1. 11.0 REACTION KINETICS
Objectives: Define reaction rate, average rate, instantaneous rate and initial rate Determine the reaction rate based on a differential equation.

2. REACTION KINETICS
Chemical kinetics is the study of the rates of chemical reactions, the factors that affect these rates, and the reaction mechanisms by which reactions occur.
Important
industrial process
-Time
Optimum yield
Optimum conditions
control over reaction,
obtain products economically,
using optimum conditions

3. Rate of reaction Reaction rate is the change in the concentration
of a reactant or a product with time.
Unit of rate (mol L-1 s-1)
rate
Example; A  B
d[A] dt -
d[A] = change in concentration of A
rate =
dt = period of time
d[B] dt
rate =
d[B] = change in concentration of B
Because [A] decreases with time, d[A] is negative.

4. A B
time
rate = -
d[A] dt
rate =
d[B]
[A] ↓
[B] ↑

5. Rate of reaction The average rate is the rate over a period of time.
The rate of reaction at a given time is called an instantaneous rate of reaction.
The instantaneous rate at the beginning of a reaction is called the initial rate of reaction.
Instantaneous rate is determined from a graph of concentration vs time by drawing a line tangent to the curve at that particular time.

6. Rate of reaction Reaction: H2O2(aq)  H2O(l) + ½ O2(g)
Reaction rates are obtained from the slopes of the straight lines;
An average rate from the purple line.
The instantaneous rate at t =300 s from the red line.
The initial rate from the blue line.
purple
blue
red

7. Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)
instantaneous rate = rate at a specific time
average rate = -
d[Br2] dt
= -
[Br2]final – [Br2]initial
tfinal - tinitial

8. The differential rate equation
A differential rate equation enables the relationship between the rate of disappearance of reactants and
the formation of products.
Consider the reaction,
aA + bB  cC + dD
Rate =
a,b,c and d are the stoichiometric coefficients

9. The differential rate equation
Example:
The formation of NH3,
N2(g) + 3H2(g)  2NH3(g)
The differential rate equation is;
Rate =
The equation means that the rate of disappearance
of N2 is 1/3 the rate of disappearance of H2 and 1/2
the rate of formation of NH3.

10. Example 1:
Consider the reaction, 2HI  H2 + I2,
determine the rate of disappearance of HI when
the rate of I2 formation is 1.8 x 10-6 M s-1.
Solution:
Rate =
= 1.8  10-6
Rate =
= 2  1.8  = 3.6  10-6 M s-1

11. EXERCISE 1:
Hydrogen gas produced nonpolluting product is water vapour when react in O2 due to this reaction has been used for fuel aboard the space shuttle, and may be used by Earth-bound engines in the near future.
2H2(g) + O2(g) H2O(g)
Express the rate in terms of changes in [H2], [O2] and [H2O] with time.
When [O2] is decreasing at 0.23 mol L-1 s-1, at what rate is [H2O] increasing?
(0.46 mol L-1 s-1)

12. Exercise 2:
Consider the reaction,
NO(g) + O2(g) 2NO2(g).
Suppose that at a particular time during the
reaction nitric oxide (NO) is reacting at the rate of
0.066 M s-1
At what rate is NO2 being formed?
At what rate is molecular oxygen reacting?

13. Exercise 3:
Consider the reaction,
N2(g) + 3H2(g)  2NH3(g)
Suppose that at a particular moment during the
reaction molecular hydrogen is reacting at the rate
of M s-1
a) At what rate is ammonia being formed?
b) At what rate is molecular nitrogen reacting?

