Chemical Kinetics

1

CHEMICAL KINETICS 25-05-2015

Contact: By: Muhammad Asif

Phone: +92-303-7807073 Lecturer (Physical Chemistry)

E-mail: [email protected] Govt. College Sahiwal, Pakistan

4.1 Introduction 02

4.2 Rate of reaction ... 02

4.3 Rate equation and rate constant.. 03

4.4 Order and molecularity of reactions ... 04

4.5 Pseudo order reactions 05

4.6 Zero order reactions ... 06

4.7 First order reactions ... 07

4.8 Second order reactions ... 09

4.9 Third order reactions .. 13

4.10 Elementary and complex reactions 23

4.10.1 Opposing reactions . 23

4.10.2 Consecutive reactions . 34

4.10.3 Parallel reactions .... 36

4.10.4 Chain reactions 37

4.11 Thermal reactions .... 37

4.12 Photochemical reactions .. 40

4.13 Effect of temperature on rate of reaction .... 44

4.14 Theories of reaction rates 48

4.14.1 Collision theory ... 48

4.14.2 Transition state theory . 52

2

4.1 INTRODUCTION

Chemical kinetics is the branch of chemistry that deals with the rates of chemical

reactions. The chemical reactions can broadly be divided into three classes:

1. Fast reactions e.g., the reaction between aqueous sodium chloride and silver nitrate.

2. Moderate reactions e.g., hydrolysis of an ester.

3. Slow reactions e.g., rusting of iron.

However, the chemical reactions can occur at a variety of rates from very fast to very slow.

For example, the paper on which this book is printed reacts extremely slowly with the

atmosphere oxygen, whereas explosion is extremely fast reaction. The two extreme classes of

reactions cannot be studied conveniently. Reactions which proceed with measureable rates

are the subjects of study by physical chemists.

In 1864 C. M. Guldberg and P. Waage pioneered the development of chemical

kinetics by formulating the law of mass action, which states that the speed of a chemical

reaction is proportional to the quantity of the reacting substances.

Experimentally it has been found that the rate of a chemical reaction depends on the

nature of the reacting species, the temperature, the pressure, the concentration of the reacting

species and the presence of catalyst.

The reactions may be classified kinetically as being either homogenous, if they take

place in one phase only, or heterogeneous if two or more phases are involved in the

processes. The reactions may be elementary or complex. An elementary reaction occurs in a

single step while a complex reaction occurs in two or more steps.

4.2 RATE OF REACTION

The rate of a chemical reaction is the change in the concentration of reactants or

products in a unit time or a given time.

Change in concentration c

RateTime period of change t

This equation gives the average velocity of the reaction during the time of observation. By

shortening the time of observation the average velocity approaches more and more closely to

the actual velocity of the reaction. So in chemical kinetics the rate of reaction is not

represented by the average velocity but by the instantaneously velocity of the reaction that is

when the time of observation approaches zero then,

d[c]

Ratedt

3

The brackets [ ] represent the concentration in moles per decimeter cube. Thus rate of

reaction is expressed in units of moles per decimeter cube per second ( 3 1mol dm sec ). For

gas phase reactions, pressure units are used in place of molar concentration.

Consider the following general reaction,

A + B C + D

The rate at which reaction proceeds can be measured in terms of the rate at which one of the

reactants disappears or one of the products appears.

dx d[A] d[B] d[C] d[D]

dt dt dt dt dt

Where [A], [B], [C] and [D] represent the concentrations of A, B, C and D in moles per

decimeter cube at time t. The negative sign indicates that the concentrations of A and B

decreases with increasing time. The plus sign indicates that the concentrations of C and D

increase with increasing time.

4.3 RATE EQUATION AND RATE CONSTANT

According to law of mass action, the rate of a chemical reaction is proportional to the

product of the molar concentrations of the reactants raised to the power equal to the number

of molecules of each species taking part in the reaction. Consider the reaction,

aA + bB cC + dD

a bdx

[A] [B]dt

a bdx

= k[A] [B]dt

Where, k is constant of proportionality known as the velocity constant or rate constant or

specific rate constant. This expression shows how the reaction rate is related to

concentration and is known as rate equation, rate expression or rate law.

If the concentration of A and B are kept unity, then

a bdx

= k[1] [1] = kdt

Thus rate constant may be defined as the rate of reaction when molar concentration of

each of the reactants is unity (1 mol dm3

). The value of k varies from reaction to reaction

and also varies with temperature for a given reaction. The value of k for a reaction does not

change with time. The units of k depend on the order of reaction.

4

Anyhow it is not essential that the powers in the rate equation are always equal to the

coefficients of balance chemical equations. So we can say that the rate equation of above

reaction may be,

m ndx

= k[A] [B]dt

The power or exponent of a concentration term, m or n in the rate equation is usually in small

whole number integers (0, 1, 2, 3) or may be fractional. Here n is called the order of reaction

with respect to A, and m is the order of the reaction with respect to B. The sum (n + m) is

called the overall order of the reaction.

Rate equation for an elementary reaction can be predicted from balanced chemical

equation but it cannot be predicted for complex reactions (involving more than one steps).

Therefore it is written from experimental facts for complex reactions. Actually rate equation

is an experimental expression. The order of the reaction is completely independent of the

reaction stoichiometry.

4.4 ORDER AND MOLECULARITY OF REACTIONS

The order of a reaction is defined as the total number of atoms, ions or molecules

whose concentration changes during a chemical reaction. It can be calculated from an

experimentally determined rate equation. It is the sum of exponents of all the

concentration terms in the differential form of rate equation. A reactant whose

concentration does not affect the rate of reaction is not included in the rate equation. The

concentration of such a reactant has the power 0 e.g., [A]0 =1. For a reaction maximum order

is three and the minimum is zero. However, the order of reaction may be in minus with

respect to a particular species in the reaction; such a species acts as an inhibitor and rate of

reaction is inversely proportional to its concentration. The reactions may be classified

according to the order. If (m + n) in the rate equation is:

m + n = 0 the reaction is zero order reaction

m + n = 1 the reaction is first order reaction

m + n = 2 the reaction is second order reaction

The order of reaction provides valuable information about the mechanism of a reaction and

the knowledge of mechanism of a given reaction allows us to control that reaction.

The order of reaction should not be confused with molecularity which is defined as

the total number of atoms, ions or molecules which take part in a reaction as given by

the balanced chemical equation. Molecularity and order are identical for elementary

reactions. Most chemical reactions are complex reactions which occur in a series of steps.

5

Each step is an elementary reaction. The stepwise sequence of elementary reactions that

convert reactants to products is called mechanism of the reaction. In any mechanism, some

of the steps will be fast, others will be slow. The rate of the reaction is determined by the

slowest step known as the rate determining step. Hence, molecularity of a reaction can

also be defined as the total number of atoms, ions or molecules taking part in the rate

determining step. The term unimolecular, bimolecular and termolecular (or trimolecular)

indicate the number of particles reacting in a single elementary process.

No Order of Reaction Molecularity

1 It may be equal to zero. It can never be equal to zero.

2 It can be in fractions. It can never be in fractions.

3 It is at most equal to three. It can be more than three.

4 It can only be determined

experimentally.

It can be obtained from a simple balanced

chemical equation.

5 It is the sum of all the

exponents of the concentration

terms in the rate equation.

It is the sum of number of molecules of the

reactants taking part in a single step chemical

reaction.

6 It helps in determining the

mechanism of a reaction.

It gives no idea about the mechanism of a reaction.

7 In case of complex reactions

order is determined by the

slowest step of the reaction.

In case of complex reactions each step of the

reaction has its own molecularity i.e., molecularity

has no significance for complex reactions.

4.5 PSEUDO ORDER REACTIONS

A reaction in which one of the reactants is present in large excess shows an order

different from the actual order. The experimental order which is not the actual one is

referred as the pseudo order. Since for elementary reactions molecularity and order are

identical, pseudo reactions may also be called pseudo molecular reactions.

Let us consider a reaction,

A + B Products

A + B Products

In which the reactant B is present in a large excess. Since it is an elementary reaction, its rate

equation can be written as;

Rate = k [A] [B]

As B is present in large excess its concentration remains practically constant in the course of

reaction. Thus the rate law can be written as;

Rate = k [A]

6

Where the new rate constant k = k [B]. Thus the actual order of the reaction is second order

but in practice it will be first order. Therefore, the reaction is said to have a pseudo first order.

Examples:

1. Hydrolysis of an ester: methyl acetate upon hydrolysis in aqueous solution using a

mineral acid as a catalyst forms acetic acid and methyl alcohol.

CH3COOCH3 + H2O CH3COOH + CH3OHH

Here a large excess of water is used and the rate law can be written as

Rate = k [CH3COOCH3] [H2O]

Rate = k [CH3COOCH3]

The reaction is actually second order but in practice it is found to be first order. Thus

it is a pseudo first order reaction.

2. Hydrolysis of sucrose: sucrose upon hydrolysis in the presence of a dilute mineral

gives glucose and fructose.

4.6 ZERO ORDER REACTION

The reaction in which rate is independent of the concentration of the reactants is

called zero order. Let us take a substance A which decomposes into products. Its initial

concentration is a moles dm3. Let after time t seconds the amount left behind is (a-x)

moles dm3

and that converted into product is x moles dm3. Then

A Products

A Products

When t = 0 a 0

When t = teq (a-x) x

Hence, the rate of reaction is given by,

dx

k(a x)dt

dx

kdt

dx kdt

Integrating,

dx k dt

x kt C

The value of integration constant C can be found by applying the initial conditions of the

reaction when t = 0, x = 0. That is

x kt (1)

7

This equation (1) is similar to the equation for a straight line. Thus, if we plot a graph

between x on y-axis and t on x-axis, we should get a straight line passing through the origin

with slope k.

