At a certain temperature, the Kp for the decomposition of H2S is 0.805

Question

H₂S (g) ---> (and the reverse arrow) H₂ (g) + S (g)

Initially, only H₂S is present at a pressure of 0.241 atm in a closed container. What is the total pressure in the container at equilibrium?

Answer 1

H₂S(g) <==> H₂(g) + S(g)

0.241............0...........0.....Initial

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

0.241-x.......x.............x......Equilibrium

Kp = (H₂)(S) / (H₂S)

0.805 = (x)(x) / 0.241 - x

x² = 0.194 - .806x

x² + 0.806x - 0.194 = 0

x = 0.194 atm

Total pressure = pressure H2S + pressure H2 + pressure S

Total pressure = 0.241 - 0.194 + 0.194 + 0.194

Total pressure = 0.047 + 0.194 + 0.194

Total pressure = 0.435 atm