OCR A Level Chemistry A: How far? (H432) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

Here you practise 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 meet each learning outcome, 5.1.2(a) to (h), from this second section of Module 5 (Physical chemistry and transition elements), all of it full A level content. Questions 9 and 10 are longer problems that combine several outcomes.

Before you start, you may want the How far? study guide or the revision notes. Every other H432 section is reachable from the course page.

Questions

1. A cylinder holds a gas mixture of 0.360 mol N₂, 0.090 mol O₂ and 0.050 mol Ar at a total pressure of 250 kPa.

(a) State what is meant by the partial pressure of a gas in a mixture. [1] (b) Calculate the partial pressures of N₂ and O₂. [2]

2. Write expressions, and give units, for:

(a) Kc for Ag₂SO₄(s) ⇌ 2Ag⁺(aq) + SO₄²⁻(aq) [2] (b) Kp for C(s) + H₂O(g) ⇌ CO(g) + H₂(g) [2]

3. Nitrogen and fluorine are heated in a sealed 2.00 dm³ vessel at a fixed temperature:

N₂(g) + 3F₂(g) ⇌ 2NF₃(g)

The vessel initially contains 0.500 mol N₂ and 1.20 mol F₂ only. At equilibrium there is 0.300 mol NF₃. Calculate Kc and give its units. [5]

4. Bromine and chlorine form an equilibrium with bromine monochloride:

Br₂(g) + Cl₂(g) ⇌ 2BrCl(g)

An equilibrium mixture contains 0.150 mol Br₂, 0.250 mol Cl₂ and 0.500 mol BrCl at a total pressure of 270 kPa. Calculate Kp. [4]

5. A student studies the hydrolysis of propyl methanoate, using hydrochloric acid as a catalyst:

HCOOCH₂CH₂CH₃(l) + H₂O(l) ⇌ HCOOH(l) + CH₃CH₂CH₂OH(l)

The flask initially contains 0.200 mol propyl methanoate and 0.600 mol water (this includes the water in the catalyst solution), with 0.00600 mol HCl. When equilibrium is reached, the mixture is poured into ice-cold water and made up to 250 cm³. Portions of 25.0 cm³ need a mean titre of 27.0 cm³ of 0.200 mol dm⁻³ NaOH.

(a) Explain why the mixture is poured into ice-cold water before titrating. [2] (b) Calculate the amount of methanoic acid at equilibrium. [2] (c) Calculate Kc. [2]

6. Use these values for the equilibrium C₂H₂(g) + H₂(g) ⇌ C₂H₄(g): Kp = 8.0 × 10⁴ kPa⁻¹ at 700 K and Kp = 1.5 × 10² kPa⁻¹ at 1000 K.

(a) Use the two values to decide the sign of the enthalpy change for the forward reaction, giving your reasoning. [2] (b) State the effect on Kp at 700 K of adding a catalyst. [1]

7. A weak acid, HA, dissociates in water: HA(aq) ⇌ H⁺(aq) + A⁻(aq). In one solution at equilibrium, [HA] = 0.0950 mol dm⁻³ and [H⁺] = [A⁻] = 5.00 × 10⁻³ mol dm⁻³.

(a) Give the Ka expression for HA. [1] (b) Work out the value of Ka, with units. [2] (c) The solution is diluted tenfold at constant temperature. Explain, using Ka, what happens to the extent of dissociation. [2]

8. Methane can be converted into ethyne at high temperature:

2CH₄(g) ⇌ C₂H₂(g) + 3H₂(g)

Explain, in terms of Kp, the effect of increasing the total pressure at constant temperature on the equilibrium yield of ethyne. [4]

9. Ethene reacts with hydrogen chloride to form chloroethane:

C₂H₄(g) + HCl(g) ⇌ C₂H₅Cl(g)

A mixture of 1.20 mol C₂H₄ and 1.00 mol HCl reaches equilibrium at a fixed temperature, by which time 60.0% of the HCl has reacted. The total pressure is 320 kPa.

