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Difficulty: EasyEquilibrium Constant Expression and Calculations

For the reversible reaction PCl5(g)⇌PCl3(g)+Cl2(g)\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g), the equilibrium concentrations of PCl5\text{PCl}_5, PCl3\text{PCl}_3, and Cl2\text{Cl}_2 in a 1.0 dm31.0\text{ dm}^3 vessel are 0.20 mol dm−30.20\text{ mol dm}^{-3}, 0.60 mol dm−30.60\text{ mol dm}^{-3}, and 0.30 mol dm−30.30\text{ mol dm}^{-3} respectively. What is the numerical value of the equilibrium constant, KcK_c?

Answer: 0.9 mol dm^-3

Answer

The equilibrium constant KcK_c is 0.90 mol dm−30.90\text{ mol dm}^{-3}.
The equilibrium constant KcK_c is calculated by substituting the equilibrium concentrations into Kc=[PCl3][Cl2][PCl5]=0.60×0.300.20=0.90 mol dm−3K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} = \frac{0.60 \times 0.30}{0.20} = 0.90\text{ mol dm}^{-3}.

Step-by-Step Solution

1
Formulate the equilibrium constant expression KcK_c for the reaction.
Kc=[PCl3][Cl2][PCl5]K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]}
The equilibrium constant expression is defined as the product of the concentrations of the reaction products divided by the product of the concentrations of the reactants, each raised to the power of their stoichiometric coefficient.
2
Substitute the equilibrium concentrations into the expression.
Kc=0.60×0.300.20K_c = \frac{0.60 \times 0.30}{0.20}
Given equilibrium values: [PCl5]=0.20 mol dm−3[\text{PCl}_5] = 0.20\text{ mol dm}^{-3}, [PCl3]=0.60 mol dm−3[\text{PCl}_3] = 0.60\text{ mol dm}^{-3}, and [Cl2]=0.30 mol dm−3[\text{Cl}_2] = 0.30\text{ mol dm}^{-3}.
3
Calculate the arithmetic result.
Kc=0.180.20=0.90 mol dm−3K_c = \frac{0.18}{0.20} = 0.90\text{ mol dm}^{-3}
Multiplying 0.60×0.300.60 \times 0.30 gives 0.180.18, and dividing by 0.200.20 yields 0.900.90.

Key Concept

Direct calculation of equilibrium constant KcK_c from equilibrium concentrations
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Topics

Equilibrium Constant Expression and Calculations