Lattice Enthalpy and Born-Haber Cycles — AQA A-level Chemistry

SPEC 3.1.8.1 • THERMODYNAMICS

Lattice Enthalpy & Born-Haber Cycles

Understanding ionic lattice stability through energy cycles and Hess's Law

Why Lattice Enthalpy?

When ionic compounds form, gaseous ions come together to create a solid lattice structure. This process releases a huge amount of energy, making ionic compounds very stable. But here's the problem: we cannot measure this energy change directly in a laboratory.

Think about it — you cannot simply combine gaseous Na⁺ ions with gaseous Cl⁻ ions in a beaker and measure the heat released. The ions don't exist in isolation under normal conditions!

The Solution

We use Born-Haber cycles — an application of Hess's Law that allows us to calculate lattice enthalpy indirectly using other enthalpy changes that CAN be measured experimentally.

This topic builds directly on your AS knowledge of enthalpy changes and Hess's Law. By the end of this lesson, you'll be able to construct Born-Haber cycles and calculate lattice enthalpies for any ionic compound.

What is Lattice Enthalpy?

Lattice enthalpy measures the strength of ionic bonding in a solid ionic compound. It can be defined in two ways — pay careful attention to which definition a question uses!

Lattice Formation

Enthalpy of lattice formation is the enthalpy change when one mole of an ionic solid is formed from its gaseous ions under standard conditions.Na⁺(g) + Cl⁻(g) → NaCl(s)Always EXOTHERMIC (negative)

Lattice Dissociation

Enthalpy of lattice dissociation is the enthalpy change when one mole of an ionic solid is separated into its gaseous ions under standard conditions.NaCl(s) → Na⁺(g) + Cl⁻(g)Always ENDOTHERMIC (positive)

Why Use Born-Haber Cycles?

Lattice enthalpy cannot be measured directly by experiment — we cannot easily create gaseous ions and measure the energy released when they form a lattice. Instead, we use multiple experimental values in a Born-Haber cycle to calculate it indirectly.

Critical Point

The two definitions have the same magnitude but opposite signs. For NaCl: lattice formation = −787 kJ mol⁻¹, lattice dissociation = +787 kJ mol⁻¹. Always check which definition your data uses!

Factors Affecting Lattice Enthalpy

The magnitude of lattice enthalpy depends on the strength of electrostatic attraction between ions. This is governed by two factors:

FactorEffect on Lattice EnthalpyExplanation

Ionic ChargeHigher charge → More exothermicGreater electrostatic attraction between ions (F ∝ q₁q₂)

Ionic RadiusSmaller radius → More exothermicIons can get closer together, stronger attraction (F ∝ 1/r²)

Why Larger Ions Give Less Exothermic Values

When ionic radius increases: (1) the charge is more spread out over the ion, reducing charge density, and (2) the ions are further apart in the lattice. Since electrostatic attraction depends on distance between ion centres, larger ions experience weaker attraction.

Comparing Compounds

MgO has a much more exothermic lattice enthalpy (−3850 kJ mol⁻¹) than NaCl (−787 kJ mol⁻¹) because:

Mg²⁺ and O²⁻ have higher charges than Na⁺ and Cl⁻ (2+ and 2− vs 1+ and 1−)

Mg²⁺ and O²⁻ are smaller ions than Na⁺ and Cl⁻

Key Enthalpy Definitions

Before constructing Born-Haber cycles, you must know these definitions precisely. All values are measured under standard conditions (298 K, 100 kPa):

TermDefinitionSign

Enthalpy of Formation (ΔH_f)Enthalpy change when 1 mole of a compound forms from its elements in their standard statesUsually −

Enthalpy of Atomisation (ΔH_at)Enthalpy change when 1 mole of gaseous atoms forms from the element in its standard stateAlways +

Ionisation Energy (IE)Energy required to remove 1 mole of electrons from 1 mole of gaseous atoms/ionsAlways +

Electron Affinity (EA)Enthalpy change when 1 mole of gaseous atoms gains 1 mole of electrons1st usually −

Bond EnthalpyEnergy required to break 1 mole of covalent bonds in gaseous moleculesAlways +

Important: Elements Have Zero Enthalpy of Formation

The enthalpy of formation of any element in its standard state is zero by definition. This is because no change occurs — the element is already in its standard state.

