Question

At a certain temperature, the Kp for the decomposition of H2S is 0.721.

Initially, only H2S is present at a pressure of 0.256 atm in a closed container. What is the total pressure in the container at equilibrium?

Answer 1

H2S (g)   <—>   H2 (g)       +   S (g)

0.256               0               0       (initial)

0.256-x           x               x       (at equilibrium)

Kp = p(H2)*p(O2) / p(H2S)

0.721 = x*x/(0.256-x)

0.1846 - 0.721*x = x^2

x^2 + 0.721*x - 0.1846 = 0

This is quadratic equation (ax^2+bx+c=0)

a = 1

b = 0.721

c = -0.1846

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 = 1.258

roots are :

x = 0.200 and x = -0.9214

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

x = 0.200

Total pressure at equilibrium = 0.256 - x + x + x

= 0.256 + x

= 0.256 + 0.200

= 0.456 atm

Answer: 0.456 atm