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JEE Main 202524 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
24 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Chemical Equilibrium

JEE Main 2025 · 24 January · Morning Shift · Chemistry · Chemical Equilibrium

37.8 ~g ~N_2 O_5 was taken in a 1 L reaction vessel and allowed to...

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}

Correct answer

0

Step-by-step solution

Initial pressure of N2O5\mathrm{N}_2 \mathrm{O}_5=37.8108×0.082×5001=14.35 bar 2N2O5⇌2N2O4+O2\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}t=014.35t=eq14.35−2P2P\begin{array}{lll}\mathrm{t}=0 & 14.35 \\ \mathrm{t}=\mathrm{eq} & 14.35-2 \mathrm{P} & 2 \mathrm{P}\end{array}PTotal  at eqb =14.35+P=18.65P=4.3PN2O5=5.75 bar PN2O4=8.6barPO2=4.3 bar kp=(8.6)2×(4.3)(5.75)2=9.619=x×10−2x=961.9≈962\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}$$
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