14. 11.1
RATE LAW

15. At the end of the lesson the students should be able to:
Objectives:
At the end of the lesson the students should be able to:
1. define rate law and write the rate equation
2. define the order of reaction and the rate constant
3. calculate the order with respect to a certain reactant
from experimental data
4. determine the overall order of a reaction from
experimental data
5. calculate the value and determine the units of rate constants

16. The Rate Law Rate = k [A]x[B]y Rate α [reactant]
The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers.
aA + bB cC + dD
Rate α [reactant]
Rate = k [A]x[B]y
reaction order is x with respect to A
reaction order is y with respect to B
Overall reaction order is (x + y)
The exponents x, y, … can be integers, fractions or decimal or negative values.
k is called rate constant

17. Rate Law The values of x and y can only be determined experimentally.
Reaction order is usually defined in terms of reactant concentrations.
The order of a reactant is not related to the stoichiometric coefficients of the reactants in the balanced chemical equation.
F2 (g) + 2ClO2 (g) FClO2 (g)
rate = k [F2][ClO2] 1

18. The units of rate constant, k
A Products
Rate, r = k [A]x
i) The reaction is zero order
Rate = k [A]0
Rate = k
unit k =
unit rate
= mol L-1 s-1 or M s-1

19. [A] k = rate ii) First order Rate = k [A]1 Unit k = M s-1 M = s-1
iii) Second order
Rate = k [A]2
[A]2
k =
rate =
M s-1 M2
= M-1 s-1
Unit k

20. S2O82- + 3I- 2SO42- + I3- Example :
The above reaction is first order with respect to iodide ions and
to thiosulphate ions.
a) Write the rate of equation for the reaction.
b) What is the unit of rate constant, k?
Solution :
a) Rate = k [S2O82-]1[I-]1
b) Rate = k [S2O82-]1[I-]1
k =
rate
[S2O82-]1[I-]1 =
Ms-1 M2 =
M-1 s-1
Unit k

21. The order of reaction For reaction A Products Rate = k [A]x
i) If x = 0
Rate = k [A]0
Rate = k
Rate is not dependent on [A]
Therefore this reaction is zero order with respect to A

22. ii) If x= 1
Rate = k [A]1
Assume [A]i = 1.0M
Rate = k (1.0M)
If the [A] is doubled from 1.0M to 2.0M,
Rate = k (2.0M)
= 2k(1.0M)
hence
Rate = 2k[A]
Doubling the [A] will double the rate of reaction.
Therefore this reaction is first order with respect to A

23. iii) If x = 2
Rate = k[A]2
Assume [A]i = 1.0 M
Rate = k (1.0 M)2
If the [A] is doubled from 1.0 M to 2.0 M,
Rate = k (2.0 M)2
= 4k(1.0 M)
hence
Rate = 4k[A]
Doubling [A], the rate will increase by a factor of 4.
Therefore the reaction is second order with respect to A

24. Determining Reaction Order from Rate Law
Example
Determining Reaction Order from Rate Law
For each of the following reactions, determine the reaction order with respect to each reactant and the overall order from the given rate law.
(a) 2NO(g) + O2(g) NO2(g); rate = k[NO]2[O2]
The reaction order respect to NO : 2
The reaction order respect to O2 : 1
overall reaction order = 3

25. CH3CHO (g) CH4(g) + CO(g); rate = k[CH3CHO]3/2
Solution:
The reaction order with respect to CH3CHO : 3/2
The reaction order (overall) : 3/2
(c) H2O2(aq) + 3I-(aq) + 2H+(aq) I3-(aq) + 2H2O(l);
rate = k[H2O2][I-]
Solution:
The reaction of order with respect to H2O2 : 1
The reaction of order with respect to I- : 1
and zero order in H+, while overall order is 2.

26. Determination of the orders of reaction rate;
O2(g) + 2NO(g) NO2(g)
Experiment
Initial Reactant Concentrations (molL-1)
Initial Rate (M s-1) O2 NO 1 2 3 4 5
1.10x10-2
1.30x10-2
3.21x10-3
3.90x10-2
28.8x10-3
2.20x10-2
3.30x10-2
2.60x10-2
6.40x10-3
12.8x10-3
9.60x10-3

27. The reaction is first order with respect to O2
Solution:
O2(g) + 2NO(g) NO2(g)
rate = k [O2]m[NO]n
Compare 2 experiments in which the concentration of one reactant varies and the concentration of the other reactant(s) remains constant.
rate2
k [O2]2m[NO]2n
k [O2]1m[NO]1n =
rate1 =
[O2]2m
[O2]1m
[O2]2
[O2]1 m
6.40x10-3Ms-1
3.21x10-3Ms-1 =
1.10x10-2mol/L
2.20x10-2mol/L m ;
2 = 2m
, m = 1
The reaction is first order with respect to O2
Do a similar calculation for the other reactant(s).

