This document discusses chemical kinetics and rate of reactions. It defines chemical kinetics as the study of reaction rates and their mechanisms. It then discusses factors that influence reaction rates such as concentration, temperature, pressure, catalysts and more. It defines rate of reaction and discusses how to determine rates. It introduces reaction orders such as zero order, first order and second order reactions. Examples of each type of reaction order are provided along with the appropriate rate equations. Pseudo-first order reactions are also discussed.

1. UNIT- 4
CHEMICAL KINETICS
BY:
ARUN GOYAL

2. • Rate of reaction
Factors influencing the rate of reaction:
Concentration
Temperature
Pressure
Solvent
Light
Catalyst
• Kinetics of various reactions
Zero order reactions
1st order reactions
• Theories of reaction rates (DELETED BY
CBSE FOR SESSION 2020-21)
Collision
Absolute reaction rate theory

3. Slow reactions Fast reactions

4. Chemical Kinetics
 Studies the rate of a chemical reactions & rate laws
a) Shows time needed for a given amount of the product
b) Shows amount of product in a given amount of time
c) Shows how to control the reaction
d) Guide towards the mechanism of a reaction
 Factors affecting the rate of a reaction
(Temperature, pressure, concentration and catalyst)
Chemical kinetics is defined as the branch of chemistry which deals with the
study of the rate of chemical reactions and their mechanism.

5. Rate of Reaction
A B
“The rate of reactions is defined as the change in concentration of
any of reactant or products per unit time”

6. Rate of Reaction
Time taken
Time taken
Rateof reaction(r)
Amount of B produced
Rateof reaction(r)
Amount of Aconsumed
dt
Rateof reaction(r)
dx

7. = =
= =
- d[ A]
dt
 d [ B ]
Rate of Reaction
A B
Rate of Reaction (r) = Rate of disappearance of A
Unit of Rate
Concentration/Time (Mole/litre)/sec mol l-1 s-1
d t
 [ B ]
dt d t
= Rate of appearance of B
-[ A]

8. Rate of Reaction
A + B C + D
Rate of Reaction (r) = Rate of disappearance of A=
= Rate of disappearance of B =
= Rate of appearance of C =
= Rate of appearance of D =
d t
- d [ A ]
d [ C ]
d t
d t
- d [ B ]
d t
 d [ D ]

9. An Example

10. Determination of Rate

11. Determination of Rate

12. nA + mB pC + qD
Rate of Reaction & Stoichiometry
= Rate of disappearance of B =
= Rate of appearance of C =
n d t
d [ A ]
Rate of Reaction (r) = Rate of disappearance of A =
 1
 1 d [ C ]
p d t
 1 d [ B ]
m d t
= Rate of appearance of C =
 1 d [ D ]
q d t

13. Determination of Rate
Zn + H2SO4 ZnSO4 + H2
rate  -
d[Zn]
 -
d[H2SO4 ]

d[ZnSO4 ]

d[H2 ]
dt dt dt dt
N2 + 3H2 2NH3
rate  -
d[N2 ]
 -
1 d[H2 ]

1 d[NH3 ]
dt 3 dt 2 dt
H2 + I2 2HI
rate  -
d[H2 ]
 -
d[I2 ]

1 d[HI]
dt dt 2 dt

14. Factors influencing the rate of reaction
• Concentration
The rate of chemical reaction is proportional to
the concentration of the reacting species taking
part in reaction. Usually increase as reactant
increases.
• Temperature
The increase in temperature increases the
reaction rate. Actually, the energy of the
reactant increases with increase of T and so
will be the no. Of collisions.
It is observed , about 10oC rise in T makes
reaction rate double.

15. • Pressure
With Increases pressure rate of reaction increases.

16. • Presence of Light: Some reactions known as photochemical reaction take
place in presence of light. So if intensity of light increases rate of reaction
increases.
• Surface Area: Increase in surface area provides more opportunity for the
reactants to come in contact or collide resulting in increased rate .

