Born-Haber Cycles

Born-Haber cycles extend the idea of Hess's law to ionic compounds. They allow us to calculate lattice energies and other thermochemical quantities that cannot be measured directly. Understanding Born-Haber cycles requires you to know all the component enthalpy changes, how they fit together in sequence, and how to extract any missing value from the cycle.

Lattice Energy

Lattice energy (ΔH°latt) can be defined in two ways — make sure you know which convention your exam board uses.

Formation convention (Edexcel): The lattice energy is the enthalpy change when one mole of an ionic lattice is formed from its gaseous ions under standard conditions.

Na⁺(g) + Cl⁻(g) → NaCl(s) ΔH°latt = −787 kJ mol⁻¹

This value is always negative (exothermic) because the electrostatic attraction between ions releases energy.

The dissociation convention (used by some other boards) is the reverse: the energy required to completely separate one mole of ionic lattice into gaseous ions. This would be +787 kJ mol⁻¹ for NaCl. Always check the sign convention being used in a question.

The Component Steps

A Born-Haber cycle for the formation of an ionic compound from its elements involves several steps. For NaCl:

Step 1: Atomisation of the metal

Na(s) → Na(g) ΔH°at = +107 kJ mol⁻¹

Converting the metal from its standard state to gaseous atoms. Always endothermic.

Step 2: Atomisation of the non-metal

½Cl₂(g) → Cl(g) ΔH°at = +121 kJ mol⁻¹

Converting the non-metal from its standard state to gaseous atoms. Always endothermic.

Step 3: First ionisation energy of the metal

Na(g) → Na⁺(g) + e⁻ IE₁ = +496 kJ mol⁻¹

Removing an electron from the gaseous metal atom. Always endothermic.

Step 4: First electron affinity of the non-metal

Cl(g) + e⁻ → Cl⁻(g) EA₁ = −349 kJ mol⁻¹

Adding an electron to the gaseous non-metal atom. Usually exothermic for the first electron (negative value).

Step 5: Lattice energy

Na⁺(g) + Cl⁻(g) → NaCl(s) ΔH°latt = −787 kJ mol⁻¹

The gaseous ions come together to form the ionic lattice. Always exothermic (formation convention).

The overall process: Enthalpy of formation

Na(s) + ½Cl₂(g) → NaCl(s) ΔH°f = −411 kJ mol⁻¹

The Born-Haber Cycle Diagram

The cycle connects the enthalpy of formation (the direct route) with all the individual steps (the indirect route). By Hess's law, the two routes must give the same overall enthalpy change.

graph TD
    A["Na(s) + ½Cl₂(g)"] -->|"ΔH°f = −411"| F["NaCl(s)"]
    A -->|"ΔH°at[Na] = +107"| B["Na(g) + ½Cl₂(g)"]
    B -->|"ΔH°at[Cl] = +121"| C["Na(g) + Cl(g)"]
    C -->|"IE₁ = +496"| D["Na⁺(g) + Cl(g)"]
    D -->|"EA₁ = −349"| E["Na⁺(g) + Cl⁻(g)"]
    E -->|"ΔH°latt = ?"| F

By Hess's law:

ΔH°f = ΔH°at(Na) + ΔH°at(Cl) + IE₁ + EA₁ + ΔH°latt

Rearranging to find lattice energy:

ΔH°latt = ΔH°f − ΔH°at(Na) − ΔH°at(Cl) − IE₁ − EA₁

ΔH°latt = (−411) − (+107) − (+121) − (+496) − (−349)

ΔH°latt = −411 − 107 − 121 − 496 + 349

ΔH°latt = −786 kJ mol⁻¹

(The small difference from the quoted value of −787 is due to rounding in the data.)

Step-by-Step Method for Constructing Born-Haber Cycles

Follow this systematic procedure:

Start with the elements in their standard states at the bottom left or top of your diagram.
Write the direct route: elements → ionic compound (this is ΔH°f).
Atomise the metal: convert the metal from its standard state to gaseous atoms.
Atomise the non-metal: convert the non-metal from its standard state to gaseous atoms.
Ionise the metal: apply first (and second, if needed) ionisation energies.
Add electron(s) to the non-metal: apply first (and second, if needed) electron affinities.
Form the lattice: combine gaseous ions to form the ionic solid.
Close the cycle by applying Hess's law: the sum around the cycle must equal zero.

Tip: When drawing energy level diagrams vertically, endothermic steps go UP and exothermic steps go DOWN. The lattice energy step is always a large downward arrow.

Born-Haber Cycle for MgCl₂: A More Complex Example

For MgCl₂, the metal forms a 2+ ion and there are two chloride ions per formula unit. This adds extra steps.

Step	Process	ΔH / kJ mol⁻¹
Atomisation of Mg	Mg(s) → Mg(g)	+148
Atomisation of Cl	Cl₂(g) → 2Cl(g) [i.e. 2 × 121]	+242
1st ionisation energy of Mg	Mg(g) → Mg⁺(g) + e⁻	+738
2nd ionisation energy of Mg	Mg⁺(g) → Mg²⁺(g) + e⁻	+1451
1st electron affinity of Cl (×2)	2Cl(g) + 2e⁻ → 2Cl⁻(g) [i.e. 2 × (−349)]	−698
Lattice energy	Mg²⁺(g) + 2Cl⁻(g) → MgCl₂(s)	?
Enthalpy of formation	Mg(s) + Cl₂(g) → MgCl₂(s)	−641

graph TD
    A["Mg(s) + Cl₂(g)"] -->|"ΔH°f = −641"| G["MgCl₂(s)"]
    A -->|"ΔH°at[Mg] = +148"| B["Mg(g) + Cl₂(g)"]
    B -->|"2×ΔH°at[Cl] = +242"| C["Mg(g) + 2Cl(g)"]
    C -->|"IE₁ = +738"| D["Mg⁺(g) + 2Cl(g)"]
    D -->|"IE₂ = +1451"| E["Mg²⁺(g) + 2Cl(g)"]
    E -->|"2×EA₁ = −698"| F["Mg²⁺(g) + 2Cl⁻(g)"]
    F -->|"ΔH°latt = ?"| G

Calculation:

ΔH°latt = ΔH°f − ΔH°at(Mg) − 2ΔH°at(Cl) − IE₁ − IE₂ − 2EA₁

ΔH°latt = (−641) − (+148) − (+242) − (+738) − (+1451) − (−698)

ΔH°latt = −641 − 148 − 242 − 738 − 1451 + 698

ΔH°latt = −2522 kJ mol⁻¹

The large magnitude reflects the 2+ and 1− charges and relatively small ionic radii.

Important Notes on Electron Affinity

First electron affinity (EA₁) is the enthalpy change when one mole of electrons is added to one mole of gaseous atoms: X(g) + e⁻ → X⁻(g). For most non-metals this is exothermic (negative).
Second electron affinity (EA₂) is the enthalpy change when a second electron is added to a 1− ion: X⁻(g) + e⁻ → X²⁻(g). This is always endothermic (positive) because you are forcing a negative electron onto an already negative ion, which requires energy to overcome the repulsion.

For oxygen:

EA₁: O(g) + e⁻ → O⁻(g) ΔH = −141 kJ mol⁻¹
EA₂: O⁻(g) + e⁻ → O²⁻(g) ΔH = +798 kJ mol⁻¹

The overall process of forming O²⁻ from O is endothermic (+657 kJ mol⁻¹). Oxide ions only exist in ionic lattices because the large exothermic lattice energy more than compensates for this endothermic step.

Common Mistakes in Born-Haber Cycles

Mistake	Correction
Forgetting to double the atomisation or EA for compounds like MgCl₂	There are 2 Cl atoms per formula unit, so multiply by 2
Using the wrong sign for EA₂	EA₂ is always endothermic (positive)
Confusing ionisation energy with electron affinity	IE removes electrons from metals; EA adds electrons to non-metals
Forgetting the 2nd ionisation energy for 2+ metals	Mg²⁺ requires both IE₁ and IE₂
Drawing the lattice energy arrow in the wrong direction	Lattice formation is exothermic (downward on an energy level diagram)

Reference Table: Lattice Energies

Compound	Lattice energy / kJ mol⁻¹	Ion charges
NaF	−930	1+, 1−
NaCl	−787	1+, 1−
NaBr	−747	1+, 1−
NaI	−704	1+, 1−
KCl	−711	1+, 1−
MgO	−3850	2+, 2−
MgCl₂	−2523	2+, 1−
CaO	−3401	2+, 2−
CaCl₂	−2258	2+, 1−

Notice how lattice energy becomes more exothermic with higher charges and smaller ions.

