melts at 25ÆC with an enthalpy change of + 6.01 kJ mol-1. Use this valueand 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 revealingthe answer.

Q18: Step 2 involves calculating a value for the entropy change in the system. Try thison paper yourself before revealing the answer.

Q19: Step 3 involves calculating a value for the total entropy change from the first twosteps. Try this yourself.

Here are two further examples to try.

Q20: Calcium carbonate, present in limestone, is stable under normal atmosphericconditions. 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 whylimestone is stable at 25ÆC but not at 1500ÆC.



CaCO3(s) � CaO(s) + CO2(g) �HÆ = +178 kJ mol-1

�SÆ = +161 J K-1 mol-1

Q21: Graphite has been converted into diamond by the use of extreme pressure andtemperature. 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 atroom temperature or 5000ÆC.

C(s)graphite � C(s)diamond �H = +2.0 kJ mol-1

The total entropy of a reaction system and its surroundings always increases for aspontaneous process and can be calculated from the temperature and the enthalpychange for the system.

7.4 Gibbs free energy�

Æ

�

�

Learning Objective

To understand how the total entropy is normally expressed in terms of Gibbs freeenergy.

To understand that the direction of spontaneous change is in the direction ofdecreasing free energy and that the standard free energy change for a reaction canbe calculated in a variety of ways.

Having to consider what happens to the surroundings in a chemical system is a bit ofa nuisance, since chemists really just wish to look at the reaction itself and be able topredict the feasibility of its occurence. This can be done by rearranging the equationwhich best represents the Second Law of Thermodynamics.

��

��'�� "�

��

+

�� "�

��

+� ��

Multiplying both sides by -T gives the following expression:

-T�SÆTOTAL = �HÆ

SYSTEM - T�SÆSYSTEM

The term -T�SÆ

TOTAL is replaced by a new symbol �GÆ called thestandard Gibbs free energy change. The expression becomes:

�GÆ = �HÆ - T�SÆ

Since �GÆ = -T�SÆTOTAL and the second law states that spontaneous changes have�SÆTOTAL positive, it follows that the direction of spontaneous change is in the direction


7.4. GIBBS FREE ENERGY 131

of decreasing free energy. Under standard conditions, �GÆ will be negative. Undernon-standard conditions, all spontaneous changes have �G negative.

Predicting spontaneity using free energy changes

15 min

�

Æ

�

�

Learning Objective

After this activity you should be able to predict the sign of �G and say whether areacton will be spontaneous or not from a consideration of temperature, enthalpy andentropy.

An on-line activity to complete a table showing a summary of variables involved indetermining the sign of �G.

Read through the four possible combinations of variables involved in calculating a signfor �G and then complete the summary table by dragging and dropping the correctentries into the frame.

The sign of the Gibbs free energy change in a reaction depends upon the positive ornegative nature of �H and �S and sometimes depends upon temperature.



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 signsof �HÆ and �SÆ.



7.4.1 Calculating �GÆ values�

Æ

�

�

Learning Objective

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 standardenthalpy and standard entropy changes.


Example Calculate a value for �GÆ and thus predict whether or not the combustion ofgraphite 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 �GÆ and is thermodynamically feasible.

In the same way that standard enthalpies of formation can be used to calculate standardenthalpy changes for a reaction (Topic 2.6 - Standard enthalpy changes), the standardfree energy change of a reaction can be calculated from the standard free energies offormation of reactants and products.

Example Calculate a value for �GÆ for this equilibrium reaction given that the standardfree energies of formation of nitrogen dioxide and dinitrogen tetroxide are +52 kJ mol -1

and +98 kJ mol-1 respectively.

2NO2 � N2O4

�-� � �-�#��$��

� �-�#��$��

�-� � � � ��

�-� � � )* ��

This reaction is therefore feasible in the direction that the equation shows (since �GÆ isnegative). This means that the equilibrium composition would favour the products overthe reactants. If the value of �GÆ had worked out at +6 kJ mol-1, the thermodynamicswould be predicting that the reaction was non-feasible. The equilibrium compositionwould favour the reactants over the products.



It follows then that when �GÆ = 0, the products and reactants will be equally favoured.This fact can be useful in calculating the temperature at which a reaction becomesfeasible. This will be the temperature around which �GÆ changes sign. The reaction willproceed 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 (thetemperature) is the only unknown in the equation.


Example Calculate the temperature at which limestone (calcium carbonate) becomesthermally unstable.

CaCO3(s) � CaO(s) + CO2(g)

�HÆ = +178 kJ mol-1 �SÆ = +161 J K-1 mol-1


� � � �� * �� + � �� * ��

�� + � ��

+ ��

��

Temperature when feasible = 1105.6 K (832.6ÆC)

Above this temperature, �GÆ is negative and decomposition occurs.



Calculations involving free energy changes

60 min

�

Æ

�

�

Learning Objective

After this activity you should be able to calculate values for �GÆ from standard freeenergy of formation data. You should also be able to calculate the temperature atwhich a reaction becomes feasible.

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 �GÆ forthese two reactions and thus predict whether or not the reaction is spontaneous .

a) 2Mg(s) + CO2(g) � 2MgO(s) + C(s)

b) 2CuO(s) + C(s) � 2Cu(s) + CO2(g)

Table 7.3:

SUBSTANCE �GÆ

FORMATION / 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 thereaction:

MgCO3(s) � MgO(s) + CO2(g)

�HÆ

f /kJ mol-1 -1113 -602 -394

SÆ/J K-1 mol-1 66 27 214

Q24: Use the data given, along with data booklet values to calculate the temperature atwhich the Haber process becomes feasible.

N2(g) + 3H2(g) � 2NH3(g)

�HÆ

f /kJ mol-1 0 0 -46.4

SÆ/J K-1 mol-1 ? ? 193.2

Q25: Given these reaction values for oxides of nitrogen:

4NO(g) � 2N2O(g) + O2(g) �GÆ = -139.56 kJ2NO(g) + O2(g) � 2NO2(g) �GÆ = -69.70 kJ

a) Calculate �GÆ for this reaction:

2N2O(g) + 3O2(g)� 4NO2(g)

b) Say whether the equilibrium position favours reactants or products.




Know about standard free energy changes for a reaction.

Understand the feasibility of a reaction and how the temperature at which a reactionbecomes feasible can be calculated from thermochemical data.

7.4.2 Free energy change under non-standard conditions�

Æ

�

�

Learning Objective

To know that a reaction will proceed spontaneously in the forward direction until thecomposition is reached where �G = zero

The values of �GÆ with which we have been working are based on the free energyvalues of reactants and products under standard conditions. However, the free energychange, �G, during the course of a reaction when a mixture of reactants and productsis 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 ofproducts rises. At the minimum point, the value of �G = 0 and this dictates theequilibrium 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, increasingentropy and decreasing free energy.

Figure 7.9 contains three graphs which show the situation for different values of the �GÆ

change. In each case, the reaction proceeds spontaneously in the forwards directionuntil the composition is reached where �G = 0.



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), �GÆ is 0. There will be equal amounts of A and B atequilibrium.

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 wellbe that the reaction has a high activation energy barrier that prevents it taking place ata measureable speed. Graph 1 (top left) could be slow to reach equilibrium and graph2 (top right) could be fast. The website animation illustrates this point.

The combustion of the chemicalsin a match is an exothermicreaction with an increase in

entropy. The free energy change istherefore negative but the high

activation energy prevents burningbefore striking.

Figure 7.10: Thermodynamics of burning a match

This is a suitable point to try the Prescribed Practical Activity on ’Verification of athermodynamic prediction.’



PPA - Verification of a thermodynamic prediction

90 min

�

Æ

�

�

Learning Objective

To be able to calculate the theoretical decomposition temperature for sodiumhydrogencarbonate and verify this experimentally.

A calculation from given values of enthalpy and entropy is carried out and an experimentundertaken to verify the results.

Consult with your tutor whether this PPA for the assessment of outcome 3 may be carriedout (Refer to SCCC Advanced Higher Chemistry Unit 2 PPA 4).

7.5 Ellingham diagrams�

Æ

�

�

Learning Objective

To be able to plot variations of free energy change with temperature and usethe subsequent graphs to predict the conditions under which reactions can occur,particularly the extraction of a metal from its ores.

The numerical value of �G alters with temperature. This is a consequence of theentropy term being temperature dependent and it is possible to calculate values of �Gfor differing temperatures.


is rearranged to give

�GÆ = -T�SÆ + �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 temperaturelike 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) + H2O(g) � CO(g) + H2(g) (7.1)

�HÆ = +131.3 kJ mol-1

�SÆ = +133.8 J mol-1

Plotting an Ellingham diagram

20 min

�

Æ

�

�

Learning Objective

After this activity you should understand how Ellingham diagrams can be used topredict the conditions under which a reaction can occur.

A data handling activity constructing Ellingham diagrams and using these to predict theconditions under which a reaction can occur.


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) + H2O(g) � CO(g) + H2(g)

This reaction can be considered as a combination of these two oxide formations:

C(s) + 1/2O2(g) � CO(g) (7.2)

H2(g) + 1/2O2(g) � H2O(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 �GÆ at varioustemperatures calculated and plotted onto the same graph.

Now return to your graph and plot the values for carbon monoxide and water formationon 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

FORMATION OF H2O(g)

Temperature / K 200 400 600 800 1000 1200

�GÆ / kJ mol-1 -223 -224 -215 -206 -197 -188

INTERPRETATION

The lines relating �G to temperature for the two equations Equation 7.2 andEquation 7.3 intersect at 981.3 K when �G = 0. The carbon and hydrogen are bothcapable of reacting with oxygen but at any temperature above 981.3 K, the carbon iscapable of winning oxygen from the water molecule and forcing the second equation(Equation 7.3) to reverse. CONSIDER 1000 K

C(s) + 1/2O2(g) � CO(g) �GÆ = -200 kJ mol-1

H2(g) + 1/2O2(g) � H2O(g) �GÆ = -197 kJ mol-1

By reversing the second equation and adding to the first, the result is:



C(s) + H2O(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) + 1/2O2(g) � CO(g) �GÆ = -182 kJ mol-1

H2(g) + 1/2O2(g) � H2O(g) �GÆ = -206 kJ mol-1

By reversing the second equation and adding to the first, the result is:

C(s) + H2O(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 thedifferent reactions and allows prediction of the conditions under which combinations ofindividual reactions become feasible.