Rearranging Equation (1) for k,

xk

t (2)

This equation (2) gives the rate constant for zero order reaction. The units of k are mole

dm3

sec1

.

Half-Life

It is defined as the time required to reduce the concentration of reactants to half of its

original value. Therefore, when t = t1/2 then x = a/2. Substitute this value in equation (2),

a2

1/2

at

k 2k

Examples

1. Thermal decomposition of HI on gold surface,

Gold Surface

2HI

Gold Surface

Gold Surface

H2 + I

2

Gold Surface

2HI

Gold Surface

Gold Surface

H2 + I

2

2. Decomposition of ammonia in the presence of tungsten,

Tungston

2NH3

Tungston

Tungston

N2 + 3H

2

Tungston

2NH3

Tungston

Tungston

N2 + 3H

2

3. Photochemical reactions are usually zero order.

4. The reactions which are catalyzed by enzymes are also zero order.

4.7 FIRST ORDER REACTIONS

The reaction in which the concentration of only one molecule is changed is called

a first order reaction. Let us take a substance A which decomposes into products. Its initial

8

concentration is a moles dm3. Let after time t seconds the amount left behind is (a-x)

moles dm3

and that converted into product is x moles dm3. Then

A Products

A Products

When t = 0 a 0

When t = teq (a-x) x


dx

k(a x)dt

dx

kdt(a x)

dx

k dt(a x)

ln(a x) kt C (1)



ln(a 0) k(0) C

lna C

Therefore Eq.(1) becomes,

ln(a x) kt lna

lna ln(a x) kt

a kt

loga x 2.303

(2)

This equation is similar to the equation for a straight line. Thus, if we plot a graph between

alog

(a x)on y-axis and t on x-axis, we should get a straight line passing through the origin

with slope k/2.303.


2.303 ak log

t a x

(3)

This equation (3) gives the rate constant k of first order reaction. The units of k are sec1

.

Half-Life

When t = t1/2 then x = a/2. Substitute this value in equation (3),

9

1/2 a2

2.303 at log

k a ( )

1/2

2.303 0.693t log2

k k

Examples

1. Thermal decomposition of hydrogen peroxide is a first order reaction,

H2O2 H2O + O

2. Decomposition of nitrogen pentoxide in CCl4 solution is a first order reaction,

N2O5 2NO2 +1/2 O2

3. The hydrolysis of an ester in the presence of a mineral acid as catalyst,

CH3COOC2H5 + H2O CH3COOH + C2H5OHH

4.8 SECOND ORDER REACTIONS

The reaction in which two molecules undergo a chemical change is called a

second order reaction. Let us consider two substances A and B which react to give the

products. The reaction can be carried out by;

a) By taking equal concentrations of A and B

b) By taking different concentrations of A and B

a) Second order reaction with equal concentrations of reactants

Let the initial concentrations in moles dm3

are a moles dm3

and after time t the

concentrations left behind are (a-x) for both A and B. Then

A + B Products

A + B Products

When t = 0 a + a 0

When t = teq (a-x) + (a-x) x


dx

k(a x)(a x)dt

2dx

k(a x)dt

2

dxkdt

(a x)

2

dxk dt

(a x)

2(a x) dx k dt

10

2 1(a x) ( 1)

kt C2 1

1

kt C(a x)

(1)



1 1

k0 C C(a 0) a

Therefore equation (1) becomes,

1 1

kt(a x) a

1 1

kt(a x) a

x

kta(a x)

x

akt(a x)

(2)


x

(a x)on y-axis and t on x-axis, we should get a straight line passing through the origin

with slope ak.


1 x

kt a(a x)

(3)

This equation gives the rate constant k for second order reaction with same initial

concentrations of the reactants. The units of k are dm3 mol

1 sec

1.

Half-Life

When t = t1/2 then x = a/2. Substitute this value in equation (3),

a2

1/2 a2

1 1t

k a(a ) ka

Examples

Saponification of an ester with a strong base is a second order reaction in solution,

CH3COOC

2H

5 + NaOH CH

3COONa + C

2H

5OH

CH3COOC

2H

5 + NaOH CH

3COONa + C

2H

5OH

11

b) Second order reaction with different concentrations of reactants

Let the initial concentrations of reactants A and B are a and b moles dm3 respectively,

and x moles dm3 of the reactants are disappeared in time t. Then

A + B Products

A + B Products

When t = 0 a + b 0

When t = teq (a-x) + (b-x) x

Hence, according to rate law,

dx

k(a x)(b x)dt

dx

kdt(a x)(b x)

dx

k dt(a x)(b x)

(1)

In order to integrate the L.H.S., we have to do the partial fractions first by using partial

fraction method,

1 A B

(a x)(b x) (a x) (b x)

(2)

1 A(b x) B(a x) (3)

To get A, put (a-x) = 0 or x = a in equation (3),

1 A(b a)

1 1

Ab a (a b)

To get B, put (b-x) = 0 or x = b in equation (3),

1 B(a b)

1

B(a b)

Substituting the value of A and B into equation (2),

1 1 1

(a x)(b x) (a b)(a x) (a b)(b x)

Now substituting this value in equation (1),

1 1

dx dx k dt(a b)(a x) (a b)(b x)

1 dx 1 dx

k dt(a b) (a x) (a b) (b x)

12

1 1

[ ln(a x)] [ ln(b x)] kt C(a b) (a b)

1 1

ln(a x) ln(b x) kt C(a b) (a b)

1

ln(a x) ln(b x) kt C(a b)

1 (a x)

ln kt C(a b) (b x)

(4)



1 (a 0)

ln k0 C(a b) (b 0)

1 a

ln C(a b) b

Therefore equation (4) becomes,

1 (a x) 1 a

ln kt ln(a b) (b x) (a b) b

1 (a x) 1 aln ln kt

(a b) (b x) (a b) b

1 (a x) a

ln ln kt(a b) (b x) b

1 b(a x)

ln kt(a b) a(b x)

2.303 b(a x)

log kt(a b) a(b x)

b(a x) (a b)k

log ta(b x) 2.303

(5)


L.H.S. of this equation (5) on y-axis and time t on x-axis, we should get a straight line

passing through the origin with slope (a b)k

2.303

.


2.303 b(a x)

k logt(a b) a(b x)

(6)

13

This equation (6) gives the rate constant k for a second order reaction when the initial

concentrations of the reactants are different. The dimensions of k are again dm3 mol

1 sec

1.

Half-Life

The half-life method cannot be used for reactions where the concentrations of A and B are

different, since A and B will have different times for half reaction.

Examples

1.

H2 + I

2 2HI

H2 + I

2 2HI

2.

2HI H2 + I

2

2HI H2 + I

2

3.

2O3 3O

2

2O3 3O

2

4.

2NO2 N

2 + 2O

2

2NO2 N

2 + 2O

2

4.9 THIRD ORDER REACTIONS

The reaction in which only three molecules undergo a chemical change is called a

third order reaction. Four different cases of 3rd

order reaction are discussed here.

Case (I): when the initial concentration of all the reactants is same

Consider the following reaction in which A, B and C reactants are converting into

products in an elementary reaction. All reactants have same initial concentration (a). After

time (t) the concentration of each reactant becomes equal to a x as given below;

A + B + C Products

A + B + C Products

When t = 0 a a a 0

When t = t ax ax ax x

Rate equation for such a reaction can be written as

dx

k[A][B][C]dt

dx

k(a x)(a x)(a x)dt

3dx

k(a x)dt

This is the differential form of rate law for a third order reaction in which concentration of

all the reactants is same.

Integrating the above equation after separating variables,

3

dxk dt

(a x)

3(a x) dx k dt

14

3 1(a x)

kt c( 3 1)( 1)

2(a x)

kt c2

2

1kt c

2(a x)

(1)

When t = 0, x = 0, putting these values in equation (1), the value of c can be calculated

2

1c

2a

Hence, equation (1) becomes

2 2

1 1kt

2(a x) 2a

2 2

1 12kt

(a x) a

(2)

This is the integrated form of rate law for third order reaction with same initial

concentrations of reactants. This equation is similar to the equation for a straight line.

Therefore if we plot a graph between 2

1

(a x)on y-axis and t on x-axis, we should get a

straight line with slope equal to 2k and intercept 21/ a .

Also from equation (2),

2 2

1 1kt

2(a x) 2a

2 2

2 2

a (a x)kt

2(a x) a

2 2 2

2 2

a (a x 2ax)kt

2(a x) a

2 2 2

2 2

a a x 2axkt

2(a x) a

2

2 2

2ax xkt

2(a x) a

2 2

x(2a x)kt

2(a x) a

2 2

1 x(2a x)k

2a t (a x)

(3)

15

This equation (3) gives the rate constant k. Units of k can be determined as follows,

3 3

3 2 3 2

1 (mol dm )(mol dm )k

(mol dm ) sec (mol dm )

2 6 1k mol dm sec

Half-Life

When t = t1/2 then x = a/2 = 0.5a. Substitute this value in equation (3),

2 2

1 x(2a x)t

2a k (a x)

1/2 2 2 2 2

1 0.5a(2a 0.5a) 3 1.5t

2a k (a 0.5a) 2ka ka

(4)

It is evident from the above equation that the half-life period is inversely proportional to the

square of the initial concentration of the reactant.

