The Equilibrium Constant, Kc - Part 2

This lesson covers: 

1. Differences between homogeneous and heterogeneous equilibria
2. Calculations involving K_c

Homogeneous versus heterogeneous equilibria

Equilibria can be categorised into homogeneous or heterogeneous systems:

• Homogeneous - All reactants and products are in the same physical state
• Heterogeneous - Reactants and products are in different physical states.

For homogeneous systems, all reactants and products are included in the K_c expression.

For heterogeneous systems, only gases and aqueous substances are included in the K_c expression. Solids and pure liquids are omitted because their concentrations stay relatively constant and do not affect the position of equilibrium.

Worked example 1 - Writing the K_c expression

Write an expression for the equilibrium constant (K_c) of the reaction:

Cu_(s) + 2Ag⁺_(aq) ^⇌ Cu²⁺_(aq) + 2Ag_(s)

Step 1: Determine if the reaction is homogeneous or heterogeneous

The reactants and products are a mixture of aqueous ions and solids, so the reaction is heterogeneous.

Step 2: Identify the substances to be included in the K_c expression

Only the aqueous substances (Ag⁺ and Cu²⁺) are included in the K_c expression.

Step 3: Write the K_c expression

Kc ​=[Ag+]2[Cu2+]​

Calculating values for K_c

There are 2 worked examples below demonstrating various equilibrium calculations involving the equilibrium constant (K_c).

If we know the equilibrium concentrations of all reactants and products, we can substitute them into the K_c expression to calculate a value for K_c.

The units for K_c vary, so we must determine the units after each calculation.

Worked example 2 - Determining the equilibrium constant (K_c)

0.25 moles of PCl₅ decomposes at 650 K in a 4.0 dm³ vessel. At equilibrium, 0.10 moles of Cl₂ is present.

Determine the equilibrium constant K_c for the reaction:

PCl_5(g) ⇌ PCl_3(g) + Cl_2(g)

Step 1: Deduce moles of PCl₃ formed at equilibrium

Moles of PCl₃ = moles of Cl₂ = 0.10 mol

Step 2: Calculate remaining moles of PCl₅ at equilibrium

Moles of PCl₅ decomposed = 0.10 mol

Moles of PCl₅ at equilibrium = 0.25 - 0.10 = 0.15 mol

Step 3: Determine equilibrium concentrations

[PCl5​] =4.00.15​=0.0375 mol dm−3

[PCl3​] =4.00.10​=0.025 mol dm−3

[Cl2​] =4.00.10​=0.025 mol dm−3

Step 4: Write the equilibrium constant (K_c) expression

Kc​=[PCl5​][PCl3​][Cl2​]​

Step 5: Substitution and correct evaluation

Kc​=0.0375(0.025×0.025)​=0.017 mol dm⁻³

If we know the equilibrium constant (K_c) value and some equilibrium concentrations, we can use the K_c expression to determine an unknown equilibrium concentration.

Worked example 3 - Determining equilibrium concentrations

When propanoic acid was allowed to reach equilibrium with ethanol at 30°C, it was found that the equilibrium mixture contained 1.8 mol dm^-3 propanoic acid and 3.0 mol dm^-3 ethanol.

If the K_c value is 3.5 at 30°C, determine the concentrations of the other components at equilibrium. The balanced equation is:

CH₃CH₂COOH_(aq) + C₂H₅OH_(aq) ⇌ CH₃CH₂COOC₂H_5(aq) + H₂O_(l)

Step 1: Write the equilibrium constant (K_c) expression

Kc​=[CH3​CH2​COOH][C2​H5​OH][CH3​CH2​COOC2​H5​][H2​O]​

Step 2: Substitute the known equilibrium concentrations into the K_c expression

3.5=1.8×3.0[CH3​CH2​COOC2​H5​][H2​O]​

Step 3: Rearrange to calculate [CH₃CH₂COOC₂H₅][H₂O]

[CH₃CH₂COOC₂H₅][H₂O] = 3.5 x 1.8 x 3.0 = 18.9

Step 4: Set [CH₃CH₂COOC₂H₅] = [H₂O] = x

From the balanced equation, [CH₃CH₂COOC₂H₅] = [H₂O] at equilibrium

x² = 18.9

x =√18.9=4.3 mol dm−3

Therefore, [CH₃CH₂COOC₂H₅]_equil = [H₂O]_equil = 4.3 mol dm^-3