Equilibrium Constant, Kc | Edexcel International A Level Chemistry Revision Notes 2017

Kc Expressions - Deduction

Equilibrium expression & constant

The equilibrium constant expression is an expression that links the equilibrium constant, K, to the concentrations of reactants and products at equilibrium taking the stoichiometry of the equation into account
So, for a given reaction:

aA + bB ⇌ cC + dD

The corresponding equilibrium constant expression is written as:

$K = \frac{{[C]}^{c} {[D]}^{d}}{{[A]}^{a} {[B]}^{b}}$

- Where:
  - [A] and [B] = equilibrium reactant concentrations (mol dm^-3)
  - [C] and [D] = equilibrium product concentrations (mol dm^-3)
  - a, b, c and d = number of moles of corresponding reactants and products
Solids are ignored in equilibrium constant expressions
The K_c of a reaction is specific to a given reaction and only changes if the temperature of the reaction changes

Worked example

Deducing equilibrium expressions

Deduce the equilibrium constant expression for the following reactions

Ag⁺ (aq) + Fe²⁺ (aq) ⇌ Ag (s) + Fe³⁺ (aq)
N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g)
2SO₂ (g) + O₂ (g) ⇌ 2SO₃ (g)

Answer 1:

$K = \frac{[{Fe}^{3 +} (aq)]}{[{Fe}^{2 +} (aq)] [{Ag}^{+} (aq)]}$

[Ag (s)] is not included in the equilibrium constant expression as it is a solid

Answer 2:

$K = \frac{[{NH}_{3} (g)]}{[N_{2} (g)] {[H_{2} (g)]}^{3}}$

Answer 3:

$K = \frac{{[{SO}_{3} (g)]}^{2}}{{[{SO}_{2} (g)]}^{2} [O_{2} (g)]}$

Kc Expressions - Calculations

Calculations involving K_c

In the equilibrium expression each figure within a square bracket represents the concentration in mol dm^-3
The units of K_c therefore depend on the form of the equilibrium expression
Some questions give the number of moles of each of the reactants and products at equilibrium together with the volume of the reaction mixture
The concentrations of the reactants and products can then be calculated from the number of moles and total volume

Equation to calculate concentration from number of moles and volume

Worked example

Calculating K_c of ethanoic acid

In the reaction:

CH₃COOH (I) + C₂H₅OH (I) ⇌ CH₃COOC₂H₅ (I) + H₂O (I)

ethanoic acid ethanol ethyl ethanoate water

500 cm³ of the reaction mixture at equilibrium contained 0.235 mol of ethanoic acid and 0.035 mol of ethanol together with 0.182 mol of ethyl ethanoate and 0.182 mol of water. Use this data to calculate a value of K_cfor this reaction.

Answer

Step 1: Calculate the concentrations of the reactants and products

- [CH₃COOH (l)] = $\frac{0.235}{0.500} = 0.470 mol {dm}^{- 3}$
- [C₂H₅OH (l)] = $\frac{0.035}{0.500} = 0.070 mol {dm}^{- 3}$
- [CH₃COOC₂H₅ (l)] = $\frac{0.182}{0.500} = 0.364 mol {dm}^{- 3}$
- [H₂O (l)] = $\frac{0.182}{0.500} = 0.364 mol {dm}^{- 3}$

Step 2: Write the equilibrium constant for this reaction in terms of concentration

- K_c = $\frac{[H_{2} O] [{CH}_{3} {COOC}_{2} H_{5}]}{[C_{2} H_{5} OH] [{CH}_{3} COOH]}$

Step 3: Substitute the equilibrium concentrations into the expression

- K_c = $\frac{[0.364] \times [0.364]}{[0.070] \times [0.470]} = 4.03$

Step 4: Deduce the correct units for K_c

- K_c = $\frac{[m o l d m^{- 3}] \times [m o l d m^{- 3}]}{[m o l d m^{- 3}] \times [m o l d m^{- 3}]}$
- All units cancel out

Therefore, K_c = 4.03

Examiner Tip

Note that the smallest number of significant figures used in the question is 3, so the final answer should also be given to 3 significant figures

Some questions give the initial and equilibrium concentrations of the reactants but not the products
An initial, change and equilibrium table should be used to determine the equilibrium concentration of the products using the molar ratio of reactants and products in the stoichiometric equation

Worked example

Calculating K_c of ethyl ethanoate

Ethyl ethanoate is hydrolysed by water:

CH₃COOC₂H₅ (I) + H₂O (I) ⇌ CH₃COOH (I) + C₂H₅OH (I)

ethyl ethanoate water ethanoic acid ethanol

0.1000 mol of ethyl ethanoate are added to 0.1000 mol of water. A little acid catalyst is added and the mixture made up to 1.00 dm³. At equilibrium, 0.0654 mol of water are present.

Use this data to calculate a value of K_cfor this reaction.

Answer

Step 1: Write out the balanced chemical equation with the concentrations of beneath each substance using an initial, change and equilibrium table

Equilibria Calculating Kc of ethyl ethanoate table

Step 2: Calculate the concentrations of the reactants and products

- [CH₃COOC₂H₅ (l)] = $\frac{0.0654}{1.00} = 0.0654 mol {dm}^{- 3}$
- [H₂O (l)] = $\frac{0.0654}{1.00} = 0.0654 mol {dm}^{- 3}$
- [CH₃COOH (l)] = $\frac{0.0346}{1.00} = 0.0346 mol {dm}^{- 3}$
- [C₂H₅OH (l)] = $\frac{0.0346}{1.00} = 0.0346 mol {dm}^{- 3}$

Step 3: Write the equilibrium constant for this reaction in terms of concentration

- K_c = $\frac{[C_{2} H_{5} OH] [{CH}_{3} COOH]}{[H_{2} O] [{CH}_{3} {COOC}_{2} H_{5}]}$

Step 4: Substitute the equilibrium concentrations into the expression

- K_c = $\frac{[0.346] \times [0.346]}{[0.654] \times [0.654]} = 0.28$

Step 4: Deduce the correct units for K_c

- K_c = $\frac{[m o l d m^{- 3}] \times [m o l d m^{- 3}]}{[m o l d m^{- 3}] \times [m o l d m^{- 3}]}$
- All units cancel out

Therefore, K_c = 0.28

Equilibrium Constant, Kc (Edexcel International A Level Chemistry)

Revision Note

Kc Expressions - Deduction

Equilibrium expression & constant

Worked example

Kc Expressions - Calculations

Calculations involving K_c

Worked example

Examiner Tip

Worked example

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Equilibrium Constant, Kc (Edexcel International A Level Chemistry)

Revision Note

Equilibrium expression & constant

Calculations involving Kc

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Author: Sonny

Calculations involving K_c