The reaction 2NO + Br_2 rightarrow 2NOBr foll | vedclass

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(C) The rate of the reaction is determined by the slow step,which is step $(II)$: $Rate = k[NOBr_2][NO]$Since $NOBr_2$ is an intermediate,we use the e

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$3$

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(C) The rate of the reaction is determined by the slow step,which is step $(II)$: $Rate = k[NOBr_2][NO]$Since $NOBr_2$ is an intermediate,we use the equilibrium from step $(I)$ to express its concentration:$K_{eq} = \frac{[NOBr_2]}{[NO][Br_2]}$$[NOBr_2] = K_{eq}[NO][Br_2]$Substituting this into the rate equation:$Rate = k \cdot K_{eq}[NO][Br_2] \cdot [NO]$$Rate = k'[NO]^2[Br_2]$The overall order of the reaction is the sum of the powers of the concentration terms in the rate law:$Order = 2 + 1 = 3$

The reaction $2NO + Br_2 \rightarrow 2NOBr$ follows the mechanism given below:
$(I)$ $NO + Br_2 \rightleftharpoons NOBr_2$ ........ Fast
$(II)$ $NOBr_2 + NO \rightarrow 2NOBr$ ......... Slow
The overall order of this reaction is

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The reaction $2NO + Br_2 \rightarrow 2NOBr$ follows the mechanism given below:$(I)$ $NO + Br_2 \rightleftharpoons NOBr_2$ ........ Fast$(II)$ $NOBr_2 + NO \rightarrow 2NOBr$ ......... SlowThe overall order of this reaction is