Miscellaneous Science;Chemical kinetics: unit P5 · Exercise 30 In the Journal of the Chemical Society for 1950, Hughes, ~ Ingold and Reed report some kinetic studies on aromatic

~~

Chemical Kinetics

II~l][llilllN24706

~ Inner London Education Authority 1984

First published 1984by John Murray (Publishers) Ltd50 Albemarle StreetLondon WlX 480

All rights reserved.Unauthorised duplicationcontravenes applicable laws

Printed and bound in Great Britain byMartinis of Berwick

British Library Cataloguing in Publication DataIndependent Learning Project for Advanced

ChemistryChemical Kinetics. - (ILPACi Unit P5)1. ScienceI. Title II. Series500 Q161.2ISBN 0 7195 4043 7

ii

CONTENTS

PREfACEAcknowledgementsKey to activity symbols and hazard symbols

INTRODUCTIONPre-knowledgePRE-TEST

LEVEL ONE

vvi

vii

1

23

5

THE RATE Of A REACTION 5The decarboxylation of 2,4,6-trinitrobenzoic acid 5Evaluating reaction rate 7RATE EQUATIONS, RATE CONSTANTS AND ORDERS Of REACTION 9Summary 11Using rate equations 12Experimental methods 13Experiment 1 - investigating the hydrolysis of benzenediazonium chloride 15Half-life of first order reactions 19The order of reaction with respect to individual reactants 21Experiment 2 - the kinetics of the reaction between iodine and

propanone in aqueous solution 24COMPUTER-AIDED REVISION 29LEVEL ONE CHECKLIST 30LEVEL ONE TEST 31

LEVEL TWO 37REACTION MECHANISMS 37Molecularity 37The rate-determining step 38Proposing a reaction mechanism 39The hydrolysis of halogenoalkanes 42ACTIVATION ENERGY AND THE EffECT Of TEMPERATURE ON REACTION RATE 44The fraction of particles with energy greater than E 46aThe Arrhenius equation 48Using the Arrhenius equation 48Experiment 3 - determining the activation energy of a reaction 50CATALYSIS 52Catalysts and activation energy 53Autocatalysis 55Experiment 4 - determining the activation energy of a catalysed reaction 56THE COLLISION THEORY 57THE TRANSITION STATE THEORY 60Energy profiles and the transition state theory 60LEVEL TWO CHECKLIST 61

END-Of-UNIT TEST 63

APPENDIX ONE 67NOTES ON TANGENTS AND THEIR SLOP~S 67Drawing tangents 67Negative and positive slopes 68A note on 'differential notation' 69INTEGRATED RATE EQUATIONS 70first order reactions 71Second order reactions 72

iii

APPENDIX TWO·CLOCK· REACTIONSExperiment 5 - a bromine ·clock' reaction

ANSWERS TO EXERCISES

iv

737374

78

PREFACEThis volume is one of twenty Units produced by ILPAC, the Independent LearningProject for Advanced Chemistry, written for students preparing for the AdvancedLevel examinations of the G.C.E. The Project has been sponsored by the InnerLondon Education Authority and the materials have been extensively tested inLondon schools and colleges. In its present revised form, however, it isintended for a wider audience; the syllabuses of all the major ExaminationBoards have been taken into account and questions set by these boards have beenincluded.

Although ILPAC was initially conceived as a way of overcoming some of thedifficulties presented by uneconomically small sixth forms, it has frequentlybeen adopted because its approach to learning has certain advantages over moretraditional teaching methods. Students assume a greater responsibility fortheir own learning and can work, to some extent, at their own pace, whileteachers can devote more time to guiding individual students and to managingresources.

By providing personal guidance, and detailed solutions to the many exercises,supported by the optional use of video-cassettes, the Project allows studentsto study A-level chemistry with less teacher-contact time than a conventionalcourse demands. The extent to which this is possible must be determinedlocally; potentially hazardous practical work must, of course, be superVised.Nevertheless, flexibility in time-tabling makes ILPAC an attractive propo-sition in situations where classes are small or suitably-qualified teachersare scarce.

In addition, ILPAC can provide at least a partial solution to other problems.Students with only limited access to labo~atories, for example, those studyingat evening classes, can·concentrate upon ILPAC practical work in the laboratory,in the confiaence that related theory can be systematically studied elsewhere.Teachers of A-level chemistry who are inexperienced, or whose main disciplineis another science, will find ILPAC very supportive. The materials can be usedeffectively where upper and lower sixth form classes are timetabled together.ILPAC can provide 'remedial' material for students in higher education.Schools operating sixth form consortia can benefit from the cohesion that ILPACcan provide in a fragmented situation. The project can be adapted for 'use inparts of the world where there is a severe shortage of qualified chemistryteachers. And so on.

A more detailed introduction to ILPAC, with specific advice both to studentsand to teachers, is included in the first volume only. Details of the ProjectTeam and Trial Schools appear inside the back cover.

LONDON 1983

v

ACKNOWLEDGEMENTSThanks are due to the following examination boards for permission toreproduce questions from past A-level papers:

Oxford Delegacy of Local Examinations;Exercise 17(1977)Level One Test 8(1980)

Southern Universities Joint Board;Exercise 45(1978)

The Associated Examining Board;End-of-Unit Test 2(1981)

University of Cambridge Local Examinations Syndicate;Exercise 20(1983)

University of London Entrance and School Examinations Council;Exercises 15(1983)~ 16(1979)~ 18(1972)~ 19(1975)~ 25(1979)1

26(1972)1 27(1975)~ 30(N1975)~ 35(N1982)1 36(N1977)Level One Test 1(N1974)1 2(1977)~ 3(N1977)~ 4(N1975). 5(N1977)1

6(N1978)~ 7(1979)~ 9(N1981)~ 10(1974)1 11(1976),12(N1976)

End-of-Unit Test 1(1976)1 5(N1980)Teacher-marked Exercise p61(1973)

Welsh Joint Education Committee;End-of-Unit Test 3(1977)1 4(1975)

Questions from papers of other examining boards appear in other Units.

Where answers to these questions are included~ they are provided by ILPACand not by the examination boards.

Exercise 9 is reproduced, by permission, from Nuffield Advanced Science:Chemistry; Students' Book II (1st edition), edited by B.J. Stokes andpublished by Longman Group Limited for the Nuffield Foundation.

Photographs are included by permission as follows:Fig. 16 (page 48): Svante Arrhenius - Geoff Cox1 by courtesy of the Royal

Society of Chemistry.Photographs of students and Fig. 17 (page 51) - Tony Langham.

vi

SYMBOLS USED IN ILPAC UNITS

&, Reading

~ Exercise~

? ~ Test~

Revealing exercise

Discussion

Computer programme

'A' Level question

'A' Level part question

'A' Level questionSpecial paper

Worked example

Teacher-marked exercise

INTERNATIONAL HAZARD SYMBOLS

~ Experiment

IC= o!1 Video programme

~ Film loop

Model-making

~Harmful

~Toxic

~Radioactive

[j] Flammable IIExplosive

~Corrosive

~Oxidising.•.~

vii

INTRODUCTIONIf you have studied Unit 03 (More Functional Groups) you will have comeacross the oxidation of ethanal by acidified dichromate(VI) ions:

3CH3CHO(aq) + Cr2072-(aq) + 8H+(aq) ~ 3CH3C02H(aq) + 2Cr3+(aq) + 4H20(I)

How does this reaction take place? Does it seem likely, for example, thatthree ethanal molecules, one dichromate(VI) ion and eight hydrogen ions wouldall collide at the right moment and burst apart again, to give three ethanoicacid moleculesl two chromium(III) ions and four water molecules?

Fig.1.

No - it's extremely unlikely that twelve particles would collide like this -even though the equation for the reaction might seem to suggest that they do.The stoichiometric equation for a reaction is the sum of all the processesthat take place. It doesn't show how they take place (the mechanism), howmany separate reaction steps there may be in the reaction, or how manymolecules are involved at each stage.

In this Unit, we aim to show how experimental studies of rates of reactionunder different conditions can be used to work out possible mechanisms forreactions. By mechanism, sometimes called reaction pathway, we mean theseries of separate steps that go to make up a reaction.

In Level One we look at the experimental side. We show you how to analysedata to get a rate equation and scan the different methods available formeasuring reaction rates.

In Level Two we show how you can work out a mechanism from a rate equation.look at the effect of temperature on chemical reactions and finally examinebriefly two theories of reaction rates - the collision theory and thetransition state theory.

1

In Appendix One, we show you various ways of constructing a tangent to acurve, and we also consider some of the mathematics which you may encounterin text-books concerned with reaction rates.There are five experiments in this Unit: two in Level One, two in Level Twoand one in Appendix Two.

There is an ILPAC video-programme designed to accompany this Unit; it isdivided into two sections. It is not essential, but you should try to see itat the appropriate times if it is available.

Reaction kinetics

PRE-KNOWLEDGEBefore you start work on this Unit, you should be able to:(1) state that a rate is a change in some quantity with time;(2) list the variety of ways in which the rate of a chemical reaction can

be measured, e.g., by reference to changes in mass, volume or concent-ration of reactant or product;

(3) find the rate of a reaction, given the slope of a graph of the measuredvariable (as in (2)) against time;

(4) state the effect of the following on the rate of a chemical reaction -(a)

(b)

(c)

(d)

PRE-TEST

change in concentration of reactantschange in surface area of reactantschange in temperaturethe presence of a catalyst.

To find out whether you are ready to start Level One, try thefollowing test which is based on the pre-knowledge items. You shouldnot spend more than thirty minutes on this test. Hand your answersto your teacher for marking.

2

~\;lmilesbe used

m min -2 (1)Eft S-lo

Some high speed trains are supposed to cruise at a speed of 125per hour (miles hr-1). Which of the following units could alsoto express the speed of a train?A m-2 B km hr-1 C miles min-1

1 •

PRE-TEST

2. Which of the following units could be used to express the rate of thereaction between magnesium and hydrochloric acid?

Mg(s) + 2HCI(aq) ~ H2(g) + MgCI2(aq)

B cm3 min-1

C mol2 dm-6 hr-1

o cm3 S-l

E mol dm-3 min-1

( 1 )

3. The following information concerns an experiment to investigate the rateof reaction between calcite (a form of calcium carbonate) and excesshydrochloric acid. 1.05 g of calcite reacted with 10 cm3 of 2 M hydro-chloric acid in a small flask. The flask was weighed at two minuteintervals to determine the loss in mass due to the evolution of carbondioxide. The results were recorded in the form of the graph below. Inthis graph the loss in mass is plotted against the time from the momentwhen the two substances were mixed. Use this graph to assist you inanswering the following questions.

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0.1

0.4

0> 0.3....•.•..enencoE.Senen.2

0.2

Fig.2. time/minutes

3

(a) What mass of carbon dioxide has been formed nine minutes fromthe start of the reaction? (1)

(b) During which one of the following two minute intervals was thereaction quickest?

(i) 0 to 2 minutes(ii) 8 to 10 minutes

(iii) 2 to 4 minutes(iv) 16 to 18 minutes (1)

(c) After how many minutes had just half of the calcite reacted? (1)(d) What was the maximum mass of carbon dioxide formed? (1)(e) What fraction of a mole of carbon dioxide is this? (1)(f) What fraction of a mole is 1.05 g of calcium carbonate? (1)(g) Write the equation for the reaction between calcium carbonate

and hydrochloric acid. (C = 12.0, 0 = 16.0, Ca = 40.0) (2)

4. 5 g of granulated zinc is put in a conical flask and covered with100 cms (an excess) of 2 M hydrochloric acid at 20°C. A fairlyslow reaction takes place.For each of the following changes state whether you would expect therate to be increased, decreased or unchanged and give a reason foryour answer.(a) 5 g of powdered zinc is used instead of granulated zinc.(b) 3 g of granulated zinc is used instead of 5 g.(c) 100 cms of 2 M ethanoic (acetic) acid is used instead of hydro-

chloric acid.(d) The temperature is raised to 40°C.(el A few drops of aqueous copper(IIl sulphate is added. (10)

(Total 20 marks)

4

LEVEL ONE

The study of chemical kinetics (reaction rates) is of importance to industrialchemists because they are concerned with making particular products aseconomically as possible. Therefore, following the rates of reaction,interpreting them and altering them have important industrial applicationsas well as improving our understanding of chemistry. These are some of thefactors which you consider in Level One.

THE RATE OF A REACTIONObjectives. When you have finished this section, you should be able to:

(1) calculate values for the rate of reaction at various times, from agraph of concentration of reactant against time;

(2) write an equation to show the relationship between rate of reactionand concentration of reactant.

The study of reacti.on kinetics relies on experimental work. We begin bypresenting you with a set of experimental results.

We have chosen a reaction where a single substance (the reactant) breaksdown. We will take you through a series of exercises on the data, to showyou how to find the rate equation for the reaction. Our purpose is toillustrate the key ideas of reaction rate, order of reaction and rateconstant.

The decarboxylation of 2~4~6-trinitrobenzoic acid2,4,6-Trinitrobenzoic acid in solution loses carbon dioxide when heated,as shown by the following equation:

C02H

02N~O· N02Y (aq)

N02

02N I6YN02

~ ~ (aq)

N02

Since one of the products is gaseous the rate of reaction could be studiedby measuring the volume of carbon dioxide produced. Alternatively, youcould follow the decrease in the amount of 2,4,6-trinitrobenzoic aciditself.

The scientists who did the experiment in 1931 (E.A. Moelwyn-Hughes andO.N. Hinshelwood) chose the second method. They set up several mixtures at90°C. After various reaction times they withdrew a sample and added alarge volume of iced water to quench (i.e. stop) the reaction. They thentitrated each mixture with 5.0 x 10-3 M barium hydroxide solution usingbromothymol blue as indicator.

5

We have taken the following figures from their results:

Table 1

Concentration of2,4,6-trinitrobenzoic

Time/min acid/mol dm-3

0 2.77 x 10-4

18 2.32 x 10-4

31 2.05 x 10-4

55 1.59 x 10-4

79 1.26 x 10-4

157 0.58 x 10-4

Infinity 0.00

The first step in analysing this sort of data is to plot a graph of concent-ration of reactant against time. You do this in the first exercise. Squarebrackets will be used to denote concentration throughout this Unit, i.e.[Xtaq)] = concentratiori of X in mol dm-3 of aqueous solution.

Exercise 1 Plot a graph of concentration of 2;4,6-trinitrobenzoic ~acid (vertical axis) against time (horizontal axis). Usea large piece of graph paper, as you will have to draw ~~tangents to the curve later and you will need plenty ofworking space for this.(Answer on page 78 )

You now have a graph showing how concentration of reactant changes with time.The rate of change per unit time (per minute in this case) is given by theslope of the graph. This is the rate of reaction.

In the next exercise you look at your graph to see how much you can learnabout the rate of reaction by inspection.

Exercise 2 Place your ruler under the curve at time 0, with the IOng~edge of the ruler hugging the curve. Notice the slope ofthe ruler. Move the ruler slowly along the curve, ~~checking how its slope changes as you go. Then answerthe following questions.(a) When is the rate of reaction greatest and when is it least?(b) Suggest one reason why the rate of reaction is changing.(Answers on page 78 )

We now show you how to get a value for the rate of reaction at an instant(instantaneous reaction rate) from a concentration/time curve.

6

Evaluating reaction rateIn Exercise 2, while moving your ruler along the graph, you were, in fact,comparing the slopes of tangents to the curve at various points.

To get a value for the slope of a graph at a particular point, draw a tangentto the curve at that point and measure the slope of the tangent. In AppendixOne (page 67) we suggest several methods which can be used to constructtangents accurately.

