Born–Haber Cycles

Spec Mapping — OCR H432 Module 5.2.1 — Born–Haber cycles, covering definitions of standard enthalpy of formation ($\Delta_f H^\ominus$Δf​H⊖), enthalpy of atomisation ($\Delta_{at} H^\ominus$Δat​H⊖), first and second ionisation energies ($\text{IE}_1$IE1​, $\text{IE}_2$IE2​), first and second electron affinities ($\text{EA}_1$EA1​, $\text{EA}_2$EA2​); construction of Born–Haber cycles for Group 1 halides, Group 2 halides, and Group 2 oxides; identification of the sign (endothermic or exothermic) of each step; and application of Hess's law to link lattice enthalpy to the experimentally measurable steps in the cycle (refer to the official OCR H432 specification document for exact wording).

A Born–Haber cycle is a particular kind of Hess cycle — specifically a thermodynamic loop that links the enthalpy of formation of an ionic compound to the lattice enthalpy via a sequence of experimentally accessible intermediate steps. It was published jointly in 1919 by Max Born (later Nobel laureate in physics, 1954, for his contributions to quantum mechanics) and Fritz Haber (Nobel laureate in chemistry, 1918, for the industrial synthesis of ammonia from N$_2$2​ and H$_2$2​ — paraphrased). The cycle solves the central problem identified in Lesson 1: lattice enthalpy cannot be measured directly, so we must find it indirectly by combining quantities that can be measured — enthalpies of formation (calorimetry), enthalpies of atomisation (vapour-pressure measurements and bond-dissociation spectroscopy), ionisation energies (photoelectron spectroscopy), and electron affinities (mass-spectrometric studies of ion-pair formation). This lesson develops the cycle qualitatively — definitions, signs, layout — for three representative compounds (NaCl, MgCl$_2$2​, MgO). Lesson 3 will then turn the cycle into a numerical calculation, including the sign-handling tricks that distinguish A from A* in this topic.

Key Equation: for an ionic compound $M_a X_b$Ma​Xb​ with metal cation $M^{n+}$Mn+ and non-metal anion $X^{m-}$Xm−, the Born–Haber cycle gives: $\Delta_f H^\ominus(M_a X_b) = a \cdot \Delta_{at} H^\ominus(M) + b \cdot \Delta_{at} H^\ominus(X) + a \sum_{k=1}^{n} \text{IE}_k(M) + b \sum_{k=1}^{m} \text{EA}_k(X) + \Delta_{LE} H^\ominus$Δf​H⊖(Ma​Xb​)=a⋅Δat​H⊖(M)+b⋅Δat​H⊖(X)+a∑k=1n​IEk​(M)+b∑k=1m​EAk​(X)+ΔLE​H⊖ where each step has its own well-defined sign. The art of Born–Haber bookkeeping is keeping every sign and every stoichiometric coefficient correct.

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Learning Objectives (OCR Spec 5.2.1)

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

• Define $\Delta_f H^\ominus$Δf​H⊖, $\Delta_{at} H^\ominus$Δat​H⊖, $\text{IE}_1$IE1​, $\text{IE}_2$IE2​, $\text{EA}_1$EA1​, and $\text{EA}_2$EA2​ with full OCR-mark-scheme precision
• Construct a Born–Haber cycle for a Group 1 halide (e.g. NaCl), a Group 2 halide (e.g. MgCl$_2$2​), and a Group 2 oxide (e.g. MgO)
• Recognise the sign of each step (endothermic or exothermic) and justify the sign physically
• Apply Hess's law to relate the steps in the cycle algebraically
• Identify where the lattice enthalpy sits in the cycle under the OCR formation convention

Why Do We Need a Cycle?

Lesson 1 established that lattice enthalpy cannot be measured directly because the gaseous-ion starting state is not experimentally accessible. The clever trick of the Born–Haber cycle is to connect the lattice-enthalpy step to a set of steps that can be measured, forming a closed thermodynamic loop. Hess's law (treated in the rates-and-equilibrium course) tells us that the total enthalpy change around any closed cycle is zero — equivalently, the enthalpy change for going A → B is independent of the route taken. If we know every step in the indirect route from elements in their standard states to the ionic solid except for the lattice-enthalpy step, we can solve for the unknown step by simple algebra.

A Born–Haber cycle is therefore a Hess-law cycle specifically designed to expose the lattice-enthalpy step. The cycle has two routes from the elements in their standard states to the ionic solid:

• Route 1 (direct): elements → ionic solid, enthalpy change $\Delta_f H^\ominus$Δf​H⊖ (measurable by combustion or solution calorimetry).
• Route 2 (indirect, via gaseous ions): elements → gaseous atoms → gaseous ions → ionic solid, with each sub-step individually measurable except for the final lattice-formation step.

Setting Route 1 = Route 2 gives a single equation whose only unknown is $\Delta_{LE} H^\ominus$ΔLE​H⊖, which we solve for.

The Required Definitions

You must be able to state each of these definitions with OCR-mark-scheme precision. Memorise them verbatim — paraphrasing usually loses one or two marks per definition.

