AQA A-Level Chemistry (7405)

## Learning Outcomes

- Construct a Born–Haber cycle for an ionic compound using correct species, state symbols, and enthalpy terms
- Calculate lattice enthalpy of formation, lattice dissociation enthalpy, and other unknown enthalpy values using Hess's law
- Distinguish between atomisation, bond dissociation, ionisation energy, and electron affinity in Born–Haber cycles
- Handle compounds with multiply charged ions, including Group 2 chlorides and oxides
- Explain why second electron affinity is often endothermic
- Compare experimental and theoretical lattice enthalpies and use the difference as evidence for covalent character

## AQA A-Level Chemistry (7405) Syllabus

For the AQA A-Level Chemistry (7405), you are expected to understand Born–Haber cycles as an application of thermodynamics and Hess's law, with a focus on the following syllabus points:

![Born–Haber cycle for sodium chloride showing enthalpy of formation, atomisation, ionisation energy, electron affinity, and lattice enthalpy steps.](https://static.pastpaperhero.com/images/aqa-a-level-chemistry-7405-thermodynamics-born-haber-cycles_00.png)

_Sodium chloride enthalpy pathway from the elements to gaseous ions and the ionic lattice, linked by Hess's law._

- Know how to construct Born–Haber cycles for ionic compounds from standard enthalpy of formation, atomisation enthalpies, ionisation energies, electron affinities, and lattice enthalpy
- Identify the difference between enthalpy of lattice formation and lattice dissociation enthalpy, including their sign conventions
- Explain why compounds with Group 2 metals need successive ionisation energies in the cycle
- Understand why oxide ions require both first and second electron affinity terms
- Calculate unknown enthalpy values from a Born–Haber cycle, not just lattice enthalpy
- Compare lattice enthalpies obtained from Born–Haber cycles with values from a perfect ionic model
- Explain how differences between theoretical and experimental values provide evidence for covalent character in ionic compounds

## Test Your Knowledge

_Attempt these questions before reading this article. If you find some difficult or cannot remember the answers, look more closely at that area during your revision._

1. What is the difference between lattice enthalpy of formation and lattice dissociation enthalpy?
2. Why does a Born–Haber cycle for MgO include a second ionisation energy and a second electron affinity?
3. How do you decide when to multiply an enthalpy value by 2 in a Born–Haber cycle?
4. Why is the second electron affinity of oxygen positive, even though the first electron affinity is negative?
5. What does it mean if the experimental lattice enthalpy is more exothermic than the theoretical value?

## Introduction

Born–Haber cycles are one of the most tested thermodynamics skills in AQA A-level Chemistry because they bring together several ideas that students often learn separately: enthalpy of formation, atomisation, bond breaking, ionisation energies, electron affinities, and Hess's law.

The key reason they matter is that lattice enthalpy cannot be measured directly. Instead, you build an indirect route from the elements in their standard states to the ionic solid, then use Hess's law to calculate the missing value. In Paper 3, this can appear as a direct calculation, an incomplete cycle to finish, or an explanation question about covalent character.

> **Key Term: Born–Haber cycle**
> An enthalpy cycle used to calculate lattice enthalpy or another unknown enthalpy change for an ionic compound by applying Hess's law.

### How a Born–Haber cycle is built

A Born–Haber cycle is just a Hess cycle for the formation of an ionic solid. The direct route is the standard enthalpy of formation. The indirect route breaks the process into realistic steps involving gaseous atoms and gaseous ions.

> **Key Term: Hess's law**
> The principle that the total enthalpy change for a reaction is the same whichever route is taken, provided the initial and final states are the same.

The standard pattern is:

1. Start with the elements in their **standard states**
2. Convert the elements into **gaseous atoms**
3. Remove electrons from the metal atom or atoms using **ionisation energies**
4. Add electrons to the non-metal atom or atoms using **electron affinities**
5. Form the ionic lattice from the gaseous ions
   > **Key Term: standard enthalpy of formation**
   > The enthalpy change when 1 mol of a compound is formed from its elements in their standard states under standard conditions.