A measured slope has a number, a sign and a unit. The number tells how fastthe concentration (or some other quantity such as volume of gas) is changingwith time. The sign indicates whether the quantity (Y), such as concentra-tion, is increasing or decreasing, as in Fig. 3 below. The unit combines theunits of the change being measured (such as concentration) and time.

Fig.3.

yY decreases with time

time

A negative slope

Y Y increases with time

time

A positive slope

Rate of reaction is usually defined as the rate of change of magnitude of theconcentration of a reactant or a product with time and can be obtained fromthe slope of the concentration/time graph. In most of the exercises ratewill have units of:

(concentration) x (time-1) e.g. mol dm-3 S-l

However, if a reaction is being followed by a physical method, involving themeasurement of, say, mass or volume, the rate can be expressed incorresponding units, such as those of mass time-1 (e.g. g S-l) or volumetime-1 (e.g. cm3 min-1).

In fact, rate can be expressed in a variety of units, as long as the quantitymeasured is proportional to the concentration or amount of either a reactantor a product. If rate does have units other than those of concentrationtime-1 it should always be clear from the context.

In the next exercise you calculate the slope and thus the rate of reactionat specified times.

7

Exercise 3 Take the graph you drew in Exercise 1 and construct ~tangents to the curve at 10, 50, 100 and 150 minutes ~(using one of the methods given in the Appendix).Calculate their slopes and complete a copy of Table 2.Bear in mind that where the slope is negative, the value of theslope, but not the sign, gives the rate.

Table 2

Time Concentration Slope RateImin Imol dm-3 Imol dm-3 min -1 Imol dm-3 min-1

10

50

100

150

(Answers on page 78 )

You now have a set of figures to show that the rate of reaction for thedecarboxylation of 2,4,6-trinitrobenzoic acid is decreasing with time. Theconcentration of reactant is also decreasing with time. What we now want toknow about this reaction is. in what way does the rate of reaction depend onthe concentration of reactant?

A standard way of finding out whether two quantities, say x and y, aredirectly related is to plot a graph of y against x. If the graph is astraight line going through zero, then y is directly proportional to x:

ory

y

<X x

mxor m

where m is a constant, equal to the slope of the line. If y is proportionalnot to x but to (x + a constant) the equation is:

y mx + c

where c is the intercept on the y axis or the value of y when xFig. 4 J.

Fig.4.

o (see

Now go on to the next Exercise, where you find the relationship between therate of the decarboxylation of 2,4,6-trinitrobenzoic acid and itsconcentration.

8

Exercise 4 Plot a graph of rate of reaction (vertical axis) against ~concentration of 2,4,6-trinitrobenzoic acid (horizontalaxis) using your values in Table 2 (Exercise 3). Now ~~answer the following questions:(a) Does your graph go through the origin? Explain why it

should.(b) Use your graph to state the relationship between rate of

reaction and concentration of reactant:(i) in words

(ii) mathematically.(Answers on page 78 )

The expression you have just worked out in Exercise 4 is called the rateequation for the reaction. We consider rate equations in more detail in thenext section.

RATE EQUATIONS~ RATE CONSTANTS AND ORDERS OF REACTIONA rate equation relates the rate of a reaction to the concentrations ofthe reacting species by means of constants called (a) the rate constantand (b) the order(s) of reaction.

Objectives. When you have finished this section, you should be able to:

(3) identify the order of reaction with respect to a named reagent;(4) determine the overall order of reaction;(5) work out a value for the rate constant of a first order reaction,

including its unit;(6) recognise a first order reaction from a graph of rate against concent-

ration of reactant.

Read about rate equations (rate expressions) and rate constants(velocity constants) in your text-book(s). Distinguish carefullybetween the order of reaction with respect to one reactant and theoverall order. Some books use the 'differential notation' for therate of reaction, e.g.

rate of reaction d[reactant]dt

You will find this easy if you study mathematics but we suggest you need notuse it in this context since it is rarely required for A-level chemistry.However, if you want some further explanation, look in Appendix One.

We will spend more time on orders of reaction and their uses later in theUnit; at the moment you should be able to apply what you have just learnedto the rate equation for the decarboxylation of 2,4,6-trinitrobenzoic acid.You do this in the next Exercise.

9

Exercise 5 Use the rate equation which you worked out in Exercise 4 ~to state:(a) the order of reaction with respect to 2,4,6-trinitro-~~

benzoic acid;(b) the overall order of reaction.(Answers on page 78 )

You should be able to work out a value for the rate constant for a firstorder reaction like this one, both graphically and by substituting into therate equation. To make sure you can use both methods, try the next twoexercises.

Exercise 6 Using both your graph and rate equation from Exercise 4, ~work out a value for the rate constant (k) of thedecarboxylation of 2,4,6-trinitrobenzoic acid at 90 °C. ~~What is the unit of the rate constant for this reaction?(Answer on page 78 )

Another method to calculate the rate constant is to substitute values of rateand concentration (from your graph) into the rate equation, rearranged togive:

k = rate/concentrationand average the results. Try this in the next exercise.

Exercise 7 (a) Use your graph from Exercise 1 to calculate a valuefor the rate of reaction at zero time (the initialratel.

(b) Calculate the rate constant (k) for each of thetimes (10, 50, 100 and 150 minutes) listed in Table 2 andfor zero time.

(c) Average your results.(Answers on page 79 )

In the next exercise; you compare the values of k which you calculated inExercises 6 and 7.

Exercise 8 Which of the two values of k that you have justcalculated is likely to be more accurate? Explain yourchoice.(Answer on page 79 )

All the exercises so far in this Unit have been based on data from thedecarboxylation of 2,4,6-trinitrobenzoic acid. The following summary willhelp you to identify the main points in these exercises.

10

Summary

So far in this Unit, we have shown you how to:(a) plot a concentration/time curve from experimental data;(b) draw tangents to this curve and hence calculate values of reaction

rate;(c) use rate/concentration data obtained from this curve to identify a

first order reaction by plotting rate versus concentration;(d) express the relationship between rate and concentration of a reactant

mathematically (a rate equation or rate expression);(e) calculate a value for the rate constant in two ways.

Now try the next exercise, where you are given rate/concentration data fora different reaction and asked to deduce the rate expression, the order ofreaction and the rate constant.

Exercise 9 Table 3 contains some data for the decomposition of ~dinitrogen pentoxide in tetrachloromethane solution:2N205(in CC14 solution) ~ 4N02(in CC14 solution) + 02(g) ~~Table 3

Rate of reaction asdecrease in concentration

Concentration of N205 per secondN205/mol dm-3 /10-5 mol dm-3 S-1

2.21 2.26

2.00 2.10

1.79 1.93

1 .51 1.57

1.23 1.20

0.92 0.95

Plot a graph of the rate of the reaction against the concent-ration of N205, and try to answer these questions:(a) What is the rate expression for the reaction?(b) What order is this reaction with respect to N205?(c) What is the value of the constant in the rate expression?

Include the correct unit.(Answers on page 79 )

In the next section we aim to extend your knowledge of rate equations andorders of reaction to more complicated reactions.

11

Using rate equationsSo far you have studied reactions in which one compound decomposes. We nowtake a brief look at reactions involving more than one reactant.

Objectives. When you have finished this section, you should be able to:

(7) use a given rate equation to identify the order of reaction withrespect to an individual reactant;

(8) use a given rate equation to identify the overall order of reaction;(9) define order of reaction with respect to an individual reactant and

overall order of reaction.

To see that you have understood the difference between overall order andorder of reaction with respect to a single reactant, try the followingthree short exercises.

Exercise 10 The reaction between mercury (II) chloride and ethane- Iidioate ions takes place according to the equation: ~

2HgC12(aq) + C2042-(aq) ~ 2Cl-(aq) + 2C02(g) + Hg2C12(s)

The rate equation for the reaction israte = k [HgC12(aqJ][C204

2-(aq)]2(a) What is the order of reaction with respect to each

reactant?(b) What is the overall order of reaction?(Answers on page 79 )

In the next exercise you may be surprised to find a fractional order in therate equation. You will not come across fractional orders very often atA-level, but you should be aware that they exist and are a sign of a morecomplex reaction mechanism than the ones you will be investigating inLevel Two.

Exercise 11 For the following reaction:C2H4(g) + I2(g) ~ C2H4I2(g)

the rate equation is:rate = k [C2H4(g)][I2(g) ]3/2

(a) What is the order of reaction with respect to eachreactant?

(b) What is the overall order of reaction?(Answers on page 79 )

12

The next exercise deals with the overall order of a hypothetical reaction.

Exercise 12 Substances P and 0 react together to form products andthe overall order of reaction is three. Which of thefollowing rate equations could not be correct?A rate k [p]Z[Q]B rate k [p]O[Q]3C rate k [p][O]Z0 rate k [pJ[OJ3E rate k [pJ[QJZ[H+Jo(Answer on page 79 )

Now do the next exercise, to check that you can work out units for the rateconstant, k, for a reaction which is not first order.

Exercise 13 The equation for the reaction between peroxodisulphateion and iodide ion is:

SzOaZ-(aql + 2I-(aq) - 2S04Z-(aq) + Iz(aq)

The rate equation is:rate = k [S20e2-(aql][I-(aq)1

If concentrations are measured in mol dm-3 and rate inmol dm-3 S-1, determine the unit of k. Show your workingclearly.(Answer on page 79 1

We now go on to consider the various ways in which reactions can be followedto collect information for kinetic studies.

Experimental methodsIn the decarboxylation of 2,4,6-trinitrobenzoic acid, the reaction wasfollowed by a titration method. This is a straightforward technique, buthas the disadvantage that the reaction mixture is destroyed in the process.

Many other methods are available for measuring the rate of reaction. Wetake a look at the more important of these physical methods which all sharethe advantage that they do not interfere with the reaction mixture in anyway.

Objectives. When you have finished this section you should be able to:

(10) choose suitable methods for measuring the rates of different types ofreaction;

(11) describe in outline each of the main methods for measuring reactionrate.

13

Read about the different methods that are commonly used to measure 0reaction rates. We suggest that you make brief notes on the followingmethods: sampling and titration; measuring gas volume; polarimetry; ~colorimetry; conductivity; dilatometry (measuring change in volume ofliquids). The video-programme mentioned below would also be useful.

For each method you should consider the type of reaction that could befollowed by it, and an outline of how it works.

At this point you should watch part 1 of the ILPAC video-programme II II'Chemical Kinetics' if it is available. This part of the tape givesdetails of all the methods of measuring reaction rates listed above. 00

Now attempt the next exercise.

Exercise 14 For each of the following reactions, decide which method~or methods (there may be more than one) you could use tomonitor its progress. ~~(a) CaCo3(s) + 2HCI(aq) + 2CaCI2(aq) + CO2 (g) + H2o(1)(b) 2Mno4 -(aq) + 5C204

2-(aq) + 16H+(aq) ~ 2Mn2+(aq) + 8H2o(l)+ 10Co2(g)

(c) (CH3)3CBr(l) + H2oCl) ~ (CH3)3CoH(aq) + H+(aq) + Br-(aq)(d) CH3CH(oC2Hs)2(aq) + H2o(1) ~ CH3CHo(aq) + 2C2HsoH(aq)(e) H202(aq) + 2I-(aqJ + 2H+(aq) ~ 2H2o(1) + 12(aq)(f) C12H22011(aq) + H2o(11 ~ C6H1206(aq) + C6H1206(aq)

sucrose(Answers on page 79 )

glucose fructose

Now attempt the following Teacher-marked Exercise. You may want to lookover your notes before starting. When you have finished, hand it to yourteacher for marking.

Teacher-markedExercise

Review the methods that are available for measuring [2]the rates of chemical reactions. For each methodyou should mention: (a) what is being measured,(b) for what type of reaction it is suitable, and(c) brief details of apparatus (no diagrams needed).

Nbw that you have looked at the various different methods ofmeasuring reaction rates, do Experiment 1, where you follow the rateof hydrolysis of benzenediazonium chloride using a physical method -measuring the volume of nitrogen produced.

14

EXPERIMENT 1An investigation of the hydrolysis ofbenzenediazonium chloride

AimThe purpose of this experiment is to deter-mine the rate equation for the reaction inwhich benzenediazonium chloride is hydrolysedand hence find the order of reaction withrespect to benzenediazonium chloride.

IntroductionBenzenediazonium chloride is an unstable substance, which decomposes whenheated above 5 °c to give phenol, nitrogen and hydrochloric acid:

C6H5N2+CI-(aq) + H20(I) ~ C6H50H(aq) + N2(g) + HCI(aq)There are several stages in the preparation of the reaction mixture, so youdo this at a temperature low enough for the rate of reaction to be negligible.When you are ready, you warm the mixture quickly to a fixed temperature andmeasure the volume (V~ of gas produced at one minute intervals for about25 minutes. You then leave the mixture until no further reaction appears tobe occurring, and measure the total volume eVoo) of gas produced since theclock was started.Because the volume of gas produced in time t is proportional to the amountof benzenediazonium chloride used up, it follows that:

Voo cr [C6H5N2+CI-(aq)] at the startVoo - Vt cr [C6H5N2+Cl-Caq)] at time t

A plot of (Voo - Vv against time will therefore have the same form as theconcentration/time graph and can be used to obtain information about therate of reaction.Note that, in this experiment, you need not attempt to judge the time whenthe reaction begins. The clock can be started at any time after the mixturehas reached a steady temperature. Even if some benzenediazonium chloridehas reacted by then, the amount remaining can be taken as giving the'initial' concentration, and this can be calculated, if necessary, from Voo•

15

Requirementssafety spectacles and protective gloveswater bath. thermostatically controlled, set between 40 and 50°Cthermometer, 0-100 °cside-arm test-tube with bung3-way tapglass syringe, 100 cm3

rubber tubing (2 short lengths)2 retort stands, bosses and clampswash-bottle of distilled watermeasuring cylinder, 10 cm3

