2017 H2 Chem P1 Q12 — Kinetics Solution

The answer is A.
The rate equation follows the slow (rate-determining) step. If a species in the slow step is an intermediate made in an earlier fast step, its concentration must be replaced using that fast step.
In A the slow step is \(2\text{NO} + \text{H}_2 \rightarrow \text{N}_2\text{O} + \text{H}_2\text{O}\): it involves two NO and one \(\text{H}_2\) and no intermediates, so rate \(= k[\text{NO}]^2[\text{H}_2]\), matching the rate equation given. The fast second step (\(\text{N}_2\text{O} + \text{H}_2\)) does not affect the rate, and the two steps add up to \(2\text{NO} + 2\text{H}_2 \rightarrow \text{N}_2 + 2\text{H}_2\text{O}\).
• B is wrong: the slow step \(\text{H}_2 \rightarrow 2\text{H}\) involves only \(\text{H}_2\), so rate \(= k[\text{H}_2]\). \(\text{I}_2\) reacts only in the fast step, so it cannot appear in the rate equation, but the given rate equation contains \([\text{I}_2]\).
• C is wrong: the slow step \(\text{HBrO} + \text{HBr}\) involves the intermediate HBrO. From the fast step, \([\text{HBrO}] \propto [\text{HBr}][\text{O}_2]^{1/2}\), so rate \(= k[\text{HBr}]^2[\text{O}_2]^{1/2}\), not \(k[\text{HBr}]^2[\text{O}_2]\).
• D is wrong: the slow step \(\text{H}_2\text{O}_2 + \text{IO}^-\) involves the intermediate \(\text{IO}^-\). From the fast step, \([\text{IO}^-] \propto [\text{H}_2\text{O}_2][\text{I}^-]\), so rate \(= k[\text{H}_2\text{O}_2]^2[\text{I}^-]\), not \(k[\text{H}_2\text{O}_2][\text{I}^-]\).