In general for a reaction of nth order, the half-life is found to be,

1/2 n 1

1t

a (where n is the order of reaction)

Thus, half-life period of any order reaction is inversely proportional to the initial

concentration raised to the power one less than the order of that reaction.

Examples:

The reaction between NO and H2 to give N2 and H2O follows the third order reaction

kinetics. The reaction is,

2NO + 2H2 N

2 + 2H

2O

2NO + 2H2 N

2 + 2H

2O

The rate law for this reaction is given by

Rate = k [H2][NO]2

Case (II): when the initial concentration of two reactants is same and that of the third

one is different

Consider the following reaction in which the reactants A and B have the same initial

concentrations equal to a but reactant C has different concentration equal to c as given

below,

A + B + C Products

A + B + C Products

When t = 0 a a c 0

When t = t ax ax cx x

Rate equation for such a reaction is given by

16

dx

k[A][B][C]dt

dx

k(a x)(a x)(c x)dt

2dx

k(a x) (c x)dt

This is the differential rate equation for a third order reaction in which concentration of two

reactants are same and that of the third one is different.


2

dxk dt

(a x) (c x)

(1)

Using partial fraction, we can write

2 2

1 A B C

(a x) (c x) (a x) (a x) (c x)

(2)

21 A(a x)(c x) B(c x) C(a x) (3)

For calculating the value of B, put (a x) = 0 or x = a that is

21 A(a a)(c a) B(c a) C(a a)

1 B(c a) or 1

B(c a)

For calculating the value of C, put (c x) = 0 or x = c that is

21 A(a c)(c c) B(c c) C(a c)

21 C(a c) or 2

1C

(a c)

For calculating the value of A, consider equation (3),

21 A(a x)(c x) B(c x) C(a x)

2 2 21 A(ac ax cx x ) B(c x) C(a x 2ax)

2 2 21 Aac Aax Acx Ax Bc Bx Ca Cx 2Cax

Comparing coefficients of x2, we get

0 A C

A C or 2

1A

(a c)

Putting the values of A, B and C in equation (2),

17

2 2 2 2

1 1 1 1

(a x) (c x) (a c) (a x) (c a)(a x) (a c) (c x)

Hence equation (1) becomes

2 2 2

1 1 1dx k dt

(a c) (a x) (c a)(a x) (a c) (c x)

2 2 2

1 dx 1 dx 1 dxk dt

(a c) (a x) (c a) (a x) (a c) (c x)

2 1

2 2

1 ln(a x) 1 (a x) 1 ln(c x)kt c

(a c) 1 (c a) ( 2 1)( 1) (a c) 1

2 2

1 1 1 1ln(a x) ln(c x) kt c

(a c) (c a) (a x) (a c)

21 1 1

ln(a x) ln(c x) kt c(a c) (c a) (a x)

2

1 (a x) 1 1ln kt c

(a c) (c x) (c a) (a x)

(4)

When t =0, x = 0, putting these values in above equation, the value of c can be calculated,

2

1 (a 0) 1 1ln k0 c

(a c) (c 0) (c a) (a 0)

2

1 a 1ln c

(a c) c a(c a)

Putting the value of c in equation (4),

2 2

1 (a x) 1 1 1 a 1ln kt ln

(a c) (c x) (c a) (a x) (a c) c a(c a)

2 2

1 (a x) 1 1 1 a 1ln ln kt

(a c) (c x) (c a) (a x) (a c) c a(c a)

2

1 (a x) a 1 1 1ln ln kt

(a c) (c x) c (c a) (a x) a(c a)

2

1 c(a x) a (a x)ln kt

(a c) a(c x) a(a x)(c a)

2

1 c(a x) a a xln kt


2

1 c(a x) xln kt


18

2

1 c(a x) xln kt

(a c) a(c x) a(a c)(a x)

2c(a x) x(a c)

ln (a c) kta(c x) a(a x)

(5)

This is the integrated rate equation for a third order reaction in which concentration of two

reactants are same and that of the third one is different. This equation is similar to the

equation for a straight line passing through the origin, from the slope of which k can be

calculated if initial concentrations are known that is,

2Slope (a c) k

2

Slopek

(a c)

Left hand side of equation (5) is a dimensionless quantity. Hence units of k comes out to be,

231 mol / dm ksec 2 61 mol dm seck

2 6 1k mol dm sec

The half-life period cannot be determined for such type of a reaction.

Case (III): when the initial concentration of all the reactants is different

Consider the following reaction in which the initial concentration of reactants A, B

and C are a, b and c respectively.

A + B + C Products

A + B + C Products

When t = 0 a b c 0

When t = t ax bx cx x

Rate equation for such a reaction can be written as

dx

k[A][B][C]dt

dx

k(a x)(b x)(c x)dt


dx

k dt(a x)(b x)(c x)

(1)

Using partial fraction, we get

1 A B C

(a x)(b x)(c x) (a x) (b x) (c x)

(2)

19

1 A(b x)(c x) B(a x)(c x) C(a x)(b x)

For calculating the value of A, put (a x) = 0 or x = a that is

1 A(b a)(c a) B(a a)(c a) C(a a)(b a)

1 A(b a)(c a)

1 1

A(b a)(c a) (a b)(c a)

For calculating the value of B, put (b x) = 0 or x = b that is

1 A(b b)(c b) B(a b)(c b) C(a b)(b b)

1 B(a b)(c b)

1 1

B(a b)(c b) (a b)(b c)

For calculating the value of C, put (c x) = 0 or x = c that is

1 A(b c)(c c) B(a c)(c c) C(a c)(b c)

1 C(a c)(b c)

1 1

C(a c)(b c) (c a)(b c)


1 1 1 1

(a x)(b x)(c x) (a b)(c a)(a x) (a b)(b c)(b x) (c a)(b c)(c x)


1 1 1dx k dt

(a b)(c a)(a x) (a b)(b c)(b x) (c a)(b c)(c x)

1 dx 1 dx 1 dx

k dt(a b)(c a) (a x) (a b)(b c) (b x) (c a)(b c) (c x)

1 1 1

ln(a x) ln(b x) ln(c x) kt c(a b)(c a) (a b)(b c) (c a)(b c)

1

(b c) ln(a x) (c a) ln(b x) (a b) ln(c x) kt c(a b)(b c)(c a)

(b c) ln(a x) (c a) ln(b x) (a b) ln(c x)

kt c(a b)(b c)(c a)

(3)


(b c) ln(a 0) (c a) ln(b 0) (a b) ln(c 0)

k0 c(a b)(b c)(c a)

20

(b c) ln a (c a) ln b (a b) ln c

c(a b)(b c)(c a)


(b c) ln(a x) (c a) ln(b x) (a b) ln(c x) (b c) ln a (c a) ln b (a b) ln ckt

(a b)(b c)(c a) (a b)(b c)(c a)


(a b)(b c)(c a) (a b)(b c)(c a)


(a b)(b c)(c a)

(a x) (b x) (c x)(b c) ln (c a) ln (a b) ln (a b)(b c)(c a)kt

a b c

(4)

This is the integrated rate equation for a third order reaction in which the concentrations of all

the reactants are different. This is the equation of a straight line passing through the origin,

where the left hand side of the equation represents the dependent variable and t is the

independent variable. The plot will give a straight line whose slope is given by

Slope (a b)(b c)(c a)k

Slope

k(a b)(b c)(c a)

Hence, the value of k can be determined if the initial concentrations of the three reactants are

known. Units of k can be determined by rearranging the equation (4),

(a x) (b x) (c x)(b c) ln (c a) ln (a b) ln

a b ck(a b)(b c)(c a)t

3

33

mol dmk

mol dm sec

2 6 1k mol dm sec

Case (IV): when only two reactants are involved

Sometime the rate of a reaction has more dependence on the concentration of one

reactant than that of the other. For example, 3rd

order reactions, rate of reaction may be

directly proportional to the square of concentration of one reactant and concentration of the

other reactants. Consider a third order reaction of the type in which two moles of reactant A

are reacting with one mole of reactant B to give products. The initial concentration of reactant

A is a and that of B is b as given below,

21

2A + B Products

2A + B Products

When t = 0 a b 0

When t = t a2x bx x


2dx

k[A] [B]dt

2dx

k(a 2x) (b x)dt


2

dxk dt

(a 2x) (b x)

(1)

Using partial fraction, we get

2 2

1 A B C

(a 2x) (b x) (a 2x) (a 2x) (b x)

(2)

21 A(a 2x)(b x) B(b x) C(a 2x) (3)

For calculating the value of C, put (b x) = 0 or x = b that is

21 A(a 2b)(b b) B(b b) C(a 2b)

21 C(a 2b)

2

1C

(a 2b)

For calculating the value of B, put (a 2x) = 0 or x = a/2 that is

2a a a a

1 A(a 2 )(b ) B(b ) C(a 2 )2 2 2 2

a 2b a

1 B(b ) B( )2 2

2 2

B(2b a) (a 2b)

For calculating the value of A, consider equation (3),

21 A(a 2x)(b x) B(b x) C(a 2x)

2 2 21 A(ab ax 2bx 2x ) B(b x) C(a 4x 4ax)

2 2 21 Aab Aax 2Abx 2Ax Bb Bx Ca 4Cx 4Cax)

Comparing coefficients of x2, we get

1 2A 4C

22

2A 4C

2

2A 2C

(a 2b)