:~~~~~:~~ trite, NaN02 - - - - - - - - - - - - - - - - - - - - -I~ II~ Ibeaker 250 cm3

crushed ice ...-1 IVIhydrochloric acid, concentrated. HCI - - - - - - - - - - - - - - l~ ftpumice or anti-bumping granules ~••__~graduated pipette. 5 cm3 or 10 cm3

~~~~~~:m:~~~e~6H5NH2- -- - - - - - - - - - - - - - - - - - - -1811~1teat-pipette ~stopclock or stopwatch

Hazard warningPhenylamine is toxic, by ingestion and by skin absorption.It is also flammable and gives off a harmful vapour.Concentrated hydrochloric acid is corrosive and gives off aharmful vapour. Therefore you MUST:WEAR SAFETY SPECTACLES AND GLOVESWORK IN A FUME CUPBOARD WHERE POSSIBLEKEEP BOTTLES AWAY FROM FLAMESKEEP STOPPERS ON BOTTLES AS MUCH AS POSSIBLE

Procedure1. Set the control on the water bath to a temperature between 40 and 50°C

and hang a thermometer in it. If possible this should be done before thelesson so that it has time to reach a steady temperature.

2. Set up the apparatus, without the chemicals, as shown in Fig. 5.

side-arm tubein thermostatat 40-50°C

3-waytap clamp

1'"'1'"'1""1""1""1""1""1""1""1'1"1'11I1""111I'1""1"1I'IT'"I""I""I'1II1o ~ w ~ ~ ~ ~ ro ~ ~ 00

syringe

Fig.5.

benzenediazonium chloride

...;.:::. anti-bumping granules

16

3. When you have connected the three-way tap between the side-arm tube andthe syringe, turn the tap so that it connects the syringe with the openair as in Fig. 6, A. Press the plunger in, so that the syringe is empt~and then turn the tap so that it connects the side-arm tube with the airas in Fig. 6, B.

open air open air

syringe syringe side-arm tube V ~

~

syringe

connectingchannel

A. Emptying syringe

Fig.6.

B. Allowing gas to escape C. Collecting gas

5.

7.

4. Using a 10 cm3 measuring cylinder, add 2.0 crn3 of distilled water intoa test-tube. Weigh 0.80 g of sodium nitrite and dissolve it in thewater. Cool this solution in a beaker of crushed ice and keep it handy.Remove the side-arm tube and, using a measuring cylinder, measureinto it 5.0 cm3 of distilled water and 2.5 cm3 of concentratedhydrochloric acid.

6. Add a few anti-bumping or pumice granules to the side-arm tube and sealits mouth with a rubber bung.Place the side-arm tube in a beaker of crushed ice and, using a ~pipette and filler, add 1.0 cm3 of phenylamine. Mix the contents ~of the tube thoroughly and allow them to cool.

8. Using a teat-pipette, add the sodium nitrite solution a few drops at atime to the phenylamine solution in the side-arm tube. Shake the tubegently to swirl its contents as you do so.

9. Check that the thermostat temperature has become constant (at about45°C), theh clamp the side-arm tube in position in the thermostat(see Fig. 5) so that the solution is completely immersed in water.

10. Wait four minutes with the tap open to the air, so that the solutionwarms to the temperature of the water-bath. Record the temperature.

11. After four m~nutes, turn the tap so that it connects the side-arm tubewit~ the syringe (Fig. 6, C), reset the clock and take a reading.Regard this as time zero.

12. Continue taking readings each minute for about 25 minutes. Enter yourresults in a larger copy of Results Table 1. The total volume of gasproduced is over 100 cm3• so as you see the volume approaching the100 cm3 mark, get ready to turn the tap to the position shown in Fig. 6, A(emptying the syringe into the air). Just as it reaches 100 cm3, turn thetap, expel the collected nitrogen. then turn the tap back to the positionshown in Fig. 6, C (connecting the side-arm tube with the syringe).Depending on how you make up your mixture, you may have to do this twiceduring the experiment.

17

13. After 25 minutes, leave the apparatus for at least another half hour forthe reaction to finish completely. This gives you the total volume ofnitrogen produced in the reaction, Voo (i.e., the volume produced atinfinite time). If you want to speed things up, immerse the side-armtube into a beaker of hot water at about 60°C. If the syringe isnearly full, empty it first, noting the volume of nitrogen expelled. Thevolume of nitrogen should reach its maximum in a few minutes. Remember toallow the syringe to cool back down to the temperature of the waterbathbefore taking your final reading (Voo).

Results Table 1Time, tlmin

Volume of N2, Vt/cm3

(V00 - V t)I cms

(Specimen results on page 80 )

Calculations1. Work out the values of (Voo-Vt) and enter them into your copy of

Results Table 1.2. Plot (Voo-Vt) (vertical axis) agalnst time (horizontal axis). Draw a

smooth curve through tile points.3. Construct tangents to your curve and measure the slope at each point.

Draw one at time 0 and at least four others evenly spaced.4. Use your graph to complete a larger copy of Results Table 2.

Results Table 2

Time Slope RateImin Icm3 min -1 Icms min -1 (V -V )/cm3

00 t

(Specimen results on page 80 15. Plot another graph of rate of reaction against lVoo - Vt), which is

proportional to the concentration of benzenediazonium chloride.

Questions1. Use the information from your second graph to write a rate equation for

the reaction.2. Use your second graph to work out a value for the rate constant, k, for

the reaction, including its units.3. Explain why the escape of gas during the first four minutes, before

recording the first volume reading, can be neglected.(Answers on page 80 )

In the next section we take a look at another aspect of first order reactions.

18

Half-life of first order reactionsIf you are asked to shBw that a reaction is of the first order you can oftendo this by calculating its half-life. As you will see, there is only onegraph to plot!

You have already come across the idea of half-life of a reaction in Unit S2(Atomic Structure), in connection with the decay of radioactive isotopes.All radioactive isotopes decay by first order kinetics. The rate equationis -

rate = k [reactant]Each radioactive isotope has its own half-life. Since the intensity of theradioactivity of a sample is proportional to the amount of the isotope itcontains, the half-life is the time taken for the radioactivity to fall tohalf its original value.

If you are not sure of the definition of half-life. look it up.read about the half-life of first order reactions.

Also 0W

Objectives. When you have finished this section I you should be able to:

(12) define the term half-life;(13) determine the half-life of a first order reaction from graphical data

of concentration versus time;(14) identify a reaction with a constant half-life as being first order.

To revise the idea of radioactive decay and half-life, try the next exercise.

605040302010

:~l , i) iT, Ii ..

; 1 : ,\ i i

~ ~·t:·'J::' -;,

i,

\ ' , ,. ;

"':" ~Hr ,

"1.:.;,

+ ','i. ;

'"\:

I,,· '-......"" "". -..

r.',

1oo

j!!·c

::::J

~ 8~:0'-~ 6co.~C 4Q)ucou 2

10

Exercise 15 (a) State what is meant by the term 'half-life'.(b) The following is a decay curve for a radioactive

element X.12

Fig.7. time/minutes'

(i) Determine the half-life of X.(ii) What is the order of the decay reaction?(Continued on page 20.)

19

(c) On a copy of the axes below, sketch the approximaterelationship between rate of decay and concent-ration of X.

Fig.8. concentrationofX/arbitraryunits


Other first order reactions, besides radioactive decay processes~ haveconstant half-lives. You use this fact in the next exercise.

Exercise 16 The following chemical reaction~ under certain conditions,0proceeds with the order shown. At::t2N205 ~ 4N02 + O2 First order(a) ~n one experiment on the reaction, the initial concent-

ration of N205 was 32 units and the half-life was 5minutes. On a copy of the graph below, plot the relationbetween the concentration of N205 and time for thisexperiment.

35en•..'i: 30j

~~ 25:eco......

20an0

N

Z.•.. 150c:0',t:; 10co~•..c:CD(,) 5c0(,)

00 5 10 15 20

Fig.9, time/min

(b) When the reaction was carried out in tetrachloromethane,it was found that an initial concentration of N205 of2.00 mol dm-3 gave an initial rate of reaction of 2.10 x10-5 mol dm-3 S-1. Calculate the rate constant for thereaction and state the units.

(Answers on page 81 1

In the next exercise you apply the half-life concept to the reaction youconsidered in Experiment 1.

20

Exercise 17 The decomposition of benzenediazonium chloride in aqueoussolution is a reaction of the first order which proceedsaccording to the equation:

C6H5N2CI ~ C6H5CI + N2(g)

A certain solution of benzenediazonium chloride containsinitially an amount of this compound which gives 80 cms ofnitrogen on complete decomposition. It is found that, at 30 °C,40 cm3 of nitrogen are evolved in 40 minutes. How long afterthe start of the decomposition will 70 cm3 of nitrogen havebeen evolved? (All volumes of nitrogen refer to the sametemperature and pressure. )(Answers on page 81)

In the next section we show you how to det8rmine the order of reaction withrespect to one reactant at a time, from experimental data, in a reactionwhich involves two or more reactants.

The order of reaction with respect to individual reactantsObjectives. When you have finished this section, you should be able to:

(15) determine the order of reaction with respect to individual reactantsfrom initial rate data;

(16) determine the rate equation for the iodination of propanone byfollowing the reaction colorimetrically.

We show you how to determine the order of reaction with respect to individualreactants in the following Worked Example.

Worked Example The data in Table 4 were obtained for the reactionbetween nitrogen monoxide and hydrogen at 700 °C.The stoichiometric equation for the reaction is:

2H2(g) + 2NO(g) ~ 2H20(g) + N2(g)Table 4

Initial concentrationfmol dm-s

Experiment Initial ratenumber H2 NO fmol dm-3 s-1

1 0.01 0.025 2.4 x 10-6

2 0.005 0.025 1.2 x 10-6

3 0.01 0.0125 0.6 x 10-6

Determine the order of reaction with respect to NO and toH2• Write the rate equation for the reaction.

21

Solution1. Consider Experiments 1 and 2 in Table 4.

Initial [NO(gJ] is constant. Therefore, any variation in initial rateis due to the variation in initial [H2(gJl.

2. Deduce the order of reaction with respect to hydrogen.HalVing initial [H2(gJ] halves the rate of reaction ..'. rate ~ [H2]1 i.e. the order with respect to H2 1

3. Consider Experiments 1 and 3 in Table 4.Initial [H2(gJ] is constant. Therefore, any variation in initial rateis due to the variation in initial [NO(gJ].

4. Deduce the order of reaction with respect to nitrogen monoxide.HalVing initial [NO(gJ] gives one quarter the rate, i.e. (1)2.

rate ~ [NO(gJ12 i.e. the order with respect to NO = 25. Write the rate equation.

rate = k[H2(gJ][No(gJ12

Use the same method for the following exercises.

Exercise 18 Two gases, A and 8, react according to the stoichiometric 0equation: ~

A(gJ + 38(gJ + A83(g) ~"A series of experiments carried out at 298 K in order todetermine the order of this reaction gave the following results.Table 5

Initial Initialconcentration concentration Initial rate ofof A of 8 formation of A83

Expt. e/mol dm-3 e/mol dm-3 /mol dm-3 min-1

1 0.100 0.100 0.00200

2 0.100 0.200 0.00798

3 0.100 0.300 0.01805

4 0.200 0.100 0.00399

5 0.300 0.100 0.00601

(aJ What is the order of the reaction between A and 8 withrespect to

(i) substance A (ii) substance 81(b) Write down a rate equation for the reaction between A and 8.(c) Using the experimental data given for Experiment 1, Table 5,

calculate the rate constant, k, for the reaction. Give theappropriate units of k.

(f\nswers on page 81 )

22

Exercise 19 A series of experiments was carried out on the reaction: ~2H2(g) + 2NO(g) ~ 2H20(g) + N2(g) ~

The initial rate of reaction at 750°C was determined bynoting the rate of formation of nitrogen. and the followingdata recorded:Table 6

Rate ofInitial concent- Initial concent- formation ofration of nitrogen ration of hydrogen nitrogen

Expt. monoxide/mol dm-3 /mol dm-3 s-l. /mol dm-3 s -l.

1 6.0 X 10-3 1.0 X 10-3 2.88 x 10-3

2 6.0 X 10-3 2.0 X 10-3 5.77 X 10-3

3 6.0 X 10-3 3.0 X 10-3 8.62 X 10-3

4 1.0 X 10-3 6.0 X 10-3 0.48 x 10-3

5 2.0 X 10-3 6.0 X 10-3 1.92 X 10-3

6 3.0 X 10-3 6.0 X 10-3 4.30 X 10-3

The rate equation for the reaction is:rate = k[H21m[Noln

Deduce the order of the reaction with respect to(a) hydrogen(b) nitrogen monoxide.(Answers on page 81 )

In the next experiment you determine a rate equation for the reaction tillbetween iodine and propane in aqueous solut~on. Since one of thereactants is coloured you will follow the reaction colorimetrically. JLIf you do not have access to a reliable colorimeter, your teacher mayrecommend that you do an alternative experiment, which you will find inAppendix Two.

23

EXPERIMENT 2The kinetics of the reaction betweeniodine and propanone in aqueous solution.

AimThe purpose of-this experiment is toobtain the rate equation for the reactionbetween iodine and propanone by deter-mining the order of reaction with respectto each reactant and to the catalyst(hydrogen ions). The equation is:I2(aq) + CH3COCH3(aq) ~

CH3COCH2I(aq) + H+(aq) + I-(aq)

IntroductionA catalyst does not necessarily appear in the stoichiometric equation (hereit appears as a product) but it can appear in the rate equation. The otherspecies which are likely to appear are the reactants and so you may assumethat the rate equation is:

rate = k[CH3COCH3]P[I2]q[H+]rYou will be determining the order of reaction with respect to each reactantby varying the concentration of each species in turn, keeping the othersconstant and following the reaction colorimetrically. As the intensity ofthe iodine colour decreases more light is transmitted through the solution(i.e. the absorbance decreases).There are three parts to the experiment:1. Choosing the right filter for the colorimeter.2. Calibrating the colorimeter so that meter readings can be converted to

concentrations of iodine.3. Obtaining values for the concentration of iodine at intervals of time

for a series of experiments with the following sets of conditions:(a) Initial [CH3COCH3] varying;(b) Initial [12] varying;(c) Initial. [H+] varying;

[121, [H+) constant[CH3COCH3], [H+J constant[CH3COCH3], [12] constant

Each set of experiments gives you the order of reaction with respect toone component. If there are three groups of students in your classworking on this experiment, then we suggest that each group assumesresponsibility for one set of conditions. Sharing your results willspeed up matters considerably.

Requirementssafety spectaclescolorimeter with a set of filtersset of optically matched test-tubes to fit colorimeter (with stoppers)wash-bottle of distilled water4 burettes with stands, filling funnels and beakersiodine solution, 0.020 M 12 (in KI(aq))propanone solution, 2.0 M CH3COCH3hydrochloric acid, 2.0 M HCIstopclock or watchthermometer, 0 to 100°C

24

Procedure1. Choose a filter. First~ switch on the

colorimeter to allow it to warm up.(Leave it switched on until you havefinished~ unless your teacher advisesotherwise.) 1deally~ the filtershould let through only light of theparticular waveleng~absorbed by thecoloured solution. So~ for a redsolution which absorbs cyan light~ youwould use a cyan filter - cyan is thecomplementary colour to red (see Fig.10). Since iodine solution is reddish~use a filter in the blue-green range.To select the best~ proceed as follows.(a) Put anyone of the suitable filters into the slot in the colorimeter

and insert a tube of distilled water, covering it to exclude straylight. Turn the adjusting knob so that the meter shows zeroabsorbance (or, on some colorimeters, 100% transmission). Mark therim of the tube so that you can replace it in the same position.

(b) Replace the tube with a tube containing the most concentrated iodinesolution you will use (see Results Table 3) and take a reading ofabsorbance (or % transmission). Mark this tube too so that you canreplace it in the same position.

(c) Repeat steps (a) and (b) above for other suitable filters in turn.(d) Choose the filter which gives the greatest absorbance (or least

transmission).2. Prepare a calibration curve, so that you can convert your meter readings

taken during the experiment to concentration of iodine.(a) To do this. prepare a series of iodine solutions as suggested in

Results Table 3. Measure and record the absorbance of each one.Most colorimeters are liable to 'drift', so you should zero themachine before each reading. by inserting the distilled water'blank' and re-setting the needle to 100% transmission.

Results Table 3Volume of Volume of Meter reading0.020 M 12 distilled [I2(aq)] (% absorbancesolution/cm3 water/cm3 Imol dm-3 or transmission)

0.0 10.0 0

