OCR A Level Chemistry A: Lattice enthalpy (H432) – Revision Notes

Use these notes for fast recall of section 5.2.1, Lattice enthalpy, in the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. They summarise outcomes 5.2.1(a) to (e) in Module 5, Physical chemistry and transition elements, with no AS or A2 split because H432 is the full A level. Fuller teaching, with longer examples, is in the Lattice enthalpy study guide.

Marked questions on this section are in the Lattice enthalpy practice questions. Browse other units from the OCR chemistry hub, tick off 5.2.1 on the printable checklist, and test the rest of the course with the free diagnostics.

5.2.1(a): Lattice enthalpy in one line

ΔLEH: the enthalpy change when 1 mol of an ionic lattice is formed from its gaseous ions, under standard conditions (100 kPa, with 298 K as the stated temperature).

Li⁺(g) + I⁻(g) → LiI(s)

• Always exothermic, because oppositely charged ions attract as they come together.
• More exothermic value = stronger ionic bonding in the giant lattice.
• Stronger ionic bonding usually means a higher melting point (link back to 2.2.2(b) and (c)).
• It cannot be measured directly; it is found from a Born–Haber cycle.

5.2.1(b): Born–Haber terms

You must define first ionisation energy and enthalpy change of formation. For the other terms you need the meaning and a correct equation.

Term	Equation pattern	Sign
ΔfH	K(s) + ½F₂(g) → KF(s)	usually −
ΔatH, metal	K(s) → K(g)	+
ΔatH, fluorine or chlorine	½F₂(g) → F(g); ½Cl₂(g) → Cl(g)	+
ΔatH, bromine	½Br₂(l) → Br(g)	+
ΔatH, iodine	½I₂(s) → I(g)	+
First ionisation energy	K(g) → K⁺(g) + e⁻	+
Second ionisation energy	Sr⁺(g) → Sr²⁺(g) + e⁻	+
First electron affinity	F(g) + e⁻ → F⁻(g)	− for halogens
Second electron affinity	O⁻(g) + e⁻ → O²⁻(g)	+
ΔLEH	K⁺(g) + F⁻(g) → KF(s)	−

Definitions to learn:

• First ionisation energy: the energy required to remove 1 mol of electrons from 1 mol of gaseous atoms, giving 1 mol of gaseous 1+ ions.
• Enthalpy change of formation: the enthalpy change when 1 mol of a compound forms, under standard conditions, from its constituent elements, each in its standard state.

Method in steps: a Born–Haber calculation

1. Write ΔfH = sum of all the other steps + ΔLEH.
2. For each halogen atom in the formula, include one ΔatH and one electron affinity. MX₂ needs two of each.
3. For a 2+ ion, include the first and second ionisation energies.
4. Put every value in with its sign, in brackets if negative.
5. Rearrange for the unknown and check its sign makes sense.

Worked reminder: lattice enthalpy of potassium fluoride

Data (kJ mol⁻¹): ΔfH(KF) = −568; ΔatH(K) = +89; first IE(K) = +419; ΔatH(F) = +79; first EA(F) = −328.

Sum of steps = 89 + 419 + 79 + (−328) = +259
−568 = 259 + ΔLEH
ΔLEH(KF) = −827 kJ mol⁻¹

Drawing the cycle

• Elements in standard states sit at the zero line.
• Endothermic arrows point up; exothermic arrows point down.
• A common order going up: atomise metal, atomise non-metal, ionise metal. Then electron affinity brings you down to the gaseous ions.
• ΔLEH runs from the gaseous ions down to the solid. ΔfH runs from the elements to the solid.
• State symbols on every species; show e⁻ on the levels where electrons are free.

5.2.1(c): Solution and hydration

Term	Meaning	Example equation	Sign
ΔsolH	1 mol of solute dissolves in water	KF(s) + aq → K⁺(aq) + F⁻(aq)	+ or −
ΔhydH	1 mol of gaseous ions dissolves in water	F⁻(g) + aq → F⁻(aq)	always −

Hydration releases energy because polar water molecules are attracted to ions: δ− oxygen towards cations, δ+ hydrogen towards anions. Infinite dilution details are not required.

ΔsolH can be measured in the lab with q = mcΔT (the 3.2.1(h) guidance includes these 5.2.1(c) enthalpy changes). ΔhydH cannot be measured directly, so it comes from a cycle.

5.2.1(d): The solution cycle

ΔsolH = −ΔLEH + ΣΔhydH

              gaseous ions
   −ΔLEH  ↗                ↘  sum of ΔhydH
 solid  ─────── ΔsolH ───────→  aqueous ions

Method in steps

1. Write the formula and count the ions: CsBr has one of each; MgI₂ has one Mg²⁺ and two I⁻.
2. Reverse the lattice enthalpy (change its sign) for lattice breaking.
3. Add the hydration enthalpy of every ion, multiplied by how many there are.
4. Rearrange for whichever term is missing.

Worked reminder: ΔsolH of caesium bromide

Data (kJ mol⁻¹): ΔLEH(CsBr) = −640; ΔhydH(Cs⁺) = −276; ΔhydH(Br⁻) = −336.

ΔsolH = +640 + (−276) + (−336) = +28 kJ mol⁻¹

Endothermic: breaking the lattice needs more energy than hydration releases.

