Question

At a certain temperature, the ?p for the decomposition of H2S is 0.813. H2S(g)↽−−⇀H2(g)+S(g) Initially, only H2S is present at a pressure of 0.201 atm in a closed container. What is the total pressure in the container at equilibrium?

Answer 1

ICE Table:

p (H2) p (H2S) p (S) initial 0.201 0 change -1x +1x +1x equilibrium +1x 0.201-1x +1x

Equilibrium constant expression is
Kp = p(H2)*p(S)/p(H2S)
0.813 = (1*x)(1*x)/((0.201-1*x))
0.813 = (1*x^2)/(0.201-1*x)
0.1634-0.813*x = 1*x^2
0.1634-0.813*x-1*x^2 = 0
This is quadratic equation (ax^2+bx+c=0)
a = -1
b = -0.813
c = 0.1634

Roots can be found by
x = {-b + sqrt(b^2-4*a*c)}/2a
x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 1.315

roots are :
x = -0.9798 and x = 0.1668

since x can't be negative, the possible value of x is
x = 0.1668

Total pressure = 0.201 - x + x + x
= 0.201 + x
= 0.201 + 0.1668
= 0.368 atm

Answer: 0.368 atm