Kc and Kp in A Level Chemistry, with Worked Steps

Kc is the equilibrium constant written with the equilibrium concentrations of the species, and Kp is the equilibrium constant written with the partial pressures of gases. Both follow the same pattern: products over reactants, each raised to the power of its coefficient in the balanced equation, at a fixed temperature. Kc applies to any equilibrium in solution or the gas phase; Kp is used for gas-phase equilibria.

This guide shows how to write each expression, work out its units and calculate it, with four worked examples. It also shows what each board asks for, because the depth differs between AQA, Cambridge International and Pearson Edexcel.

What each syllabus asks for

The full wording is in AQA's physical chemistry specification and on Cambridge's Chemistry 9701 page, which links to the current syllabus. Read your own board's version before an exam: it decides, for instance, whether you will meet heterogeneous equilibria at all.

Writing a Kc expression

Kc is the product of the equilibrium concentrations of the products, each raised to the power of its coefficient in the balanced equation, divided by the same product for the reactants. Square brackets mean the concentration at equilibrium, in mol dm⁻³. For the Haber process, N₂(g) + 3H₂(g) ⇌ 2NH₃(g):

Kc = [NH₃]² / ([N₂][H₂]³)

Three rules prevent most lost marks:

• Products go on top and reactants underneath, always for the equation exactly as written. Reverse the equation and the new Kc is the reciprocal of the old one.
• Each concentration is raised to the power of its coefficient in the balanced equation, not to a number taken from the data.
• Only equilibrium values go into the expression, never starting amounts.

Kp is written the same way, with the partial pressure of each gas, p(X), in place of its concentration: for the Haber process, Kp = p(NH₃)² / (p(N₂) × p(H₂)³).

Working out the units

Substitute mol dm⁻³ for each bracket and cancel. For the Haber process:

(mol dm⁻³)² / ((mol dm⁻³)(mol dm⁻³)³) = (mol dm⁻³)⁻² = mol⁻² dm⁶

If the powers on the top and the bottom add up to the same number, everything cancels and Kc has no units. The same method works for Kp with a pressure unit in place of mol dm⁻³.

Worked example 1: Kc when the volume cancels

Ethanoic acid reacts with ethanol to form ethyl ethanoate and water, all in the same liquid mixture:

CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O

A mixture of 0.500 mol of ethanoic acid and 0.800 mol of ethanol, with an acid catalyst, reaches equilibrium. It then contains 0.360 mol of ethyl ethanoate. Calculate Kc.

1. Write the expression: Kc = [CH₃COOC₂H₅][H₂O] / ([CH₃COOH][C₂H₅OH]).
2. Each concentration is the amount divided by the same total volume, V. With two species on top and two underneath, V cancels, so the equilibrium amounts can go straight in.
3. Kc = (0.360 × 0.360) / (0.140 × 0.440) = 0.1296 / 0.0616 = 2.10, with no units.

The shortcut in step 2 works only because the powers on the top and the bottom are equal. That is the trap in the next example.

Worked example 2: Kc when the volume does not cancel

For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), 0.80 mol of SO₂ and 0.50 mol of O₂ are sealed in a 2.00 dm³ vessel. At equilibrium the vessel contains 0.60 mol of SO₃. Calculate Kc.

1. Use the equation to find the changes: forming 0.60 mol of SO₃ uses 0.60 mol of SO₂ and 0.30 mol of O₂.
2. Equilibrium amounts: SO₂ = 0.80 − 0.60 = 0.20 mol; O₂ = 0.50 − 0.30 = 0.20 mol; SO₃ = 0.60 mol.
3. Convert to concentrations by dividing by 2.00 dm³: [SO₂] = 0.10, [O₂] = 0.10 and [SO₃] = 0.30 mol dm⁻³.
4. Kc = [SO₃]² / ([SO₂]²[O₂]) = 0.30² / (0.10² × 0.10) = 0.090 / 0.0010 = 90.
5. Units: (mol dm⁻³)² / ((mol dm⁻³)² × mol dm⁻³) = mol⁻¹ dm³.

