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?
ICE Table:
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
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