OCR A Level Chemistry A: Lattice enthalpy (H432)

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, is taught from scratch on this page. The section is part of Module 5, Physical chemistry and transition elements, and each of its outcomes, 5.2.1(a) to (e), has its own heading below. Because H432 is the whole A level, no content here carries an AS or A2 label. According to the specification’s assessment overview, Module 5 is examined in the Periodic table, elements and physical chemistry component (01), and Unified chemistry (03) draws on every module.

Find other H432 units on the OCR chemistry hub, record what you have covered on the printable checklist, and spot gaps elsewhere in the course with the free diagnostics.

Outcomes in 5.2.1

Outcome	Skill tested	Specification guidance
5.2.1(a)	Explain the term lattice enthalpy, ΔLEH, and use it as a measure of the strength of ionic bonding in a giant ionic lattice	Definition required; see also 2.2.2(b) and (c)
5.2.1(b)	Use the lattice enthalpy of a simple ionic solid and other energy terms to (i) construct Born–Haber cycles and (ii) carry out related calculations	Terms: enthalpy change of formation, ionisation energy, enthalpy change of atomisation, electron affinity. Definitions required for first ionisation energy and enthalpy change of formation only. M2.2, M2.3, M2.4, M3.1; HSW2
5.2.1(c)	Explain and use the terms enthalpy change of solution, ΔsolH, and enthalpy change of hydration, ΔhydH	Definitions required; details of infinite dilution not required
5.2.1(d)	Use the enthalpy change of solution, hydration enthalpies and lattice enthalpy to (i) construct enthalpy cycles and (ii) carry out related calculations	M2.2, M2.3, M2.4, M3.1; HSW2
5.2.1(e)	Explain qualitatively how ionic charge and ionic radius affect the exothermic value of lattice enthalpy and hydration enthalpy	None given

No PAG is listed against 5.2.1 itself. The guidance for 3.2.1(h), on techniques for finding enthalpy changes directly and indirectly, says it includes the enthalpy changes in 5.2.1(c).

5.2.1(a): What lattice enthalpy means

Lattice enthalpy, ΔLEH, is the enthalpy change for the formation of 1 mol of an ionic lattice from its gaseous ions, under standard conditions.

For lithium bromide:

Li⁺(g) + Br⁻(g) → LiBr(s) ΔLEH = −817 kJ mol⁻¹

Three details matter:

• The ions start as separate gaseous ions, far apart.
• The product is one mole of the solid, so write the equation for one formula unit.
• It is lattice formation, so the value is always negative: opposite charges attract, and bringing them together releases energy.

The specification says standard conditions can be taken as 100 kPa and a stated temperature, 298 K.

Lattice enthalpy as a measure of ionic bond strength

In 2.2.2 you met the giant ionic lattice: positive and negative ions packed so that each is pulled strongly by its oppositely charged neighbours on every side. Lattice enthalpy puts a number on that attraction. A more exothermic (more negative) value means more energy is released as the lattice forms. The same amount would be needed to pull it apart again, so the ionic bonding is stronger.

A very exothermic lattice enthalpy usually goes with a high melting point. Gaseous ions cannot be combined in a calorimeter, so lattice enthalpy is found indirectly, from an enthalpy cycle.

5.2.1(b): Born–Haber cycles

A Born–Haber cycle applies Hess’ law to an ionic compound. One route goes straight from the elements to the solid: the enthalpy change of formation. The other route goes the long way round, through gaseous atoms and then gaseous ions. Conservation of energy means both routes give the same total (HSW2).

The energy terms you need

Term	What it describes	Example equation	Sign
Enthalpy change of formation, ΔfH	1 mol of a compound is made from its elements, each in its standard state	Na(s) + ½Br₂(l) → NaBr(s)	Usually negative
Enthalpy change of atomisation, ΔatH	The element, in its standard state, is turned into 1 mol of gaseous atoms	Na(s) → Na(g); ½Br₂(l) → Br(g)	Always positive
First ionisation energy	1 mol of electrons is removed from 1 mol of gaseous atoms	Na(g) → Na⁺(g) + e⁻	Always positive
Second ionisation energy	1 mol of electrons is removed from 1 mol of gaseous 1+ ions	Mg⁺(g) → Mg²⁺(g) + e⁻	Always positive
Electron affinity (first)	1 mol of electrons is added to 1 mol of gaseous atoms	Br(g) + e⁻ → Br⁻(g)	Negative for halogens
Lattice enthalpy, ΔLEH	1 mol of ionic lattice forms from gaseous ions	Na⁺(g) + Br⁻(g) → NaBr(s)	Always negative

Only first ionisation energy and enthalpy change of formation need formal definitions, but you must use every term and write its equation correctly. Two points are easy to miss:

• Atomisation is per mole of atoms formed. For bromine, the standard state is a liquid, so the equation is ½Br₂(l) → Br(g).
• For a compound such as MgBr₂, you need two moles of bromine atoms and two electron affinities.