(a) Calculate the amount of each gas at equilibrium. [2] (b) Find p(C₂H₄), p(HCl) and p(C₂H₅Cl). [2] (c) Write an expression for Kp and calculate its value, with units. [3] (d) The total pressure is suddenly doubled at constant temperature. Use your answer to (c) to explain which way the equilibrium shifts. [2]

10. Solid ammonium carbamate is placed in an evacuated flask at a fixed temperature:

NH₂COONH₄(s) ⇌ 2NH₃(g) + CO₂(g)

At equilibrium the total pressure is 60.0 kPa.

(a) Write an expression for Kp and explain why ammonium carbamate does not appear in it. [2] (b) Calculate Kp, with units. [2] (c) Extra NH₃ is pumped in at the same temperature. At the new equilibrium, p(NH₃) = 80.0 kPa. Calculate p(CO₂). [2] (d) State the effect on p(CO₂) of adding more solid ammonium carbamate to the original equilibrium. [1]

11. This question uses your results from questions 3 and 4.

(a) State the value of Kp for 2BrCl(g) ⇌ Br₂(g) + Cl₂(g) at the temperature in question 4. [1] (b) More F₂ is pumped into the vessel in question 3 at constant temperature. Explain why Kc stays the same but the amount of NF₃ increases. [2]

Answers

1. (a) Its share of the total pressure: mole fraction of that gas multiplied by the total pressure of the mixture. [1] (b) Total amount = 0.500 mol; x(N₂) = 0.720 and x(O₂) = 0.180 [1]. p(N₂) = 180 kPa; p(O₂) = 45.0 kPa [1]. [3] Examiner insight: Argon still counts in the total amount of gas even though it takes no part in any reaction; leaving it out gives wrong mole fractions throughout.

2. (a) Kc = [Ag⁺]²[SO₄²⁻] [1]; units mol³ dm⁻⁹ [1]. (b) Kp = p(CO) × p(H₂) / p(H₂O) [1]; units kPa [1]. [4] Examiner insight: An expression that includes Ag₂SO₄(s) or C(s) is wrong even if the rest is right, so check the state symbols before writing anything.

3. N₂ reacted = 0.300 ÷ 2 = 0.150 mol, so at equilibrium N₂ = 0.350 mol, F₂ = 1.20 − 3(0.150) = 0.750 mol, NF₃ = 0.300 mol [1]. Concentrations: 0.175, 0.375 and 0.150 mol dm⁻³ [1]. Kc = [NF₃]² / ([N₂][F₂]³) [1] = 0.150² / (0.175 × 0.375³) = 2.44 [1] dm⁶ mol⁻² [1]. [5] Examiner insight: The 1:3:2 ratio is where slips happen, especially taking 0.300 mol as the N₂ used; an initial / change / equilibrium table keeps your method visible.

4. Total amount = 0.900 mol [1]. Partial pressures: p(Br₂) = 45.0 kPa, p(Cl₂) = 75.0 kPa, p(BrCl) = 150 kPa [1]. Kp = p(BrCl)² / (p(Br₂) × p(Cl₂)) = 150² / (45.0 × 75.0) = 6.67 [1]; no units, because the kPa terms cancel [1]. [4] Examiner insight: Writing “kPa” here is a units error; when the number of gas moles is equal on both sides, say “no units” explicitly rather than leaving the space blank.

5. (a) Cooling and dilution slow the reaction almost to a stop [1], so the equilibrium does not shift to replace the acid being removed by NaOH and the titre reflects the equilibrium composition [1]. (b) Acid per portion = 0.0270 × 0.200 = 5.40 × 10⁻³ mol, so 0.0540 mol in 250 cm³ [1]. Methanoic acid = 0.0540 − 0.00600 = 0.0480 mol [1]. (c) Ester = 0.152 mol, water = 0.552 mol, propan-1-ol = 0.0480 mol [1]. Kc = (0.0480 × 0.0480) / (0.152 × 0.552) = 0.0275 (no units, volumes cancel) [1]. [6] Examiner insight: Answers that forget to subtract the HCl catalyst, or forget the factor of ten for the 25.0 cm³ portion, are the easiest ways to reach a wrong value; state each step in words.

6. (a) Kp decreases as temperature increases [1], so the forward reaction is exothermic and ΔH is negative [1]. (b) No effect; a catalyst does not change Kp [1]. [3] Examiner insight: A bare “negative” or “exothermic” earns little; tie it to the direction Kp moves as temperature rises, quoting the data.

