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QJEEM22Chceb1: JEE Main 2022 Chemistry - Class 11 JEEM22Chceb1 Solution: At of are mixed with of . The reaction occurring as above comes …

At 600 K,2 mol600 \mathrm{~K}, 2 \mathrm{~mol} of NO\mathrm{NO} are mixed with 1 mol1 \mathrm{~mol} of O2\mathrm{O}_{2} .

2NO(g)+O2(g)⇄2NO2(g)2 \mathrm{NO}_{(\mathrm{g})}+\mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{NO}_{2}(\mathrm{g})

The reaction occurring as above comes to equilibrium under a total pressure of 1 atm. Analysis of the system shows that 0.6 mol0.6 \mathrm{~mol} of oxygen are present at equilibrium. The equilibrium constant for the reaction is ________. (Nearest integer)

Step-by-Step Verified Solution

JEE Main 2022 (Online) 28th July Evening Shift Chemistry - Chemical Equilibrium Question 30 English Explanation
Partial pressure of NO(g)=1.22.6×1\mathrm{NO}(\mathrm{g})=\frac{1.2}{2.6} \times 1

Partial pressure of O2( g)=0.62.6\mathrm{O}_{2}(\mathrm{~g})=\frac{0.6}{2.6}

Partial pressure of NO2( g)=0.82.6\mathrm{NO}_{2}(\mathrm{~g})=\frac{0.8}{2.6}

Kp=(PNO2)2(PNO)2(PO2)amp;=0.8×0.8×2.61.2×1.2×0.6amp;=1.925amp;≈2\begin{aligned}\mathrm{K}_{\mathrm{p}}=\frac{\left(\mathrm{P}_{\mathrm{NO}_{2}}\right)^{2}}{\left(\mathrm{P}_{\mathrm{NO}}\right)^{2}\left(\mathrm{P}_{\mathrm{O}_{2}}\right)} &=\frac{0.8 \times 0.8 \times 2.6}{1.2 \times 1.2 \times 0.6} \\\\&=1.925 \\\\& \approx 2\end{aligned}

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