AQA A-Level Chemistry: Equilibrium constant Kp for homogeneous systems (7405) – Revision Notes

For full explanations and longer worked examples, read the Kp study guide first. These notes are for fast recall in the final weeks.

They follow section 3.1.10 of the AQA AS and A-level Chemistry specification (7404/7405), version 1.2 (July 2026), for AS and A-level exams from June 2016 onwards. The whole section is A-level only: it is not in the AS course. The A-level assessment overview lists section 3.1.10 under Paper 1, and Paper 3 can test any content. There is no required practical in this section.

Use the AQA A-Level Chemistry hub for every topic and the printable checklist to track progress. Kp builds on Kc and Le Chatelier’s principle, condensed in the Kc revision notes.

Definitions (3.1.10)

• Kp: the equilibrium constant calculated from partial pressures for a system at constant temperature. It is deduced from the equation for a reversible reaction in the gas phase.
• Homogeneous system: all species are in the same phase. For Kp in this section, all species are gases.
• Mole fraction, x(A): moles of A divided by the total moles of gas in the mixture. No units.
• Partial pressure, p(A): the pressure gas A would exert if it alone filled the container. Same unit as the total pressure.

Formulas

Quantity	Formula	Notes
Mole fraction	x(A) = n(A) / n(total)	all mole fractions add up to 1
Partial pressure	p(A) = x(A) × P(total)	all partial pressures add up to P(total)
Kp for aA + bB ⇌ cC + dD	Kp = p(C)^c × p(D)^d / (p(A)^a × p(B)^b)	equilibrium values only
Units of Kp	(pressure unit)^Δn	Δn = gas moles on right − gas moles on left

Unit patterns

Δn (right − left)	Units with kPa	Example
0	none	H₂(g) + Cl₂(g) ⇌ 2HCl(g)
+1	kPa	N₂O₄(g) ⇌ 2NO₂(g)
+2	kPa²	CH₄(g) + H₂O(g) ⇌ CO(g) + 3H₂(g)
−1	kPa⁻¹	2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
−2	kPa⁻²	N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

If the question uses Pa or MPa, the same powers apply to that unit.

Method in steps: Kp from initial moles

1. Write the balanced equation.
2. Draw a moles table: initial, change, equilibrium.
3. Fill in the change row using the equation’s mole ratios.
4. Add up the equilibrium moles of all gases.
5. Mole fraction of each gas = its equilibrium moles ÷ total.
6. Partial pressure = mole fraction × total pressure.
7. Substitute into Kp, then work out the units.
8. Round only at the end, to the s.f. of the least accurate data.

Small worked reminder

An equilibrium mixture contains 0.900 mol A, 0.300 mol B and 0.600 mol C at a total pressure of 180 kPa.

total = 0.900 + 0.300 + 0.600 = 1.800 mol
x(A) = 0.500   p(A) = 90.0 kPa
x(B) = 0.167   p(B) = 30.0 kPa   (0.300/1.800 x 180 = 30.0)
x(C) = 0.333   p(C) = 60.0 kPa
check: 90.0 + 30.0 + 60.0 = 180 kPa

Tip: n(A)/n(total) × P can be done in one step on the calculator. This keeps full accuracy.

Method in steps: finding an unknown

1. Write the Kp expression.
2. Rearrange so the unknown partial pressure is on its own.
3. Substitute the known partial pressures and Kp.
4. Take a square root if the unknown is squared.
5. If asked for the total pressure, add all the partial pressures.

Effects of changing conditions

Change	Position of equilibrium	Value of Kp
Increase total pressure (constant T)	shifts to side with fewer moles of gas	no change
Decrease total pressure (constant T)	shifts to side with more moles of gas	no change
Equal gas moles on both sides, change pressure	no change	no change
Increase T, forward exothermic	shifts left	decreases
Increase T, forward endothermic	shifts right	increases
Add a catalyst	no change; equilibrium reached faster	no change

Why pressure moves the position but not Kp. Each p = x × P. For a reaction where Δn is not zero, a factor of P^Δn appears in the expression. When P changes, the mole fractions must change to keep Kp constant. That change in mole fractions is the shift in position.

Working backwards. If Kp falls as temperature rises, the forward reaction is exothermic. If Kp rises as temperature rises, it is endothermic.

