In: Chemistry
At a certain temperature, the Kp for the decomposition of H2S is 0.785. H2S(g) to H2(g) + S(g) Initially, only H2S is present at a pressure of 0.142 atm in a closed container. What is the total pressure in the container at equilibrium
H2S (g) <—> H2 (g) + S (g)
0.142 0 0 (initial)
0.142-x x x (at equilibrium)
Kp = p(H2)*p(O2) / p(H2S)
0.785 = x*x/(0.142-x)
0.11147 - 0.785*x = x^2
x^2 + 0.785*x - 0.11147 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 0.785
c = -0.1115
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 = 1.062
roots are :
x = 0.123 and x = -0.9078
since x can't be negative, the possible value of x is
x = 0.123
Total pressure at equilibrium = 0.142 - x + x + x
= 0.142 + x
= 0.142 + 0.123
= 0.265 atm
Answer: 0.265 atm