For a reaction, $\mathrm{N}_2 \mathrm{O}_{5(\mathrm{~g})} \rightarrow 2…

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
  1. 5 times of initial pressure
  2. $5 / 2$ times of initial pressure
  3. $7 / 2$ times of initial pressure
  4. $7 / 4$ times of initial pressure

Solution


$\begin{gathered}x=\frac{P_0}{2} \\ P_{\text {total }}=P_0-\frac{P_0}{2}+P_0+\frac{P_0}{4}=\frac{7}{4} P_0\end{gathered}$

Asked in: JEE Main 2025 (24 Jan Shift 1)

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