$37.8 \mathrm{~g} \mathrm{~N}_2 \mathrm{O}_5$ was taken in a 1 L reaction vessel and allowed to undergo the…
$2 \mathrm{~N}_2 \mathrm{O}_{5(\mathrm{~g})} \rightleftharpoons 2 \mathrm{~N}_2 \mathrm{O}_{4(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}$
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 $\mathrm{L} \mathrm{mol}^{-1} \mathrm{~K}^{-1}$
Solution
$\begin{aligned}
& =\frac{\frac{37.8}{108} \times 0.082 \times 500}{1}=14.35 \text { bar } \\
& 2 \mathrm{~N}_2 \mathrm{O}_5 \rightleftharpoons 2 \mathrm{~N}_2 \mathrm{O}_4+\mathrm{O}_2
\end{aligned}$
$\begin{array}{lll}\mathrm{t}=0 & 14.35 \\ \mathrm{t}=\mathrm{eq} & 14.35-2 \mathrm{P} & 2 \mathrm{P}\end{array}$
$\begin{aligned} & \mathrm{P}_{\text {Total }} \text { at eqb }=14.35+\mathrm{P}=18.65 \\ & \mathrm{P}=4.3 \\ & \mathrm{P}_{\mathrm{N}_2 \mathrm{O}_5}=5.75 \text { bar } \\ & \mathrm{P}_{\mathrm{N}_2 \mathrm{O}_4}=8.6 \mathrm{bar} \\ & \mathrm{P}_{\mathrm{O}_2}=4.3 \text { bar } \\ & \mathrm{k}_{\mathrm{p}}=\frac{(8.6)^2 \times(4.3)}{(5.75)^2}=9.619=\mathrm{x} \times 10^{-2} \\ & \mathrm{x}=961.9 \approx 962\end{aligned}$
Asked in: JEE Main 2025 (24 Jan Shift 1)