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The following mechanism has been proposed for the reaction of $NO$ with ${Br}_{2}$ to form $NOBr$ .
$NO (g) + {Br}_{2} (g) ⇌ {NOBr}_{2} (g)$
${NOBr}_{2} (g) + NO (g) ⟶ 2 NOBr (g)$ (slow step)
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 D.

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Detailed Solution

Given, the rate determining step,

$Rate = k [{NOBr}_{2}] [NO]$

$K_{eq} = \frac{[{NOBr}_{2}]}{[NO] [{Br}_{2}]}$

substitute this value in equation 1, we get,

$\begin{array}{l} r = k [NO] [{NOBr}_{2}] \\ Also, [{NOBr}_{2}] = k_{eq} [NO] [{Br}_{2}] \\ \Rightarrow r = k^{'} [{NO}^{2} [{Br}_{2}] \end{array}$

$Rate \propto [NO]^{2} {[{Br}_{2}]}^{1}$

The order of reaction with respect to NO is 2.

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