Module
7 :
Theories of Reaction Rates
Lecture 33 : Transition State Theory
Example
For the K + Br
2
KBr + Br reaction at 300 K, find the thermodynamic activation parameters
G
,
H
and
S
of the transition state theory.
Solution :
The Arrhenius parameters for the above reaction are (from Table 32.1 ) A = 1.0 * 10
12
m
-1
s
-1
and E
a
= 0 kJ/ mol
S
= R ln { h A
/ N
A
e
2
k
2
B
T
2
}
Here h = 6.65 * 10
-34
Js,
= 1 bar = 10
5
J/m
3
, N
A
= 6.02 * 10
23
mol
-1
k
B
= 1.38 * 10
-23
J/ K, A = 10
12
M
-1
s
-1
= mol
-1
dm
+3
s
-1
Substitute all these values and show that
h A
/ N
A
e
2
k
2
B
T
2
= 7.812 * 10
-11
(A/M
-1
s
-1
) / (T/K)
2
Substituting the above value in the equation for
S
, we have
S
= R ln { 7.812 * 10
-11
* 10
12
* 300
2
}
= - 58 JK
-1
mol
-1
H
= E
a
- 2 RT = 0 - 2* 8.314 * 300 J / mol
= -2.5 kJ / mol
G
=
H
- T
S
= -2.5 - 300 * (-58) / 10
-3
kJ / mol
= 15 kJ / mol
1
2
3
4
5
6
7
8
9
10
11
12
13
14
KBr + Br reaction at 300 K, find the thermodynamic activation parameters 
G
, 
/ NA e2 k2BT2 } 
