Chemistry - Chemical Kinetics Question with Solution | TestHub

The rate of reaction is given by:

$\text{Rate}=-\frac{1}{3}\frac{d[{\text{I}}^{-}]}{dt}=\frac{1}{2}\frac{d[{\text{SO}}_4^{2-}]}{dt}$

The rate of consumption of ${\text{I}}^{-}$ is $-\frac{d[{\text{I}}^{-}]}{dt}$.

$-\frac{d[{\text{I}}^{-}]}{dt}=\frac{3}{2}\frac{d[{\text{SO}}_4^{2-}]}{dt}$

Substituting the given rate of formation of ${\text{SO}}_4^{2-}$:

$-\frac{d[{\text{I}}^{-}]}{dt}=\frac{3}{2}\times 2.2\times 10^{-2}$

$-\frac{d[{\text{I}}^{-}]}{dt}=3.3\times 10^{-2}{\text{ mol dm}}^{-3}{\text{ s}}^{-1}$

Answer: $3.3\times 10^{-2}{\text{mol dm}}^{-3}{\text{s}}^{-1}$