Question

For the reaction H2 (g) + CO2 (g) ⇌ H2O(g) + CO(g), Kc = 1.60. Exactly 1.00 mole of each gas is added simultaneously to a 10.0 L flask. Calculate all equilibrium concentrations.

Answer 1

The equilbirium expression

Kc = [H2O][CO]/([H2][CO2])

Kc = 1.6

initially

[H2O][CO]= 1/10 = 0.1

[H2][CO2]= 0.1

in equilibrium (expect more product sto form due to KC value)

[H2O][CO]= 0.1 + x

[H2][CO2]= 0.1 - x

substitute

1.60 = (0.1 + x)^2 / (0.1 - x)^2

sqrt(1.6) = (0.1+x) / (0.1-x)

1.2649*(0.1-x) = 0.1+x

0.12649 - 1.2649x = 0.1 + x

-2.2649x = 0.1 - 0.12649

x = 0.02649 / 2.2649

x = 0.0116958

so

[H2O][CO]= 0.1 + 0.0116958 = 0.1116958

[H2][CO2]= 0.1 - 0.0116958 = 0.0883042

proof:

Q = [H2O][CO]/([H2][CO2])

Q = (0.1116958 *0.1116958 ) / (0.08830422^2) = 1.5999 which is approx to 1.6

answered by: anonymous