In any Ellingham diagram, the reaction with the most negative �G value at a chosentemperature will operate as written. Any reaction with an equation above this value canbe 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 linesdrawn goes ’as written’.

7.5.1 Extraction of metals�

Æ

�

�

Learning Objective

To show that Ellingham diagrams can be used to predict the conditions required toextract a metal from its oxide

Ellingham diagrams plot values of �GÆ against temperature. If the lines are drawn formetal oxide formation reactions, these can be used to predict the conditions required toextract a metal from its oxide. This requires the formation of the metal oxide processto be reversed. Any chemical used to aid the reversing of this process must provideenough free energy to supply this reversal. It is normal to write all reactions that are onthe graph to involve one mole of oxygen (so that oxygen is removed when two equationsare combined).

Interpreting Ellingham diagrams

20 min

�

Æ

�

�

Learning Objective

After this activity you should be able to use Ellingham diagrams to predict theconditions required to extract a metal from its ore.

An interpretation exercise using an Ellingham diagram and answering questions thatlead to an understanding of metal ore extraction.

Answer these questions on paper before displaying the explanation. In each case referto the Ellingham diagram in Figure 7.11


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 byheat alone?

Q29: Use the graph Figure 7.11 to calculate the �GÆ value for the reaction in whichcarbon reduces zinc oxide at:

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 thereaction feasible or not?

Q30: At what temperature does the reduction of zinc oxide by carbon become feasible?


Try the next two questions yourself, on paper.

Q31: This Ellingham diagram, Figure 7.12, shows the reactions involved in the blastfurnace reduction of iron(II) oxide with carbon.



��

��

��

��

��

��

��

��

��

��

��

��

�

�

�

�

�

�

�

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 1500K.

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 toreduce iron(II) oxide?

e) Can you suggest a reason (apart from the temperature) why carbon monoxide mightbe more efficient than carbon at reducing iron(II) oxide?

Q32: Although magnesium ores are very abundant in the Earth’s crust, the very highreactivity of magnesium makes it difficult to extract the metal. During the Second WorldWar, 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?


7.6. SUMMARY 143

b) Describe two problems that the operation of the process at this temperature wouldpresent.

c) Use the Ellingham diagram to calculate �GÆ for the production of magnesium in thefollowing:

2MgO + Si� SiO2 + 2Mg at 1500 K

In industry, the extraction of magnesium from magnesium oxide using silicon involvestwo modifications.

(i) A mixture of calcium oxide and magnesium oxide is used, and the calcium oxidereacts 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:

CaO + 2MgO + Si � CaSiO3 + 2Mg 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. Standardentropies (SÆ) are quoted at 298 K and 1 atmosphere pressure. A perfect crystalhas zero entropy at zero Kelvin.

• Calculations of the standard entropy change for a reaction can be made from thestandard entropy values for reactants and products:

�So � �SoPRODUCTS � �So

REACTANTS

• Spontaneous changes involve an increase in the total entropy for a process thatis 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 � � �HT

• The free energy change (Gibbs free energy change) �G provides a way ofaccounting for the surroundings as well as the system.

�Go � �Ho � T�So

Calculations of the standard free energy change for a reaction can be made byusing standard enthalpy and standard entropy values in this equation.



• The standard free energy change in a reaction can also be calculated from aknowledge of the standard free energy of formation of the reactants and productsin this equation:

�Go � �GoPRODUCTS � �Go

REACTANTS

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 bychanging the reaction conditions, particularly temperature. The temperature atwhich a reaction becomes feasible can be calculated from values of �HÆ and�SÆ.

• Reactions proceed spontaneously in a forward direction until the compositionwhere �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 Ellinghamdiagrams. These graphs can be used to predict the conditions under whichreactions can occur, and in particular, to predict conditions necessary to extractmetals from their ores.

7.7 Resources• Chemistry in Context: Hill and Holman, Nelson, ISBN 0-17-438401-7



• Chemistry - Advanced Higher: Unit 2: Principles of Chemical ReactionsHigher Still Support, ISBN 1-85955-874-7



7.8 End of topic test



145

Topic 8

Electrochemistry

Contents

8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147

8.2 Electrochemical cells . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147

8.2.1 Shorthand cell conventions for electrochemical cells . . . . . . . . . . . 151

8.3 Standard Electrode Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154

8.3.1 The Standard Hydrogen Electrode . . . . . . . . . . . . . . . . . . . . . 155

8.3.2 The Electrochemical Series and associated calculations . . . . . . . . . 158

8.4 EÆ values and the standard free energy change . . . . . . . . . . . . . . . . . . 162

8.5 Fuel cells . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166

8.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169

8.7 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170

8.8 End of Topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170



• 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 H2Omolecules; (Higher, Unit 1)

• define the terms reducing agent and oxidising agent and identify these in redoxreactions; (Higher, Unit 1)

• understand what is meant by the standard free energy change (�GÆ). (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 itsmagnitude;

146 TOPIC 8. ELECTROCHEMISTRY

• explain what is meant by the standard electrode potential and explain how suchvalues are measured;

• calculate the e.m.f. of a cell under standard conditions (EÆ) and relate this to thestandard free energy change (�GÆ);

• explain how fuel cells are different from electrochemical cells and give somepossible benefits of their use.



8.1 Introduction

Electrochemistry is concerned with chemical reactions in which electrons are transferredfrom one reactant to another (redox reactions) and with the relationships between theelectron transfer and the electron currents generated or used in the process. This Topicconcentrates on reactions which generate electricity. These reactions spontaneouslyrelease 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 tinysilver oxide or lithium batteries that power heart pacemakers, and from the humble torchbattery 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 processor the power source in environmentally friendly cars.

8.2 Electrochemical cells�

Æ

�

�

Learning Objective

To explain how an electrode potential arises and how two electrodes can be combinedto produce an electrical cell



When zinc metal is placed in contactwith a solution containing Zn2+(aq)ions, a few of the metal atoms ionise.This leaves electrons on the metalthat can be accepted by other Zn2+

ions.

Figure 8.1: Zinc metal electrode

An equilibrium is set up: Zn(s)� Zn2+(aq) + 2e-

For zinc, this equilibrium lies slightlytowards the ions. Thus there isan increase in the positive chargein the solution and an increase inthe negative charge on the metal.In other words, the metal becomesslightly negative with respect to thesolution. There is also a very slightdecrease 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 possibleand an equilibrium is established:

Cu(s) � Cu2+(aq) + 2e-


8.2. ELECTROCHEMICAL CELLS 149

In this case however, the equilibrium lies slightly towards the atoms. Electrons areremoved 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 losemass?

a) gainb) lose

Q3: Which of the two metals is more reactive?

a) zincb) copper

Q4: Before revealing the answer, write on paper a general statement linking thereactivity 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 soa 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 withits ions in solution.

If zinc metal is placed in a solution containing copper(II) ions, a displacement reactiontakes place because zinc atoms have a tendency to lose electrons and copper(II) ionshave 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 theenergy is not released as heat but instead mostly as electricity. This is because thetwo reactants are kept separate and electron transfer takes place through an externalconnecting wire.



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-coppercell 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 copperelectrode.

Q10: What is moving as the electric current in the wire?

Q11: What is the purpose of the salt bridge?


An electrochemical cell is composed of two half-cells between which electrical contactis made by an electrolyte, often in the form of a salt bridge (ion bridge). Such a cell isable to produce an electric current because there is a potential difference between thetwo electrodes.

The electrode at which oxidation occurs is called the anode. In the above cell, the zincelectrode is the anode because the reaction that occurs is:

Zn(s) � Zn2+(aq) + 2e-



The electrode at which reduction occurs is called the cathode. In the above cell, thecopper 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 electrodesis 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 abilityto push electrons around the external circuit.

8.2.1 Shorthand cell conventions for electrochemical cells�

Æ

�

�

Learning Objective

To use the cell convention. To work out the overall cell reaction and to describe a cellusing 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) � � Cu2+(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 twodifferent states, i.e. solid and solution).

• The double vertical line represents the junction between the two solutions (i.e. thesalt bridge).

When written in this way, the electrons flow from the electrode on the left to the electrodeon the right. This is the spontaneous direction for the cell and the e.m.f. will always bepositive for this direction.

�� !��"!��"!�� !

�#� ��$��%

&��' ()*

+��)�,��$�#�� ,�)�,��$�#��

�,� �'��)� -

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 contactbetween the external circuit and the solution, e.g.



Figure 8.5

Figure 8.5 represents a half-cell in which hydrogengas is oxidised to hydrogen ions on a carbon rod.

Figure 8.6

Figure 8.6 represents a half-cell in which chlorinegas is reduced to chloride ions on a platinumelectrode.

Figure 8.7

Figure 8.7 represents a half-cell in which iron(II)ions are oxidised to iron(III) ions at a platinumelectrode.

In Figure 8.5 and Figure 8.6, note the extra phase boundary between the solution andgas. In Figure 8.7, note the comma between Fe2+(aq) and Fe3+(aq) since these are bothin the same phase (both in solution).

A word of warning: Sometimes a cell diagram may be drawn with the anode to the rightof the diagram. However, no matter which way round the diagram is, the cell notationmust have the anode on the left. A useful memory aid is that the anode and cathode arein 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 oution-electron half equations for each electrode process.

Oxidation half equation

Oxidation occurs at the anode. Anode is always to the left hand side.

i.e. Mg(s) � Mg2+(aq)

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:



2Ag+(aq) + 2e- � 2Ag(s)

Mg(s) � Mg2+(aq) + 2e-

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 thecorrect answer.

Write balanced equations for the following electrochemical cells:

Q12: Mg � Mg2+(aq) � � Zn2+(aq)� Zn(s)

Q13: Cr(s) � Cr3+(aq) � � Cu+(aq) � Cu(s)

Q14: Pt(s) � H2(g)� H+(aq) � � Cl2(g) � Cl-(aq) � Pt(s)

Q15: Ni(s)� Ni2+(aq)� � H2(g)� H+(aq) � Pt(s)


Example : Writing cell notations

Write the cell notation for a cell containing copper in a solution of copper (II) ionsconnected 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:

Cu(s)� Cu2+(aq)

2. Add the salt bridge: Cu(s) � Cu2+(aq) � �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.



Q16: Write the cell notation for a cell in which zinc metal in zinc(II) sulphate solution isconnected to magnesium metal in magnesium sulphate solution.