1.0 9.0 0.0020

2.0 8.0 0.0050

3.0 7.0 0.0060

4.0 6.0 0.0080

5.0 5.0 0.010

(b) Plot a graph of meter reading against concentration of iodine andkeep this to use in analysing your results.


25

3. Decide which series of experiments you will do from Table 7 below andnote their letters. The figures in heavy type will help you choose. Ifno other students are working on this investigation then you will needto do all the experiments.

Table 7

Experiment

a b c d e f g

Volume of 2 Mpropanone/cm3 2 4 6 2 2 2 2

Volume of 0.02 Miodine/cm3 2 2 2 4 1 2 2

Volume of 2 M HelIcm3 2 2 2 2 2 4 6

Volume of waterIcm3 4 2 0 2 5 2 0

[Propanone]Imol dm-3 0.4 0.8 1 .2 0.4 0.4 0.4 0.4

[I2J/mol dm-3 0.004 0.004 0.004 0.008 0.002 0.004 0.004

[H+]/mol dm-3 0.4 0.4 0.4 0.4 0.4 0.8 1 .2

4. For your first mixture, measure out iodine solution, acid and water fromburettes into a test-tube. Wipe the tube clean and handle only at the top.

5. Measure the propanone solution into another test-tube, again using aburette. Keep the outside of all tubes clean and dry.

6. Adjust the colorimeter to zero against the tube of distilled water.7. Add the propanone solution to the first mixture and start the clock.

Quickly stopper the test-tube and invert it six or seven times to mixthe contents thoroughly.

B. Put the tube in the colorimeter in time to take a reading~at 30 secondsafter the clock was started.

9. Take further readings at 30 second intervals for 6 minutes, or until thecolour disappears, if this occurs sooner. Record your results in a copyof Results Table 4.

10. Record the temperature of the room and the temperature of the mixtureafter the final reading. If they differ by more than 2 or 3 °c you maybe advised to repeat the experiment, modifying the procedure as in step 12.

11. Repeat steps 4 to 10 for the other two mixtures in your set ofexperiments.

12. If time permits, repeat your measurements. Some colorimeters may givebetter results if you remove the tube after each reading, and reset themeter to zero just before re-inserting the reaction tube in time for thenext reading experiments.

26

Results Table 4

Time/min 0 1 1 1~ 2 2~ 3 31 4 4~ 5 5~ 62 2

Meter reading

[12 (aq) J/10-3 mol dm-3


Analysis of results1. Use your calibration curve to convert the meter readings to iodine

concentrations and enter these values in Results Table 4. Calculate avalue of initial concentration from the data in Table 7.

2. Plot graphs of concentration of iodine (vertical axis) against time(horizontal axis) for each mixture in the set of experiments you havedone. Plot all three on the same sheet of graph paper.

3. From these graphs obtain values for the initial rate of reaction. Notethat, under the conditions used in this experiment, the graphs willprobably be straight lines. In this case the initial rate is simply theslope of the line. If the graphs are curves, simply draw a tangent tothe curve at time zero and measure its slope.

4. Determine the order of reaction with respect to the component you havebeen varying, by comparing the initial rates at different concentrations.

Questions1. Study the graphs plotted by other groups of students for varying the

concentrations of the other two components. Work out the order ofreaction with respect to each of these components by comparing theinitial rates at different concentrations. Use all the information youhave collected to write the rate equation for the reaction.

2. What is the overall order of the reaction?3. Compare the rate equation with the stoichiometric equation for the

reaction. What is the main diffe~ence between the two? How do youexplain this difference?

4. Calculate the rate constant for the overall reaction using each of thethree initial rates from your set of experiments and average the results.Compare this with values obtained from the other sets of experiments.

5. What are the main sources of error in this experiment?(Answers on page 83)

Computer simulation of the experimentIf you have access to the ILPAC computer program 'Chemical Kinetics', r-.you can use part of the program to compare your experimental results ~with results calculated from the accepted rate equation. Note thatthe other part of the program is more suitable for use later in theUnit.

27

The kinetics of the iodination of propanone can be investigated equally wellby a sampling method such as that described in the next exercise.

Exercise 20 Under conditions of acid catalysis, propanone reactswith iodine as follows:

(CH3)2CO(aq) + I2(aq) ~ CH2ICOCH3(aq) + HI(aq)50 cm3 of 0.02 M I2(aq) and 50 cm3 of acidified 0.25 Mpropanone were mixed together. 10 cm3 portions of the reactionmixture were removed at 5 minute intervals and rapidly added toan excess of 0.5 M NaHC03(aq). The iodine remaining wastitrated against aqueous sodium thiosulphate.

C')

E(.) 20.......~c.g

19::J'0In.sco 18.l:a.:;In0£ 17-0Q)

E 16::J'0>

10 20 30Fig.11. time,t/min

The graph (Fig. 11) records the volume of aqueous sodiumthiosulphate required to react with the iodine remainingat different times after mixing the reactants.(a) Why are the 10 cm3 portions of the reaction mixture added

to aqueous sodium hydrogencarbonate before titration withsodium thiosulphate?

(b) How does the rate of change of iodine concentration varyduring the experiment?

(c) What is the rate of reaction in terms of cm3 of sodiumthiosulphate per minute?

(d) What is the order of the reaction with respect to iodine?(el Suppose the reaction is first order with respect to

propanone. What would be the rate of reaction (in cm3 ofsodium thiosulphate min-1) if 0.50 M propanone were usedinstead of 0.25 M propanone?

(f) Indicate by means of a sketch how the volume of aqueoussodium thiosulphate used would vary with time if nocatalytic acid was present in the reacting mixture ofpropanone and iodine.

(g) Explain your answer to (fl.(Answers on page 83 1

28

Finally, in this section, we present an exercise where you are giventhe rate equation and the rate constant. for a reaction, at aparticular temperature, and asked to calculate the rates at differentinitial concentrations of reactants.

Exercise 21 Hydrogen and iodine react together to produce hydrogeniodide:

The rate equation for the reaction is:rate = k [Ha(g)l[Ia(g)]

)and the rate constant at 374°C is 8.58 X 10-5 mol-1 dm3 S-l

(a) Work out the rate of reaction at each of the followinginitial concentrations of hydrog~n and iodine6 inexperiments A6 Band C.Table 8

Initial [H2] Initial [12]Expt. Imol dm-3 Imol dm-3

A 0.010 0.050

B 0.020 0.050

C 0.020 0.10

(b) From your answers to (a), what can you say about the effectof concentration of reactants on reaction rate?

(Answers on page 83 1

COMPUTER-AIDED REVISIONAt this pointl if it is available, you should use the ILPAC computerprogram 'Chemical Kinetics'.

The program allows you to recall and practice much of what you have learnedin Level One, including graphical techniques, finding orders of reaction,rate equations and rate constants. It should be good revision before you doyour Level One Test. However, you should not attempt the questions aboutactivation energy until you have done the relevant work in Level Two.

You do not need to know anything about. computing but, if you havenot used a program before, you will need a few minutes of instructionon getting started and using the keyboard.

29

LEVEL ONE CHECKLISTYou have now reached the end of Level One of this Unit. The following is asummary of the objectives in Level One. Read carefully through them andcheck that you have adequate notes.

At this stage you should be able to:

(1) calculate values for the rate of reaction at various timesl from agraph of concentration of reactant against time;

(2) write an equation to show the relationship between rate of reaction andconcentration of reactant;

(3) identify the order of reaction with respect to a named reactant;(4) determine the overall order of reaction;(5) work out a value for the rate constant of a first order reactionl

including its units;(6) recognise a first order reaction from a graph of rate against concent-

ration of reactant.(7) use a given rate equation to identify the order of reaction with

respect to an individual reactant;(8) use a given rate equation to identify the overall order of reaction;(9) define order of reaction with respect to an individual reactant and

overall order of reaction.(10) choose suitable methods for measuring the rates of different types of

reaction;(11) describe in outline each of the main methods for measuring reaction

rate.(12) define the term half-life;(13) determine the half-life of a first order reaction from graphical data

of concentration versus time;(14) identify a reaction with a constant half-life as being first order.(15) determine the order of reaction with respect to individual reactants

from initial rate data;(16) determine the rate equation for the iodination of propanone by

following the reactions colorimetrically.

LEVEL ONE TESTTo find out how well you have learned the material in Level Onel try ~.the test which follows. Read the notes below before starting. •

~\\\1. You should spend about 1~ hours on this test.2. You will need a sheet of graph paper and a ruler for this test.3. Hand your answers to your teacher for marking.

30

LEVEL ONE TEST ~\;gIn Questions 1, 2 and 3, one or more of the suggested responses is correct.Answer as follows:A if only 1, 2 and 3 are correct8 if only 1 and 3 are correctC if only 2 and 4 are correct0 if only 4 is correctE if some other response or combination is correct.

1. The extent of a reaction may be determined without interferingwith any of the reacting materials by1 measuring the optical rotation2 measuring the electrical conductance3 collecting any gas evolved in a gas syringe4 titrating against a suitable standard solution.

2. Hydrogen peroxide oxidizes hydriodic acid according toH202(aq) + 2H+(aq) + 2I-(aq) ~ I2(aq) + 2H20(1)

The rate of the reaction could be investigated-by measuringchanges in the1 volume of the system2 intensity of colour of the system3 optical rotation of the system4 electrical conductivity of the system.

( 1 )

(1)

3. The rate of the reaction between sodium hydroxide solution andethyl propanoate

C2H5C02C2H5(1) + OH-(aq) ~ C2H5C02-(aq) + C2H50H(aq)is first order with respect to both ethyl propanoate and sodiumhydroxide. It is true to say that1 halVing the concentration of either of the two reactants independ-

ently will halve the rate of reaction2 the reaction is first order overall3 the rate of reaction is proportional to the concentration of ethyl

propanoate4 halVing the concentrations of both reactants simultaneously will

halve the rate of reaction. (1)

31

4. The rate of reaction between X and Y is third order overall.Which of the following rate equations must be INCORRECT?A Rate k[XJ[y]3

B Rate k[XJ[Y]2

C Rate k[X]2[y]

0 Rate k[X]2[y][Z]O

E Rate k[X]O[y]3

5. The rate of decomposition of a compound X in aqueous solution isgiven by: rate = k[x12• The units of k areA mol dm-3 S-l

B mol -1 dm3 s -1

C mol -1 dm6 s -2

0 mol -2 dm6 s -2

E mol -2 dm6 s -1

6. For the reaction X + Y ~ Z. the rate expression is1Rate = k[X]2[Y]2

If the concentrations of X and Yare both increased by a factorof 4, by what factor will the rate increaserA 4

B 8

C 16

o 32E 64

7. The kinetics of the reaction between Mn04- ions and C2042- ions

in acidic solution is to be investigated.2Mn04-Caq) + 16H+Caq) + 5C204

2-Caq)~ 2Mn2+Caq) + 8H20Cl) + 10C02Cg)

In addition to a stop clock, which piece of apparatus would enablethe rate of reaction to be followed most easily and accurately?A DilatometerB Gas syringeC Polarimetero Standard electrode and valve voltmeterE Colorimeter.

32

C 1)

C 1)

C11

C 1 )

8. The reaction between potassium iodide and potassium peroxodi-sulphate(VI), K2S20e, in aqueous solution proceeds accordingto the overall equation

2KI(aq) + K2S20e(aq) = 2K2S0~(aq) + 12The rate of this reaction is found by experiment to be directlyproportional to the concentration of the potassium iodide, and directlyproportional to that of the potassium peroxodisulphate(VI).(a) Write the above equation in ionic form. (1)(b) What is the overall order of the reaction? (1)(c) Show the meaning of the term velocity constant by writing an

equation for the rate of this reaction, such that the concent-ration of the peroxodisulphate(VI) ions decreases with time. (2)

(d) With an initial concentration of potasslum iodide of1.0 x 10-2 mol dm-3 and of potassium peroxodisulphate(VI) of5.0 x 10-~ mol dm-3, it is found that at 298 K the initial rateof disappearance of the peroxodisulphate(VIl ions is1.02 x 10-e mol dm-3 S-l. What is the velocity constant of thereaction at this temperature? (2)

9. When a dilute aqueous solution of a benzenediazonium salt ismaintained at 35°C, it slowly decomposes evolving nitrogen.The equation for the reaction is

C6HsN2+(aq) + H20(I) ~ C6HsOH(aq) + H+(aq) + N2(g)The course of the reaction may be followed by collecting the gas evolvedand measuring its volume at known times. Three of the substancesrequired for the preparation of the solution of the benzenediazoniumsalt are phenylamine, sodium nitrite and water.(a) Name an additional reagent which would be required for the

preparation of the benzenediazonium salt solution. (1)-(b) State one condition which is necessary for the preparation of

the benzenediazonium salt solution. (1)(c) In order to follow the rate of decomposition, why is it

necessary to stir well while the nitrogen gas is being evolved? (1)(d) Sketch, using the axes given below, the shape of the graph

obtained by plotting the volume of nitrogen gas collected againsttime since the reaction began.

otime

(1 )

(e) Describe how a set of results for such an experiment, showingvolumes of gas obtained at various times since the reactionbegan, could be used to show that the hydrolysis reaction isfirst order with respect to benzenediazonium ion. (3)

33

10. In an experiment designed to investigate the effect of acidconcentration on the reaction between an acid and a metal, alarge excess of hydrochloric acid of concentrations varying from0.2 M to 1.0 M reacts with a fixed length of magnesium ribbon(5 cm in each case). The reaction times required fo~ the evol-ution of 15 cm3 of hydrogen gas are determined at a constant pressureof 1 atm and at 300 K.The following table summarises the results obtained by a group of sixth-form students.

Time required for theConcentration of evolution of 15 cm3 ofhydrochloric acid hydrogen at 1 atm and/mol dm-3 300 K/seconds

0.25 160

0.35 80

0.50 40

0.70 20

1.00 10

(a) Suggest a reason why the experiments were carried out with alarge excess of hydrochloric acid. (1)

(b) Sketch out an assembly of apparatus which might have been usedfor carrying out the above experiments. (2)

(c) (i) Consider the data in the table. Give a relationshipbetween reaction time t and the concentration of hydro-chloric acid. ( 1)

(iil What is the order of the reaction with respect to hydro-chloric acid? (1)

(d) Since hydrochloric acid is a strong acid, it contains hydrogenions and chloride ions only. Also, as a result of this

Let n denote the overall order of the reaction between magnesiumand hydrochloric acid.

Three rate equations are possible:(i) Rate

(ii) Rate(iii) Rate

k[H+ Ink [ H + In -p [ C1 - JPk[Cl-ln

Devise a set of investigations which would enable you to decidewhich one of the above rate equations is correct. (5)

34

11 . Dinitrogen pentoxide (N20S) decomposes according to the equation2N20S(g) -+ 2N204(g) + °2(g)

The reaction proceeds at a rate which is conveniently measurableat a temperature of 45°C; at this temperature the followingresults were obtained.

Rate of disappearance[N20s1/mol dm-3 of N20s/mol dm-3 S-l

22.3 x 10-3 11.65 x 10-6

17.4 x 10-3 8.67 x 10-6

13.2 x 10-3 6.63 x 10-6

9.5 x 10-3 4.65 x 10-6

4.7 x 10-3 2.35 x 10-6

(a) Plot these results on a suitable graph.(b) The rate law for the reaction can be expressed in the form

rate k[N20SJx(i) What is the value of x as deduced from the graph?

( 4 )

( 1)

(ii) What is the value of k at 45°C? In what units is itexpressed? (2)

(c) It can be calculated from the figures given that the concent-ration would fall from 2 x 10-2 mol dm-3 to 1 x 10-2 mol dm3

"in 1386 seconds. How much longer would it take to fal) to2 .5 x 10-3 mo I dm -3? ( 2 )

(d) It will be seen from the equation that two N20S molecules giverise to three molecules of products. (You may neglect dissocia-tion of N204 at this temperature.)(i) Indicate how in principle you would use this fact to

