Introduction to Reaction Kinetics, Hazle Cox

8/14/2019 Introduction to Reaction Kinetics, Hazle Cox

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INTRODUCTION TO

REACTION KINETICS

Hazle Cox


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Introduction to Reaction Kinetics

INTRODUCTION TO REACTION KINETICS

Aims

The aim of this course is to introduce the student to the principles of

reaction kinetics, covering the necessary mathematics within the course.

In addition, the student will learn how to present the data from reaction

kinetics measurements by the plotting of graphs.

Syllabus

1. Introduction to kinetics: a general definition of rate, stoichiometry,

types of reaction (elementary and composite), energy profiles.

2. Rate in chemical kinetics: kinetic reaction profiles, rate of a chemical

reaction.

3. Factors determining rate: theoretical rate equation: a simple collision

model; experimental rate equation: rate constants, order of reaction.

4. Determining experimental rate equations at fixed temperature:

a. Reactions involving a single reactant: half-life check, the

differential method: a check for second-order behaviour and

general approach (taking logarithms) to determine partial order.

The integration method: integrated rate equation for first order,

second order reactions, to determine the rate constant.


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b. Reactions involving several reactants: the isolation method,

pseudo-order rate equation, the initial rate method.

5. The effect of temperature on the rate of a chemical reaction: The

Arrhenius equation, determining the Arrhenius parameters, the

magnitude of the activation energy.

Assessment

The final mark will consist of continuous assessment (30%) and an

unseen end of course exam (70%).

Reading List

The Molecular World: Chemical kinetics and mechanism edited by M.

Mortimer and P. Taylor (The Open University and RSC) 2002.Physical Chemistry by P.W. Atkins (Oxford University Press).

Name and Location of Course Lecturer

Dr Hazel Cox, Room 3R514, Chichester Building

Tel: 8972, Email [email protected]


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Introduction


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INTRODUCTION

There are two main driving forces behind chemical reactions are:

1. THERMODYNAMICS: the study of energy and entropy.

2. KINETICS: the study of the rate at which chemical reactions

occur.

Thermodynamics considers the relationship between the system the

reaction, process or organism under study and the surroundings the

rest of the universe. Since energy is either released or taken in by all

chemical and biochemical processes, thermodynamics enables the

prediction of whether a reaction may occur or not without need to

consider the nature of matter itself. In chemical systems, it allows

determination of the feasibility, direction and equilibrium position of

reactions.

This course is concerned only with chemical kinetics. Kinetics is the

study of movement; in particular, it is concerned with the measurement

and interpretation of the rates of chemical reactions. It is an area quite

distinct from that of chemical thermodynamics, which is concerned only

with the initial states of the reactants (before the reaction begins) and the

final state of the system when equilibrium is reached (so that there is no

longer any net change).

What happens between these initial and final states of reaction and

exactly how, and how quickly, the transition from one to the other occurs

is the province of chemical kinetics.


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At the molecular level chemical kinetics seeks to describe the behaviour

of molecules as they collide, are transformed into new species, and move

apart again. Whatever the process, information about how quickly it

occurs and how it is affected by external factors is of key importance.

Without such knowledge, for example, we would be less well equipped to

generate products in the chemical industry at an economically acceptable

rate, or design appropriate drugs, or understand the processes that occur

within our atmosphere.

Empirical Chemical kinetics

This is an approach to chemical kinetics in which the aim is to describe

the progress of a chemical reaction with time in the simplest possible

mathematical way.

Experimental methods in kinetics measure change in the composition of areaction mixture with time, either continuously as the reaction progresses,

or at fixed intervals after the reactants have come together. The

techniques applied vary depending on the timescale of the reaction and

the chemical species under study. Additional kinetic information is

obtained by varying experimental parameters such as the initial

concentration of reactant(s) or the temperature of the mixture.

A General Definition Of Rate

The rate of a reaction of a designated species is the rate of change of

concentration of that species with time. For a physical quantity that

changes linearly with time, we can take as a definition:


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unitstypicalinintervaltime

unitstypicalinquantityphysicalinchangechangeofrate =

For time, typical units are seconds, minutes, hours, and so on. If, for

example, the physical quantity was distance then typical units could be

metres and the rate of change would correspond to speed measured in,

say, metres per second (ms-1

). Since the physical quantity changes

linearly with time this means that the change in any one-time interval is

exactly the same as that in any other time interval. In order words a plot

of physical quantity versus time will be a straight tine and there is a

uniform, or constant, rate of change.

The equation above can be written in a more compact notation. If the

physical quantity is represented byy, then it will change by an amount y

during a time interval tand we can write

t

yy

=ofchangeofrate

This rate of change, y/t, corresponds mathematically to the slope (or

gradient) of the straight line, and, as already stated has a constant value.

A very important situation arises when a rate of change itself varies with

time. For example a car accelerating; as time progresses, the car goesfaster and faster. In this case a plot of physical quantity versus time is no

longer a straight line. It is a curve. At any particular time, the rate of

change is often referred to as the instantaneous rate of change. It is

measured as the slope of the tangent to the curve at that particular time

and is represented by the expression:


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dt

dyy =ofchangeofrateousinstantane

A rate of reaction is therefore equal to the gradient of the curve of species

concentration against time evaluated at the time of interest. The steeper

the gradient, the greater the rate of reaction. A species that is being

consumed has a negative gradient, whilst a species that is being formed

has a positive gradient. The units of rate of reaction always have

dimensions ofconcentration per time.

However, the best curve that can be computed will always depend on thequality of the experimental data; for example in a chemical kinetic

investigation on how well concentrations can be measured at specific

times. The uncertainty in the value of the tangent that is computed at any

point will reflect these factors.

Chemical Reactions

Reactants and products may be consumed and formed at different rates

according to the particular reaction stoichiometry (the number of

molecules of reagent and products in the balanced equation). For

example, at any point in the reaction between hydrogen and nitrogen to

form ammonia:

N2 (g) + 3H2 (g) = 2NH3 (g)

The rate of consumption of hydrogen is three times the rate of

consumption of nitrogen, whilst the rate of production of ammonia is

twice the rate of consumption of nitrogen but only two-thirds the rate of

consumption of hydrogen.


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Important as it may be, knowing the stoichiometry of a reaction still

leaves open a number of fundamental questions:

v Does the reaction occur in a single step, as might be implied by a

balanced chemical equation, or does it involve a number of

sequential steps?

v In any step, are bonds broken, or made, or both? Furthermore,

which bonds are involved?

v In what way do changes in the relative positions of the various

atoms, as reflected in the stereochemistry of the final products,

come about?

v What energy changes are involved in the reaction?

The key information required to answer these question is embodied in the

reaction mechanism for a given reaction. This refers to a molecular

description of how the reactants are converted into products during thereaction. However, a reaction mechanism is only as good as the

information on which it is based. Essentially, it is a proposal of how a

reaction is thought to proceed and its plausibility is always subject to

testing by new experiments.

A powerful means of gaining information about the mechanism of a

chemical reaction is via experimental investigations of the way in which

the reaction rate varies, for example, with the concentration of species in

the reaction mixture, or with temperature. There is thus a strong link

between, on the one hand, experimental study and, on the other, the

development of models at the molecular level.


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Types of Reactions

If a reaction occurs in a single step, it is referred to as an elementary

reaction. Therefore, the balanced chemical equation does actually convey

the essential one-step nature of the process.

However, if a reaction occurs via more than one elementary step, it is

referred to as a composite reaction.

Examples:

If we consider the reaction between bromomethane (CH3CH2Br) and

sodium hydroxide in a mixture of ethanol and water at 25 C then the

stoichiometry is represented by the following equation

CH3CH2Br (aq) + OH-(aq) = CH3CH2OH (aq) + Br

-(aq)

It is well established that this reaction occurs in a single step. Therefore,

it is elementary and the balanced chemical equation does actually convey

the essential one-step nature of the process. The reaction mechanism,

although consisting of only one step, is written as

CH3CH2Br + OH- CH3CH2OH + Br

-

The arrow sign () is used to indicate that the reaction is known (or

postulated) to be elementary and, by convention, the states of the species

involved are not included.

