OCR A Level Chemistry A: Lattice enthalpy (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.

Practise section 5.2.1, Lattice enthalpy, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. The questions below test outcomes 5.2.1(a) to (e) from Module 5, Physical chemistry and transition elements, and question 10 uses the technique for measuring an enthalpy change of solution that the 3.2.1(h) guidance points to. H432 is the whole A level, so the questions carry no AS or A2 labels. Each question supplies the energy values it needs; use those values.

Before you start, work through the Lattice enthalpy study guide or skim the revision notes. Other units sit on the OCR chemistry hub; log your progress on the printable checklist, and try the free diagnostics to test other parts of the course.

Questions

1. Explain the term lattice enthalpy, and write the equation, with state symbols, for the lattice enthalpy of strontium fluoride. [3]

2. Write equations, with state symbols, for each of these.

(a) The enthalpy change of atomisation of iodine. [1] (b) The first ionisation energy of caesium. [1] (c) The second ionisation energy of calcium. [1] (d) The enthalpy change of formation of caesium iodide. [1]

3. Calculate the lattice enthalpy of caesium iodide, CsI. Use these values, in kJ mol⁻¹: ΔfH(CsI) = −347; ΔatH(I) = +107; first electron affinity of I = −295; ΔatH(Cs) = +76; first ionisation energy of Cs = +376. [3]

4. Calculate the first electron affinity of iodine using the Born–Haber cycle for sodium iodide, NaI. Use these values, in kJ mol⁻¹: ΔfH(NaI) = −287; ΔatH(Na) = +107; first ionisation energy of Na = +496; ΔatH(I) = +107; ΔLEH(NaI) = −702. [3]

5. This question is about the Born–Haber cycle for calcium iodide, CaI₂.

(a) Starting from Ca(s) + I₂(s), give the species, with state symbols, on each level reached after: atomising calcium; atomising iodine; the first ionisation of calcium; the second ionisation of calcium; adding electrons to iodine. [3] (b) Calculate ΔLEH(CaI₂). Use these values, in kJ mol⁻¹: ΔfH(CaI₂) = −533; ΔatH(Ca) = +178; first ionisation energy of Ca = +590; second ionisation energy of Ca = +1145; ΔatH(I) = +107; first electron affinity of I = −295. [3] (c) The lattice enthalpy of MgI₂ is −2327 kJ mol⁻¹. Explain why ΔLEH(CaI₂) is less exothermic. [2]

6. Lithium iodide dissolves readily in water.

(a) State what is meant by the enthalpy change of hydration. [1] (b) Give the equation for the enthalpy change of solution of lithium iodide, including state symbols. [1] (c) Write the equation, with state symbols, for the enthalpy change of hydration of the iodide ion. [1] (d) Explain why the hydration of an iodide ion is exothermic. [2]

7. Calculate the enthalpy change of solution of potassium iodide, KI, and state whether a solution gets warmer or cooler as KI dissolves. Use these values, in kJ mol⁻¹: ΔLEH(KI) = −649; ΔhydH(K⁺) = −322; ΔhydH(I⁻) = −307. [2]

8. Calculate the enthalpy change of hydration of the lithium ion, Li⁺. Use these values, in kJ mol⁻¹: ΔsolH(LiI) = −63; ΔLEH(LiI) = −763; ΔhydH(I⁻) = −307. [3]

9. This question is about the factors that affect lattice and hydration enthalpies.

(a) Put CsI, NaF and NaI in order of increasingly exothermic lattice enthalpy. Explain your order. [3] (b) Predict whether Al³⁺ or Mg²⁺ has the more exothermic enthalpy change of hydration. Explain your answer. [2]

10. A student finds the enthalpy change of solution of anhydrous lithium iodide. He dissolves 6.69 g of LiI in 50.0 g of water in a polystyrene cup. The temperature rises by 14.2 K. Use Ar(Li) = 6.9 and Ar(I) = 126.9, c = 4.18 J g⁻¹ K⁻¹, and assume the solution has a mass of 50.0 g.

