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Difficulty: Very hardEquilibrium Constant Expression and Calculations
Excess solid carbon, C(s)\text{C}(s), and 4.0 mol4.0\text{ mol} of carbon dioxide gas, CO2(g)\text{CO}_2(g), are introduced into an evacuated 2.0 dm32.0\text{ dm}^3 rigid reaction vessel at a constant temperature. The system reaches equilibrium according to the reaction equation:
C(s)+CO2(g)⇌2CO(g)\text{C}(s) + \text{CO}_2(g) \rightleftharpoons 2\text{CO}(g)
If the equilibrium concentration of carbon monoxide gas, CO(g)\text{CO}(g), is determined to be 0.80 mol dm−30.80\text{ mol dm}^{-3}, calculate the numerical value of the equilibrium constant, KcK_c, in mol dm−3\text{mol dm}^{-3}.

Answer: 0.4 mol dm^-3

Answer

The numerical value of the equilibrium constant, KcK_c, is 0.40.4 (or 0.400.40).
To find KcK_c, first convert moles of CO2\text{CO}_2 to initial concentration: [CO2]0=4.0 mol2.0 dm3=2.00 mol dm−3[\text{CO}_2]_0 = \frac{4.0\text{ mol}}{2.0\text{ dm}^3} = 2.00\text{ mol dm}^{-3}. From stoichiometry (C(s)+CO2(g)⇌2CO(g)\text{C}(s) + \text{CO}_2(g) \rightleftharpoons 2\text{CO}(g)), creating 0.80 mol dm−30.80\text{ mol dm}^{-3} of CO\text{CO} consumes 0.802=0.40 mol dm−3\frac{0.80}{2} = 0.40\text{ mol dm}^{-3} of CO2\text{CO}_2. At equilibrium, [CO2]eq=2.00−0.40=1.60 mol dm−3[\text{CO}_2]_{eq} = 2.00 - 0.40 = 1.60\text{ mol dm}^{-3}. Omitting the solid carbon C(s)\text{C}(s) from the expression gives Kc=[CO]2[CO2]=(0.80)21.60=0.40 mol dm−3K_c = \frac{[\text{CO}]^2}{[\text{CO}_2]} = \frac{(0.80)^2}{1.60} = 0.40\text{ mol dm}^{-3}.

Step-by-Step Solution

1
Calculate initial concentration of reactant gas
[CO2]0=2.00 mol dm−3[\text{CO}_2]_0 = 2.00\text{ mol dm}^{-3}
Concentration must be calculated in mol dm−3\text{mol dm}^{-3} by dividing moles by vessel volume (2.0 dm32.0\text{ dm}^3).
2
Determine equilibrium concentrations using stoichiometric ratios
[CO2]eq=1.60 mol dm−3[\text{CO}_2]_{eq} = 1.60\text{ mol dm}^{-3} and [CO]eq=0.80 mol dm−3[\text{CO}]_{eq} = 0.80\text{ mol dm}^{-3}
The stoichiometric mole ratio of CO2\text{CO}_2 to CO\text{CO} is 1:21:2. Thus, consuming 0.40 mol dm−30.40\text{ mol dm}^{-3} of CO2\text{CO}_2 yields 0.80 mol dm−30.80\text{ mol dm}^{-3} of CO\text{CO}.
3
Formulate the equilibrium constant expression for the heterogeneous reaction
Kc=[CO]2[CO2]K_c = \frac{[\text{CO}]^2}{[\text{CO}_2]}
Pure solids such as C(s)\text{C}(s) have constant concentration and are omitted from the equilibrium constant expression.
4
Calculate the value of KcK_c
Kc=(0.80)21.60=0.40K_c = \frac{(0.80)^2}{1.60} = 0.40
Substitute equilibrium concentrations into the KcK_c expression and solve.

Key Concept

Equilibrium constant expression for heterogeneous equilibria and ICE table calculations.
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Topics

Equilibrium Constant Expression and Calculations