OxfordAQA A-Level Chemistry: Equilibrium constant Kp for homogeneous systems (9620) – Revision Notes

This page condenses the Kp study guide, which has the full teaching and longer examples. Use these notes for fast recall close to the exam.

They follow section 3.1.12 (subsection 3.1.12.1) of the OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards. Being International A2 physical chemistry, the whole section is International A-level only. Unit 4 (Organic 2 and Physical 2) examines it; Unit 5 (Practical and synoptic) covers the whole specification, so it may appear there too. The section names no required practical.

Other 9620 topics are on the course hub, and a checklist to print keeps track of outcomes. Check weaker areas with the free 10-minute diagnostics, then drill with the Kp practice questions. The AS groundwork on Kc and Le Chatelier’s principle has its own summary page.

Definitions (3.1.12.1)

• Kp: the equilibrium constant worked out from partial pressures, for a system held at constant temperature. You deduce its expression from the equation of a gas-phase reversible reaction.
• Homogeneous system: every species is in the same phase. Here, all are gases, so all appear in Kp.
• Mole fraction: amount of one gas ÷ total amount of gas. No units; never more than 1.
• Partial pressure: the pressure that one gas would produce on its own in the same container. Its unit matches the total pressure’s.

Key relationships

What	Relationship	Check or note
Mole fraction of A	n(A) ÷ n(total)	all mole fractions sum to 1
Partial pressure of A	mole fraction of A × total pressure	all partial pressures sum to P
Kp for wW + xX ⇌ yY + zZ	p(Y)^y × p(Z)^z ÷ (p(W)^w × p(X)^x)	use equilibrium partial pressures
Kp units	(pressure unit)^Δn	Δn = product gas moles − reactant gas moles

Units at a glance

Δn	Example equation	Units if pressures are in kPa
+3	2N₂O₅(g) ⇌ 4NO₂(g) + O₂(g)	kPa³
+1	2NO₂(g) ⇌ 2NO(g) + O₂(g)	kPa
0	butane(g) ⇌ 2-methylpropane(g)	none
−1	2CO(g) + O₂(g) ⇌ 2CO₂(g)	kPa⁻¹
−2	CO₂(g) + 4H₂(g) ⇌ CH₄(g) + 2H₂O(g)	kPa⁻²

Changing the pressure unit. 1 kPa = 1000 Pa. Multiply a Kp value by 1000^Δn to go from kPa-based to Pa-based units. For Δn = −2, a Kp of 3.30 × 10⁻⁵ kPa⁻² becomes 3.30 × 10⁻¹¹ Pa⁻².

Method in steps: Kp from starting amounts

1. Balanced equation first.
2. Table of amounts: initial, change, equilibrium (all in mol).
3. Change row from the mole ratio in the equation.
4. Total every gas at equilibrium, including any that started at zero.
5. Mole fraction of each gas, then partial pressure (fraction × P).
6. Substitute into Kp. Work out Δn for the units.
7. Round only at the end, to the significant figures the data support.

Shortcut that keeps accuracy: key (moles ÷ total) × P in one go for each gas.

Method in steps: an unknown from Kp

1. Write Kp in terms of partial pressures.
2. Rearrange so the unknown stands alone.
3. Substitute, then take a root if the unknown carries a power.
4. If the question then asks for a total pressure, add all the partial pressures.

When Δn = 0 you can work in moles directly: the total pressure and the total amount cancel. Use this to save time, but say so in your working so the method is clear.

Condition changes at a glance

You change	Equilibrium position	Kp
Raise total pressure (constant T)	towards fewer moles of gas	unchanged
Lower total pressure (constant T)	towards more moles of gas	unchanged
Pressure change, Δn = 0	no shift	unchanged
Raise T, forward reaction exothermic	moves left	decreases
Raise T, forward reaction endothermic	moves right	increases
Add a catalyst	no shift; equilibrium reached sooner	unchanged

Why pressure shifts the position but not Kp: substituting (mole fraction × P) for every p leaves P^Δn multiplying the mole-fraction terms. A new P therefore forces new mole fractions if Kp is to stay the same, and that adjustment is the shift.

Working from data: a Kp that grows on heating means an endothermic forward reaction; a Kp that shrinks on heating means an exothermic one.

Explaining a temperature effect in full: state the direction the equilibrium moves (towards the endothermic side), say whether that raises or lowers the product partial pressures relative to the reactant ones, then state what happens to Kp. For an exothermic forward reaction heated up, that reads: moves left, product partial pressures fall, Kp decreases.

