UNIT

CHEMICAL KINETICS

SYLLABUS
Rate of a reaction (average and instantaneous), Factors affecting rate of reaction; Concentration, Temperature,
Catalyst; Order and Molecularity of a reaction; Rate law and specific rate constant, Integrated rate equations
and Half life (only for zero and first order reactions); Concept of collision theory (elementary idea, No mathematical
treatment)

KEY CONCEPTS
4.1 RATE OF REACTION
Rate of reaction defined as the change in concentration of reactant or product per unit time. It is always a
positive quantity.
Total Change in concentration of reactant or product
dc
Rate of reaction = ; v=±
Change in time (in sec) dt
where dc = change in concentration in a small interval dt

Average Rate of Reaction : The rate of reaction over a certain measurable period of time during the course
of reaction, is called as average rate of reaction. It is denoted by R .
For a reaction A  B

  A 2 –  A 1   A
R  
 t 2 – t1  t

where [A]1 = Concentration of reactant A at time t1, [A]2 = Concentration of reactant A at time t2.

Instantaneous Rate of Reaction : The rate of reaction at any particular instant during the course of reaction
is called as instantaneous rate of reaction.
Mathematically, Instantaneous rate = (Average rate )t 0

   A    B  d  A  d  B
Rt = – = or Rt = – 
 t   t  dt dt
t  0 t  0

Hence, Slope of the tangent at any point of curve indicates instantaneous rate of reaction.

1
Rate of reaction in the form of stoichiometry of a chemical reaction :
Let us consider a reaction :
m1A + m2B + -------  n1P + n2Q + n3R + ---------

1 d  A 1 d  B 1 d  P  1 d  Q 
Rate of reaction = –   
m1 dt m 2 dt n1 dt n 2 dt
Unit of Reaction Rate = mol L–1 time–1 i.e. (mol L–1 s–1 or mol L–1 min–1 or mol L–1 h–1)

Example 1 :
For a reaction A B the concentration of reactant decreases from 0.04 M to 0.03 M in 20 minutes. Calculate
the average rate of the reaction.
c
Sol. Average rate = ±
t
Here the decrease in concentration of reactant takes place therefore,
c (C  C1 ) (0.03  0.04) 0.01
Average rate = – = 2   = 0.5 × 10–3
t t 20 20
Average rate = 0.5 × 10–3 mole litre–1 minute–1.

Example 2 :
In the reaction A + 2B 3C the rate of disappearance of A is 10–2 mole lit–1 sec–1. Calculate for the reaction.
(i) Rate of disappearance of B. (ii) Rate of formation of C.
Sol. From the stoichiometry of the reaction
dCA 1 dCB 1 dC d [A] 1 d [B] 1 d [C]
Rate of reaction =      
dt 2 dt 3 dt dt 2 dt 3 dt
Hence the rate of disappearance of B = 2 × rate of disappearance A = 2 × 10–2 mole litre–1 sec–1
Rate of formation of C = 3 × Rate of disappearance of A = 3 × 10–2 = 3 × 10–2 mole litre–1 sec–1

4.2 RATE LAW

(a) The rate law represents the experimentally observed rate of reaction which depend upon the slowest step of
the reaction.
(b) Rate law can not be deduce from the equation for a given reaction. It can be find by experiments only.
(c) Rate of a reaction is directly proportional to the product of the concentration of reactants.
Suppose n1 A + n2 B  Product
R   A  n1  B n 2

R = k  A  n1  B n 2
where k is called as specific reaction rate or velocity constant or velocity coefficient.
Unit of Rate Constant : R = k  A  n1  B n 2

STUDY MATERIAL: XII CHEMISTRY 2 CHEMICAL KINETICS

  mole  n
 mole  1  mole  2
n
 litre sec.  k    
litre   litre 

1(n1 n 2 )
 mole 
 (mole)  1 2  (litre) 1 2  sec sec1 .
1– n  n n  n –1 k= 
 litre 
k= –1 or

Example 3 :
For the reaction, A + B  C the following data were obtained.

Initial rate
S.N. [A] moles litre 1 [B] moles litre 1
[moles litre1 min 1 ]
1 0.10 0.10 1.0  104
2 0.10 0.30 9.0  104
3 0.30 0.30 2.7  103

(a) Write the rate law for this reaction.

(b) Calculate the value of specific rate constant for this reaction.
Sol. (a) Let the orde w.r.t. to A be x and with respect to B by y.
In the first experiment, r = K [0.10]x × [0.10]y = 1.0 × 10–4 mol –1 min–1 ....... (i)
In the second experiment, r = K [0.10]x × [0.30]y = 9.0 × 10–4 mol –1 min–1 ....... (ii)
In the third experiment, r = K [0.30]x × [0.30]y = 2.7 × 10–3 mol –1 min–1 ....... (iii)
Dividing eq. (ii) by (i) and (iii) by (ii), we have

K [0.10]x  [0.30]y 9.0  104 mol  1 min 1


K [0.10]x  [0.10]y 1.0  104 mol  1 min 1

K [0.30]x  [0.30]y 2.7  103 mol  1 min 1

and 
K [0.10]x  [0.30]y 9.0  104 mol  1 min 1
or [3]y = 9 = [3]2 and [3]x = 3 = [3]1 or y = 2 and x = 1
Rate law will be, r = K [A] [B]2
(b) r = K [A] [B]2

