Born–Haber Cycles with Worked Examples

A Born–Haber cycle is an energy cycle, based on Hess’s law, that links the formation of an ionic compound from its elements to a series of individual steps: turning elements into gaseous atoms, turning atoms into ions, and bringing ions together into a lattice. Because lattice energy can’t be measured directly, Born–Haber cycles are the standard way to find it. This article explains each step, gives a reliable method, and works through four examples.

The idea

Take sodium chloride. There are two routes from the elements to the solid:

Direct route: Na(s) + ½Cl₂(g) → NaCl(s), with enthalpy change ΔH_f (the enthalpy of formation, which can be measured by calorimetry).

Indirect route, through gaseous ions:

1. Turn solid sodium into gaseous sodium atoms.
2. Turn chlorine molecules into gaseous chlorine atoms.
3. Remove an electron from each sodium atom.
4. Add an electron to each chlorine atom.
5. Bring the gaseous ions together to form the lattice.

By Hess’s law, the total enthalpy change is the same by both routes. If we know every term except one, we can calculate the unknown — usually the lattice energy.

The terms you need

Term	Definition	Sign
Enthalpy of formation, ΔH_f	1 mol of compound formed from its elements in their standard states	Usually −
Enthalpy of atomisation, ΔH_at	1 mol of gaseous atoms formed from the element in its standard state	Always +
First ionisation energy, IE₁	1 mol of electrons removed from 1 mol of gaseous atoms: M(g) → M⁺(g) + e⁻	Always +
Second ionisation energy, IE₂	M⁺(g) → M²⁺(g) + e⁻	Always +, larger than IE₁
First electron affinity, EA₁	1 mol of electrons added to 1 mol of gaseous atoms: X(g) + e⁻ → X⁻(g)	Usually −
Second electron affinity, EA₂	X⁻(g) + e⁻ → X²⁻(g)	Always + (adding an electron to a negative ion)
Lattice formation enthalpy, ΔH_latt	1 mol of solid formed from gaseous ions	Always −

A common trap: the atomisation of chlorine is for forming one mole of Cl atoms, so it’s half the Cl–Cl bond enthalpy: ½ × 243 = +122 kJ mol⁻¹. Similarly for oxygen: ½ × 498 = +249 kJ mol⁻¹.

The method

1. Write the formation equation and the lattice equation.
2. List every step of the indirect route, with the right number of moles of each (e.g. two chlorine atoms for CaCl₂).
3. Apply Hess’s law in this form:

ΔH_f = (sum of all steps to make gaseous ions) + ΔH_latt

4. Rearrange for the unknown.
5. Check the sign makes sense: lattice formation enthalpies are always negative.

If you draw the cycle as an energy-level diagram, endothermic steps go up and exothermic steps go down.

Example 1: sodium chloride

Data (kJ mol⁻¹): ΔH_f(NaCl) = −411; ΔH_at(Na) = +107; IE₁(Na) = +496; ΔH_at(Cl) = +122; EA₁(Cl) = −349.

Steps to gaseous ions:

Step	Equation	ΔH
Atomise Na	Na(s) → Na(g)	+107
Ionise Na	Na(g) → Na⁺(g) + e⁻	+496
Atomise Cl	½Cl₂(g) → Cl(g)	+122
Electron affinity	Cl(g) + e⁻ → Cl⁻(g)	−349
Sum		+376

Hess’s law: −411 = +376 + ΔH_latt

ΔH_latt = −411 − 376 = −787 kJ mol⁻¹

Notice: making the gaseous ions costs +376 kJ mol⁻¹ overall. The reaction is still strongly exothermic only because the lattice energy releases so much more (see ionic bonding explained).

Example 2: potassium chloride

Data: ΔH_f(KCl) = −437; ΔH_at(K) = +89; IE₁(K) = +419; ΔH_at(Cl) = +122; EA₁(Cl) = −349.

• Sum of steps: 89 + 419 + 122 − 349 = +281
• ΔH_latt = −437 − 281 = −718 kJ mol⁻¹

Compare with NaCl (−787): KCl’s lattice energy is smaller in magnitude, because K⁺ is larger than Na⁺ (see lattice energy).

