Equilibrum Expressions and the Equilibrium Constant, Kc

This lesson covers: 

  1. What the equilibrium constant (Kc) is
  2. How Kc indicates the position of equilibrium
  3. How to write an expression for Kc
  4. Calculations involving Kc
  5. The effect of changing conditions on the equilibrium constant (Kc)

Introducing the equilibrium constant

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

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

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


A Kc 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 Kc

For the general equilibrium reaction:

aA + bB ⇌ dD + eE

The equilibrium constant Kc is given by:

Kc =AaBbDdEe


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


For example, for the reaction:

H2(g) + I2(g) ⇌ 2HI(g)

The Kc expression would be:

Kc =[H2][I2][HI]2

Calculating values for Kc

There are 3 worked examples below demonstrating various equilibrium calculations involving the equilibrium constant (Kc).

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

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

Worked example 1 - Determining the equilibrium constant (Kc)

For the reaction H2(g) + I2(g) ⇌ 2HI(g) at 580 K, the equilibrium concentrations are:

[HI] = 0.60 mol dm−3

[H2] = 0.20 mol dm−3

[I2] = 0.20 mol dm−3

Determine the equilibrium constant (Kc).


Step 1: Write the equilibrium constant (Kc) expression

Kc=[H2][I2][HI]2


Step 2: Substitution and correct evaluation

Kc=0.20×0.20(0.60)2=9


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

Worked example 2 - Determining the equilibrium constant (Kc)

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 Kc for the reaction:

PCl5(g) ⇌ PCl3(g) + Cl2(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 (Kc) 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 (Kc) value and some equilibrium concentrations, we can use the Kc 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 Kc value is 3.5 at 30°C, determine the concentrations of the other components at equilibrium. The balanced equation is:

CH3CH2COOH(aq) + C2H5OH(aq) ⇌ CH3CH2COOC2H5(aq) + H2O(l)


Step 1: Write the equilibrium constant (Kc) expression

Kc=[CH3CH2COOH][C2H5OH][CH3CH2COOC2H5][H2O]


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

3.5=1.8×3.0[CH3CH2COOC2H5][H2O]


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

Temperature changes affect Kc

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

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

  • A decrease in the amount of product at equilibrium causes Kc to decrease.
  • An increase in the amount of product at equilibrium causes Kc to increase.

This ensures that the value of Kc aligns with the new position of equilibrium.


Example:

2SO2(g) + O2(g) ⇌ 2SO3(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, Kc (the ratio of products to reactants) decreases.

The impact of temperature changes on exothermic and endothermic reactions is summarised in the table below:

Type of reactionTemperature changeImpact on equilibrium positionImpact on Kc
Exothermic reactionIncreaseShifts leftDecreases
Exothermic reactionDecreaseShifts rightIncreases
Endothermic reactionIncreaseShifts rightIncreases
Endothermic reactionDecreaseShifts leftDecreases

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, Kc remains unchanged at a specific temperature.


For example:

N2(g) + 3H2(g) ⇌ 2NH3(g)

  • Adding more N2 leads to an increased formation of NH3 at the new equilibrium. However, the equilibrium ratio of [NH3]2 to [N2][H2]3 remains equal to the constant value of Kc.
  • Thus, changes in concentration and pressure do not directly influence Kc - 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 Kc.

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

Factor changedImpact on equilibrium positionImpact on Kc
ConcentrationChangesRemains the same
PressureChangesRemains the same
TemperatureChangesChanges
CatalystRemains the sameRemains the same