r 1.0  104 mol  1 min 1 1.0  104 min 1

K  
[A] [B]2 [0.1 mol  1 ] [0.1 mol  1 ]2 0.1  (0.1) 2 mol2  2

1.0  104 min 1

or K 1 2 2 2
= 0.1 mol–2 2 min–1.
1.0  10  1.0  10 mol 

STUDY MATERIAL: XII CHEMISTRY 3 CHEMICAL KINETICS

4.2.1 Factors affecting rate of reaction:
(1) Nature of reactants: In a chemical reaction some bonds are broken some new bonds are formed. Hence
rate of reaction depends on the type of bonding i.e. on the nature of the reactants. Rate of the reaction is
higher if the strength of breaking bonds is less or the number of breaking bonds is less.
For example, 2NO +O2  2NO2 ......... (i)
CH4 + O2  CO2 + 2 H2O ......... (ii)
2CO + O2  2CO2 ......... (iii)
At same temperature reaction (i) is faster in comparision to reaction (ii) Because in reaction number (i)
only one bond is broken up whereas in reaction (ii) Four C-H bonds are broken up.
Similarly reaction (i) is faster in comparision to reaction (iii) because the bond NO is weak in comparision
to bond CO.
If the reactants are in ionic or free radical state then the reaction proceeds faster in comparision to molecular
reactions.
(2) Concentration of reactants: According to the law of mass action the rate of reaction is directly proportional
to the product of the concentration of the reactants.Hence by increasing the concentration of reactants the
rate of reaction increases. For example the rusting of iron increases in rainy season due to increase in water
vapour in atmosphere.
(3) Temperature: By increasing temperature the rate of most of the reactions increases. The rate of most of
the reaction is doubled or triple by increasing the temperature 10°C. By increasing temperature the kinetic
energy increases hence the fraction of the effective collisions increases and rate increases.
(4) Catalyst or inhibitors : A catalyst is a substance which influences the rate of a reaction without undergoing
any chemical change itself. For example when a small amount of MnO2 is added to KClO3 the rate of
dissociation of KClO3 increases and yield of O2 increases considerably. For the hydrogenation of unsaturated
hydrocarbon nickel is used as catalyst. Similarly by adding H3PO4 to H2O2 its dissociation decreases
hence it is used as inhibitor.
(5) Nature of medium : Reaction rate of ionic or polar reactants is fast in water or polar solvents which
slower in non-polar solvents. The reaction rate is slower of non-polar reactants in polar solvents.
(6) Effect of pressure : The reactions in which volume of the products is less than the volume of the reactants,
rate of reaction increases with increase in pressure.
For example: For the reaction N2 + 3H2  2NH3 by increasing pressure rate of reaction increases and
for reaction PCl5  PCl3 + Cl2 by decreasing pressure rate of reaction increases. There is no effect of
pressure on H2+ I2  2HI.
(7) Surface area of the reactants: In some homogeneous reactions rate of reaction depends on the contact
surface of the reactants. The rate of reaction is proportional to the surface area of the reactants.
For example small pieces of a coal burn rapidly in comparision to big pieces of coal although they have
same mass.

STUDY MATERIAL: XII CHEMISTRY 4 CHEMICAL KINETICS

4.2.2 Order of reaction :
The order of a reaction is the sum of the powers to which all concentration terms are raised in the experimentally
determined rate equation. For example,
For general reaction, aA + bB  product
dx
The reaction rate,  k [A]a [B]b  k CaA C bB
dt
Hence total order of the reaction = n = (a + b)
Order of any reaction is an experimental quantity. It is not necessary that order of a reaction is a whole number,
it can be fractional or even zero.
Examples of different order reactions
(i) Zero order reactions :
Photolysis of HCl : H2(g) + Cl2(g)  2HCl (g) ; Rate of reaction = k0
(ii) First order reactions :
1
Dissociation of H2O2 : H2O2  H2O + O2 ; Rate of reaction = k1 [H2O2]
2
(iii) Second order reactions :
Dissociation of HI : 2HI  H2 + I2 ; Rate of reaction = k2 [HI]2
(iv) Third order reactions :
2NO + O2  2NO2 ; Rate of reaction = k3 [NO]2 [O2]

4.2.3 Molecularity
The total number of molecules, atoms, ions or radicals taking part in the fundamental chemical reaction is
known as molecularity of reaction. If one, two, three molecules take part in the reaction then the reactions are
known as unimolecular, bimolecular, tri-molecular respectively. The possibility of collisions between four or
more moleculars is very less, so the reactions of high molecularity are very less.
Examples -
H2O2  H2O + O unimolecular
NH4NO2  N2 + 2H2O unimolecular
H2 + I2  2HI bimolecular
2FeCl3 + SnCl2  2FeCl2 + SnCl4 tri-molecular
2NO + O2  2NO2 tri-molecular

4.2.4 Rate determining step : When a chemical reaction occurs in a series of steps, one of the steps is much slower
than all the other steps. Such a slowest step in the reaction mechanism is called a rate determining step.
For example, the reaction, 2NO2Cl (g)  2NO2 (g) + Cl2 (g)
Take place in two steps : (i) NO2Cl (g) 
k1
 NO2 (g) + Cl (g) (slow, unimolecular)

(ii) NO2Cl (g) + Cl (g) 

k2
NO2 (g) + Cl2 (g) (fast, bimolecular)
–––––––––––––––––––––––––––––––––––
2NO2Cl (g)  2NO2 (g) + Cl2 (g) (Overall reaction)
The first step being slower than the second is the rate determining step. The rate law predicted by the slow step
is rate = k1 [NO2Cl]. It can be seen that Cl is formed in the first step and consumed in the second. Hence, it
is the reaction intermediate.
STUDY MATERIAL: XII CHEMISTRY 5 CHEMICAL KINETICS
4.2.5 Pseudo-Unimolecular Reaction
Such reactions whose order is one but the molecularity is two, are called pseudo-unimolecular reaction.
Example :

H
(a) CH3 COOC2H5 + H2O 
 CH3COOH + C2H5OH

(b) C12H22O11 + H2O 
H
 C6H12O6 + C6H12O6
(c) C6H5N2Cl + H2O C6H5OH + N2 + HCl
(d) R–X + H2O ROH + HX

Example 4 :
For the reaction Cl2 (g) + 2NO (g) 2NOCl (g), the rate law is expressed as : rate = k [Cl2] [NO]2. What
is the overall order of this reaction ?
Sol. Order = 3

Example 5 :
The rate of reaction X  Y becomes 8 times when the concentration of the reactant X is doubled. Write the
rate law of the reaction.
dx
Ans.  k [X]3
dt

Example 6 :
What is meant by elementary step in a reaction ?
Ans. Each step of a complex reaction is called elementary step.

