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The reaction 2H2S(g)⇌2H2(g)+S2(g) Kc=1.67×10−7 at 800∘C is carried out with the following initial concentrations: [H2S] =...

The reaction 2H2S(g)⇌2H2(g)+S2(g) Kc=1.67×10−7 at 800∘C is carried out with the following initial concentrations: [H2S] = 0.325 M , [H2] =0.125 M , and [S2] = 0.00 M. Find the equilibrium concentration of [S2].

Solutions

Expert Solution

Kc = [H2]²[S2]/[H2S]²
In going to equilibrium, H2S loses 2x in concentration, H2 gains 2x in concentration, and S2 gains x in concentration, based on the stoichiometry of the reaction. The equation becomes
1.67×10^−7 = (0.325+2x)²(x)/(0.125-2x)².
This seems like a formidable equation to solve, be we can take a short cut. Since Kc is very small, x is very small, and we can make the assumption that
0.325 >> 2x
and therefore make the approximations
0.325+2x ≈ 0.325.

(0.125-2x) ≈ 0.125
The equation becomes
1.67×10^−7 = 0.350²(x)/0.125².
                  = 0.1225 (x) /0.0156

                   = 7.852 (x)
x ≈ 2.12×10^−8 = [S2] at equilibrium.
If Kc were about 0.1 or so larger, this assumption fails, and we would have to solve the Kc equation for x, either explicitly or using Newton's method.

queen_honey_blossom answered 4 hours ago
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