The reaction between phenylchloromethane (C6H5CH2Cl) and sodium

hydroxide in water at 25C


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C6H5CH2Cl (aq) + OH-(aq) = C6H5CH2OH (aq) + Cl

-(aq)

Here, all of the available experimental evidence suggests that the reaction

does not occur in a single elementary step. The most likely mechanism

involves two steps

C6H5CH2Cl [C6H5CH2]+

+ Cl-

[C6H5CH2]+

+ OH- C6H5CH2OH

This is a composite reaction as it proceeds via more than one elementary

step. The corresponding reaction mechanism given above is referred to as

a composite reaction mechanism, or just composite mechanism. In

general, for any composite reaction, the number and nature of the steps in

the mechanism cannot be deduced from the stoichiometry.

The species [C6H5CH2]+

in the mechanism is known as a reaction

intermediate. All mechanisms with more than a single step will involve

intermediate species and these will be formed in one step and consumed,

in some way, in another step.

In general, for most composite mechanisms the sum of the various steps

should add up to give the overall balanced chemical equation. (An

important exception is a radical chain mechanism.)

Energy Profiles For A Chemical Reaction

Chemical reactions are not instantaneous. Even explosions, although

extremely rapid, require a finite time for completion. This resistance to

change implies that at the molecular level individual steps in a

mechanism require energy in order to take place. For a given step, the

energy requirement will depend on the species involved.


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An energy profile can be used to depict the energy changes that occur

during an elementary reaction schematically.

An energy profile such as above can be interpreted in two distinct ways: As representing the energy changes that occur when individual

molecular species interact with one another in a single event, or

As representing what happens on a macroscopic scale, in which

case some form of average has to be taken over many billions of

reactions.

From a molecular viewpoint, the energy profile shows the energy changes

that occur when a single bromomethane molecule encounters, and reacts

with, a single hydroxide ion in solution. As these species come closer and

closer together they interact and, as a consequence, chemical bonds

become distorted and the overall potential energy (PE) increases. At

distances typical of chemical bond lengths, the reactant species become

Reactants

Products

Energy

Barrier

Transition State

Potential

energy

Reaction coordinate


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partially bonded together and new chemical bonds being to form. At this

point the PE reaches a maximum and any further distortion then favours

the formation of product species and a corresponding fall in PE.

The PE maximum is referred to as the transition state. The molecular

species that is present at this energy maximum is one in which old bonds

are breaking and new ones are forming: it is called the activated

complex. This complex is a transient species and not a reaction

intermediate.

It is clear from the figure above that there is an energy barrier to reaction.

So, for example, for a bromomethane molecule to react with a hydroxide

ion, energy must be supplied to overcome this barrier. The source for this

energy is the kinetic energy of collision between the two species in

solution; so the more violent the collision process, the more likely the

reaction will occur.

From a macroscopic interpretation of the energy profile, we can stillretain the ideas of a transition state and an activated complex. The energy

barrier to reaction is now a very complex average over many molecular

events but, as we shall see later, it can still be related to a quantity that is

measured experimentally.

From a thermodynamic viewpoint, the energy difference between the

products and the reactants can be taken to a good approximation to be

equal to the enthalpy change for the elementary reaction. The difference

in potential energy (products reactants) is negative, thus the enthalpy

change will be negative and the elementary reaction is exothermic.


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Self-test Questions:

1. What is the stoichiometry of the reaction?

2. In general terms, what does a reaction mechanism provide?

3. What is an elementary reaction?

4. What sort of equations will involve intermediate species?

5. For composite mechanism (except for radical chain mechanisms)

how can you obtain the overall chemical equation?

6. What is a transition state, and what is the name given to the species

at this point?

Answers:

1. Information concerning the relative amounts of reactants and

products taking part in a chemical reaction is known as the

stoichiometry of the reaction.2. In general terms, a reaction mechanism provides a molecular

description of how reactants are thought to be converted into

products during a chemical reaction.

3. An elementary reaction is one that takes place in a single step, does

not involve the formation of any intermediate species, and which

passes through a single transition state.

4. All reactions with more than one step will involve intermediate

species. These are formed in one step and consumed, in some way,

in another.

5. The sum of various steps on a composite mechanism gives the

overall balanced chemical equation.


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6. The transition state lies at the top of the energy barrier to reaction;

the species at the top of this barrier is transient and is called the

activated complex.

Question 7: Does the following energy profile represent an exothermic or

endothermic change and what is the significance of the point marked X?

Answers:

(i) From a thermodynamic viewpoint, it is only the initial and final states

of a composite reaction that need to be considered. Overall the reaction is

exothermic.

(ii) It is a local minimum that corresponds to the formation of the reaction

intermediate.

Reaction coordinate

Reactant

TS for step 1TS for step 2

X

Products

PE


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Rates in Chemical

Kinetics


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Rates in Chemical Kinetics

In the following reaction between hypochlorite ions ( ClO ) and bromide

ions in aqueous solution at room temperature

ClO (aq) + Br (aq) = BrO (aq) + Cl (aq)

The stoichiometry is such that one mole of each reactant is converted into

one mole of each product, i.e., it has a 1:1 stoichiometry. (The

stoichiometry is an essential preliminary step in any kinetic study.) This

reaction will be used in the following discussions.

Kinetic Reaction Profiles

A kinetic study involves following a reaction as a function of time. This

can be achieved by using a suitable analytical technique to measure the

concentration of reactants, or products, or both, at different times during

the progress of the reaction. To avoid changes in reaction rate due to

temperature changes, it is essential that measurements are made under

isothermal, i.e., constant temperature, conditions.

A typical set of results obtained for the above reaction at a constant

temperature of 25 C is shown below. This type of plot is called a kinetic

reaction profile.

It should be clear that the initial conditions of the experiment were

selected so that the initial concentration of ClO was greater than that of

Br ; we can say that ClO was in excess. In more concise terms,

0

]ClO[ - >0

]Br[ , where the subscript zero has been used to indicate


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initial concentration. Experimentally, these initial concentrations were ,

0- ]ClO[ = 3.230 10-3 mol dm-3 and 0]Br[

= 2.508 10-3 mol dm-3.

Question 1: By how much have the concentrations of ClO , Br ,

BrO and Cl changed after 2000 s of reaction?

Answer: The changes in concentration for ClO and Br correspond, in

each case, to a decrease of about 1.95 10-3 mol dm-3 compared to their

initial concentrations. For BrO and Cl there is an increase in

concentration from zero to about 1.95 10-3 mol dm-3.

Thus, after 2000 s of reaction the magnitudes of the changes in the

concentrations of reactants and products are the same, although there is a

decrease for reactants and increase for products. In fact, this result would

have been obtained irrespective of the time period selected. This means

that the stoichiometry of the reaction applied throughout the whole course

of reaction; that is it has time-independent stoichiometry.

This does not mean that intermediates are not present for a reaction that

has time-independent stoichiometry, just that, within the accuracy of the

chemical analyses used, intermediates cannot be detected and so they do

not affect the stoichiometric relationship between reactants and products.

In fact, the above reaction is thought to be composite with a three-step

mechanism in which case intermediates mustbe involved.

Finally, the concentration of Br , which is the reactant not in excess,

appears to be progressing towards zero. In fact, after 10 hours in an


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extended experiment it was found that [ Br ]= 0.007 10-3 mol dm-3.

Thus, nearly all the Br has been consumed to give product. From a

kinetic viewpoint, it is reasonable to say in these circumstances that the

reaction has gone to completion, that is, had the reactants been initially

present with equal concentrations, they would both have been virtually

completely converted into products since there is greater than 99 %

reaction.

