2H2S (g) 2H2 (g) + S2 (g)

Kc = 1.67 * 10 ^-7 at 800 C

A 0.500 L reaction vessel initially contains 0.0125 mol of H2S at 800 C. Find the equilibrium concentrations of H2 and S2.

Chemical reaction:

2H2S (g) ⇄ 2H2 (g) + S2 (g)

The equilibrium constant is given by: Kc = [H2]^2 * [S2] / [H2S]^2 = 1.67 * 10^ - 7

The initial concentration is 0.0125 mol / 0.500 L = 0.0250 M

Make a table showing the initial concentrations, the change and the final concentrations of each species

2H2S (g) ⇄ 2H2 (g) + S2 (g)

start 0.0250M 0 0

change - 2x + 2x + x

end 0.0250 - 2x 2x x

Kc = (2x) ^2 (x) / (0.0250 - 2x) ^2

Kc = 4x^3 / (0.0250 - 2x) ^2

To solve that equation in an easy way you can assume that 2x is << 0.0250, which leads to

Kc = 4x^3 / (0.0250) ^2 = 1.67 * 10^ - 7

=> x^3 = 1.67 * 10^ - 7 * 0.0250 / 4 = 2.6 * 10 ^-11

=> x = 2.97 * 10^ - 4 M

With this you can check that your assumption that x << 0.0250 is good and continue.

From the table you know that the concentrations at equilibrium are:

[H2] = 2x = 2 * 2.97 * 10 ^ - 4 M = 5.94 * 10 ^ - 4

[S2] = x = 2.97 * 10^ - 4 M