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:

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

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 1 - 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 2 - 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 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.

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

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

Temperature changes affect Kc

The equilibrium constant (K_c_) is influenced solely by temperature. Changes in concentration or pressure have no impact on K_c.

When the position of equilibrium is altered due to a temperature shift:

• A decrease in the amount of product at equilibrium causes K_c_ to decrease.
• An increase in the amount of product at equilibrium causes K_c_ to increase. This ensures that the value of K_c_ aligns with the new position of equilibrium.

Example:

2SO_2(g) + O_2(g) ⇌ 2SO_3(g), ΔH = $-$197 kJ mol^-1

Raising the temperature causes the equilibrium to shift left towards the reactants, absorbing heat and resulting in less SO₃ at the new equilibrium position.

With a decrease in SO₃:

• The proportion of products reduces.
• Consequently, K_c_ (the ratio of products to reactants) decreases. The impact of temperature changes on exothermic and endothermic reactions is summarised in the table below:

Concentration and pressure changes do not affect Kc

While shifts in concentration or pressure can move the equilibrium position, altering the quantities present to counteract the change, K_c_ remains unchanged at a specific temperature.

For example:

N_2(g) + 3H_2(g) ⇌ 2NH_3(g)

• Adding more N_2_ leads to an increased formation of NH_3_ at the new equilibrium. However, the equilibrium ratio of [NH_3_]^2^ to [N_2_][H_2_]^3^ still remains equal to keep value of K_c_ constant.
• Thus, changes in concentration and pressure do not directly influence K_c_ - its value is only determined by temperature.

Catalysts do not change Kc or equilibrium position

Catalysts increase the rate of both the forward and reverse reactions, reducing the time needed to reach equilibrium following a change.

However, catalysts neither alter the equilibrium position nor the value of K_c_.

The table below summarises the effects of changing reaction conditions on the position of equilibrium and the value of K_c_: