For H2(g) + I2(g) 2HI(g) at equilibrium in a rigid vessel, [H2] = [I2] = 0.050 M and [HI] = 0.400 M. Then 0.050 M of H2 is added at constant temperature. Which statements are correct?

Immediately after the addition, Q_c = 32

At the new equilibrium, [HI] is greater than 0.400 M

Immediately after the addition, Q_c = 32

The net reaction after the addition is backward

At the new equilibrium, [HI] is greater than 0.400 M

At the new equilibrium, [H2] equals 0.100 M

JEE Advanced levelOne or more correct optionsDifficulty 3 of 3ce055

For $H X_{2} (g) + I X_{2} (g) 2 HI (g)$ at equilibrium in a rigid vessel, $[H X_{2}] = [I X_{2}] = 0.050 M$ and $[HI] = 0.400 M$ . Then 0.050 M of $H X_{2}$ is added at constant temperature. Which statements are correct?

K_{c} = (0.400)^{2} / (0.050)^{2} = 64

. After adding

H X_{2}

Q_{c} = (0.400)^{2} / ((0.100) (0.050)) = 32 < K_{c}

, so the reaction goes forward (A true, B false). Solving

(0.400 + 2 y)^{2} = 64 (0.100 - y) (0.050 - y)

gives

y = 0.0156

, so

[HI] = 0.431

M, above 0.400 M (C true), while

[H X_{2}] = 0.084

M, below 0.100 M (D false).