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

JEE Main · 2006 · 30 Apr 2006 · Offline · Chemical Kinetics

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)

If the second step is the rate determining step, the order of the reaction with respect to

NO (g)

Answer: (D)

The slow second step,

NOB r_{2} + NO \to 2NOBr

, fixes the rate as

r = k [NOB r_{2}] [NO]

, but

NOB r_{2}

is an intermediate and must be replaced using the fast pre-equilibrium of the first step. From that equilibrium

[NOB r_{2}] = K [NO] [B r_{2}]

, and substituting gives

r = k K [NO]^{2} [B r_{2}]

. The reaction is therefore second order in

NO

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option