JEE Main 2026 Chemistry question

ChemistryEquilibriumJEE Main 2026

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)⇌B(g). 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)

Answer: (A)

Solution

=x/(1−x)=4 gives x=0.8 mol of B. =nRT/V=1×0.082×400/10=3.28 atm; =0.8×0.082×400/10=2.624 atm.

Question ID VK-015691