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Difficulty: MediumEquilibrium Constant Expression and Calculations
In the industrial synthesis of methanol, carbon(II) oxide gas reacts with hydrogen gas according to the equation:
CO(g)+2H2(g)⇌CH3OH(g)\text{CO}(g) + 2\text{H}_2(g) \rightleftharpoons \text{CH}_3\text{OH}(g)

At a given temperature, an equilibrium mixture in a 2.0 dm32.0\text{ dm}^3 sealed container contains 0.40 mol0.40\text{ mol} of CO(g)\text{CO}(g), 0.40 mol0.40\text{ mol} of H2(g)\text{H}_2(g), and 0.16 mol0.16\text{ mol} of CH3OH(g)\text{CH}_3\text{OH}(g). What is the numerical value of the equilibrium constant, KcK_c, for this reaction?

  1. 10.0 dm6 mol−210.0\text{ dm}^6\text{ mol}^{-2}Answer
  2. B
    2.5 dm6 mol−22.5\text{ dm}^6\text{ mol}^{-2}
  3. C
    2.0 dm6 mol−22.0\text{ dm}^6\text{ mol}^{-2}
  4. D
    0.10 dm6 mol−20.10\text{ dm}^6\text{ mol}^{-2}

Answer

10.0 dm6 mol−210.0\text{ dm}^6\text{ mol}^{-2}
To find KcK_c, first divide the moles of each gas at equilibrium by the volume of the vessel (2.0 dm32.0\text{ dm}^3) to obtain their equilibrium concentrations: [CO]=0.20 mol dm−3[\text{CO}] = 0.20\text{ mol dm}^{-3}, [H2]=0.20 mol dm−3[\text{H}_2] = 0.20\text{ mol dm}^{-3}, and [CH3OH]=0.08 mol dm−3[\text{CH}_3\text{OH}] = 0.08\text{ mol dm}^{-3}. Then substitute these values into the equilibrium expression Kc=[CH3OH][CO][H2]2K_c = \frac{[\text{CH}_3\text{OH}]}{[\text{CO}][\text{H}_2]^2}, giving 0.080.20×(0.20)2=10.0 dm6 mol−2\frac{0.08}{0.20 \times (0.20)^2} = 10.0\text{ dm}^6\text{ mol}^{-2}.

Step-by-Step Solution

1
Calculate the equilibrium concentration of each species
[CO]=0.40 mol2.0 dm3=0.20 mol dm−3[\text{CO}] = \frac{0.40\text{ mol}}{2.0\text{ dm}^3} = 0.20\text{ mol dm}^{-3}, [H2]=0.40 mol2.0 dm3=0.20 mol dm−3[\text{H}_2] = \frac{0.40\text{ mol}}{2.0\text{ dm}^3} = 0.20\text{ mol dm}^{-3}, [CH3OH]=0.16 mol2.0 dm3=0.08 mol dm−3[\text{CH}_3\text{OH}] = \frac{0.16\text{ mol}}{2.0\text{ dm}^3} = 0.08\text{ mol dm}^{-3}
Equilibrium constant KcK_c requires molar concentrations in mol dm−3\text{mol dm}^{-3}, calculated using C=nVC = \frac{n}{V}.
2
Write the equilibrium constant expression
Kc=[CH3OH][CO][H2]2K_c = \frac{[\text{CH}_3\text{OH}]}{[\text{CO}][\text{H}_2]^2}
Products are placed in the numerator and reactants in the denominator, each raised to the power of its stoichiometric coefficient.
3
Substitute concentrations into the expression and solve
Kc=0.08(0.20)×(0.20)2=0.080.20×0.04=0.080.008=10.0 dm6 mol−2K_c = \frac{0.08}{(0.20) \times (0.20)^2} = \frac{0.08}{0.20 \times 0.04} = \frac{0.08}{0.008} = 10.0\text{ dm}^6\text{ mol}^{-2}
Accurate substitution and exponent evaluation.

Key Concept

Calculation of equilibrium constant (KcK_c) from equilibrium amounts and container volume
Estimated Time:1m 30s
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Topics

Equilibrium Constant Expression and Calculations