Asked by Caroline
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.
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.
Answers
Answered by
DrBob222
(H2S) = mols/L = 0.46 mol/4.6L = 0.1M
............2H2S ==> 2H2 + S2
I............0.1......0.....0
C............-2x......2x....x
E..........0.1-2x.....2x....x
Substitute the E line into Kc expression and solve for x, then 2x = (H2)
............2H2S ==> 2H2 + S2
I............0.1......0.....0
C............-2x......2x....x
E..........0.1-2x.....2x....x
Substitute the E line into Kc expression and solve for x, then 2x = (H2)