OCR A Level Chemistry A: How far? (H432)

This study guide teaches 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. As the second section of Module 5 (Physical chemistry and transition elements), it lists eight lettered learning outcomes, 5.1.2(a) to (h); each is taught below with worked examples. The content belongs to the full A level, H432. In the written exams, Module 5 content can appear on paper 01, Periodic table, elements and physical chemistry, and on paper 03, Unified chemistry, the paper that spans every module. The specification links no PAG to 5.1.2.

It builds on 3.2.3, Chemical equilibrium, where you met dynamic equilibrium, le Chatelier’s principle and Kc without units. Other H432 sections are listed on our course page for H432 Chemistry A, and each outcome can be ticked off on an H432 checklist you can print. Condensed How far? revision notes help with recall, and the 5.1.2 practice set tests it. Find your gaps with a free diagnostic.

Outcomes in 5.1.2

Outcome	Skill you need	Additional guidance in the specification
5.1.2(a)	Use the terms mole fraction and partial pressure	See also 3.2.3
5.1.2(b)	Calculate quantities present at equilibrium from given data	M0.2
5.1.2(c)	Describe techniques and procedures to find quantities present at equilibrium	Not for Kp; HSW4
5.1.2(d)	Write Kc and Kp expressions for homogeneous and heterogeneous equilibria	Liquid and solid concentrations are constant and are omitted in heterogeneous Kc and Kp expressions
5.1.2(e)	Calculate Kc and Kp, or related quantities, including their units	You will not be required to solve quadratic equations
5.1.2(f)	(i) Effect of temperature on K for exothermic and endothermic reactions; (ii) K stays constant when concentration, pressure or a catalyst changes	M0.3
5.1.2(g)	Explain how K controls the position of equilibrium when concentration, pressure or temperature change	M0.3
5.1.2(h)	Apply the same principles to other equilibrium constants, such as Ka in 5.1.3	Where appropriate

5.1.2(a): Mole fraction and partial pressure

For a mixture of gases:

• Mole fraction of A, written x(A) = (amount of A) / (total amount of gas). Mole fractions have no units and add up to 1.
• Partial pressure of A, written p(A) = x(A) × total pressure, P. It is the pressure A would exert if it alone filled the container.

The partial pressures of all the gases add up to the total pressure. This checks your arithmetic.

Worked example 1: partial pressures

Cyclohexane vapour dehydrogenates over a catalyst at a fixed temperature:

C₆H₁₂(g) ⇌ C₆H₆(g) + 3H₂(g)

A sample starting as 0.400 mol C₆H₁₂ reaches equilibrium after 0.250 mol has reacted. The total pressure is 115 kPa. Find each partial pressure.

1. Equilibrium amounts: C₆H₁₂ = 0.400 − 0.250 = 0.150 mol; C₆H₆ = 0.250 mol; H₂ = 3 × 0.250 = 0.750 mol.
2. Total amount = 0.150 + 0.250 + 0.750 = 1.150 mol.
3. Mole fractions: x(C₆H₁₂) = 0.150/1.150 = 0.130; x(C₆H₆) = 0.217; x(H₂) = 0.652.
4. Partial pressures (use unrounded fractions): p(C₆H₁₂) = 15.0 kPa; p(C₆H₆) = 25.0 kPa; p(H₂) = 75.0 kPa.
5. Check: 15.0 + 25.0 + 75.0 = 115 kPa.

5.1.2(b): Quantities present at equilibrium

Most problems give you starting amounts and one equilibrium amount. A three-row table keeps the working clear: initial, change, equilibrium. Fill in the change row using the coefficients of the balanced equation. Then convert to whatever K needs: concentrations (divide by volume) for Kc, or partial pressures for Kp.

Worked example 2: an esterification

A student mixes 0.500 mol propanoic acid with 0.600 mol methanol and a few drops of acid catalyst. No ester or water is present at the start. At equilibrium, 0.180 mol of propanoic acid remains.

