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ChemistryRates and equilibrium

The Kc equation

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The Kc equation is the equilibrium constant expression for a reversible reaction. For aA + bB ⇌ cC + dD, Kc = [C]ᶜ[D]ᵈ ÷ ([A]ᵃ[B]ᵇ): the equilibrium concentrations of the products, each raised to the power of its number in the balanced equation, divided by those of the reactants. Square brackets mean concentration in mol dm⁻³. A large Kc means the equilibrium lies to the right, with mostly products; a small Kc means mostly reactants.

How to write a Kc expression

K_c = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}For aA + bB ⇌ cC + dD
  1. Write the balanced equation.
  2. Put the products on top and the reactants underneath.
  3. Use each species' number from the balanced equation as the power on its concentration.
  4. Multiply the terms together; never add them.

AQA's A-level specification says Kc is deduced from the equation for a reversible reaction, with [X] meaning the concentration of X in mol dm⁻³, and asks students to construct an expression for Kc for a homogeneous system, where everything is in the same phase.

Example 1: the Haber process

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g):

K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}

Units of Kc

Kc's units depend on the expression, so work them out each time by substituting mol dm⁻³ for every concentration and cancelling.

  • For Example 1: (mol dm⁻³)² ÷ (mol dm⁻³)⁴ = (mol dm⁻³)⁻² = mol⁻² dm⁶.
  • For H₂ + I₂ ⇌ 2HI: (mol dm⁻³)² ÷ (mol dm⁻³)² cancels, so Kc has no units.
  • For 2SO₂ + O₂ ⇌ 2SO₃: (mol dm⁻³)² ÷ (mol dm⁻³)³ = (mol dm⁻³)⁻¹ = mol⁻¹ dm³.

Calculating Kc

Example 2: from equilibrium concentrations

For H₂(g) + I₂(g) ⇌ 2HI(g), the equilibrium concentrations are [H₂] = 0.020 mol dm⁻³, [I₂] = 0.015 mol dm⁻³ and [HI] = 0.12 mol dm⁻³. Calculate Kc.

K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = \frac{0.12^2}{0.020 \times 0.015} = \frac{0.0144}{0.000300} = 48

Kc = 48, with no units.

Example 3: from starting amounts, using a table

2.0 dm³ of a mixture starts with 0.80 mol of SO₂ and 0.60 mol of O₂. At equilibrium, 0.50 mol of SO₃ has formed: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g). Calculate Kc.

Use the equation's ratio, 2 : 1 : 2, to find how much of each reactant was used, then divide moles by the volume:

SO₂O₂SO₃
Start (mol)0.800.600
Change (mol)−0.50−0.25+0.50
Equilibrium (mol)0.300.350.50
Equilibrium (mol dm⁻³)0.150.1750.25
K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2[\text{O}_2]} = \frac{0.25^2}{0.15^2 \times 0.175} = 15.9\ \text{mol}^{-1}\ \text{dm}^3

Example 4: an esterification, where volume cancels

Ethanoic acid reacts with ethanol: CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O. At equilibrium the mixture contains 0.33 mol of acid, 0.33 mol of ethanol, 0.67 mol of ester and 0.67 mol of water. Calculate Kc.

Two moles of particles go in and two come out, so every volume term cancels and the moles can go straight into the expression:

K_c = \frac{0.67 \times 0.67}{0.33 \times 0.33} = 4.1

Example 5: finding an unknown concentration

For the hydrogen iodide equilibrium with Kc = 48, the equilibrium concentrations of H₂ and I₂ are both 0.010 mol dm⁻³. Find [HI].

[\text{HI}]^2 = 48 \times 0.010 \times 0.010 = 0.0048 \quad \Rightarrow \quad [\text{HI}] = 0.069\ \text{mol dm}^{-3}

Now try a question on reversible reactions and equilibrium in the course, then press Mark to check it there.

Tap the time on a concentration graph when equilibrium is reached.

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What changes Kc, and what does not

AQA's specification states that the value of the equilibrium constant is not affected by changes in concentration or by adding a catalyst, and asks students to predict the qualitative effect of temperature on Kc.

ChangeEffect on KcEffect on the position of equilibrium
ConcentrationNoneShifts to oppose the change, so Kc stays the same
CatalystNoneNone: equilibrium is reached faster
Temperature, exothermic forward reactionRaising T makes Kc smallerShifts to the left
Temperature, endothermic forward reactionRaising T makes Kc biggerShifts to the right

Only temperature changes Kc. This is why a Kc value is always quoted at a stated temperature. For example, if an exothermic reaction has Kc = 48 at one temperature, its Kc will be smaller than 48 at a higher temperature, and the equilibrium mixture will contain a smaller proportion of products.

Common mistakes

  • Using moles instead of concentrations when the moles on each side are not equal. Divide by the volume first.
  • Forgetting the powers. 2NH₃ gives [NH₃]², not 2[NH₃].
  • Products on the bottom. Products go on top.
  • Using starting amounts. Kc uses equilibrium concentrations. Build a table to find them.
  • Missing units, or giving units when there are none. Work them out from the expression each time.
  • Saying concentration or a catalyst changes Kc. Only temperature does.

Kc equation questions with answers

1. Write the Kc expression for 2NO₂(g) ⇌ N₂O₄(g).
Kc = [N₂O₄] ÷ [NO₂]²
2. Give the units of Kc for 2NO₂ ⇌ N₂O₄.
(mol dm⁻³) ÷ (mol dm⁻³)² = mol⁻¹ dm³
3. For A + B ⇌ 2C, [A] = 0.20, [B] = 0.10 and [C] = 0.40 mol dm⁻³ at equilibrium. Find Kc.
Kc = 0.40² ÷ (0.20 × 0.10) = 8.0 (no units)
4. The forward reaction of an equilibrium is exothermic. What happens to Kc if the temperature rises?
Kc decreases: the equilibrium shifts towards the reactants.
5. Does adding a catalyst change Kc?
No. A catalyst speeds up the forward and backward reactions by the same factor. The mixture reaches equilibrium sooner, but with the same composition and the same Kc.

Practise equilibria in a short timed paper

A-level Chemistry: Year 1 gives you a short timed paper every day, with 8 questions worth 25 to 35 marks across the course, so equilibrium calculations turn up regularly but not in every paper.

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Common questions

What is the Kc equation?
Kc = [products]^(their coefficients) ÷ [reactants]^(their coefficients), using equilibrium concentrations in mol dm⁻³.
How do you calculate Kc?
Find the equilibrium concentration of every species (using a table if you start from initial amounts), substitute them into the Kc expression, and work out the units.
What does a large Kc mean?
The equilibrium lies far to the right: at equilibrium there are many more products than reactants.
Does Kc have units?
Sometimes. It depends on the expression: if the powers on top and bottom add to the same total, the units cancel.
What affects the value of Kc?
Only temperature. Concentration, pressure and catalysts do not change Kc.
What is a homogeneous equilibrium?
One where all the reactants and products are in the same phase, such as all gases or all in solution.

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