JEE Main 2026 · 6 Apr · Shift 1 · Chemistry Q57

ChemistrySingle correct+4 / -1Question 57 of 75

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.

$\ce{A(g) <=> B(g)}$

$K_c$ 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 $^{-1}$ )

3.28, 2.624

2.624, 3.28

3.28, 0.656

0.656, 6.56

Answer and worked solution

Official answer: A

Equilibrium. $K_c=\dfrac{[\mathrm B]}{[\mathrm A]}=4$ , and A + B is 1 mol in total, so B = 0.8 mol and A = 0.2 mol.

Partial pressures ( $\tfrac{RT}V=\tfrac{0.082\times400}{10}=3.28$ atm per mol):

(A)

Topics:Law of mass action and Kc, Kp

JEE Main 2026 · 6 Apr · Shift 1, Chemistry question 57