28. To find the order with respect to NO, we compare experiment 3 and 1, in which [O2] is held constant and [NO] is doubled:
[NO]3 =
[NO]1 n =
k [O2]3m
[NO]3n
k [O2]1m
[NO]1n
Rate 3
Rate 1
12.8 x 10-3Ms-1
3.21 x 10-3Ms-1 =
2.60 x 10-2mol/L
1.30 x 10-2mol/L n
4 = 2
n = 2 ;
The reaction is second order with respect to NO
Thus the rate law is :
Rate =
k[O2][NO]2

29. Exercise: ClO2(aq) + 2OH- (aq)  products
The results of the kinetic studies are given below.
exp
[ClO2] M
[OH-]
Initial rate,
Ms-1 1
0.0421
0.0185
8.21 1 0-3 2
0.0522
1.26 1 0-2 3
0.0285
Explain what is meant by the order of reaction.
Reffering to the data determine
(i) rate law /rate equation
(ii) rate constant, k
(iii) the reaction rate if the concentration of both ClO2
and OH- = 0.05 M

30. Exercise:
Write rate law for this equation,
A + B  C
When [A] is doubled, rate also doubles. But doubling the [B] has no effect on rate.
When [A] is increased 3x, rate increases 3x, and increasing of [B] 3x causes the rate to increase 9x.
Reducing [A] by half has no effect on the rate, but reducing [B] by half causes the rate to be half the value of the initial rate.

31. Many gaseous reactions occur in a car engine and exhaust
Exercise:
Many gaseous reactions occur in a car engine and exhaust
system. One of the gas reaction is given below.
NO2(g) + CO(g)
NO(g) + CO2(g)
Rate =
k [NO2]m[CO]n
Use the following data to determine the individual and overall reaction orders:
Experiment
Initial Rate(Ms-1)
Initial [NO2](M)
Initial [CO](M) 1
0.0050
0.10 2
0.080
0.40 3
0.20

32. Integrated Rate Law Objectives:
Write the rate law for zero order, 1st order and 2nd order reaction
2. Define half-life.
3. Draw the respective graphs for the different order reactions
4. Solve quantitative problems.

33. Integrated rate equations
Zero Order Reaction
A zero order reaction is a reaction independent of the concentration of reactant.
A product
The rate law is given by
rate = k[A]0
rate = k
rate
[A] M

34. - d[A] = k dt
Using calculus,
- d[A] = kdt
- ∫d [A] = k ∫dt
- [A] = kt + c
substituting t=0, [A] = [A]0
- [A]0 = k(0)+c
c = - [A]0
[A]0 -[A] = kt
Unit of k for zero order reaction M s-1

35. [A]o – [A] t [A] rate [A]o t [A] Graphs for zero order reaction
[A]o – [A] = kt
y = mx + c t
[A] t
[A]o
rate
[A] = -k t + [A]o
y = m x + c
[A]

36. Half-life (t½)
Half life (t½) is the time required for the concentration of a reactant to decrease to half of its initial value.
zero order reaction
Substituting t = t1/2 , and [A] = [A]0 into the zero
order reaction, gives
[A]0 - [A] = kt
[A]0 – [A]0 = kt1/2 2
Solving for t1/2 gives
t1/2 = [A]0 2k

37. First Order Reactions
A first order reaction is a reaction whereby its rate depends on the concentration of reactant raised to the first power.
From the rate law,
rate = k[A]
To obtain the units of k
k = rate
[A]
unit k = M s-1 M
= s-1

