OCR A Level Chemistry A: Enthalpy changes (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.

This set covers section 3.2.1, Enthalpy changes, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. The questions test every learning outcome, 3.2.1(a) to (h), from Module 3, Periodic table and energy. H432 is the full A level. Questions 11 and 12 are longer problems.

Revise first with the Enthalpy changes study guide and the Enthalpy changes revision notes. Other topics are on the OCR chemistry hub and the printable checklist.

Use c = 4.18 J g⁻¹ K⁻¹ and take 1.00 cm³ of solution as 1.00 g.

Questions

1. This question is about enthalpy definitions.

(a) State the conditions meant by standard conditions. [1] (b) Write the equation, with state symbols, for the standard enthalpy change of formation of ethanal, CH₃CHO(l). [2] (c) Write the equation, with state symbols, for the standard enthalpy change of combustion of cyclohexane, C₆H₁₂(l). [2]

2. The reaction N₂(g) + O₂(g) → 2NO(g) is endothermic.

(a) Sketch a labelled enthalpy profile diagram for this reaction. Show ∆H and the activation energy. [3] (b) Use ideas about the bonds broken and formed to explain why ∆H is positive. [2]

3. Magnesium ribbon burns exothermically in air, but only after it is held in a flame. Explain why. [2]

4. A student burns propanone (M = 58.0 g mol⁻¹) to heat 125 g of water. The spirit burner’s mass falls by 0.290 g and the water warms by 10.4 K.

(a) Calculate the enthalpy change of combustion of propanone from these results. Give your answer to 3 significant figures. [3] (b) The data value is −1790 kJ mol⁻¹. Suggest two reasons why the student’s value is less exothermic. [2]

5. 25.0 cm³ of 0.800 mol dm⁻³ H₂SO₄(aq) is mixed with 50.0 cm³ of 0.800 mol dm⁻³ NaOH(aq). The temperature rises by 7.2 K. Calculate the enthalpy change of neutralisation. [4]

6. 5.95 g of potassium bromide, KBr (M = 119.0 g mol⁻¹), is dissolved in 50.0 cm³ of water in a polystyrene cup. The temperature falls by 4.8 K.

(a) Calculate the enthalpy change for dissolving 1 mol of KBr. Include a sign. [3] (b) Suggest why this value is likely to be smaller than the true value. [1]

7. Methanal burns in oxygen: HCHO(g) + O₂(g) → CO₂(g) + H₂O(g). Use these values (kJ mol⁻¹): C–H 413; C=O in methanal 736; O=O 498; C=O in CO₂ 805; O–H 464.

(a) Calculate ∆H for this reaction using the bond enthalpies. [3] (b) Methanal’s standard enthalpy change of combustion is −571 kJ mol⁻¹. Suggest two reasons why your answer to (a) differs from it. [2]

8. For Cl₂(g) + F₂(g) → 2ClF(g), ∆H = −101 kJ mol⁻¹. The Cl–Cl bond enthalpy is 243 kJ mol⁻¹ and the F–F bond enthalpy is 158 kJ mol⁻¹. Calculate the Cl–F bond enthalpy. [3]

9. Use these standard enthalpy changes of combustion (kJ mol⁻¹) to calculate the standard enthalpy change of formation of cyclohexane, C₆H₁₂(l): C(s) −393.5; H₂(g) −285.8; C₆H₁₂(l) −3919.6. [3]

10. Phosphorus(V) chloride reacts with water: PCl₅(s) + 4H₂O(l) → H₃PO₄(s) + 5HCl(g). Use these standard enthalpy changes of formation (kJ mol⁻¹): PCl₅(s) −444; H₂O(l) −286; H₃PO₄(s) −1279; HCl(g) −92. Calculate ∆rH. [3]

11. A student finds ∆H for the thermal decomposition ZnCO₃(s) → ZnO(s) + CO₂(g) indirectly.

Experiment 1: 3.76 g of ZnCO₃ (M = 125.4 g mol⁻¹) reacts with 50.0 cm³ of 2.00 mol dm⁻³ HCl. The temperature rises by 2.9 K. ZnCO₃(s) + 2HCl(aq) → ZnCl₂(aq) + H₂O(l) + CO₂(g) ∆H₁

