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

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) <===> H2(g) + S(g)

0.186.................0............0......Initial..

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

0.186 - x............x.............x....Equilibrium..

K = 0.709 = (P_H2)(P_S)/(P_H2S)

0.709 = (x)(x)/0.186-x

x2 = 0.132 - 0.709x

x = 0.153

Equilibrium Pressures:

H2S = 0.186 - 0.153 = 0.033 atm

H2 = 0.153 atm

S = 0.153 atm

TOTAL Pressure = 0.339 atm