JEE Main 2023 — Chemical Kinetics Question with Solution

${\mathrm{KClO}}_3 + 6{\mathrm{FeSO}}_4 + 3{\mathrm{H}}_2{\mathrm{SO}}_4 \to \mathrm{KCl} + 3{\mathrm{Fe}}_2({\mathrm{SO}}_4{)}_3 + 3{\mathrm{H}}_2\mathrm{O}$

From the above equation, rate of decomposition ${\mathrm{FeSO}}_4$ and rate of production of ${\mathrm{Fe}}_2{\left({\mathrm{SO}}_4\right)}_3$ can be written as follows,

$\frac{1}{6}\times \mathrm{Rate} \mathrm{of} {\mathrm{decomposition}}_{{\mathrm{FeSO}}_4}=\frac{{\displaystyle 1}}{{\displaystyle 3}}\times \mathrm{Rate} \mathrm{of} {\mathrm{production}}_{{\mathrm{Fe}}_2{\left({\mathrm{SO}}_4\right)}_3}$

Rate of decomposition of ${\mathrm{FeSO}}_4$

$\frac{\mathrm{Change} \mathrm{in} \mathrm{concentration}}{\mathrm{time} \mathrm{in} \mathrm{seconds}}=\frac{(10-8.8)}{30\times 60}$

$=\frac{1.2}{1800} \mathrm{mol} {\mathrm{L}}^{-1} {\mathrm{s}}^{-1}$

Rate of production of ${\mathrm{Fe}}_2{\left({\mathrm{SO}}_4\right)}_3$

$=\frac{0.6}{1800}=3.33\times {10}^{-4}$

$=333.3\times {10}^{-6} {\mathrm{mol}}^{-1} {\mathrm{s}}^{-1}$