38. Rate = k[A]
y = mx + c
Rate
Ms-1
[A] ,M

39. For first order reaction,
Rate = k[A]
- d[A] dt =
substituting
t = 0, [A]=[A]0
k[A]
- d[A]
[A] =
k dt
- ln [A]0 = k(0) + c
c =  ln[A]0
-ln [A] = kt – ln[A]0
- d[A]
[A] =
k dt ∫
- ln [A] = kt + c ln
[A]0
[A] = kt

40. [A] t ln[A] ln[A]o [A] ln[A]o t t
Characteristic graphs of 1st order reaction
ln[A]o – ln[A] =kt
[A] t
ln[A] = - kt +ln[A]o ln
[A]0
------
[A] = kt
ln[A]
ln[A]o
[A]
ln[A]o t t

41. Example
The reaction 2A B is first order with respect A with a rate constant of 2.8 x 10-2 s-1 at 800C. How long will it take for A to decrease from 0.88 M to 0.14 M ?
[A]0 = 0.88 M
ln[A]o – ln[A] = kt
[A] = 0.14 M
kt = ln[A]0 – ln[A] ln
[A]0
[A] k = ln
0.88 M
0.14 M
2.8 x 10-2 s-1 =
ln[A]0 – ln[A] k
t =
= 66 s

42. Example
Decomposition of H2O2 (aq) is first order, given that k = 3.66 x 10-3 s-1 and [H2O2 ]o = M, determine
a) the time at which [H2O2] = M
b) the [H2O2 ] after 225 s.
Solution : kt ln
[H2O2]0
[H2O2] = a) ln
0.882
0.600
3.66 x 10-3 s-1 x t =

43. ln = x 10-3 s-1 x t
t =
ln 1.47
3.66 x 10-3
= s b) ln
[H2O2]0
[H2O2] = kt ln
0.882
[H2O2] =
3.66 x 10-3s-1x 225 s
[H2O2] = M

44. Exercise,
The conversion of cyclopropane to propene in the gas phase is a first order reaction with a rate constant of 6.7 X 10-4 s-1 at 500°C.
CH2
CH CH CH3-CH=CH2
If the initial concentration of cyclopropane was 0.25 M, what is the concentration after 8.8 minute. (0.18 M)
How long will it take for the concentration of cyclopropane to decrease from 0.25 M to 0.15 M? (13 min)
How long will it take to convert 74 percent of the starting material? (33 min)

45. How do you know decomposition is first order?
Half-life of a first-order reaction
The half-life, t½, is the time required for the concentration of a reactant to decrease to half of its initial concentration.
t=t½ when [A] = [A]0 2 ln
[A]0
[A]0/2 k = t½
ln2 k =
0.693 k =
What is the half-life of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1? t½
ln2 k =
0.693
5.7 x 10-4 s-1 =
= 1200 s = 20 minutes
How do you know decomposition is first order?
units of k (s-1)

46. First-order reaction
A product
No. of
half-lives
[A]o = 8 M 1 4 2 2 3 1 4
1/2 t½
ln2 k =

47. Example
The decomposition of ethane C2H6 to methyl radicals is a first order reaction with a rate constant of 5.36 x 10-4 s-1 at 700o C.
C2H6 (g) → CH3 (g)
Calculate the half life of the reaction in minutes.
Solution
t1/2 =
ln 2 k
0.693
5.36 x 10-4 = =
1.29 x 103 s =
21.5 min

48. Problem 2
What is the half-life of a compound if 75% of a given sample of the compound decomposes in 60 min? Assume it is first-order reactions kinetics.
(t1/2=30 min)

49. Problem 3
The decomposition of SO2Cl2 is a first-order reaction.
SO2Cl2(g) SO2 (g) + Cl2 (g)
Write the rate differential equation for the reaction.
Calculate the value of rate constant, k at 500 K if
5.00 % SO2Cl2 decomposed in 6.75 min.
Ans (7.6X10-3)
iii) Specify the half-life for the decomposition reaction. Ans (91.20 )

