Calculating Concentration Using the Equilibrium Constant

Question

At a given temperature, Kc = 3.24 for the reaction H2 (g) + CO2(g) ⇌ H2O (g) + CO (g) Of 0.800 mole of both H2 and CO2 (g) are placed in a 1.0 L container at this temperature, what is the concentration of CO (g) when the system comes to equilibrium?

Answer 1

@Jacques D's answer is perfectly acceptable, I thought I'd add the typical ICE table, in case the reader isn't sure of how the concentrations were arrived at.

H₂(g) + CO₂(g) <==> H₂O(g) + CO(g)

0.8.........0.8.....................0................0.........Initial

-x...........-x......................+x...............+x.......Change

0.8-x......0.8-x...................x.................x........Equilibrium

K = [H₂O][CO] / [H₂][CO₂]

3.24 = (x)(x) / (0.8-x)(0.8-x)

Use the quadratic formula and solve for x which will be the equilibrium concentration of CO and H₂O

0.8 - x will be the equilibrium concentrations of both H₂ and CO₂.

Answer 2

if x is the amount of reaction that happens in molarity (volume is constant, so molarity and moles are proportional) the equilibrium concentrations of each species will be: (Note also that the volume is 1 liter so the moles of reactants .8/1 liter or .8 Molar)

.H₂ : .8M - x CO₂: .8 M -x H₂O: x CO: x Each product adds x times the stoichiometric coefficient and each reactant subtracts x times the stoichiometric coefficient)

The K expression is [CO][H₂O]/([H₂][CO₂]) or x²/(.8-x)² = 3.24 for the given conditions.

Solve for x using the quadratic formula once you multiply by (8-x)² and rearrange into the quadratic form. x is the concentration of CO.