Verified Solution • 2025–26 Session moderate

QJEEM25Chf273: JEE Main 2025 Chemistry - Class 11 JEEM25Chf273 Solution: was taken in a 1 L reaction vessel and allowed to undergo the…

37.8 g N2O537.8 \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 N2O5( g)⇌2 N2O4( g)+O2( g)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, Kp=\mathrm{Kp}= _________ ×10−2\times 10^{-2} [nearest integer]

Assume N2O5\mathrm{N}_2 \mathrm{O}_5 to behave ideally under these conditions.

Given: R=0.082\mathrm{R}=0.082 bar Lmol−1 K−1\mathrm{L} \mathrm{mol}^{-1} \mathrm{~K}^{-1}

Step-by-Step Verified Solution

amp; Initial pressure of N2O5amp;amp;=37.8108×0.082×5001=14.35baramp;2 N2O5⇌2 N2O4+O2\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}

amp;t=014.35amp;t=eq14.35−2P2PPamp;PTotal  at eqb =14.35+P=18.65amp;P=4.3amp;PN2Os=5.75baramp;PN2O4=8.6baramp;PO2=4.3baramp;kp=(8.6)2×(4.3)(5.75)2=9.619=x×10−2amp;x=961.9≈962\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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