Question

At a certain temperature, the KpKp for the decomposition of H2SH2S is 0.769.0.769.

H2S(g)−⇀↽−H2(g)+S(g)H2S(g)↽−−⇀H2(g)+S(g)

Initially, only H2SH2S is present at a pressure of 0.240 atm0.240 atm in a closed container. What is the total pressure in the container at equilibrium?

Ptotal=

Answer 1

         H2S(g) <----> H2(g) + S(g)

initial 0.24 atm       0 atm   o atm

change     -x             +x       +x

equil     0.24-x          x        x

Kp = pH2*pS/pH2S

0.769 = x^2/(0.24-x)

X = 0.192 atm

At equilibrium,

partial pressure of H2S = 0.24-x

                                    = 0.24 - 0.192

          = 0.048 atm

partial pressure of H2 = X = 0.192 atm

partial pressure of S = X = 0.192 atm

Total pressure at equilibrium = pH2S+pH2+pS

                              = 0.048+0.192+0.192

                              = 0.432 atm