Reaction : 2Br^{-} + H_2O_2 + 2H^{+} to Br_2 | vedclass

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(A) For a multi-step reaction,the rate of the overall reaction is determined by the slowest step,which is known as the rate-determining step.In the gi

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$3$

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(A) For a multi-step reaction,the rate of the overall reaction is determined by the slowest step,which is known as the rate-determining step.In the given mechanism,step $(a)$ is the slow step:$Br^{-} + H^{+} + H_2O_2 \xrightarrow{slow} HOBr + H_2O$The rate law for this reaction is given by the rate of the slow step:$Rate = k[Br^{-}][H^{+}][H_2O_2]$The order of the reaction is the sum of the powers of the concentration terms in the rate law expression:$Order = 1 + 1 + 1 = 3$Therefore,the order of the reaction is $3$.

Reaction : $2Br^{-} + H_2O_2 + 2H^{+} \to Br_2 + 2H_2O$
takes place in two steps :
$(a)$ $Br^{-} + H^{+} + H_2O_2 \xrightarrow{slow} HOBr + H_2O$
$(b)$ $HOBr + Br^{-} + H^{+} \xrightarrow{fast} H_2O + Br_2$
The order of the reaction is

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Reaction : $2Br^{-} + H_2O_2 + 2H^{+} \to Br_2 + 2H_2O$takes place in two steps :$(a)$ $Br^{-} + H^{+} + H_2O_2 \xrightarrow{slow} HOBr + H_2O$$(b)$ $HOBr + Br^{-} + H^{+} \xrightarrow{fast} H_2O + Br_2$The order of the reaction is