CH₃CH₂COOH(l) + CH₃OH(l) ⇌ CH₃CH₂COOCH₃(l) + H₂O(l)

	Acid	Methanol	Ester	Water
Initial / mol	0.500	0.600	0	0
Change / mol	−0.320	−0.320	+0.320	+0.320
Equilibrium / mol	0.180	0.280	0.320	0.320

Kc = [ester][water] / ([acid][methanol]). Every concentration equals amount ÷ V. With two terms in the numerator and two in the denominator, V cancels out:

Kc = (0.320 × 0.320) / (0.180 × 0.280) = 2.03 (no units)

Water is included: it is mixed into one liquid phase, not present as a separate pure liquid, so this equilibrium is homogeneous.

5.1.2(c): Finding equilibrium quantities by experiment

The specification asks for the techniques for Kc, not Kp. A typical method for an esterification or ester hydrolysis:

1. Mix known amounts of the reagents with a small, known amount of acid catalyst in a sealed flask.
2. Leave the flask at a constant temperature for long enough to reach equilibrium, often several days.
3. Quench: pour the mixture into a large volume of ice-cold water. Cooling and dilution slow the reaction so much that the composition is effectively frozen while you analyse it.
4. Titrate the acid present with standard sodium hydroxide, using phenolphthalein.
5. Run a separate titration of the catalyst alone and subtract it. What remains is the organic acid at equilibrium.
6. Deduce the other equilibrium amounts from the equation, then Kc.

Why quench? The titration removes acid. Without quenching, le Chatelier’s principle says the equilibrium would shift to replace it, and you would measure too much. A second flask analysed a day later should give the same titre if equilibrium was reached. Titration technique is revised in this H432 practical page.

5.1.2(d): Writing Kc and Kp expressions

For aA + bB ⇌ cC + dD (all in one phase):

• Kc = [C]ᶜ[D]ᵈ / ([A]ᵃ[B]ᵇ), with square brackets meaning equilibrium concentration in mol dm⁻³.
• Kp = p(C)ᶜ p(D)ᵈ / (p(A)ᵃ p(B)ᵇ), for gases only.

Products go on top; each term is raised to its coefficient.

Homogeneous means every species is in the same phase. Heterogeneous means more than one phase is present. The specification’s note is the key rule: the concentration of a pure solid or pure liquid is constant, so it is left out of both Kc and Kp.

Equilibrium	Kc	Kp
H₂O(l) ⇌ H₂O(g)	[H₂O(g)]	p(H₂O)
Ni(s) + 4CO(g) ⇌ Ni(CO)₄(g)	[Ni(CO)₄] / [CO]⁴	p(Ni(CO)₄) / p(CO)⁴

Worked example 3: a heterogeneous Kp

For Ni(s) + 4CO(g) ⇌ Ni(CO)₄(g) at one temperature, the gas phase at equilibrium holds 0.200 mol Ni(CO)₄ and 0.800 mol CO. The total pressure is 50.0 kPa.

• Only gases count in the total: 1.000 mol. So x(Ni(CO)₄) = 0.200 and x(CO) = 0.800.
• p(Ni(CO)₄) = 0.200 × 50.0 = 10.0 kPa; p(CO) = 40.0 kPa.
• Kp = 10.0 / 40.0⁴ = 3.91 × 10⁻⁶ kPa⁻³ (kPa ÷ kPa⁴).

Nickel does not appear. Adding more nickel does not move this equilibrium.

5.1.2(e): Calculating K and its units

Units method. Replace each term with its unit (mol dm⁻³ for Kc, kPa or Pa or atm for Kp), then cancel. Equivalently, count (moles of terms on top) − (moles of terms below) = n:

• n = 0: no units.
• n = 1: mol dm⁻³ (or kPa).
• n = −2: dm⁶ mol⁻² (or kPa⁻²).

Worked example 4: Kc with units

Ethanol vapour is heated in a sealed 2.50 dm³ vessel:

C₂H₅OH(g) ⇌ CH₃CHO(g) + H₂(g)

The vessel starts with 1.20 mol ethanol. At equilibrium there is 0.450 mol H₂.

