Answer to Question #300862 in General Chemistry for ash

Question

The reaction: 2 H2S(g) ⇌ 2 H2(g) + S2(g) K = 5.13 X 10-6 (at 873 K)

is carried out with the following initial partial pressures: P (H2S) = 1.00 X 10-3 bar, P(H2) = 1.00 X 10^-5 bar, and P(S2) = 0.100 bar. Find the partial pressures of each of the reactants and products at equilibrium. Also show the % error calculation with an explanation of your reasoning and show the ICE table and the calculation for Q.

Answer 1

2H₂S(g) ⇌ 2H₂(g) + S₂(g)

K = [H₂]²[S₂]/[H₂S]²

2H₂S(g) ⇌ 2H₂(g) + S₂(g)

I 1 x 10^-3 1 x 10^-5 0.100

C - 2x + 2x +x

E 1 x 10^-3 - 2x 1 x 10^-5 + 2x 0.100 + x

5.13 x 10^-6 = (1 x 10^-5 + 2x)²(0.100 + x)/(1 x 10^-3 - 2x)²

x = 1.7 x 10^-4

[H₂] = 1 x 10^-5 + 2x = 1 x 10^-5 + 2 x 1.7 x 10^-4 = 0.00035

[S₂] = 0.100 + x = 0.100 + 1.7 x 10^-4 = 0.1002

[H₂S] = 1 x 10^-3 - 2x = 1 x 10^-3 - 2 x 1.7 x 10^-4 = 0.00066