Q17: Write the cell notation for a cell in which the following reactions occur:

Ni(s) � Ni2+(aq) + 2e-

Pb2+(aq) + 2e- � Pb(s)

Q18: Write the cell notation for a cell in which copper is oxidised to Cu2+ ions by chlorinegas 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)


8.3 Standard Electrode Potentials�

Æ

�

�

Learning Objective

To outline the use of the standard hydrogen electrode in determining absolute valuesof electrode potentials.

The cell e.m.f. (E) is the electrical potential difference between the two electrodes whenno 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 takingplace.

The individual electrode potentials depend on the position of the equilibrium betweenthe metal and its ions. i.e.

Zn(s) � Zn2+(aq) + 2e-

Cu(s) � Cu2+(aq) + 2e-

The position of these equilibria, as with all equilibria, will depend on the temperatureand 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 madeunder the same conditions. A set of standard conditions have been defined for themeasurement of a cell e.m.f.

Pressure of any gas - 1 atmosphere

Concentration of solutions - 1 mol �-1

Temperature - 298 K (25ÆC)


8.3. STANDARD ELECTRODE POTENTIALS 155

These are the same standard conditions used for the measurements of thermodynamicquantities in Topics 6 and 7 Unit 2.

The absolute value of the electrode potential of a half cell cannot be determinedexperimentally since by its definition redox reactions require two half reactions. It isimpossible to have an oxidation without a reduction. It is only possible to measure theoverall 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 ofzero, then a number of cells can be set up using this standard electrode. The voltagemeasured 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 Kand 1 atmospherepressure.

B A glass tube with holesto allow hydrogen gas toescape.

C Platinum foil coated withplatinum black (very finepowdered platinum).

D A solution containingH+(aq) ions ofconcentration 1 mol�-1

The half-cell reaction is:

2H+(aq) + 2e-� H2(g)

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 copperelectrode, a voltage of +0.34 V is recorded (Figure 8.9). This is measured using a highresistance voltmeter to ensure that only a tiny current is drawn.



Figure 8.9: Copper-hydrogen cell

This voltage can be assigned as the standard electrode potential for the reactionoccurring 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 fromthe hydrogen, through the meter, to the copper.

Q20: Which electrode is the anode?

a) Copper electrodeb) 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 standardconditions, we can say that the standard reduction potential of copper is +0.34 V.By convention, electrode potentials are quoted as standard reduction potentials andare given the symbol, EÆ.



So, Cu2+(aq) + 2e- � Cu(s) EÆ = +0.34 V

If a similar cell is set up using a zinc electrode (Figure 8.10), the electron flow is fromzinc 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 electrodeb) 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.


Summarising: Zn2+(aq) + 2e-� Zn(s) EÆ = -0.76 V

2H+(aq) + 2e-� H2(g) EÆ = 0.00 V

Cu2+(aq) + 2e-� Cu(s) EÆ = +0.34 V



8.3.2 The Electrochemical Series and associated calculations�

Æ

�

�

Learning Objective

To use the Electrochemical Series to estimate the relative strengths of oxidisingagents and reducing agents. To use standard reduction potentials to calculate celle.m.f. and predict the direction of a redox reaction

The standard reduction potentials of a very large number of half-cells have beenmeasured against the standard hydrogen electrode. These values have been organisedinto a list called the Electrochemical Series, part of which is shown on page 11 of theData Booklet.

Usually, the reduction potentials are arranged in numerical order, with the lowest (mostnegative) 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 highnegative 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 theother will go in the opposite direction (the oxidation half). Since an electrode potentialcan 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) �� Cu2+(aq) � Cu(s)

Anode - oxidation:

Zn(s) � Zn2+(aq) + 2e-

Reduction Potential (EÆ) = -0.76 V

Reversing the sign gives +0.76 V � the zinc contribution is +0.76 V.

Cathode - reduction:

Cu2+(aq) + 2e-� Cu(s)

Reduction Potential (EÆ) = +0.34 V

Reaction goes as written � 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)



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, calculatethe e.m.f. in volts, assuming standard conditions (include the sign but not the units).

Q28: Mg(s) �Mg2+(aq) �� Zn2+(aq) � Zn(s)

Q29: Cr(s) � Cr3+(aq)�� Cu2+(aq) � Cu(s)

Q30: Pt(s) � H2(g) � H+(aq)�� Cl2(g) � Cl-(aq) � Pt(s)

Q31: Ni(s)� Ni2+(aq) ��Br2(aq), Br-(aq)� Pt(s)


The table of standard reduction potentials (EÆ values) on page 11 of the Data Bookletcan also be used to estimate the relative strengths of reducing agents and oxidisingagents. It must be emphasised that EÆ values refer to standard conditions. If theconditions are significantly different, then any predictions based on EÆ values may bewrong.

Oxidising agents and reducing agents

20 min

A series of questions on oxidising and reducing agents designed to familiarise studentswith the Electrochemical Series and enable them to make generalisations about therelative strengths of oxidising agents and reducing agents.

Look at the Electrochemical Series on page 11 of the Data Book and use the informationto answer the following questions. Write your answers on paper before displaying theanswer.

Remember that the more positive the value for a standard reduction potential, the morelikely the reaction is to go as a reduction.

Q32: Which of the following reductions is the easiest?

a) Na+(aq) + e- � Na(s)b) F2(g) + 2e- � 2F-(aq)c) Br2(g) + 2e- � 2Br-(aq)d) Cu2+(aq) + 2e- � Cu(s)

Q33: In the Electrochemical Series shown, where will the easiest reduction be found?

a) at the top of the tableb) at the bottom of the table

Q34: What is meant by an oxidising agent?

Q35:

In the reduction,



F2(g) + 2e- � 2F-

is F2 an oxidising agent or a reducing agent?


Remember that the more negative the value for a standard reduction potential, the morelikely the reaction is to go as an oxidation.

Q36: Where in the Electrochemical Series in the data booklet will you find the reactionwhich is most likely to go in the opposite direction?

a) top of the tableb) 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) Top leftb) Top rightc) Bottom leftd) Bottom right


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 lefthand 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.



Figure 8.11: Redox summary

It can be used in the following set of questions which will give you more practice inidentifying and comparing oxidising agents and reducing agents.

If you feel comfortable with this concept, you may only need to do the first two or threequestions. If not, do them all.

Q40: From the grid Figure 8.12, enter the letter of the box containing the best oxidisingagent.

Figure 8.12: Answer grid

Q41: From the grid Figure 8.12, enter the letter of the box containing the best reducingagent.

Q42: From the grid Figure 8.12, enter the letter of the substance that could be used asan oxidising agent or a reducing agent.







It also follows that a half-reaction will be able to force any half-reaction above it in theseries to go in reverse (i.e. as an oxidation). This can be described as the ’AnticlockwiseRule’.

Figure 8.14: The ’anticlockwise’ rule

Any pair of half-reactions combined in this way will produce a positive value for the celle.m.f. Under standard conditions, the reaction will be spontaneous in this direction. EÆ

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. Thestandard free change (�GÆ) has to be negative for a reaction to do this. These twoquantities are connected.

8.4 EÆ values and the standard free energy change�

Æ

�

�

Learning Objective

To define the �GÆ in terms of the cell e.m.f. and use cell e.m.f to calculate �GÆ


8.4. EÆ VALUES AND THE STANDARD FREE ENERGY CHANGE 163

Consider a zinc-copper cell under standard conditions. The cell e.m.f. (EÆ) when nocurrent is drawn is +1.10 V. The positive sign means that the reaction is spontaneous inthe direction:

Zn(s) + Cu2+(aq)� Zn2+(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 availableis going to depend upon the EÆvalue 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 inthe 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 ofelectrons 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 zincreacted in the cell under thermodynamically reversible conditions. For this reaction,

�GÆ = -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 freeenergy for the cell reaction is given by the expression:

Figure 8.15: Relationship between �G and EÆ

Example : Calculation of �GÆ from cell e.m.f.

Calculate �GÆ for the nickel-silver cell.



1) Calculate the cell e.m.f.

Write the two half-equations with their EÆ values.

Ni2+(aq) + 2e-� Ni(s) EÆ = -0.23 V

Ag+(aq) + e-� Ag(s) EÆ = +0.80 V

Using the Anticlockwise rule, the silver half-reaction will be able to force the nickel half-reaction 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.

Oxidation Ni(s) � Ni2+(aq) + 2e-

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 cellreaction, you must not double the EÆ 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.

20 min

A set of questions to give practice at calculating �GÆ from the cell e.m.f.

Calculate the standard free energy change in each of the following questions. Give youranswer 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. Thecell reaction is:

Zn(s) + Ag2O(s) � ZnO(s) + 2Ag(s)


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 anodereaction 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 andhigh voltage.

Q48: Ni(s) � Ni2+(aq) �� Cu2+(aq) � Cu(s)


Calculating standard entropy changes.

20 min

Description of an experiment which enables �HÆ to be calculated. Calculation of �GÆ

allows �SÆ for the reaction to be calculated.

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: mass =0.635 g

4) Final Temperature = 39ÆC

2) Excess silver (I) nitrate:Volume of solution = 25 cm3

5) Blue solution containingCu2+(aq) ions.

3) Initial Temperature = 25ÆC 6) Precipitate of silver metal.

Q49: What is the relationship between the enthalpy change and the temperature rise?



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, �HÆ inkJ mol-1

Q52: The same reaction can be carried out in an electrochemical cell. Calculate thecell e.m.f. under standard conditions, in volts. Remember to include the sign.


8.5 Fuel cells�

Æ

�

�

Learning Objective

To be able to describe similarities and also the main difference between fuel cells andconventional electrochemical cells

Like batteries, fuels use redox reactions to generate electricity. In a battery, the chemicalenergy is stored in the materials that make up the battery. When one of the reactants isused up, the battery stops working and must be recharged or replaced. In a fuel cell, thereactants 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 producingelectricity.

The simplest fuel cells use hydrogen as the fuel but methane, ethanol and even petrolor diesel fuel cells are also possible.

Figure 8.17 shows a diagram of one type of fuel cell that uses hydrogen as the fuel.


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 migratethrough the electrolyte to the cathode where they react with oxygen and electrons onthe platinum electrode. Comparison of the fuel cell (Figure 8.17) with the zinc-coppercell ( 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 signor units)

Q56: Write the overall cell equation.




Table 8.2: Fuel cells for transport

Courtesy of Ballard Power Systems andBreakthrough Technologies Institute, Inc.

Courtesy of Mazda and BreakthroughTechnologies Institute, Inc.