Rate of change of concentration of a reactant or product with time

The concentration-versus-time profiles for each of the reactants and

products in the given figure are curved. This means that the rate of

change of concentration with time for each of these species is not

constant; in each case it will vary continuously as the reaction progresses.

Question 2: How would you determine the rate of change of

concentration with time for BrO at 1500 s?

Answer: The rate will be equal to the slope of the tangent drawn to the

curve at 1 500 s. This will measure the instantaneous rate of change and

will be represented by dt

d ]BrO[

.

Question 3: If the coordinates for two point on the tangent are (t= 0 s,

[ BrO ] = 1.14 10-3 mol dm-3) and (t= 4000 s, [ BrO ] = 2.82 10-3

mol dm-3

), what is the value ofdt

d ]BrO[ at 1 500 s?


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Answer: the slope of the tangent is calculated as follows:

1-7-3

3-3

3-33-3

sdmmol1020.4

s4000

dmmol1068.1

s0s4000

dmmol1014.1dmmol1082.2

slope

=

=

=

This value, which is an estimate that depends on how well the tangent has

been drawn, is the rate of change of concentration of BrO with time,

dt

d ]BrO[ at 1 500 s.

Clearly,dt

d ]BrO[ and

dt

d ]Cl[ will be equal in value since the kinetic

reaction profiles for BrO and are identical. They are also both positive

quantities since they represent the formation of product species. Thus, if

we represent the rate of the reaction at any time by the symbol J, then one

possible definition would be

dt

d

dt

dJ

]Cl[]BrO[ ==

However, we could equally well have considered reaction profiles for

ClO and Br and determineddt

d ]ClO[ and

dt

d ]Br[ . The rates of

change of concentration of ClO and Br at any time in the reaction are:

Negative, because they represent the consumption of reactant

species


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Equal to one another, because according to the stoichiometry if a

ClO ion reacts then so musta Br ion.

Equal in magnitude, but opposite in sign, to dt

d ]BrO[

and

dt

d ]Cl[ , because according to the stoichiometry, the reaction of a

ClO ion with a Br ion must produce one each of the two

product ions.

Note, these points will hold no matter which reactant species is in excess,

and irrespective of the amount of excess.

Question 4: If at 1500 s,dt

d ]ClO[ = 4.18 10-7 mol dm-3 s-1 what will

be the value ofdt

d ]ClO[ at this time?

Answer:

1-3-7

1-3-7

sdmmol1018.4

)sdmmol1018.4(]ClO[

=

=dt

d

The fact thatdt

d ]ClO[ and

dt

d ]Br[ are positive quantities puts us in

a position to give a final definition ofJfor reaction (*):

dt

d

dt

d

dt

d

dt

dJ

]Cl[]BrO[]Br[]ClO[ ====


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Defined in this way, J, irrespective of whether it is expressed in terms of

a reactant or a product in Reaction (*), always has a single positive value

at any time in the reaction.

One special case of the rate of reaction is that corresponding to the start

of the reaction. This is referred to as the initial rate of reaction and is

represented byJ0. The figure below is a plot of the kinetic reaction profile

for Cl and shows the tangent drawn to the curve so that the initial rate

of change of concentration of Cl can be determined.

Question 5: If the tangent drawn above passes through a point with

coordinates (t = 500 s, [

Cl ] = 1.6

10

-3

mol dm

-3

), what is the initialrate of reaction?

Experimentally there are two important factors that must be taken into

account when measuring initial rates of reaction:

1. It is preferable to measure an initial rate by observing the

appearance of product rather than the disappearance of reactant.

Cl

0

0.5

1

1.5

2

2.5

0 1000 2000 3000 4000 5000 6000

time / s

[Cl-]x10-3/moldm-3 Initial

tangent


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2. It is essential to make measurements in the very early stages, say

the first 5 %, of a reaction in order to obtain accurate values of the

initial rate.

A general definition of the rate of a chemical reaction

The discussion that resulted in the definition of J above can be applied to

any chemical reaction that has a time-independent stoichiometry. For

example, nitrogen dioxide (NO2) decomposes in the gas phase at

temperatures in the region of 300 C to give nitric oxide (NO) and

oxygen:

2 NO2 (g) = 2 NO (g) + O2 (g)

If the progress of this gas-phase reaction is monitored in a closed reaction

vessel then concentrations can simply be expressed in terms of mol dm-3

.

At any time in the decomposition, the rate of decrease in the

concentration of NO2 will be directly related to the rates of increase in the

concentrations of NO and O2, respectively.

Question: What is the relationship between these quantities?

Answer: If we considerdt

d ]NO[ 2 , which is a positive quantity, then

dt

d

dt

d

dt

d ]O[2

]NO[]NO[ 22 ==


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These relationships are consistent with the fact that according to the

stoichiometry, the rate of increase in the concentration of O2 is only equal

to one-half of that for NO.

Of course, we could equally well have written,

dt

d

dt

d

dt

d ]O[]NO[

2

1]NO[

2

1 22 ==

It is this form that is conventionally used to define the rate of reaction and

so

dt

d

dt

d

dt

dJ

]O[]NO[

2

1]NO[

2

1 22 ===

It is common practise in chemical kinetics (as well as in chemical

thermodynamics) to use chemical reaction written in an alphabetical

form to help to express a definition in a general way. Thus, we could

write a chemical reaction with known stoichiometry as

KK ++=++ QPBA qpba

where A, B and so on, represent reactants and P, Q and so on, represent

products. The numbers Kba, and Kqp, ensure that the equation is

balanced and so are known as balancing coefficients. In practise they are

usually chosen to have their smallest possible integer values, and they

must be positive. Writing a chemical reaction in this way allows a

quantity called the stoichiometric number to be introduced. It is given

the symbol Y (pronounced nu Y) where the sbscript Y represents a

given species (reactant or product) in the reaction. The stoichiometric

number is then defined so that for

reactant A, a=A , reactant B, b=B , reactant P, p+=P , and

reactant Q, q+=Q


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dt

dJ

]Y[1

Y=

Strictly, this definition assumes constant volume conditions during the

course of a reaction. For solution reactions this is a reasonable

approximation. It is also valid for gas-phase reactions carried out in

sealed containers.

Question: For the following two reactions express the rate of reaction, J,

in terms of the rate of change of concentrations of each reactant and each

product. (Assume both reactions have time-independent stoichiometry.)

(a) 2 H2 (g) + 2 NO (g) = 2 H2O (g) + N2 (g)

(b) S2O82-

(aq) + 3 I-(aq) = 2 SO4

2-(aq) + I3

-(aq)

Answer:

(a) 1,2and,2,2222 NOHNOH

+=+=== , therefore,

dt

d

dt

d

dt

d

dt

dJ

]N[]OH[

2

1]NO[

2

1]H[

2

1 222 ====

(b)

dtd

dtd

dtd

dtdJ -- ]I[]SO[

21]I[

31]OS[ 324282 ====


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Factors Determining The Rate Of A Chemical Reaction

The rate of a reaction depends on

1. The concentration of the reactants,

2. The temperature of the reactants,

3. The surface area of the reactants,

4. Whether or not a catalyst is present

Theoretical Model: A simple collision model

The essential theoretical picture is chemical kinetics is that for a step in a

reaction mechanism to occur, two things must happen:

Reactant species involved in the step must collide with one

another, and

Colliding particles must have sufficient energy to overcome the

energy barrier separating reactants from products.

The constituent particles in gases or solutions are in constant, random

motion. We can envisage, therefore, that collisions occur continuously

and this suggests that the more frequently reactant species collide, then

the faster the consequent reaction.

Concentration

The more concentrated the particles are, i.e., the more particles there are

per unit volume, the more collisions there will be and hence the rate of

reaction will increase (i.e., the number of collisions per unit volume per

unit time).


