Question

At a certain temperature, the Kp for the decomposition of H2S is .0711 H2S(g) image from custom entry tool H2(g) + S(g) Initially, only H2S is present at a pressure of 0.171 atm in a closed container. What is the total pressure in the container at equilibrium?

Answer 1

Kp = 0.0711

H2 +S = H2S

The Kp expression:

Kp = P-H2S /(P-H2)(P-S)

initial conditions:

P-H2S = 0.171 atm

P-S = 0

P-H2 = 0

in equilibrium, and due to stoichiometry

P-H2S = 0.171 -x

P-S = 0 + x

P-H2 = 0 + x

in the Kp expression:

Kp = P-H2S /(P-H2)(P-S)

Kp = (0.171-x) /(x*x)

0.0711 = (0.171-x) /(x^2)

0.0711*x^2 = 0.171-x

0.0711*x^2 + x - 0.171 = 0

solve for x

x = 0.1689

then

P-H2S = 0.171 -x = 0.171-0.1689 = 0.0021

P-S = 0 + x = 0.1689

P-H2 = 0 + x = 0.1689

Total P = PH2S + P-S P-H2 = 0.0021 +0.1689+0.1689 = 0.3399 atm