Question

At a certain temperature, the Kp for the decomposition of H2S is 0.709. Initially, only H2S is present at a pressure of 0.181 atm in a closed container. What is the total pressure in the container at equilibrium?

H2S(g)<---->H2(g)+S(g)

Answer 1

         H2S(g) <----> H2(g) + S(g)

initial 0.181 atm       0 atm   o atm

change     -x             +x       +x

equil     0.181-x          x        x

   Kp = pH2*pS/pH2S

    0.709 = X^2/(0.181-X)

X = 0.15 atm

At equilibrium,

partial pressure of H2S = 0.181-x

                = 0.181 - 0.15

       = 0.131 atm

partial pressure of H2 = X = 0.15 atm

partial pressure of S = X = 0.15 atm

at equilibrium, total pressure = 0.131+0.15+0.15 = 0.431 atm