Chemistry - Chemical Kinetics Question with Solution | TestHub

The given reaction is:

${\mathrm{CH}}_3{\mathrm{Br}}_{(\mathrm{aq})}+{\mathrm{OH}}_{(\mathrm{aq})}^{-}\to {\mathrm{CH}}_3{\mathrm{OH}}_{(\mathrm{aq})}+{\mathrm{Br}}_{(\mathrm{aq})}^{-}$

The rate of reaction can be expressed as:

Rate $=-\frac{d[{\mathrm{CH}}_3\mathrm{Br}]}{dt}=-\frac{d[{\mathrm{OH}}^{-}]}{dt}=\frac{d[{\mathrm{CH}}_3\mathrm{OH}]}{dt}=\frac{d[{\mathrm{Br}}^{-}]}{dt}$

Given that the rate of consumption of ${\mathrm{OH}}^{-}$ is $x{\text{ mol dm}}^{-3}{\text{ s}}^{-1}$, we have:

$-\frac{d[{\mathrm{OH}}^{-}]}{dt}=x{\text{ mol dm}}^{-3}{\text{ s}}^{-1}$

From the rate expression, the rate of formation of ${\mathrm{Br}}^{-}$ is equal to the rate of consumption of ${\mathrm{OH}}^{-}$:

$\frac{d[{\mathrm{Br}}^{-}]}{dt}=x{\text{ mol dm}}^{-3}{\text{ s}}^{-1}$

Answer: $\mathrm{x}\mathrm{mol}\mathrm{dm}{}^{-3}{\mathrm{s}}^{-1}$