JEE MAIN - Chemistry (2025 - 24th January Morning Shift - No. 1)
For a reaction, $\mathrm{N}_2 \mathrm{O}_{5(\mathrm{~g})} \rightarrow 2 \mathrm{NO}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})}$ in a constant volume container, no products were present initially. The final pressure of the system when $50 \%$ of reaction gets completed is
$5 / 2$ times of initial pressure
$7 / 2$ times of initial pressure
$7 / 4$ times of initial pressure
5 times of initial pressure
Explanation
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$$\begin{aligned} & \quad x=\frac{P_0}{2} \\ & P_{\text {total }}=P_0-\frac{P_0}{2}+P_0+\frac{P_0}{4}=\frac{7}{4} P_0 \\ & \text { Option (4) } \end{aligned}$$
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