What is the total pressure in the container at equilibrium?

Question

At a certain temperature the Kp for the decomposition of H2S is 0.709

H2S(g) (double arrows) H2(g)+S(g)

Initially only H2S is present at a pressure of 0.186 atm in a closed container. What is the total pressure in the container at equilibrium?

Answer 1

H₂S_(g) ↔ H_2(g) + S_(g) k_p = 0.709

I 0.186 0 0

C -x + x + x

E 0.186 - x x x

k_p = [H₂][S][H₂S]^-1

0.709 = [x][x][0.186 - x]^-1

[0.186 - x]0.709 = [x²][0.186 - x]^-1

0.131874 - 0.709x = x²

0 = x² + 0.709x - 0.131874 ⇒ x = (−b ± √(b² − 4ac))/(2a) ⇒ x = -0.861985 or 0.152985

Since x represents the equilibrium partial pressures of H_2(g) and S_(g), we only consider the positive x value since pressure cannot be negative.

Thus, the equilibrium partial pressure of H₂S_(g) is (0.186 − 0.152985) = 0.030015 atm and 0.152985 atm, both for H_2(g) and S_(g).

The total pressure in the container is then equal to partial pressure H₂S_(g) + partial pressure H_2(g) + partial pressure S_(g). So,

P_total = (0.030015)atm + (0.152985)atm + (0.152985)atm = 0.335985 ⇒ 0.336 atm

-Dan