Question

7. Calculate the equilibrium concentrations for the following reaction when the 0.200 mol of H2S is placed in an empty 1.00 L container. (This one will be hard to solve by hand: use an online tool once you set up the expression.) 2 H2S (g) ⇌ 2 H2 (g) + S2 (g) K = 4.2 x 10-6 at 1100 K

8. Consider the reaction 2 CO2(g) ⇄ 2 CO (g) + O2 (g). In one equilibrium mixture [CO2] = 0.18 M, [CO] = 0.35 M, [O2] = 0.29 M. If the concentrations are changed so that [CO2] = 0.30 M, [CO] = 0.18 M, [O2] = 0.29 M, what will be the new equilibrium concentrations? (This one will be hard to solve by hand: use an online tool)

Answer 1

7)

2H2S(g) -------- 2H2(g) + S2(g)

Initial 0.200M 0 0

Final (0.200-2x) 2x x

Solving the equation

[S2] = 3.34 * 10^(-4) M

[H2] = 6.68 * 10^(-4) M

[H2S] = 0.200 - 2(3.34 * 10^(-4)) = 0.19932

8)

2CO2(g) --------- 2CO(g) + O2(g)

K = [Products]/[Reactants] = [CO]^2[O2]/[CO2]^2 = (0.35)^2 * (0.29)/(0.18)^2 = 1.09645

2CO2(g) --------- 2CO(g) + O2(g)

Initial 0.30 0.18 0.29

Final (0.30-2x) (0.18+2x) (0.29+x)

Solving the equation

x = 0.063

[O2] = 0.29 + 0.063

[CO] = 0.18 + 2(0.063)

[CO2] = 0.30 - 2(0.063)