OCR A Level Chemistry A: How far? (H432) – Revision Notes

These revision notes cover section 5.1.2, How far?, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. You will find every lettered outcome, 5.1.2(a) to (h), from this second section of Module 5 (Physical chemistry and transition elements). The content belongs to the full A level, and the specification lists Module 5 as assessed in components 01 and 03. If any point here feels thin, go back to the How far? study guide, which explains each idea in full with longer examples.

Course links: all H432 sections, outcome checklist to print, practice questions on this section and our free diagnostics.

Key terms, 5.1.2(a)

• Mole fraction: x(A) = moles of A ÷ total moles of gas present. It has no units, and the values for one mixture sum to 1.
• Partial pressure: x(A) multiplied by the total pressure. Summing every gas’s partial pressure gives back the total pressure.
• Homogeneous: every species in the equation shares one phase.
• Heterogeneous: two or more phases are present.
• Kc: built from equilibrium concentrations in mol dm⁻³.
• Kp: built from equilibrium partial pressures; gases only.

Expressions and units, 5.1.2(d) and (e)

Item	Rule
Kc for aA + bB ⇌ cC + dD	[C]ᶜ[D]ᵈ / ([A]ᵃ[B]ᵇ)
Kp for the same, all gases	p(C)ᶜ p(D)ᵈ / (p(A)ᵃ p(B)ᵇ)
Heterogeneous	Leave out pure solids and pure liquids (their concentration is constant)
Units	Substitute units for every term and cancel
n = moles of terms on top − moles below	n = 0: none; n = 1: mol dm⁻³ or kPa; n = −1: dm³ mol⁻¹ or kPa⁻¹; n = 2: mol² dm⁻⁶ or kPa²
Kp pressure units	kPa, Pa or atm; use whichever the data give and keep it consistent

Examples of heterogeneous expressions:

• CaO(s) + SO₂(g) ⇌ CaSO₃(s): Kp = 1 / p(SO₂), units kPa⁻¹.
• Fe(s) + 2Fe³⁺(aq) ⇌ 3Fe²⁺(aq): Kc = [Fe²⁺]³ / [Fe³⁺]², units mol dm⁻³.

Method in steps: Kc from starting amounts, 5.1.2(b) and (e)

1. Write the balanced equation.
2. Draw an initial / change / equilibrium table in moles.
3. Use the mole ratio for the change row.
4. Divide each equilibrium amount by the volume.
5. Substitute into Kc; work out units.

Worked reminder. J(g) + 3K(g) ⇌ 2L(g) in 3.00 dm³. At equilibrium: 0.30 mol J, 0.90 mol K, 0.60 mol L. Concentrations: 0.100, 0.300 and 0.200 mol dm⁻³. Kc = 0.200² / (0.100 × 0.300³) = 14.8 dm⁶ mol⁻² (n = 2 − 4 = −2).

Method in steps: Kp, 5.1.2(a) and (e)

1. Find equilibrium amounts (table as above).
2. Add them to get the total amount of gas only.
3. Mole fraction of each gas.
4. Multiply by total pressure to get each partial pressure.
5. Substitute into Kp; work out units.

Worked reminder. 3M(g) ⇌ N(g). At equilibrium 0.60 mol M and 0.20 mol N; total pressure 200 kPa. x(M) = 0.60/0.80 = 0.750, so p(M) = 150 kPa; p(N) = 50.0 kPa. Kp = 50.0 / 150³ = 1.48 × 10⁻⁵ kPa⁻² (n = 1 − 3 = −2).

Finding equilibrium quantities by experiment, 5.1.2(c)

For Kc only (the specification says not for Kp). Typical esterification or ester hydrolysis method:

• Seal known amounts plus a known amount of acid catalyst in a flask at constant temperature.
• Wait until equilibrium is reached (repeat analysis later gives the same result).
• Quench in ice-cold water: cooling and dilution slow the reaction so the composition stays fixed during analysis.
• Titrate the acid with standard NaOH; subtract the titre for the catalyst alone.
• Use the equation to deduce the other equilibrium amounts.

What changes K, 5.1.2(f)

Change	Effect on K	Effect on position
Raise temperature, exothermic forward	K decreases	Shifts left
Raise temperature, endothermic forward	K increases	Shifts right
Add a reactant	None	Shifts right
Increase total pressure	None	Shifts towards fewer gas moles
Add a catalyst	None	None; equilibrium reached sooner

Method in steps: reading K against temperature, 5.1.2(f)

1. Check that every K value in the data is quoted at a stated temperature. A value of K means nothing without one.
2. See which way K moves as the temperature rises.
3. K rises with temperature: the forward reaction is endothermic.
4. K falls with temperature: the forward reaction is exothermic.
5. Link to yield: a larger K means a greater proportion of products at equilibrium.

