Question

Hydrogen sulfide decomposes according to the following reaction, for which Kc = 9.30 10-8 at 700°C. 2 H2S(g) 2 H2(g) + S2(g) If 0.33 mol H2S is placed in a 2.6 L container, what is the equilibrium concentration of H2(g) at 700°C?

Answer 1

initial concnetration of H2S = number of mol/volume
= 0.33/2.6
= 0.127 M
2 H2S(g) <------->2 H2(g) + S2(g)
0.127 0 0 (initial)
0.127-2x 2x x (at equilibrium)

Kc = [H2]^2 [S2] / [H2S]^2
9.30*10^-8 = (2x)^2 * x / (0.127-2x)
since Kc is small, x will be small and can be ignored as compared to 0.127
above expression thus becomes,
9.30*10^-8 = (2x)^2 * x / (0.127)
4x^2*x = 1.1804*10^-8
4x^3 = 1.1804*10^-8
x = 1.434*10^-3 M
[H2] = 2x = 2*1.434*10^-3 = 2.868*10^-3 M
Answer: 2.868*10^-3 M