For an oxide, the oxide ion needs a second electron added to O⁻(g). That second electron affinity is endothermic, because the electron is pushed onto an ion that is already negative.

Building the cycle

Draw energy levels with endothermic steps going up and exothermic steps going down. Start from the elements at zero.

1. Write the elements in their standard states as your baseline.
2. Atomise the metal, then atomise the non-metal (each arrow goes up).
3. Ionise the metal atom, once or twice as needed (up).
4. Add electrons to the non-metal atoms (down for a first electron affinity of a halogen).
5. You now have gaseous ions at the top level. The lattice enthalpy arrow goes down to the solid.
6. The formation arrow goes from the elements to the solid.

Every species needs a state symbol, and free electrons appear on the levels between ionisation and electron affinity.

Worked example 1: lattice enthalpy of lithium bromide

Use these values (kJ mol⁻¹): ΔfH(LiBr) = −351; ΔatH(Li) = +159; first ionisation energy of Li = +520; ΔatH(Br) = +112; first electron affinity of Br = −325.

The cycle, with level enthalpies relative to the elements:

Li⁺(g) + e⁻ + Br(g)    (+791)
   ↑ first IE of Li, +520     ↓ first EA of Br, −325
   │                        Li⁺(g) + Br⁻(g)    (+466)
Li(g) + Br(g)          (+271)    │
   ↑ ΔatH(Br), +112              │
Li(g) + ½Br₂(l)        (+159)    │ ΔLEH(LiBr) = ?
   ↑ ΔatH(Li), +159              │
Li(s) + ½Br₂(l)        (zero)    │
   ↓ ΔfH(LiBr), −351             ↓
LiBr(s)                (−351) ←──┘

Hess’ law: ΔfH = ΔatH(Li) + ΔatH(Br) + IE + EA + ΔLEH

−351 = 159 + 112 + 520 + (−325) + ΔLEH
−351 = 466 + ΔLEH
ΔLEH = −351 − 466 = −817 kJ mol⁻¹

The answer is negative, as expected.

Worked example 2: a compound with a 2+ ion

Find ΔLEH for magnesium bromide, MgBr₂. Use these values (kJ mol⁻¹): ΔfH(MgBr₂) = −524; ΔatH(Mg) = +148; first ionisation energy of Mg = +738; second ionisation energy of Mg = +1451; ΔatH(Br) = +112; first electron affinity of Br = −325.

Step 1: list the long-route steps, doubling for bromine.

Step	Value / kJ mol⁻¹
Mg(s) → Mg(g)	+148
Br₂(l) → 2Br(g), which is 2 × ΔatH(Br)	+224
Mg(g) → Mg⁺(g) + e⁻	+738
Mg⁺(g) → Mg²⁺(g) + e⁻	+1451
2Br(g) + 2e⁻ → 2Br⁻(g), which is 2 × EA	−650
Total before the lattice forms	+1911

Step 2: apply Hess’ law.

ΔfH = (sum of the steps) + ΔLEH
−524 = 1911 + ΔLEH
ΔLEH = −2435 kJ mol⁻¹

Rearrange the same cycle to find any one missing term, such as an electron affinity.

5.2.1(c): Enthalpy changes of solution and hydration

Enthalpy change of solution, ΔsolH: the enthalpy change for dissolving 1 mol of a solute in water.

RbBr(s) + aq → Rb⁺(aq) + Br⁻(aq)

Enthalpy change of hydration, ΔhydH: the enthalpy change for dissolving 1 mol of gaseous ions in water.

Rb⁺(g) + aq → Rb⁺(aq) Br⁻(g) + aq → Br⁻(aq)

Hydration is always exothermic. Water molecules are polar, so the δ− oxygen atoms are attracted to a positive ion, and the δ+ hydrogen atoms are attracted to a negative ion. Forming these attractions releases energy. Enthalpy changes of solution can be either sign, because they are a balance between two large terms. The specification does not require details of infinite dilution.

