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(B) Step $(i)$: $NO_{(g)} + Br_{2(g)} \rightleftharpoons NOBr_{2(g)}$ (Fast equilibrium)Step $(ii)$: $NOBr_{2(g)} + NO_{(g)} \longrightarrow 2NOBr_{(g

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$2$

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(B) Step $(i)$: $NO_{(g)} + Br_{2(g)} ightleftharpoons NOBr_{2(g)}$ (Fast equilibrium)Step $(ii)$: $NOBr_{2(g)} + NO_{(g)} \longrightarrow 2NOBr_{(g)}$ (Slow,rate-determining step)The rate law is determined by the slow step: $	ext{Rate} = k[NOBr_2][NO]$.Since $NOBr_2$ is an intermediate,we express its concentration using the equilibrium constant $K_C$ from step $(i)$: $K_C = \frac{[NOBr_2]}{[NO][Br_2]}$.Therefore,$[NOBr_2] = K_C[NO][Br_2]$.Substituting this into the rate law: $	ext{Rate} = k \cdot K_C[NO][Br_2][NO] = k'[NO]^2[Br_2]$.The order of the reaction with respect to $NO_{(g)}$ is $2$.

$[A] \ (mol/L)$	$[B] \ (mol/L)$	Rate $(mol/L \cdot s)$
$0.05$	$0.05$	$1.2 \times 10^{-3}$
$0.10$	$0.05$	$2.4 \times 10^{-3}$
$0.05$	$0.10$	$1.2 \times 10^{-3}$

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