Physical Chemistry Third Edition

26.4 The Activated Complex Theory of Bimolecular Chemical Reaction Rates in Dilute Gases 1113

z‡rot,HHBr′′ 

8 π^2 IekBT

h^2



8 π^2 (1. 73 × 10 −^46 kg m^2 )(1. 3807 × 10 −^23 JK−^1 )(500.0K)

(6. 6261 × 10 −^34 Js)^2

 214. 3

Normal mode number 1 is the symmetric stretch:



hc ̃ν 1

kBT



(6. 6261 × 10 −^34 J s)(2. 9979 × 1010 ms−^1 )(2340 cm−^1 )

(1. 3807 × 10 −^23 JK−^1 )(500.0K)

 6. 73

Normal mode number 2 is the bend (there are two of these)



hcν ̃ 2

kBT



(6. 6261 × 10 −^34 J s)(2. 9979 × 1010 cm s−^1 )(460 cm−^1 )

(1. 3807 × 10 −^23 JK−^1 )(500.0K)

 1. 32

zvib,HHBr

1

1 −e−x^1

(

1

1 −e−x^2

) 2



1

1 −e−^6.^73

(

1

1 −e−^1.^32

) 2

(1.0011)(1.3645)^2  1. 864

z‡HHBr(1. 558 × 1033 m−^3 )(214.3)(1.864) 6. 22 × 1035 m−^3

k

kBT

h

e−∆ε

‡

0 /kBT

z‡

′′

HHBr

z′′Hz′′HBr

NAv



(1. 3807 × 10 −^23 JK−^1 )(500.0K)

6. 6261 × 10 −^34 Js

exp

(

− 8. 3 × 10 −^21 J

(1. 3807 × 10 −^23 JK−^1 )(500.0K)

)

×

6. 22 × 1035 m−^3

(4. 254 × 1030 m−^3 )(6. 161 × 1034 m−^3 )

(6. 02214 × 1023 mol−^1 )

 4. 475 × 106 m^3 mol−^1 s−^1

 4. 475 × 109 L mol−^1 s−^1

Exercise 26.12

a.Calculate the rate constant at 500.0 K for the reverse of the reaction in the previous example:

H 2 +Br→H+HBr

Assume the same activated complex as in the forward reaction so that∆ε‡ 0  1. 24 × 10 −^19 J

for the reverse reaction.

b.Use the values of the forward and reverse rate constants to calculate the equilibrium constant

for the reaction. If you have already studied Part I of this textbook, compare your result with

the result of a calculation from thermodynamic data.

Equation (26.4-15) can be written in a “thermodynamic” form, using Eq. (7.1-20):

k

kBT

h

1

c◦

K

‡

c

kBT

h

1

c◦

e−∆G

‡◦/RT

(26.4-16)