Example 7 :
For the reaction 2X  X2, the rate of reaction becomes three times when the concentration of X is increased
27 times. What is the order of the reaction ?
Ans. Order of the reaction = 1/3

Example 8 :
A reaction occurs by the following mechanism:
(i) NO2 (g) + F2 (g)  NO2F (g) + F (g) (slow bimolecular)
(ii) F (g) + NO2 (g)  NO2F (g) (Fast, bimolecular)
(a) What is the molecularity of each of the elementary steps?
(b) What rate law is predicted by the mechanism?
(c) Identify the intermediate.
Sol. (a) The step (i) and (ii) both involve two reactant molecules each. Hence, both the steps are bimolecular.
(b) The rate law is predicted by the stoichiometric equation of the slow, rate determining step (i).
The predicted rate law is, rate = k [NO2] [F2]
(c) The species F is a reaction intermediate because it is produced in step (i) and consumed in step (ii).

STUDY MATERIAL: XII CHEMISTRY 6 CHEMICAL KINETICS

 LEARNING CHECK-1

Q.1 Define the term order of reaction for chemical reactions.

Q.2 Express the rate of the following reaction in terms of disappearance of hydrogen in the reaction :
3H2 (g) + N2 (g)  2NH3 (g)
Q.3 What is meant by order of a reaction being zero ?
Q.4 For the reaction A  B, the rate of reaction becomes twenty-seven times, when the concentration of A is
increased three times. What is the order of the reaction ?
Q.5 State the rate law for chemical reactions.
Q.6 For the reaction N2 (g) + 3H2 (g)  2NH3 (g), express the rate of reaction in terms of formation of ammonia.
Q.7 The rate law for the reaction, A + B  prodcuts is rate = k [A] [B]2. The rate of the reaction at 25°C is found
to be 0.25 M/s when [A] = 1.0 M and [B] = 0.2 M. Calculate the rate constant k at this temperature.

4.3 DIFFERENT ORDER OF REACTIONS

4.3.1 Zero order reactions : The chemical reactions in which rate of reactions is proportional to the zeroth power
of the concentration of reactant or does not depend on the concentration of the reactants are known as zero
order reactions.
For zero order reaction, [A]  product
dx
Rate of this reaction at time t is  [A]0
dt
where x is amount of reactant decomposed at time t or product formed at time t.
dx
 k 0 [A]0 , where k0 is known as rate constant for zero order reaction
dt

dx
or  k0 [ [A]0  1]
dt
This is rate expression for zero order reaction. Thus, rate of zero order reaction remains constant.
or dx = k0 dt

On integration both sides,  dx   k0dt

x = k0t + C ......... (1) Where C is integration constant.
If t = 0 then x = 0  C=0
Putting value of C in eq. (1), x = k0 t
k0 = x/t ......... (2)
This is integrated rate expression or integrated form of zero order reaction.
Characteristics of zero order reaction :
x Concentration
(i) The units of rate constant k0 : k 0   = conc. time–1
t Time
If unit of concentration is mole per litre and time in second.
Then unit of k0 = mole litre–1 second–1 (mole dm–1 sec–1)

STUDY MATERIAL: XII CHEMISTRY 7 CHEMICAL KINETICS

 (ii) On plotting a graph between (concentration of product) versus time a straight line passing through origin is
obtained.
(iii) Half life period (t1/2): Suppose in t1/2 time half reaction is complete
a
If t = t1/2 then x = a/2 , t1/2  or t1/2 a
2k 0
(iv) Graphical representation :

Zero Order

Rate Conc. t 1/2

Conc. t Conc.

Example :
(a) H2 (g) + Cl2 (g) 
h
 2HCl (g) (b) 2 NH3 (g) 
h
 N2 (g) + 3H2 (g)
(c) Reaction between Acetone and Bromine. (d) Dissociation of HI on gold surface.
(e) Decomposition of gas on surface of catalyst

4.3.2 First Order Reaction : Reaction in which the rate of reaction depends only on one concentration term of
reactant. All natural and artificial radioactive decay of unstable nuclei take place by first order kinetics.
Reaction : A  Product.
t=0 a 0
After time t (a – x) x
Here, 'a' be the concentration of A at the start and (a – x) is the concentration of A after time taken t. i.e. x part
of A has been changed in the product. So, the rate of reaction after time t is equal to.
dx dx dx
 (a  x) or  k (a  x) or = k dt
dt dt (a  x)
where, k is specific reaction rate for the first order reaction, 'a' is initial concentration of A and x moles/lit of A
has been decomposed up to the time t.

 a  x   k dt
dx
Upon integration of above equation, or – n (a – x) = kt + C

At t = 0, x = 0, C = – n a
Put the value of 'C'
 n(a – x) Slope = –x
– n (a – x) = kt – n a
 a  2.303  a 
or kt = n   or k= log  
a–x t a – x
a a–x a–x Ct t
kt = n ; –kt = n ; = n–kt ; C = n–kt
a–x a a 0
Ct = C0n–kt (Willhalpi equation)
Here C0  Concentration on time t = 0 ; Ct Concentration on time t
STUDY MATERIAL: XII CHEMISTRY 8 CHEMICAL KINETICS
 Characteristics of first order reaction.
Unit of rate constant : k = sec–1
Half-life period : Putting a – x = a/2 and t = t1/2 in the integrated equation we get
n a n2 0.693
t1/2 = ; t1/2 = ; t1/2 =
k a/2 k k
Half-life period for first order reaction is independent from the concentration of reactant.
Graphical representation :

Rate log [A] t1/2

Conc. t t

A few examples of 1st order and their formulae for rate constant are as under :
1. CH3COOC2H5 + H2O  CH3COOH + C2H5OH
2.303 v – v0
k= log 
t v – v t
where v0, vt and v are the volumes of NaOH solution used for titrating a definite volume of the reaction
mixture of zero time, after time t and after time infinite respectively.

