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12 Chemical Reaction Mechanisms I: Rate Laws and Mechanisms
E X A M P L E 12.16
The stoichiometry of a reaction is
B + 2F + H − → D + other products
and its rate law is
rate k app
[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
rate k 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).
E X A M P L E 12.17
For the gaseous reaction
4HNO 3 − → 4NO 2 + 2H 2 O + O 2
(12.4-42)
the rate law is found to be
rate k app [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.