JEE Chemistry 2025 Question with Solution

Given: The reaction is $N_2O_5(g) \rightarrow 2NO_2(g) + \frac{1}{2} O_2(g)$ in a constant volume container. Initially, only $N_2O_5$ is present with initial pressure $P_0$.

Find: The final pressure when $50\%$ of the reaction is completed.

Assume the initial moles of $N_2O_5$ are $1$.

When $50\%$ of the reaction is completed, moles of $N_2O_5$ left are $0.5$.

From stoichiometry, decomposition of $0.5$ mole of $N_2O_5$ forms:

• $NO_2 = 2 \times 0.5 = 1$
• $O_2 = \frac{1}{2} \times 0.5 = 0.25$

Total moles at this stage are

$$0.5 + 1 + 0.25 = 1.75$$

At constant volume and temperature, pressure is proportional to total moles. Therefore,

$$P_{\text{final}} = \frac{1.75}{1} P_0 = \frac{7}{4} P_0$$

Therefore, the final pressure is $\frac{7}{4}$ times of initial pressure, so the correct option is D.

Given: Initial pressure of $N_2O_5$ is $P_0$.

Find: Final pressure at $50\%$ completion.

At $50\%$ completion:

• Remaining $N_2O_5$ contributes $\frac{P_0}{2}$
• Formed $NO_2$ contributes $P_0$
• Formed $O_2$ contributes $\frac{P_0}{4}$

So,

$$P_{\text{final}} = \frac{P_0}{2} + P_0 + \frac{P_0}{4} = \frac{7P_0}{4}$$

This shortcut works because at constant volume and temperature, partial pressure is directly proportional to moles. Therefore, the final pressure is $\frac{7}{4}$ of the initial pressure, and the correct option is D.