Introducing the equilibrium constant

When a reversible reaction reaches a state of dynamic equilibrium, we can calculate a value called the equilibrium constant (K_c_) using the molar concentrations of the reactants and products at equilibrium.

K_c_ gives us a quantitative measure of where the equilibrium lies - whether there are more products or more reactants present at equilibrium.

• A large K_c_ value indicates the equilibrium position favours the products.
• A small K_c_ value indicates the equilibrium position favours the reactants.

A K_c_ value of 1 indicates that the reaction is at equilibrium, and the concentrations of reactants and products are equal when raised to their respective stoichiometric coefficients.

Writing an expression for K_c_

For the general equilibrium reaction:

aA + bB ⇌ dD + eE

The equilibrium constant K_c_ is given by:

${\text{K}}_{\text{c }}=\frac{D^d{E}^e}{A^a{B}^b}$

Where the lower case letters represent the coefficients in the balanced chemical equation.

For example, for the reaction:

H_2(g)_ + I_2(g)_ ⇌ 2HI_(g)_

The K_c_ expression would be:

${\text{K}}_{\text{c }}=\frac{{\text{[HI]}}^2}{{\text{[H}}_2{\text{][I}}_2\text{]}}$

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 Cu^2+^) are included in the K_c_ expression.

Step 3: Write the K_c_ expression

${\text{K}}_{\text{c }}=\frac{{\text{[Cu}}^{\text{2+}}\text{]}}{{\text{[Ag}}^{\text{+}}{\text{]}}^{\text{2}}}$

Calculating values for K_c_

There are 3 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_)

For the reaction H_2(g)_ + I_2(g)_ ⇌ 2HI_(g)_ at 580 K, the equilibrium concentrations are:

[HI] = 0.60 mol dm^−3^

[H_2_] = 0.20 mol dm^−3^

[I_2_] = 0.20 mol dm^−3^

Determine the equilibrium constant (K_c_).

Step 1: Write the equilibrium constant (K_c_) expression

$K_c=\frac{{[HI]}^2}{[H_2][I_2]}$

Step 2: Substitution and correct evaluation

$K_c=\frac{{(0.60)}^2}{0.20\times 0.20}=9$

In some cases, we may need to calculate some equilibrium concentrations before we can find K_c_.

Worked example 3 - Determining the equilibrium constant (K_c_)

0.25 moles of PCl_5_ decomposes at 650 K in a 4.0 dm^3^ vessel. At equilibrium, 0.10 moles of Cl_2_ 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 PCl_3_ formed at equilibrium

Moles of PCl_3_ = moles of Cl_2_ = 0.10 mol

Step 2: Calculate remaining moles of PCl_5_ at equilibrium

Moles of PCl_5_ decomposed = 0.10 mol

Moles of PCl_5_ at equilibrium = 0.25 - 0.10 = 0.15 mol

Step 3: Determine equilibrium concentrations

${\text{[PCl}}_5\text{] }=\frac{0.15}{4.0}=0.0375{\text{ mol dm}}^{\text{-3}}$

${\text{[PCl}}_3\text{] }=\frac{0.10}{4.0}=0.025{\text{ mol dm}}^{\text{-3}}$

${\text{[Cl}}_2\text{] }=\frac{0.10}{4.0}=0.025{\text{ mol dm}}^{\text{-3}}$

Step 4:Write the equilibrium constant (K_c_) expression

$K_c=\frac{[PC{l}_3][C{l}_2]}{[PC{l}_5]}$

Step 5: Substitution and correct evaluation

$K_c=\frac{(0.025\times 0.025)}{0.0375}=0.017$ 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 4 - 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.

CH_3_CH_2_COOH_(aq)_ + C_2_H_5_OH_(aq)_ ⇌ CH_3_CH_2_COOC_2_H_5(aq)_ + H_2_O_(l)_

If the K_c_ value is 0.35 at 30°C, determine the concentration of ethyl propanoate (CH_3_CH_2_COOC_2_H_5_) at equilibrium. Give your answer to 2 significant figures.

Step 1:Write the equilibrium constant (K_c_) expression

because water is a pure liquid (activity ≈ 1), it must be omitted from the K_c_ expression

$K_c=\frac{[C{H}_{3}C{H}_{2}COO{C}_2{H}_5]}{[C{H}_{3}C{H}_{2}COOH][C_2{H}_{5}OH]}$

Step 2: Rearrange expression

$[C{H}_{3}C{H}_{2}COO{C}_2{H}_5]=K_c\times [C{H}_{3}C{H}_{2}COOH][C_2{H}_{5}OH]$

Step 3:Substitution and correct evaluation

$[C{H}_{3}C{H}_{2}COO{C}_2{H}_5]=0.35\times 1.8\times 3.0=1.9{\text{ mol dm}}^{-3}$