Summary

A Born-Haber cycle is a Hess's law cycle applied to the formation of an ionic compound.
It includes atomisation, ionisation, electron affinity, and lattice energy steps.
Any one unknown value can be calculated if all others are known.
For compounds with 2+ cations, include both IE₁ and IE₂.
For compounds with two halide ions per formula unit, double the atomisation and EA values.
Always check your signs carefully at each step.

A-Level Deep Dive: Born-Haber Cycles

Spec mapping

Edexcel 9CH0 specification Topic 13 requires construction and use of Born–Haber cycles to calculate lattice enthalpies (formation or dissociation) for ionic compounds, using ΔH°_at, ΣIE, ΣEA and ΔH°_f as inputs (refer to the official specification document for exact wording). The cycle: ΔH°_f = ΔH°_at(metal) + ½n·ΔH°_at(non-metal) + ΣIE + ΣEA + ΔH°_lat(formation), with sign and stoichiometry handled carefully. Born–Haber questions are characteristic Paper 1 quantitative items (8–10 marks) and form a backbone of Paper 3 synoptic analysis. Candidates must distinguish the experimental (Born–Haber) lattice energy from the theoretical (Madelung-style, perfect-ionic) value, and recognise that discrepancies indicate degree of covalent character. The Edexcel data booklet provides ionisation energies, electron affinities, and atomisation enthalpies for selected elements; lattice energies for common compounds are also tabulated.

Worked example with full mark scheme

Question (10 marks):

Use the data below to calculate the experimental lattice enthalpy of formation of MgO(s).

ΔH°_f[MgO(s)] = −602 kJ mol⁻¹ ΔH°_at[Mg(s)] = +148 kJ mol⁻¹ First ionisation energy of Mg = +738 kJ mol⁻¹ Second ionisation energy of Mg = +1451 kJ mol⁻¹ ΔH°_at[½O₂(g)] = +249 kJ mol⁻¹ First electron affinity of O = −141 kJ mol⁻¹ Second electron affinity of O = +798 kJ mol⁻¹

(a) Construct a labelled Born–Haber cycle. (4) (b) Calculate ΔH°_lat(formation) for MgO(s). (4) (c) Identify whether the second EA is endothermic or exothermic and explain why. (2)

Solution with mark scheme:

(a) Cycle (top to bottom): Mg(s) + ½O₂(g) → Mg²⁺(g) + O²⁻(g) → MgO(s); alternatively bottom: Mg(s) + ½O₂(g) → MgO(s) directly via ΔH°_f.

M1 — atomisation steps shown for both Mg and O. M1 — both ionisation energies of Mg shown stepwise. M1 — both electron affinities of O shown stepwise (note: 2nd EA is positive). M1 — lattice formation arrow Mg²⁺(g) + O²⁻(g) → MgO(s) labelled correctly.

(b) Apply Hess: ΔH°_f = ΔH°_at[Mg] + ΔH°_at[½O₂] + IE₁ + IE₂ + EA₁ + EA₂ + ΔH°_lat(form).

Substitute: −602 = 148 + 249 + 738 + 1451 + (−141) + 798 + ΔH°_lat(form).

M1 — correct cycle equation. M1 — correct substitution including signs.

ΔH°_lat(form) = −602 − [148 + 249 + 738 + 1451 − 141 + 798] = −602 − 3243 = −3845 kJ mol⁻¹

M1 — correct arithmetic of the bracketed sum. A1 — final answer with sign and units.

(c) B1 — Second EA is endothermic (+798 kJ mol⁻¹). B1 — adding a second electron to O⁻(g) requires energy because the negative ion repels the incoming electron — electrostatic repulsion outweighs the attraction to the nuclear charge after one electron has already been gained.

Total: 10 marks (M7 A1 B2).