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

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answer is B.

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Detailed Solution

$(i) NO (g) {+ Br}_{2} (g) ⇌ {NOBr}_{2} (g)$

$(ii) {NOBr}_{2} (g) + NO (g) \to 2NOBr (g)$

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

${But NOBr}_{2}$ is intermediate and must not appear in the rate law equation From 1^st step $K_{C} = \frac{[{NOBr}_{2}]}{[NO] [{Br}_{2}]}$

$∴ [{NOBr}_{2}] {= K}_{c} [NO] [{Br}_{2}]$

$∴ {Rate law equation = K. K}_{C} {[NO]}^{2} [{Br}_{2}]$

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