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For example, we can consider an elementary reaction between two

difference species A and B (which could be molecules, fragments of

molecules, atoms or ions) in the gas phase

A + B products

The number of collisions between species A and species B that occur in a

fixed volume in unit time (say, 1 s) is a measure of the collision rate

between A and B. For example, doubling the concentration of B means

that the number of targets for individual A species in a given volume is

increased by a factor of two; hence the rate at which A species collide

with B species is doubled. Thus, overall, the collision rate between A and

B species is directly proportional to their concentrations multiplied

together, so that

Collision rate [A][B]

or

Collision rate = c [A][B]

c is a constant of proportionality. In fact, the form of this constant can be

calculated for any gas-phase elementary reaction using a theory of

collisions in the gas phase that was first put forwards in the 1920s.

If every collision between species A and B resulted in chemical

transformation to products, then the rate of reaction (J) would be identical

to the collision rate. For many elementary reactions, however, this is not

the case.

Question: Can you suggest a reason for this?


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Of all the collisions that occur between reactant species A and B, only a

fraction,f, will be successful. We can therefore write the rate of reaction

as

J=f (collision rate)

i.e.,

J=fc [A][B]

In this discussion so far we have implicitly assumed that the temperature

is fixed. This being the case, the quantityfc can be replaced by a single

constant, ktheory, so that

J= ktheory [A][B]

This equation is an example of a rate equation and, more explicitly, it is

the theoretical rate equation for the elementary equation

A + B products

The quantity ktheory is the theoretical rate constant for the elementary

reaction, it has a value that is independent of the concentrations of

reactants A and B.

Temperature

The theoretical rate constant, although called a constant, does depend on

temperature. Increasing the temperature increases, in most circumstances,

the magnitude of ktheory . So carrying out a reaction at a higher

temperature, but with the same initial concentrations of A and B, will be

expected to result in an increase in the rate of reaction. This behaviour


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can be understood in a qualitative way in terms of a simple collision

model.

In any gas, at a particular instant, the particles will be moving about with

a wide distribution of speeds. It is shown in a schematic form for a gas

(consisting of molecules) at two different temperatures. Note, that the

area under each curve is the same and is a constant for a given sample

since it represents the total number of molecules in that sample.

Increasing the temperature clearly results in an increase in the number of

more rapidly moving molecules, and the distribution becomes flatter and

wider. Furthermore, the peak of the distribution, which corresponds to the

most probable speed, moves to a higher value.

For an elementary reaction, raising the temperature has two distinct

consequences.

There is an increase in the fraction of rapidly moving species for

both reactants. In turn this means that the fraction of collisions, f,

with a kinetic energy sufficient to overcome the energy barrier to

Number of

molecules

Speed


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reaction also increases. The effect on the rate of reaction can be

quite significant.

There is a general increase in speed for both reactant species and

this results in an increased collision rate; in other words, the

constant c is temperature dependent. This effect, however, is

relatively small and for increases of temperature over a range of,

say, 100 C, it would be difficult to detect experimentally.

Question: What would be the effect of an increase in temperature for an

elementary reaction with an energy barrier to reaction that was very close

to zero?

Answer: In these circumstances, most collisions will lead to chemical

reaction, i.e., the fraction of successful collisions will be effectively

constant and independent of temperature. Thus, experimentally, it would

be difficult to detect any change in the rate of reaction with increasing

temperature.

The above discussion has been based on simple qualitative ideas about

how an elementary reaction may occur. The way to test this picture is to

see if rates of reaction measured experimentally, using difference

concentrations of each reactant and at different temperatures, show thesame predicted behaviour. From comparison of the experimental rate

constant with the theory we can conclude that for elementary reactions

there is good agreement between theory and experiment.


34/83


Summary

The simple collision theory assumes that

Reaction occurs when and only when collisions between the

reactants occur, and

Temperature is a measure of the velocity of the particles.

Factors that seem to be explained by this theory:

The rate of reaction increases with concentration of reactants. The more

concentrated the particles are the more collisions per unit volume per unit

time and hence a greater rate of reaction.

The rate of reaction increases with increasing temperature. As the

temperature of the reactants increases, we in effect mean that the velocity

of the particles increases. Obviously this will increase the number of

collisions per unit volume per unit time, and thus the rate of reaction willincrease.

The rate of reaction increases will surface area. If we think here of a

reaction such as that between calcium carbonate and dilute hydrochloric

acid, the bigger the surface area of the chalk, the more collisions between

the chalk and the acid, and therefore the greater the rate of reaction.

The role of the catalyst: if the catalyst is a solid, then a possibleexplanation of the increased rate of reaction could be that the reactants

group together on the surface of the catalyst, and this increased

concentration of reactants will lead to a higher rate. (This does not

explain the catalytic effect of dissolved substances. The activation energy

idea is useful in explaining catalytic effectssee later.)


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The use of a simple collision model to predict the behaviour of

elementary reactions involving two reactant species is instructive but

nonetheless limited in scope. To extend such a model to chemical

reactions in general would be difficult because the vast majority of these

are composite. To make progress in understanding the rates of chemical

reactions it is necessary to adopt an experimental approach .

An experimental approach

Experimental investigation, under isothermal conditions, for a wide range

of chemical reactions that can be represented in a general form as

KK +++=+++ RQPCBA rqpcba

has shown in many cases, both in the gas and solution phase, that the

experimental rate equation takes the form

K

]C[]B[]A[RkJ=

Here Rk is the experimental rate constant. Two key points about this

relationship are

One concentration term appears for each reactant

Each concentration term is raised to a particular power: ,, ,

and so on.

It is important to emphasize that the relationship is empirical in that it

represents a generalization of the simplest mathematical way of


36/83


representing the experimental rate equations for each of the reactions

studied. The relationship is based simply on the results of observation and

experiment.

The powers to which the concentration terms are raised are known as

partial orders of reaction. Thus is the partial order with respect to

reactant A, is the partial order of reaction with respect to B, is the

partial order of reaction with respect to reactant C, and so on. The overall

order of reaction (n) is defined by the sum of partial orders

K+++= n

It is often, but not always, the case that the partial orders of reaction turn

out to be small integers. If the partial order for a reactant is either 1 or 2,

then the reaction is referred to as being first-order or second-order in the

particular reactant.

The most frequently observed values of overall order n are also 1 or 2 and

the corresponding reactions are then referred to as being, respectively,

first- and second-order processes. An overall order of reaction can only

be defined for a reaction that has an experimental rate equation

corresponding to the general form give above, i.e.,

K

]C[]B[]A[RkJ=

A few selected examples of reactions with experimental rate equations of

this form are given below:


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Reaction Experimental rate

equation

(a) S2O82-

(aq) + 3 I-

(aq) = 2 SO42-

(aq) + I3-

(aq) J = kR [S2O82-

][I-

]

(b) 3 ClO - (aq) = ClO3- (aq) + 2 Cl- (aq) J = kR [ClO

-]2

(c) BrO3-(aq) + 5Br-(aq) + 6H+(aq) = 3Br2(aq)+ 3H2O(l) J = kR [BrO3

-][Br-][H+]2

(d) (CH3)3CBr(aq) + OH-(aq) = (CH3)3COH(aq) + Br

-(aq) J = kR [(CH3)3CBr]

(e) CO(g) + Cl2(g) = COCl2(g) J = kR [CO][Cl2]1.5

Question: Reaction (a) is that between persulfate ion and iodide ion in

aqueous solution. What are the partial orders with respect to these

reactants and the overall order of reaction?

Answer: The experimental rate equation is

J = kR [S2O82-

][I-]

In this equation the concentration of the persulfate is raised to the power

of 1: the partial order with respect to S2O82-

is therefore 1 and the reaction

is first-order in S2O82-

. Similarly, the partial order with respect to the

iodide is 1 and the reaction is therefore also first-order in this reactant.

The overall order is n = 1 + 1 = 2, that is, second-order overall.