Fuel cells hold a great deal of promise for the future, although many problems still remainto be solved. This type of fuel cell would be the ideal solution for cutting down, if noteliminating, car and bus exhaust emissions (Table 8.2).

If hydrogen is to be used as the fuel, the obvious source would be water. Extractinghydrogen from water is the first problem. Electrolysis, even using renewable sources ofenergy, still remains expensive. Photochemical processes and biological methods arebeing investigated. Storage of the hydrogen also remains a safety problem. Liquefactionof hydrogen is difficult due to the very low boiling point. Absorption onto metals asmetal 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 asmethanol is promising, although CO2

will still be a product. Indeed anumber of car manufacturers aredeveloping cars powered by amethanol fuel cell with performanceapproaching that of petrol-enginedcars.

Courtesy of Opel and BreakthroughTechnologies Institute, Inc.


8.6. SUMMARY 169

Table 8.4: Power generation by fuel cells

Other types of fuel cell, althoughunsuitable for vehicles, are in use forpower generation in hospitals, office

buildings and factories.

Courtesy of Ballard Power Systems andBreakthrough 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 inthe external circuit. The overall cell voltage (e.m.f.) is the sum of the electrodepotentials 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) � � Cu2+(aq) � Cu(s)

Anode Cathode

• Individual electrode potentials cannot be measured but values are obtained bycomparison with the standard hydrogen electrode.

• Standard Reduction Potentials (EÆ 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: �GÆ = -nFEÆ.

• Fuel cells operate like electrochemical cells, the only difference being that the fuelfor the reaction is continually provided from an external source.









Websites:

http://chemistry.about.com/science/chemistry/msub26.htm

http://www.fuelcells.org/




http://chemistry.about.com/science/chemistry/msub26.htm

http://www.fuelcells.org/

171

Topic 9

Kinetics

Contents

9.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172

9.2 Measuring reaction rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172

9.3 Rate equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175

9.3.1 Order of reaction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177

9.4 Reaction mechanisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181

9.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187

9.6 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189

9.7 End of Topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189



• 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


• explain what is meant by the overall order of a chemical reaction and the orderwith 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 chemicalprocess 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 mixedwith 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 theway.

In industry, ’time is money’. It is obviously very important that chemicals are producedas quickly and as efficiently as possible.

In this Topic, your knowledge of the factors which affect the rate of a reaction will beput on a more mathematical basis. In particular, the Topic will investigate the effects ofchanging the concentrations of reactants on the reaction rate. The information obtainedwill be used to gain an insight into reaction mechanisms.

9.2 Measuring reaction rates�

Æ

�

�

Learning Objective

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 decreaseand the concentrations of products will increase.

The rate of a reaction can therefore be defined either as the rate of consumption of areactant or as the rate of formation of a product, i.e.

0� �!��

�� or

0� �� %��!��

�� or

0� � ��

� ��

or

0� � ��

� �� %��!��


9.2. MEASURING REACTION RATES 173

The method used to measure the change in concentration of reactant or productdepends on the reaction being studied.

The most convenient methods involve monitoring a physical quantity which changesduring 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 presentchanges)

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 ofthis 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 ofthe reaction between iodine and propanone.

I2(aq) + CH3COCH3(aq) CH3COCH2I(aq) + H+(aq) + I-(aq)

propanone iodopropanone

Change in concentration with time

15 min

In this activity, experimental data are used to find out what happens to the concentrationsof 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 ofcarbon dioxide produced can also be measured. The following data were obtained froman experiment.



Table 9.2: Concentration and volume against time - sample data

Time / s Concentration of Br2 / mol �-1 Volume of CO2 / cm3

0 12.0 x 10-3 0

30 11.0x 10-3 22

60 10.1 x 10-3 42

90 9.3 x 10-3 59

120 8.6 x 10-3 75

240 6.4 x 10-3 123

360 4.8 x 10-3 158

600 3.7 x 10-3 191

Use the data in Table 9.2 to answer the following questions. Where appropriate, answeron paper before revealing the answer.

Q3: What is the average rate, in cm3 per second, of CO2 production over the first 30seconds?

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 cm3 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 sametime period.


The reaction rate decreases as the reaction proceeds because the concentration ofreactants decreases.

It is difficult to measure precisely the rate of reaction at any given time (the instantaneousrate), since the rate is constantly changing. It is possible to measure the average rateover 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 simplereaction.


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 = t1.

The instantaneous rate at time = t1 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 initialconcentration of a reactant or product is accurately known, and since the reaction isfastest at the start, initial rates of reaction are often used in kinetic studies.

9.3 Rate equations�

Æ

�

�

Learning Objective

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 theconcentration is altered.

The following graphs (Figure 9.2) show the results obtained for the catalyticdecomposition of hydrogen peroxide.

2H2O2(aq) � 2H2O(�) + O2(g)

Only the concentrations of H2O2 were changed. All other variables were kept constant.



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 first10 seconds. However, it is normal to determine initial rates of reaction (the gradient attime = 0 secs) and use these values in subsequent calculations. (Table 9.3)

Table 9.3: Rates of oxygen evolution

Expt [H2O2] / mol �-1 Average rate over10 s / cm3 s-1

Initial rate / cm3 s-1

A 0.40 0.50 0.51B 0.32 0.38 0.40C 0.16 0.20 0.21D 0.08 0.10 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 ofhydrogen peroxide.

If the enzyme catalase is used, the rate of reaction will depend also on the concentrationof catalase, Further experiments can be done to show that the rate of reaction is alsodirectly proportional to the catalase, i.e.

Rate [H2O2] and Rate [catalase]



Combining these gives:

Rate [H2O2] [catalase]

or0� � ) �"�$�� (9.1)

This expression (Equation 9.1) is known as the rate equation and k is known as therate constant.

Learning Point

It is important to note that the rate equation and the rate constant can only bedetermined experimentally. They cannot be obtained from the stoichiometric equation.

9.3.1 Order of reaction�

Æ

�

�

Learning Objective

To be able work out the order of reaction with respect to individual reactants, theoverall 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 theconcentration 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 concentrationsof 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 theorder 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 finalgraph of initial rate against concentration. In simple cases, an inspection of the initialrate data can be enough.

For example, look at Table 9.3 (shown again below) which shows the initial rate data for



the catalytic decomposition of hydrogen peroxide.

Expt [H2O2] / mol �-1 Average rate over10 s / cm3 s-1

Initial rate / cm3 s-1

A 0.40 0.50 0.51B 0.32 0.38 0.40C 0.16 0.20 0.21D 0.08 0.10 0.11

Compare experiments B and C. The initial concentration used in experiment B is twicethat used in C. You will notice that the initial rate in experiment B is twice that of C, withinexperimental error.

Similarly, the initial concentration in C is twice that in D and again the initial rate in C istwice that in D.

The rate is directly proportional to [H2O2] and the reaction is first order with respect toH2O2.

Example : Rate equations and constants

Consider the reaction between nitrogen monoxide (nitric oxide) and hydrogen.

2NO(g) + 2H2(g) � N2(g) + 2H2O(g)

Use the following data to work out the rate equation and calculate the rate constant. Allconcentrations are initial concentrations.

Experiment [NO] / mol �-1 [H2] / mol �-1 Initial rate /

mol �-1 s-1

1 1 x 10-3 1 x 10-3 6.0 x 10-5

2 1 x 10-3 2 x 10-3 1.2 x 10-4

3 2 x 10-3 1 x 10-3 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 [H2] has been doubled. The initial rate has alsobeen doubled.

Rate [H2] i.e. n = 1

So the reaction is first order with respect to H2.

Compare experiments 1 and 3. The [NO] has been doubled. In this case, the initial ratehas 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 22 = 4. The reaction is therefore secondorder with respect to NO, and m = 2. The rate equation is therefore:



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 evaluatek. 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 rateequation. To work out the units of k for the previous example (a third order reaction):

�� ) � ��

� ) � ��

�� ) ��

��

� ��

So k = 6.0 x 104 mol-2 l2 s-1

The units of k depend on the overall order of the reaction.

Rate equation Overall order Units of k

Rate = k[A]0 0 mol �-1 s-1

Rate = k[A]1 1 s-1

Rate = k[A]2 2 mol-1 � s-1

Rate = k[A][B] 2 mol-1 � s-1

Rate = k[A]2[B] 3 mol-2 �2 s-1

It is important to point out again that the rate equation can only be derivedexperimentally. It cannot be obtained from the balancing numbers in the balancedequation. In the reaction in the previous example, two moles of NO react with twomoles of H2. Equal numbers of moles react but the experimentally determined order ofreaction with respect to each reactant is different.

Orders and rate constants

20 minA series of questions to give practice at working out orders of reaction, rate equationsand calculating rate constants.

For any questions which are not marked by the computer, answer first on paper beforerevealing the correct answer.

The next two questions refer to the following reaction:

Bromide ions are oxidised by bromate ions (BrO3-) in acidic solution according to the

equation:



5Br-(aq) + BrO3-(aq) + 6H+(aq) � 3Br2(aq) + 3H2O( � )

By experiment, the reaction is found to be first order with respect to both bromide andbromate 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, N2O5:

2N2O5(g) � 4NO2(g) + O2(g)

Experiments were carried out in which the initial concentration was changed and theinitial rate of reaction was measured. The following data were obtained.

[N2O5] / mol �-1 Initial Rate / mol �-1 s-1

0.05 2.2 x 10-5

0.10 4.4 x 10-5

0.20 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 decimalform, 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 inmol �-1 s-1. (Again, answer in normal decimal form, and do not include units.)


9.4. REACTION MECHANISMS 181

The next three questions refer to the following reaction:

Iodide ions are oxidised in acidic solution to triiodide ions, I3-, by hydrogen peroxide.

H2O2(aq) + 3I-(aq) + 2H+(aq) � I3-(aq) + 2H2O( � )

The following initial rate data were obtained:

Experiment Initial Concentrations / mol �-1 Initial Rate /mol �-1 s-1

[H2O2] [I-] [H+]

1 0.02 0.02 0.001 9.2 x 10-6

2 0.04 0.02 0.001 1.84 x 10-5

3 0.02 0.04 0.001 1.84 x 10-5

4 0.02 0.02 0.002 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�

Æ

�

�

Learning Objective

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 wereaffected by changes in concentration, temperature and surface area (if one of thereactants 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 sufficientenergy and the correct orientation. A transition state forms which breaks down to givethe products.