Worked reminder. For an equilibrium, Kc = 4.6 at 400 K and 1.9 at 500 K. Kc falls as temperature rises, so the forward reaction is exothermic, and the equilibrium yield of products is lower at 500 K. A catalyst at 500 K would give the same Kc of 1.9; it would just reach it sooner.

How K controls position, 5.1.2(g)

Work out the same expression with the new, non-equilibrium values (the quotient) and compare with K:

• Quotient < K: reaction goes forwards until they match.
• Quotient > K: reaction goes backwards until they match.
• Concentration or pressure change: K fixed, quotient moves away from it, position shifts.
• Temperature change: K moves, the old mixture no longer matches, position shifts.

Pressure reminder. Doubling total pressure doubles every partial pressure. If the top of Kp has two more gas moles than the bottom, the quotient becomes 2² = 4 times Kp, so the mixture shifts left.

Other constants, 5.1.2(h)

The same rules carry over to Ka (5.1.3) and any other K:

• Write it from the equation; leave out water as solvent.
• Only temperature changes it.
• HA ⇌ H⁺ + A⁻, Ka = [H⁺][A⁻]/[HA]: diluting lowers the quotient below Ka, so more HA dissociates.

Must-know distinctions

• Position of equilibrium vs value of K. Concentration and pressure changes can move the position. A catalyst moves neither. Only temperature changes K.
• Kc vs Kp. Kc uses concentrations in square brackets; Kp uses p( ) and applies to gases.
• Homogeneous vs heterogeneous. In a homogeneous liquid mixture, such as an esterification, water is included. Water is left out only as a pure liquid phase or as the solvent.
• Mole fraction vs partial pressure. A mole fraction is a pure number; a partial pressure carries pressure units.

Quick self-test

1. A gas mixture contains 0.20 mol A, 0.30 mol B and 0.50 mol C at 300 kPa. Find p(C).
2. Ethanoic acid vapour dimerises: 2CH₃COOH(g) ⇌ (CH₃COOH)₂(g). At equilibrium p(CH₃COOH) = 40.0 kPa and p((CH₃COOH)₂) = 60.0 kPa. Calculate Kp with units.
3. Give the units of Kc for 2NO(g) + 2CO(g) ⇌ N₂(g) + 2CO₂(g).
4. Sodium sulfate decahydrate loses water to dry air: Na₂SO₄·10H₂O(s) ⇌ Na₂SO₄(s) + 10H₂O(g). Write Kp and state its units.
5. For X(g) ⇌ Y(g) + Z(g), Kc = 0.0450 mol dm⁻³. At equilibrium [Y] = [Z] = 0.0600 mol dm⁻³. Find [X].
6. P(g) + Q(g) ⇌ 2R(g) in 0.500 dm³: 0.40 mol P, 0.10 mol Q, 0.24 mol R at equilibrium. Find Kc.
7. The forward reaction is exothermic. What happens to Kp when the temperature rises?
8. What does adding a catalyst do to Kc?
9. Why is an equilibrium mixture poured into ice-cold water before titration?
10. Cl₂(g) + F₂(g) ⇌ 2ClF(g): what happens to the position when the total pressure is increased? Explain using Kp.
11. A solution of a weak acid is diluted. What happens to Ka and to the extent of dissociation?

Answers

1. Total 1.00 mol, x(C) = 0.50, p(C) = 150 kPa.
2. Kp = 60.0 / 40.0² = 0.0375 kPa⁻¹.
3. n = (1 + 2) − (2 + 2) = −1, so dm³ mol⁻¹.
4. Kp = p(H₂O)¹⁰, units kPa¹⁰ (both solids left out).
5. The equilibrium concentration is [X] = 0.0600² / 0.0450 = 0.0800 mol dm⁻³.
6. Concentrations 0.80, 0.20, 0.48 mol dm⁻³; Kc = 0.48² / (0.80 × 0.20) = 1.44, no units.
7. Kp decreases.
8. Nothing; it only shortens the time to reach equilibrium.
9. Cooling and dilution slow the reaction, so the composition does not shift while acid is removed by the titration.
10. No change. Two gas moles on each side, so every term scales equally; the quotient still equals Kp.
11. Ka is unchanged; the quotient falls below Ka, so the extent of dissociation increases.

Where marks are usually lost

• Using initial amounts, not equilibrium amounts, in the expression.
• Forgetting to divide by volume when the number of moles on each side differs.
• Adding the amount of a solid into the total moles when finding mole fractions.
• Rounding mole fractions early, then giving Kp to too few significant figures.
• Including solids or a pure liquid in a heterogeneous Kc or Kp.
• Writing units by guesswork; substitute and cancel every time.
• Claiming Kc changes when a concentration or pressure changes.
• Explaining a pressure effect only with le Chatelier when the question says “in terms of Kp”.
• Forgetting to subtract the catalyst when using titration data.

Official syllabus

Source document: OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017 (OCR). These notes follow its section 5.1.2, How far?, from Module 5.