Question

At a certain temperature, the Kp for the decomposition of H. S is 0.752. HS(g) =H2(g) + S(g) Initially, only H, S is present at a pressure of 0.127 bar in a closed container. What is the total pressure in the container at equilibrium? Protal = bar

Answer 1

initial partial pressure of H2S = 0.127 atm

equilibrium constant, Kp = 0.752

The balanced reaction at equilibrium is,

H2S(g) = H2(g) + S(g)

I     0.127 0      0

     C        -x        x      x

     E   0.127-x     x      x

The equation for the equilibrium constant, Kp, obtained during the equilibrium phase is,

Kp = P_H2P_S/P_H2S

0.752 = x*x / 0.127−x

x² + 0.752x - 0.096 = 0

x = 0.111 atm

The equilibrium partial pressures of the reaction are -

P_H2S = 0.127−x = 0.127−0.111 = 0.016 atm

P_H2 = P_S = x = 0.111atm

The total pressure P is the sum of the partial pressures at equilibrium.

P = P_H2S + P_H2 + P_S = 0.016 + 0.111 + 0.111 = 0.238 atm