Standard enthalpy of formation

$\Delta_f H^\ominus$Δf​H⊖ is the enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions.

Example: $\text{Na}(s) + \tfrac{1}{2}\,\text{Cl}_2(g) \rightarrow \text{NaCl}(s) \quad \Delta_f H^\ominus = -411\ \text{kJ mol}^{-1}$Na(s)+21​Cl2​(g)→NaCl(s)Δf​H⊖=−411 kJ mol−1

Enthalpy of atomisation

$\Delta_{at} H^\ominus$Δat​H⊖ is the enthalpy change when one mole of gaseous atoms is formed from the element in its standard state under standard conditions.

Examples:

• $\text{Na}(s) \rightarrow \text{Na}(g) \quad \Delta_{at} H^\ominus = +108\ \text{kJ mol}^{-1}$Na(s)→Na(g)Δat​H⊖=+108 kJ mol−1
• $\tfrac{1}{2}\,\text{Cl}_2(g) \rightarrow \text{Cl}(g) \quad \Delta_{at} H^\ominus = +122\ \text{kJ mol}^{-1}$21​Cl2​(g)→Cl(g)Δat​H⊖=+122 kJ mol−1
• $\tfrac{1}{2}\,\text{O}_2(g) \rightarrow \text{O}(g) \quad \Delta_{at} H^\ominus = +249\ \text{kJ mol}^{-1}$21​O2​(g)→O(g)Δat​H⊖=+249 kJ mol−1

Note that the atomisation of a diatomic gas starts from $\tfrac{1}{2}$21​ mole of molecules, because the definition requires one mole of atoms. Atomisation is always endothermic because bonds must be broken (in diatomic gases) or atoms must be freed from a metallic lattice (in solid metals). For Cl$_2$2​(g) the relationship to bond enthalpy is $\Delta_{at} H^\ominus(\text{Cl}) = \tfrac{1}{2} E(\text{Cl–Cl})$Δat​H⊖(Cl)=21​E(Cl–Cl).

First ionisation energy

$\text{IE}_1$IE1​ is the enthalpy change when one mole of gaseous 1+ ions is formed from one mole of gaseous atoms.

Example: $\text{Na}(g) \rightarrow \text{Na}^+(g) + \text{e}^- \quad \text{IE}_1 = +496\ \text{kJ mol}^{-1}$Na(g)→Na+(g)+e−IE1​=+496 kJ mol−1

Second ionisation energy

$\text{IE}_2$IE2​ is the enthalpy change when one mole of gaseous 2+ ions is formed from one mole of gaseous 1+ ions.

Example: $\text{Mg}^+(g) \rightarrow \text{Mg}^{2+}(g) + \text{e}^- \quad \text{IE}_2 = +1451\ \text{kJ mol}^{-1}$Mg+(g)→Mg2+(g)+e−IE2​=+1451 kJ mol−1

Ionisation energies are always endothermic because energy is needed to overcome the attraction between the nucleus and the outgoing electron. Successive ionisation energies increase (IE$_2$2​ > IE$_1$1​) because the electron is being removed from an increasingly positively charged species.

First electron affinity

$\text{EA}_1$EA1​ is the enthalpy change when one mole of gaseous 1− ions is formed from one mole of gaseous atoms.

Example: $\text{Cl}(g) + \text{e}^- \rightarrow \text{Cl}^-(g) \quad \text{EA}_1 = -349\ \text{kJ mol}^{-1}$Cl(g)+e−→Cl−(g)EA1​=−349 kJ mol−1

The first electron affinity is usually exothermic because the incoming electron is attracted to the positively charged nucleus of the neutral atom — energy is released when the system descends to a lower potential-energy well.

Second electron affinity

$\text{EA}_2$EA2​ is the enthalpy change when one mole of gaseous 2− ions is formed from one mole of gaseous 1− ions.

Example: $\text{O}^-(g) + \text{e}^- \rightarrow \text{O}^{2-}(g) \quad \text{EA}_2 = +798\ \text{kJ mol}^{-1}$O−(g)+e−→O2−(g)EA2​=+798 kJ mol−1

Crucially, the second electron affinity is endothermic (positive). This is because the electron is being forced onto a species that is already negatively charged, so there is electrostatic repulsion to overcome. The energy cost of this repulsion outweighs the attraction to the nucleus. Recognising this sign is a classic OCR examination discriminator — candidates who treat EA$_2$2​ as exothermic lose the MgO and CaO calculations entirely.

