At a certain temp, Kp for decomposition of H2S is 0.718. H2S-->H2+S

Question

Initially only H2S is present at a pressure of 0.140 bar in a closed container. What is total pressure in the container at equilibrium?

Answer 1

Use an ICE table to help with this problem.

H₂S ===> H₂ + S

0.140..........0.......0.......Initital

0.140-x......+x.....+x......Change

0.140-x.......x.......x......Equilibrium

Kp = 0.718 = (H₂)(S) / (H₂S) = (x)(x) / (0.140 - x)

Solve for x and that will be the pressure of both H₂ and S. Subtract x from 0.140 and that will be the pressure of H₂S. Add them tall to get the total pressure at equilibrium