$$\mathrm{KClO}_{3}+6 \mathrm{FeSO}_{4}+3 \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{KCl}+3 \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}+3 \mathrm{H}_{2} \mathrm{O}$$
The above reaction was studied at $300 \mathrm{~K}$ by monitoring the concentration of $\mathrm{FeSO}_{4}$ in which initial concentration was $10 \mathrm{M}$ and after half an hour became 8.8 M. The rate of production of $\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}$ is _________ $\times 10^{-6} \mathrm{~mol} \mathrm{~L} \mathrm{~s}^{-1}$ (Nearest integer)
Solution
The Rate of Reaction (ROR) for a reaction can be calculated based on the changes in concentrations of the reactants or the products over time. In the given reaction,
<br/><br/>$$\mathrm{KClO}_{3}+6 \mathrm{FeSO}_{4}+3 \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{KCl}+3 \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}+3 \mathrm{H}_{2} \mathrm{O}$$
<br/><br/>The rates of reaction can be written as :
<br/><br/>$$
\begin{aligned}
& \mathrm{ROR}=-\frac{\Delta\left[\mathrm{KClO}_3\right]}{\Delta \mathrm{t}}=\frac{-1}{6} \frac{\Delta\left[\mathrm{FeSO}_4\right]}{\Delta \mathrm{t}} \\\\
&=\frac{+1}{3} \frac{\Delta\left[\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3\right]}{\Delta \mathrm{t}} \\
\end{aligned}
$$
<br/><br/>From this, we can express the rate of formation of $\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3$ in terms of the change in concentration of $\mathrm{FeSO}_4$ :
<br/><br/>$$ \frac{\Delta\left[\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3\right]}{\Delta \mathrm{t}}=\frac{1}{2} \frac{-\Delta\left[\mathrm{FeSO}_4\right]}{\Delta \mathrm{t}} $$
<br/><br/>We know that the initial concentration of $\mathrm{FeSO}_4$ was 10 M and after 30 minutes (or 1800 seconds), it became 8.8 M. Substituting these values in, we get :
<br/><br/>$=\frac{1}{2} \frac{(10-8.8)}{30 \times 60} = 0.333 \times 10^{-3}$
<br/><br/>To express the rate in terms of $10^{-6} \, \mathrm{mol \, L^{-1} \, s^{-1}}$, we multiply the rate by $10^{3}$ :
<br/><br/>$=333 \times 10^{-6} \, \mathrm{mol \, L^{-1} \, s^{-1}}$
<br/><br/>Therefore, the rate of production of $\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}$ is $333 \times 10^{-6} \, \mathrm{mol \, L^{-1} \, s^{-1}}$.
About this question
Subject: Chemistry · Chapter: Chemical Kinetics · Topic: Rate of Reaction
This question is part of PrepWiser's free JEE Main question bank. 96 more solved questions on Chemical Kinetics are available — start with the harder ones if your accuracy is >70%.