Example 3: magnesium oxide (two ionisations, two electron affinities)

Data: ΔH_f(MgO) = −602; ΔH_at(Mg) = +148; IE₁(Mg) = +738; IE₂(Mg) = +1,451; ΔH_at(O) = +249; EA₁(O) = −141; EA₂(O) = +798.

Steps:

Step	ΔH
Mg(s) → Mg(g)	+148
Mg(g) → Mg⁺(g) + e⁻	+738
Mg⁺(g) → Mg²⁺(g) + e⁻	+1,451
½O₂(g) → O(g)	+249
O(g) + e⁻ → O⁻(g)	−141
O⁻(g) + e⁻ → O²⁻(g)	+798
Sum	+3,243

ΔH_latt = −602 − 3,243 = −3,845 kJ mol⁻¹

Two things stand out:

• Making Mg²⁺ and O²⁻ costs a huge +3,243 kJ mol⁻¹ — including the endothermic second electron affinity, because an electron is being forced onto an ion that’s already negative.
• The lattice energy is about five times that of NaCl, because both ions are doubly charged. That’s what makes the whole process favourable.

The second electron affinity of oxygen can’t be measured directly and is itself usually worked out from Born–Haber cycles, so published values vary (and published MgO lattice energies range from about −3,790 to −3,890 kJ mol⁻¹ depending on the data used). Always use the numbers given in the question.

Example 4: calcium chloride (two chlorine atoms)

Data: ΔH_f(CaCl₂) = −796; ΔH_at(Ca) = +178; IE₁(Ca) = +590; IE₂(Ca) = +1,145; ΔH_at(Cl) = +122; EA₁(Cl) = −349.

Because there are two chloride ions, the chlorine steps are doubled:

Step	ΔH
Ca(s) → Ca(g)	+178
Ca(g) → Ca²⁺(g) + 2e⁻ (IE₁ + IE₂)	+590 + 1,145 = +1,735
Cl₂(g) → 2Cl(g) (2 × 122)	+244
2Cl(g) + 2e⁻ → 2Cl⁻(g) (2 × −349)	−698
Sum	+1,459

ΔH_latt = −796 − 1,459 = −2,255 kJ mol⁻¹

The most common error in this type of question is forgetting to double the chlorine atomisation and electron affinity.

Example 5: finding a different unknown

Born–Haber cycles can be rearranged to find any one missing quantity. Suppose for NaCl you’re given the lattice formation enthalpy (−787) but not the electron affinity of chlorine:

−411 = 107 + 496 + 122 + EA₁ + (−787) −411 = −62 + EA₁ EA₁ = −349 kJ mol⁻¹ ✓

This is how many second electron affinities (such as oxygen’s EA₂) are obtained in the first place.

Why Born–Haber cycles matter

• They give experimental lattice energies, which can be compared with values calculated from a purely ionic model. A big difference reveals covalent character (see lattice energy).
• They explain why compounds exist in the formulas they do — for example, why magnesium forms MgO and MgCl₂ with Mg²⁺, and why sodium doesn’t form NaCl₂.
• They show that ionic compounds form because of lattice energy, not because electron transfer itself is favourable.

Common mistakes

• Using the full bond enthalpy for atomising chlorine or oxygen instead of half.
• Forgetting to double terms for compounds like CaCl₂ or Na₂O.
• Getting the sign of EA₂ wrong — it’s always positive.
• Mixing up lattice formation and dissociation signs.
• Arithmetic sign errors — write each term with its sign and add carefully.

Practice

Use the data to find the lattice formation enthalpy of sodium oxide, Na₂O: ΔH_f = −414; ΔH_at(Na) = +107; IE₁(Na) = +496; ΔH_at(O) = +249; EA₁(O) = −141; EA₂(O) = +798.

Answer: Sum = 2(107) + 2(496) + 249 − 141 + 798 = 214 + 992 + 249 − 141 + 798 = +2,112. ΔH_latt = −414 − 2,112 = −2,526 kJ mol⁻¹.

Key takeaways

• A Born–Haber cycle applies Hess’s law: ΔH_f = (steps to make gaseous ions) + ΔH_latt.
• Steps: atomise both elements, ionise the metal, add electrons to the non-metal (electron affinity), then form the lattice.
• Use half the bond enthalpy for diatomic non-metals, and multiply terms by the number of ions.
• EA₂ is positive; lattice formation enthalpies are negative.
• The cycle shows that lattice energy drives ionic compound formation.