One mole each of He and A ( g ) are taken in a 10 L closed flask and heated to 400 K to establish the following equilibrium. A ( g ) rightleftharpoons B ( g ) K C for this reaction at 400 K is 4.0. The partial pressures (in atm ) of He and B ( g ) are respectively (at equilibrium). (Assume He , A ( g ) and B ( g ) behave as ideal gases) (Given : R =0.082 L atm K -1 mol -1 )

Chemistry · Chemical Equilibrium

JEE Main 2026 — 6 April, Morning Shift — Question 51

One mole each of He and $\mathrm{A}(\mathrm{g})$ are taken in a 10 L closed flask and heated to 400 K to establish the following equilibrium. $\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})$ $\mathrm{K}_{\mathrm{C}}$ for this reaction at 400 K is 4.0. The partial pressures (in atm ) of He and $\mathrm{B}(\mathrm{g})$ are respectively (at equilibrium). (Assume $\mathrm{He}, \mathrm{A}(\mathrm{g})$ and $\mathrm{B}(\mathrm{g})$ behave as ideal gases) (Given : $\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ )

Option A:
$3.28,2.624$
Correct
Option B:
$2.624,3.28$
Option C:
$3.28,0.656$
Option D:
$0.656,6.56$

Answer: A

Step-by-step solution

\begin{aligned} & \mathrm{A} \rightleftharpoons \mathrm{~B} & \mathrm{t}=0 \quad 1 \mathrm{~mol}- & \mathrm{t}_{\mathrm{eq}^{\mathrm{m}}} \quad 1-\mathrm{x} \quad \mathrm{x} & \mathrm{~K}_{\mathrm{C}}=\frac{\mathrm{x} / 10}{\frac{1-\mathrm{x}}{10}}=4 & \mathrm{x}=0.8 & \mathrm{P}_{\mathrm{He}}=\frac{1 \times 0.082 \times 400}{10}=3.28 \mathrm{~atm} & \mathrm{P}_{\mathrm{B}}=\frac{0.8 \times 0.082 \times 400}{10}=2.624 \mathrm{~atm} \end{aligned}

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 6 April, Morning Shift
Subject: Chemistry
Chapter: Chemical Equilibrium
Topic: Analysis of Chemical Equilibrium, Equilibrium Constant and Reaction Quotient