2 H2 (g) + S2 (g) >>> 2 h2 S(g)?

At a particular temparture a 2.0 liter flask contains 2.0 mol of h2S, 0.40 mol of H2, and 0.80 mol of S2. Caculate the equilibrium constant for the reaction.

2 H2 (g) + S2 (g) >>> 2 h2 S(g)

Answer:
[ H2S] = 2.0 mol / 2.0 L = 1.0 M
[ H2 ] = 0.40 mol / 2.0 L = 0.20 M
[ S2 ] = 0.80 mol / 2.0 L = 0.40 M
K = [H2S]^2 / [ H2]^ [S2 ] = 1 / 0.04 x 0.4 = 62.5

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