JEE MAIN - Chemistry (2023 - 11th April Morning Shift - No. 19)
$$\mathrm{KClO}_{3}+6 \mathrm{FeSO}_{4}+3 \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{KCl}+3 \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}+3 \mathrm{H}_{2} \mathrm{O}$$
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)
Answer
333
Explanation
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,
$$\mathrm{KClO}_{3}+6 \mathrm{FeSO}_{4}+3 \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{KCl}+3 \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}+3 \mathrm{H}_{2} \mathrm{O}$$
The rates of reaction can be written as :
$$ \begin{aligned} & \mathrm{ROR}=-\frac{\Delta\left[\mathrm{KClO}_3\right]}{\Delta \mathrm{t}}=\frac{-1}{6} \frac{\Delta\left[\mathrm{FeSO}_4\right]}{\Delta \mathrm{t}} \\\\ &=\frac{+1}{3} \frac{\Delta\left[\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3\right]}{\Delta \mathrm{t}} \\ \end{aligned} $$
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 $\mathrm{FeSO}_4$ :
$$ \frac{\Delta\left[\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3\right]}{\Delta \mathrm{t}}=\frac{1}{2} \frac{-\Delta\left[\mathrm{FeSO}_4\right]}{\Delta \mathrm{t}} $$
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 :
$$=\frac{1}{2} \frac{(10-8.8)}{30 \times 60} = 0.333 \times 10^{-3} $$
To express the rate in terms of $10^{-6} \, \mathrm{mol \, L^{-1} \, s^{-1}}$, we multiply the rate by $10^{3}$ :
$$ =333 \times 10^{-6} \, \mathrm{mol \, L^{-1} \, s^{-1}}$$
Therefore, the rate of production of $\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}$ is $333 \times 10^{-6} \, \mathrm{mol \, L^{-1} \, s^{-1}}$.
$$\mathrm{KClO}_{3}+6 \mathrm{FeSO}_{4}+3 \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{KCl}+3 \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}+3 \mathrm{H}_{2} \mathrm{O}$$
The rates of reaction can be written as :
$$ \begin{aligned} & \mathrm{ROR}=-\frac{\Delta\left[\mathrm{KClO}_3\right]}{\Delta \mathrm{t}}=\frac{-1}{6} \frac{\Delta\left[\mathrm{FeSO}_4\right]}{\Delta \mathrm{t}} \\\\ &=\frac{+1}{3} \frac{\Delta\left[\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3\right]}{\Delta \mathrm{t}} \\ \end{aligned} $$
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 $\mathrm{FeSO}_4$ :
$$ \frac{\Delta\left[\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3\right]}{\Delta \mathrm{t}}=\frac{1}{2} \frac{-\Delta\left[\mathrm{FeSO}_4\right]}{\Delta \mathrm{t}} $$
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 :
$$=\frac{1}{2} \frac{(10-8.8)}{30 \times 60} = 0.333 \times 10^{-3} $$
To express the rate in terms of $10^{-6} \, \mathrm{mol \, L^{-1} \, s^{-1}}$, we multiply the rate by $10^{3}$ :
$$ =333 \times 10^{-6} \, \mathrm{mol \, L^{-1} \, s^{-1}}$$
Therefore, the rate of production of $\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}$ is $333 \times 10^{-6} \, \mathrm{mol \, L^{-1} \, s^{-1}}$.
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