2. C12H22O11 + H2O 
H
 C6H12O6 + C6H12O6 ;

2.303 r0 – r
k= log r – r
t t 

where r0, rt and r are the polarimetric readings at zero time, after time t and after time infinitely respectively.
3. 2N2O5  4NO2 + O2 and NH4NO2  2H2O + N2

2.303 v
k= log v – v
t  t

where vt and v are the volume of O2 or N2 gas collected after time t and after time infinite respectively.
1
4. H2O2  H2O + O
2 2

2.303 v0
k= log v
t t

where v0 and vt are the volumes of KMnO4 solution used for titrating a definite volume of the reaction mixture
in the beginning (at t = 0) and at time t respectively.

STUDY MATERIAL: XII CHEMISTRY 9 CHEMICAL KINETICS

 Radio active disintegration: The disintegration of radioactive elements obeys first order kinetics.
A (radio active element)  Product
Let at time t = 0 the number of atoms of A is N0 and at any time t the number of atom be Nt.
Then a = N0; (a – x) = Nt and  = decay constant
2.303 N
  t
log 0
Nt ....... (1)

Half life: Time in which radioactive elements is half of its initial concentration.
Hence t = t1/2, Nt = N0/2
By putting the value in equation (1),
2.303 N0 2.303 0.693
 log t1/2  log 2 
t1/2 N0 / 2 ;  

4.4 PSEUDO FIRST ORDER REACTION

A second order (or of higher order) reactions can be converted into a first order reaction if the other reactant
is taken in large excess. Such first order reactions are known as pseudo first order reactions.
For Example : Inversion of cane sugar is pseudo first order reaction.

H
C12H22O11 + H2O   C6H12O6 + C6H12O6
Cane sugar Glucose Fructose
Rate = k [C12H22O11]

Example 9 :
A first order reaction takes 20 minutes for 90% completion. Calculate the rate constant for this reaction.
Sol. We know the rate constant for first order reaction is
2.303 a
k1  log Given, t = 20 minute, a = 1, x = 0.9
t ax

2.303 1 2.303 1 2.303

k1  log  log  log10
20.0 1  0.9 20.0 0.1 20.0

2.303
k1   0.1151 minute 1 [log 10 = 1]
20.0
Example 10 :
A first order reaction is 30% completed in 50 minutes. Calculate the value of rate constant.
70
Sol. 30% completed means [A] = [A]0 (70% left)
100
2.303 [A]0 2.303 [A]0 2.303 2.303
k log  log or k  [log10  log 7]  [1.000  0.8451]
t [A] 50 70 50 50
[A]0
100
= 7.13 × 10–3 min–1.
STUDY MATERIAL: XII CHEMISTRY 10 CHEMICAL KINETICS
Example 11 :
Show that in case of first order reaction, the time required for 99.9% of the reaction to complete is 10 times that
required for half of the reaction to take place. [log 2 = 0.3010]
2.303 [A]0
Sol. Using the equation for first order, t  log
k [A]
Substituting the values for 99.9% change, we get
2.303 [A]0 2.303 6.909
t 99.9%  log  log1000 
k 0.001  [A]0 k k

0.693 t 99.9% 6.909 k

t1/2  or    10
k t1/2 k 0.693

Example 12 :
Express the relation between the half life period of a reaction and its initial concentration if the reaction involved
is of second order.
Sol. General expression for half life period of nth order :
1
t1/2  [A0]1-n or t1/2 
[A0 ]n 1

1
For second order, t1/2  where [A0] is initial concentration.
[A0 ]

Example 13 :
A substance with initial concentration a follows zero order kinetics with rate constant k mol L–1 s–1.
In how much time will the reaction go to completion ?
a a a
Sol. k or t  ; In time the reaction will be completed.
t k k

Example 14:
A first order decomposition reaction takes 40 minutes for 30% decomposition. Calculate t1/2 value.
Sol. 30% decomposition means (a – x) = (a – 0.3) = 0.7a
2.303 a
Apply the following relation, k  log
t ax
Substituting the values, we get
2.303 a 2.303 2.303
k log = log1.4286   0.1551 = 0.00893 min–1
40 0.7a 40 40

0.693 0.693
and t1/2   = 77.6 minutes
k 0.00893
STUDY MATERIAL: XII CHEMISTRY 11 CHEMICAL KINETICS
 LEARNING CHECK-2

Q.1 State any one condition under which a bimolecular reaction may be kinetically first order.
Q.2 What is meant by ‘rate constant k’ of reaction ? If the concentration be expressed in mol L–1 units and time in
seconds, what would be the units for k (i) for a zero order reaction and (ii) for a first order reaction ?
Q.3 The rate constant for the first order decomposition of N2O5 at 45°C is 3.00 × 10–2 min–1. If the initial
concentration of N2O5 is 2.00 × 10–3 mol L–1, how long will it take for the concentration to drop to
5.00 × 10–4 mol L-1?