50. Second Order Reactions
A second order reaction is a reaction which rate depends on the concentration of one reactant raised to the second power or on the concentration of two different reactants each raised to the first power.
Example
A product
Where
Rate = -
d[A] dt
= k[A]2

51. To obtain the units of k
rate k =
[A]2
M/s
Unit k = M2
= M-1 s-1
Using calculus, the following expression can be obtained 1 1 -
[A]0 =
+ kt
[A]

52. Characteristic graphs for second order reaction
Rate = k [A]2
rate
[A]
rate
[A]2

53. [A] t 1/[A] – 1/[A]o 1/[A] M-1 1/[A]o t t
Graphs for second order reaction
[A] 1
+ kt =
[A]o
[A] t
1/[A] M-1
1/[A] – 1/[A]o
1/[A]o t t

54. 2nd –order reaction, r = k[A]2
If [A] doubles, r2 = k (2[A])2 = k ( 4 [A]2 ) = 4 k [A]2 = 4 r R will increase by 4 times if [A] doubles

55. Half life of a second order reaction1
[A]
[A]0 =
+ kt
[A]= [A]o
Substituting t = t1/2 2 1 1
[A]0
[A]0
+ kt1/2 = 2 1
t1/2=
k[A]0

56. Detemination of half-life using graph for second order reaction1
k[A]0
[A]0/2
[A]0/4
[A]0/8 x 2x 4x t

57. Example
Iodine atoms combine to form molecular iodine in
the gaseous phase
I (g) + I (g)  I2(g)
This reaction is a second order reaction , with the rate constant of 7.0 x 109 M-1 s-1
If the initial concentration of iodine was M,
i) calculate it’s concentration after 2 min.
ii) calculate the half life of the reaction if the
initial concentration of iodine is 0.06 M and
0.42 M respectively.

58. Solution : 1
[A]
[A]0 =
+ kt i) 1
[A] 1
[0.086]
+ (7.0 x 109 x 2 x 60 ) =
= 8.4 x 1011
[A] = x M
ii)
[I2] = 0.06 M 1
t1/2= 1
k[A]0 =
7.0 x 109 x 0.060 =
2.4 x s

59. [I2] = 0.42 M
t1/2= 1
k[A]0 1 =
7.0 x 109 x 0.042
= 3.4 x s

60. Using graph
Example,
The following results were obtained from an experimental investigation on dissociation of dinitrogen pentoxide at 45oC
N2O5(g)  2 NO2(g) + ½ O2(g)
time, t/min
[N2O5] x 10-4 M
Plot graph of [N2O5] vs time, determine
The order of the reaction
the rate constant k

61. Solution :
180
160
140
120
[N2O5] x 10-4 /M
100 80 60 40 20 10 20 30 40 50 60 70 80
Time ( min)

62. i)
Based on the above graph,
Time taken for concentration of N2O5 to change from 176 x 10-4 M to 88 x 10-4 M is 20 min
Time taken for concentration of N2O5 to change from 88 x 10-4 M to 44 x 10-4 M is also 20 min
The half life for the reaction is a constant and does not depend on the initial concentration of N2O5
Thus, the above reaction is first order
ln2
ii)
= 0.03 min-1
k =
20 min

63. Concentration-Time Equation Half-Life
Summary of the Kinetics of Zero-Order, First-Order
and Second-Order Reactions
Order
Rate Law
Concentration-Time Equation
Half-Life
t½ =
[A]0 2k
rate = k
[A] = [A]0 - kt t½
ln2 k = 1
rate = k [A]
ln[A] = ln[A]0 - kt 1
[A] =
[A]0
+ kt
t½ = 1
k[A]0 2
rate = k [A]2

