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 0.36 mol H2S is placed in a 3.4 L container, what is the equilibrium concentration of H2(g) at 700°C?
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initial [H2S] = 0.36 mol / 3.4 L
initial [H2S] = 0.1059M
Kc = [H2]² [S2] / [H2S]²
9.30e-8 = (2x)² (x) / (0.1059M - 2x)²
Assuming 2x << 0.1059M (and we can check this assumption later),
9.30e-8 ≈ (2x)² (x) / (0.1059M)²
1.0426e-9 = 4x³
x = 6.39e-4 M
final [H2] = 2x
final [H2] ≈ 2(6.39e-4 M)
final [H2] ≈ 1.3e-3 M (and 2x is about 1/100 of 0.1059)
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