17. • Catalyst
Catalysts are the chemical substances which
increases rate of reaction without undergoing any
physical or chemical change. Presence of positive
catalyst also increases rate of reaction. Speed
chemical reactions
Catalyst and activation energy..
 Toincrease the rate of a reaction you need to
increase the number of successful collisions.
One possible way of doing this is to provide
an alternative way for the reaction to happen
which has a lower activation energy.
 Adding a catalyst has exactly this effect of
shifting the activation energy. A catalyst
provides an alternative route for the reaction.
That alternative route has a lower activation
energy.

18. Law Mass Action & Rate of Reaction
Law of mass action, first proposed by Guldberg & Waage
in 1867 and states that the rate at which a substance
reacts is proportional to its active mass, i.e., molar
concentration and the rate of a chemical reaction is
directly proportional to the product of the active masses or
molar concentrations of the reactants.
Rate = k [A]n [B]m
nA + mB Product
Rate  [A]n [B]m
K is the rate constant
(Velocity constant, Velocity co-efficient or Specific reaction rate)

19. Rate Laws/Rate Equation
It is an expression showing the relationship between the
reaction rate and the concentrations of reactants
Rate = k [A]n [B]m
nA + mB Product
Rate  [A]n [B]m
K is the rate constant
(Velocity constant, Velocity co-efficient or Specific reaction rate)

20. Rate Constant
Rate = k [A] [B]
A + B Product
Rate  [A] [B]
If [A] = [B] = 1, then
Rate = k  1  1 = k
Thus, rate constant of a reaction may be defind as the rate of
reaction when the concentration of each of the reactants is
unity at a given temperature
Characteristics of k
• Different value for different reactions.
• A measure of rate of reaction.
• Independent of reactant concentration.
• Varies with change in temperature.

21. More Examples
Rate k[CO][NO2 ]
Rate  k[N2O5 ]
Rate  k[H2 ][I2 ]
Rate  k[NO ]2
2
Rate  k[NO] [H2 ]2
CO  NO2  CO2  NO
2N2O5  4NO2  O2
H2  I2  2HI
2NO2  2NO O2
2NO  2H2  N2  2H2O

22. Order of a Reaction
The sum of the powers of concentrations in the rate law
Rate = k [A]m [B]n
nA + mB Product
Rate  [A]m [B]n
Order of the reaction = m + n
m + n = 0, a zero order reaction
m + n = 1, a first order reaction
m + n = 2, a second order reaction
m + n = 3, a third order reaction

23. Molecularity of a Reaction
The number of reactant molecules involved in a reaction
A ProductUnimolecular reactions
Bimolecular reactions A + B Product
Termolecular reactions A + B + C Product
The decomposition of N2O5
N2O5
N2O5 + NO3
2N2O5
NO2 + NO3
3NO2 + O2
4NO2 + O2
(Slow)
(Fast)

24. Order of a reaction Molecularity of a reaction
Sum of the power of the
concentration terms in the rate
low.
Number of reacting species
involved in a simple reaction.
Experimentally determined. A theoretical concept
Can have fractional value. Always a whole number
Can have zero value. Do not have zero value
Can be changed with reaction
conditions.
Can not be changed with
reaction conditions.
For a complex reaction, the
slowest step gives the order of
the reaction
For a complex reaction, each
step has its own molecularity.

25. Zero Order Reaction
The rate is independent of reactant concentrations.
A Product
Initial conc. a 0
Final conc. a-x x
dt
- d[ A]
 k[ A]0
R a t e 
dt
 k
d x

- d ( a  x)
 k ( a  x)0
dt
dx kdt
t
x = kt or k 
x
Unit of k = concentration per unit time

26. Examples of Zero Order Reaction
22 2
2
N O Pt
N 
1
O
Photochemical reactions:
H2 (g)  Cl2 (g)  2HCl(g)Sunlight
Heterogeneous reactions:
2HI Au
H  I
2 2
2NH3 N2 3H2Pt