Figure 9.3: Formation of a transition state

If there is insufficient energy or the wrong orientation, the molecules simply bounceapart.

This is a simple, one-step reaction. If the concentration of either reactant is doubled, thenumber of successful collisions will be doubled. In other words, the reaction will be firstorder 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 processcan the stoichiometry be used to work out the rate equation.

Most reactions are not so simple, e.g.

2NO(s) + 2H2(g) � N2(g) + 2H2O(g)

The balanced equation shows two molecules of NO reacting with two moles of H2.

If this were to occur by a simple one-step reaction, then two molecules of NO wouldhave to collide simultaneously with two molecules of H2. This simultaneous collision offour 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 bya series of simple steps, each involving one or two molecules or ions. Such a series ofsimple reaction steps is known as a reaction mechanism. The sum of all the individualsteps must equal the stoichiometric (balanced) equation. The reaction mechanism mustalso fit the experimentally determined rate equation.

This poses a problem. Do we need to know how fast each step is in order to work outthe overall rate?

As an analogy, consider this production line in the bottling plant in a distillery.



Figure 9.4: A whisky production line

There are three parts and the maximum capacity of each is:

Part 1 Filler 2 bottles filled per minute.

Part 2 Capper 120 bottles capped per minute.

Part 3 Labeller 60 bottles labelled per minute.

The production line is switched on. After 30 seconds a bottle has been filled and ispassed to the Capper which caps it in 0.5 seconds and passes it to the Labeller whichtakes 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 isgoverned 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 thanthe others and this step will determine the overall reaction rate, i.e. it will bethe 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.



A possible mechanism is as follows:

2NO2 � NO3 + NO slow step

NO3 + CO � NO2 + CO2 very fast step

The first step very slowly produces the intermediate, NO3. As soon as it is formed, step2 occurs. The NO3 reacts with a CO molecule to form the product, CO2. The rate offormation of CO2 does not depend on the rate of reaction between the NO3 and CO.Instead it depends on the rate of formation of NO3. 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 NO2. The reaction issecond order with respect to NO2 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 themechanism. 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 RDSand only gets involved in a fast step which takes place after the RDS. Changes in theconcentration of CO will not affect the reaction rate.

It must be emphasised that this reaction mechanism is a suggestion and there may beother possible mechanisms which also fit the rate equation. If further evidence could befound for the existence of the intermediate, NO3, this would strengthen the case for thismechanism.

We can now explain the apparent discrepancy between the balancing numbers in thestoichiometric equation and the orders with respect to the individual reactants. Theorder 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 notfeature in the RDS.

It is also possible for a substance to appear in the rate equation but not in thestoichiometric equation, e.g. a catalyst is not a reactant but must feature in the rate-determining 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 (CH3COCH3) and alsofirst order with respect to H+ ions which act as a catalyst.

To find out the order with respect to iodine, try the PPA.



PPA - Kinetics of the acid-catalysed propanone / iodine reaction.

Consult with your tutor whether the practical work for this PPA should be be carried outat 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 ionswhich catalyse the reaction. The order with respect to iodine can be determinedby taking a reaction mixture in which the initial concentration of propanone andhydrogen ions are very much larger than that of iodine. With such conditions, onlythe concentration of iodine will vary significantly during the reaction and this will allowus 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 andanalysing them for iodine.

This simulated practical is not like a normal simulation, where you press thebutton and the work is done for you, in this simulation you must prepare thesolutions, 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 programto 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 goodidea to have a look at them all before you start. As you might expect, some things laterin the experiment won’t work unless you’ve actually run the experiment.

If you get stuck, use the button at the top right of each screen.



Questions on reaction mechanisms

20 minA number of questions to give practice at interpreting rate data and relating these toreaction mechanisms.

For any questions which are not marked by the computer, answer first on paper beforerevealing the correct answer.

Q29: Which of the following reactions is most likely to occur by a simple one-stepprocess?

a) 4HBr + O2 � 2H2O + 2Br2

b) H2S + Cl2 � S + 2HClc) 2NO + O2 � 2NO2

d) 2H2 + O2 � 2H2O

Q30: Which of the following reactions is least likely to occur by a simple one-stepprocess?

a) 4HBr + O2 � 2H2O + 2Br2

b) H2S + Cl2 � S + 2HClc) 2NO + O2 � 2NO2

d) 2H2 + O2 � 2H2O

The next six questions refer to the reaction between propanone and bromine in alkalinesolution.

The balanced equation is:

CH3COCH3(aq) + Br2(aq) + OH-(aq) � CH3COCH2Br(aq) + H2O( � ) + Br-(aq)

The experimentally determined rate equation is:

Rate = k [CH3COCH3][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) Trueb) False

Q32: The reaction involves a simple one-step process.

a) Trueb) False

Q33: The reaction is second order overall.

a) Trueb) False

Q34: The rate determining step involves one molecule of propanone and one moleculeof bromine.


9.5. SUMMARY 187

a) Trueb) False


The next five questions refer to a reaction involving hydrogen peroxide and bromide ionsin aqueous solution.

H2O2 + Br- � BrO- + H2O Step 1

H2O2 + BrO- � Br- + H2O + 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) Rate = k [H2O2]b) Rate = k [H2O2][Br-]c) Rate = k [H2O2]2

d) Rate = k [H2O2]2 [Br-]

Most reactions occur by a series of simple steps - a reaction mechanism. Therate determining step (RDS) is the slowest step in the mechanism. Experimentallydetermined rate equations give information about the RDS and can be used to suggestpossible 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 ofone 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 �-1.

• 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



• The order of the reaction with respect to A is n and the order with respect to B ism.

• The overall order of reaction is n + m.

• The rate equation, and hence value for k, is always derived experimentally, usuallyfrom 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 ratedetermining step.

• Rate equations can provide evidence for a proposed reaction mechanism butcannot provide proof as other possible reaction mechanisms may also give thesame rate equation.


9.6. RESOURCES 189






• A-level Chemistry: E.N. Ramsden, Stanley Thorne Publishers, ISBN 0-85950-154-X




191

Topic 10

End of Unit 2 Test (NAB)

Contents

192 TOPIC 10. END OF UNIT 2 TEST (NAB)



GLOSSARY 193

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 inother 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 andproducts.

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 amountsof acid or base are added

cathode

The electrode at which reduction occurs

chromatography

Chromatography is an analytical method where mixtures are separated intotheir components by partitioning between a stationary and mobile phase. Thestationary/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 processesbecome equal, so that no net change results

electrode potential

The difference in electric potential between the electrode (metal) and the ions insolution


194 GLOSSARY

electromotive force

The electric potential difference between the electrodes in a cell when no currentis 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 completelyhydrated, 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 inwater

entropy

The entropy of a system is the degree of disorder of the system. The greater thedisorder, the greater the entropy. Low entropy is associated with strongly orderedsubstances

equivalence point

the equivalence point in a titration experiment is reached when the reactionbetween the titrant (added from the burette) and the titrate (in the flask) is justcomplete.

equivalence point.

the equivalence point in a titration experiment is reached when the reactionbetween the titrant (added from the burette) and the titrate (in the flask) is justcomplete.

excess

Excess reactants are present in greater than stoichiometric amounts and will notbe 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 reactionenthalpies of each step of the reaction.

heterogeneous reaction

Heterogeneous reactions have reactants and/or products in more than one phase


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 ina subsequent step

limiting

Limiting reactants are present in smallest stoichiometric amounts and will becompletely used to produce products whose amounts will be determined by theinitial quantity of limiting reactant.

mean molar bond enthalpy

An average value that is quoted for a bond that can occur in different molecularenvironments

molar bond enthalpy

The molar bond enthalpy for a diatomic molecule X-Y is the energy required tobreak 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 rateequation

overall order of a reaction

The sum of the powers to which the concentrations of all reactants in the rateequation are raised in the rate equation

partition coefficient

The ratio of the concentrations of a solute in two immiscible liquids is called thepartition coefficient

primary standard

A primary standard is a stable, soluble solid which can be accurately weighed anddissolved 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 toratios 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 particulartemperature


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 eachreactant

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-liquidchromatographic column after the injection time

second Law of Thermodynamics

The total entropy of a reaction system and its surroundings always increases for aspontaneous change

stable equilibrium

A stable equilibrium state is one which is regained after a small disturbance fromthis state has occurred

standard conditions

Conditions at a pressure of one atmosphere and a specific temperature (298K)(25ÆC).

standard enthalpy change

The enthalpy change for a reaction in which reactants and products are consideredto 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 standardenthalpy and entropy changes by


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 fromthe 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 inoxygen under standard conditions.

standard molar enthalpy of formation

The enthalpy change that occurs when one mole of a substance is produced fromits elements in their standard states


GLOSSARY 197

standard reduction potential

The electrode potential for a reduction half-cell measured under standardconditions (relative to the standard hydrogen electrode)

standard solution

A standard solution is one with an accurately known concentration. It can beprepared by weighing a primary standard and dissolving it in a known volume ofsolution, 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 exactwhole numbers of particles (atoms, ions or molecules) in fixed, characteristicratios.

third Law of Thermodynamics

The Third Law of Thermodynamics states that the entropy of a perfect crystal at 0K 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 areaction


198 FURTHER QUESTIONS

Further questions

Topic 1: Stoichiometry

Questions continued from page 8.

Q61: Ethanoic acid is prepared industrially by reacting liquid methanol with carbonmonoxide gas.

CH3OH(�) + CO(g) � CH3CO2H(�)

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 thereaction above?


Q63: It must be coloured.

a) Trueb) False

Q64: It must have a low boiling point.

a) Trueb) False

Q65: It must be a liquid.

a) Trueb) False

Q66: It must have a relatively high formula mass.

a) Trueb) False


Q67: What volume (in ml to 2 decimal places) of 2 mol � -1 sodium carbonate would berequired to neutralise 1 ml of concentrated ’syrupy’ phosphoric acid (15 mol �-1)?


Q68: When sodium carbonate solution is added to aqueous copper(II) ions, theprecipitate in fact consists of a mixture of copper(II) carbonate and hydroxide. Wouldthis affect the analysis above? Give an explanation for your answer.


FURTHER QUESTIONS 199


Q69: When plaster of Paris [(CaSO4)2.H2O] sets after it has been mixed with water, itforms gypsum (CaSO4.2H2O). How many grams of water (to 1 decimal place) shouldbe added to 580 g of plaster of Paris to ensure that it is converted completely to gypsumwith 27 g of water remaining?

