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(D) The rate-determining step is the second reaction: $NOBr_{2(g)} + NO_{(g)} \rightarrow 2 NOBr_{(g)}$.The rate law for this step is: $Rate = k[NOBr_

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$2$

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(D) The rate-determining step is the second reaction: $NOBr_{2(g)} + NO_{(g)} ightarrow 2 NOBr_{(g)}$.The rate law for this step is: $Rate = k[NOBr_2][NO]$.Since $NOBr_2$ is an intermediate,we express its concentration using the equilibrium constant $(K_c)$ of the first step: $NO_{(g)} + Br_{2(g)} ightleftharpoons NOBr_{2(g)}$.$K_c = \frac{[NOBr_2]}{[NO][Br_2]} \implies [NOBr_2] = K_c[NO][Br_2]$.Substituting this into the rate law: $Rate = k 	imes K_c[NO][Br_2] 	imes [NO] = k'[NO]^2[Br_2]$.Thus,the order of the reaction with respect to $NO_{(g)}$ is $2$.

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