The following mechanism has been proposed for the reaction of NO with ${Br}_{2}$ to form $NOBr.$
$\begin{array}{l} NO (g) + {Br}_{2} (g) ⇌ {NOBr}_{2} (g) \\ {NOBr}_{2} (g) + NO (g) ⇌ 2 NOBr (g) \end{array}$
If the second step is the rate determining step, the order of the reaction with respect to $NO (g)$ is

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answer is A.

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Detailed Solution

The steps are

$\begin{array}{ll} NO (g) + {Br}_{2} (g) ⇌ {NOBr}_{2} (g); K_{eq} = \frac{[{NOBr}_{2}]}{[{NO}_{2}] [{Br}_{2}]} \\ {NOBr}_{2} + NO (g) \overset{k}{⟶} 2 {NOBr}^{2} (g); (slow step) \end{array}$

The rate expression is

$rate = k [{NOBr}_{2}] [NO] = k \{K_{eq} [NO] [{Br}_{2}]\} [NO] = k K_{eq} [NO]^{2} [{Br}_{2}]$

Hence, the order of the reaction with respect to NO is 2.

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