For H2(g) + I2(g) 2HI(g) the equilibrium constant is K_c = 50 at a given temperature. A mixture has [H2] = 0.20 mol L^-1, [I2] = 0.10 mol L^-1 and [HI] = 0.80 mol L^-1. What will happen?

The reaction proceeds in the forward direction

The reaction proceeds in the backward direction

The reaction proceeds in the forward direction

The mixture is already at equilibrium

No prediction is possible without knowing the volume

NEET levelSingle correct optionDifficulty 1 of 3ce111

For $H X_{2} (g) + I X_{2} (g) 2 HI (g)$ the equilibrium constant is $K_{c} = 50$ at a given temperature. A mixture has $[H X_{2}] = 0.20 mol L^{- 1}$ , $[I X_{2}] = 0.10 mol L^{- 1}$ and $[HI] = 0.80 mol L^{- 1}$ . What will happen?

Q_{c} = [HI]^{2} / ([H X_{2}] [I X_{2}]) = (0.80)^{2} / (0.20 \times 0.10) = 32.0

. Since

Q_{c} < K_{c} = 50

, the mixture has too little product, so the net reaction goes forward.