Question

Initially, only H2S is present at a pressure of 0.245 atm in a closed container. What is the total pressure in the container at equilibrium?

Answer 1

The Kp for the decomposition of H2S is 0.842 (from literature value).

Kp = (P_{H2, eq} P_S,eq) / P_H2S,eq = 0.842

All the coefficients were 1 so there are no exponents to worry about. We know that P_{H2S , i} = 0.245 atm .
An ICE table gives:

   H2S(g)    ​    H2(g) +    S(g)
IC:    0.245    0 0
​C:    -x    +x +x
​EC:    0.245 - x x x

We define x as the partial pressure of H₂(g) or S(g) at equilibrium, i.e. P_{H2, eq} = P_{S, eq}.

This becomes:

0.842 = (x) (x) / (0.245 - x)
​or, 0.842 = x² / (0.245 - x) [The KP is not small, so we'll have to solve this quadratic equation in full.]
or, 0.842 (0.245 - x) ​ = x²
or, 0.20629 - 0.842x ​ = x²
​or, x² + 0.842x - 0.20629 = 0

Solving the quadratic equation, we get;
x = 0.198 an x = -1.040​

Discarding negative value of x, we get;
​x = 0.198

So, x = 0.198 = P_{H2, eq} = P_{S, eq}.

PTotal = P_{H2S, eq} + P_{H2, eq} + P_{S, eq}.
​ = (0.245 - x) + x + x
   = 0.245 + x
           = 0.245 + 0.198
           = 0.443 atm