64. Zero order 1st order 2nd order A  product A  product A  product
r = k [A]1
r = k [A]0
r = k [A]2
Unit k = s-1
Unit k = M-1 s-1
Unit k = M s-1
[A] r
[A] r
[A]2 r
[A] r
Integrated rate law
Integrated rate law
Integrated rate law
[A]0 – [A] = kt
ln([A]0 / [A]) = kt
1/[A] – 1/[A]0 = kt t
[A]
[A]0 t
ln[A]
ln[A]0 t
[A]
t1/2 = [A]0/2k t
1/[A]
1/[A]0 t
[A] t
[A]0 - [A] t
ln([A]0 / [A]) t
1/[A] – 1/[A]0
t1/2 = ln2/k
t1/2 = 1/k[A]0

65. 11.2 Collision Theory Objectives
At the end of the lesson the students should be able to:
1. explain reaction rates in terms of collision theory.
2. identify factors affecting the effectiveness of collision.
3. define activation energy.
4. Define and state the characteristics of an activated complex

66. Collision Theory
Collision Theory is the theory to explain the rate of chemical reactions. It is based on;
1- molecule must collide to react
2- molecules must possess a certain minimum kinetic energy (activation energy) to initiate the chemical reaction.
Rate 
Number of effective collisions
time

67. 3- molecule must collide in the right orientation in order for the reaction to occur.
Only effective collisions cause formation of product; collisions of molecules with Ea and at correct orientation.
The activation energy (Ea) is the minimum energy that must be supplied or required by collisions for a reaction to occur.

68. The activation energy (Ea)
…..is the minimum energy is required to initiate the chemical reaction.

69. Importance of Orientation
Orientation is unimportant
Orientation is important

70. Importance of Orientation
Orientation is important

71. Transition State Theory
The configuration of the atoms of the colliding species at the time of the collision is called the transition state.
Species formed at transition state is called activated complex.

72. Characteristics of Activated Complex
Very unstable i.e. It has a short half-life.
Its potential energy is greater than reactants or products.
The activated complex and the reactants are in chemical equilibrium.
It decomposes to form products or reactants.

73. Potential energy
- A reaction profile shows potential energy plotted as a function of the progress of the reaction. - The difference in potential energies between the products and the reactants is -DH for the reaction. - Reactant molecules must have enough energy to overcome an energy barrier separating products from reactants, Ea. Ea
Reactant DH
product
Progress of reaction

74. A Reaction Profile: exothermic reaction
Activated complex
Transition state Ea
(Forward reaction)
Ea (reverse reaction) DH
CO(g) + NO2(g) CO2(g) + NO(g)

75. A Reaction Profile for endothermic process
activated complex. Ea
(reverse reaction) Ea
(forward reaction)
Product H
Reactant

76. A Reaction profile: endothermic reactionDH
2NOCl → 2NO + Cl2

77. Example:
1. For the reaction A + B C + D , the enthalphy change of the forward reaction is + 21 kJ/mol. The activation energy of the forward reaction is 84 kJ/mol.
a) What is activation energy of the reverse reaction?
b) Sketch the reaction profile of this reaction
2. Draw a potential energy diagram for an exorthermic reaction. Indicate on the drawing:
a) Potential energy of the reactants and the products
b) The activation energy for the forward and the reverse reaction
c) The heat of the reaction

78. Factors affecting rate of reaction
CONCENTRATIONS OF REACTANTS:
Reaction rates generally increase as the concentrations of the reactants are increased.
TEMPERATURE:
Reaction rates generally increase rapidly as the temperature is increased.
CATALYSTS:
Catalysts speed up reactions.
PARTICLE SIZE:
The rate increases as the smaller the size of reacting particles .

79. A) CONCENTRATIONS OF REACTANTS
Reaction rate 
collision
time
The frequency of collision increases increases with the concentration
4 particle system
(2 and 2)  4 collision

80. A) CONCENTRATIONS OF REACTANTS
A concentration of reactants increases, the frequency of collision increases.
This would also result in the increase in the quantity of effective collision. Thus the reaction rate increases.
5 particle system
(3 and 2)  6 collision

81. A) CONCENTRATIONS OF REACTANTS
This observation correlates with the RATE LAW that has been previously discussed…
Reaction rate = k [ A ]x [ B ]y …
(A & B = reactants)
(x & y = rate order)
Based on this equation,
Reaction rate  concentration of reactants
(depending on its rate order)
REMINDER!
Only in zero order reactions, the rate of reaction is not dependant upon the concentration of the reactants.
Reaction rate = k [ A ]0 = k (constant)

