The mechanism of the reaction 2NO_{(g)} + Br_ | vedclass

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(C) The rate of the reaction is determined by the slow step:$Rate = k[NOBr_2][NO]$Since $NOBr_2$ is an intermediate,we express its concentration using

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$2$

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(C) The rate of the reaction is determined by the slow step:$Rate = k[NOBr_2][NO]$Since $NOBr_2$ is an intermediate,we express its concentration using the equilibrium constant $K_{eq}$ from the first step:$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]^1$The order of reaction with respect to $NO_{(g)}$ is the exponent of $[NO]$,which is $2$.

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