5.2.1(d): Enthalpy cycles for dissolving

Dissolving can be imagined in two stages. First, break the lattice into gaseous ions; this needs the reverse of the lattice enthalpy, so it is +|ΔLEH|. Second, hydrate those ions. Hess’ law then gives:

ΔsolH = −ΔLEH + ΣΔhydH

The sum covers every ion in the formula, so a 1:2 compound counts its anion twice.

            M⁺(g) + X⁻(g)
   −ΔLEH ↗               ↘ ΔhydH(M⁺) + ΔhydH(X⁻)
  MX(s)  ──── ΔsolH ────→  M⁺(aq) + X⁻(aq)

Worked example 3: enthalpy change of solution of rubidium bromide

Use these values (kJ mol⁻¹): ΔLEH(RbBr) = −659; ΔhydH(Rb⁺) = −301; ΔhydH(Br⁻) = −336.

ΔsolH = −(−659) + (−301) + (−336)
      = 659 − 637
      = +22 kJ mol⁻¹

Dissolving is endothermic: breaking the lattice needs slightly more energy than hydration releases, so the solution cools.

Worked example 4: finding a hydration enthalpy

Find ΔhydH(Mg²⁺). Use these values (kJ mol⁻¹): ΔsolH(MgBr₂) = −185; ΔLEH(MgBr₂) = −2435; ΔhydH(Br⁻) = −336.

ΔsolH = −ΔLEH + ΔhydH(Mg²⁺) + 2ΔhydH(Br⁻)
−185 = 2435 + ΔhydH(Mg²⁺) + 2(−336)
−185 = 1763 + ΔhydH(Mg²⁺)
ΔhydH(Mg²⁺) = −1948 kJ mol⁻¹

The factor of 2 for bromide is the step most often dropped.

Linking to practical work

The guidance for 3.2.1(h) includes measuring the enthalpy changes in 5.2.1(c). For a solution enthalpy, dissolve a weighed mass of solid in a measured volume of water in an insulated cup, record the temperature change, use q = mcΔT and divide by the moles of solid. A temperature rise means ΔsolH is negative; a fall means it is positive. Hydration enthalpies come from cycles instead.

5.2.1(e): Effect of ionic charge and ionic radius

Both lattice enthalpy and hydration enthalpy depend on the attraction between charges, so the same two factors control them.

Ionic radius. Smaller ions can get closer together. The attraction between them is stronger, so more energy is released. The lattice enthalpy, or the hydration enthalpy, is more exothermic.

Ionic charge. Ions with a larger charge attract oppositely charged ions (or the polar ends of water molecules) more strongly. More energy is released, so the value is more exothermic.

Applying this to the compounds above:

• MgBr₂ (−2435) is far more exothermic than LiBr (−817). Mg²⁺ and Li⁺ are similar in size, so the larger charge on Mg²⁺ is the main factor.
• Take ΔfH(KBr) = −394, ΔatH(K) = +89 and first ionisation energy of K = +419, with the bromine values from worked example 1. The same cycle gives ΔLEH(KBr) = −689 kJ mol⁻¹, less exothermic than LiBr. Both cations are 1+, but K⁺ is bigger, so its centre sits further from each Br⁻ ion.
• Mg²⁺ (−1948) has a far more exothermic hydration enthalpy than Rb⁺ (−301). Mg²⁺ carries twice the charge and is much smaller, so it pulls the δ− oxygen of water in more strongly.

In an explanation, name the ions, say which factor differs (charge, radius or both), and link it to attraction and energy released.

Common errors

• Defining lattice enthalpy as breaking the lattice. In H432 it is formation from gaseous ions, so it is negative.
• Using ½Br₂(g) as the standard state of bromine, or forgetting that ΔatH is per mole of atoms.
• For MgBr₂-type compounds, forgetting to double the atomisation and electron affinity of the halogen.
• Missing the second ionisation energy of a 2+ ion.
• Mixing up the sign when rearranging: lattice dissociation is −ΔLEH in a solution cycle.
• Explaining charge and radius effects without linking them to attraction.

Where to go next

Condensed tables and a quick self-test are in the Lattice enthalpy revision notes. Then try the marked Lattice enthalpy practice questions. Giant ionic lattices are introduced in Bonding and structure, and electron shells and sub-shells in Electron structure.

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.