27. First Order Reaction
Rate is determined by the change of only one concentration term
A Product
Initial conc. a 0
Final conc. a-x x
dt
- d[ A]
 k[ A]R a t e 
d x

- d ( a  x)
 k ( a  x)
dt dt
 k d t
a  x
d x

28. -ln (a-x) = kt + I, the constant of integration
a x

dx
 kdt
If t = 0 and x = 0
I = -ln a
a
 kt
a x
ln
First Order Reaction
k 
1
ln
a
t a x
k 
2.303
log
a
t a x
Unit of k = (time)-1

29. Examples of First Order Reaction
tV
10
Vt
Decomposition of N2O5 in CCl4 solution:
N2O5  2NO2  1
2 O2
k 
2.303
log
V
Problem:
From the following data for the decomposition of N2O5 in CCl4
solution at 48 C, show that it is a first order reaction.
t (mins)
VCO2
10 15 20 25 
6.30 8.95 11.40 13.5 34.75

30. Examples of First Order Reaction
0 5 15 25 45
37 29.8 19.6 12.3 5.0
Decomposition of H2O2 in aqueoussolution:
H O Pt
H O O
2 2 2
k 
2.303
log
a
t 10
a x
Problem:
The catalysed decomposition of H2O2 in aqueous solution is
followed by titrating equal volume of sample solutions with
KMnO4 solution at different time interval give the following
results. Show that the reaction is a first order reaction.
t (mins)
VKMnO4

31. Acid Hydrolysis of a Ester:
t 10
V -Vt
3

V -V0
2 5 2 3 2 5
k 
2.303
log
CH COOC H  H O A
cid
CH COOH  C H OH
Examples of First Order Reaction
Problem:
The following data was obtained on hydrolysis of methyl
acetate at 25 C in 0.2N HCl. Show that it is a first order
reaction.
t (mins)
Valkali
0 75 119 180 
19.24 24.20 26.60 29.32 42.03

32. Inversion of Cane sugar:

 C H O
tt 10
r r
6 12 6
C H O  H O A
cid
C H O
12 22 11 2 6 12 6
k 
2.303
log
r0  r
Examples of First Order Reaction
Problem:
The optical rotation of sucrose in 0.9N HCl at different time
interval is given in the following table. Show that it is first order
reaction.
t (mins) 0 7.18 18 27.05 
Rotation +24.09 +21.4 +17.7 +15 -10.74

33. Pseudo-Order Reaction
The experimental order which is not the actual one observed.
A + B (excess) Product
R a t e  k '
[ A ]
R a t e  k [ A ] [ B ]
Where,k'
 k[B]
Acid Hydrolysis of a Ester:
Acid
CH 3COOC 2H5  H2O  CH3COOH  C2H5OH
Inversion of Cane sugar:
6 12 6 C H OC H O  H O A
cid
C H O
12 22 11 2 6 12 6

34. Second Order Reaction
Rate is determined by the change of two concentration term
2A Product
Initial conc. a 0
Final conc. a-x x
dt
- d[ A]
 k[ A]2
R a t e 
d t
d x
=
dt
- d[a  x] = k(a-x)2
= kdt
d x
( a  x ) 2

35. = kt + I
a  x
1
 (a  x)2 = kdt
dx
If t = 0 and x = 0
I = 1/a = kt +
a  x
1 1
a
x
t a(a  x)
k 
1
Unit of k = (conc.)-1 (time)-1
Second Order Reaction