4.5 TEMPERATURE DEPENDENCE OF THE RATE OF A REACTION

Arrhenius related rate constant k of a chemical reaction with the absolute temperature T by the following
equation: k  Ae Ea /RT .......... (1)
Where A is pre exponential factor, Ea activation energy T absolute temperature and R is gas constant
(The value of R = 1.987 cal K–1 mole–1 or 8.314 J K–1mole–1)
Equation (1) is known as Arrhenius Equation.
Taking log on both side in eq. (1)
Ea
logek = logeA –
RT Ea
Slope =
log k 2.303
Ea 10
2.303 log10k = 2.303 log10A –
RT
1/T
–E a
A plot between log k and 1/T is a straight line of slope =
2.303R
Hence Ea = 2.303 × R × slope

Main postulates of Arrhenius theory -

(i) All the molecules present in reaction do not take part in the reaction but only those molecules having a
particular amount of energy or more are able to react and form products. These molecules are known as
active molecules and the molecules which do not take part are known as inactive molecules.
(ii) Energy of active molecules is higher than a average energy. The minimum extra energy possessed by
reactant molecules which is required for a chemical reaction is known as energy of activation.
Activation energy = [Threshold energy] – [Average energy of the reactants]
Ea = Et – ER

STUDY MATERIAL: XII CHEMISTRY 12 CHEMICAL KINETICS

 The energy changes during endothermic and exothermic reactions versus the process of reaction are
shown :
Activated complex

Activation energy
Threshold energy

H
H endothermic
Average energy
of A + B exothermic

where Ea = Activation energy

For fast reaction : Activation energies are low.
For slow reaction : Activation energies are high.
(iii) The excess energy possessed by the reactant molecules is due to collisions taking place between the
molecules.
(iv) As low is the value of activation energy, more molecules will be active and reaction rate will be as high.
(v) There is an equilibrium between active molecules and normal (passive) molecules.
(vi) On an increase in temperature there is significant increase in the number of active molecules hence reaction
rate increases.
(vii) When active molecules collide with each other they form activated complex, which is high energy species
or whose potential energy is maximum. It is unstable and it decomposed rapidly and give prodcuts. Thus
rate of reaction depends on number of active molecules.

MAXWELL-BOLTZMANN DISTRIBUTION CURVE

* The distribution of kinetic energy may be described by plotting the fraction of molecules (NE/NT) with a given
kinetic energy (E) vs kinetic energy (Fig. a). Here, NE is the number of molecules with energy E and NT is total
number of molecules.

t
Fraction of molecules
Fraction of molecules

Energy of
(t+10) activation This area shows
fraction of additional
This area
molecules which
shows fraction
react at (t+10)
Most probable of molecules
reacting at t
kinetic energy
Kinetic energy
Kinetic energy
Figure (a) : Distribution curve showing Figure (b) : Distribution curve showing temperature
energies among gaseous molecules dependence of rate of a reaction
* The peak of the curve corresponds to the most probable kinetic energy, i.e., kinetic energy of maximum
fraction of molecules. There are decreasing number of molecules with energies higher or lower than this value.
* When the temperature is raised, the maximum of the curve moves to the higher energy value (Fig. b) and the
curve broadens out, i.e., spreads to the right such that there is a greater proportion of molecules with much
higher energies.
STUDY MATERIAL: XII CHEMISTRY 13 CHEMICAL KINETICS
* The area under the curve must be constant since total probability must be one at all times.
* Increasing the temperature of the substance increases the fraction of molecules, which collide with energies
greater than Ea. It is clear from the diagram that in the curve at (t + 10), the area showing the fraction of
molecules having energy equal to or greater than activation energy gets doubled leading to doubling the rate of
a reaction.
k2 E a T2 – T1
* Arrhenius equation at two different temperatures k  2.303R T T
1 1 2

k (t 10ºC)
Temperature Coefficient: Temperature Coefficient =  between 2 & 3.
k tº
(i) The ratio of the rate constant of a reaction at two temperature separated by 10°C (usually 25°C and
35°C) is known as temperature coefficient.
(ii) The temperature coefficient for most chemical reactions lies between 2 and 3.
(iii) Reaction between NO and O2 to form NO2 exhibit a small negative temperature coefficient and so the
rate of such reactions decreases with rise in temperature.

Example 11:
The rate constant of a reaction is 1.5 × 107 s–1 at 50°C and 4.5 × 107 s–1 at 100°C. Calculate the value of
activation energy, Ea for the reaction.
Sol. Applying Arrhenius equation and substituting the values, we get

k2 Ea  1 1 
log  
k1 2.303R  T1 T2 

4.5  107 s 1 Ea  1 1  Ea 50
or log     or log 3  
7 1
1.5  10 s 2.303R  8.314 323 373  19.147 323  373
0.4771  19.147  323  373 1100579.725
or Ea   = 22011.595 J mol–1
50 50
or Ea = 22.01 kJ mol–1
Reaction path
4.5.1 Effect of Catalyst : without catalyst
Energy of
* The catalyst provides an alternate pathway or activation
Energy of
reaction mechanism by reducing the activation
Potential energy

Reaction activation without

Reactants
path with catalyst
energy between reactants and products and catalyst
with
catalyst
hence lowering the potential energy barrier
* Positive catalyst increase the rate of reaction Products

while the negative catalyst decreases.

* By Using catalyst, the rate constant is always increased. Reaction coordinate
* A catalyst does not alter Gibbs energy, G of a reaction. Figure : Effect of catalyst on activation energy
It catalyses the spontaneous reactions but does not catalyse non-spontaneous reactions.

STUDY MATERIAL: XII CHEMISTRY 14 CHEMICAL KINETICS

* It is also found that a catalyst does not change the equilibrium constant of a reaction rather, it helps in attaining
the equilibrium faster, that is, it catalyses the forward as well as the backward reactions to the same extent so
that the equilibrium state remains same but is reached earlier.

4.6 COLLISION THEORY

The important points of this theory are :
(i) If two molecules are to react together they must collide together.
(ii) Threshold energy : (Energy Barrier) The minimum energy which the molecule should possess so that their
mutual collision result in chemical reaction is called threshold energy.
(iii) Effective collisions : The collisions in which molecules collide with sufficient kinetic energy (called threshold
energy) and proper orientation, so as to facilitate breaking of bonds between reacting species and formation of
new bonds to form products are called as effective collisions.
For example, formation of methanol from bromoethane depends upon the orientation of reactant molecules as
shown in Fig. The proper orientation of reactant molecules lead to bond formation whereas improper orientation
makes them simply bounce back and no products are formed.

