Question

At a certain temperature, the ?p for the decomposition of H2S is 0.823. H2S(g)↽−−⇀H2(g)+S(g) Initially, only H2S is present at a pressure of 0.103 atm  in a closed container. What is the total pressure in the container at equilibrium? atm

Answer 1

ICE Table:

                    p(H2S)              p(H2)               p(S)              

initial             0.103               0                   0                 

change              -1x                 +1x                 +1x               

equilibrium         0.103-1x            +1x                 +1x               

Equilibrium constant expression is
Kp = p(H2)*p(S)/p(H2S)
0.823 = (1*x)(1*x)/((0.103-1*x))
0.823 = (1*x^2)/(0.103-1*x)
8.477*10^-2-0.823*x = 1*x^2
8.477*10^-2-0.823*x-1*x^2 = 0
This is quadratic equation (ax^2+bx+c=0)
a = -1
b = -0.823
c = 8.477*10^-2

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 = 1.016

roots are :
x = -0.9156 and x = 9.258*10^-2

since x can't be negative, the possible value of x is
x = 9.258*10^-2

Total P = p(H2S) + p(H2) + p(S)
= 0.103 - x + x + x
= 0.103 + x
= 0.103 + 9.258*10^-2
= 0.196 atm

Answer: 0.196 atm