Q70: There are only 5 g of copper(II) sulphate (CuSO4.5H2O) left in a bottle. Whatvolume of 0.2 mol �-1 solution could be made from this weight? Give your answer to thenearest unit. ml

Q71: The insecticide DDT is known to increase in concentration in living tissues asit moves along a food chain. Its concentration in the water of Lake Michigan is 2 x10-5 mg �-1. If it increases in concentration 300,000 times in fish tissues, estimate theconcentration (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 theincrease in concentration in going from fish to these birds?

Q73: An indigestion remedy (antacid) contains an aqueous suspension of magnesiumhydroxide (Mg(OH)2). This suspension was diluted 50 times and 10.0 ml of this reactedcompletely with 24.0 ml of 0.15 mol �-1hydrochloric acid. What is the concentration ofmagnesium hydroxide in the original suspension in mol � -1, to 1 decimal place?

Q74: What is the concentration of the magnesium hydroxide in g � -1? Give your answerto 1 decimal place.

Q75: Alcohol levels in blood can be determined by a redox titration with potassiumdichromate according to the balanced equation:

C2H5OH(aq) + 2Cr2O72-(aq) + 16H+(aq) � 2CO2(g) + 4Cr3+(aq) + 11H2O(l)

What is the blood alcohol level in mol �-1 (to 3 decimal places) if 8.76 ml of 0.050 mol �-1

K2Cr2O7 is required for titration of a 10.026 ml sample of blood?

Q76: It is more usual to express blood alcohol levels in mg alcohol / 100 ml blood. Whatis the alcohol level in the question above in these units? Express your answer to 3significant figures.

Q77: Titration with solutions of potassium bromate (KBrO3) can be used to determinethe concentration of As(III). What is the molar concentration of As(III) in a solution if22.35 ml of 0.100 mol �-1 KBrO3 is needed to titrate 50.00 ml of the As(III) solution? Thebalanced equation is:

3H3AsO3 + BrO3- � Br- + 3H3AsO4

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+ reactedcompletely with 37.5 ml of 0.20 mol �-1 ethylene diamine. How many ethylene diaminemolecules 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 of0.0233 mol � -1 EDTA. Calculate the percentage (to 3 decimal places) of zinc in thissample.



Q80: 17.870 g of chlorophyll were heated in an oxygen atmosphere to destroy theorganic part and leave a residue of magnesium oxide. If this residue weighed 0.806g, what is the percentage of magnesium in chlorophyll? Give your answer to 2 decimalplaces.

Q81: The formula mass of chlorophyll is 893.5. How many atoms of magnesium arethere in each molecule? Enter your answer as a number.

Q82: Gold has compounds containing gold(I) ion or gold(III) ion. A compound of goldand chlorine was treated with a solution of silver nitrate, AgNO3, to convert the chlorideion in the compound to a precipitate of AgCl. A 162.7 mg sample of the gold compoundgave 100.3 mg AgCl. Calculate the percentage of chlorine in the gold compound, to 2decimal places.

Q83: Decide whether the formula is AuCl or AuCl3.

Topic 2: Chemical Equilibrium


Q34: Would the process to manufacture SO3 be more productive at 636ÆC or 856ÆC?


Q35: Around lightning, the temperature of the air is raised to approximately 2400 ÆC.At this temperature, Kp for the previous reaction is increased to 2.44 x 10-3. What isthe equilibrium partial pressure of NO around a lightning strike? Assume that the initialpartial pressures of nitrogen and oxygen are 0.78 and 0.21 respectively.


Q36: 3Fe(s) + 4H2O(g)� Fe3O4(s) + 4H2(g)


Q37: 2NO(g) + Cl2(g) � 2NOCl(g)


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 thishad on the partition coefficient?



a) increases the partition coefficientb) decreases the partition coefficientc) no effect


Q27: Both the blue and dark blue spots from both the original inks have moved similardistances. What might you conclude from this?

Topic 4: Acid/base Equilibria


Q52: What type of bond is formed?

Q53: What type of substance is ammonium chloride?

a) acidb) basec) 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-�?


Q57: What is the pH of a solution of sulphurous acid of molar concentration0.05 mol � -1?

Q58: What is the pH of a solution of carbonic acid of molar concentration0.001 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


Q24: Which of the following expressions represents the acid dissociation constant ofHIn in Figure 5.1.

a)

� ��"&�� "�$�

�"�$�� &��



b)

� ��"&��

�"�$�� &��

c)

� ��"�$

�� &��

�"&�� "�$�

d)

� ��"�$

�� &��

�"&��

pH titration (page 81)

Q25: Which of these statements is true?

a) The alkali is more concentrated than the acid.b) The pH rises rapidly at the beginning.c) The alkali is less concentrated than the acid.d) The pH changes rapidly only around the equivalence point.


Q26: Using the graphs in Table 5.2 and your previous answers, write a generalstatement about the pH at the equivalence point in an acid/ alkali titration. Then displaythe answer.

Buffer calculations (page 93)

Q27: To prepare 1 litre of a buffer solution which would maintain a pH 5.5, 0.6g ofethanoic acid was used. What mass in grams of sodium ethanoate should the solutioncontain? Answer to one decimal place.

Q28: One of the systems which maintains the pH of blood at 7.40 involves the acidH2PO4

- and the salt containing HPO42-. Calculate the ratio of salt to acid in blood (Ka

and pKa of H2PO4- 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 isnegative, 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-lineexample before displaying the answer to check your working.

Calculate the enthalpy change involved in this reaction:

2H2(g) + O2(g) � 2H2O(g)



Hess’s Law questions (page 105)

Q43: Solid magnesium chloride exists in two forms: anhydrous (MgCl2) and hydrated(MgCl2.6H2O).

Calculate the enthalpy change for converting anhydrous magnesium chloride tohydrated magnesium chloride, given the following enthalpies of formation:

�HÆ

f (hydrated magnesium chloride) = -2500 kJ mol-1

�HÆ

f (anhydrous magnesium chloride) = -642 kJ mol-1

�HÆ

f (water) = -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 enthalpyof combustion of ethanol? (again: remember the sign).

Q45: The difference could be due to heat losses or the non-standard conditions, butassuming the conditions to be standard, calculate the percentage efficiency of thisexperiment.

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) � 2NH3(g)

Q39: Zn(s) + Cu2+(aq) � 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 pointof any liquid, the gas and liquid are in equilibrium. Use this information to calculate aboiling point for chloroform.

CHCl3(l) � CHCl3(g) �HÆ = 31.4 kJ mol-1

�SÆ = 94.2 J K-1 mol-1



Interpreting Ellingham diagrams (page 140)

Q41: The data booklet gives the melting point of zinc as approximately 700 K. Whathappens to the entropy and what effect does it have on the gradient of the graph? (pointA)

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? Giveexamples of suitable substances.

Q58: Why does the current stop flowing?


Q59: Al(s) � Al3+(aq) � � Pb2+(aq)� Pb(s)


Q60: Write the cell notation for a cell in which this cell reaction occurs:

Fe2+(aq) + Ag+(aq) � Fe3+(aq) + Ag(s)


Q61: What will be the value of the standard reduction potential (EÆ) of zinc? (Rememberto show the sign and units).


Q62: Al(s) � Al3+(aq) �� Pb2+(aq) � Pb(s)

Q63: Pt(s) � Sn2+(aq), Sn4+(aq) �� Fe3+(aq), Fe2+(aq) � Pt(s)

Oxidising agents and reducing agents (page 159)

Q64: Where in the table will the best oxidising agents be found?

a) Top leftb) Top rightc) Bottom leftd) Bottom right

Q65: Draw a diagram to summarise this information.




Q66: From the grid Figure 8.13, enter the letter of the substance that could be used asan oxidising agent or a reducing agent.




Calculation of standard free energy change. (page 164)

Q69: Al(s) + 3/2 I2(s) � Al3+(aq) + 3 I-(aq)

Q70: Fe(s) � Fe2+(aq) �� Ag+(aq) � Ag(s)

Calculating standard entropy changes. (page 165)

Q71: Use this answer to calculate the standard free energy change, �GÆ, in kJ mol-1 toone decimal place.

Q72: What is the relationship between free energy, enthalpy and entropy?

Q73: Using this relationship and your answers above, calculate the standard entropychange at 298K (25ÆC), in J K-1 mol-1.

Q74: By considering the overall redox equation, comment on the sign of �SÆ.


Q75: Apart from generating electricity, what advantage would the use of such a cellhave 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?



Questions on reaction mechanisms (page 186)

Q40: The following mechanism fits the rate equation.

a) Trueb) False

Q41: The following mechanism fits the rate equation.

a) Trueb) False

Topic 10: End of Unit 2 Test (NAB)


ANSWERS: TOPIC 1 207

Answers to questions and activities

1 Stoichiometry

Answers from page 3.

Q1: CaF2

Q2: Fe(NO3)3

Q3: Ba3(PO4)2

Q4: K2Cr2O7


Q5: Calcium sulphate

Q6: Copper(II) chloride

Q7: Potassium permanganate or potassium manganate(VII)

Q8: Phosphorus(V) oxide


Q9: 56.0

Q10: 342.3

Q11: 152.0

Q12: 58.0


Q13: c) statements 1, 2 and 3.

Q14: c) statement 3


Q15: 11.0

Q16: 300.0

Q17: 21.17

Q18: 5.3


208 ANSWERS: TOPIC 1


Q19: 6.0

Q20: 3.0

Q21: 2.0

Q22: 9.0


Q23: 1.02

Q24: 0.75

Q25: ethanol

Q26: 44.0


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 FeCl3.6H2O 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 � -1

solution.

Q30: 12.41


Q31: 270.3

Q32: 1.6

Q33: 20.0

Q34: 260




Q35: a) True

Q36: a) True

Q37: a) True

Q38: b) False


Q39: 0.3125

Q40: 0.017

Q41:

Q42: 22.5


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 orginalperoxide solution.

• Molar concentration is 1.7056 mol �-1

• The molar mass of H2O2 is 34, so multiplying this by the molar concentration gives58.0 g �-1. The hydrogen peroxide concentration is therefore 5.8 g/100ml (or 5.8%).




Q45: First, use the thiosulphate titration information to calculate the number of moles ofiodine released.

• The equation for reduction of iodine by thiosulphate is:

• I2 + 2S2O32- � 2I- + S4O6

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 I2) = 25.38 mg

Q46:

• From the equation for ozone and iodide calculate the weight of ozone given theweight of iodine produced.