2 2

1 A B C

(a 2x) (b x) (a 2x) (a 2x) (b x)

2 2 2 2

1 2 2 1

(a 2x) (b x) (a 2b) (a 2x) (a 2b)(a 2x) (a 2b) (b x)


2 2 2

2 2 1dx k dt

(a 2b) (a 2x) (a 2b)(a 2x) (a 2b) (b x)

2 2 2

1 2dx 1 2dx 1 dxk dt

(a 2b) (a 2x) (a 2b) (a 2x) (a 2b) (b x)

2 1

2 2

1 1 2 (a 2x)ln(a 2x) ln(b x) k dt

(a 2b) (a 2b) (a 2b) ( 2 1)( 2)

2

1 (a 2x) 1ln kt c

(a 2b) (b x) (a 2b)(a 2x)

2

1 (a 2x) (a 2b)ln kt c

(a 2b) (b x) (a 2x)

(4)


2

1 (a 2 0) (a 2b)ln k0 c

(a 2b) (b 0) (a 2 0)

2

1 a (a 2b)ln c

(a 2b) b a


2 2

1 (a 2x) (a 2b) 1 a (a 2b)ln ln kt

(a 2b) (b x) (a 2x) (a 2b) b a

2

1 (a 2x) (a 2b) a (a 2b)ln ln kt

(a 2b) (b x) (a 2x) b a

2b(a 2x) (a 2b) (a 2b)

ln (a 2b) kta(b x) (a 2x) a

2b(a 2x) 1 1

ln (a 2b) (a 2b) kta(b x) (a 2x) a

23

2b(a 2x) a (a 2x)

ln (a 2b) (a 2b) kta(b x) a(a 2x)

2b(a 2x) 2x

ln (a 2b) (a 2b) kta(b x) a(a 2x)

2b(a 2x) 2x(a 2b)

ln (a 2b) kta(b x) a(a 2x)

(5)

This is the equation of straight line passing through the origin. The value of k can be found

from the slope if the initial concentrations are known.

2Slope (a 2b) k

2

Slopek

(a 2b)

Units of k can be found as follows,

1 n

1

1 n

moleUnits of k time

litre

2 2 1Units of k mol litre time

4.10 ELEMENTARY AND COMPLEX REACTIONS

An elementary reaction is a simple reaction which occurs in a single step. A complex

reaction is that reaction which occurs in two or more steps. Chemistry is comprised of

billions of reactions. The majority of reactions with which the chemist deals are not

elementary instead they involve two or more elementary steps and are complex. The complex

reactions are also known as composite reactions. There are various types of complex

reactions which may involve:

1. Reversible or opposing reactions

2. Consecutive reactions

3. Simultaneous or parallel or side reactions

4. Chain reactions

4.10.1 REVERSIBLE OR OPPOSING REACTIONS

These reactions may be of the following types:

a) First order opposed by first order reaction

b) First order opposed by second order reaction

c) Second order opposed by first order reaction

d) Second order opposed by second order reaction

24

a) First order opposed by first order reaction

In such reactions, the forward as well as the backward reaction is first order. This is

the simplest case of opposing reactions. Consider the following reaction

A B

k1

k-1

When t = 0 a 0

When t = t ax x

When t = teq axe xe


1 1

dxk (a x) k x

dt (1)

If xe is the equilibrium concentration then at the state of dynamic equilibrium,

Rf = Rb and x = xe

1 e 1 ek (a x ) k x

1 e1

e

k (a x )k

x

(2)

Putting the value of k-1 from equation (2) in equation (1),

1 e1

e

k (a x )dxk (a x) x

dt x

e1e

(a x )xdxk (a x)

dt x

e e1e

x (a x) (a x )xdxk

dt x

e e e1e

ax xx ax x xdxk

dt x

e1e

a(x x)dxk

dt x

1e

e

k adx(x x)

dt x


1

e e

k adxdt

(x x) x

25

1e

e

k aln(x x) t c

x (3)

When t =0, x = 0, putting these values in above equation, the value of c comes out to be,

ec ln x


1e e

e

k aln(x x) t ln x

x

1e e

e

k aln x ln(x x) t

x

e 1

e e

x k aln t

(x x) x

e e1

e

x xk ln

at (x x)

(4)

This is the integrated rate equation for opposing reaction of the type first order opposed by

first order.

b) First order opposed by Second order reaction

In such reactions, the forward reaction is first order and backward reaction is second

order. Consider the following reaction,

A B + C

k1

k-1

When t = 0 a 0 0

When t = t ax x x

When t = teq axe xe xe


21 1

dxk (a x) k x

dt (1)


Rf = Rb and x = xe

21 e 1 ek (a x ) k (x )

1 e1 2

e

k (a x )k

(x )

(2)


26

21 e1 2

e

k (a x )dxk (a x) x

dt (x )

21 e1 2

e

k (a x )dxk (a x) x

dt (x )

2 2

1 e 1 e

2

e

k (a x)x k (a x )xdx

dt x

2 2 2e 1 e 1 e

dxx k (a x)x k (a x )x

dt

2 2 2 2 2e 1 e 1 e 1 1 e

dxx k ax k xx k ax k x x

dt

2 2 2 2 2e 1 e 1 1 e 1 e

dxx k ax k ax k x x k xx

dt

2 2 2e 1 e 1 e e

dxx k a(x x ) k xx (x x)

dt

2e 1 e e 1 e e

dxx k a(x x)(x x) k xx (x x)

dt

2e 1 e e edx

x k (x x) a(x x) xxdt

2e 1 e e edx

x k (x x) ax ax xxdt

12

e e e e

k dtdx

(x x)(ax ax xx ) x

Integrating the above equation,

12

e e e e

kdxdt

(x x)(ax ax xx ) x

(3)

Using partial fraction method, we get

e e e e e e

1 A B

(x x)(ax ax xx ) (x x) (ax ax xx )

(4)

e e e1 A(ax ax xx ) B(x x) (5)

For calculating the value of A, put (xe x) = 0 or x = xe that is

e e e e e e1 A(ax ax x x ) B(x x )

2e e1 A(2ax x )

e e1 A x (2a x )

27

e e

1A

x (2a x )

For calculating the value of B, consider equation (5),

e e e1 A(ax ax xx ) B(x x)

e e e1 Aax Aax Axx Bx Bx

Comparing coefficients of x, we get

e0 Aa Ax B

eB Aa Ax

eB A(a x )

Putting the value of A,

e

e e

(a x )B

x (2a x )

Putting the values of A and B in equation (4),

e

e e e e e e e e e e

(a x )1 1

(x x)(ax ax xx ) x (2a x )(x x) x (2a x )(ax ax xx )


e 12

e e e e e e e e

(a x )dx kdxdt

x (2a x )(x x) x (2a x )(ax ax xx ) x

e 12

e e e e e e

(a x )dx k1 dxdt

x (2a x ) (x x) (ax ax xx ) x

1e e e 2e e e

k1ln(x x) ln(ax ax xx ) t c

x (2a x ) x

e e 12

e e e e

(ax ax xx ) k1ln t c

x (2a x ) (x x) x

(6)

When t =0, x = 0, putting these values in above equation, the value of c comes out to be,

e e 12

e e e e

(ax a0 0x ) k1ln 0 c

x (2a x ) (x 0) x

e e

1c ln a

x (2a x )


e e 12

e e e e e e

(ax ax xx ) k1 1ln t ln a

x (2a x ) (x x) x x (2a x )

28

e e 12

e e e e e e

(ax ax xx ) k1 1ln ln a t

x (2a x ) (x x) x (2a x ) x

e e 12

e e e e

(ax ax xx ) k1ln t

x (2a x ) (x x)a x

e e e1

e e

x ax x(a x )k ln

t(2a x ) (x x)a

(7)

This is the integrated rate equation for opposing reaction of the type first order opposed by

second order.

c) Second order opposed by first order reaction

In such reactions, the forward reaction is second order and backward reaction is first

order. Consider the following reaction,

A + B Ck1

k-1

When t = 0 a a 0

When t = t ax ax x

When t = teq axe axe xe


21 1

dxk (a x) k x

dt (1)


Rf = Rb and x = xe

21 e 1 ek (a x ) k x

2

1 e1

e

k (a x )k

x

(2)


2

2 1 e1

e

k (a x )dxk (a x) x

dt x

2 2

1 e 1 e

e

k (a x) x k (a x ) xdx

dt x

2 21e e

e

kdx(a x) x (a x ) x

dt x

2 2 2 21e e e

e

kdx(a x 2ax)x (a x 2ax )x

dt x

29

2 2 2 21e e e e e

e

kdxa x x x 2axx (a x x x 2ax x)

dt x

2 2 2 21e e e e e

e

kdxa x x x 2axx a x x x 2ax x

dt x

2 2 2 21e e e

e

kdxa x x x a x x x

dt x

2 2 2 21e e e

e

kdxa x a x x x x x

dt x

21e e e

e

kdxa (x x) xx (x x )

dt x

21e e e

e

kdxa (x x) xx (x x)

dt x

21e e

e

kdx(x x)(a xx )

dt x

12

e e e

k dtdx

(x x)(a xx ) x


12

e e e

kdxdt

(x x)(a xx ) x

(3)

Using partial fraction method, we get

2 2

e e e e

1 A B

(x x)(a xx ) (x x) (a xx )

(4)

2e e1 A(a xx ) B(x x) (5)


2 e e e e1 A(a x x ) B(x x )

2 2e1 A(a x )

2 2

e

1A

(a x )


2 e e1 A(a xx ) B(x x)