– –
CH3Br + OH CH3OH + Br

improper H + – –
H Br OH No products
H + – orientation C
– H
H C Br OH
H H
H Proper – + – –
HO C Br HO C H + Br
orientation
H H H
Intermediate
Expression of reaction rate constant and collision theory :
Consider the following bimolecular reaction.

A  A 
 Products A  B 
 Products ........ (ii)
k ........ (i) or k

d [P] d [P]
Thus the rate of reaction for reaction (i),  k [A]2 . For reaction (ii),  k [A] [B]
dt dt

d [P]
If the concentration of each reactant A and B one mole per litre then rate of reaction  k.
dt

d [P]
But according to collision theory rate of reaction  k  Zq ......... (1)
dt
= (number molecules collide per second per litre) × (fraction of effective collision)
Where q = fraction of the total molecules which are activated,
Z = number of collisions per second per litre between reactant molecules i.e. collision frequency or collision
number (ZAA or ZAB).
According to Maxwell Boltzmann law of molecular velocities, the fraction of the molecules q, whose energy is
equal or more than activation energy can be given as

STUDY MATERIAL: XII CHEMISTRY 15 CHEMICAL KINETICS

 n * Number of activated molecules
q   e Ea /RT ......... (2)
n Number of total molecules
Put the value of q in equation (1) we get.
k  Ze  Ea /RT ......... (3)
On comparing the value of rate constant k obtained and experimental values it was found that both the values
are nearly same for so many reactions. But for complex reactions there was much difference between experimental
and theoretical values.
By considering the above fact the collision equation (3) was modified as follows

k  pZe Ea /RT ......... (4)

Here p is probability factor or steric factor or orientation factor, which is related to geometry of the molecules.
On the basis of this equation the calculated values are almost similar to practical values. Hence the equation (4)
is known as modified collision theory equation.
By comparision of equation (4) with Arhemius equation, A = pZ
Thus the expression of reaction rate derived from collision theory is similar to Arrhenious equation if A = pZ.

LEARNING CHECK-3

Q.1 Define activation energy of a reaction.

Q.2 Why is it that rate of most of the reactions increase, when the temperature of reaction mixture is increased ? In
what units is the rate of reaction expressed ?
Q.3 The reaction 2H2 (g) + O2 (g)  2H2O () is thermodynamically feasible. How is it that a mixture of hydrogen
and oxygen kept at room temperature shows no tendency to form water ?
Q.4 The decompositioin of phosphine 4PH3 P4 + 6H2 (g)
has rate law, rate = k [PH3]. The rate constant is 6.0 × 10–4 s–1 at 300 K and Ea is 3.05 × 105 J mol–1. What
is the value of rate constant at 310 K ?

ADDITIONAL EXAMPLES
Example 1 :
d [H 2 ] d [NH3 ]
For the reaction 3H2(g) + N2 (g)  2NH3 (g), how is the rate of reaction expressions – and
dt dt
interrelated ?
1 d [H 2 ] 1 d [NH3 ]
Sol.  
3 dt 2 dt

Example 2 :
For the reaction : A + H2O B, rate [A].
What is its (i) molecularity (ii) order of the reaction ?
Sol. Molecularity = 2, Order of reaction = 1

STUDY MATERIAL: XII CHEMISTRY 16 CHEMICAL KINETICS

Example 3 :
What are photochemical reactions ? Explain the mechanism of the photochemical reaction occuring between
hydrogen and chlorine gas.
Sol. Photochemical reactions are those reactions which take place in presence of sunlight, e.g.,

H 2 (g)  Cl 2 (g) 

 2HCl(g)
sunlight

Mechanism :
h
(i) Chain initiation step Cl  Cl  2Cl 
Chlorine free
radical

(ii) Chain propagation step H2• + Cl•  H• + HCl

H• + Cl2  HCl + Cl•
(iii) Chain termination step H• + Cl•  H – Cl
Cl• + Cl•  Cl2
Example 4 :
The possible mechanism for the reaction : 2NO (g) + O2 (g)  2NO2 (g) is
(i) NO + O2  k
  NO3 (fast)
 (ii) NO3 + NO 
k2
  NO2 + NO2 (g) (slow)

Write the rate law and order for the reaction.
dx
Sol. Rate law is derived from slow step,  k 2 [NO3 ] [NO]
dt

[NO3 ]
From (i), k  or [NO3] = k [NO] [O2]
[NO] [O2 ]
dx
Susbtituting value of [NO3] in the above rate law, we get  k 2  k [NO]2 [O 2 ]1
dt
It is rate law. The order of reaction is 2 + 1 = 3.

Example 5 :
What will be the initial rate of reaction it its rate constant is 10–3 s–1 and the concentration of the reactant is
0.2 mol L–1 ? What fraction of the reactant will be converted into the products in 200 seconds ?
Sol. Initial rate = k × [A]
Substituting the values, we get, Initial rate = 10–3 s–1 × 0.2 mol L–1 = 2 × 10–4 mol L–1 s–1.
To obtain fraction of reactants that will be converted into products after 200s, proceed as follows :

2.303 1 1 200s  103 s1 2  101

t log or log    0.086
k N N 2.303 2.303

1 1
or = Antilog of 0.086 = 1.219 or N  0.820 or 82%
N 1.219
Percent that will change into products = 100 – 82 = 18%

STUDY MATERIAL: XII CHEMISTRY 17 CHEMICAL KINETICS

Example 6 :
A first order reaction is 20% complete in 20 minutes. Calculate the time it will take the reaction to complete
80%.
Sol. Use the equation for first order. 20% complete means [A] = 0.8 [A]0
2.303 [A]0 2.303 [A]0 2.303 5 2.303
k log  log or k log  (log 5  log 4)
t [A] 20 0.8 [A]0 20 4 20

2.303 2.303
= (0.6990 – 0.6021) = × 0.0969 min–1
20 20
Time for 80% completion : 80% complete means [A] = 0.2 [A]0
2.303 [A]0 2.303 [A]0
t log  log
k [A] k 0.2 [A]0

2.303 20 20
or t  log 5   0.6990 = 144.2 min.
2.303 0.0969 0.0969
Example 7 :
 d[A]
Calculate the rate of reaction from the following rate law :  k [A]1[B]2
dt
When the concentrations of A and B are 0.01 M and 0.02 M respectively and k = 5.1 × 10–3 L2 mol–2 s–1.
 d[A]
Ans.  k [A]1[B]2
dt
Substituting the values of k, [A] and [B], we get
Rate of reaction = 5.1 × 10–3 × (0.01) (0.02)2 = 5.1 × 10–3 × 10–2 × 4 × 10–4
= 20.4 × 10–9 = 2.04 × 10–8 mol L–1 s–1.