• 1 mol of ozone produces 1 mol of iodine, so 0.0001 mol of ozone will produce0.0001 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


Q48: 0.024

Q49: 92

Q50: 80


Q51: 0.0025

Q52: 0.0016


Q53: 11.33

Q54: 0.50



Q55:

Al2(SO4)3 + 6OH- 2Al(OH)3 + 3SO42-

2Al(OH)3 Al2O3 + 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 GIVEN

GIVEN REQUIRED

Al2O3 Al2(SO4)3

3 x 16.0 = 48.02 x 27.0 = 54.0

12 x 16.0 = 192.0 3 x 32.1 = 96.3 2 x 27.0 = 54.01 mol = 102.0 g

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


Q57: 151.7

Q58: 18.0

Q59: 9.0

Q60: 3.74

Further answers


Q61: carbon monoxide

Q62: 21.43


Q63: b) False



Q64: b) False

Q65: b) False

Q66: a) True


Q67: 11.25


Q68: It would not affect the result because the precipitated hydroxide would alsoproduce oxide (like the carbonate) when heated.


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.



2 Chemical Equilibrium


Q1: At t = 15 there are 5, 5 and 2, respectively. At t = 30 and t = 70 there is no changewith four of each.


Q2: There are four of each at both times.

Q3: They are the same.


Q4:

Kc =[Fe2 + ]2 [I -

3 ]

[Fe3 + ]2 [I - ]3

Q5:

Kc =[H + ]2 [HPO2 -

4 ][H3PO4]


Q6: c) hydrogen iodide


Q7: a) phosphorus(V) chloride

Q8: b) Mg

Q9: d)

2SO2(g) + O2(g) � 2SO3(g) Kc at 636ÆC = 3343

Q10: The value increases from 21 to 3343 as the temperature drops.


Q11:

Kp =p2

NOpCl2

p2NOCl

Q12:

Kp =p�SO3

pO22p2SO2




Q13: c)

Kc =[Zn2 + ]

�Cu��

Q14: c)Kp = p2

H2OpN2O


Q15: 0.0035

Q16: 0.050


Q17:

x = 0.094 or 0.229 but only 0.094 is a sensible solution.

� [H2] = 0.1 - x = 0.006 mol � -1

[I2] = 0.2 - x = 0.106 mol � -1

[HI] = 2x = 0.188 mol � -1

Q18:

x = 0.01

� [N2O4] = 0.05 - x = 0.04 mol � -1

[NO2] = 2x = 0.02 mol � -1

Q19:

x = -0.0092

This is a negative number, which means that in reaching equilibrium, the backwardreaction occurs, increasing the concentration of N2O4.

so [N2O4] = 0.02 - x = 0.0292

[NO2] = 0.03 + 2x = 0.0116

Q20: 2.76 x 10-16


Q21: 0.059

Q22: 0.470




Q23: b) right to left

Q24: c) no change

Q25: a) left to right

Q26: b) right to left


Q27: absorb

Q28: right

Q29:

Q30: increase


Q31: Endothermic, because energy always has to be supplied to break bonds.

Q32: a) increase


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 thetemperature); addition of SO2(adding reactant will push equilibrium towards products);addition of oxygen.

Further answers


Q34: 636ÆC


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




Q36: c) no change

Q37: a) left to right



3 Phase Equilibria


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


Q6: temperature

Q7: yes

Q8: 1

Q9: yes


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 realitymillions of particles are involved. Reaching true 100% extraction is very difficult.


Q12: Partition coefficient = 3.0. This comes about since the concentration of acid in theether 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 cm3 there would be 3 g of soluteand in the second batch 1.2 g giving a total of 4.2 g.


Q14: non-polar

Q15: non-polar

Q16: decaffeinated


Q17: b) black



Q18: e) green

Q19: a) has the highest solvent/water partition coefficient

Q20: c) dark blue


Q21: b) pentane

Q22: a) r.t. increases as m increases

Q23: c) xylene(C8H10)

Further answers


Q24: 4

Q25: yes

Q26: c) no effect


Q27: They are probably the same dyes in both cases since the Rf values would bethe same. If they were different materials, they would probably have moved differentdistances.



4 Acid/base Equilibria


Q1: ammonia

Q2: hydrogen chloride

Q3: ammonium ion

Q4: chloride ion


Q5: 68


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.


Q9: b) a base

Q10: b) the conjugate base of H2O

Q11: b) The equilibrium lies well to the left.

Q12:

� ��"�$

�� $"��

�"�$��


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. So the forwardreaction must be endothermic and therefore �HÆ is positive.




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


Q28:

� ��"�$

��

�"�� "�$�


Q29: b) HIO3

Q30: a) HF




Q31: b) IO3-

Q32: a) F-

Q33:

The higher the value of Ka the stronger the acid.

Strong acids will have Ka values very much greater than 1. (These are not normallyquoted.)

Weak acids have Ka values less than 1.


Q34: c) high Ka, low pKa

Q35: sulphurous acid (pKa 1.8)

Q36: d) low Ka, high pKa

Q37: the hydrogen phosphate ion, HPO42- (pKa 12.7)


Q38: From page 12 of the data booklet,

%�� and � � ��

� ��

%" ��

�%��

��

��

��

��

��

�

� ��

��

� ��

Q39: 5.4

Q40: 0.025

Q41: From the data booklet,%��



%" ��

�%��

��

rearranging �

��

�

�%�� %"

doubling �� %�� %"

� ��

� � ��

� � antilog ��

� ��

Q42: 2.4

Q43: 3.3


Q44: ammonium ion

Q45: 9.3

Q46:

�� 1"�� "

��1"�

�

�or

�� 1"�� "�$

��1"�

�

�Q47: a) weak


Q48: 8.2

Q49: piperidine

Q50: codeine

Q51:

The more methyl groups the stronger the base becomes.

The Ka value decreases (and pKa increases), meaning that the conjugate acid is gettingweaker. Hence the base is getting stronger.

Further answers


Q52: A dative covalent bond (or coordinate bond) is formed since both electrons in thebond originate from the same atom.

Q53: c) salt




Q54: �OH-� is 10-13.

Q55: 1 x 10 -14.

Q56: 1 x 10 -14.


Q57: 1.6

Q58: 4.7

Q59: 0.016

Q60: 0.10



5 Indicators and buffers


Q1: a) left

Q2: red

Q3: b) right

Q4: blue


Q5: 0.0000001


Q6: 100

Q7: blue

Q8: 1

Q9: green


Q10: 7

Q11: At the equivalence point, the exact amount of alkali has been added to neutralisethe acid; no more, no less.

Q12: 0.1

Q13: 3


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 whichphenolphthalein changes colour. So there will be a sharp endpoint. Both the other



indicators will change colour gradually. For methyl orange, the colour change takesplace long before the equivalence point.

Q19: For both methyl orange and bromothymol blue, the pH of the solution is changingrapidly over the indicators pH range. So there will be a sharp endpoint even althoughthe equivalence point falls in neither range.


Q20: 5.47

Q21: 1.58

Q22: 1.79 x 10-5

Q23: 4.6

Further answers


Q24: d)

� ��"�$

�� &��

�"&��


Q25: d) The pH changes rapidly only around the equivalence point.


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


Q27: 4.1

Q28: 1.58



6 Thermochemistry


Q1: -894

Q2: Activation Energy

Q3: 137

Q4: The reaction shown in Figure 6.2 has a lower reaction vessel temperaturebecause 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 energyto break than a double bond, as in oxygen.


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



Q17:

You might like to compare this value with the data booklet value and suggest a reasonfor the difference (refer to Figure 6.5 for a hint).


Q18:

�H(2) = �H(1) + �H(3)

�H(3) = �H(2) - �H(1)

�H(3) = +33.2 - 90.2 kJ mol-1

�H(3) = -57 kJ mol-1

Q19: -849

Q20:

By using the algebraic method, you should reach this point:-



�HÆ

f (methane) = �HÆ

c (carbon) + 2�HÆ

c (hydrogen) + ( - �HÆ

c (methane) )

�HÆ

f (methane) = - 75 kJ mol-1

Q21: a) Using �HÆ = ��HÆ


f(reactants)

The reaction: B2H6(g) + 3O2(g) � B2O3(s) + 3H2O(l)

Equation B2H6 + 3O2� B2O3 +3H2O

given �Hf (kJmol-1)

+41.0 0 -1225.0 -286.0

Multiply bymoles present

+41.0 0 -1225.0 -858.0

�HÆ = (-1225.0 - 858.0) - (+41.0 + 0) kJ

= -2083.0 - 41.0�H Æ = - 2124.0 kJ

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 mostenergy. (Try working out whether one gram of each would still leave diborane as thebetter fuel in terms of heat released per g of fuel.)


Q22: 10.30

Q23: 5.74

Q24: 59.12

Q25: -1359.76


Q26:

�HÆ

LATT= �HÆ

f - (�HÆ

AT + �HÆ

I.E. + 1/2 �HÆ

BOND + �HÆ

E.A.)

�HÆ

LATT=

-411 - ( +109 + 502 + (1/2 x 243) + (-348.7))

�HÆ

LATT=

-411 - (+383.8)

�HÆ

LATT=

-794.8 kJ mol-1



A Born-Haber cycle for lithium fluoride (page 111)

Q27:

�HÆ

f=

�HÆ

AT + �HÆ

I.E. + 1/2 �HÆ

BOND + �HÆ

E.A. + �HÆ

LATT

�HÆ

f=

+159 + 526 + (1/2 x 155) + (-3282) + (-1030)

�HÆ

f = +762.5 - 1358.2

�HÆ

f=

- 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 havingelectrons added.

Q31: -170.4


Q32: There are two values, -418 and -338 kJ because there are two types of ion beinghydrated.