It is very important to recognize in this example that

You cannot equate the partial orders of reaction for S2O82-

and I-

to their

balancing coefficients in the chemical equation. Partial orders of

There is no simple link between the stoichiometry of the

reaction and the form of the experimental rate equation.


38/83


reaction can be determined only from experimental measurements of

the kinetics of a process. In the case of S2O82-

it turns out by coincidence

that the partial order has the same value as the balancing coefficient. For

I-, the partial order and the balancing coefficient (equal to 3) are very

different.

Question: For reactions (b) and (c) what are the partial orders of reaction

with respect to the individual reactants and the overall order of reaction in

each case?

For reaction (d), the experimental rate equation does not depend on the

concentration of one of the reactants in the chemical equation, that is the

hydroxide ion, OH-. In this case the reaction is said to be zero-order in

OH-.

Question: Can you suggest why the term zero-order is used?

Answer: If the experimental rate equation is written in the general form

suggested earlier, then it would be

J = kR [(CH3)3CBr][OH

-]

Where = 1. Since the rate of reaction does not depend on [OH-] then the

only possible value for is zero; since [OH-

]

0

= 1. Thus, the reaction iszero-order in OH

-and a change of concentration of this reactant does not

affect the rate of reaction.

Reaction (e) demonstrates that a partial order of reaction may be

fractional; the partial order with respect to Cl2 is 1.5. This type of


39/83


behaviour is often found for gas-phase reactions that have a particular

type of mechanism.

The most important conclusion to be drawn from the above discussion is

that:

There is no systematic relationship between the stoichiometry of a

reaction and the partial orders of reaction that are determined by

experiment.

The only exception to this general conclusion is in the case of reactions

which, according to all available evidence, are elementary. This is

discussed more later. However, for now it can be noted that a simple

collision theory can predict the form of the experimental rate equation for

an elementary reaction involving two species. For reactions, which are

not elementary, no such theoretical approach is available. Indeed, if it

were, there would be no need for the large area of experimental chemical

kinetics that is currently in existence.

The rate equation for a chemical reaction, which providesinformation on the partial orders of reaction and the rate constant,

has to be determined experimentally.

Partial orders of reaction are of more interest that the overall order.

Essentially, the overall order of reaction provides a convenient means of

categorizing reactions, but otherwise is of little importance. We will see

later that it is the values of partial orders of reaction, together with the

value of the rate constant and the way in which it varies with temperature

that enables us to propose detailed mechanisms for reactions.


40/83

Determiningexperimental rate

equations at a fixed

temperature


41/83


Determining experimental rate equations at a fixed temperature

A strategy

To establish the form of an experimental rate equation it is necessary to

determine the values of both the partial orders of reaction and the

experimental rate constant. These is no definitive set of rules for carrying

out this process but the flow diagram below outlines a common strategy.

It is assumed here that the stoichiometry of the reaction has been

established and that kinetic reaction profiles are to be measured at a fixed

temperature.

One key strategic point is to distinguish between reactions that involve

either a single reactant or several reactants. We will start with reactions

involving only a single reactant.

NO

NO

YES

YES

Isolate in turnthe

contribution ofeach reactant

Determine the value of the experimental rate constant using ara hical method the inte ration method

Is there more thanone reactant?

Check for first-order

behaviour (half-life check)

Check for second-orderbehaviours, or use a more general

approach for determining thepartial order of reaction (the


42/83


Reactions involving a single reactant

A preliminary half-life check

The thermal decomposition of dinitrogen pentoxide (N2O5) in the gas

phase has time-independent stoichiometry

2 N2O5 (g) = 4 NO2 (g) + O2 (g)

A kinetic profile for N2O5 measured at 63.3 C is shown below:

It would be convenient if the kinetic reaction profile above could be used

directly without the need for further data processing, to obtain

information about the experimental rate equation for the decomposition

of N2O5. In fact, a preliminary check can be carried out using a method

based on the idea ofreaction half-life, which is denoted by 2/1t .

The half-life for a chemical reaction involving a single reactant can be

defined as:

0

0.5

1

1.5

2

2.5

3

3.5

4

0 100 200 300 400 500 600 700 800 900

time / s

[N2O5]x10^(-3)/moldm^(-


43/83


The time it takes for the concentration of the reactant to fall to one-half of

its initial value.

Question: What is the half-life for the decomposition of N2O5 at 63.3 C?

Answer: To determine 2/1t , the time taken for the initial concentration of

N2O5 (4.0 10-3

) to fall to one-half of its value, (i.e., 2.0 10-3).

According to the above figure this time is about 165 s; so 2/1t = 165 s.

Successive half-lives can be defined on the same kinetic reaction profile

of a reactant A with initial concentration [A0]:

The first half-life, 2/1t (1), corresponds to the time taken for the

initial concentration to fall to [A0],

The second half-life, 2/1t (2), corresponds to the time taken for the

concentration to fall from [A0] to [A0]..

The third half-life, 2/1t (3), corresponds to the time taken for the

concentration to fall from [A0] to1/8[A0].

There is no need to consider the fourth half-life, and so on.

Question: What are the successive half-lives for the decomposition of

N2O5 at 63.3 C ?

Answer: The successive half-lives are equal to one another, each having a

value close to 165 s.

The observation that successive half-lives are equal to one another is

important because such behaviour is unique to first-order reactions. The


44/83


experimental rate equation for the decomposition of N2O5 at 63.3 C must

therefore be

]ON[ 52RkJ=

We will come back to the determination of the experimental rate constant

later.

To summarize: It is useful to check whether successive half-lives for a

reaction are equal to one another, since, if this is the case, then it is safe to

conclude that the reaction is first-order. In order for the check to be

carried out, data must be available over at least two, preferably three,

half-lives.

The differential method

This method has been used for many years since it was first suggested in

1884 by the first Nobel prizewinner in Chemistry (1901) vant Hoff. The

method is sometimes named after him. Alternatively, as here, it is

described by the term differential, which reflects the fact that rate

equation are differential equations. However, this is not to imply that the

method involves calculus but simply to indicate that it seeks directly to

determine the form of experimental rate equations without changing them

into another form (as we will do later).

A reaction involving a single reactant is the gas-phase decomposition of

NO2 at 300 C (the homework question):

2 NO2 (g) = 2 NO (g) + O2 (g)


45/83


A preliminary half-life check of the kinetic reaction profile for NO 2

shows that the reaction is not first-order. (You should check this when

you get your homeworks back.)

To use the differential method it is necessary to propose a plausible rate

equation and to do this the simplestproposal is made.

Question: Can you suggest what this might be?

Answer: The simplest proposal would be that

]NO[ 2RkJ=

so that the rate of reaction depends only on the concentration of the

reactant NO2 raised to the power of the partial order .

It is important to recognize that the plausible rate equation is only a

suggestion. Nonetheless it is relatively simple and has the form that is

found experimentally for many chemical reactions.

The form of the proposed rate equation indicates that analysis should

focus on the relationship between the rate of reaction, J, and the

concentration of the reactant, [NO2].

Question: How isJdefined for the thermal decomposition of NO2 ?

Answer:

The stoichiometric numbers are 1and,2,222 ONONO

+=+== ;

hence

dt

d

dt

d

dt

dJ

]O[]NO[

2

1]NO[

2

1 22 =+==


46/83


The rate of reaction can therefore by determined from the kinetic reaction

profiles for NO2, NO or O2 and you were asked to find values ofJat

different times for the decomposition reaction in the homework. These

values are repeated in the Table below which, also includes the initial rate

of reaction and two further determinations at 250 s and 750 s. You should

note that this table includes the values of d[NO2]/dt from which J is

calculated (although values of d[NO]/dt or d[O2]/dtcould equally well

have been used for this purpose). In addition the table gives the values of

[NO2] at the selected times; these values are simply taken from the kinetic

reaction profile for NO2.