Example 8 :
What is meant by ‘mechanism of a reaction’? Suggest a mechanism for the decomposition of ozone, O3 into
O2.
Sol. The exact path followed by a reaction to give the products is called mechanism of reaction.
The mechanism of the reaction 2O3 3O2 is as follows :
k
 O2  [O] (fast) ; [O]  O3  2O 2 (slow)
k1
O3 


Example 9 :
NO2 (g) + CO (g) –––– CO2 (g) + NO (g), the experimentally determined rate expression below 400 K
is : rate = k [NO2]2. What mechanism can be proposed for above reaction ?
Sol. As the rate is dependennt on [NO]2, there are two [NO] terms in the slow step of the reaction
NO2 (g) + NO2 (g) –––– NO (g) + NO3 (g) (slow)
NO3 (g) + CO (g) –––– CO2 (g) + NO2 (g) (fast)
Rate = k [NO2]2.

STUDY MATERIAL: XII CHEMISTRY 18 CHEMICAL KINETICS

Example 10 :
The decomposition of N2O5 in CCl4 solution follows the first order rate law. The concentrations of N2O5
measured at different time intervals are given below :

Time in
second (t) 0 80 160 410 600 1130 1740

[N 2O5 ] molL1 5.5 5.0 4.8 4.0 3.4 2.4 1.6

Calculate its rate constant at t = 410 s and t = 1130 s. What do these results show ?
Sol. [N2O5] at zero time [A]0 ; [N2O5] at the time [A]
2.303 [A]0
Applying the equation for first order, k  log
t [A]

2.303 5.5 2.303

At t = 410s, k  log  [log 5.5  log 4]
410 4.0 410

2.303 2.303
 log [0.7404  0.6021] =  0.1383 = 7.76 × 10–4 s–1
410 410

2.303 5.5 2.303 2.303

At t = 1130s, k log  [0.7404  0.3802]   0.3602 = 7.34 × 10–4 s–1.
1130 2.4 1130 1130
Since k is almost constant, therefore reaction is of first order.

Example 11 :
A first order reaction takes 69.3 minutes for 50% completion. Set up an equation determining the time needed
for 80% completion of this reaction.
Sol. Calculate rate constant k of the reaction as follows :

0.693 0.693 69.3  102

k   = 1 × 10–2 min–1
t1/2 69.3min 69.3 min
Write the equation for first order and substitute the values for 80% completion.
2.303 [A]0 2.303 [A]0
k log or t  log [80% is complete, 20% is left]
t [A] 1  10 2 20
[A]0
100

2.303 2.303
or t 2
log 5   0.6990 or t = 1.6097 × 102 min = 160.97 min.
1  10 102

STUDY MATERIAL: XII CHEMISTRY 19 CHEMICAL KINETICS

 CONCEPT MAP
Rate of chemical reaction : It is destroyed when performing this independent of the initial
=
defined as the change in function : Catalysts don't "cause" a concentrations of the reactants. At
concentration of a reactant (or a reaction, they just speed it up. Rate law and rate constant : An a fixed temperature, k is constant
product) in a particular time (d) The surface area of the reactants expression which relates the rate of characteristic of the reaction. Larger
interval. Type of rate of reaction (i) or catalyst : Reactions that involve a reaction to the concentration of value of k indicates fast reaction
Average (ii) Instantaneous solids often proceed faster if the the reactants is called the Rate and small k indicates slow
Factor affecting reaction rate : solid is a fine powder instead of big Equation or Rate Law. Rate  [A]a. reactions.
(a) Concentrations of reactants : chunky bits. The physical [B]b or Rate = k [A]a [B]b. The Molecularity : The minimum

STUDY MATERIAL: XII CHEMISTRY

Generally speaking, the higher the difference between a fine powder, constant of proportionality, k is number of molecules, atoms or ions
concentration of reactants, the and big chunky bits, will be the known as the Rate Constant of reactants required for an
faster the rate of the reaction. surface area (specific reaction rate) and may be elementary reaction to occur is
(b) The temperature : The higher Expression of the rate : defined as the rate at unit indicated by the sum of the
the temperature, the faster the rate For a reaction, : m1A + m2B + ------ concentrations of the reactants. k stoichiometric coefficients of the
of the reaction. n1P + n2Q + n3R + --------- depends on the temperature and is reactant(s) in the chemical
(c) The presence of a catalyst : A Rate of reaction equation, is known molecularity of
catalyst is able to increase the rate the reaction.
of a reaction, although the catalyst Molecularity of a reaction is :

