At 400ºC, K_c = 64 for the equilibrium H₂(g) + I₂(g) (⇌) 2HI(g). If 3.00 mol H₂ and 3.00 mol I₂ are introduced into an empty 4.0 L vessel, find the equilibrium concentration of HI at 400ºC.

Question

Quizplus · Accepted Answer

The initial concentrations of H2 and I2 are both 0.75 M (3.00 mol / 4.0 L). At equilibrium, let the concentration of HI be 2x. The equilibrium concentrations of H2 and I2 will be 0.75 - x. The equilibrium expression is Kc = [HI]^2 / ([H2][I2]) = 64. Solving 64 = (2x)^2 / ((0.75-x)(0.75-x)) gives x = 0.6, so [HI] = 2x = 1.2 M.

At 400ºC, Kc = 64 for the Equilibrium H2(g)

At 400ºC, K_c = 64 for the Equilibrium H₂(g)