2 e e1 Aa Axx Bx Bx


30

e0 Ax B

eB Ax


e2 2

e

xB

(a x )


e2 2 2 2 2 2

e e e e e e

x1 1

(x x)(a xx ) (a x )(x x) (a x )(a xx )


e 12 2 2 2 2

e e e e e

x dx kdxdt

(a x )(x x) (a x )(a xx ) x

e 12 2 2

e e e e

x dx k1 dxdt

(a x ) (x x) (a xx ) x

2 1e e2 2e e

k1ln(x x) ln(a xx ) t c

(a x ) x

2

e 1

2 2

e e e

(a xx ) k1ln t c

(a x ) (x x) x

(6)

When t =0, x = 0, therefore the value of c comes out to be,

2

e 1

2 2

e e e

(a 0x ) k1ln 0 c

(a x ) (x 0) x

2

2 2

e e

1 aln c

(a x ) x


2 2

e 1

2 2 2 2

e e e e e

(a xx ) k1 1 aln t ln

(a x ) (x x) x (a x ) x

2 2

e 1

2 2 2 2

e e e e e

(a xx ) k1 1 aln ln t

(a x ) (x x) (a x ) x x

2

e e 1

2 2 2

e e e

(a xx )x k1ln t

(a x ) (x x)a x

2

e e e1 2 2 2

e e

x (a xx )xk ln

t(a x ) (x x)a

(7)

31

This is the integrated rate equation for opposing reaction of the type second order opposed by

first order.

d) Second order opposed by second order reaction

In such reactions, the forward reaction is second order and backward reaction is also

second order. Consider the following reaction,

A + Bk1

k-1

C + D

When t = 0 a a 0 0

When t = t ax ax x x

When t = teq axe axe xe xe


2 21 1

dxk (a x) k x

dt (1)


Rf = Rb and x = xe

2 21 e 1 ek (a x ) k x

2

1 e1 2

e

k (a x )k

x

(2)


2

2 21 e1 2

e

k (a x )dxk (a x) x

dt x

2 2 2 2

1 e 1 e

2

e

k (a x) x k (a x ) xdx

dt x

2 2 2 21e e2

e

kdx(a x) x (a x ) x

dt x

2 2 2 2 2 21e e e2

e

kdx(a x 2ax)x (a x 2ax )x

dt x

2 2 2 2 2 2 2 2 2 21 e e e e e2e

kdxa x x x 2axx a x x x 2ax x

dt x

2 2 2 2 2 21e e e2

e

kdxa x 2axx a x 2ax x

dt x

2 2 2 2 2 21e e e2

e

kdxa x a x 2axx 2ax x

dt x

32

2 2 21e e e2

e

kdxa (x x ) 2axx (x x)

dt x

21e e e e2

e

kdxa (x x)(x x) 2axx (x x)

dt x

21 e e e2e

kdx(x x) a (x x) 2axx

dt x

1

22ee e e

kdxdt

x(x x) a (x x) 2axx


1

22ee e e

kdxdt

x(x x) a (x x) 2axx

(3)

Using partial fraction method we get

22 e e ee e e

1 A B

(x x) a (x x) 2axx(x x) a (x x) 2axx

(4)

2 e e e1 A a (x x) 2axx B(x x) (5)


2 e e e e e e1 A a (x x ) 2ax x B(x x )

2 2e e1 A(2a x 2ax )

e e

1A

2ax (a x )


2 e e e1 A a (x x) 2axx B(x x)

2 2

e e e1 Aa x Aa x 2Aaxx Bx Bx


2e0 Aa 2Aax B

2 eB A(a 2ax )


2

e

e e

(a 2ax )B

2ax (a x )


33

2

e

22e e e e e e ee e e

(a 2ax )1 1

2ax (a x )(x x) [2ax (a x )][a (x x) 2axx ](x x) a (x x) 2axx


2

e 1

2 2 2

e e e e e e

(a 2ax )dx k1 dxdt

2ax (a x ) (x x) a x a x 2axx x

2 2 1e e e 2e e e

k1ln(x x) ln(a x a x 2axx ) t c

2ax (a x ) x

2 2

e e 1

2

e e e e

(a x a x 2axx ) k1ln t c

2ax (a x ) (x x) x

(6)

When t =0, x = 0, therefore the value of c comes out to be,

2 2

e e 1

2

e e e e

(a x a 0 2a0x ) k1ln 0 c

2ax (a x ) (x 0) x

2

e

e e e

a x1ln c

2ax (a x ) x

2

e e

1c ln a

2ax (a x )


2 2

2e e 1

2

e e e e e e

(a x a x 2axx ) k1 1ln t ln a

2ax (a x ) (x x) x 2ax (a x )

2 2

2e e 1

2

e e e e e e

(a x a x 2axx ) k1 1ln ln a t

2ax (a x ) (x x) 2ax (a x ) x

2 2

e e 1

2 2

e e e e

(a x a x 2axx ) k1ln t

2ax (a x ) (x x)a x

2

e e e1 2

e e e

x a(ax ax 2xx )k ln

2atx (a x ) (x x)a

e e e1e e

x ax ax 2xxk ln

2at(a x ) (x x)a

e e e1e e

x x(a 2x ) axk ln

2at(a x ) (x x)a

(7)

This is the integrated rate equation for opposing reaction of the type second order opposed by

second order.

34

4.10.2 CONSECUTIVE OR SEQUENTIAL REACTIONS

The reactions which proceed from reactants to products in one or more steps

through intermediates are known as consecutive reactions. Each step involved here has its

own rate constant. Consider the simplest consecutive reaction as follows,

k2A B C

k1

When t = 0 a 0 0

When t = t [A] [B] [C]

According to law of mass action, rate of decomposition of A is given by,

1

d[A]k [A]

dt

Integrating this equation after separating variables,

1d[A]

k dt[A]

1ln[A] k t c (1)

When t =0, [A] = a, therefore the value of c comes out to be,

1ln a k 0 c

ln a c


1ln[A] k t ln a

1ln[A] ln a k t

1

[A]ln k t

a

Taking antilog on both sides,

1k t[A]

ea

1k t[A] ae (2)

Now rate of formation of B,

1 2

d[B]k [A] k [B]

dt

2 1d[B]

k [B] k [A]dt

1k t2 1

d[B]k [B] k ae

dt

(By equation (2))

35

Multiplying both sides by 2k te

2 1 2k t k t k t2 1

d[B]k [B] e (k ae )e

dt

2 2 2 1k t k t (k k )t2 1

d[B]e [B]k e k ae

dt

2 2 1k t (k k )t1d

[B]e k aedt

2 2 1k t (k k )t1d [B]e k ae dt

2 2 1k t (k k )t1d [B]e k a e dt

2 1

2

(k k )tk t

1

2 1

e[B]e k a D

k k

(3)

When t =0, [B] = 0, therefore the value of D comes out to be,

2 1

2

(k k )0k 0

1

2 1

e[0]e k a D

k k

1

2 1

k aD

k k

Putting the value of D in equation (3),

2 1

2

(k k )tk t 1

1

2 1 2 1

k ae[B]e k a

k k k k

2 2 1k t (k k )t12 1

k a[B]e e 1

k k

2 2 1k t k t k t12 1

k a[B]e e e 1

k k

2

1

2 2

k tk t1

k t k t

2 1

k a e 1[B] e

k k e e

1 2k t k t12 1

k a[B] e e

k k

(4)

The concentration of C can be found as,

a [A] [B] [C]

[C] a [A] [B]

1 1 2k t k t k t12 1

k a[C] a ae e e

k k

36

1 2

1

k t k tk t 1 1

2 1

(k ae k ae )[C] a ae

k k

1 1 2k t k t k t

2 1 2 1 1 1

2 1

a(k k ) ae (k k ) k ae k ae[C]

k k

1 1 1 2k t k t k t k t

2 1 2 1 1 1

2 1

a(k k ) k ae k ae k ae k ae[C]

k k

1 2k t k t

2 1 2 1

2 1

a(k k ) a(k e k e )[C]

k k

1 2k t k t

2 1

2 1

(k e k e )[C] a 1

k k

(5)

Radioactive decay is an example of such a process. Other examples include chemical

processes such as polymerization, thermal cracking and chlorination of hydrocarbons.

4.10.3 SIMULTANEOUS OR PARALLEL OR SIDE REACTIONS

The reactions in which reactants undergo two or more independent reactions

simultaneously are called as parallel reactions. These are also known as side reactions.

The reaction in which maximum yield of the product is obtained is called main or major

reaction while the others are called side reactions. Consider a general parallel reaction,

A

B

C

k1

k2when t = 0 awhen t = t a - x

0 at t = 0x at t = t

0 at t = 0x at t = t

Rate equation for such a reaction is given by,

1 2

dxk (a x) k (a x)

dt

1 2

dx(k k )(a x)

dt


1 2dx

(k k ) dt(a x)

1 2ln(a x) (k k )t c (1)

When t =0, x =0, therefore the value of c comes out to be,

1 2ln(a 0) (k k )0 c

37

ln a c


1 2ln(a x) (k k )t ln a

1 2(k k )t ln a ln(a x)

1 2

1 a(k k ) ln

t a x

(2)

The equation (2) is the rate constant expression for a parallel reaction. It is similar to the

equation of a first order reaction, (k1 + k2) being the sum of the specific rates of the two

simultaneous reactions. If k1 > k2 then AB is main reaction and AC is side reaction.