Atomisation vs Bond Enthalpy

For diatomic elements like Cl₂: atomisation produces 1 mole of atoms (½Cl₂ → Cl), while bond enthalpy breaks 1 mole of bonds (Cl₂ → 2Cl). So: ΔH_at = ½ × Bond Enthalpy. If you need 2 Cl atoms, double the atomisation value!

Understanding Electron Affinity

Electron affinity deserves special attention because its sign can vary:

First Electron Affinity

The enthalpy change when 1 mole of gaseous atoms each gain one electron to form 1 mole of gaseous ions with a 1− charge.Cl(g) + e⁻ → Cl⁻(g) EA₁ = −349 kJ mol⁻¹Usually exothermic — the atom "wants" the electron to complete its outer shell.

Second Electron Affinity

The enthalpy change when 1 mole of gaseous 1− ions each gain one electron to form 1 mole of gaseous ions with a 2− charge.O⁻(g) + e⁻ → O²⁻(g) EA₂ = +798 kJ mol⁻¹Always endothermic — energy is needed to overcome repulsion between the negative ion and incoming electron.

Common Mistake

Students often assume all electron affinities are exothermic. Remember: adding an electron to an already negative ion requires energy to overcome electrostatic repulsion!

Constructing Born-Haber Cycles

A Born-Haber cycle applies Hess's Law: the total enthalpy change is independent of the route taken. We create two routes from elements to ionic compound:

Two Routes

Direct route: Elements → Ionic compound (enthalpy of formation)

Indirect route: Elements → Gaseous atoms → Gaseous ions → Ionic compound

Steps in the indirect route (for NaCl):

1

Atomisation of metal: Na(s) → Na(g)

2

Ionisation of metal: Na(g) → Na⁺(g) + e⁻

3

Atomisation of non-metal: ½Cl₂(g) → Cl(g)

4

Electron affinity: Cl(g) + e⁻ → Cl⁻(g)

5

Lattice formation: Na⁺(g) + Cl⁻(g) → NaCl(s)

Diagram Conventions

Energy axis: Runs vertically — higher positions = higher energy. Endothermic processes (atomisation, ionisation) go UP. Exothermic processes (electron affinity, lattice formation) go DOWN. Arrow length is proportional to energy change.

Hess's Law Equation

ΔH_f = ΔH_at(Na) + IE₁(Na) + ΔH_at(Cl) + EA₁(Cl) + ΔH_LE

Born-Haber Cycle: NaCl

01Worked Example

Use the following data to calculate the lattice enthalpy of formation of sodium chloride (NaCl).

Enthalpy ChangeValue / kJ mol⁻¹

Enthalpy of formation of NaCl−411

Enthalpy of atomisation of Na+107

First ionisation energy of Na+496

Enthalpy of atomisation of Cl+122

First electron affinity of Cl−349

1

Apply Hess's Law: ΔH_f = ΔH_at(Na) + IE₁(Na) + ΔH_at(Cl) + EA₁(Cl) + ΔH_LE

2

Substitute values: −411 = (+107) + (+496) + (+122) + (−349) + ΔH_LE

3

Simplify: −411 = +376 + ΔH_LE

4

Rearrange: ΔH_LE = −411 − 376 = −787 kJ mol⁻¹

Answer: Lattice enthalpy of formation = −787 kJ mol⁻¹

Born-Haber Cycles: Group 2 Compounds

For compounds like MgCl₂ or CaCl₂, the cycle requires additional steps because Group 2 metals form 2+ ions:

Adaptations for MgCl₂

Two ionisation energies: Mg → Mg⁺ → Mg²⁺ (include both IE₁ and IE₂)

Double the chlorine: 2 × ΔH_at(Cl) and 2 × EA₁(Cl)

The Hess's Law equation becomes:

Formula for MgCl₂

ΔH_f = ΔH_at(Mg) + IE₁(Mg) + IE₂(Mg) + 2×ΔH_at(Cl) + 2×EA₁(Cl) + ΔH_LE

Remember

The formula of the compound tells you what to multiply: MgCl₂ means 2 chlorine atoms, so double all chlorine-related terms. CaBr₂ would mean 2 bromine atoms.

Born-Haber Cycle: MgCl₂

02Worked Example

Calculate the lattice enthalpy of formation of magnesium chloride (MgCl₂).

