On analysis, an equilibrium mixture for the reaction 2h2s(g) 2h2(g) + s2(g) was found to contain 1.0 mol h2s, 4.0 mol h2, and 0.80 mol s2 in a 4.0 l vessel. calculate the equilibrium constant, kc, for this reaction.

Kc = 3.2

The chemical equation is

2H2S ⇌ 2H2 + S2

The equilibrium constant expression is

Kc ={[H2]^2[S2]}/[H2S]^2

1.Calculate the equilibrium concentrations of each component

[H2] = (4.0 mol)/(4.0 L) = 1.0 mol/L

[S2] = (0.80 mol)/(4.0 L) = 0.20 mol/L

[H2S] = (1.0 mol)/(4.0 L) = 0.25 mol/L

2.Calculate the value of the equilibrium constant

Kc = (1.0^2 x 0.20)/0.25^2 = 3.2

the answer is a and have a great day