36. Second Order Reaction
A + B Product
Initial conc. a b 0
Final conc. a-x b-x x
dx
dt
 kdt
(a  x)(b  x)

dx
 k(a  x)(b  x)

d[A]
 
d[B]

dx
 k[A][B]
dt dt dt
Integrating,
ab  x)
2.303
log
b(a  x)
t(a b)
k 

37. A + B Product
Initial conc. a a 0
Final conc. a-x a-x x
dx
dt
 kdt
(a x)2

dx
 k(a  x)(a  x)  k(a  x)2

d[A]
 
d[B]

dx
 k[A][B]
dt dt dt
aa  x
k 
2.303
log
x
t
Integrating,
Second Order Reaction

38. Examples of Second Order Reaction
Alkali Hydrolysis of Ester:
x
t a(a  x)
k 
1
CH3COONa  C2H5OHCH3COOC2 H5  NaOH
Problem:
A gram mole of ethyl acetate was hydrolysed with a gram
mole of NaOH and was studied by titrating 25ml of the
reaction mixture at different time interval against a standard
acid. Show that the reaction is of the second order.
t (mins)
Vacid
0 4 6 10 15 20
8.04 5.3 4.58 3.5 2.74 2.22

39. Examples of Second Order Reaction
Thermal decompostion of Acetaldehyde:
t P0(2P0  Pt)
Pt P0
k 
1

2CH CHO 5
20
C
2CH  2CO
3 4

CH CHO 5
20
C
CH  CO
3 4
Problem:
The thermal decompostion of acetaldehyde was studied at
518 C. Starting with the initial pressure of 363 mm of Hg, the
following results were obtained at different time interval. Show
that the reaction is of second order.
t (sec) 42 73 105 190
p (mm) 34 54 74 114
x
t a(a  x)
k 
1

40. Half-Life of a Reaction (t1/2 ort0.5)
Half-life is another expression for reaction rate and is defined
as the time required for the concentration of a reactant to
decrease to half its initial value

41. Half-Life of a Zero-Order Reaction
As, [A] at t1/2 is one-half of the original [A],
[A]t = 0.5 [A]0.
k
k k
t
k
2k
0.5
0
0.5

[A]0
 t

[ A]0  0.5[A]0
 t
[ A]t
 t 
x

[A]0

x

42. Half-Life of a First-Order Reaction
NOTE: For a first-order process, the half-life does not depend on [A]0 and
is inversely proportional to k.
k
t1/2
1/2
0.5[A]0t1/2
k k

0.693
t 
2.303
log 2 
2.303
0.3010
k 
2.303
log2
t1/ 2
As, [A] at t1/2 is one-half of the original [A],
[A]t = 0.5 [A]0.
k 
2.303
log
a
t a x
k 
2.303
log
[A]0
t [A]
k 
2.303
log
[A]0

43. Half-Life of a Second-Order Reaction
0
1/2
00 00
1/2
1
11 21
1
k[A]
[A] [A] [A]0.5[A]
[A] [A]0[A]0[A]
x
t 
 kt 

x

[A]0 [A]

1
a(a x)
kt 
t a(a  x)
k 
1
As, [A] at t1/2 is one-half of the original [A],
[A]t = 0.5 [A]0.
NOTE: For a second-order process, the half-life is inversely
proportional to both k and[A]0.

44. nA  Product
rate k[A]n
Order Rate Law Integrated Rate Law Half-Life Straight line Plot
0 r = k[A]0
[A]   kt[A]0
(a x)   kt  a
t 
[A]0
0.5
2k
t 
a
0.5
2k
[ A] vs t
(a  x) vs t
1 r = k[A]1
ln [A]   kt  ln[A]0
ln(a  x)   kt  ln a
t 
0.693
0.5
k
ln [ A] vst
ln(a  x) vs t
2 r = k[A]2
1
 kt 
1
[A] [A]0
1
 kt 
1
(a  x)
a
t 
1
0.5
k[A]
0
t 
1
0.5
ka
1
vs t
[A]
1
vs t
(a x)
Summarising the all……….
At Time = 0, Concentration of reactant = [A]0 or a
At Time = t, Concentration of reactant = [A] or (a-x)

45. Determination of the Order of a Reaction
 Using integrated rate expression
• A hit-and-trial method
• Calculating the constant k
 Using half-life period
 
 
1
2
log 
log 
 A2 
 Graphical method
Linear Fitting of rate expression:
• Zero-order, [A] vs t
• 1st-order, ln[A] vs t
• 2nd-order, 1/[A] vs t
A
t
n 1 t1 t  , nth order
1
[A]n1