(a) Calculate the energy released, q, in J. [1] (b) Calculate ΔsolH(LiI) in kJ mol⁻¹, giving your answer to three significant figures. [3] (c) The value used in question 8 is ΔsolH(LiI) = −63 kJ mol⁻¹. Suggest one reason why the student’s value is less exothermic. [1]

11. This question is about strontium iodide, SrI₂.

(a) Calculate ΔfH(SrI₂). Use these values, in kJ mol⁻¹: first and second ionisation energies of Sr = +550 and +1064; ΔatH(Sr) = +164; ΔatH(I) = +107; first electron affinity of I = −295; ΔLEH(SrI₂) = −1960. [3] (b) The enthalpy change of solution of SrI₂ is −99 kJ mol⁻¹ and ΔhydH(I⁻) = −307 kJ mol⁻¹. Draw a labelled enthalpy cycle linking SrI₂(s), its gaseous ions and its aqueous ions, and use it to calculate ΔhydH(Sr²⁺). [4]

Answers

1. The enthalpy change when 1 mol of an ionic lattice is formed [1] from its gaseous ions, under standard conditions [1]. Sr²⁺(g) + 2F⁻(g) → SrF₂(s), with correct state symbols [1] Examiner insight: A definition describing the lattice being broken apart into ions describes the reverse process and does not match the H432 definition, so it is unlikely to gain the first mark.

2. (a) ½I₂(s) → I(g) [1] (b) Cs(g) → Cs⁺(g) + e⁻ [1] (c) Ca⁺(g) → Ca²⁺(g) + e⁻ [1] (d) Cs(s) + ½I₂(s) → CsI(s) [1] Examiner insight: State symbols are part of each equation here; iodine as I₂(g) in (a) or (d), or Ca(g) as the starting species in (c), would not describe the defined process.

3. Sum of steps before the lattice forms = 76 + 376 + 107 + (−295) = +264 kJ mol⁻¹ [1]; ΔfH = 264 + ΔLEH, so ΔLEH = −347 − 264 [1] = −611 kJ mol⁻¹ [1] Examiner insight: A positive answer for a lattice enthalpy is a signal to recheck signs; it usually comes from subtracting the electron affinity instead of adding it.

4. −287 = 107 + 496 + 107 + EA + (−702) [1]; EA = −287 − (107 + 496 + 107) + 702 = −287 − 710 + 702 [1] = −295 kJ mol⁻¹ [1] Examiner insight: Writing the full Hess’ law expression first shows the method clearly, so a slip in the arithmetic still leaves the method visible to the marker.

5. (a) Ca(g) + I₂(s), then Ca(g) + 2I(g) [1]; Ca⁺(g) + e⁻ + 2I(g), then Ca²⁺(g) + 2e⁻ + 2I(g) [1]; Ca²⁺(g) + 2I⁻(g) [1] (b) Steps: 178 + 590 + 1145 + 2(107) + 2(−295) [1] = +1537 kJ mol⁻¹ [1]; ΔLEH = −533 − 1537 = −2070 kJ mol⁻¹ [1] (c) Ca²⁺ has the same charge as Mg²⁺ but a larger ionic radius [1], so Ca²⁺ and I⁻ ions sit further apart, attract each other less strongly, and less energy is given out as the lattice forms [1] Examiner insight: In (b), using the iodine values once instead of twice is the usual source of a wrong answer; in (c), comparing charge alone cannot explain a difference between two 2+ ions.

6. (a) The enthalpy change when 1 mol of gaseous ions is dissolved in water [1] (b) LiI(s) + aq → Li⁺(aq) + I⁻(aq) [1] (c) I⁻(g) + aq → I⁻(aq) [1] (d) Water molecules are polar; the δ+ hydrogen atoms are attracted to the negative iodide ion [1]. Forming these attractions releases energy [1] Examiner insight: For an anion, say that the hydrogen end of water is attracted; describing the oxygen end around a negative ion contradicts the charges.

7. ΔsolH = −ΔLEH + ΔhydH(K⁺) + ΔhydH(I⁻) = 649 + (−322) + (−307) = +20 kJ mol⁻¹ [1]; the value is positive (endothermic), so the solution gets cooler [1] Examiner insight: The lattice enthalpy must be reversed for the lattice-breaking step; using −649 gives −1278, which is far too large for a solution enthalpy and has the wrong sign.