7. (a) Ka = [H⁺][A⁻] / [HA] [1] (b) Ka = (5.00 × 10⁻³)² / 0.0950 = 2.63 × 10⁻⁴ [1] mol dm⁻³ [1]. (c) Immediately after dilution, the quotient is (5.00 × 10⁻⁴)² / 0.00950 = 2.63 × 10⁻⁵, which is less than Ka [1]. Ka is unchanged at constant temperature, so more HA dissociates until the ratio equals Ka again: the extent of dissociation increases [1]. [5] Examiner insight: Saying “Ka decreases on dilution” contradicts the whole idea of an equilibrium constant; the constant stays fixed and the position moves.

8. Kp stays the same because the temperature is constant [1]. Increasing total pressure raises every partial pressure by the same factor; there are 4 gas moles on top of the Kp expression and 2 below, so the quotient increases [1]. The quotient is now greater than Kp [1]. The equilibrium shifts left until the quotient equals Kp again, so the yield of ethyne decreases [1]. [4] Examiner insight: A le Chatelier answer (“shifts to fewer moles”) does not answer “in terms of Kp”; the comparison of quotient and Kp needs to appear.

9. (a) HCl reacted = 0.600 mol [1]. At equilibrium: C₂H₄ = 0.600 mol, HCl = 0.400 mol, C₂H₅Cl = 0.600 mol [1]. (b) Total = 1.600 mol; mole fractions 0.375, 0.250, 0.375 [1]. p(C₂H₄) = 120 kPa, p(HCl) = 80.0 kPa, p(C₂H₅Cl) = 120 kPa [1]. (c) Kp = p(C₂H₅Cl) / (p(C₂H₄) × p(HCl)) [1] = 120 / (120 × 80.0) = 0.0125 [1] kPa⁻¹ [1]. (d) Each partial pressure doubles, so the quotient becomes 240 / (240 × 160) = 0.00625, which is less than Kp [1]. The equilibrium shifts right, towards chloroethane, until the quotient equals Kp [1]. [9] Examiner insight: Using the percentage on C₂H₄ rather than HCl is the usual misread; underline which reactant the percentage refers to before starting.

10. (a) Kp = p(NH₃)² × p(CO₂) [1]. Ammonium carbamate is a solid; its concentration is constant, so it is omitted [1]. (b) The gases form in a 2:1 ratio, so p(NH₃) = ⅔ × 60.0 = 40.0 kPa and p(CO₂) = 20.0 kPa [1]. Kp = 40.0² × 20.0 = 3.20 × 10⁴ kPa³ [1]. (c) p(CO₂) = Kp / p(NH₃)² [1] = 32 000 / 80.0² = 5.00 kPa [1]. (d) No effect on p(CO₂) [1]. [7] Examiner insight: Splitting the total pressure 50:50 instead of 2:1, or not squaring p(NH₃), are the usual errors; note the mole ratio first.

11. (a) Kp = 1 / 6.67 = 0.150 [1]. (b) Kc depends only on temperature, which has not changed [1]. Adding F₂ makes the quotient smaller than Kc, so the forward reaction occurs until the ratio equals Kc again, forming more NF₃ [1]. [3] Examiner insight: Reversing an equation inverts the constant; use the unrounded Kp (20/3), which gives exactly 0.150.

Where marks are usually lost

• Leaving an inert gas, such as argon, out of the total amount of gas.
• Including solids in heterogeneous Kc or Kp expressions.
• Using starting amounts, not equilibrium amounts, in K.
• Forgetting the dilution factor or the catalyst blank in titration problems.
• Giving units by guesswork, or giving units when there are none.
• Saying K changes with concentration, pressure or a catalyst.
• Answering “in terms of Kp” questions with le Chatelier’s principle alone.

Next steps

• Recap with the How far? revision notes.
• Revisit worked examples in the How far? study guide.
• For a teacher’s help with Kc and Kp, book a free trial class.
• Spot gaps with the free 10-minute diagnostics.
• Find the next section on the H432 course page and record progress on the outcome checklist.

Official syllabus

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