46. 0th Order n=0
Rate = k[A]0
[A] = - kt + [A]o
1st Order n=1
Rate = k[A]1
ln[A] = - kt + ln[A]o
2th Order n=2
Rate = k[A]2
1/[A] = kt + 1/[A]o
[A]
Time, t
slope = k
nA  Product
rate k[A]n
ln[A]
Time, t
slope = k
1/[A]
Time, t
slope = k

47. Determination of the Order of a Reaction
 Van’t Hoff’s Differential Method
1 2
21
1 2
21
2
2
1
1
2
dCdC
dCdC
dCdC
dtdt
log C  log C 
dt





 log 
dt

 




log 
n 
 nlog C  log C 
dt






 log 
dt






log 
 log k  nlogC
dt






log  log k  nlogC
dt






log 
 kC n dC2
dt
dC1
 kC1
n
dC
 kCn
 Ost wald’s Isolation Method
A B  C  Pr oduct
n  nA  nB nC

48. Problems
P (mm Hg)
t0.5
707
84
79
84
3.5
84
Q4. For the reaction between gaseous chlorine and nitric oxide, it is found that
doubling the concentration of both reactants increases the rate eight times, but
doubling the chlorine concentrationalone doubles the rate. What is the order of
reaction with respect to nitric oxide and chlorine?
2NO + Cl2 = 2NOCl
Q1. In the reduction of nitric oxide, 50% of reaction was completed in 108 sec
when initial pressure was 336 mm Hg and in 147 sec initial pressure was 288 mm
Hg. Find the order of the reaction.
2NO  2H2  N2  2H2O
Q2. In the thermal decomposition of a gaseous substance, the time taken for the
decomposition of half of the reactant was 105 min when the the initial pressure was
750 mm and 950 min when the initial pressure was 250mm. Find the order of the
reaction.
Q3. The half-life for the thermal decompostion of phosphine at three different
pressures are given below: Find the order of the reaction.

49. Problems
Q5. The half-life of a chemical reaction at a particular concentration is 50 min.
When the concentration is doubled, the half-life become 100 mins. Find out the
order of the reaction.
Q6. The half-life of a chemical reaction at a particular concentration is 50 min.
When the concentration is doubled, the half-life remains 50 mins. Find out the
order of the reaction.
Q7. The half-life of a chemical reaction at a particular concentration is 50 min.
When the concentration is doubled, the half-life become 25 mins. Find out the
order of the reaction.
Q8. Compound A decomposes to form B and C is a first order reaction . At 25 C
the rate constant for the reaction is 0.45 s-1 . What is the half-life of A at this
temperature.
Q9. The half-life of a susbstance in a first order reaction is15 minutes. Calculate
the rate constant.
Q10. For a certain first order reaction, half-life is 100 sec. How long will it take for
the reaction to be completed 75% ?
Q11. 50% of a first order reaction is completed in 23 min. Calculate the time
required to complete 90% of the reaction.

50. Temperature Effect on Reaction Rate
• A measure of the average kinetic energy of the molecules in a sample.
• At any temperature there is a wide distribution of kinetic energies.
• At higher temperatures, a larger population of molecules has higher energy.
• As the temperature increases, the fraction of molecules that can overcome the
activation energy barrier increases.
• As a result, the reaction rate increases.

51. The ratio of rate constants of a reaction at two different
temperatures differing by 10 degree is known as
temperature coefficient. The value of temperature coefficient
is generally 2 to 3.
Temperature Effect on Reaction Rate
Temperature Coefficient =
298
308
25
35
k
k
k
k
 2 to 3
 Increases the rate of a reaction
 Initiate a reaction
Temperature Coefficient:

52. Arrhenius Equation
In 1889 Arrhenius developed a mathematical relationship
between k, T and Ea, which is known as Arrhenius equation.
where, A is an experimentally determined quantity
Ea is the activation energy
R is the gas constant
T is temperature in Kelvin
Taking the natural logarithm of both sides,
y = mx + b k is determined experimentally
at several temperatures.

