Question

At a certain temperature, the equilibrium constant, Kc, for this reaction is 53.3.

H₂(g) + I₂(g) <-------> 2 HI(g) K_c=53.3

At this temperature, 0.600 mol of H2 and 0.600 mol of I2 were placed in a 1.00-L container to react. What concentration of HI is present at equilibrium?

Answer 1

[H2] = 0.600 mol / 1 L = 0.600 M
[I2] = 0.600 mol / 1 L = 0.600 M

Set up an ICE table. I is the initial concentration. C is the change. E is the equilibrium equation, which is just I + C.

In the reaction, the reactants are forming products. Therefore, the change would be -nx for reactants and +nx for products, where n is the stoichiometric amount.

. . . .H2(g) + I2(g) <-----> 2HI(g)
I. . 0.600. . .0.600. . . . . . 0
C. . .-x. . . . . -x. . . . . . . +2x
E.0.600-x. .0.600-x. . . . ..2x

Kc = [HI]^2 / [I2][H2]

53.3 = (2x)^2 / (0.600-x)^2

x = 0.471 and 0.826

[I2] = 0.600 - x = 0.600 - 0.471 = 0.129 M
[I2] = 0.600 - x = 0.600 - 0.826 = -0.226 M

0.826 gives you a negative value for [I2] and [H2], so it is not the answer. Therefore, the answer is x = 0.471.

Plug that value back into the equilibrium for [HI].

[HI] = 2x = 2*0.471 = 0.942 M