Consider the reaction:Cl_{2(aq)} + H_2S_{(aq) | vedclass

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(D) The rate-determining step $(RDS)$ is the slow step of the reaction mechanism.For mechanism $A$:The slow step is $Cl_2 + H_2S \rightarrow H^{+} + C

Vedclass · Accepted Answer

$A$ only

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(D) The rate-determining step $(RDS)$ is the slow step of the reaction mechanism.For mechanism $A$:The slow step is $Cl_2 + H_2S ightarrow H^{+} + Cl^{-} + Cl^{+} + HS^{-}$.The rate law derived from this step is $	ext{rate} = k[Cl_2][H_2S]$,which matches the given rate equation.For mechanism $B$:The slow step is $Cl_2 + HS^{-} ightarrow 2Cl^{-} + H^{+} + S$.The rate law is $	ext{rate} = k[Cl_2][HS^{-}]$.From the fast equilibrium $H_2S ightleftharpoons H^{+} + HS^{-}$,the equilibrium constant is $K = \frac{[H^{+}][HS^{-}]}{[H_2S]}$,so $[HS^{-}] = \frac{K[H_2S]}{[H^{+}]}$.Substituting this into the rate law gives $	ext{rate} = k' \frac{[Cl_2][H_2S]}{[H^{+}]}$,which does not match the given rate equation.Therefore,only mechanism $A$ is consistent.

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