53. Arrhenius Equation
Calculation of Ea:
Ea can be calculated from the slope of a plot of (ln k) vs (1/T).
RT
Ea
2.303RT
 log Alog k  
 ln Aln k  
Ea If k1 and k2 corresponds to T1 andT2,

 1 2 1
T2
k 2.303R  TT
log
k2

Ea T1
R 1
T
Slope 
 ln k
 
Ea

54. Theories of Reaction Rates
Collision Theory
Absolute Reaction Rate Theory
Or
(Transition State Theory)

55. Collision Theory
According to this theory, a chemical reaction takes place only by
collisions between the reacting molecules.
 The molecules must collide with sufficient kinetic energy.
Energy barrier
Ea = ActivationEnergy

56.  The molecules must collide with correct orientation.
Collision Theory
Thus, only molecules colloid with kinetic energy
greater than activation energy and with correct
orientation can cause reaction.

57. Collision Theory & Reaction Rates:
A + B Product
Rate  f  p z
Where, f = fraction molecules having sufficient kinetic energy
p = probable fraction of molecules with effective orientation
z = collision frequency

58. Limitation of the Collision Theory
• Applicable to simple gaseous reaction
• Good for simple bimolecular reaction
• Only kinetic energy is considered
• Silent for bond cleavage and bond formation
• No method to calculate Probability factor or Steric factor

59. Absolute Reaction Rate Theory
The transition state represents the point of
highest free energy for a reaction step.
Developed by Henry Erying in 1935 and postulated that during collision, the
reactant molecules form a transition state or activated complex which
decomposes to give the products.
AB  C  A BC  A  B- C
In summary,
• Kinetic energy is converted into potential energy
• Interpenetration of electron clouds
• Rearrangement of valence electrons
• Formation of an activated complex or transition state

60.  B-C
Potentialenergy
Ea
A B  C  ABC  A
Activated Complex or
Transition state
A BC
AB  C
Reactant
E
A  B- C
Product
Reaction coordinate
Reaction Energy Diagram for an Exothermic Reaction

61. Potentialenergy
Ea
E
AB  C
Reactant
Reaction coordinate
A  B- C
Product
 B-CA B  C  ABC  A
Activated Complex or
Transition state
A BC
Reaction Energy Diagram for an Endothermic Reaction

62. Potentialenergy
AB  C  A BC  A  B-C
Activated Complex or
Transition state
A BC
AB  C
Reactant
E
A  B- C
Product
Reaction coordinate
Ea
Ecat
Activation Energy and Catalysis

63. Some Problems
Q1. The value of rate constant for the decomposition of N2O5 were
determined at several temperatures. A plot of ln k vs 1/T gave a
straight line of which the slope was found to be -1.2  104 K. Calculate
the activation energy of the reaction.
Q2. The value of rate constant for the decomposition of ethyl iodide is
1.6  10-5 s-1 at 327 ºC and 6.36  10-3 s-1 at 427 ºC. Calculate the
activation energy for the reaction. (R = 8.314 JK-1 mol-1)
Q3. The rate of a particular reaction quadruples when the temperature
changes from 293K to 313K. Calculate the activation energy for the
same reaction.
Q4. The activation energy of a non-catalyzed reaction at 37 ºC is 200
kcal mol-1 and the activation energy of the same reaction when
catalyzed by an enzyme is 6 kcal mol-1. Calculate the ratio of the rate
constant of the catalysed and the non-catalysed reaction. Assume
frequency factor to be same in both cases. (R = 1.987cal).

