Build a cycle from correctly balanced changes, then use Hess’s law to find lattice enthalpy or another missing value.
Define the change, not just its name
Every enthalpy definition describes a particular reaction amount and physical states. Standard changes use a pressure of 100 kPa and a stated temperature, often 298 K. The standard state is the most stable physical form under those conditions. A gaseous atom is not interchangeable with an aqueous ion.
| Change | What happens for one mole | Sign or caution |
|---|---|---|
| Standard formation | One mole of compound forms from its elements in their standard states. | Can be positive or negative. |
| Atomisation of an element | One mole of gaseous atoms forms from the element in its standard state. | Usually positive; ½Cl₂(g) → Cl(g). |
| First ionisation | One mole of gaseous atoms loses one mole of electrons to form gaseous 1+ ions. | Positive. |
| Second ionisation | One mole of gaseous 1+ ions loses one mole of electrons to form gaseous 2+ ions. | Positive; starts with ions, not atoms. |
| First electron affinity | One mole of gaseous atoms gains one mole of electrons to form gaseous 1− ions. | For the common halogen/oxygen examples, negative. |
| Second electron affinity | One mole of gaseous 1− ions gains one mole of electrons to form gaseous 2− ions. | Positive: an incoming electron is repelled. |
| Bond dissociation | One mole of a specified gaseous covalent bond is broken homolytically. | Positive; distinguish a particular bond from a mean bond enthalpy. |
| Lattice formation | One mole of solid ionic compound forms from separated gaseous ions. | Negative. |
| Lattice dissociation | One mole of solid ionic compound separates into gaseous ions. | Positive; opposite of lattice formation. |
Write the particles before inserting numbers
For an oxide, both electron affinities of oxygen are needed. For a chloride containing two chloride ions, use twice the first electron affinity of chlorine; a second electron affinity of chlorine would make Cl²− and describe the wrong substance. Likewise Mg²+ needs IE₁ + IE₂, not 2IE₁.
Construct a cycle that conserves matter
Start with the formation equation. Follow an alternative route: atomise the elements, ionise the metal, add electrons to the non-metal atoms, then form the solid lattice. Carry unused particles and electrons through each energy level so that the whole level has the same atoms and net charge. Upward arrows represent positive enthalpy changes in an energy-level diagram. Reverse any arrow and its sign reverses.
For MX₂ formed from M(s) and X₂(g), one mole of X₂ bonds is broken. If the data give atomisation per mole of X atoms, multiply that value by two. If they give X–X bond dissociation, use it once.
Diagram placeholder
Born–Haber energy-level diagram to add
Labels to include:
- Vertical axis: enthalpy
- M(s) + X₂(g) and MX₂(s)
- M(g) + 2X(g)
- M²⁺(g) + 2X(g) + 2e⁻
- M²⁺(g) + 2X⁻(g)
- Signed arrows and stoichiometric multipliers
Draw the formation arrow from the elements to MX₂(s). The alternative route climbs through atomisation and ionisation, falls through two first electron affinities, then falls through lattice formation. Separate the two ionisation steps when these are required. The direct and indirect changes must have equal sums.
Worked example: a fictional MX₂ dataset
Use these illustrative values, all in kJ mol⁻¹: formation −680; metal atomisation +160; IE₁ +520; IE₂ +1040; X₂ bond dissociation +220; first electron affinity −330 per mole of X atoms. These are constructed practice data, not a measured compound dataset.
The non-lattice sum is 160 + 520 + 1040 + 220 − 660 = 1280. Therefore −680 = 1280 + ΔlattH° and lattice formation is −1960 kJ mol⁻¹. Lattice dissociation would be +1960 kJ mol⁻¹.
To find a missing IE₂ instead, rearrange the same signed equation before substituting. Do not treat all arrows as positive and subtract a remembered list.
Quick checks
Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official AQA mark allocations.
Q1. Define lattice dissociation for MgO and give its equation.Show answer
The enthalpy change when one mole of solid MgO separates into its constituent gaseous ions. MgO(s) → Mg²⁺(g) + O²⁻(g). Under standard conditions this is the standard lattice dissociation enthalpy.
Q2. Why is oxygen’s second electron affinity positive?Show answer
An electron is added to O⁻(g). Energy is needed to overcome repulsion between the negative ion and the incoming electron.
Q3. A Br–Br bond dissociation enthalpy is 194 kJ mol⁻¹. Is bromine atomisation automatically +97?Show answer
Not from Br₂(l), bromine’s standard state at 298 K. Half the vaporisation enthalpy must also be included. Half the bond value only describes ½Br₂(g) → Br(g).
Q4. For fictional MX, formation = −410, atomisation terms = +90 and +110, IE₁ = +480 and EA₁ = −320. Find lattice formation.Show answer
Non-lattice sum = 90 + 110 + 480 − 320 = 360. Lattice formation = −410 − 360 = −770 kJ mol⁻¹.
Q5. In the worked MX₂ dataset, recover IE₂ using lattice formation −1960.Show answer
−680 = 160 + 520 + IE₂ + 220 − 660 − 1960. The known terms total −1720, so IE₂ = +1040 kJ mol⁻¹.
Sources
Sources and examiner guidance (reviewed 2 October 2026)
- AQA 7405 physical chemistry specification — 3.1.8 coverage and required skills.
- Chemrevise: Thermodynamics — Coverage checklist; explanations, data exercises and quick checks on this page are original Finesse material.
- AQA June 2023 Paper 1 mark scheme — Q05, pp18–19: solution and Born–Haber cycles; report Q05, p4. Definitions and examples are independently written.
- AQA June 2023 Paper 1 examiner report — Read alongside the question-specific marking guidance; not a universal wording checklist.
Finesse Tuition is not endorsed by AQA or Chemrevise. All explanations and examples here are our own.