0.51 mol/3 L = (H2S) in mols/L = M
.........2H2S --> 2H2 + S2
I........0.17......0.....0
C.........-2x.....2x.....x
E......0.17-2x....2x.....x
Substitute the E line into the Kq expression and solve for x. Then evaluate each specie.
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?
1 answer