64. Consecutive Reactions
At t = 0,
At t = t,
A k1

[A]0
[A]
B k2
 C
0 0
[B] [C]
dt
dt
dt
-d[A]
2
1 2
d[C]
 k [B]
d[B]
 k [A] - k [B]
 k1[A]
[A]0 [A] [B] [C]
[A]
[B]
[C]

65. Parallel or Side Reactions
dt
dt
1 2
1 2 1 2
 d[A]
k'
[A]
dt
 d[A]
 (k  k )[A]
 d[A]
 r  r  k [A]  k [A]
r1 k1[A]
r2  k2[A]
r1

k1
r2 k2
r1 k1[A]
r2  k2[A]
r1

k1[A]

k1
r2 k2[A] k2

66. Riversible or Opposing Reactions
kf
A B
kb
At t = 0,
At t = t,
[A]0
[A]
0
[B]
eq
b f
eq
f 0 b
f b
f b
x
dt
dt
dt
dt dt dt
 k  k
0  kf ([A]0  xeq )  kb xeq
 kb xeq  kf ([A]0  xeq )
([A]0  xeq )
At equliberium,
dx
 0 & x x
dx
 k ([A]  x)  k x
dx
 k [A]  k[B]
 d[A]

d[B]

dx
 k [A]  k [B]

67. Chemical Kinetics & Catalysis
A catalyst is a substance which alters the rate of a chemical reaction,
itself remaining chemically unchanged at the end of the reaction.
The process is called catalysis.

68. Positive Catalyst Negative Catalyst
Homogeneous Catalysis Heterogeneous Catalystis
Enzyme Catalysis
Catalyst
Catalysis

69. Homogeneous Catalysis
In homogeneous catalysis, the catalyst is in the same phase
as the reactants and is evenly distributed throughout.
-
C12H22O11
 -  -
CH3COOC2H5  H2O  [H /OH ]  CH3COOH  C2H5OH  [H /OH ]
2H2O2  [I ]  2H2O  O2
Catalysis in Gas Phase
2SO2  O2  [NO]  2SO3  [NO]
2CO  O2  [NO]  2CO2  [NO]
CH3CHO  [I2 ]  CH4  CO  [I2 ]
Catalysis in Solution Phase
 H2O  [H2SO4 ]  C6H12O6  C6H12O6  [H2SO4 ]

70. Heterogeneous Catalyst
In heterogeneous catalysis, the catalyst is in different phase
from the reactants. Also known as Contact Catalysis.
Surface area
Promoter action
Activation energy
Catalytic poisons
Catalysis with Gaseous Reactants
2SO2  O2  [Pt]  2SO3  [Pt]
N2  3H2  [Fe]  2NH3  [Fe]
CH2  CH2  H2  [Ni]  CH3 -CH3  [Ni]
Catalysis with Liquid Reactants
2H2O2  [Pt]  2H2O  O2  [Pt]
C6H6  CH3COCl  [AlCl3 ]  C6H5COCH3  HCl  [AlCl3 ]
Catalysis with Solid Reactants
KClO3  [MnO2 ]  2KCl 3O2  [MnO2 ]
Some Features:

71. Characteristic of Catalytic Reactions
 It does not affect the equilibrium position
 Depends on temperature
P
tbl
ack
2H O2H2 O2
 Remains unchanged in mass and chemical composition
 A small quantity is generally needed
 More effective when finely divided
 It is very specific
Al O
C2H5OH 2
3
 CH2  CH2  H2O
Cu
C2H5OH CH3CHO  H2
 In general, it cannot initiate a reaction
H2  O2 No reactionRT

72. Catalytic Promoter Catalytic Poison
A substance which, though itself
not a catalyst, promotes the activity
of a catalyst
A substance which, destroy the
activity of a catalyst to accelerate
a reaction
3
N  3H Fe
 2NH
2 2
Promoter: Mo orAl2O3
• Change of lattice spacing
• Increase of peaks and cracks
3
Fe
N2  3H2  2NH
• Preferential adsorption
• Chemical reaction
CO  2H 2 2
3
 CH OHZnO & CrO
3
Poison: H2S
SO 2  O2 2SO 3Pt
Poison: As2O3