Enthalpy ChangeValue / kJ mol⁻¹

Enthalpy of formation of MgCl₂−641

Enthalpy of atomisation of Mg+148

First ionisation energy of Mg+738

Second ionisation energy of Mg+1451

Enthalpy of atomisation of Cl+122

First electron affinity of Cl−349

1

ΔH_f = ΔH_at(Mg) + IE₁ + IE₂ + 2×ΔH_at(Cl) + 2×EA₁(Cl) + ΔH_LE

2

−641 = (+148) + (+738) + (+1451) + 2(+122) + 2(−349) + ΔH_LE

3

−641 = 148 + 738 + 1451 + 244 − 698 + ΔH_LE

4

−641 = 1883 + ΔH_LE

5

ΔH_LE = −641 − 1883 = −2524 kJ mol⁻¹

Answer: Lattice enthalpy of formation = −2524 kJ mol⁻¹

Compounds Containing O²⁻ Ions

For oxides like Na₂O or MgO, we need to consider that oxygen forms a 2− ion, requiring two electron affinities:

First EA of Oxygen

O(g) + e⁻ → O⁻(g)EA₁ = −141 kJ mol⁻¹

Exothermic

Second EA of Oxygen

O⁻(g) + e⁻ → O²⁻(g)EA₂ = +798 kJ mol⁻¹

Endothermic!

Adaptations for Na₂O

Double the sodium: 2 × ΔH_at(Na) and 2 × IE₁(Na)

Both electron affinities: EA₁(O) + EA₂(O)

Why is EA₂ Endothermic?

Adding an electron to O⁻ means forcing a negative electron towards an already negative ion. Energy must be supplied to overcome this electrostatic repulsion.

03Worked Example

Use the Born-Haber cycle data below to calculate the first electron affinity of chlorine.

Enthalpy ChangeValue / kJ mol⁻¹

Enthalpy of formation of KCl−437

Enthalpy of atomisation of K+89

First ionisation energy of K+419

Enthalpy of atomisation of Cl+122

Lattice enthalpy of formation of KCl−718

1

ΔH_f = ΔH_at(K) + IE₁(K) + ΔH_at(Cl) + EA₁(Cl) + ΔH_LE

2

−437 = (+89) + (+419) + (+122) + EA₁(Cl) + (−718)

3

−437 = 630 − 718 + EA₁(Cl)

4

−437 = −88 + EA₁(Cl)

5

EA₁(Cl) = −437 + 88 = −349 kJ mol⁻¹

Answer: First electron affinity of chlorine = −349 kJ mol⁻¹

Theoretical vs Experimental Lattice Enthalpies

Lattice enthalpies can be calculated in two ways, and comparing them reveals important information about the nature of bonding:

Experimental Value

Calculated from a Born-Haber cycle using measured enthalpy data.

Based on real experimental measurements.

Theoretical Value

Calculated using the perfect ionic model — assumes ions are perfect spheres with purely electrostatic attraction.

Based on mathematical calculations.

The Perfect Ionic Model Assumes

Ions are perfect spheres with uniform charge distribution

Bonding is 100% ionic (purely electrostatic)

No covalent character or electron sharing

Comparing Lattice Enthalpies

The difference between experimental and theoretical values reveals the extent of ionic bonding:

CompoundExperimental / kJ mol⁻¹Theoretical / kJ mol⁻¹Difference

NaCl−787−76126 (3.4%)

MgCl₂−2524−2326198 (8.5%)

AgCl−905−770135 (17.5%)

AgI−876−736140 (19.0%)

Interpreting the Data

NaCl: Close agreement → nearly purely ionic

MgCl₂: Larger difference → Mg²⁺ has higher charge density, polarises Cl⁻

AgCl → AgI: Increasing difference → I⁻ is larger and more easily polarised than Cl⁻

Key Exam Point

A larger difference indicates more covalent character. Experimental values are more exothermic because covalent bonding adds extra bond strength beyond pure ionic attraction.

Ion Polarisation

The difference between experimental and theoretical lattice enthalpies arises due to ion polarisation.

Polarisation

The distortion of the electron cloud of an anion by a nearby cation. The cation's positive charge attracts the anion's electrons, pulling them towards it and distorting the spherical shape.

How Polarisation Creates Covalent Character

When the anion's electron cloud is distorted, electron density becomes shared between the two ions. This sharing of electrons is the defining feature of covalent bonding. The greater the polarisation, the more covalent character the bond has — and the stronger the overall bonding becomes.

