In the formation of HBr from H2 and Br2, following mechanism is observed:I. Br2⇌2Br∙ (Equilibrium step)II. H2+Br∙→HBr+H∙ (Slow step)III. H∙+ HBr + H∙ (Fast step)The rate law for the above reaction is

Br2⇌keq2Br⋅⇒Br∘=keq1/2Br21/2 and H2+Br⋅⟶k1HBr+H∙⇒ Rate =k1H2Br∗=k1keq⏟Rate constant H2Br21/2=keff H2Br21/2

In the formation of HBr from H2 and Br2, following mechanism is observed:I. Br2⇌2Br∙ (Equilibrium step)II. H2+Br∙→HBr+H∙ (Slow step)III. H∙+ HBr + H∙ (Fast step)The rate law for the above reaction is

Br2⇌keq2Br⋅⇒Br∘=keq1/2Br21/2 and H2+Br⋅⟶k1HBr+H∙⇒ Rate =k1H2Br∗=k1keq⏟Rate constant H2Br21/2=keff H2Br21/2

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In the formation of HBr from H₂ and Br₂, following mechanism is observed:
I. ${Br}_{2} ⇌ 2 {Br}^{∙}$ (Equilibrium step)
II. $H_{2} + {Br}^{∙} \to HBr + H ∙$ (Slow step)
III. $H ∙ + HBr + H ∙$ (Fast step)
The rate law for the above reaction is

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$r = k {[H_{2}]}^{1 / 2} [Br]^{1 / 2}$

$r = k {[H_{2}]}^{1 / 2} [{Br}_{2}]$

$r = k [H_{2}] [{Br}_{2}]$

$r = k [H_{2}] {[{Br}_{2}]}^{1 / 2}$

answer is B.

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Detailed Solution

$\begin{array}{l} {Br}_{2} \overset{k_{eq}}{⇌} 2 {Br}^{\cdot} \\ \Rightarrow [{Br}^{\circ}] = k_{eq}^{1 / 2} {[{Br}_{2}]}^{1 / 2} \\ and H_{2} + {Br}^{\cdot} \overset{k_{1}}{⟶} HBr + H^{∙} \\ \Rightarrow Rate = k_{1} [H_{2}] [{Br}^{*}] \\ = \underset{Rate constant}{\underset{⏟}{k_{1} \sqrt{k_{eq}}}} [H_{2}] {[{Br}_{2}]}^{1 / 2} = k_{eff} [H_{2}] {[{Br}_{2}]}^{1 / 2} \end{array}$