follow the extent of decomposition of N20S with time. (2)(ii) Sketch a simple apparatus that could be used for this

purpose. (2)

(Continued on page 36.)

35

12. The table shows the results of an investigation into the ratereaction between hydroxide ions. OH-, and p~osphinate ions,PH202-, at 80°C. The overall equation for the reaction is:

PH202-(aq) + OH-(aq) ~ PH032-(aq) + H2(g)

Initial concentrations Initial rateof hydrogen

[PH202 -(aq) J [OH -(aq) ] productionExperiment Imol dm-3 Imol dm-3 jcm3 min-1

1 0.6 1 .0 2.42 0.6 2.0 9.63 0.6 3.0 21.54 0.1 6.0 14.45 0.2 6.0 28.86 0.3 6.0 43.2

(a) How does the initial rate of this reaction depend on theconcentration of hydroxide ions?

(b) How does the initial rate of this reaction depend on theconcentration of phosphinate ions?

(c) What is the rate expression for this reaction?(d) In what units will the rate constant be expressed?(e) Is the hydroxide ion acting as a base in this reaction?

Justify your answer.

of ~

(1)

( 1 )

(2)(1 )

(2)(Total 50 marks)

36

LEVEL TWOIn the next section we show how the rate equation for a chemical reactioncan be used to work out a possible mechanism. Proposing a mechanism involvessuggestingl at a molecular level, the route by which the reaction occurs.

REACTION MECHANISMSObjectives. When you have finished this section you should be able to:

(17) explain why a reaction mechanism can be worked out only from the rateequation for a reaction, not directly from the stoichiometric equation;

(18) state that most chemical reactions take place in a series of steps,each called an elementary step or reaction step;

(19) explain the meaning of the term molecularity;(20) distinguish between the molecularity of a reaction step and the order

of reaction as shown by the rate equation;(21) identify the slowest step in a reaction as the rate-determining step;(22) suggest a simple mechanism for a reaction from the rate equation and

vice versa.

Molecularitystart by reading about molecularity and rate-determining step in yourtext book(s). LooK in the index under 'reaction mechanism' or'mechanism of a reaction'. Make sure that you understand thedifference between molecularity and order of reaction and that theterm molecularity can strictly be applied only to a reaction stepand not to the overall reaction.

oCO

A reaction with more than three reacting species is unlikely to take placein a single step. The probability of even three particles colliding andreacting instantaneously is very small. We assume that most of thesereactions which proceed at a measurable rate occur in two or more separatesteps. The series of reaction steps, taken together, is the reactionmechanism, or reaction pathway as it is sometimes called.

In the next exercise you consider the mechanism for the iodine-propanonereaction whose rate equation you have determined experimentally and foundto be: Rate = k[CH3COCH3][H+l

37

step 2 FastH-C==C-CH + H t

I I 3

H O-H

A suggested mechanism for the iodination of propanone is:~lS:

H H ~~I I @

H-C-C-CH + H+ -+ H-C-C-CH step 1 Slow~ g -. ~ ~ 6-H 3

HI,@H-C-C-CH

I I 3

H O-H

Exercise 22

H-C==C-CH + 12 -+I I 3

H O-H

T f'H-C-C-CHI I 3

H O~H

I II IH-C-C-CHI I 3

H O-HIIH-C-C-CH + H+ + I-I II 3H 0

step 3 Fast

step 4 Fast

(a) Write down the molecularity of each step.(b) Does the mechanism show H+(aq) acting as a catalyst?(Answers on page 84 )

Notice that in the above mechanism the first step is slow whilst the othersare fast. This slow step is called the rate-determining step andinformation concerning it is obtained from the rate equation.

The rate-determining stepWhenever a process takes place in several steps. the overall rate of theprocess depends on the speed of the slowest step. There are lots ofsituations like this; for example, driving along a busy motorway whichconverges into a single lane and then opens up again into three lanes.Consider Fig. 12 below and imagine you want to travel from A to O.

~

r:07l~-- -

- --+-0

B

~v-ffj)~

~ 0- .[2J.D~~~.--

A

+- -

Fig.12.

If at A your car is travelling at 70 m.p.h. you will reach point B quicklybut getting past the lane closure can take a long time. Once your car hasreached C it can accelerate again and arrive at 0 shortly afterwards·. In thisanalogy the stages correspond to the elementary steps in a reaction mechanism.

In a more realistic comparison with a three-step chemical reaction yourspeed from A to B would be so great that the distance A - B is immaterial,and it would take several days to get past the lane closure. Thus, gettingpast the lane closure is the rate-determining step, i.e. it determines theoverall rate of travel from A to D.

38

You may disagree with this analogy. It, may not represent all the steps; itcan only be suggested, then tested. The same is true for mechanisms ofchemical reactions. Keep thinking about the idea of rate-determining stepsin odd moments over the next few days. Can you suggest a better analogythan the one we describe above?

Now we look at the ways in which we decide what might be the rate-determiningstep in a chemical reaction.

Proposing a reaction mechanism'In any reaction mechanism of the type that you are likely to meet, there isusually one step which acts as a bottleneck, the rate-determining step forthe reaction. The only way to find out which molecules and how many of themare involved in the rate-determining step is to look at the rate equationfor the reaction.

We now take you through a Worked Example to show you how to identify the rate-determining step and to suggest a mechanism for a reaction.

Worked Example. Hydrogen and iodine monochloride react as shown bythe equation:

2ICICg) + H2Cg) ~ 2HCICg) + I2Cg)The rate equation for the reaction is:

rate =. k[IClCg)][H2Cg) 1

Suggest a possible mechanism for the reaction.

Solution1. Decide which molecules Cspecies) are involved in the rate-determining

step:The rate equation shows that the reaction is of the first order withrespect to ICI and first order with respect to H2• The rate-determiningstep is likely to involve ICI and H2 in a collision.

IClCg) + H2Cg) ~ products2. Suggest possible products from the reactants you have chosen:

CThese products may include species which do not appear as products inthe stoichiometric equation.)The most probable products for the reaction between ICI and H2 are HCland HI. Thus we can write the equation for the rate-determining step:

IClCg) + H2(g) ~ HClCg) + HICg) SLOW3. Compare the rate-determining step with the stoichiometric equation to

see how many molecules are still to be accounted for:We have to account for another molecule of ICI. HI is not a finalproduct so a second step might be:

ICICg) + HICgJ ~ HCICg) + I2Cg) FAST

39

4. Add up the reaction steps:Intermediate products and reactants should cancel, so that the steps,taken together, contain the same amounts of reactants and products givenin the stoichiometric equation. If they do not add up to the stoichio-metric equation your mechanism can't possibly work!Adding up the steps for this reactionl we have:~

ICl(g) + H2(g)ICl(g) + HI(g)

2ICl(g) + H2(g)

~ HCl(g) + HI(g)~ HCl(g) + I2(g)

slowfast

(overall reaction)

The separate steps do add up to give the same overall amounts of reactantsand products as the stoichiometric equation, and so this is a possiblemechanism. It must be accepted as a possibility unless and until it isdisproved as the result of further experimental work. For example, if thereaction mixture were shown to contain traces of chlorine, the above mechan-ism would not fit all the facts and a new one would have to be found.

Now try the next exercise, where we give alternative mechanisms for theabove reaction.

Exercise 23 The following mechanisms have been proposed for the ~reaction between iodine monochloride and hydrogenl but ~none of them is satisfactory. Identify the main fault(slin each mechanism. Assume all species are gaseous.(a) ICI + H2 -+ Hel + HI slow

HCl + ICI -+ HI + C12 fastC12 + HI -+ HCl + leI fast

(b) ICI + ICI -+ 12 + C12 slowC12 + H2 -+ 2HCl fast

(c) ICI -+ I- + Cl- slowI- + H2 -+ HI + H- fastH· + ICI -+ HI + Cl· fastCl- + Cl- -+ C12 fastHI + HI -+ 12 + H2 fastC,12 + H2 -+ 2HCl fast


In the next exercise you refer back to the proposed mechanism for theiodination of propanone which appeared in Exercise 22.

40

Exercise 24 study the mechanism for the iodination of propanone ~given in Exercise 22 (page 38) and answer the followingquestions. ~~(a) Identify the rate-determining step.(b) What is the relationship between the molecularity of the

rate-determining step and the rate equation?(c) Add up the reaction steps and show that the overall

reaction agrees with the stoichiometric equation.(Answers on page 84 )

In Exercises 16, 18 and 19 in Level One, which were part A-level questions,you worked out rate equations for three reactions from data. We now addanother part of each question, which asks you to suggest a mechanism foreach reaction.

Exercise 25

Exercise 26

Given that the reaction: 2N20S ~ 4N02 + 02is first order, suggest a mechanism to account for this.(Answer on page 84 )

For the reaction: A + 3B ~ ABsthe rate equation is: rate = k[A][B]2Suggest a possible mechanism for the reaction betweenA and B, leading to the formation of ABs.(Answer on page 84 )

Exercise 27 For the reaction 2H2Cg) + 2NOCg) ~ 2H20(g) + N2(g)the rate equation is: rate = k[H2J[No12

Suggest a possible mechanism which is consistent withthe rate equation.(Answer on page 84

Now that you have had some practice in working out mechanisms for reactions,we go on to consider one example in .more detail, the hydrolysis of halogeno-alkanes.

41

The hydrolysis of halogenoalkanesYou have already studied the hydrolysis of primary, secondary and tertiaryhalogenoalkanes in Unit 02 (Some Functional Groups) and considered themechanism for the hydrolysis of a primary halogenoalkane.

Fig.13.

H", 0" 0-

H-/'C -X -H

H H\'/I

HO··· ·C···· XI

Hactivated complex

H

-HO-C~H + X-H

We study this mechanism in the light of its rate equation and consider themechanisms for the hydrolyses of secondary and tertiary halogenoalkanes.


(23) use the rate equation for the hydrolysis of a halogenoalkane to workout a mechanism for the reaction;

(24) explain the effect of the carbon skeleton on the rate of hydrolysis ofhalogenoalkanes in terms of the stability of carbonium ions.

Many experiments have been carried out to investigate the hydrolysisreactions of halogenoalkanes. In the next exercise we present initial-ratedata from a series of experiments carried out on bromomethane and 2-bromo-2-methylpropane. Use the data to work out the rate equation for the hydrolysisof each compound.

Exercise 28 Samples of bromomethane and 2-bromo-2-methylpropane were~dissolved in dilute aqueous ethanol and reacted withsodium hydroxide solution. Several experiments were ~~carried out, at constant temperature. using differentinitial concentrations of each bromoalkane and hydroxideions. The initial rate of reaction was determined in each case.Use the data to establish a rate equation for each reaction.Table 9 Data for the hydrolysis of bromomethane, CH3Br

[CH3Brl [OH-] RateExperiment Imol dm-3 Imol dm-3 Imol dm-3 s -1

A 0.010 0.0050 0.107

B 0.010 0.010 0.216

C 0.010 0.020 0.425

D 0.020 0.020 0.856

E 0.040 0.020 1 .72

F 0.080 0.020 3.42

42

Table 10 Data for the hydrolysis of 2-bromo-2-methylpropane,(CH313CBr

[(CH3)3CBr] [OH -1 RateExperiment Imol dm-3 Imol dm-3 Imol dm-3 S-l

A 0.020 0.010 20.2

B 0.020 0.020 20.1

C 0.020 0.030 20.2

D 0.040 0.030 40.1

E 0.060 0.030 60.2

F 0.080 0.030 80.0

(a) Use the data to establish a rate equation for each reaction.(b) Identify the rate-determining step in the mechanism for the

hydrolysis of bromomethane shown in Fig.13 .(c) Propose a mechanism for the hydrolysis of 2-bromo-2-

methylpropane. (Hint: the mechanism involves the form-ation of a carbonium ion.)

(d) Why do you think the mechanism for the primary halogeno-alkane is labelled SN2 and the mechanism for the tertiaryhalogenoalkane is labelled SN1?


You should now make a space-filling model of the tertiary halide and ?use it to do the next short exercise. ~

Exercise 29 What feature of the structure of (CH3)3CBr makes abimolecular mechanism for its hydrolysis unlikely?(Answer on page 84 )

In Unit 01 (Hydrocarbons) you learned that tertiary carbonium ions are morestable than primary carbonium ions. We can therefore conclude that tertiaryhalogenoalkanes hydrolyse via the SN2 mechanism partly because of the stericproblems associated with the 'activated complex' and partly due to thegreater stability of the tertiary carbonium ion compared to that of theprimary carbonium ion.

The kinetics of hydrolysis of secondary halogenoalkanes are intermediatebetween those for primary and tertiary halogenoalkanes. The rate equationis more complex than the examples you have met so far and you are notexpected to know it for most A-level syllabuses. We give it below forcompleteness and draw certain conclusions from it.

Rate = k1[RX] + k2[RX][OH-Jwhere RX is the halogenoalkane. The relative importance of the two termsk1[RXJ and k2[RX][OH-] depends on the reaction conditions and the structureof the halogenoalkane. Thus secondary halides may react according to eitherSN1 or SN2 mechanisms depending on the conditions.

43

If it is available, you should now watch part 2 of the ILPAC video- It IIprogramme 'Chemical' Kinetics' which deals with mechanisms, includingthose of the halogenoalkane hydrolyses. 00

The next exercise is an A-level question based on a different reaction. Youare given data and asked to use these to determine the order of reaction andcomment on its mechanism.

Exercise 30 In the Journal of the Chemical Society for 1950, Hughes, ~Ingold and Reed report some kinetic studies on aromatic Anitration. In one experiment ethanoic acid containing ~0.2% water was used as solvent and pure nitric acid wasadded to make a 7 M solution.The kinetics of the nitration of ethylbenzene, C6H5C2H5, werestudied with the following result at 20°C.Table 11

Time Concentration of/min ethylbenzene/mol dm-3

0 0.090

8.0 0.063

11.0 0.053

13.0 0.049

16.0 0.037

21.0 0.024

25.0 0.009

(a) Determine an order of reaction from these results.(b) What does this suggest about the mechanism of the reaction?(c) What products are likely?(Answers on page 85 )

We now go on to consider the effect of temperature on the rate of reaction,and introduce the important concept of activation energy.

ACTIVATION ENERGY AND THE EFFECT OF TEMPERATURE ON REACTION RATEThe rate of many gaseous and aqueous reactions often doubles for a temper-ature rise of only 10 K. Why is it that such a small rise in temperaturecan cause a large percentage increase in the reaction rate? We consider thequestion in the Tollowing section, which also helps to explain why somereactions, which we might expect to proceed on the basis of values of ~ or6c&, do not appear to occur.

44


(25) state the meaning of the term activation energy and identify it on anenergy profile diagram for a reaction;

(26) explain the effect of increased temperature on the rate of reaction interms of the fraction of molecules with energy greater than theactivation energy;

(27) state the Arrhenius equation~which shows how the rate constant~ k~changes with temperature;

(28) use a graphical method to obtain values for the activation energy fromthe Arrhenius eguation.

Read the section on activation energy in your text-book(s). Look for 0energy profile diagrams indicating activation energy. This will help ~Iyou to answer the exercises which follow. ~I

Consider the following reaction:CeH18(1) + 12~02(g) ~ 8C02(g) + 9H20(g); 6H = -5498 kJ mol-1

This reaction is highly exothermic and can occur with explosive force underthe right conditions. However~ it does not take place at room temperature!

There are many other examples like this, where there is an 'energy barrier'to be overcome before reaction can take place. This barrier is called theactivation energy for the reaction. Since collisions between reactantparticles are always occurring it seems reasonable to assume that particlesdo not always react when they collide. A reaction occurs as a result ofcollisions between particles which possess more than a certain minimumamount of energy - the activation energy~ Ea (sometimes EA).

The next exercise is concerned with the energy barrier and the forces whichneed to be overcome before a reaction can occur.

Exercise 31 Suggest two reasons why molecules must reach a certainenergy before reaction between them is possible.(Answers on page 85 )