20
itself is neither created nor (i) Always a whole number (not
zero) and never a fraction.
CHEMICAL KINETICS
(ii) The value of molecularity of a rate equation (or Rate law) is Bromine. (50%) of its initial value is called half-
simple or one step reaction does not Rate  [A]n = k [A]n. (d) Dissociation of HI on gold life time of the reaction. The half life
exceed 3. In a multi-step complex reaction, the surface. period for a zero order reaction is
Order of reaction : It is defined as order of the reaction depends on (e) Decomposition of gas on directly proportional to the initial
the sum of the exponents (powers) the slowest step. surface of catalyst concentration of the reactants.
of the molar concentrations of the Zero Order Reaction : Reaction in Differential rate equation : t1/2 = a/2k
reactants in the experimentally which the concentration of reactant First order reaction : A reaction is
determined rate equations. do not change with time are said to = k0[A]0 or = k0[A]0 said to be of the first order if the
If rate of reaction  [A]p [B]q [C]r or be of zero order reaction. Examples: rate of the reaction depends upon
Rate of reaction = k [A]p [B]q [C]r where x is molar concentration of only on concentration term only.
(a) H2 (g) + Cl2 (g) 2HCl (g) product at time 't'.
Order of reaction = p + q + r & the Generally radioactivity
order w.r.t. A, B & C are p, q & r (b) 2NH3 (g) N2(g)+3H2 (g) Half-life period : The time period phenomenon shows first order
respectively. (c) Reaction between Acetone and during which the concentration of reaction.
For a “Reaction of nth order”, the a reactant gets reduced to one-half
order of the reaction is n and the

CHEMICAL KINETICS
 CONCEPT MAP
Differential rate equation : where x1 and x2 be the amounts (ii) Inversion of cane-sugar between reactants and products is
decomposed up to the time t1 and called Activation energy. It's depend
t2 respectively. on nature of reactant and is free form
Psuedo first order reaction : A Threshold energy (Energy effect of temperature.
second order (or of higher order) Barrier): The minimum energy Activation energy = [Threshold
, where [A]0 is reactions can be converted into a which the molecule should possess energy] – [Average energy of the
first order reaction if the other so that their mutual collision result reactants] ; Ea = Et – ER
initial concentration and [A] is reactant is taken in large excess. in chemical reaction is called The energy changes during

STUDY MATERIAL: XII CHEMISTRY

concentration after time t. Such first order reactions are known threshold energy. endothermic and exothermic
as psuedo first order reactions. Activation Energy (E a ) : The reactions versus the process of
Half-life period : t1/2 = Examples : minimum extra energy over and reaction are shown :
(i) Hydrolysis of ethyl acetate above the average potential energy where, Ea = Activation energy
Half-life period for first order of the reactants which must be For fast reaction:
reaction is independent from the supplied to the reactants to enable Activation energies are low.
concentration of reactant. them to cross over the energy barrier For slow reaction :
Interval formula :
Activation energies are high.

21
Temperature Coefficient : The ratio
k= log ;
of the rate constant of a reaction at
two temperature separated by 10°C
(usually 25°C and 35°C) is known as
temperature coefficient.
CHEMICAL KINETICS
Temperature Coefficient
Graphical representation : substances are also known as in- i.e. the expression has no
= between 2 & 3. hibitors or negative catalyst. By dependence CO(g) concentration.
using catalyst, the rate constant is The reason is that the reaction
Arrhenius equation : k = Ae–Ea/RT always increased. occurs by a series of elementary
where A = Frequency factor or the Mechanism of reactions : The path steps.
pre-exponential factor. way which reactants are converted The sequence of elementary
A & E a are collectively called into the products is called the processes leading to the overall
Arrhenius parameters of the Positive Catalysis : reaction mechanism. For example for stoichiometry is known as the
reaction. The phenomenon in which the reaction : “Mechanism of the reaction”.
The Logarithmic expressions (Two- presence of catalyst accelerates the NO2(g) + CO(g) CO2(g) + NO(g), In a sequence of reaction leading to
temperature form) : rate of reaction. the rate expression is rate the formation of products from
Negative catalysis : The phenom- reactants, the slowest step is the rate
log enon in which presence of catalyst = determining step.
retards the rate of reaction. Such

CHEMICAL KINETICS
 ANSWERS TO LEARNING CHECK  1
(1) Sum of the powers of concentration of the reactants in the rate law expression is called the order of that
chemical reaction.
1 d [H 2 ]
(2) 
3 dt
(3) The rate of reaction does not vary with concentrations of the reaction and the reaction is very fast.
(4) Order = 3
(5) The relation between the rate of reaction and concentration of the reactants is called rate law.
1 d [NH3 ]
(6) 
2 dt
(7) Rate = k [A] [B]2 ; Rate = 0.25 Ms–1, [A] = 1.0 M, [B] = 0.2 M
0.25 (Ms 1)
0.25 (Ms–1) = k × 1.0 (M) × (0.2)2 (M2) = 0.04 × k (M3). Hence, k  3
 6.25 M 2s 1
0.04 (M )
ANSWERS TO LEARNING CHECK  2
(1) If one of the reactants is in excess, its concentration does not change. Therefore, the reaction may be first order.
(2) Rate constant is the rate of reaction when the concentration of reactants are unity.
Units of k : For zero order reaction : mol L–1 s–1.
For first order reaction : s–1
2.303 [A]0 2.303 [A]0
(3) Applying the equation of first order, k  log or t log
t [A] k [A]

2.303 2  103 2.303  0.6021

Substituting the values, we get, t  2
log  = 46.2 min.
3  10 5  104 3  10 2

ANSWERS TO LEARNING CHECK  3

(1) It is the extra energy which must be supplied to the reactants so that they can change into the products after
crossing the activated complex.
(2) When temperature is increased, the kinetic energy of molecules increases, therefore, number of molecules
possessing threshold energy increases which results in more effective collisions. Hence, rate of reaction increases.
Units of rate of reaction is mol L–1 s–1.
(3) They require activation energy to give the product.
E a  T2  T1 
(4) Applying Arrhenius equation, log k 2  Substituting the value, we get
k1 2.303R  T1T2 

k2 3.05  105  310  300  3.05  106 3.05  103 3050

log    = =   1.7128
6  104 2.303  8.314 300  310 19.147  93000 19.147  93 1780.671
k2
or 4
= Antilog 1.7128 = 51.62 or k2 = 3.097 × 10–2 s–1.
6  10
STUDY MATERIAL: XII CHEMISTRY 22 CHEMICAL KINETICS