8. ΔsolH = −ΔLEH + ΔhydH(Li⁺) + ΔhydH(I⁻) [1]; −63 = +763 + ΔhydH(Li⁺) + (−307) [1]; ΔhydH(Li⁺) = −63 − 763 + 307 = −519 kJ mol⁻¹ [1] Examiner insight: A hydration enthalpy must come out negative; an answer of +1007 shows that the lattice enthalpy was not reversed for the lattice-breaking step.

9. (a) CsI < NaI < NaF (least to most exothermic) [1]. Na⁺ is smaller than Cs⁺, so NaI is more exothermic than CsI [1]. F⁻ is smaller than I⁻, so the ions in NaF are closest and attract most strongly, releasing the most energy [1] (b) Al³⁺ [1]. It is more highly charged and smaller than Mg²⁺, so the δ− oxygen atoms of surrounding water molecules are drawn in more strongly and more energy is released [1] Examiner insight: An order on its own gains only the first mark in (a); each comparison needs the relevant ion named and linked to the strength of attraction.

10. (a) q = 50.0 × 4.18 × 14.2 = 2968 J (2970 J to three significant figures) [1] (b) n(LiI) = 6.69 ÷ 133.8 = 0.0500 mol [1]; 2.968 kJ ÷ 0.0500 = 59.4 kJ mol⁻¹ [1]; temperature rose, so ΔsolH = −59.4 kJ mol⁻¹ [1] (c) Energy was lost to the surroundings, so the measured temperature rise was smaller than it should have been [1] Examiner insight: Converting q from J to kJ before dividing and adding the negative sign are separate steps; an answer of 59.4 without a sign, or −59 400, does not complete the calculation.

11. (a) Steps: 164 + 550 + 1064 + 2(107) + 2(−295) = +1402 kJ mol⁻¹ [1]; ΔfH = 1402 + (−1960) [1] = −558 kJ mol⁻¹ [1] (b) Cycle: SrI₂(s) → Sr²⁺(g) + 2I⁻(g) labelled −ΔLEH; Sr²⁺(g) + 2I⁻(g) → Sr²⁺(aq) + 2I⁻(aq) labelled ΔhydH(Sr²⁺) + 2ΔhydH(I⁻); SrI₂(s) → Sr²⁺(aq) + 2I⁻(aq) labelled ΔsolH [1]. −99 = 1960 + ΔhydH(Sr²⁺) + 2(−307) [1]; ΔhydH(Sr²⁺) = −99 − 1960 + 614 [1] = −1445 kJ mol⁻¹ [1] Examiner insight: In (b), the cycle should show both ions with state symbols on every level; a cycle that shows only one I⁻ will carry the same error into the calculation.

Where marks are usually lost

• Defining lattice enthalpy as lattice breaking, which reverses its sign under the H432 definition.
• Writing iodine or bromine in the wrong standard state in atomisation and formation equations.
• Using one halogen atomisation and one electron affinity for an MX₂ compound (questions 5 and 11).
• Leaving out the second ionisation energy for a Group 2 cation.
• Not reversing ΔLEH when it is used in a solution cycle (questions 7, 8 and 11).
• Counting the hydration enthalpy of a doubled anion only once.
• Explaining hydration of an anion with the oxygen end of water pointing at it.
• Giving a trend order without explaining it in terms of ionic size or charge and attraction.
• Leaving the sign off an experimental ΔsolH, or forgetting to convert J to kJ.

Next steps

• Definitions, tables and a self-test: Lattice enthalpy revision notes
• Teaching from scratch, with worked cycles: Lattice enthalpy study guide
• Find weak spots in other topics: all free 10-minute diagnostics
• Ionic lattices first met in Module 2: Bonding and structure
• Hess’ law and measuring enthalpy changes: Enthalpy changes
• Entropy and feasibility, the next section: Enthalpy and entropy
• Every H432 unit in one place: OCR A Level Chemistry hub
• Track outcomes on paper: 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 5.2.1 Lattice enthalpy, learning outcomes (a) to (e). Published by OCR.