Must-know distinctions

• Position vs value. Pressure and catalysts can never change Kp. Pressure can change the position; temperature changes both.
• Mole fraction vs partial pressure. Mole fraction has no units and is at most 1. Partial pressure has pressure units.
• Kp vs Kc. Kp uses partial pressures p(X); Kc uses concentrations [X] in mol dm⁻³. Never put square brackets into a Kp expression.
• Rate vs equilibrium. A catalyst changes how fast equilibrium is reached, not where it lies.
• Initial vs equilibrium moles. Only equilibrium moles go into the mole fractions.

Quick self-test

1. A gas mixture contains 0.600 mol of oxygen in 2.40 mol of gas. The total pressure is 80.0 kPa. Calculate the mole fraction and partial pressure of oxygen.
2. What do the partial pressures in a gas mixture add up to?
3. Write the Kp expression for CH₄(g) + H₂O(g) ⇌ CO(g) + 3H₂(g) and give its units in kPa.
4. For N₂O₄(g) ⇌ 2NO₂(g), p(N₂O₄) = 40.0 kPa and p(NO₂) = 60.0 kPa at equilibrium. Calculate Kp.
5. For H₂(g) + Cl₂(g) ⇌ 2HCl(g), p(H₂) = 25.0 kPa, p(Cl₂) = 40.0 kPa and p(HCl) = 150 kPa at equilibrium. Calculate Kp.
6. For N₂O₄(g) ⇌ 2NO₂(g) at the same temperature as question 4, Kp = 90.0 kPa and p(N₂O₄) = 10.0 kPa. Calculate p(NO₂).
7. The total pressure on the equilibrium in question 4 is increased at constant temperature. State the effect on the position of equilibrium and on Kp.
8. The forward reaction of 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) is exothermic. State the effect of raising the temperature on Kp.
9. State the effect of adding a catalyst on the value of Kp.
10. Data are given to 2 and 3 significant figures. To how many significant figures should Kp be quoted?
11. For a gas-phase reaction, Kp is 4.5 × 10⁻³ kPa⁻¹ at 300 K and 2.0 × 10⁻⁴ kPa⁻¹ at 400 K. Is the forward reaction exothermic or endothermic?
12. Give two quick checks you can apply to a set of mole fractions and partial pressures.

Answers

1. x(O₂) = 0.600 / 2.40 = 0.250; p(O₂) = 0.250 × 80.0 = 20.0 kPa.
2. The total pressure of the mixture.
3. Kp = p(CO) × p(H₂)³ / (p(CH₄) × p(H₂O)); units kPa² (Δn = 4 − 2 = +2).
4. Kp = 60.0² / 40.0 = 90.0 kPa.
5. Kp = 150² / (25.0 × 40.0) = 22 500 / 1000 = 22.5 (no units).
6. p(NO₂)² = 90.0 × 10.0 = 900, so p(NO₂) = 30.0 kPa.
7. Position shifts left, to the side with fewer moles of gas; Kp unchanged.
8. Kp decreases (the equilibrium shifts left, in the endothermic direction).
9. No effect; equilibrium is just reached faster.
10. 2 significant figures, the limit of the least accurate data.
11. Kp falls as the temperature rises, so the equilibrium shifts left on heating. The reverse reaction is endothermic, so the forward reaction is exothermic.
12. The mole fractions add up to 1, and the partial pressures add up to the total pressure.

Where marks are usually lost

• Using initial moles, not equilibrium moles, to find the total moles of gas.
• Forgetting that a product which starts at zero still counts in the total at equilibrium.
• Missing the power on a partial pressure, such as p(H₂)³.
• Writing [X] instead of p(X) inside a Kp expression.
• Wrong or missing units, or giving units when Δn = 0.
• Mixing Pa, kPa and MPa within one calculation.
• Rounding partial pressures early, so the final Kp drifts in the third figure.
• Stating that higher pressure increases Kp. Only temperature changes Kp.
• Saying a catalyst shifts the equilibrium or increases the yield.
• Explaining a temperature effect without naming the endothermic or exothermic direction.
• Forgetting to square-root at the end when the unknown partial pressure is squared in the expression.

Next steps

Test yourself fully with the Kp practice questions, and check paper formats in the exam preparation guide. For a quick check of weaker areas, try the free diagnostics.

Official syllabus

AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards, published by AQA. Section 3.1.10: Equilibrium constant Kp for homogeneous systems (A-level only).