Physical Chemistry Third Edition

552 12 Chemical Reaction Mechanisms I: Rate Laws and Mechanisms

EXAMPLE12.16

The stoichiometry of a reaction is

B+2F+H−→D+other products

and its rate law is

ratekapp

[B][F]^2 [H]

[D]

Propose a mechanism.

Solution

The first step cannot be rate-limiting because of the presence of [D] in the denominator of the

rate law. The following three-step mechanism corresponds to this rate law if the third step is

rate-limiting:

(1) B+F
A+D (fast)

(2) F+H
G (fast)

(3) A+G−→products (slow)

(12.4-40)

The rate law corresponding to this mechanism is

ratek 3

(

K 1 [B][F][D]−^1

)

(K 2 [F][H])k 3 K 1 K 2

[B][F]^2 [H]

[D]

(12.4-41)

in agreement with the given rate law.

Exercise 12.15

Verify Eq. (12.4-41).

EXAMPLE12.17

For the gaseous reaction

4HNO 3 −→4NO 2 +2H 2 O+O 2 (12.4-42)

the rate law is found to be

ratekapp[HNO 3 ]^2 [NO 2 ]−^1 (12.4-43)

Propose a possible mechanism.

Solution

We must have NO 2 on the right-hand side of some step prior to a rate-limiting step. One

possibility is

(1) HNO 3 
HO+NO 2 (fast)

(2) HO+HNO 3 −→further intermediates/products (slow) (12.4-44)

(3) further steps (no effect on rate law)

with step 2 assumed to be rate-limiting. The steps following step 2 do not affect the rate law.