1. Equilibrium amounts: CH₃CHO = 0.450 mol; C₂H₅OH = 1.20 − 0.450 = 0.750 mol.
2. Concentrations: [C₂H₅OH] = 0.300; [CH₃CHO] = 0.180; [H₂] = 0.180 mol dm⁻³.
3. Kc = (0.180 × 0.180) / 0.300 = 0.108 mol dm⁻³.
4. Units: (mol dm⁻³ × mol dm⁻³) / mol dm⁻³ = mol dm⁻³.

Worked example 5: Kp with units

Using worked example 1, Kp = p(C₆H₆) × p(H₂)³ / p(C₆H₁₂) = 25.0 × 75.0³ / 15.0 = 703 125, so Kp = 7.03 × 10⁵ kPa³. Units: kPa × kPa³ / kPa = kPa³.

Related quantities

“Related quantities” means rearranging K to find something else. If Kc for the ethanol reaction is still 0.108 mol dm⁻³ at the same temperature, and a new equilibrium mixture has [C₂H₅OH] = 0.500 and [CH₃CHO] = 0.150 mol dm⁻³, then [H₂] = 0.108 × 0.500 / 0.150 = 0.360 mol dm⁻³. No quadratics are required, so expect data that allow a direct rearrangement like this.

5.1.2(f): What changes K, and what does not

(i) Temperature is the only factor that changes K.

• Exothermic forward reaction: raising the temperature decreases K.
• Endothermic forward reaction: raising the temperature increases K.

You can run this backwards. If a table shows Kp rising as temperature rises, the forward reaction is endothermic.

(ii) K is constant when you change concentration, pressure, or add a catalyst (at constant temperature). A catalyst raises the rates of both directions by the same factor. Equilibrium arrives sooner; the final composition and the value of K do not change.

5.1.2(g): How K controls the position of equilibrium

Think of the expression as a ratio the mixture must return to. Call the value of the same expression, worked out with the current (non-equilibrium) values, the quotient.

• Concentration. Add a reactant: the denominator grows, so the quotient falls below Kc. The system reacts forwards, making more products, until the ratio equals Kc again. K has not changed; the position has.
• Pressure. Increase the total pressure at constant temperature and every partial pressure rises by the same factor. The quotient changes whenever the two sides have different numbers of gas moles. The mixture then shifts until the ratio equals Kp again.
• Temperature. Here K itself changes. The old mixture no longer fits the new K, so the position shifts until it does.

Worked example 6: pressure explained with Kp

Take the equilibrium in worked example 1 and suddenly double the total pressure at constant temperature. Each partial pressure doubles at that instant:

quotient = (2 × 25.0)(2 × 75.0)³ / (2 × 15.0) = 2³ × Kp = 8 × Kp

The quotient is now larger than Kp. To bring it back down, the numerator must fall and the denominator must rise, so the equilibrium shifts to the left, towards fewer gas moles. Le Chatelier’s principle predicts the same, but 5.1.2(g) wants the argument through the fixed value of Kp.

5.1.2(h): Other equilibrium constants

The same rules apply to every equilibrium constant you meet later, such as the acid dissociation constant, Ka, in 5.1.3. For a weak acid HA:

HA(aq) ⇌ H⁺(aq) + A⁻(aq) Ka = [H⁺][A⁻] / [HA]

• The expression is written from the equation in the same way. Water as solvent is left out.
• Ka changes only with temperature.
• Diluting the acid lowers every concentration. The quotient drops below Ka (two terms on top, one below), so more HA dissociates until the ratio equals Ka again. The extent of dissociation increases; Ka does not change.

For acid basics, revisit OCR acids.

Common errors

• Using rounded mole fractions to get partial pressures, then losing accuracy in Kp.
• Writing initial amounts into Kc instead of equilibrium amounts.
• Forgetting to divide by volume when Kc has units. Volume only cancels when n = 0.
• Including a solid such as nickel in a heterogeneous expression.
• Saying “Kc increases” when you add a reactant. Only temperature changes K.
• Quoting le Chatelier’s principle for a pressure change when the question wants the argument made through Kp.
• Leaving units off, or giving kPa when the expression gives kPa³.

Official syllabus

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