Graphically this can be represented as shown in figure (4.1). For example, nitration of

phenols gives ortho-nitrophenol and para-nitrophenol in two simultaneous side reactions

Figure (4.2).

Fig(4.1): Graphical representation Fig(4.2): Simultaneous side reactions

4.10.4 CHAIN REACTIONS

The chemical reactions which take place in a series of successive processes

involving the formation of free atoms and radicals are known as chain reactions. The

kinetic laws for such reactions are considerably complex. A well-known example of these

types of reactions is the hydrogenbromine reaction. Chain reactions usually proceed very

rapidly. Many explosive reactions occur by a chain mechanism; atomic fission and atomic

fusion are of this type of reactions.

4.11 THERMAL REACTIONS

A well known example of a thermal reaction is the hydrogen bromine reaction. The

stoichiometry for this reaction is,

H2 + Br2 2HBr

But it does not follow the simple rate expression

2 2d[HBr]

k[H ][Br ]dt

38

It follows the following empirical rate equation given by Bodenstein and Lind in 1906:

1/2

2 2

2

k[Br ] [H ]d[HBr]

[HBr]dt1 k

[Br ]

The complexity of this equation could not be explained until 1919 when Christiansen,

Herzfeld and Polanyi independently and almost simultaneously solved the problem. They

proposed a chain of reactions with the following steps,

(1) k1

BrBr2 2 Chain initiation

(2) k2

Br + H2 HHBr + Chain propagation

(3) k3

+ Br2 HBr +H Br Chain propagation

(4) k4

BrH2 +H + HBr Chain inhibition

(5) k5

Br Br22 Chain termination

According to this mechanism HBr is formed in reactions (2) and (3) and removed in reaction

(4). Consequently, the rate of formation of HBr is given by

2 2 3 2 4d[HBr]

k [Br ][H ] k [H ][Br ] k [H ][HBr]dt

2 2 3 2 4d[HBr]

k [Br ][H ] k [Br ] k [HBr] [H ]dt

(1)

The concentration of [Br ] and [H ] can be determined using steady state approximation.

The steady state approximation is also occasionally called stationary state approximation.

This approximation assumes that the concentration of reaction intermediates remains constant

throughout the reaction after an initial buildup. This approximation can only be applied to

short lived or very reactive species. Therefore mathematically we can write,

21 2 2 2 3 2 4 5d[Br ]

k [Br ] k [Br ][H ] k [H ][Br ] k [H ][HBr] k [Br ] 0dt

(2)

2 2 3 2 4d[H ]

k [Br ][H ] k [H ][Br ] k [H ][HBr] 0dt

(3)

Solving equation (3),

2 2 3 2 4k [Br ][H ] k [H ][Br ] k [H ][HBr] 0

39

2 2 3 2 4k [Br ][H ] k [Br ] k [HBr] [H ] 0

3 2 4 2 2k [Br ] k [HBr] [H ] k [Br ][H ] (4)

2 2

3 2 4

k [Br ][H ][H ]

k [Br ] k [HBr]

(5)

Now solve equation (2),

21 2 2 2 3 2 4 5k [Br ] k [Br ][H ] k [H ][Br ] k [H ][HBr] k [Br ] 0

21 2 2 2 3 2 4 5k [Br ] k [Br ][H ] k [Br ] k [HBr] [H ] k [Br ] 0

21 2 2 2 2 2 5k [Br ] k [Br ][H ] k [Br ][H ] k [Br ] 0

(By equation (4))

21 2 5k [Br ] k [Br ] 0

25 1 2k [Br ] k [Br ]

1/2

1 2

5

k [Br ][Br ]

k

(6)

Now putting the value of [Br ] from equation (6) in equation (5),

1/2

1 22 2

5

3 2 4

k [Br ]k [H ]

k[H ]

k [Br ] k [HBr]

(7)

Putting the values of [Br ] and [H ] from Eq.(6) and Eq.(7) in Eq.(1) we get

2 2 3 2 4d[HBr]


1/2

1 21/2 2 2

51 22 2 3 2 4

5 3 2 4

k [Br ]k [H ]

kk [Br ]d[HBr]k [H ] k [Br ] k [HBr]

dt k k [Br ] k [HBr]

1/2

3 2 41 22 2

5 3 2 4

k [Br ] k [HBr]k [Br ]d[HBr]k [H ] 1


1/2

3 2 4 3 2 41 22 2

5 3 2 4

k [Br ] k [HBr] k [Br ] k [HBr]k [Br ]d[HBr]k [H ]


1

5

1/2k

2 2 2 3 2k

3 2 4

k [Br ] [H ]2k [Br ]d[HBr]

dt k [Br ] k [HBr]

40

1

5

1/2k 1/2

2 2 2k

4

3 2

2k [Br ] [H ]d[HBr]

k [HBr]dt1

k [Br ]

1/2

2 2

2

k[Br ] [H ]d[HBr]

[HBr]dt1 k

[Br ]

(8)

Where 15

k

2 kk 2k and 4

3

k

kk are constants. This equation (8) is in agreement with the rate

equation given by Bodenstein and Lind. The appearance of the term [HBr] in the

denominator implies that the velocity of the reaction is decreased by the product HBr and this

product acts an inhibitor of the reaction. This is an example of self-inhibition.

4.12 PHOTOCHEMICAL REACTIONS

Photochemical reaction is a chemical reaction initiated by the absorption of

energy in the form of light. In photochemical reactions free radicals are produced which

initiate the chain reactions. An example is the photosynthesis of HCl gas. All photochemical

reactions are of zero order. These reactions are independent of concentration of reactants.

Change in concentration of reactants has no effect on rate of photochemical reactions.

a) HydrogenBromine Reaction

Consider the following photochemical reaction in which H2 and Br2 are going to

produce HBr,

H2 + Br2 2HBrh

The mechanism of this reaction consists of following five elementary steps,

(1) Br2 + h 2k1

Br Chain initiation

(2) k2

Br + H2 HHBr + Chain propagation

(3) k3

+ Br2 HBr +H Br Chain propagation

(4) k4

BrH2 +H + HBr Chain inhibition

(5) k5

Br Br22 Chain termination

According to this mechanism HBr is formed in reactions (2) and (3) and removed in reaction

(4). Consequently, the rate of formation of HBr is given by

41

2 2 3 2 4d[HBr]

k [Br ][H ] k [H ][Br ] k [H ][HBr]dt

2 2 3 2 4d[HBr]


(1)

The concentration of [Br ] and [H ] can be determined using steady state approximation. This

approximation assumed that at steady state the rate of formation of reaction intermediates can

be considered to be equal to their rate of disappearance. Mathematically we can write,

21 a 2 2 3 2 4 5d[Br ]

k I k [Br ][H ] k [H ][Br ] k [H ][HBr] k [Br ] 0dt

(2)

2 2 3 2 4d[H ]

k [Br ][H ] k [H ][Br ] k [H ][HBr] 0dt

(3)


2 2 3 2 4k [Br ][H ] k [H ][Br ] k [H ][HBr] 0

2 2 3 2 4k [Br ][H ] k [Br ] k [HBr] [H ] 0

3 2 4 2 2k [Br ] k [HBr] [H ] k [Br ][H ] (4)

2 2

3 2 4

k [Br ][H ][H ]

k [Br ] k [HBr]

(5)


21 a 2 2 3 2 4 5k I k [Br ][H ] k [H ][Br ] k [H ][HBr] k [Br ] 0

21 a 2 2 3 2 4 5k I k [Br ][H ] k [Br ] k [HBr] [H ] k [Br ] 0

21 a 2 2 2 2 5k I k [Br ][H ] k [Br ][H ] k [Br ] 0

(By equation (4))

21 a 5k I k [Br ] 0

25 1 ak [Br ] k I

1/2

1a

5

k[Br ] I

k

(6)

Now putting the value of [Br ] from equation (6) in equation (5),

1/2

12 a 2

5

3 2 4

kk I [H ]

k[H ]

k [Br ] k [HBr]

(7)

Putting the values of [Br ] and [H ] from Eq.(6) and Eq.(7) in Eq.(1) we get

42

2 2 3 2 4d[HBr]


1/2

11/2 2 a 2

512 a 2 3 2 4

5 3 2 4

kk I [H ]

kkd[HBr]k I [H ] k [Br ] k [HBr]


1/2

3 2 412 a 2

5 3 2 4

k [Br ] k [HBr]kd[HBr]k I [H ] 1


1/2

3 2 4 3 2 412 a 2

5 3 2 4

k [Br ] k [HBr] k [Br ] k [HBr]kd[HBr]k I [H ]


1

5

1/2k

2 a 2 3 2k

3 2 4

k I [H ]2k [Br ]d[HBr]

dt k [Br ] k [HBr]

1

5

1/2k 1/2

2 a 2k

4

3 2

2k I [H ]d[HBr]

k [HBr]dt1

k [Br ]

1/2

a 2

2

kI [H ]d[HBr]

[HBr]dt1 k

[Br ]

(8)

Where 15

k

2 kk 2k and 4

3

k

kk are constants. The equation (8) is the final expression.

b) HydrogenChlorine Reaction

The chemical reaction between hydrogen and chlorine in the presence of light

produces hydrogen chloride with explosion. In 1930 Bodenstein and Hanger investigated the

kinetics of this photochemical reaction. The rate law expression for the reaction on the basis

of their experimental observations is given below,

H2 + Cl2 2HClh

a 2d[HCl]

kI [H ]dt

Where k is constant and Ia is the intensity of absorbed light (expressed in Einstein dm-3

sec-1

).

