2018 H2 Chem P1 Q11 — Kinetics Solution

The answer is C.
A correct mechanism must (1) have a slow (rate-determining) step whose reactants match the rate equation, and (2) have steps that add up to the overall equation.
Rate \(= k[\text{NO}_2]^2\), so the slow step must involve two \(\text{NO}_2\) molecules and no CO.
In C the slow step is \(\text{NO}_2 + \text{NO}_2 \rightarrow \text{NO} + \text{NO}_3\), giving rate \(= k[\text{NO}_2]^2\). Adding the steps: \(2\text{NO}_2 + \text{NO}_3 + \text{CO} \rightarrow \text{NO} + \text{NO}_3 + \text{CO}_2 + \text{NO}_2\), which simplifies to \(\text{CO} + \text{NO}_2 \rightarrow \text{CO}_2 + \text{NO}\), the overall equation. \(\text{NO}_3\) is an intermediate.
• A is wrong: the slow step involves only one \(\text{NO}_2\), which gives rate \(= k[\text{NO}_2]\), first order in \(\text{NO}_2\).
• B is wrong: the slow step fits the rate equation, but the steps add up to \(2\text{NO}_2 + \text{CO} \rightarrow \text{CO}_2 + 2\text{NO} + \text{O}\), not the overall equation (an O atom is left over and two \(\text{NO}_2\) are used).
• D is wrong: CO takes part in the slow step, which gives rate \(= k[\text{NO}_2][\text{CO}]\), first order in each reactant.