Deriving the Rate Law for a Mechanism with a Fast Initial

Chapter 6, Problem 15PE

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Deriving the Rate Law for a Mechanism with a Fast Initial Step

Show that the following mechanism for Equation 14.24 also produces a rate law consistent with the experimentally observed one:

Step 1: \(\mathrm{NO}(g)+\mathrm{NO}(g) \underset{k_{-1}}{\stackrel{k_{1}}{\rightleftarrows}} \mathrm{N}_{2} \mathrm{O}_{2}(g)\)      (fast, equilibrium)

Step 2: \(\mathrm{N}_{2} \mathrm{O}_{2}(g)+\mathrm{Br}_{2}(g) \stackrel{k_{2}}{\longrightarrow} 2 \mathrm{NOBr}(g)\)     (slow)

The first step of a mechanism involving the reaction of bromine is

\(\mathrm{Br}_{2}(g) \underset{k_{-1}}{\stackrel{k_{1}}{\rightleftharpoons}} 2 \mathrm{Br}(g)\)     (fast, equilibrium)

What is the expression relating the concentration of \(\mathrm{Br}(g)\) to that of \(\mathrm{Br}_{2}(g)\)?

Equation Transcription:

NO(g)+NO(g)   N2O2(g)

N2O2(g)+Br2(g)  2NOBr(g)

Br2(g)    2Br⁡(g)

Br(g)

Br2(g)

Text Transcription:

NO(g)+NO(g)⇌^k_1 N2O2(g)_k_-1

N_2O_2(g)+Br_2(g)⟶^k_2 2NOBr(g)

Br2(g)⇄^k_1 _k_-1 2Br⁡(g)

Br(g)

Br2(g)