Experiment 2: 2.44 g of ZnO (M = 81.4 g mol⁻¹) reacts with a fresh 50.0 cm³ of the same acid. The temperature rises by 12.8 K. ZnO(s) + 2HCl(aq) → ZnCl₂(aq) + H₂O(l) ∆H₂

(a) Suggest why ∆H for the decomposition cannot easily be measured directly. [1] (b) Show that the acid is in excess in Experiment 1. [2] (c) Calculate ∆H₁. [3] (d) Calculate ∆H₂. [2] (e) Use a Hess cycle to calculate ∆H for the decomposition of ZnCO₃. [2]

12. 30.0 cm³ of 1.50 mol dm⁻³ HCl in a polystyrene cup stays at 19.6 °C for 3 minutes. At 4 minutes, 30.0 cm³ of 1.60 mol dm⁻³ NaOH at 19.6 °C is added and stirred. Later readings:

Time / min	5	6	7	8
Temperature / °C	29.4	29.0	28.6	28.2

(a) Extrapolate the cooling readings back to 4 minutes. State the temperature at 4 minutes and the temperature change. [2] (b) Calculate the enthalpy change of neutralisation. [3] (c) State how the answer to (b) would change if the student had used the highest recorded temperature instead. [1] (d) Suggest two changes to the method that would improve the accuracy of the result. [2]

Answers

1. (a) 100 kPa and a stated temperature, usually 298 K [1] (b) 2C(s) + 2H₂(g) + ½O₂(g) → CH₃CHO(l): correct formulae and balancing for 1 mol of ethanal [1]; correct state symbols [1] (c) C₆H₁₂(l) + 9O₂(g) → 6CO₂(g) + 6H₂O(l): balanced for 1 mol of cyclohexane [1]; state symbols, with water as a liquid [1] Examiner insight: Keep the ½O₂ in (b); doubling the equation to clear the fraction makes 2 mol of ethanal.

2. (a) Product level (2NO) drawn above reactant level (N₂ + O₂), with axes labelled [1]; ∆H arrow pointing up from reactants to products [1]; Ea arrow drawn upwards from the reactant line to the peak [1] (b) More energy is taken in breaking bonds than is released making new bonds [1]; breaking the very strong N≡N triple bond needs a large amount of energy [1] Examiner insight: An Ea arrow drawn from the product level, or a ∆H line with no arrowhead, costs credit on a sketch.

3. The reaction has an activation energy, a minimum energy needed before it can take place [1]. The flame supplies this energy; once started, the energy released keeps the reaction going [1] Examiner insight: Use the phrase “minimum energy”; it matches the specification’s wording.

4. (a) q = 125 × 4.18 × 10.4 = 5434 J = 5.434 kJ [1]; n = 0.290 ÷ 58.0 = 0.00500 mol [1]; ∆cH = −5.434 ÷ 0.00500 = −1087, so −1090 kJ mol⁻¹ [1] (b) Heat lost to the air or absorbed by the can [1]; incomplete combustion, or propanone evaporating from the wick [1] Examiner insight: m is the 125 g of water, not the 0.290 g of fuel.

5. n(H₂SO₄) = 0.0250 × 0.800 = 0.0200 mol, so n(H⁺) = 0.0400 mol [1]; n(OH⁻) = 0.0500 × 0.800 = 0.0400 mol, so 0.0400 mol of water forms [1]; q = 75.0 × 4.18 × 7.2 = 2257 J = 2.257 kJ [1]; ∆neutH = −2.257 ÷ 0.0400 = −56.4 kJ mol⁻¹ [1] Examiner insight: Dividing by 0.0200 mol of acid gives about double the answer; the definition is per mole of water.

6. (a) q = 50.0 × 4.18 × 4.8 = 1003 J = 1.003 kJ [1]; n = 5.95 ÷ 119.0 = 0.0500 mol [1]; temperature falls, so ∆H = +1.003 ÷ 0.0500 = +20.1 kJ mol⁻¹ [1] (b) Heat flows in from the surroundings, so the measured fall is too small [1] Examiner insight: Write the + sign; 20.1 alone does not show the direction of the change.

7. (a) Broken: 2(413) + 736 + 498 = 2060 kJ [1]; made: 2(805) + 2(464) = 2538 kJ [1]; ∆H = 2060 − 2538 = −478 kJ mol⁻¹ [1] (b) The standard value has H₂O(l), so it includes energy released on condensing [1]; C–H and O–H values are averages, not the actual bonds in these molecules [1] Examiner insight: “Bond enthalpies are not accurate” is too vague; say that they are averages from many compounds.