They proposed the following mechanism for the reaction,

(1) Cl2 + h 2k1

Cl Initiation step

43

(2) k2

Cl + H2 HHCl + Propagation step

(3) k3

+ Cl2 HCl +H Cl Propagation step

(4) k4

Cl Cl21/2 Termination step

HCl is more stable than HBr. Bond between HCl is strong so [H ] radical do not react with

HCl molecule as in case of HBr. The net rate of reaction in terms of rate of formation of HCl

according to above mechanism can be written as,

2 2 3 2d[HCl]

k [Cl ][H ] k [H ][Cl ]dt

(1)

The concentration of [Cl ] and [H ] can be determined applying steady state approximation.

This approximation assumed that at steady state the net change in concentration of reaction

intermediates is zero. Mathematically,

1 a 2 2 3 2 4d[Cl ]

k I k [Cl ][H ] k [H ][Cl ] k [Cl ] 0dt

(2)

2 2 3 2d[H ]

k [Cl ][H ] k [H ][Cl ] 0dt

(3)


2 2 3 2k [Cl ][H ] k [H ][Cl ] 0

2 2 3 2k [Cl ][H ] k [H ][Cl ]

(4)

2 2

3 2

k [Cl ][H ][H ]

k [Cl ]

(5)


1 a 2 2 3 2 4k I k [Cl ][H ] k [H ][Cl ] k [Cl ] 0

1 a 3 2 3 2 4k I k [H ][Cl ] k [H ][Cl ] k [Cl ] 0

(By equation (4))

1 a 4k I k [Cl ] 0

1 a

4

k I[Cl ]

k

(6)

Now putting the value of [Cl ] from equation (6) in equation (5),

2 1 a 2

3 4 2

k k I [H ][H ]

k k [Cl ]

(7)

44

Putting the values of [Cl ] and [H ] from Eq.(6) and Eq.(7) in Eq.(1) we get

2 2 3 2

d[HCl]k [Cl ][H ] k [H ][Cl ]

dt

1 a 2 1 a 22 2 3 2

4 3 4 2

k I k k I [H ]d[HCl]k [H ] k [Cl ]

dt k k k [Cl ]

2 1 a 2 1 a 22

4 4

k k I k k I [H ]d[HCl][H ]

dt k k

2 1a 2

4

2k kd[HCl]I [H ]

dt k

a 2

d[HCl]kI [H ]

dt (8)

Where 2 1 4k 2k k / k is a constant. The expression (8) is same as was given by Bodenstein

and Hanger.

4.13 EFFECT OF TEMPERATURE ON RATE OF REACTION

Arrhenius Equation (1889)

It is well established that the rate of reaction increases to an appreciable extent with

the rise in temperature. Experimentally, it has been found that for a 10 C rise in temperature,

the velocity of the reaction is doubled or trebled.

The first empirical equation to show the variation of rate constant k of a reaction with

temperature T was suggested by Hood. His equation is,

B

log k AT

(1)

Where A and B are positive empirical constants and T is the absolute temperature. The

verification of this equation lies in the fact that a plot of log k versus 1/T is nearly a straight

line with negative slope for most of the reactions. This equation was theoretically justified by

Vant Hoff in 1884. His arguments were based on the variation of equilibrium constant with

temperature.

Svante Arrhenius extended his idea and suggested a similar equation to show the

variation of rate constant with temperature in the year 1889. The differential form of

Arrhenius equation is given by,

a2

Ed ln k

dT RT (2)

Where, Ea = Energy of activation, T = Absolute temperature and R = Gas constant.

Integrating equation (2) we get,

45

a2

E dTd ln k

R T

aE

ln k ln A(constant)RT

aE

ln k ln ART

(3)

aEln k ln ART

aEk

lnA RT

Taking antilog on both sides,

aE

RTk

eA

aE

RTk Ae

(4)

The constants A and Ea are related to Aand B in equation (1). The factor A is known as

Arrhenius pre-exponential factor, also called frequency factor, have same units as that of

specific rate constant. The frequency factor is related to collision frequency and steric factor.

The steric (or orientation) effect, also called probability factor, is the fraction of collisions in

which the molecules have proper orientation favourable for reaction. Equation (3) and

equation (4) are the alternate forms of the Arrhenius equation. Arrhenius equation shows that

the value of k for a reaction is directly related to A. The value of A increases if collision

frequency or steric factor increases.

The second factor in Arrhenius equation Ea/RT depends on the value of Ea. This term

usually has a positive value because Ea is usually positive. Since this term is subtracted from

lnA in equation (3), we see that the value of k decreases as the value of Ea/RT increases. In

other words, an increase in the value of Ea decreases the rate of the reaction. Conversely, an

increase in temperature (which is in the denominator) reduces the value of the term Ea/RT

and thus increases the value of k. This increase is consistent with the observation that the

rates of almost all reactions increase as the temperature rises at the same concentration.

Rewriting equation (3) as follows,

aE

log k log A2.303RT

(5)

This equation is identical with the empirical equation (1). To test the validity of this equation;

a plot of log k against 1/T should be a straight line having slope Ea/2.303R and intercept logA

46

as shown in figure. Knowing the slope, the value of constant Ea which is the characteristic

of the reaction can be calculated.

Alternatively, Ea can also be determined if the rate constants are known at two

different temperatures. For this purpose Arrhenius equation (2) can be integrated between the

limits k1 (the rate constant at T1) and k2 (the rate constant at T2),

2 2

1 1

k T

a

2

k T

E dTd ln k

R T

22

1 1

Tk a 1Tk T

Eln k

R

a2 12 1

E 1 1ln k ln k

R T T

a2 2 1

1 1 2

Ek T Tln

k R T T

a2 2 1

1 1 2

Ek T Tln

k 2.303R T T

(6)

Equation (6) is another form of Arrhenius equation that is useful for calculating the value of

activation energy of a reaction when the value of rate constant at two different temperatures

is known for that reaction. The Arrhenius equation correlates rate constant with temperature.

It gives the idea of energy of activation.

Energy of Activation and Activated Complex

The concept of energy of activation was developed by Arrhenius in 1888 which

constitute the backbone of all the modern theories. According to Arrhenius, the molecules

must acquire a discrete minimum amount of energy before the end products are formed.

These activated molecules will then collide and lead to the reaction. Collisions between

molecules which are not activated will be of no use and no reaction will take place. Thus the

reactants must pass through an energy rich or activated state before they can react. The

47

minimum amount of energy required by the molecules to overcome the activated state

or energy barrier before the reaction takes place is known as energy of activation.

It follows from the concept of activation that the reactants are not directly converted

into products. The molecules first acquire the energy to form an activated complex and this

activated complex is then decomposed into products.

Reactants Activated Complex Products

In other words, there exists an energy barrier between the reactants and the products. If the

reactant molecules can cross this energy barrier, they will be converted into products.

The energy of activation is usually expressed in J mol-1

and kJ mol-1

and is denoted by

the symbol Ea. The relationship between enthalpy of reaction and energy of activation for an

exothermic reaction and an endothermic reaction is shown in figure (4.3) and (4.4).

Fig (4.3): Exothermic Reaction Fig (4.4): Endothermic Reaction

In an exothermic reaction products are at lower energy level than the reactants and in

an endothermic reaction the products are at higher energy level than the reactants. With both

types of reactions the activation energy is an energy barrier which must be overcome before

products are formed. Surmounting this barrier is similar to carrying a ball to the top of a hill

and then rolling down the either side. But if the ball cannot be carried to the top, it will roll

back. Similarly if Ea is not supplied, the reaction will not start and the reactants will never get

converted into the products.

According to the arguments given above, a general reaction of the type

A2 + B

2 2AB

A2 + B

2 2AB

takes place as follows;

It must be kept in mind that the energy of activation is always positive whether the reaction is

exothermic or endothermic.

48

4.14 THEORIES OF REACTION RATES

There are two main theories of reaction rates,

1) Collision theory

2) Transition state theory or activated complex theory

4.14.1 Collision Theory

This theory is based upon the kinetic theory of gases according to which the

molecules of a gas are continuously moving and hence colliding with each other. Since for a

collision to takes place, at least two molecules must be involved, therefore the simple case to

consider is that of bimolecular reactions. Hence in discussion that follows, we shall first

discuss the collision theory for bimolecular reactions and then for unimolecular reactions.

1) Collision Theory for Bimolecular Reactions

Bimolecular collision theory for the reactions consisting of two reactant molecules

was proposed independently by Max Trautz in 1916 and William Lewis in 1918. This

theory is based on following assumptions:

1. For a chemical reaction to takes place reactants must collide with each other.

2. Not all the collisions occurring lead to reaction but only those which form activated

molecules.

3. The molecules must be suitably oriented at the time of collisions.

Consider a gasphase bimolecular elementary reaction,

A + A Products

A + A Products e.g.,

2HI H2 + I

2

2HI H2 + I

2

A + B Products

A + B Products e.g.,

H2 + I

2 2HI

H2 + I

2 2HI

From reaction kinetics, the rate of reaction is given by,

2 2ARate k[A] kn (1)

A BRate k[A][B] kn n (2)

Where nA and nB are the number of molecules per dm3 of the reactants.

The rate of reaction can also be written according to collision theory as follows,

Rate Z q (3)

Where Z is number of binary collisions per second per dm3 of the reaction mixture and q is

fraction of activated molecules.