Worked reminder: a 1:2 salt, magnesium iodide

Data (kJ mol⁻¹): ΔLEH(MgI₂) = −2327; ΔhydH(Mg²⁺) = −1926; ΔhydH(I⁻) = −307.

ΔsolH = +2327 + (−1926) + 2(−307)
      = 2327 − 1926 − 614
      = −213 kJ mol⁻¹

The iodide term is doubled because MgI₂ releases two iodide ions per formula unit. Leaving out the factor of 2 would give +94, which is wrong in both size and sign.

Reading a cycle you have not seen before

Questions sometimes give a part-drawn Born–Haber or solution cycle with one arrow left blank. Work out what the blank arrow does by comparing the species above and below it: a change from atoms to ions is ionisation or electron affinity; from element to atoms is atomisation; from gaseous ions to aqueous ions is hydration. Then add up the arrows on one route, in the direction of travel, and set the total equal to the other route.

5.2.1(e): Charge and radius

Change	Effect on attraction	Lattice enthalpy	Hydration enthalpy
Smaller ionic radius	ions (or water dipoles) get closer, stronger attraction	more exothermic	more exothermic
Larger ionic charge	stronger attraction	more exothermic	more exothermic

Must-know comparisons

• Down Group 1 with the same anion (Li⁺ → Cs⁺): radius increases, lattice enthalpy becomes less exothermic.
• Na⁺ compared with Mg²⁺: Mg²⁺ is both smaller and more highly charged, so its lattice enthalpies and hydration enthalpy are much more exothermic.
• F⁻ compared with I⁻: the smaller fluoride ion gives more exothermic values.

Must-know distinctions

• Lattice enthalpy vs lattice breaking. In H432, ΔLEH is formation from gaseous ions (negative). Breaking the lattice is −ΔLEH (positive).
• Atomisation vs bond dissociation. ΔatH of chlorine is per mole of Cl atoms: ½Cl₂(g) → Cl(g). It is half the Cl–Cl bond enthalpy.
• Hydration vs solution. Hydration starts from gaseous ions; solution starts from the solid.
• First vs second electron affinity. The first is exothermic for halogens; the second (onto a negative ion, such as O⁻) is endothermic.

Quick self-test

1. Write the equation for the lattice enthalpy of calcium chloride, with state symbols.
2. State the sign of every lattice enthalpy, and explain why.
3. Write the equation for the enthalpy change of atomisation of iodine.
4. Which two Born–Haber terms does the specification require you to define?
5. Which has the more exothermic lattice enthalpy, LiCl or RbCl? Explain.
6. Use these values to find the first ionisation energy of potassium: ΔfH(KBr) = −394; ΔatH(K) = +89; ΔatH(Br) = +112; first EA(Br) = −325; ΔLEH(KBr) = −689 (all kJ mol⁻¹).
7. How many electron affinity terms appear in the Born–Haber cycle for SrBr₂?
8. Write the equation for the enthalpy change of solution of strontium chloride.
9. ΔsolH(KF) = −18, ΔLEH(KF) = −827 and ΔhydH(K⁺) = −322 kJ mol⁻¹. Calculate ΔhydH(F⁻).
10. Explain why ΔhydH(Mg²⁺) is more exothermic than ΔhydH(Na⁺).
11. Why can ΔsolH be measured in a lab but ΔhydH cannot?

Answers

1. Ca²⁺(g) + 2Cl⁻(g) → CaCl₂(s)
2. Always negative. The lattice forms from gaseous ions, and the attraction between oppositely charged ions releases energy.
3. ½I₂(s) → I(g)
4. First ionisation energy and enthalpy change of formation.
5. LiCl. Li⁺ is smaller than Rb⁺ with the same charge, so the ions are closer and attract more strongly.
6. −394 = 89 + IE + 112 + (−325) + (−689), so IE = −394 − 89 − 112 + 325 + 689 = +419 kJ mol⁻¹.
7. Two: one for each bromine atom (two Br(g) + 2e⁻ → two Br⁻(g)).
8. SrCl₂(s) + aq → Sr²⁺(aq) + 2Cl⁻(aq)
9. −18 = +827 + (−322) + ΔhydH(F⁻), so ΔhydH(F⁻) = −18 − 827 + 322 = −523 kJ mol⁻¹.
10. Mg²⁺ carries a 2+ charge and is smaller than Na⁺, so water’s δ− oxygen atoms are pulled closer and held more strongly, and more energy is released.
11. A solid can be weighed and dissolved in water in a calorimeter. Gaseous ions cannot be produced and added to water, so ΔhydH comes from a cycle.

Where marks are usually lost

• Writing lattice enthalpy as the breaking of a lattice, which gives the wrong sign under the H432 definition.
• Using Br₂(g) or Cl₂(l) as the standard state in an atomisation equation.
• Writing ΔatH for one mole of Cl₂ molecules instead of one mole of Cl atoms.
• Forgetting to double the halogen atomisation and electron affinity for an MX₂ compound.
• Leaving out the second ionisation energy for a Group 2 metal.
• Not reversing ΔLEH in a solution cycle.
• Counting the hydration enthalpy of a doubled ion only once.
• Explaining trends with “bigger ion, bigger value” instead of linking size or charge to attraction and energy released.
• Missing state symbols, or missing e⁻, on Born–Haber levels.

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.