Kc = 90 mol⁻¹ dm³. Putting the amounts in without dividing by the volume gives 45, half the correct value, because here three powers sit underneath and only two on top.

Worked example 3: Kp from mole fractions

Kp uses partial pressures. The partial pressure of a gas is its mole fraction multiplied by the total pressure, and the mole fraction is the amount of that gas divided by the total amount of gas.

For N₂O₄(g) ⇌ 2NO₂(g), 1.00 mol of N₂O₄ is heated, and at equilibrium 0.400 mol of it has dissociated. The total pressure is 200 kPa. Calculate Kp.

1. Equilibrium amounts: N₂O₄ = 1.00 − 0.400 = 0.600 mol; NO₂ = 2 × 0.400 = 0.800 mol; total = 1.400 mol.
2. Mole fractions: N₂O₄ = 0.600 / 1.400 = 0.4286; NO₂ = 0.800 / 1.400 = 0.5714. They should add up to 1, and they do.
3. Partial pressures: p(N₂O₄) = 0.4286 × 200 = 85.71 kPa; p(NO₂) = 0.5714 × 200 = 114.29 kPa.
4. Kp = p(NO₂)² / p(N₂O₄) = 114.29² / 85.71 = 152 kPa, to three significant figures.

Keep the unrounded values on your calculator until the last line; rounding the mole fractions to two figures early moves the answer. Pearson International A Level students should note that their specification uses partial pressures in atm: with a total pressure of 2.00 atm and the same mole fractions, Kp = 1.52 atm. The chemistry is identical; only the unit, and therefore the number, changes.

Worked example 4: using Kp to find a partial pressure

Exam questions also run the calculation backwards. Using illustrative figures, suppose that at a certain temperature Kp = 0.040 kPa⁻¹ for 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), with p(SO₂) = 20 kPa and p(O₂) = 50 kPa at equilibrium. Find p(SO₃).

1. Write the expression: Kp = p(SO₃)² / (p(SO₂)² × p(O₂)).
2. Rearrange: p(SO₃)² = Kp × p(SO₂)² × p(O₂) = 0.040 × 20² × 50 = 800 kPa².
3. Take the square root: p(SO₃) = 28.3 kPa.

Checking the units is a useful test of the rearrangement: kPa⁻¹ × kPa² × kPa gives kPa², and its square root is a pressure, as it should be.

What changes Kc and Kp, and what does not

Pearson's specification puts it plainly: the value of the equilibrium constant is not affected by changes in concentration or pressure, or by adding a catalyst, and AQA and Cambridge test the same idea. A catalyst only helps the system reach equilibrium faster. Temperature is the one factor that changes K:

• for an exothermic forward reaction, raising the temperature lowers Kc or Kp, and the position of equilibrium moves to the left;
• for an endothermic forward reaction, raising the temperature increases Kc or Kp, and the position moves to the right.

When the pressure on a gas equilibrium is increased, the position can shift, as Le Chatelier's principle predicts, but the new partial pressures still give the same Kp at the same temperature.

Converting between Kc and Kp

For gas equilibria the two constants are linked: Kp equals Kc multiplied by RT raised to the power Δn, where Δn is the change in the number of moles of gas from left to right, and the units of pressure, concentration and R must be consistent. None of these three syllabuses asks you to use the relationship: Cambridge 9701 says so explicitly, and AQA and Pearson do not list it. In the exam, answer in the constant the question asks for, using the data it gives.

Five mistakes that cost marks

• Using starting amounts instead of equilibrium amounts.
• Leaving out the volume when the powers on the top and the bottom differ.
• Using the total pressure in place of a partial pressure.
• Giving no units, or units copied from a different expression.
• Rounding at every step instead of at the end, then quoting too many or too few significant figures.

Equilibrium calculations reward a fixed layout: equation, expression, amounts, concentrations or partial pressures, substitution, units. If you would like to practise it with your own board's past papers, our A Level Chemistry tutoring page shows how each board examines the subject, and the A Level tutoring overview lists the other A Level subjects.

Frequently asked questions