82. As temperature increases, kinetic energy, of molecules increases
B) TEMPERATURE
As temperature increases, kinetic energy, of molecules increases
So, more collisions occur in a given time
Furthermore, the higher the kinetic energy, the higher the energy of the effective collisions.
So more molecules will have energy greater than activation energy, Ea
Thus, the rate of reaction increases

83. Distribution of Kinetic Energies of Molecules
Represent total number of molecules
with kinetic energy greater than Ea

84. k = A e-Ea∕RT ARRHENIUS EQUATION
B) TEMPERATURE
ARRHENIUS EQUATION
In 1889, Svante Arrhenius proposed the following mathematical expression for the effect of temperature on the rate constant, k:
k = A e-Ea∕RT
Where…
k = rate constant
A = constant known as the collision frequency factor
e = natural log exponent
Ea = activation energy for the reaction
R = universal gas constant (8.314 J mol-1 K-1)
T = absolute temperature

85. k = A e-Ea∕RT ARRHENIUS EQUATION
B) TEMPERATURE
ARRHENIUS EQUATION
The relation ship between the rate constant, k and temperature can be seen in the k vs T graph:
k = A e-Ea∕RT
1/T (K-1)

86. y = m x + C ARRHENIUS EQUATION - DERIVATION
B) TEMPERATURE
ARRHENIUS EQUATION - DERIVATION
The relationship between k and T is clearer when we further derive the Arrhenius Equation…
Natural log both ends…
(But ln e = 1)
Thus…
y =
m x
+ C
See the linear relationship…?

87. Graph Representation Of The Arrhenius Equation
B) TEMPERATURE
Graph Representation Of The Arrhenius Equation
Plotting a ln k vs 1/T graph would show a clearer relationship between k (Rate constant) and temperature
Where,
Ea = Activation Energy
R = Jmol-1K-1
T = Absolute Temp
A = Collision freq. factor

88. If the value of A (collision frequency factor) is not known and the same reaction conducted at two different temperatures.The Arrhenius equation at each temperature can be written and combined to formed the equation shown in the box.
B) TEMPERATURE
and
Rearranging the equations would give…
Since “A” is a constant…

89. 2 HI (g)  H2(g) + I2(g) Exercise: the Activation energy
The decomposition of hydrogen iodide,
2 HI (g)  H2(g) + I2(g)
has rate constants of 9.51 x 10-9 L mol-1 s-1 at 500 K and
1.10x10-5 L mol-1 s-1 at 600 K. Find Ea.
DATA:
k1 = 9.51 x 10-9 L mol-1 s T1 = 500K
k2 = 1.10 x 10-5 L mol-1 s T1 = 600K
SOLUTION:
Ea = x 105 J/mol = 176 kJ/mol

90. C) CATALYST
A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed.
Addition of a catalyst increases the reaction rate by increasing the frequency of effective collision. That is by
Decreasing the Ea, and
Correct orientation

92. Addition of a catalyst changes the value of k (rate constant) .
C) CATALYST
Addition of a catalyst changes the value of k (rate constant) .
Reaction rate = k [ A ]x [ B ]y
(A & B = reactants)
(x & y = rate order)
The catalyst reacts by reducing the Ea and increasing A, thus increasing the k.

93. Ea > E’a rateuncatalyzed < ratecatalyzed
C) CATALYST
When Ea decreases, k increases,
REACTION RATE increases
uncatalyzed
catalyzed
rateuncatalyzed < ratecatalyzed
Ea > E’a
13.6

94. Reaction pathway

95. D) PARTICLE SIZE
The smaller the size of reacting particles, the greater is the total surface area exposed for reaction and consequently the faster the reaction. In the case of heterogeneous systems, in which the reactants are in different phases, the area of contact between the reacting substances will influence the reaction rate considerably.