For the reaction,CH_3Br_{(aq)} + OH_{(aq)}^{- | vedclass

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(B) The rate of reaction is determined by the stoichiometry of the balanced chemical equation.For the reaction: $CH_3Br_{(aq)} + OH_{(aq)}^{-} \longri

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$x$

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(B) The rate of reaction is determined by the stoichiometry of the balanced chemical equation.For the reaction: $CH_3Br_{(aq)} + OH_{(aq)}^{-} \longrightarrow CH_3OH_{(aq)} + Br_{(aq)}^{-}$The rate expression is: $-\frac{d[CH_3Br]}{dt} = -\frac{d[OH^{-}]}{dt} = \frac{d[CH_3OH]}{dt} = \frac{d[Br^{-}]}{dt}$Given that the rate of consumption of $OH_{(aq)}^{-}$ is $x \ mol \ dm^{-3} \ s^{-1}$,i.e.,$-\frac{d[OH^{-}]}{dt} = x$.Since the stoichiometric coefficients of $OH^{-}$ and $Br^{-}$ are both $1$,the rate of formation of $Br_{(aq)}^{-}$ is equal to the rate of consumption of $OH_{(aq)}^{-}$.Therefore,the rate of formation of $Br_{(aq)}^{-}$ is $x \ mol \ dm^{-3} \ s^{-1}$.

For the reaction,$CH_3Br_{(aq)} + OH_{(aq)}^{-} \longrightarrow CH_3OH_{(aq)} + Br_{(aq)}^{-}$,the rate of consumption of $OH_{(aq)}^{-}$ is $x \ mol \ dm^{-3} \ s^{-1}$. What is the rate of formation of $Br_{(aq)}^{-}$ (in $mol \ dm^{-3} \ s^{-1}$)?

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