Time

s

[NO2] 10-3

mol dm-3

d[NO2]/dt 10-6

mol dm-3

s-1

J 10-6

Mol dm-3

s-1

0 4.00 -16.64 8.32

250 1.96 -3.79 1.90

500 1.30 -1.77 0.88

750 0.97 -0.97 0.48

1000 0.78 -0.61 0.30

1500 0.55 -0.32 0.16

This table provides all of the information necessary to proceed with the

differential method. There are two distinct ways of using the differentialmethod.

A check for second-order behaviour


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A preliminary half-life check has shown that the decomposition is not

first-order. It seems reasonable therefore that the next step should be to

check whether it is second-order, that is

22 ]NO[RkJ=

This equation can be rearranged, so that

RkJ

=2

2 ]NO[

In other words, the quantity 22 ]NO/[J should be constant throughout the

reaction. Values of this quantity taken at three different times in the

reaction, is given below.

Time / s 22 ]NO/[J / dm

3mol

-1s

-1

250 0.49

750 0.51

1500 0.53

Within the uncertainties inherent in the method, the values can reasonably

be taken to be constant. It can thus be concluded that the thermal

decomposition of NO2 at 300 C does have an experimental rate equation

of the form postulated:

]NO[ 2RkJ=

The partial order with respect to NO2 is 2 and, overall, the reaction is

second-order.

You will have noticed that the ratio 22]NO/[J is equal to the

experimental rate constant. One way of determining this rate constant,

therefore, would be to calculate the average value of this ratio using all of


48/83


the available date. This is an acceptable approach but, on balance, it is

better to use a graphical method which we will discuss later.

A general Approach

This is an approach for the determination of partial order that does not

involve any checking of data and which can be used to determine partial

orders ofany value, integral or otherwise.

If we continue with the previous example, then as already discussed, a

plausible rate equation is

]NO[ 2RkJ=

The quantity of interest in this equation is and it appears as an

exponent. In order to get at the exponent we need to take logarithms of

both sides of the equation, such that

)]NO[ln(ln 2

RkJ= In this case we have taken logarithms to the base e (ln), but it would have

been equally valid to select logarithms to the base 10 (log); as long as the

same thing is done to both sides of the equation, there is no reason to

prefer one type of logarithm to the other. We can now use the log rules to

simplify, giving

R

R

kJ

kJ

ln]NOln[ln

]NOln[lnln

2

2

+=

+=

This is in the same form as an equation for a straight line, so that

A plot of ln J versus ln([NO2]) should be linear if the assumed

form of the rate equation is correct


49/83


The slope of the straight line will be given by , thereby providing

a value for the partial order of reaction with respect to NO2.

A plot of ln(J / mol dm-3

s-1

) versus ln([NO2] / mol dm-3

) is a straight line

(you should prove this for yourselves). The fact that it is a straight line

confirms that the experimental rate equation is of the form proposed .

The numerical value of the slope of the straight line can be reasonably

taken to equal to 2, and so (as expected) the partial order with respect to

NO2 is 2.

The intercept of the ln J axis, which occurs when ln ([NO2]) = 0, will

provide a value of ln kR, and hence a value of the experimental rate

constant. However, the data points are located in a region of the graph

lying far from the intercept and to determine this value a long

extrapolation is required. The accuracy of this extrapolation will depend

critically on the quality of the experimental data.

Any small uncertainty in the slope of the line will result in a much greateruncertainty in the intercept. In general, the determination of an

experimental rate constant from the value of an intercept calculated from

such a plot can introduce fairly significant uncertainties and, for this

reason, it is not recommended.

To summarize.We have discussed two approaches for determining partial order using the

differential method. One is very specific and simply checks for second-

order behaviour; in essence, it is a trial-and-error method. The other is a

general graphical method that can be used to determine any partial order

of reaction, whether it be an integer or a fraction.


50/83


The Integration Method

A reaction involving a single reactant A

aA = products

May have an experimental rate equation of the form

]A[

]A[1Rk

dt

d

aJ == (*)

Mathematically, this rate equation can be described as being a differential

equation. By the application of a set of rules, it can be integrated, which

means that it is converted into an alternative form. Specifically,integrating the above equation (*) has the effect of changing the

relationship between rate and concentration to one between concentration

and time, and this provides a new way of analysing the kinetic data.

In the case of the above equation (*), the result of integration depends on

the numerical value of the partial order . The equations that result from

integration are referred to as integrated rate equations and they are

particularly important in the case of first order (= 1) and second-order

(= 2) reactions.

Reminder:

Integration as the inverse of differentiation: The Indefinite Integral

Integration is the inverse process to differentiation. When you

differentiate a function, say )(tf , you obtain its rate of change )(tf .

With integration you go the opposite way - you are starting with a

function representing the rate of change, and you are interested in

determining the cumulative change.


51/83


Some standard integrals are given below:

Special cases:

The integral of a constant kis

+= ckxkdx

The integral of 1, written simply as dx, not 1 dx, is

+= cxdx

Differential Equations

A differential equation is an equation that contains derivatives. That

integration is the inverse of differentiation is useful because the

mathematical description of physical processes is often given in the form

of differential equations, which must be undifferentiated or integrated in

order to obtain a useful solution.

caxa

axdx

caxa

axdx

ca

dx

caxdxax

aca

xdxx

axax

aa

+ =

+ =

+ =

++=+

++

=+

sin1

cos

cos1sin

e1

e

ln1

1for,1

1


52/83


The integration of the rate equation for first-order chemical reactions

Returning to the above example, the kinetics of a first orderchemical

reaction of the sort involving a single reactant A

aA = products

will have a first-order experimental rate equation of the form

]A[]A[1

Rkdt

d

aJ ==

Formal integration of this equation would proceed according the

following steps. The differential equation is rearranged so that thefunction of the variable [A] and the differential d[A] appear on one side,

and functions of the variable tand the differential dtappear on the other.

dtakA

AdR

][

][=

This equation is now integrated:

CtakA

dtakA

Ad

R

R

+=

=

]ln[

][

][

The value of the constant C is determined by noting that when the

reaction begins at the time t = 0, the concentration of A is equal to its

initial value, which will be denoted by [A]0. Putting tequal to zero then

gives

CA =0]ln[

so that

0]ln[]ln[ AtakA R +=


53/83


54/83


Question: If the slope of the straight line is 4.18 10-3 s-1, what is the

value of the experimental rate constant?

Answer: According to the integrated rate equation

)]ONln([2])ONln([ 05252 += tkR

the slope of a plot of ln([N2O5]) versus time will be equal to 2kR. Thus

13

13

s1009.2

s1018.42

-R

-R

k

k

=

=

Thus, the first order integrated rate equation provides a convenientgraphical method that uses all of the experimental data, for determining

the value of the experimental rate constant. In general, depending on the

quality of the experimental data that are available from a kinetic reaction

profile, the uncertainty in the computed value for a given experimental

rate constant can be quite small. For example, a fuller statistical analysis

of the data used above in the plot gives13

s10)02.009.2(-

Rk

= .

The integrated rate equation for a second-order reaction involving a

single reactant

For a second-order experimental rate equation of the form

2]A[

]A[1Rk

dt

d

aJ ==

Integration of this equation would proceed in the same way as for first-

order. The differential equation is rearranged so that the function of the

variable [A] and the differential d[A] appear on one side, and functions of

the variable tand the differential dtappear on the other.

dtakA

AdR

][

][2

=


55/83


This equation is now integrated:

CtakA

CtakA

CtakA

dtakAdA

R

R

R

R

+=

+=

+=+

=

+

][

1

1

][

12][

][][

1

12

2

The value of the constant Cis again determined by noting that when the

reaction begins at the time t = 0, the concentration of A is equal to its

initial value, denoted by [A]0. Putting tequal to zero then gives

CA

=0][

1

so that

0][

1

][

1

Atak

AR +=

Once again, this equation represents a straight line.