8. Bonds broken: 243 + 158 = 401 kJ [1]; −101 = 401 − 2x [1]; x = 502 ÷ 2 = 251 kJ mol⁻¹ [1] Examiner insight: Two moles of Cl–F bonds form in the equation; forgetting to halve gives 502.

9. Equation: 6C(s) + 6H₂(g) → C₆H₁₂(l); ∆fH = Σ∆cH(reactants) − Σ∆cH(products) [1]; = 6(−393.5) + 6(−285.8) − (−3919.6) = −4075.8 + 3919.6 [1]; = −156.2 kJ mol⁻¹ [1] Examiner insight: Multiply both the carbon and hydrogen values by six.

10. Σ∆fH(products) = −1279 + 5(−92) = −1739 kJ [1]; Σ∆fH(reactants) = −444 + 4(−286) = −1588 kJ [1]; ∆rH = −1739 − (−1588) = −151 kJ mol⁻¹ [1] Examiner insight: Keep brackets around negative values when you subtract.

11. (a) It needs strong, continuous heating, so no temperature change can be measured in a calorimeter [1] (b) n(HCl) = 0.0500 × 2.00 = 0.100 mol [1]; n(ZnCO₃) = 3.76 ÷ 125.4 = 0.0300 mol, which needs 0.0600 mol of HCl, less than 0.100 mol [1] (c) q = 50.0 × 4.18 × 2.9 = 606 J = 0.606 kJ [1]; n = 0.0300 mol [1]; ∆H₁ = −0.606 ÷ 0.0300 = −20.2 kJ mol⁻¹ [1] (d) q = 50.0 × 4.18 × 12.8 = 2675 J and n(ZnO) = 2.44 ÷ 81.4 = 0.0300 mol [1]; ∆H₂ = −2.675 ÷ 0.0300 = −89.2 kJ mol⁻¹ [1] (e) Both reactions give ZnCl₂(aq) + H₂O(l) + CO₂(g), so ∆H = ∆H₁ − ∆H₂ [1]; = −20.2 − (−89.2) = +69.0 kJ mol⁻¹ [1] Examiner insight: In (e), write ∆H₁ − ∆H₂ before substituting, so your method is visible if the arithmetic slips.

12. (a) The readings fall by 0.4 °C per minute, so the temperature at 4 minutes is 29.8 °C [1]; ∆T = 29.8 − 19.6 = 10.2 K [1] (b) n(HCl) = 0.0300 × 1.50 = 0.0450 mol, n(NaOH) = 0.0480 mol, so 0.0450 mol of water forms [1]; q = 60.0 × 4.18 × 10.2 = 2558 J = 2.558 kJ [1]; ∆neutH = −2.558 ÷ 0.0450 = −56.8 kJ mol⁻¹ [1] (c) Less exothermic (about −54.6 kJ mol⁻¹), because ∆T of 9.8 K ignores heat lost before the reading [1] (d) Put a lid on the cup to cut heat loss [1]; take readings every 30 seconds, or measure volumes with a pipette [1] Examiner insight: Use the limiting reagent, here HCl, for the moles of water; using NaOH overstates the amount formed.

Where marks are usually lost

• Putting the fuel’s or solid’s mass into q = mc∆T (questions 4 and 6).
• Leaving q in joules when ∆H is wanted in kJ mol⁻¹.
• Dividing by moles of acid for a diprotic acid (question 5).
• Missing the + sign for an endothermic change.
• Doubling a definition equation to remove ½, which breaks the 1 mol rule.
• Running a ∆cH cycle as products minus reactants.
• Writing ∆H₂ − ∆H₁ in an indirect cycle without checking the direction of each arrow.

Next steps

• Recap: Enthalpy changes revision notes
• Full explanations: Enthalpy changes study guide
• Check the whole course: all free 10-minute diagnostics
• Mole calculations: Amount of substance practice
• More OCR chemistry topics: course hub
• Printable checklist: OCR A Level Chemistry checklist
• Book a free trial class: /trial/

Official syllabus

OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 3.2.1 Enthalpy changes, learning outcomes (a) to (h). Published by OCR.