Initially, only H2S is present at a pressure of 0.169 atm in a closed container. What is the total pressure in the container at equilibrium?

Question

At a certain temperature, the Kp for the decomposition of H2S is 0.756.

H2S(g)↽−−⇀H2(g)+S(g).

Answer 1

At a certain temperature, the Kp for the decomposition of H2S is 0.756.

H2S(g) ↽−−⇀ H2(g) + S(g).

initial 0.169 atm 0 0

change -x +x +x

equilibrium 0.169 -x x x

Kp= (P_H2) (P_S) 0.756 = x² [quadratic formula to solve for x]

(P_H2S) (0.169 - x) x = 0.142 atm

P_H2S= 0.027 atm

P_H2= 0.142 atm

P_S = 0.142 atm

P_total = P_H2S + P_H2 + P_S = 0.027 atm + 0.142 atm + 0.142 atm = 0.311 atm