An energy profile diagram is a way of showing how the energy of reactingmolecules changes. To ensu~e that you can distinguish between the activationenergy and the enthalpy change for a reaction, try the next exercise.

(a)Exercise 32 Draw an energy profile to represent the reaction: ~2N20(g) ~ 2N2(g) + 02(g); ~~ = -164.0 kJ mol-1

~\The activation energy, Ea, for the forward reactionis 250 kJ mol-1• Use graph paper and let 1 cm represent50 kJ mol-J..

(b) Use your energy profile diagram to calculate the activationenergy for the back reaction.


45

Next we go on to consider what proportion of particles in a gaseous reactionmixture reach the activation energy and how the proportion is affected bytemperature.

The fraction of particles with energy greater than Ea

In Unit P1 (The Gaseous state) you learned about the Maxwell distributioncurves for a sample of gas at different temperatures. You then used thesecurves to calculate the fraction of molecules with kinetic energies in acertain range at a particular temperature. For our purposes, in this Unit,it would be useful to know the fraction of molecules with energy greaterthan the energy of activation, Ea.

Consider Fig. 14 below, which shows the distribution of kinetic energies ofparticles in the gas phase and the activation energy, Ea.

>-e>Q)cQ)

cQ)>'tj)co£'§(/)Q)

:;uQ)

"0E'0co

'~

~-Fig.14,

kinetic energy ..

m fraction of molecules withenergy> E a at a particular

temperature

number of particles with E > Eatotal number of particles

The area beneath the curve is proportional to the total number of particlesinvolved and the shaded area under the curve is proportional to the numberof particles with energy greater than Ea. Hence the fraction of particleswith energy greater than E is given by the ratio:a

shaded area under curvetotal area under curve

Maxwell and Boltzmann derived a useful mathematical expression for thisfraction in terms of the activation energy, Ea, the gas constant, R, and theabsolute temperature, T:

Fraction of particles with energy> Ea -Ea/RTe

The following Worked Example shows you how to use this expression tocalculate the number of molecules with energy greater than a given energy.

Worked Example. Calculate the number of molecules in 1.0 mol of gasat 25°C with energy greater than 55.0 kJ mol-1•

46

Solution1. Calculate the fraction of molecules with energy greater than

55.0 kJ mo 1-1•

Fraction of molecules with E > 55.0 kJ mol-1 = e-E/RT whereR = 8.31 J K-1 mol, T = 298 K and E 55000 J mol-1•

-(55000 J mol-1)/(8.31 J K-1 mol-1)(298 K)Fraction of molecules e= 2.28 X 10-10

2. Calculate the total number of molecules with energy greater than55.0 kJ mol-1•

The total number of molecules present in 1.0 mol of gas is given by:L = 6.02 X 1023 mol-1 where L is the Avogadro constant.

with E > 55.0 kJ mol-1

of moleculesnumber of molecules

total numberThen number with E > 55.0 kJ mol-1 L e-E/RT

6 .02 x 102• ma 1 -1 X 2.28 x 10 -1 0 ~ 1-1-.-3-7-X-1-0-J.-jj.-m-o-l---1-1

Since fraction of moleculeswith E > 55.0 kJ mol-1

Now you do a similar calculation at a different temperature.

Exercise 33 (a)

(b)

Calculate the number of molecules in 1.0 mol of gas~at 35 °c with energy greater than 55.0 kJ mol-1•

Compare this value to that worked out at 298 K. ~~(Answers on page 85 1

You have just calculated that for a rise in temperature of only 10 K twiceas many molecules exceed the energy barrier of 55.0 kJ mol-1• Thus, on theMaxwell distribution curves the shaded area under the (T + 10) K curve willbe twice that on the T K curve (where T is the absolute temperature) asshown in Fig. 15.

>0>(j)CQ)

CQ)>'0,ro

-;:;'~

CIlQ)

:i(.)Q)

"0E-oco

'~

~-Fig.15.

TK

kinetic energy •

fraction of molecules withenergy> E a at TK

fraction of molecules withenergy> E a at (T + 10)K

We now go on to consider the relationship between the rate constant, k, fora reaction and the fraction of particles with energy greater than Ea·

47

THE ARRHENIUS EQUATIONFor particles to react~ their energy must exceed Ea for that reaction. Thissuggests that at a given temperature

-E IRTRate ~ e a ...•......... (1)

Since the rate changes during the progress of a reaction~ it is more usefulto use k (rate constant).

Consider the general reactionA(g) + B(g) ~ productsRate = k[A]X[BJY at a room temperature T~

If the experiment is repeated using the same concentrations of A and B at ahigher temperature~ T2~ the rate increases. Since [A]X and [B]X have the sameinitial value then k must increase. Therefore~ we can say that the rateconstant is a general measure of the reaction rate at a particular temperature.Equation (1) becomes:

k ~ e-EaIRT

-EaIRTk Ae ••••••••••••••• ( 2 )

A is a constant~ sometimes called theArrhenius constant or pre-exponentialconstant. Equation (2) is oftencalled the Arrhenium equation~ namedafter the Swedish chemist~ SvanteArrhenius~ who formulated it in 1880.

Using the Arrhenius equation Fig.16. S.A. Arrhenius (1859-1927)

log )e(In

The equation provides an extremely useful means of getting values for theactivation energy and the pre-exponential factor for a reaction. It isusually changed to a logarithmic form to make it more manageable:

k = Ae -EaIRT

Ink lnAe -EaIRT

1 nA + 1 ne -.EaIRT

InA EalRT

Changing this to the log~o form:Ealog ~0k = log ~oA - 2.3 RT.. • • • • • • • • • • • • • ( 3 )

Equation (3) is often quoted for you in A-level questions.

The next exercise deals with equation (3).

48

(bJ

(a JExercise 34 Which of the terms in equation (3) are variables ~for a particular reaction?Compare equation (3) to the equation for a straight ~~line, i.e. y = mx + c. Which of the terms inequation (3) are analogous to y, m, x and c?

(c) Sketch the shape of the graph you would expect to obtainby plotting log 10k against 1/T.

(d) Show how you could calculate A and Ea from your graph.(Answers on page 85 )

Use the method you have just devised in the next exercise.

Exercise 35 When gaseous hydrogen iodide decomposes in accordancewith the equation:

2HI(g) ? H2(g) + I2(gJthe reaction is found to be second order with respect tohydrogen iodide. In a series of experimentsl the rate constantfor this reaction was determined at several different tempe-ratures. The results obtained are shown in the table below.Table 12

Temperature Rate constant (kJ(T)/K /dm3 mol-1 S-1

633 1 .78 X 10-5

666 1.07 x 10-4

697 5.01 x 10-4

715 1.05 x 10-3

781 1 .51 x 10-2

Arrhenius deduced that for a reaction:

logk E2.3 RT + logA

where R is the gas constant and A is a constant.(a) Use the data given to determine the activation energy E

for the reaction.(b) Determine by what factor the rate increases when the

temperature rises from 300 K to 310 K.(You may wish to use the alternative form of the Arrheniusequation

k c A 10 -E /2 •3RT )

(c) The kinetic energy of a fixed mass of gas is directlyproportional to its temperature mea~ured on the Kelvinscale. Calculate the ratio of the kinetic energies of afixed mass of gas at 310 K and at 300 K.

(d) Compare this ratio with the increase in the rate ofreaction determined from experimental data. Discuss thesignificance of this result.


49

In the next experiment you determine the activation energy for theoxidation of iodide ions by peroxodisulphateCVI) Cpersulphate) ions.The experiment is an example of a 'clock' reaction, in which youmeasure the time taken for a reaction to reach a certain stage.You will find more details on 'clock'reactions in Appendix Two of this Unit.

EXPERIMENT 3Determining the activation energyof a reaction

AimThe purpose of this experiment is todetermine the activation energy, Ea,for the reduction of peroxo-disulphate(VI) ions, S20a2-Caq), byiodide ions, I-Caq), using a'clock' reaction.

IntroductionThe equation for the reduction ofperoxodisulphate(VI) ions by iodide ions is:

S20a2-Caq) + 2I-Caq) ~ 2S042-Caq) + I2Caq)

A small, known amount ofthiosulphate ions is added to the reaction mixture,which also contains some starch indicator. The thiosulphate reacts with theiodine formed in the above reaction as in the following equation:

2S2032-Caq) + I2Caq) ~ S4062-Caq) + 2I-Caq)

At the instant that all the thiosulphate has reacted, free iodine isproduced in the solution and its presence is shown by the appearance of theblue-black colour of the iodine-starch complex, i.e. the thiosulphate ionsact as a 'monitor' indicating the point at which a certain amount of iodinehas been formed. For this reason the reaction is often referred to as aniodine 'clock'reaction. In general, for a 'clock' reaction:Rate of reaction oc 1/t wheret is the time taken to reach a specified stage.You carry out the experiment at five different temperatures between about20°C and 50 °C. You then find the activation energy for the reaction byplotting a graph of log C1/t) against 1/T CT is the absolute temperature)

Requirementssafety spectaclesbeaker, 400 cm3

2 thermometers, 0-100 °cBunsen burner, tripod, gauze and mat4 burettes2 boiling-tubesclamp and standpotassium peroxodisulphate(VI) solution, 0.020 M K2S20apotassium iodide solution, 0.50 M KIsodium thiosulphate solution, 0.010 M Na2S203starch solution, 0.2%stopclock or watch

50

Procedure1. Half-fill the beaker with water and heat it to between 49°C and 51 °C.

This will be used as a water-bath.2. Using a burette, measure out 10 cm3 of potassium peroxodisulphateCVI)

solution into the first boiling-tube. Clamp this in the water-bath andplace a thermometer in the solution in the boiling-tube.

3. Using burettes, measure out 5 cm3 each of the potassium iodide and sodiumthiosulphate solutions and 2.5 cm3 of starch solution into the secondboiling-tube. Place another thermometer in this solution and stand itin the water-bath.

4. When the temperatures of the two solutions are equal and constant (towithin! 1°C), pour the contents of the second boiling-tube into thefirst, shake to mix, and start the clock.

5. When the blue colour of the starch-iodine complex appears, stop theclock and write down the time in a copy of Results Table 5.

6. Repeat the experiment at temperatures close to 45°C, 40 °c, 35°C, 30 °C.(The temperatures you use may differ from those by a few degrees but must,of course, be recorded carefully.)

Results Table 5Temperature/oC

Temperature, T/K

Time, tis

log10C1/tJ

¥K-1


Calculations1. Plot a graph of log1o (1/t) (vertical axis) against 1/T (horizontal axis).2. Use your graph to calculate a value for the activation energy.(Answers on page 86 )

The experiment you have justperformed can be automatedusing a micro-computer(see Fig. 17). Ask yourteacher for details, if youare interested in finding outhow this is done.

Fig.17.

51

In the next exercise we present data for a reaction between peroxo-disulphate(VI) ions and cobalt metal. In this case, you take the units ofk as loss in mass of metal per minute and work out a value for the activationenergy for the reaction.

Exercise 36 When strips of cobalt foil are rotated rapidly at aconstant rate in sodium peroxodisulphate(VI) solution,there is a slow reaction and the cobalt dissolves. Thereaction can be followed by removing and weighing thefoil at intervals.

Table 13Experiment 1 Experiment 2 Experiment 3at 0.5 °c at 13.5 °c at 25°C

Time Mass Time Mass Time Mass/min /mg /min /mg /min /mg

0 130 0 130 0 130

20 120 10 115 4 116

60 98 15 106 8 103

80 86 30 86 12 87

Determine a rate constant k at each temperature as a loss inmass per minute (mg min-1), and then determine the activationenergy E of the reaction using the relationship

k cc 10 -E /2 • 3RT

What difference in the results would you predict if the cobaltfoil were NOT rotated?(Answers on page 87 )

The ILPAC computer program 'Chemical Kinetics' contains a section onactivation energy. If it is available, and if you have time, youcould now run this part of the program in order to test your under-standing of the topic. Ask your teacher.

We turn now to another factor affecting the rate of chemical reactions, thepresence of catalysts.

CATALYSISA catalyst can be defined as a substance which alters the rate of a reactionwithout itself undergoing any permanent chemical change. In this sectionwe consider the effect of a catalyst on the activation energy of thereaction and the mechanism of catalysis.

52


(29) describe the difference between homogeneous and heterogeneouscatalysis and give examples of each type;

(30) explain the action of a catalyst in terms of activation energy andreaction pathways;

(31) explain what happens in an autocatalytic reaction and give one exampleof this type of reaction.

Start by reading the section on catalysis in your text-book(s). At <=2this stage, read rapidly through the whole section, keeping the above ~Iobjectives in mind as a guide. You will need to read some parts in ~more detail later.

You have already met several reactions which are catalysed. In thenext exercise you recall some examples of catalysed reactions and classifythe catalysts involved as homogeneous or heterogeneous. If the reactantsare in a different physical state from the catalyst then the catalyst isclassed as a heterogeneous catalyst. If the reactants and catalyst are inthe same physical state then it is classed as a homogeneous catalyst. Usethis information to answer the following question.

Exercise 37 Choose six reactions that you have studied so far in ~your course which are catalysed. Write your answer inthe form of a table with three columns, classifying your~~catalyst as homogeneous or heterogeneous in the thirdcolumn.(Answers on page 87 )

In the iodination of propanone you learned that the H+ ions which catalysethe reaction do not appear as reactants in the stoichiometric equation butthey do appear in the rate equation. This is generally the case if thecatalyst is involved in the rate-determining step of the reaction. However,catalysts' are not always involved in the rate-determining step and in suchcases do not usually appear as concentration terms in the rate equation.

In the next section, we consider the way in which a catalyst can affect therate of reaction without necessarily being involved in the rate-determiningstep.

Catalysts and activation energy

Most catalysts are thought to operate by providing an alternative reactionpathway with a lower activation energy. The catalysed reaction proceeds bya different mechanism, which may involve more reaction steps than theuncatalysed one, but it takes place more rapidly because each step has a loweractivation energy than that of the uncatalysed reaction.

53

It's a bit like crossing the Alps from Italy to France. One way is on foot- climbing up glaciers with ropes and crampons, and this requires a hugeamount of energy. This is the uncatalysed route. The other way is to gounder Mont Blanc in your car, using the road tunnel - a different route,which is faster.

Fig.18. -

Chamonix

In the next exercise, you interpret an energy profile diagram for the decom-position of hydrogen iodide, both catalysed and uncatalysed.

Exercise 38 The decomposition of hydrogen iodide is catalysed by ~platinum metal. The following diagram, Fig.19,represents the energy changes that take place during the ~~course of the uncatalysed reaction, with the changesduring the catalysed reaction superimposed.

,

'T, , .•..

, ".• T:.

·:·:r::·.; -I··! .,;I .1..;.

"........ '

:i

: ;

:.;.. :. 1-: i··c.:·1

Fig.19. progress of reaction __

Use the scale on Fig. 19 to work out the following for both theforward and back reactions:(a) the activation energies (uncatalysed),(b) the activation energies (catalysedJ,(c) the enthalpy changes.(Answers on page 87 )

54

The energy profile (Fig. 19) shows that~ by lowering the activation energy~the catalyst speeds up both the forward and back reactions.

We consider more fully the mechanism of catalysis in Unit IS (TransitionElements). HoweVer~ before ending this section~ we take a brief look atreactions where the catalyst is also a product of the reaction.

AutocatalysisReactions in which a product acts as a catalyst are called 'autocatalytic'.Once the reaction startsl more catalyst is produced~ so that the rate ofreaction increases with time for at least part of its duration. You havealready come across one example of a reaction like this in Level One - theacid-catalysed iodination of propanone:

CH3COCH3(aq) + Ia(aq) ~ CHaICOCH3(aq) + H+(aq) + I-(aq)

The ILPAC computer program 'Chemical Kinetics' has a section whichillustrates the autocatalytic nature of this reaction. Ask yourteacher if it is available.

The next exercise is about the autocatalytic oxidation of manganate(VII) ions~Mn04: by ethanedioate ions~ Ca04a-~ catalysed by manganese(II) ions~ Mna+.

