At a certain temperature, the Kp for the decomposition of H2S is 0.890.

Question

Initially, only H2S is present at a pressure of 0.176 atm in a closed container. What is the total pressure in the container at equilibrium?

 

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

 

Answer 1

H2S(g) <==> H2(g) + S(g) ... Kp = 0.890

0.176 atm......0...........0...Initial

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

0.176-x..........x............x..Equilibrium

Kp = 0.890 = (H2)(S) / (H2S)

0.890 = (x)(x) / (0.176-x)

0.157 - 0.889x = x²

x² + 0.889x - 0.157 = 0

x = 0.151 atm

At equilibrium we have the following partial pressures:

P_H2S = 0.176 - 0.151 = 0.025 atm

P_H2 = 0.151 atm

P_S = 0.151 atm

TOTAL PRESSURE = 0.025 atm + 0.151 atm + 0.151 atm = 0.327 atm