Consider the reaction mechanism:A2⇌keq2A (fast) (where A is the intermediate.)A+B⟶k1P(slow)The rate law for the reaction is

A2⇌keq2A (fast) ⇒keq=[A]2/A2⇒[A]=keqA2A+B⟶k1P( slow )→RDS⇒RateRDS=k1[A][B]=k1keqA2⋅[B]

Consider the reaction mechanism:A2⇌keq2A (fast) (where A is the intermediate.)A+B⟶k1P(slow)The rate law for the reaction is

A2⇌keq2A (fast) ⇒keq=[A]2/A2⇒[A]=keqA2A+B⟶k1P( slow )→RDS⇒RateRDS=k1[A][B]=k1keqA2⋅[B]

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Consider the reaction mechanism:
$A_{2} \overset{k_{eq}}{⇌} 2 A (fast)$ (where A is the intermediate.)
$A + B \underset{k_{1}}{⟶} P (slow)$
The rate law for the reaction is

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$k_{1} [A] [B]$

$k_{1} k^{1 / 2} {[A_{2}]}^{1 / 2} [B]$

$k_{1} k^{1 / 2} [A] [B]$

$k_{1} k^{1 / 2} [A]^{2} [B]$

answer is B.

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Detailed Solution

$\begin{array}{l} A_{2} \overset{k_{eq}}{⇌} 2 A (fast) \\ \Rightarrow k_{eq} = [A]^{2} / [A_{2}] \Rightarrow [A] = \sqrt{k_{eq} [A_{2}]} \\ A + B \overset{k_{1}}{⟶} P (slow) \to RDS \\ \Rightarrow {Rate}_{RDS} = k_{1} [A] [B] = k_{1} \sqrt{k_{eq}} \sqrt{[A_{2}]} \cdot [B] \end{array}$