2Mn04-(aq) + 16H+(aq) + 2Ca04a-(aq) ~ 4COa(g) + 2Mna+(aq) + 8HaO(1)

Exercise 39 The graph in Fig. 20 represents the reaction betweenmanganate(VII) ions and ethanedioate ions.

3.50

3.00

M

IE 2.501::l<5E 2.00vI0•.......•..

1.500-~Iv0 1.00c~

0.50

Fig.20.20 40 60 80 100 120 140 160

time/s

(a) Estimate from the graph the rates of reaction at thebeginning of the process and after 50 seconds. How do youexplain the difference in these values?

(b) The rate of reaction suddenly drops again after about 90seconds. Give a -reason for this.


55

In the next experiment, which you are expected to plan in advance, tillyou determine the activation energy for the catalysed reaction youstudied in Experiment 3. If you have studied Unit IS you will have 1discovered that this experiment is catalysed by iron(III) ions, wbichyou use in this experiment.

EXPERIMENT 4Determining the activation energyof a catalysed reaction

AimThe purpose of this experiment is todetermine the activation energy for theoxidation of iodide ions by peroxo-disulphate(VI) ions in the presence ofiron(III) ions. You then compare thevalue obtained with that for theuncatalysed reaction, determined inExperiment 3.

IntroductionBecause this is a planning experiment, we give fewer details and instructionsthan you have been used to. It is, of course, very similar to Experiment 3but you should consider carefully which concentrations of solutions to usesince the reaction is catalysed. You should also consider the effect oftemperature on reaction rate.

RequirementsMake a list of requirements including the masses and amounts needed: discussthe list with YDur teacher or technician at least a day before you want to dothe experiment.

ProcedureWork this out for yoUrse If and -keep an accurate record.

Results1. Tabulate your results in an appropriate form.2. Calculate a value for the activation energy of this reaction.(Your teacher has a set of specimen results)

QuestionCompare your result with the activation energy you calculated inExperiment 3. Comment on the two values, and discuss them with your teacher.

You should now attempt the following Teacher-marked Exercise which is ~concerned with activation energy and catalysis. ~

56


When ethanal is heated to about 800 K. it decomposesby a second order reaction according to the equation

CH3CHO(g) ~ CH~(g) + CO(g)If. however. the ethanal is heated to the sametemperature with iodine vapour, the iodine catalysesthe reaction; this latter reaction is first order withrespect to both iodine and ethanal and has a loweractivation energy than the reaction with ethanal alone.

(a) Explain the meaning of the terms which are underlined.(b) Suggest a set of experiments by which you could

confirm that the uncatalysed reaction is second order.(c) Indicate briefly what experiments would be necessary

to show that the reaction with iodine present has alower activation energy than the decomposition ofethanal by itself.

In the last two sections of this Unit, we take a look at two theories whichattempt to explain the observations obtained in chemical kinetics: (a) thecollision theory, and (b) the transition state theory.

THE COLLISION THEORYWe have been assuming a simplistic version of the collision theory through-out the Unit: before a reaction can take place, molecules must collide.

You also know, from the section on activation energy and the effect oftemperature of the rate of chemical reaction, that the colliding moleculesmust possess at least a certain energy - the activation energy - before theycan react.

In this section, we go on to consider a third factor - the orientation ofthe colliding molecules.


(32) state that, according to the collision theory, molecules must collidebefore they can react;

(33) state that before colliding molecules can react they must have anenergy equal to, or greater than, the activation energy for thereaction;

(34) recognise that it is possible, using the collision theory. to calculatethe number of 'fruitful' collisions occurring per unit time, understated conditions, for a reaction;

(35) explain the significance of the steric factor, P.

Start by reading about the collision theory in your textbook(sJ. 0Look out for terms such as collision frequency (or collision number) ~Iand collision geometry (the steric factor concerned with the ~orientation of particles).

57

You know from Level One that, for a general reaction:A + B ~ productsrate = k[AJrn[BJn

This means that, provided we choose reactions of" the same order, comparingrate constants is a way of comparing the rates of those reactions at agiven temperature.

You also know, from the section on the effect of temperature on reactionrates, that k is related to the activation energy by the Arrhenius equation:

k = Ae -Ea/RT

Arrhenius derived this equation from a study of experimental results - it isan empirical equation. The collision theory produces an almost identicalequation by means of calculations based on the theoretical model of gasconsisting of spherical particles of a certain mass and size collidingtogether. For a second order gas reaction, the collision theory gives theexpression:

k = ZO e -E a/RT ••••••.••.••.•• ( 4 )

where ZO is called the collision number and is related to the collisionfrequency of the particles. The collision frequency can be calculated froma knowledge of the number of particles in a given volume, their averagespeed and their size.

In the next exercise we give you values of the rate constants bothexperimentally determined and calculated from the collision theory for aseries of reactions for which agreement is very good. The calculated valuesare obtained from the calculated value of ZO and the experimentally deter-mined energy of activation. study the table below and answer the exercisewhich follo\.AJs.

Table 14

kCobserved] kCcalculated)Reaction T/K Imol dm-3 S-l Imol dm-3 S-l kCobs)/kCcalc)

2HI(g) ~H2Cg) + I2Cg) 556 3.5 x 10-7 5.2 X 10-7

H2Cg) + 12(g) ..•2HICg) 700 6.4 x 10-2 14.0 x 10-2

N02Cg) + CoCg) ~NoCg) + C02Cg) 500 0.55 1 .7

2NoClCg) -+

2NoCg1 + C12Cg) 300 1.5 x 10-5 g.o x 10-5

Exercise 40 Use the information in Table 14 to compare the observed ~and calculated values of k. Calculate the ratio ofkCobserved)/kCcalculated) CkCobs)/kCcalc) and enter the ~~values in the last column of the table.CAnswerson page 88 )

58

As you have found in each case, the exp~rimentally determined value issmaller than the calculated value. So how reliable is the collision theory?For many reactions, including some that take place in solution, the agreementis very poor. Agreement can be improved by refining the collision theory alittle further. As well as needing enough energy to react, molecules mustcollide with the correct orientation. For example, in the reaction betweenCO and NO,:

CO(g) + NO,(g) -+ CO,(g) + ~O(gJit appears that the carbon atom of the CO must collide with one of the oxygenatoms on the NO, molecules so that a new bond may start to form between thecarbon atom and an oxygen atom from NO,:

,~--~Fig.21.

If the carbon atom of CO were to collide with, s~, the nitrogen atom of NO"there would be no reaction. The molecules would simply bounce apart again.

~--~Fig.22. -::::-

This orientation correction is incorporated into equation (4) by an arbitraryconstant P often referred to as the steric factor. Thus we can write:

If we compare this equation with the Arrhenius equation, the constant A isequal to PZo and we can therefore obtain values for P for some reactions.In some cases, it has even been possible to see why reacting molecules ofdifferent structure give different values of P. However, since the collisiontheory does not provide an independent method of estimating P or Ea, it haslittle predictive value. Its strength lies in the simple pictorial modelwhich it provides.

We end this section on the collision theory with an exercise concerned with- the orientation of molecules.

Exercise 41 Consid~r the reaction ~2AB(g) 4> A,(g) + B,(g) ~

Assuming that the reaction occurs directly by means of simplecollisions between AB molecules (i.e. bimolecular). drawdiagrams to show the likely orientation of molecules:(a) for a 'successful' collision which will lead to reaction;(b) for an 'unsuccessful' collision ..(Answers on page 88 )

We now consider another theory of reaction kinetics: the transition statetheory. The main point ab~ut this theory is that it focuses attention onwhat happens to the reacting molecules after collision.

59

THE TRANSITION STATE THEORYObjectives. When you have finished this section you should be able to:

(36) outline the transition state theory;(37) apply the theory to a simple bimolecular reaction such as the

hydrolysis of 1-iodobutane.

The transition state theory is covered in most physical chemis~rytext-books. You should have no difficulty in finding it. Read thesection in at least one book.

In the transition state theory, attention is focused on the species existingat the point where the reactants are just about to transform into products.A species of this type is called an activated complex.

The basic assumption of the transition state theory is that the activatedcomplex is in equilibrium with the reactant molecules, i.e.:

A + B ~ AB # ~ productswhere the symbol# refers to the activated complex in the transition state.The activated complex AB*can decompose either into products or reactantsagain.

We now ]ink the transition state theory to work you have already done onenergy profiles.

Energy profiles and the transition state theoryTry the following exercise, in which you apply the transition state theoryto the hydrolysis of 1-iodobutane.

(a)Exercise 42 Draw a simple energy profile diagram to illustrate ~the mechanism of hydrolysis of l-iodobutane. Yourdiagram should not be to scale, but you can assume ~~that the reaction is exothermic. You should includeon the diagram the activated complex and theactivation energy for the reaction.

(b) What would the activated complex be for the hydrolysis of2-bromo-2-methylpropane?


The transition state theory formulates an equation which is analogous to theArrhenius equation. The advantages of the transition state theory over thecollision theory are that it automatically takes the steric factor intoaccount and it allows direct calculation of activation energies in somecases. However, this theory is also unreliable for many reasons, but you arenot expected to study it in any depth at A-level.

60

To help you focus your ideas on these two theories, we suggest youattempt the following Teacher-marked Exercise.


The collision theory and the transition state theoryrepresent two different scientific models which thechemist uses in order to interpret experimentalreaction rates. Give a concise account of each ofthese theories, and discuss the major similaritiesand/or differences that exist between them.

LEVEL TWO CHECKLISTYou have now reached the end of this Unit. In addition to what was listedat the end of Level Onel you should now be able to:

(17) explain why a reaction mechanism can only be worked out from the rateequation for a reaction, not directly from the stoichiometric equation;

(18) state that most chemical reactions take place in a series of steps,each called an elementary step or reaction step;

(19) explain the meaning of the term molecularity;(20) distinguish between the molecularity of a reaction step and the order

of reaction as shown by the rate equation;(21) identify the slowest step in a reaction as the rate-determining step;(22) suggest a simple mechanism for a reaction from the rate equation and

vice versa;(23) use the rate equation for the hydrolysis of a halogenoalkane to work

out a mechanism for the reaction;(24) explain the effect of the carbon skeleton on the rate of hydrolysis of

halogenoalkanes in terms of the stability of carbonium ions;(25) state the meaning of the term activation energy and identify it on an

energy profile diagram for a reaction;(26) explain the effect of increased temperature on the rate of reaction in

terms of the fraction of molecules with energy greater than theactivation energy;

(27) state the Arrhenius equation which shows how the rate constant, k,changes with temperature;

(28) use a graphical method to obtain values for the activation energy fromthe Arrhenius equation;

(29) describe the difference between homogeneous and heterogeneouscatalysis and give examples of each type;

(30) explain the action of a catalyst in terms of activation energy andreaction pathways;

(31) explain what happens in an autocatalytic reaction and give one exampleof this type of reaction;

(32) state that, according to the collision theory, molecules must collidebefore they can react;

61

(33) state that before colliding molecules can react they must have anenergy equal to, or greater than, the activation energy for thereaction;

(34) recognise that it is possible, using the collision theory, to calculatethe number of 'fruitful' collisions occurring per unit time, understated conditions, for a reaction;

(35) explain the significance of the steric factor, Pj

(36) outline the transition state theory;(37) apply the theory to a simple bimolecular reaction such as the

hydrolysis of 1-iodobutane.

Check that you have adequate notes before going on to the End-of-Unit Test.

END-OF-UNIT TESTTo find out how well you have learned the material in this Unit, trY~.the test which follows. Read the notes below before starting.1. You should spend about 1~ hours on this test. ~~2. Hand your answers to your teacher for marking.

62

END-Of-UNIT TEST1 .

~\;lThe rate of the reaction between peroxodisulphate ions and iodide ~ions in aqueous solution may be studied by measuring the amountof iodine formed after different reaction times. The stoichio- ~metric equation is:

S20a2-CaqJ + 2I-CaqJ ~ 2S042-Caq) + I2Caq)The graphs :show the results of a kinetic investigation of this reaction.

I I I I

f~ Concentrationofiodinej A:arbitrary units B

ITTT, f

••• c-

I'"

I~I"'"

.JII .J

•

0 1 2. 3 4 5 6.,I f

Reaction timet min

The experimental conditions were at a fixed temperature:

Initial concentration of Initial concentration ofS20a2-CaqJ/mol dm-3 I-Caq)/mol dm-3

Curve A 0.01 0.3

Curve B 0.01- 0.2

Curve C 0.01 0.1

(a) Evaluate the initial reaction rates (given as 6[I2J/6tJ for thethree curves. (3)

Cb) Ci) With respect to which reactant concentration do thesereaction rates vary?

(ii) .What is the order of the reaction with respect to thisreactant? (2)

63

(c) With reference to experimental curve A find the times required forthe completion of:

(i) one half of the reaction.(ii) three quarters of the reaction.In the light of these results what is the order of the reaction withrespect to the second reactant? (4)

(d) (i) From your answers to parts (b) and (c). suggest an overall (2)rate equation for the reaction.

(ii) What is likely to be the rate-determining step in thisreaction? (2)

(e) Investigation of the kinetics of this reaction requires iodineconcentrations to be determined. Outline a method by which thiscould be done. (3)

2. The Arrhenius equation which involves the activation energy of areaction. E. may be given in the forms

either k = Ae-E/RT or In k = In A - EIRTFor the reaction

H2(g) + I2(g) ~ 2HI(g).the values of k at 590 K and 700 K are 1.4 x 10-3 dm3 mol-1 S-l and6.4 x 10-2 dm3 mol-1 S-l respectively. (R = 8.3 J K-1 mol-1)

(a) What do the symbols k. A. R. T represent? (2)(b) Explain the meaning of the term activation energy, (2)(c) Use an Arrhenius equation and the above data to calculate the

activation energy for the reaction( 51

3. The diagram shows the distribution of the velocities of molecules 0in a gas at temperature T. 1\

J;:i

B

A

(a) The axes are labelled A and B. State what A and B represent. (2)(b) Draw another curve on (a copy of) the diagram. showing a likely

distribution at a higher temperature Pl' (2)(c) How does the diagram now afford an explanation for the effect of

increased temperature on the rate of a reaction? (2)

64

4. (a)

( b )

( c)

Homogeneous catalysts for reac~ions may be found. Define eachof the italicised words. (2)Give one example of a homogeneous catalyst and specify areaction which it catalyses. (2)

>C>Q)c:Q)

reaction co-ordinate

In the diagram above. curve 1 shows the normal path for a part-icular reaction, while curve 2 shows the path in the presence ofa catalyst. Define the energies marked I, II and III. (3)