Q33: +13

Q34: b) endothermic

Q35: a) go down


Q36: exothermic

Q37: hydration

Enthalpy of solution (page 115)

Q38: +16

Q39: b) endothermic

Q40: a) go down



Further answers


Q41: Since:

BOND BREAKING = +675 kJBOND MAKING = -856 kJOVERALL ENTHALPY CHANGE = -181 kJ

Q42:

BOND BREAKING (ALWAYS POSITIVE) BOND MAKING (ALWAYS NEGATIVE)

2 mole of H-H bonds =864 4 moles of H-Cl bonds =4x458

1 mole of O=O bonds =497ENERGY IN =+1361 ENERGY OUT =-1832

OVERALL ENERGY CHANGE = - 471 kJ


Q43: a) Using �HÆ = ��HÆ


f(reactants)

The reaction: MgCl2(s) + 6H2O(l) � MgCl2.6H2O(s)

Equation MgCl2(s) + 6H2O(l) � MgCl2.6H2O(s)

given �Hf (kJmol-1)

-642 -286 -2500

Multiply bymoles present

-642 -1716 -2500

�HÆ = ��HÆ


f(reactants)

�HÆ = (-2500) - (-642 + (-1716)) kJ

= (-2500 + 2358)

�H Æ = - 142 kJ


Q44: -1367

Q45: 99.47



7 Reaction Feasibility


Q1: 0

Q2: melting

Q3: positive

Q4: increase


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 = 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 3moles of gas reactants become two moles of products.

Q14:


- 99.5 = 2 x SÆAMMONIA - (191.6 + 3 x 130.7)

2 x S Æ

AMMONIA = - 99.5 + 583.7

S Æ

AMMONIA = + 242.1 J K-1mol-1

Q15: �SÆ = - 313.5 J K-1 mol-1

Q16: -156.3



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 25ÆC (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 25ÆC (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.


Q22: a) 2Mg(s) + CO2(g) � 2MgO(s) + C(s)

�-� � �-�!��"�� -��"��

� ��

� � �� )* ��

The reaction where magnesium reduces carbon dioxide is feasible under standardconditions.

b) 2CuO(s) + C(s) � 2Cu(s) + CO2(g)



�-� � �-�!��"�� -��"��

� ��

� � �� )* ��

The reaction where carbon reduces copper(II) oxide is feasible under standardconditions.

Q23:

�HÆ = +117 kJ mol-1

�SÆ = +175 J K-1 mol-1

Since �GÆ = �HÆ - T�SÆ

At 400 K �GÆ

400 = +47 kJ mol-1

At 1000 K �GÆ

1000 = -58 kJ mol-1

This reaction is feasible only at higher temperatures.

Q24:�"� � � �� * ��

�� * ��

�� -� � �

+ ��"�

��

+ � ��

Q25:

a) �GÆ = +0.16 kJ

b) Equilibrium position favours the reactants.


Q26:

�-� � �"� � +��

� ��'� �%��, �- � �

��, � � �� + � � ��

+ ��

�� + � ��

This reaction becomes feasible at 708.4ÆC and above.


Q27: Silver(I) oxide. At 1000 K, �G is +60 kJ mol-1 on the graph. Even with no otherchemical involved, this reverses to breakdown silver(I) oxide with �GÆ = -60 kJ mol-1



Q28: Above 2200 K approximately. This would allow �G to be negative for:

2ZnO � 2Zn + O2

Q29:

a)

(i) At 1000 K the target equation is:

2C(s) + 2ZnO(s) � 2Zn(s) + 2CO(g)

(ii) 2C(s) + O2(g) � 2CO(g) �GÆ = -400 kJ mol-1

2Zn(s) + O2(g) � 2ZnO(s) �GÆ = -500 kJ mol-1

(iii) Reverse the zinc equation:

2ZnO(s) � 2Zn(s) + O2(g) �GÆ = +500 kJ mol-1

(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:

2C(s) + 2ZnO(s) � 2Zn(s) + 2CO(g)

(ii) 2C(s) + O2(g) � 2CO(g) �GÆ = -500 kJ mol-1

2Zn(s) + O2(g) � 2ZnO(s) �GÆ = -300 kJ mol-1

(iii) Reverse the zinc equation:

2ZnO(s) � 2Zn(s) + O2(g) �GÆ = +300 kJ mol-1

(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, �GÆ = 0. Above this temperature, the reaction isfeasible.

Remember: The lower of the two lines operates as written and the upper line will bereversed.




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 thistemperature.

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� SiO2 + 2Mg

Further answers


Q33: boiling

Q34: increase

Q35: positive

Q36:

The degree of disorder (entropy) of the water molecules as a gas above 373 K increasesrapidly as the temperature increases. The molecules also vibrate and move much morefreely. Below 273 K, the water molecules in ice vibrate more as they are heated but theirmovement is still restricted and thus their entropy only increases slowly.


Q37: Butane is a gas with a high degree of disorder, and although pentane is a morecomplex molecule, it is a liquid under standard conditions and therefore has a moreordered 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 otherway by the same amount.




Q40: Boiling point = 333 K or 60ÆC


Q41: As the zinc melts, the disorder (entropy) increases. Since the gradient is given by-�S (from the straight line �G = -T�S + �H), the slope of the line changes.

Q42: Zinc vaporises at 1180 K with an increase in entropy and a subsequent change inthe gradient of the line on the Ellingham diagram.



8 Electrochemistry


Q1:

Copper will be positive withrespect 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.


Q5: a) reduced?

Q6: a) an oxidising agent?

Q7: Zinc atoms lose electrons and copper(II) ions gain them, i.e. electrons aretransferred from zinc to copper (II) ions.


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 twosolutions. For this reason, it is often known as an ion bridge.




Q12:

Oxidation is Mg(s)�Mg2+(aq) + 2e-

Reduction is Zn2+(aq) + 2e-�Zn(s)

Mg(s) + Zn2+(aq) �Mg2+(aq) + Zn(s)

Q13:

Oxidation is Cr(s)�Cr3+(aq) + 3e-

Reduction is Cu+(aq) + e-�Cu(s)

Reduction must be multiplied by 3

Cr(s) + 3Cu+(aq) �Cr3+(aq) + 3Cu(s)

Q14: H2(g) + Cl2(g) �2H+(aq) + 2Cl-(aq)

Q15: Ni(s) + 2H+(aq) �Ni2+(aq) + H2(g)


Q16: Magnesium more reactive � e- flow from Mg � Zn.

Mg(s) �Mg2+(aq)��Zn2+(aq)� Zn(s)

Q17: Ni(s)�Ni2+(aq)��Pb2+(aq)� Pb(s)

Q18: Cu(s) �Cu2+(aq) ��Cl2(g)�Cl-(aq) �Pt(s)

Q19:

Pt(s)�H2(g)�H+(aq)�� Hg2+(aq) �Hg(l)

Or

Pt(s)�H2(g) �H+(aq) ��Hg2+(aq) �Hg(l)�Pt(s)


Q20: b) Hydrogen electrode

Q21: H2(g) � 2H+(aq) + 2e-

Q22: Cu2+(aq) + 2e- � Cu(s)

Q23:

Pt(s) � H2(g) � H+(aq) �� Cu2+(aq) �Cu(s)

Note that in this case, the diagram (Figure 8.9) has the hydrogen electrode on the righthand side. So we cannot work out the cell notation by looking at the diagram.




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)


Q28: +1.61

Q29: +1.08

Q30: +1.36

Q31: +1.30


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. Ifthe 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 othersubstance is reduced, the reducing agent must be oxidised.

Q38: right

Q39: b) Top right


Q40: D

Q41: C

Q42: E

Q43: B

Q44: A




Q45: -308.8

Q46: -371.3

Q47: -289.5

Q48: -110.0


Q49:

�H = -cm�TWhere: �H is the enthalpy change in kJ.

c is the specific heat capacity of water (4.18 kJ kg-1 ÆC-1).

m is the mass of substance being heated in kg.

�T is the change in temperature in ÆC.

(If the temperature rises, �T is positive and hence �H is negative - an exothermicprocess.)

Q50:�" � ��

� �� *� �� )*

Q51:�� %%� � � ��

�� %%� � ��

�"� � ��

��)* ��

�"� � �� )* ��

Q52: +0.46


Q53: proton exchange membrane

Q54: Pt(s) � H2(g)� H+(aq)� � H+(aq)� O2(g) � Pt(s)

Q55: 1.23

Q56: 2H2(g) + O2(g) � 2H2O(l)



Further answers


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.


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)


Q60: Pt(s) �Fe2+(aq), Fe3+(aq) ��Ag+(aq) �Ag(s)


Q61: EÆ = -0.76 V

The voltmeter reading of +0.76 V is assigned to the half-cell reaction occurring at thezinc electrode, since the value for hydrogen is zero. In the cell, zinc atoms are oxidisedto zinc(II) ions. So the oxidation potential is +0.76 V. The sign must be reversed to getthe reduction potential.


Q62: +1.55

Q63: +0.62


Q64: c) Bottom left



Q65: A possible diagram is:

Figure 10.2: Summary of oxidising/reducing agents


Q66: C

Q67: D

Q68: F


Q69: -642.7

Q70: -239.3


Q71: -88.8

Q72: �GÆ = �HÆ - T�SÆ



Q73:�-� � �"� � +��

��

+�� "� ��-�

�� "� ��-�

+

�-� � � � )* �� * ��

�"� � �� )* �� * ��

��

�

� ��

� � �� * ��

�� )� ��

�2 .�� )* � *�

�2 -�� ! ��

Q74: The negative sign means a local decrease in entropy, i.e. the system has becomemore 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 lowerentropy.


Q75: Production of water for drinking. The cell would also produce heat which could beuseful.



9 Kinetics


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 massof the apparatus will decrease as the gas escapes. You will probably have used thesemethods 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 asthe 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 thereaction proceeds. The change in colour can be monitored using a colorimeter.


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


Q7:

• Concentration of catalyst if homogeneous



• Mass and particle size of catalyst if heterogeneous• Temperature• Pressure


Q8: A straight line graph is obtained.

Q9:

The rate is directly proportional to [H2O2]

i.e. Rate [H2O2]

or Rate = constant x [H2O2]


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 [N2O5]

Q20: s-1

Q21: 0.00044

Q22: 0.0000308

Q23:

Rate = k [H2O2][I-]

or

Rate = k [H2O2][I-][H+]0

Q24: mol-1 � s-1

Q25: 0.023




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 x30 s and a further 1.5 s will be needed to cap and label it.


Q28: Rate = k [NO2][CO]


Q29: b) H2S + Cl2 � S + 2HCl

Q30: a) 4HBr + O2 � 2H2O + 2Br2

Q31: b) False

Q32: b) False

Q33: a) True

Q34: b) False

Q35: 2H2O2 � 2H2O + O2

Q36: catalyst

Q37: intermediate

Q38: b) Rate = k [H2O2][Br-]

Further answers


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 ofbromine with time and the change in volume of CO2 with time.


Q40: b) False

Q41: a) True


scholar study guide sqa advanced higher chemistry unit · pdf filescholar study guide sqa...

Documents