Born-Haber Cycles

Learning Objectives (OCR Spec 5.2.1)

By the end of this lesson you should be able to:

• Define enthalpy of atomisation, first ionisation energy, first electron affinity, second electron affinity and standard enthalpy of formation
• Construct a Born-Haber cycle for a Group 1 halide and a Group 2 oxide
• Recognise the sign of each step (endothermic or exothermic) and explain why
• Apply Hess's law to relate the steps in the cycle
• Identify where the lattice enthalpy sits in the cycle under the OCR convention

Why Do We Need a Cycle?

Lesson 1 showed that the lattice enthalpy cannot be measured directly. The clever trick is to connect the lattice enthalpy to a set of steps that can be measured, forming a closed loop. Hess's law tells us that the total enthalpy change around a closed loop is zero, so if we know all the steps except one we can calculate the missing one.

A Born-Haber cycle is exactly this - a thermodynamic cycle that links the lattice enthalpy of an ionic compound to quantities such as enthalpy of formation, enthalpy of atomisation, ionisation energies and electron affinities.

The Required Definitions

You must be able to state each of these definitions precisely for full marks in OCR exams.

Standard enthalpy of formation (ΔH°_f): the enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions.

Example: Na(s) + 1/2 Cl2(g) -> NaCl(s) ΔH°_f = -411 kJ mol^-1

Enthalpy of atomisation (ΔH°_at): the enthalpy change when one mole of gaseous atoms is formed from the element in its standard state under standard conditions.

Examples:

• Na(s) -> Na(g) ΔH°_at = +108 kJ mol^-1
• 1/2 Cl2(g) -> Cl(g) ΔH°_at = +122 kJ mol^-1

Note that the atomisation of a diatomic gas starts from 1/2 mole of molecules, because the definition requires one mole of atoms. Atomisation is always endothermic because bonds must be broken or atoms freed from a lattice.

First ionisation energy (IE_1): the enthalpy change when one mole of gaseous 1+ ions is formed from one mole of gaseous atoms.

Example: Na(g) -> Na+(g) + e- IE_1 = +496 kJ mol^-1

Second ionisation energy (IE_2): the enthalpy change when one mole of gaseous 2+ ions is formed from one mole of gaseous 1+ ions.

Example: Mg+(g) -> Mg^2+(g) + e- IE_2 = +1451 kJ mol^-1

Ionisation energies are always endothermic because energy is needed to overcome the attraction between the nucleus and the outgoing electron.

First electron affinity (EA_1): the enthalpy change when one mole of gaseous 1- ions is formed from one mole of gaseous atoms.

Example: Cl(g) + e- -> Cl-(g) EA_1 = -349 kJ mol^-1

The first electron affinity is usually exothermic because the incoming electron is attracted to the positively charged nucleus.

Second electron affinity (EA_2): the enthalpy change when one mole of gaseous 2- ions is formed from one mole of gaseous 1- ions.

Example: O-(g) + e- -> O^2-(g) EA_2 = +798 kJ mol^-1

Crucially, the second electron affinity is endothermic (positive) because the electron is being forced onto a species that is already negatively charged, so there is electrostatic repulsion. Understanding this is a classic exam point.

Signs of the Steps: A Summary

Step	Typical sign	Why
Atomisation	+	bond-breaking / lattice-breaking
First IE	+	removing electron from neutral atom
Second IE	+	removing from positive ion (larger)
First EA	-	attraction to nucleus
Second EA	+	repulsion from existing negative charge
Lattice enthalpy (OCR)	-	ions attract to form lattice
Enthalpy of formation	usually -	net stability

Constructing a Born-Haber Cycle - NaCl

We want to link ΔH°_f(NaCl) to the lattice enthalpy of NaCl. The cycle has two routes from the elements in their standard states to the ionic solid:

Route 1 (direct): Na(s) + 1/2 Cl2(g) -> NaCl(s) with enthalpy change ΔH°_f

Route 2 (indirect, via gaseous ions):

1. Atomise Na: Na(s) -> Na(g), ΔH°_at(Na)
2. Atomise Cl: 1/2 Cl2(g) -> Cl(g), ΔH°_at(Cl)
3. Ionise Na: Na(g) -> Na+(g) + e-, IE_1(Na)
4. Add electron to Cl: Cl(g) + e- -> Cl-(g), EA_1(Cl)
5. Combine ions: Na+(g) + Cl-(g) -> NaCl(s), ΔH°_LE