5. An experiment was carried out to investigate the rate of reactionof an organic chloride of molecular formula C4H9CI with hydroxideions. The reaction was carried out in solution in a mixture ofpropanone and water with the following result:

Time elapsed Concentration of Concentration of/sec C4HgCl/mol dm-3 hydroxide ions/mol dm-3

0 0.0100 0.0300

294 0.0050 0.0250

595 0.0025 0.0225

(a) Write a balanced equation for the reaction of an organicchloride of formula C4HgCI with hydroxide ions. (2)

(b) Suggest a reason why water alone is not used as the solvent inthe experiment. (1)

(c) [i) From the result given deduce an order of reaction.Explain your answer.

[ii) Write a rate expression for the overall reaction.(iii) Suggest a mechanism for the reaction that would be

consistent with your rate expression.

(4)(1 )

(2)(d) Suggest a practical method by which it should be possible to

follow the extent of the reaction. (2)(Total 50 marks)

65

APPENDIX ONE

NOTES ON TANGENTS AND THEIR SLOPES

Drawing tangentsIt might look easy to line your ruler up against the curve and draw a tangent'by eye-, but it is unlikely that you would get accurate tangents this way.We suggest two better methods in this Appendix to help you construct tangentsaccurately. You will need the following items: a ·sharp pencil and a rulerjeither a small plane mirror or a glass rod; either a pair of compasses or aprotractor.

The first step in drawing a tangent at a particular point on a curve is todraw a normal (perpendicular) at that point. We show you two methods - choosewhichever is more convenient.

1. Drawing a normal to a curve using a mirror

Fig.23. ------,IPlace a mirror acrossthe curve so that yousee the reflection ofthe curve in themirror.

_IMove the mirror fromside to side untilthe reflection of thecurve in the mirrorappears to be in acontinuous line withthe curve itself.

Draw a line along theedge of the mirror atthis point. This isthe normal to thecurve.

2. Drawing a normal to a curve using a glass rod

Fig.24.

Place the glass rodacross the curve.

Move the rod sidewaysuntil the part of thecurve seen throughthe rod appears tobe in line with theother two parts.

67

Draw a line along theglass rod. This is thenormal to thecurve.

Having drawn a normal to the curve, you can easily draw a tangent at the pointof intersection. Once again~ we show you two methods.

1. Drawing a tangent using compasses

Fig.25. A B c

Rest the point of the~ompasses on theintersection of thenormal and the curve.Using a convenientradius~ draw an arcthat cuts the normalon either side.

Open the compassesuntil the radius isabout 1~ times as bigas it was. Rest thepoint on each arc inturn where it cuts thenormal and draw fourshort arcs to inter-sect near the curve.

Using a ruler~ drawa straight linejoining the twocrosses. This lineis a tangent to thecurve.

2. Drawing a tangent using a protractor (or a set-square)

Line up the mark at the base ofthe protractor with the point wherethe normal crosses the curve.Make a mark opposite 90°. Dothis each side~ and join thetwo marks to get a tangent.

It is important to draw tangents accurately because we can obtain usefulinformation from measurements of their slopes. Slopes of lines can bepositive or negative in sign as we now show.

Negative and positive slopesThe sign of a measured slope of a tangent indicates whether the change~ y~is increasing or decreasing with time~ t. We show you how this sign isdeduced with the aid of diagrams.

68

Figs. 27 and 28 show the two curves of the type you often meet in reactionkinetics showing the change in some quantity, Y, with time, t. A tangentis drawn in each case, and its slope is required in order to determine therate of reaction.

t,• •

Fig.27. A positive slope Fig.28. A negative slope

The slope is obtained by taking two points along the tangent at times t1 andt2, and noting the corresponding values of the measured quantity, Y1 and Y2•

6.YThe slope is defined as 6.t

In Fig. 27, Y (e.g. the volume of a gaseous product) is increasing with time,so that Y2 > Y1 and Y2 - Y1 is positive. The slope, therefore, is alsopositive.

In Fig. 28, Y (e.g. the concentration of a reactant) is decreasing with time,so that Y2 < Y1 and Y2 - Y1 is negative. The slope, therefore, is alsonegative.

Since the rate of reaction is always a positive quantity, it is necessary tochange the sign of a negative slope when calculating rate from it. Thisconcern about signs is particularly important when using the differentialnotation, as we show in the next section.

A note on 'differential notation'd

In your reading you may have seen the symbol dt written in front of aconcentration term. For example, the rate equation for the decomposition ofdinitrogen pentoxide could be written as

- d[N205J = keN ° ]dt 2 ~

This is a mathematical notation meaning Ithe rate of change of [N2051 withtime'. Thus, the equation says that the rate of change of concentration ofN205 is proportional to the concentration of N205• The negative signindicates that the concentration is decreasing.

69

In practical terms, d[N205J is the slope of the concentration/time curve.dt

When rates are expressed in this shorthand notation, much information can beexpressed with brevity. For example, for the reaction

2N205(g) ~ 4N02(g) + °2(g)the rate of formation of nitrogen dioxide is four times the rate of formationof oxygen and twice as fast as the disappearance of dinitrogen pentoxide, i.e.

and

The negative sign is necessary because [N02J is increasing while [N205] isdecreasing.To see that you have understood this idea, try the following short exercise.

Exercise 43 Phosphine decomposes when heated~ as follows: ~4PH3Cg) ~ P4Cg) + 6H2Cg)

~At a given instant the rate at which phosphine decomposesis 2.4 x 10-3 mol dm-3 s-~.(a) Express the rate of reaction in three different ways, using

differential notation, and show the relationships betweenthem.

(b) What is the rate of formation of


The great value of expressing rates of reaction in differential notation isthat it enables us to express rate equations in another very useful formby means of the mathematical technique called integration. We give a briefsurvey of this in the next section - ask your teacher whether you need itfor your syllabus.

INTEGRATED RATE EQUATIONSWe give here a very brief account, which you should fill out byreading the relevant sections in your textbookCs). We suggest youonly do this if you have dealt with integration in mathematics.


(38) write an integrated rate equation for a first order reaction;(39) obtain the rate constant for a first order reaction from a plot of

log(concentration) against time;(40) show that the half-life of a first order reaction is constant;(41) write an integrated rate equation for a second order reaction;(42) obtain the rate constant for a second order reaction from a plot of

time against 1/concentration.

70

First order reactionsYou have already written a rate equation for the decomposition of dinitrogenpentoxide:

rate = k[N20sJUsing differential notation, and writing c for [N20S]' this becomes

Integrating gives

dcdtInc

kc

-kt + I (I = integration constant.)When t = 0, c = Co (the initial concentration) and I = lnco. The equationthus becomes

Inc = -kt + lnco or logc = -ktI2.3 + logco

This is an equation for a straight line and' enables us to obtain the rateconstant directly from concentration/time data by plotting logc against tand measuring the slope.

In the following exercises you can see how much easier this is than the methodyou learned in Level One, which involved two graphs and several intermediatecalculations.

Exercise 44 Use data in Results Table 1 (page 80) to show that the ~hydrolysis of 2,4.6-trinitrobenzoic acid is a first 'order reaction. Calculate the rate constant. ~~(Answer on page 88 )

Exercise 45 The decomposition of dinitrogen pentoxide (N20S)dissolved in tetrachloromethane (carbon tetrachloride) at45 0 C is first order. Using the concentrations ofdinitrogen pentoxide and the times given, estimate by agraphical method the rate constant for the decomposition,stating the unit in which it is expressed.Table 15

Time, tis 0 250 500 750 1000 1500 2000 2500

clmol dm-3 2.33 1 .95 1 .68 1.42 1 .25 0.95 0.70 0.50


In Level One you learned that first order reactions have a constant half-life. In the next exercise you confirm this mathematically by using theintegrated rate equation.

Exercise 46 From the integrated rate equation for a first orderreaction, obtain an expression for the time taken forthe concentration to fall to half its initial value.How does this show that the half-life is constant?(Answer on page 89 )

71

Finally~ we take a brief look at an integrated rate equation forsecond order reactions.

Second order reactionsWe limit our discussion to two simple types:

A + A ~ productsA + 8 ~ products

[A] = e[AJ = [8] = e

Each of these gives the same rate expression:dedt

Integrating givesWhen t = 0, e = Co

kt 1/e - IIi/co and the equation becomes

1 1kt = - --e Coor 1 1

t = ke - ""fC;

This is an equation for a straight line and, once again, we can obtain therate constant directly from concentration/time data. Try this in the nextexercise - you should be able to see from the equation what graph to draw.

Exercise 47 Methyl ethanoate reacts with aqueous alkali according to ~the following equation:CH3C02CH3(lJ + OH-(aq) ~ CH3C02-(aq) + CH30H(aq) ~~The reaction may be followed by withdrawing samples at intervalsand titrating with standard acid solution. The followingresults were obtained:Table 16

Time/min [OH-( aq) J/mol dm-3

3 7.40 x 10-3

5 6.34 X 10-3

7 5.50 X 10-3

10 4.64 X 10-3

15 3.63 x 10-3

21 2.88 x 10-3

25 2.54 x 10-3

Determine a value for k, the rate constant for this reaction.(Assume that the initial concentrations of methyl ethanoateand hydroxide ions were equal.)(Answers on page 89 J

72

APPENDIX TWO

'CLOCK' REACTIONS

In this appendix we deal with the theory of Iclockl reactions and we providean alternative experiment to the iodination of propanone (Experiment 2).

The theory of 'clock' reactionsIn a typical reaction, the early part of a concentration against time curveapproximates to a straight .line, as shown in Fig. 29 below:

Fig.29.

c:o'';:;~cQ)(,)

c:o(,)

o t,

time

If we choose any value of c (say c1) which lies on this straight line, theinitial rate of reaction is given by c1/t1. However, for a value of c (sayC2) beyond the straight part of the curvel the initial rate is greater thanC2/t2•

In clock reactions, the initial rates of a series of reactions are obtainedvery simply by choosing a fixed value of c for all the reactions so thatthe rate = cit. As long as c is constantl it need not necessarily be quanti-fied, and we can take lit as a measure of the rate of reaction.

If you are doing the next experiment as an alternative to Experiment 2 tilland have not, therefore, .done Experiment 3, we suggest that you read ~the introduction to Experiment 3 (page 50). This contains some useful Jlinformation about clock reactions.

73

EXPERIMENT 5A bromine Iclockl reaction

AimThe purpose of this experiment is todetermine the rate equation for thereaction between bromide and bromate(V)ions in aqueous solution.

IntroductionBromine and bromate(V) ions in acid solution react according to the equation:

5Br-(aq) + Br03-(aq) + 6H+(aq) ~ 3Br2(aq) + 3H20(I) •........•......... (1)

In order to follow the reaction, two other substances are added to thereaction mixture.(a) A precisely known, small amount of phenol. This reacts immediately

with the bromine produced, removing it from solution:3Br2(aq) + C6H50H(aq) ~ C6H2Br30H(aq) + 3H+(aq) + 3Br-(aq) (2)

(b) Methyl orange solution, which is bleached colourless by free bromine:Br2(aq) + methyl orange ~ bleached methyl orange .........•........ (3J

(acid form: pink) (colourless)As soon as all the phenol has reacted with bromine produced inreaction (1), free bromine will appear in solution and bleach the methylorange. If the time taken for the methyl orange solution to bebleached is t, then the rate of reaction (1) is proportional to 1/t.

In the experiment you study the effect on the rate of reaction (1) of varyingthe concentration of bromide ions, bromate (V) ions, and hydrogen ions in turn,with the concentrations of the others held constant. To save time, you couldcooperate with other students and pool your results.

Requirementssafety spectaclesphenol solution, 0.00001 M C6H50Hwash-bottle of distilled water3 burettes and standsmeasuring cylinder, 25 cm3potassium bromide solution, 0.010 M KBrpotassium bromate(V) solution, 0.0050 M KBr03acidified methyl orange solution, labelled C, 0.001%white tilethermometer 0-100 °c2 beakers, 100 cm3

stopclock or watchsulphuric acid, 0.01 M H2S04potassium bromate(V) solution, 0.20 M KBr03methyl orange solution, labelled 0, 0.001% in 0.40 M KBr

74

ProcedureA. Varying the concentration of bromide ions

1. Prepare the first pair of mixtures in two beakers~ as specified inTable 17. Use burettes to measure the potassium bromide and phenolsolutions and the water; use a measuring cylinder for the others.Table 17

Beaker X Beakei' Y

Volume of Volume of Volume of Volume of Volume of0.01 M KBr H2O 0.005 M solution C 0.00001 M/cm3 /cm3 KBr03/cm3 /cm3 phenol/cm3

10.0 0 10.0 15.0 5.0

8.0 2.0 10.0 15.0 5.0

6.0 4.0 10.0 15.0 5.0

5.0 5.0 10.0 15.0 5.0

4.0 6.0 10.0 15.0 5.0

3.0 7.0 10.0 15.0 5.0

2. Have ready a copy of Results Table 6.Results Table 6

Volume of Br-( aq)/cm3 10.0 8.0 6.0 5.0 4.0 3.0

Time~ tis

1/10-2 s-J.tTemperature/oC

Average temperature of solutions °c(Specimen results on page 90 )

3. Pour the contents of beaker X into beaker Y and start the clock.Mix the solutions by pouring from one beaker to the other~ twice~and place the beaker containing the mixture on the white tile.

4. When the pink colour disappears~ stop the clock and record the timein a copy of Results Table 6. (As you~are looking for a disappearanceof colour~ this may need a little practice. If in doubt~ repeat thereaction once or twice until you get consistent times.) Record thetemperature of the solution.

5. Work through the rest of the mixtures in Table 17 in the same way~recording each result as you go.

B. Varying the concentration of bromate(V) ions6. Follow a similar procedure to that for part A~ but keep the volume of

bromide solution constant at 10.0 cm3 and vary the volume ofbromate(V) solution as shown in Table 18.

75

Table 18

Beaker X Beaker Y

Volume of Volume of Volume of Volume of Volume of0.005 M H2O 0.01 M KBr solution C 0.00001 MKBr03/cm3 /cm3 /cm3 /cm3 phenol/cm3

10.0 0 10.0 15.0 5.0

8.0 2.0 10.0 15.0 5.0

6.0 4.0 10.0 15.0 5.0

5.0 5.0 10.0 15.0 5.0

4.0 6.0 10.0 15.0 5.0

3.0 7.0 10.0 15.0 5.0

Record your results in a copy of Results Table 7.Results Table 7

Volume of Br03 -Caq )/cm3 10.0 8.0 6.0 5.0 4.0 3.0

Time~ tis

2";10-2 S-1 .tTemperature/oC


C. Varying the concentration of hydrogen ion7. For this part of the experiment you need different solutions~ as

stated in Table 19. Follow a similar procedure to that for part A.

Table 19

Beaker X Beaker Y

Volume of Volume of Volume of Volume of Volume of0.01 M H2O 0.20 M solution 0 0.00001 MH2S04/cm3 /cm3 KBr03/cm3 /cm3 phenol/cm3

10.0 0 10.0 15.0 5.0

8.0 2.0 10.0 15.0 5.0

6.0 4.0 10.0 15.0 5.0

5.0 5.0 10.0 15.0 5.0

4.0 6.0 10.0 15.0 5.0

3.0 7.0 10.0 15.0 5.0

76

Record your results in a copy of Results Table 8.Results Table 8

Volume of acid/cm3 10.0 8.0 6.0 5.0 4.0 3.0

Time~ tis

¥10-2 s-J.

Temperature/oC


Treatment of results1. For each part of the experiment~ plot a graph of 1/t against volume of

the reactant under consideration. 1/t is proportional to the rate ofreaction and the volume of reactant is proportional to the concentration~since the total volume is constant.

2. Deduce from each graph whether or not the reaction is first order withrespect to the reactant under consideration.

3. If you think the reaction is not first order~ plot another graph~ asexplained below.Suppose that~ for a reactant A~ rate k1[AJn

Under the conditions of this experimentl it follows that(V is the initial volume of reactant)1/t = k2vn

log(l/tJPlotting log(1/tJwith slope n.

nlog V + logk2

against logV should therefore give a straight line

Questions1. Write the rate equation for the reaction.2. Qualitatively compare the rates of the three reactions stated in the

introduction.3. Why does the phenol solution need to be very dilute?(Answers on page 91 )

77

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Miscellaneous Science;Chemical kinetics: unit P5 · Exercise 30 In the Journal of the Chemical Society for 1950, Hughes, ~ Ingold and Reed report some kinetic studies on aromatic