Must-know distinctions

• Position vs value. Pressure can change the position, never Kp. Temperature changes both.
• Rate vs extent. A catalyst changes how quickly equilibrium is reached, not where it lies.
• Kp vs Kc. Kp uses p(X) from partial pressures; Kc uses [X] in mol dm⁻³.
• Mole fraction vs partial pressure. One is a unitless fraction; the other is a pressure.
• Initial vs equilibrium amounts. Only equilibrium amounts give mole fractions for Kp.

Small worked reminder

Equilibrium partial pressures for 2NO₂(g) ⇌ 2NO(g) + O₂(g) at a fixed temperature are listed below.

p(NO2) = 60.0 kPa   p(NO) = 30.0 kPa   p(O2) = 15.0 kPa

Kp = p(NO)^2 x p(O2) / p(NO2)^2
   = 30.0^2 x 15.0 / 60.0^2
   = 13 500 / 3600 = 3.75 kPa       (Δn = 3 − 2 = +1)

Quick self-test

1. A gas mixture holds 0.270 mol of one gas in 0.900 mol of gas in total, at 250 kPa. Calculate the mole fraction and the partial pressure of that gas.
2. Write the Kp expression for 2N₂O₅(g) ⇌ 4NO₂(g) + O₂(g) and give its units in MPa.
3. In the worked reminder above, what would Kp be if every pressure were given in Pa?
4. At the same temperature as the worked reminder, p(NO₂) = 40.0 kPa and p(O₂) = 15.0 kPa at equilibrium. Calculate p(NO).
5. An equilibrium mixture for butane(g) ⇌ 2-methylpropane(g) contains 0.300 mol butane and 0.750 mol 2-methylpropane at 120 kPa. Calculate Kp.
6. The total pressure on the mixture in question 5 is doubled at constant temperature. What happens to the position of equilibrium?
7. The total pressure on 2N₂O₅(g) ⇌ 4NO₂(g) + O₂(g) is raised while the temperature stays fixed. What happens to the equilibrium position, and what happens to Kp?
8. For a gas-phase reaction, Kp increases when the temperature increases. Is the forward reaction exothermic or endothermic?
9. State the effect of a catalyst on the time to reach equilibrium and on Kp.
10. Three gases in an equilibrium mixture have partial pressures of 30.0, 45.0 and 75.0 kPa. Give the total pressure and the mole fraction of the third gas.

Answers

1. Mole fraction = 0.270 ÷ 0.900 = 0.300; partial pressure = 0.300 × 250 = 75.0 kPa.
2. Kp = p(NO₂)⁴ × p(O₂) ÷ p(N₂O₅)²; Δn = 5 − 2 = +3, so the units are MPa³.
3. Δn = +1, so multiply by 1000: 3.75 × 10³ Pa.
4. p(NO)² = Kp × p(NO₂)² ÷ p(O₂) = 3.75 × 1600 ÷ 15.0 = 400, so p(NO) = 20.0 kPa.
5. Δn = 0, so Kp = 0.750 ÷ 0.300 = 2.50, with no units. The total pressure is not needed.
6. No shift: there is 1 mole of gas on each side, so pressure cancels from Kp.
7. The position moves left, towards the 2 moles of gas on the reactant side rather than the 5 on the product side. Kp is unchanged.
8. Endothermic: a rise in temperature moved the equilibrium to the right, the endothermic direction.
9. Equilibrium is reached in less time; Kp is unchanged.
10. Total = 150 kPa; mole fraction = 75.0 ÷ 150 = 0.500.

Where marks are usually lost

• Using the starting amounts, not the equilibrium amounts, to find the total moles of gas.
• Missing a product that began at zero when adding up the equilibrium total.
• Using a 1 : 1 change when the equation’s ratio is 1 : 4 or 2 : 1.
• Writing square brackets inside a Kp expression.
• Missing a power, such as the 4 on p(NO₂) in the N₂O₅ expression.
• Units with the wrong sign of power when Δn is negative, or any units at all when Δn = 0.
• Mixing Pa and kPa in one substitution.
• Forgetting the square root when the unknown partial pressure is squared.
• Saying Kp changes with pressure, or that a catalyst increases the equilibrium yield.
• Explaining a temperature change without naming the endothermic or exothermic direction.

Official syllabus

OxfordAQA, International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards: section 3.1.12 Equilibrium constant Kp for homogeneous systems (International A2), subsection 3.1.12.1.