73. Autocatalyst
One of the product itself acts as a catalyst for the same reaction
CH3COOH
 -
 C2H5OH  [H /OH ]
Negative Catalyst/Inhibitor
A substance, which reduces the rate of a reaction
Inhibitor: 2% ethanol
2MnSO4  K2SO4  8H2O 10CO 2
 -
CH3COOC 2H5  H2O  [H /OH ] 
2KMnO4  5H 2C2O4  H2SO4 
2As  3H22AsH3 
Inhibitor: Glycerol
4CHCl3  3O2  4COCl 2  2H 2O  2Cl 2
2H2O2  2H 2O  O2
Combustionof fuel Inhibitor: Pb(C2H5)4

74. Potentialenergy
Activated Complex or
Transition state
A BC
AB  C
Reactant
E
A  B- C
Product
Reaction coordinate
Ea
Ecat
Activation Energy and Catalysis

75. Theories of Catalysis
1. Intermediate Compound Formation Theory
A  B C
AB
A  C AC
AC  B  AB C
AlCl
3 6 5 3
322
3
C6H6  CH Cl  C H CH  HCl
2NO  O2  2NO2 (Intermediate)
NO2  SO2  SO3  NO
NO
2SO2SO  O
 
CH3Cl  AlCl3  [CH3 ] [AlCl 4 ] (Intermediate)
 
C6H6  [CH3 ] [AlCl 4 ]  C6H5CH3  AlCl 3  HCl

76. 2. Adsorption Theory
This theory explains the mechanism of a reaction catalyzed by a solid catalyst.
The catalyst function by adsorption of the reacting molecules on its surface
Step I: Adsorption of reactant molecules.
Step II: Formation of activated complex.
Step III: Decomposition of activated complex.
Step IV: Desorption of products.
Theories of Catalysis

77. Theories of Catalysis
2. Adsorption Theory
 H2  CH  CHNi
3 3CH 2  CH2
Step I: Adsorption of Hydrogen molecules.
Step II: Broken of H-H bond.
Step III: Formation of activated complex.
Step IV: Decompostion of the activated
complex and Desorption of
ethane molecules.

78. Active Centers on Catalyst surface

79. Adsorption Theory & Catalytic Activity
 Metals in a state of fine subdivision or colloidal
form are rich in free valence bonds and hence they
are more efficient catalysts than the metal in lumps
 Catalytic poisoning occurs because the so-called
poison blocks the free valence bonds on its
surface by preferential adsorption or by chemical
combination.
 A promoter increases the valence bonds on the
catalyst surface by changing the crystal lattice and
thereby increasing the active centers.

80. Acid-Base Catalysis
Homogeneous catalytic reaction catalyzed by acids or bases,
or both acids and bases are known as acid-base catalysts.
Inversionof Cane sugar
6 12 6
Hydrolysis of Ester
CH3COOC2H5  H2O CH3COOH  C2H5OHAcid/Base
Decomposition of Nitramide
NH2 NO2   N2O  H2OBase
 C H OC H O  H O A
cid
C H O
12 22 11 2 6 12 6

81. Enzyme Catalysis
• Enzymes are catalysts in biological systems.
• The substrate fits into the active site of the
enzyme much like a key fits into a lock.
The catalysis brought about by enzymes is known as
enzyme catalysis
E + S ES P + E

82. Characteristics of Enzyme Catalysis
• Most efficient catalysts known
• Marked by absolute specificity
• Maximum at optimum temperature
• Maximum at optimum pH
• Greatly affected by inhibitors
• Greatly enhanced by Activator or Coenzyme
Inversion of cane sugar by Invertase present in yeast
6 12 6 6 12 6 C H OI
nve
r
tase
 C H OC12 H 22O11  H2O
Conversion of glucose into ethanol by Zymase present in yeast
C H O Zym
ase
2C H OH  2CO
6 12 6 2 5 2
Hydrolysis of urea by Urease present in soya bean
H N  CO  NH  H
Ure
ase
 2NH  CO
2 2 3 2

83. Thanks
References:
NCERT textbook of Class XII
Images and some matter taken from internet sources.