JEE Chemistry 2023 Question with Solution

Given:

• Reaction: $\mathrm{KClO_3 + 6FeSO_4 + 3H_2SO_4 \rightarrow KCl + 3Fe_2(SO_4)_3 + 3H_2O}$
• Initial concentration of $\mathrm{FeSO_4}$ = $10 \, \text{M}$
• Concentration after $30$ minutes = $8.8 \, \text{M}$
• Time interval = $30 \times 60 = 1800 \, \text{s}$

Find: Rate of production of $\mathrm{Fe_2(SO_4)_3}$.

Rate of disappearance of $\mathrm{FeSO_4}$:

$$\frac{-\Delta [\mathrm{FeSO_4}]}{\Delta t} = \frac{10 - 8.8}{1800} = \frac{1.2}{1800} = 6.67 \times 10^{-4} \, \text{mol L}^{-1} \, \text{s}^{-1}$$

From the stoichiometry,

$$6\,\mathrm{FeSO_4} \rightarrow 3\,\mathrm{Fe_2(SO_4)_3}$$

So, the rate of production of $\mathrm{Fe_2(SO_4)_3}$ is:

$$\frac{3}{6} \times 6.67 \times 10^{-4} = 3.33 \times 10^{-4} \, \text{mol L}^{-1} \, \text{s}^{-1}$$

Expressing this as $\times 10^{-6}$:

$$3.33 \times 10^{-4} = 333.33 \times 10^{-6}$$

Therefore, the rate of production of $\mathrm{Fe_2(SO_4)_3}$ is $333 \times 10^{-6} \, \text{mol L}^{-1} \, \text{s}^{-1}$. The correct option is A.

The concentration drop of $\mathrm{FeSO_4}$ in $1800 \, \text{s}$ is $1.2 \, \text{M}$.

$$\text{Rate of disappearance of } \mathrm{FeSO_4} = \frac{1.2}{1800} = 6.67 \times 10^{-4}$$

In the balanced equation, $6$ moles of $\mathrm{FeSO_4}$ produce $3$ moles of $\mathrm{Fe_2(SO_4)_3}$. Hence product formation rate is half of the reactant disappearance rate:

$$\text{Rate of formation of } \mathrm{Fe_2(SO_4)_3} = \frac{3}{6} \times 6.67 \times 10^{-4}$$ $$= 3.33 \times 10^{-4} = 333.33 \times 10^{-6} \, \text{mol L}^{-1} \, \text{s}^{-1}$$

Hence, the required value is $333$.