JEE MAIN - Chemistry (2025 - 24th January Morning Shift - No. 21)
$37.8 \mathrm{~g} \mathrm{~N}_2 \mathrm{O}_5$ was taken in a 1 L reaction vessel and allowed to undergo the following reaction at 500 K
$$2 \mathrm{~N}_2 \mathrm{O}_{5(\mathrm{~g})} \rightleftharpoons 2 \mathrm{~N}_2 \mathrm{O}_{4(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}$$
The total pressure at equilibrium was found to be 18.65 bar.
Then, $\mathrm{Kp}=$ _________ $\times 10^{-2}$ [nearest integer]
Assume $\mathrm{N}_2 \mathrm{O}_5$ to behave ideally under these conditions.
Given: $\mathrm{R}=0.082$ bar $\mathrm{L} \mathrm{mol}^{-1} \mathrm{~K}^{-1}$
Explanation
$$\begin{aligned} & \text { Initial pressure of } \mathrm{N}_2 \mathrm{O}_5 \\ & \qquad \\ & \qquad=\frac{\frac{37.8}{108} \times 0.082 \times 500}{1}=14.35 \mathrm{bar} \\ & 2 \mathrm{~N}_2 \mathrm{O}_5 \rightleftharpoons 2 \mathrm{~N}_2 \mathrm{O}_4+\mathrm{O}_2 \end{aligned}$$
$$\begin{aligned} & \mathrm{t}=0 \quad 14.35 \\ & \mathrm{t}=\mathrm{eq} \quad 14.35-2 \mathrm{P} \quad 2 \mathrm{P} \quad \mathrm{P} \\ & \mathrm{P}_{\text {Total }} \text { at eqb }=14.35+\mathrm{P}=18.65 \\ & \mathrm{P}=4.3 \\ & \mathrm{P}_{\mathrm{N}_2 \mathrm{O}_{\mathrm{s}}}=5.75 \mathrm{bar} \\ & \mathrm{P}_{\mathrm{N}_2 \mathrm{O}_4}=8.6 \mathrm{bar} \\ & \mathrm{P}_{\mathrm{O}_2}=4.3 \mathrm{bar} \\ & \mathrm{k}_{\mathrm{p}}=\frac{(8.6)^2 \times(4.3)}{(5.75)^2}=9.619=\mathrm{x} \times 10^{-2} \\ & \mathrm{x}=961.9 \approx 962 \end{aligned}$$
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