From kinetic theory of gases, number of collisions per second per dm3 between like

molecules of a gas in the reaction mixture is given by,

2 2AA A A1

Z n u2

(4)

49

Where A is the collision diameter of the reactant molecules, nA is the number of molecules

per dm3 of reactants, u is the average velocity of molecules. Now the value of u is equal to,

8RT

uM

Where R is the general gas constant, T is absolute temperature and M is the molar mass.

Putting this value in equation (4),

2 2AA A A

1 8RTZ n

M2

2 2

AA A A

RTZ 2 n

M

(5)

When the two molecules are not alike then collision frequency (collision number) is given by,

2AB A B AB

8 RTZ n n

(6)

Where A BAB

2

is the mean collision diameter of the molecules; A B

A B

M M

M M

is the

reduced mass; nA and nB are the number of molecules per dm3 of the reactants; MA and MB

are the molecular weights of two gaseous reactants.

The fraction of molecules q having the necessary energy of activation is given by

Boltzmann law as follows,

aE /RTn

q en

(7)

Where n is the number of activated molecules having energy equal to or greater than the

energy of activation Ea, and n is the total number of molecules per dm3.

Thus the expression for the rate of reaction involving identical molecules is given by,

aE /RT2 2

AA A A

RTRate Z q 2 n e

M

(8) (By Eq. (3), (5) & (7))

And for different reactant molecules

aE /RT2AB A B AB8 RT

Rate Z q n n e

(9) (By Eq. (3), (6) & (7))

Hence the rate constant expressions for bimolecular reactions comes out to be,

For like molecules, from equation (1) and (8)

aE /RT2 2 2

A A A

RTkn 2 n e

M

50

aE /RT2

A

RTk 2 e

M

(10)

For unlike molecules, from equation (2) and (9)

aE /RT2A B A B AB

8 RTkn n n n e

aE /RT2AB

8 RTk e

(11)

Now the rate constant estimated by Arrhenius equation is equal to,

aE

RTk Ae

(12)

Comparing Eq.(10) and Eq.(11) with Eq.(12) we find that the frequency factor A in

Arrhenius equation is proportional to ZAA or ZAB and depends on the square root of

temperature. Hence the expression for the rate constant becomes,

aE

RTk Ze

(13)

Where 2

AZ 2 RT / M (for like molecules) and 2

ABZ 8 RT / (for unlike

molecules). It has been found experimentally that for reactions between simple molecules Z

agrees well with the Arrhenius factor A. Therefore, collision theory successfully accounts for

the rates of simple reactions but in many cases the rates differ considerably. For this reason

the equation is further modified as

aE

RTk PZe

(14)

Where, P is called the probability factor or steric factor. The factor P is a measure of the

geometrical requirements that must be met when activated colliding molecules are

interacting. This factor, the steric factor, limits the successful collisions to the ones in which

molecules are favourably oriented. In other words, P is the fraction of collisions in which the

molecules have an orientation favourable for reaction. Equation (14) is the equation of the

collision theory.

Limitations of Collision theory

The collision theory in the form of equation (14) is not applicable to the following cases,

1. When the complex molecules are involved in the chemical reactions.

2. When the chain mechanism is to be obeyed by the chemical reactions.

3. When the reaction is surface catalyzed.

It is applicable only to gas phase reactions and cannot be applied to kinetic studies in

solutions. It does not give any explanation for abnormally high rates of reactions.

51

2) Collision Theory for Unimolecular Reactions (Lindemanns Mechanism)

In unimolecular reactions, only one molecule takes part in the reaction so

consequently the question arises: How do molecules in unimolecular reactions attain their

energy of activation? If the energy of activation is attained in a bimolecular collision, then the

reaction should be of second order. But the fact is that the unimolecular reactions are first

order except in gas phase at low pressure. Explanation of how a bimolecular collision can

give rise to first order kinetics was given by F. A. Lindemann in 1922.

Lindemann pointed out that the reacting molecules acquire activation energy through

collisions with other molecules and as a result some molecules are activated. Further he

postulated that the activated molecules do not decompose immediately, but remain in the

activated form for a definite period. Meanwhile the added excess energy may increase the

amplitude of vibration and thus can lead to rupture the bond, or may be robbed by a less

energetic molecule in the ensuing collision. These possibilities can be represented by a

mechanism consisting of following series of elementary steps,

A + Ak1 A + A*(1) (Activation)

A* + Ak2 A + A(2) (Dectivation)

A*k3

Products(3) (Decomposition)

Where, A represents a normal molecule and A

* an activated molecule. k1, k2 and k3 are the

rate constants of three steps. According to this mechanism rate of formation of product is

given by,

*3

dxk [A ]

dt (1)

The concentration of A*, the reaction intermediate, can be determined by using steady state

approximation. That is

*2 * *

1 2 3

d[A ]k [A] k [A ][A] k [A ] 0

dt

* * 2

2 3 1k [A ][A] k [A ] k [A]

* 2

2 3 1[A ](k [A] k ) k [A]

2

* 1

2 3

k [A][A ]

k [A] k

(2)

52

Purring the value of A* from equation (2) in equation (1), we get

2

3 1

2 3

k k [A]dx

dt k [A] k

(3)

Equation (3) predicts two limiting possibilities for gaseous reactions.

Case (1): At high pressure the concentration of reacting molecules is relatively high. The

activated molecules will have a greater chance of colliding with inactive molecules and thus

become deactivated. Under such conditions, the rate of bimolecular deactivation will be much

faster than that of unimolecular decomposition then k2 [A] >> k3 and equation (3) reduces to

2

3 1 3 1

2 2

k k [A] k k [A]dxk[A]

dt k [A] k

Where, 3 1 2k k k / k . The equation predicts a first order reaction.

Case (2): At low pressure the concentration of A decreases and hence the number of

collisions decreases. Consequently, the rate of decomposition will be much faster than its rate

of deactivation. Under such conditions k3 >> k2 [A] and equation (3) reduces to

223 1

1

3

k k [A]dxk [A]

dt k

i.e., the reaction should be second order. Therefore it is the pressure which governs the order

of reaction. This conclusion was experimentally verified by Hinshelwood.

Examples:

Such changes from first order kinetics in gaseous reactions at higher pressures to

second order at lower pressures actually have been observed in many gaseous reactions. For

example, the thermal decomposition of nitrogen pentoxide and of azomethane at high

pressure are the first order reactions,

N2O5 N2O4 +

1/2 O2

(CH3)2N2 C2H6 + N2

whereas at very low pressures these reactions become second order. Such observations

confirm the Lindemanns mechanism.

4.14.2 Transition State Theory

This theory was first formulated in 1935 by Henry Eyring who was awarded Nobel

prize for this valuable work. This theory is also known as absolute rate theory because with

the help of it is possible to get the absolute value of the rate constant. The main postulates of

this theory are:

53

1. All reactions proceed through an activated or transition state which is higher in energy

than both reactants and products.

2. In the transition state the reactants are combined in a species called the activated

complex.

3. In activated complex the bonds between the atoms are in the process of being formed

and broken.

4. A true thermodynamic equilibrium exists between the reactant molecules and the

activated complex species even though the overall chemical reaction is irreversible.

5. The rate of the reaction is equal to the concentration of activated complex species

times the frequency at which the complex intermediates dissociate into products.

In the light of above mentioned postulates, the reaction between two reactants A and B can be

represented as follows

A + B [AB]* Products

k

Where [AB]* is the activated complex and k is the rate constant.

On the basis of these ideas, Eyring derived an equation for the rate constant k of any reaction,

given by

*

A

RTk K

N h (1)

In this equation, R is the gas constant, NA is Avogadros number, h is Planks constant, T is

Kelvin temperature and K* is equilibrium constant for the formation of activated complex

from the reactants.

Now from thermodynamics equilibrium constant for the first step reversible process is related

with the change in Gibbs free energy according to following expression,

* *G RTln K

*

* Gln KRT

*G

* RTK e

(2) (Taking antilog on both sides)

Where G* is the change in free energy of activation and is obtained by subtracting the free

energy of the activated complex from that of reactants.

Putting this value of K* from equation (2) in equation (1) we get

*G

RT

A

RTk e

N h

(3)

54

Also from fundamental relations of thermodynamics we know that

* * *G H T S

Where G*, H* and S* represents the free energy, enthalpy and entropy of activation

respectively and T is the absolute temperature. Putting this value of G* equation (3)

becomes,

* *( H T S )

RT

A

RTk e

N h

* *S H

R RT

A

RTk e e

N h

*aES

R RT

A

RTk e e

N h

(4) ( H* = Ea)

Equation (4) is the equation of transition state theory and is known as Eyring equation. This

equation gives valuable information about the reaction.

A comparison of equation (4) with Arrhenius equation shows that the pre-exponential

factor A is related to the entropy of activation by the expression,

*S

R

A

RTA e

N h

The entropy of activation introduced by the transition state theory is analogous to the steric

factor from collision theory.

Transition state theory is applicable to all types of reactions. This is due to the

presence of S* factor in the theory. It has been found that the value of S* is negative. The

negative value of S* indicates that the formation of activated complex (which is more

ordered than the molecules of the reactants) from the reactants is accompanied by a large

decrease of entropy. Larger the complexity of the reacting molecules, greater will be the

decrease in the value of entropy. This results in the smaller value of the quantity *S /Re and

since ART / N h is same for all reactions at same temperature, therefore, PZ value in collision

theory would be much less than the interpreted value.

Date post:	24-Sep-2015
Category:	Documents
Upload:	muhammad-asif
View:	48 times
Download:	6 times

Chemical Kinetics

Documents