Signs of the Steps — Comparison Table

Step	Typical sign	Magnitude (kJ mol⁻¹)	Physical reason
Atomisation of metal	+	80–200	breaking metallic bonding
Atomisation of halogen (½ X₂)	+	75–150	breaking ½ X–X bond
Atomisation of O₂ (½ O₂)	+	249	breaking ½ O=O bond
First ionisation energy IE₁	+	380–900	electron removed from neutral atom
Second ionisation energy IE₂	+	1100–2000	from a 1+ ion (more energy needed)
Third ionisation energy IE₃	+	2700–7700	from a 2+ ion
First electron affinity EA₁	− (usually)	−70 to −350	attraction to nucleus of neutral atom
Second electron affinity EA₂	+	+600 to +800	electron repelled by existing − charge
Lattice enthalpy (OCR)	−	−700 to −16,000	ions form crystal lattice (exothermic)
Enthalpy of formation	− (usually)	−100 to −1700	net stability of the compound

Constructing the Born–Haber Cycle — NaCl

We want to link $\Delta_f H^\ominus(\text{NaCl})$Δf​H⊖(NaCl) to $\Delta_{LE} H^\ominus(\text{NaCl})$ΔLE​H⊖(NaCl). The cycle has two routes from Na(s) + ½Cl$_2$2​(g) to NaCl(s):

Route 1 (direct): $\text{Na}(s) + \tfrac{1}{2}\,\text{Cl}_2(g) \xrightarrow{\Delta_f H^\ominus} \text{NaCl}(s)$Na(s)+21​Cl2​(g)Δf​H⊖​NaCl(s)

Route 2 (indirect, via gaseous atoms then gaseous ions):

1. Atomise Na: $\text{Na}(s) \rightarrow \text{Na}(g)$Na(s)→Na(g), enthalpy change $\Delta_{at} H^\ominus(\text{Na}) = +108$Δat​H⊖(Na)=+108 kJ mol$^{-1}$−1
2. Atomise Cl: $\tfrac{1}{2}\,\text{Cl}_2(g) \rightarrow \text{Cl}(g)$21​Cl2​(g)→Cl(g), enthalpy change $\Delta_{at} H^\ominus(\text{Cl}) = +122$Δat​H⊖(Cl)=+122 kJ mol$^{-1}$−1
3. Ionise Na: $\text{Na}(g) \rightarrow \text{Na}^+(g) + \text{e}^-$Na(g)→Na+(g)+e−, enthalpy change $\text{IE}_1(\text{Na}) = +496$IE1​(Na)=+496 kJ mol$^{-1}$−1
4. Add electron to Cl: $\text{Cl}(g) + \text{e}^- \rightarrow \text{Cl}^-(g)$Cl(g)+e−→Cl−(g), enthalpy change $\text{EA}_1(\text{Cl}) = -349$EA1​(Cl)=−349 kJ mol$^{-1}$−1
5. Lattice formation: $\text{Na}^+(g) + \text{Cl}^-(g) \rightarrow \text{NaCl}(s)$Na+(g)+Cl−(g)→NaCl(s), enthalpy change $\Delta_{LE} H^\ominus$ΔLE​H⊖

By Hess's law (Route 1 = Route 2):

$\Delta_f H^\ominus(\text{NaCl}) = \Delta_{at} H^\ominus(\text{Na}) + \Delta_{at} H^\ominus(\text{Cl}) + \text{IE}_1(\text{Na}) + \text{EA}_1(\text{Cl}) + \Delta_{LE} H^\ominus$Δf​H⊖(NaCl)=Δat​H⊖(Na)+Δat​H⊖(Cl)+IE1​(Na)+EA1​(Cl)+ΔLE​H⊖

This single equation is the cycle in algebraic form. Lesson 3 will rearrange it to find $\Delta_{LE} H^\ominus$ΔLE​H⊖ numerically.

Born–Haber Cycle Diagram for NaCl

The conventional pictorial layout stacks energy levels vertically: endothermic steps go upward, exothermic steps go downward.

graph TD
  E["Na(s) + ½Cl₂(g)<br/>elements, ΔH = 0"] -->|+Δat H Na<br/>+108| D["Na(g) + ½Cl₂(g)"]
  D -->|+Δat H Cl<br/>+122| C["Na(g) + Cl(g)"]
  C -->|+IE₁ Na<br/>+496| A["Na⁺(g) + Cl(g) + e⁻"]
  A -->|+EA₁ Cl<br/>−349| B["Na⁺(g) + Cl⁻(g)<br/>gaseous ions"]
  B -->|+ΔLE H<br/>−787| F["NaCl(s)<br/>ionic solid"]
  E -->|ΔfH<br/>−411| F

The lattice-enthalpy arrow is the longest downward arrow in the diagram, ending at the ionic solid. All other arrows except $\text{EA}_1$EA1​ point upward (endothermic).

Constructing the Cycle for MgCl$_2$2​

MgCl$_2$2​ is slightly more involved because Mg must be doubly ionised (IE$_1$1​ and IE$_2$2​ both needed) and two Cl atoms are required per formula unit (the atomisation and electron-affinity steps for Cl each appear with a stoichiometric coefficient of 2).

Direct route: $\text{Mg}(s) + \text{Cl}_2(g) \rightarrow \text{MgCl}_2(s)$Mg(s)+Cl2​(g)→MgCl2​(s), $\Delta_f H^\ominus(\text{MgCl}_2) = -641$Δf​H⊖(MgCl2​)=−641 kJ mol$^{-1}$−1

Indirect route (six steps):