Why Experimental Values Are More Exothermic

Covalent bonding adds extra bond strength beyond what purely ionic attraction provides. The theoretical model (assuming 100% ionic) underestimates the true bond strength, so experimental values are more negative.

Factors Affecting Ion Polarisation

Polarising Power of Cation

Small, highly charged cations have greater polarising power:

Smaller radius = higher charge density

Higher charge = stronger attraction for electrons

Example: Mg²⁺ polarises more than Na⁺

Polarisability of Anion

Large anions with high charge are more easily polarised:

Larger radius = electrons further from nucleus, more easily distorted

Higher charge = more electrons available to be pulled

Example: I⁻ is more polarisable than Cl⁻

Key Trends to Remember

Across Period 3: Na⁺ → Mg²⁺ → Al³⁺ — polarising power increases (smaller size, higher charge)

Down Group 7: F⁻ → Cl⁻ → Br⁻ → I⁻ — polarisability increases (larger radius)

Silver halides: AgF → AgCl → AgBr → AgI — covalent character increases

The spectrum from ionic to covalent:

Practice

Question 1:

ΔH_f(KBr) = −394 kJ mol⁻¹

ΔH_at(K) = +89 kJ mol⁻¹

IE₁(K) = +419 kJ mol⁻¹

ΔH_at(Br) = +112 kJ mol⁻¹

EA₁(Br) = −325 kJ mol⁻¹

1

ΔH_f = ΔH_at(K) + IE₁(K) + ΔH_at(Br) + EA₁(Br) + ΔH_LE

2

−394 = (+89) + (+419) + (+112) + (−325) + ΔH_LE

3

−394 = 295 + ΔH_LE

4

ΔH_LE = −394 − 295 = −689 kJ mol⁻¹

Practice

Question 2:

ΔH_f(CaCl₂) = −795 kJ mol⁻¹

ΔH_at(Ca) = +178 kJ mol⁻¹

IE₁(Ca) = +590 kJ mol⁻¹

IE₂(Ca) = +1145 kJ mol⁻¹

ΔH_at(Cl) = +122 kJ mol⁻¹

EA₁(Cl) = −349 kJ mol⁻¹

1

ΔH_f = ΔH_at(Ca) + IE₁ + IE₂ + 2×ΔH_at(Cl) + 2×EA₁(Cl) + ΔH_LE

2

−795 = (+178) + (+590) + (+1145) + 2(+122) + 2(−349) + ΔH_LE

3

−795 = 178 + 590 + 1145 + 244 − 698 + ΔH_LE

4

−795 = 1459 + ΔH_LE

5

ΔH_LE = −795 − 1459 = −2254 kJ mol⁻¹

Practice

Question 3:

ΔH_f(Na₂O) = −414 kJ mol⁻¹

ΔH_at(Na) = +107 kJ mol⁻¹

IE₁(Na) = +496 kJ mol⁻¹

ΔH_at(O) = +249 kJ mol⁻¹

EA₂(O) = +798 kJ mol⁻¹

ΔH_LE(Na₂O) = −2478 kJ mol⁻¹

1

For Na₂O: ΔH_f = 2×ΔH_at(Na) + 2×IE₁(Na) + ΔH_at(O) + EA₁(O) + EA₂(O) + ΔH_LE

2

−414 = 2(+107) + 2(+496) + (+249) + EA₁ + (+798) + (−2478)

3

−414 = 214 + 992 + 249 + EA₁ + 798 − 2478

4

−414 = −225 + EA₁

5

EA₁ = −414 + 225 = −189 kJ mol⁻¹

Summary

Key Definitions

TermDefinition

Lattice enthalpy (formation)Energy released when 1 mol ionic solid forms from gaseous ions

Lattice enthalpy (dissociation)Energy required to separate 1 mol ionic solid into gaseous ions

Ion polarisationDistortion of anion electron cloud by cation's positive charge

Key Exam Skills

Construct Born-Haber cycles for Group 1 and Group 2 compounds

Calculate any unknown value using Hess's Law

Adapt cycles for compounds with multiple atoms (MgCl₂, Na₂O)

Compare experimental vs theoretical values to identify covalent character

Explain polarisation: small, highly charged cations + large anions → covalent character