So a second-order integrated rate equation also provides a convenient

graphical means of determining the value of an experimental rate

constant.

Question: You showed in homework 2 that the gas-phase decomposition

of NO2 at 300 C

2 NO2 (g) = 2 NO (g) + O2 (g)

has a second-order experimental rate equation


56/83


22 ]NO[RkJ=

Use the information given in the homework to determine the value of the

experimental rate constant for this decomposition.

Determining reaction order

The most important use of integrated rate equations is in the

determination of values for experimental rate constants. However, they

are sometimes used in a trial-and-error procedure to determine whether a

reaction is first- or second-order. The essence of the method is to see

whether a good straight-line plot is obtained.

Thus, without any prior knowledge of the form of the experimental rate

equation a straight-line plot would be very strong evidence that the

reaction was the order used to define the straight-line equation.

There is, however, one important point to note:

To use first- and second-order integrated rate equations in a trial-and-

error procedure to determine order, it is essential that the data analysed

should extend to at least 50 % of complete reaction, and preferably more.

To summarize:

Integrated rate equations provide a convenient graphical means, taking

into account all of the experimental data, for the determination of the

values of experimental rate constants. First- and second-order integrated

rate equations can be used on a trial-and-error basis to establish order but

the experimental data must extend to at least 50 % of complete reaction.


57/83

Reactions involving

several reactants


58/83


Reactions involving several reactants

The majority of chemical reactions involve not one but several reactants

and so it is important to consider how to establish the form of an

experimental rate equation in these circumstances,

If we consider a general reaction between two reactants A and B

aA + bB = products

then we need to be able to establish whether the rate equation is of the

form

]B[]A[RkJ=

The problem is that the rate of reaction now depends upon the

concentration of both reactants and as a consequence, it is difficult to

disentangle the effect of one from the other. If there is a third reactant

then the situation becomes even more complex. The solution to the

problem is to arrange matters experimentally so that the analysis of the

kinetic data can be simplified. There are two ways to achieve this:

1. The isolation method this is quite general

2. The initial rate method this is more restricted and applies only to

the initial states of reaction.

The isolation method

Earlier in the course we used the reaction between hypochlorite ion and

bromide ion in aqueous solution


59/83


ClO (aq) + (aq) = (aq) + Cl (aq)Br BrO

Question: What would be a plausible rate equation for this reaction?

Answer: It would be of the form

]Br[]ClO[ -RkJ=

The kinetic reaction profile we discussed for this reaction was one for

which the initial concentrations of [ and [ were relatively

small and similar in magnitude (3.230 10

0]ClO

-

0]Br

-3mol dm

-3and = 2.508 10-3

mol dm-3

, respectively). However, the reaction can be investigated over a

much wide range of reactant concentrations.

The initial concentration of ClO is [ = 100 100]ClO-

]Br[

3 mol dm-3

. It is

considerably in excess of that for Br: = 2.00 1003 mol dm

-3. In

fact,

50][

]ClO[

0

0 =Br

-

and so ClO

is referred to as being in fifty-fold excess. The consequence

of this large excess can clearly be seen; during the course of the reaction

Br

is nearly all consumed, where as most of the ClO

remains unreacted.

This being the case it is a reasonable approximation to treat the

concentrations of ClO as remaining constant as its initial value, [ ,

throughout the course of the reaction. Under these circumstances, the

proposed experimental rate equation becomes

0]ClO-


60/83


]Br[]ClO[ 0-

RkJ=

and the term , which is effectively a constant, is referred to

as a pseudo-order rate constant. It is represented by , so that

0]ClO[

Rk

'Rk

0

' ]ClO[ = RR kk

The proposed experimental rate equation therefore becomes

]Br[' -RkJ=

The crucial point about this equation is that the rate of reaction now

depends only on the concentration of Br; in other words the kinetic

contribution of this reactant has been isolated by arranging for the

concentration of ClO

to be in large excess.

The form of the expression is exactly the same as that which would be

proposed for a reaction involving a single reactant. Hence, the various

methods available for determining the partial order for a single-reactant

process can be applied in an identical way in order to determine .

Question: Values of [Br] as a function of time are given below. By

plotting the kinetic reaction profile for [Br], determine a value for.

Time / s [Br] / mol dm-3

0 2.00 10-3

15 1.57 10-3

28 1.27 10-3

44 0.98 10-3

70 0.64 10-3

91 0.46 10-3

130 0.25 10-3


61/83


Answer: The first three successive half-lives are very close in value

(approximately 42 s in each case) which is typical of a first order

reaction; thus = 1.

The above equation is therefore referred to as a pseudo-

order rate equation in fact it is pseudo-first-order.

]Br[' -RkJ=

Question: Use the information taken from the kinetic reaction profile for

Br

to determine a value for the pseudo-order rate constant in

.

'

Rk

]Br[' -RkJ=

The kinetic reaction profile data consists of bromide concentration and

time. We can therefore, either determine J by taking the gradient of a

series of tangents to the curve and then plotting lnJagainst ln [Br] or by

using the integration method as shown below:

We know from the stoichiometric equation that

ClO (aq) + (aq) = (aq) + Cl (aq)Br BrO

and so

]Br[]Br[

1

1 ' -R

-

kdt

dJ ==

Therefore, we can integrate this to give

Ctk

dtkd

R-

R-

-

+=

=

'

'

]Brln[

]Br[]Br[

1


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From the initial conditions (t = 0, [Br] = [Br]0) we can determine the

integration constant Cand write the first-order integrated rate equation as

0' ]Brln[]Brln[ -R

- tk +=

A plot of against time is shown below, and from the gradient we

can determine that .

]Brln[ -

12' 1060.1 = skR

y = -0.016x - 6.2199

-8.500

-8.000

-7.500

-7.000

-6.500

-6.000

0 20 40 60 80 100 120 14

time / s

ln[Br-]

0

To determine the partial order with respect to ClO

it is necessary to carry

out a second experiment but this time with Br in large excess, that is

[Br]0 >> [ClO]0

In this case, the concentration of Br

is treated as remaining constantthroughout the course of the reaction so that

]ClO['' = RkJ

where is another pseudo-order rate constant. We have

already established that = 1. In fact, although we shall not go into

0

'' ]Br[ -RR kk =


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detail, it turns out that = 1 and so the experimental rate equation for the

reaction between ClO and Br in aqueous solution is

]Br][ClO[ = RkJ

One option for determining the value of the experimental second-order

rate constant, k , is to return the equation .

Rearranging this equation gives

R

0' ]ClO[ = RR kk

0

'

]ClO[ = RR

kk

Values of , [ and are know and so, therefore'Rk 0]ClO-

113

3-

12

smoldm16.0

dmmol1.0

s1060.1

=

=

-

Rk

To summarize:

The isolation method provides a very valuable means of investigating the

chemical kinetics of reactions that involve two, or more, reactants. In

essence, it involves isolatingin turn the contributions of each reactant by

arranging (experimentally) that all of the other reactants are in large

excess, such that their concentrations remain virtually unchanged during

the course of reaction. Normally this means at least a ten-fold, but more

preferably forty-fold or more, excess in concentration compared with the

initial concentration of the reaction to be isolated.


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It is worth noting that some types of reaction automatically have one

reactant in large excess. Thus, for example, benzenediazonium chloride

(C6H5N2Cl) decomposes in water at 40 C to liberate nitrogen gas

C6H5N2Cl (aq) + H2O (l) = C6H5OH (aq) + N2 (g) + HCl (aq)

In this case water is the solvent, as well as a reactant, and it remains in

large excess throughout the reaction. In this, and similar circumstances,

the experimental rate equation is often written in a way that does not

explicitly take the water into account. So for the above reaction, the

experimental rate equation would be proposed to be

]ClNHC[ 256RkJ=

where it is understood that k is a pseudo-order rate constant.R

The initial rate method

This method, as its name implies, focuses on the initial stages of a

reaction and it is initial rates of reaction that are measured. If a reaction

has time-independent stoichiometry then it is reasonable to assume that

the kinetic information obtained from investigating the initial states of the

reaction will apply throughout its whole course.

