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 0.46 mol H2S is placed in a 4.6 L container, what is the equilibrium concentration of H2(g) at 700°C?
So I thought that by using ice H2 would be 0.46 mol and that it would be 0.1 M but that was not right.
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 0.51 mol H2S is placed in a 3.0 L container, what is the equilibrium concentration of H2(g) at 700°C?
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