Alternatively, if a reaction forms products that decompose, or interfere in

some way with the progress of a reaction, then the initial rate method may

be the only viable means of obtaining useful kinetic information. The

method is not included in the flow chart since it is essentially a stand-

alone approach.

If we return to a general reaction between two reactants


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67/83


'

0

0

]ClO[R-

kJ

=

i.e., the fraction should, within experimental uncertainty, equal a constant

value. Thus is the case of Experiment 1:

1-4

3-3-

1-3-6

0

0

s1088.9

dmmol103.230

sdmmol1019.3

]ClO[

=

=

-

J

the corresponding value for Experiment2 and 3 are 9.85 10-4 s-1 and

9.88 10-4 s-1, respectively. Thus in all three experiments the value of

0

0

]ClO[ -J

is essentially constant.

If the partial order with respect to ClO-

had no been equal to 1 then it

would have been necessary to proceed by a trial-and-error method in

order to find the value of that made

0

0

]ClO[ -J

equal to the same

constant for all three experiments.

To summarize:

The initial method is essentially an isolation technique but it does notrequire that any reactants have to be in large excess. In general, for a

reaction involving two or more reactants, arranging that the initial

concentrations of the others be held at fixed values during a series of

experiments isolates one of these. The main application of the method is

for the determination of partial order. Values of pseudo-order rate


68/83


constants can be determined but with an accuracy that, in turn, depends

on how accurately initial rates of reaction can be measured.


69/83

The Effect of the

Temperature on the

Rate of a Chemical

Reaction


70/83


The Effect Of Temperature On The Rate Of A Chemical Reaction

The Arrhenius Equation

As we have seen, in order to determine the form of the experimental rate

equation for a chemical reaction it is necessary to carry out experiments

at a fixed temperature. This is to avoid any complications due to the rate

of reaction changing as a function of temperature.

In general, it is the rate constant, k , for a chemical reaction that is

temperature-dependent and this is illustrated below for the reaction

between iodomethane (CH

R

3I) and ethoxide ion (C2H5O) in a solution of

ethanol

CH3I + C2H5O

= C2H5OCH3 + I

0

0.001

0.002

0.003

0.004

0.005

0.006

280 285 290 295 300 305 310 315

temperature / K

k_

R/dm^3mol^-1s^-1

This reaction is second-order overall, that is the experimental rate

equation is of the form

]OHC][ICH[ 523= RkJ


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and so, as shown in the figure above the units of are dmRk-3

mol-1

s-1

. It

is very important to note in the figure that temperatures are expressed on

the Kelvin scale, rather than the Celsius scale.

For a period of over 60 years, spanning 1850 to 1910, in spite of

considerable experimental effort, there was much uncertainty and, in

some cases, controversy as to how to describe the temperature-dependent

behaviour of the kinetics of a chemical reaction. However, from about

1910 onwards the equation that gained general acceptance was the one

used by Svante Arrhenius in a paper entitled, On the reaction velocity of

the inversion of cane sugar by acids, published in 1889. The equation,

which is now referred to as the Arrhenius equation, takes the form

=

RT

EAk aR exp

so that an exponential dependence is involved in relating the experimental

rate constant to temperature (T).

The parameter is known as the Arrhenius activation energy but

usually is just referred to as the activation energy. The parameterA is the

ArrheniusA-factor, but again usually shortened to justA-factor; the terms

pre-exponential factor or frequency factor are sometimes also used.

aE

Together the two parameters A and are known as the Arrhenius

parameters. The quantityR is the gas constant; R = 8.31441 J K

aE

-1mol

-1.

An important reason for the widespread acceptance of the Arrhenius

equation was that the parameters A and could be given a physicalaE


72/83


meaning. A very important practical aspect of the Arrhenius equation is

that it accounts very well for the temperature-dependent behaviour of a

large number of chemical reactions. In this context the Arrhenius

parameters can be treated simply as experimental quantities. Even so,

they provide very useful information about a given reaction, particularly

in terms of the magnitude of its activation energy.

Determining The Arrhenius Parameters

In order to determine the Arrhenius parameters for a reaction it is

necessary to determine values of the experimental rate constant as a

function of temperature. This set of data is then fitted to the Arrhenius

equation using a graphical procedure.

For many reactions, particularly in solution involving organic solvents,

Arrhenius studies are restricted to differences between the melting and

boiling temperatures or the solvent and this limits the number of datapoints that can be collected.

Given the Arrhenius equation we can take logarithms to the base e, i.e.,

RT

EAk

RTEAk

RT

EAk

aR

aR

aR

=

=

=

lnln

explnln

exp

If the order of the terms on the right-hand side are changed then

ART

Ek aR lnln +=


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This equation is in the form of a straight line so that a plot of ln kRversus

1/T should be linear. This assumes that both A and are constants,

independent of temperature.

aE

This is a reasonable assumption for most reactions when studies over a

limited range of temperature, say of the order of 100 K. A graph of ln kR

versus 1/T is referred to as an Arrhenius plot, and it involves reciprocal

temperature or inverse temperature.

Question: What is a suitable label for a column heading in a table of

reciprocal temperature?

Answer:

Suppose a reaction is studied at a temperature such that T= 297 K. The

reciprocal or inverse temperature will be

13 K10367.3

K297

11

=

=

T

To plot a graph, only pure numbers can be used. If

13 K10367.31 =T

then, multiplying both sides by K, gives

10367.3K 3=T

Thus a label for a column heading in a table could be K/T. However, it is

more convenient to include the power of ten in the label. If we return to

the previous equation and multiply both sides by 103, then


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( )

367.3

K10

therefore,

10367.310K

10

3

333

=

=

T

T

We shall use the label 103

K / Tfor the horizontal axes of Arrhenius plots.

The figure below shows an Arrhenius plot for the reaction between

iodomethane and ethoxide ion in a solution of ethanol based on the data

used to plot the previous graph.

he Arrhenius plot is a good straight line.

uestion

-9

-8.5

-8

-7.5

-7

-6.5

-6

-5.5

-5

3.2 3.25 3.3 3.35 3.4 3.45 3.5 3.55 3.6

10^3 K / T

ln(k_

R/d

m^-3mol^-1s^-1)

T

Q : If the coordinates for two points on this line are (1/T = 3.25

nswer

10-3

K-1

, ln(kR / dm3

mol-1

s-1

) = - 5.70) and (1/T = 3.50 10-3 K-1, ln(kR/

dm3

mol-1

s-1

) = - 8.15), what is the value of the slope?

A : The slope is calculated as follows


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K1080.9

K1025.0

45.2

K)1025.31050.3(

)70.5

3

13

133

+

=

=

(15.8slope

=

uestionQ : What is the activation energy?

nswer

A : According the equation used to plot the graph

ART

kR lnln +=Ea

and so the slope is equal toR

Ea

=

, so it follows that

uestion

( ) ( )

1

13

113

molkJ5.81

molJ105.81

molJ K314.8K1080.9

slope

=

=

=

REa

Q : How is the intercept related to theA-factor?

nswer

A : The straight line will intercept the ln(kR) axis when 1/Tis zero.

he computed value of the intercept (taken from Excel) is 26.21. Since

The intercept, therefore, is equal to lnA.

T

the A-factor has the same units as kR, then


76/83


11311

11113

113

113

smoldm1042.2

1042.2smoldm

)21.26exp(smoldm

21.26smoldm

ln

=

=

=

=

A

A

A

A

herefore, the value of the A-factor is calculated to be 2.42 1011 dm3

e noted that, as shown by the extended Arrhenius plot below

ence of the relatively long extr

Introduction to Reaction Kinetics, Hazle Cox

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Transcript of Introduction to Reaction Kinetics, Hazle Cox