For sodium chloride, the sequence is:

- Na(s) → Na(g) : atomisation of sodium
- 1/2Cl₂(g) → Cl(g) : atomisation of chlorine
- Na(g) → Na⁺(g) + e⁻ : first ionisation energy
- Cl(g) + e⁻ → Cl⁻(g) : first electron affinity
- Na⁺(g) + Cl⁻(g) → NaCl(s) : lattice formation enthalpy

The direct route is:

- Na(s) + 1/2Cl₂(g) → NaCl(s) : standard enthalpy of formation

So the relationship is:

ΔHf = atomisation + ionisation + electron affinity + lattice formation

Rearranging:

lattice formation enthalpy = ΔHf − all the other steps

This is where sign errors happen. Do not just memorise one rearranged equation without understanding the route.

> **Key Term: enthalpy of lattice formation**
> The enthalpy change when 1 mol of an ionic solid is formed from its gaseous ions under standard conditions.
>
> **Key Term: lattice dissociation enthalpy**
> The enthalpy change when 1 mol of an ionic solid is completely separated into its gaseous ions.

These two values have the same size but opposite signs:

- lattice formation is usually **negative** because bond formation releases energy
- lattice dissociation is **positive** because separating ions needs energy

A strong exam habit is to label arrows with equations, not just numbers. That forces you to think about what each step represents.

> **Test Tip:** If the exam asks for lattice dissociation but you calculate lattice formation, do not throw away the answer. Reverse the sign at the end.

### Worked examples and how to avoid common calculation mistakes

The biggest skill is turning a formula into the correct set of steps. The chemistry decides the cycle, not the diagram shape.

Take magnesium chloride, MgCl₂.

Formation equation:

Mg(s) + Cl₂(g) → MgCl₂(s)

Indirect steps:

- Mg(s) → Mg(g)
- Mg(g) → Mg⁺(g) + e⁻
- Mg⁺(g) → Mg²⁺(g) + e⁻
- Cl₂(g) → 2Cl(g)
- 2Cl(g) + 2e⁻ → 2Cl⁻(g)
- Mg²⁺(g) + 2Cl⁻(g) → MgCl₂(s)

Why are there more steps than for NaCl?

Because magnesium forms **Mg²⁺**, so you need both the first and second ionisation energies. Also, there are two chloride ions, so chlorine atomisation and chlorine electron affinity are both multiplied by 2.

Worked example using the values:

- ΔHf[MgCl₂(s)] = −642 kJ mol⁻¹
- atomisation of Mg = +148
- atomisation of Cl = +122
- first IE of Mg = +738
- second IE of Mg = +1451
- first EA of Cl = −349
- lattice formation = ?

Use:

lattice formation = ΔHf − [atomisation of Mg + 2(atomisation of Cl) + IE1 + IE2 + 2(EA of Cl)]

Substitute:

= −642 − [148 + 2(122) + 738 + 1451 + 2(−349)]

Now simplify carefully:

2(122) = 244  
2(−349) = −698

So bracket becomes:

148 + 244 + 738 + 1451 − 698 = 1883

Then:

lattice formation = −642 − 1883 = −2525 kJ mol⁻¹

That large negative value makes sense because a 2+ ion and two 1− ions form a strong ionic lattice.

Now compare that with MgO. Students often think oxide is the same pattern as chloride with “double electron affinity”. It is not.

For magnesium oxide, the indirect steps include:

- Mg(s) → Mg(g)
- Mg(g) → Mg⁺(g) + e⁻
- Mg⁺(g) → Mg²⁺(g) + e⁻
- 1/2O₂(g) → O(g)
- O(g) + e⁻ → O⁻(g)
- O⁻(g) + e⁻ → O²⁻(g)
- Mg²⁺(g) + O²⁻(g) → MgO(s)

The key detail is the oxygen part. You must use:

- first electron affinity of oxygen
- second electron affinity of oxygen

You do **not** use 2 × first electron affinity, because the second electron is being added to an already negative ion, not to a neutral atom.

> **Key Term: first electron affinity**
> The enthalpy change when 1 mol of gaseous atoms each gains 1 electron to form 1 mol of gaseous 1− ions.
>
> **Key Term: second electron affinity**
> The enthalpy change when 1 mol of gaseous 1− ions each gains 1 electron to form 1 mol of gaseous 2− ions.

The first electron affinity is often exothermic because an electron is attracted to a neutral atom. The second electron affinity is often endothermic because the electron is forced onto an already negative ion, so energy is needed to overcome repulsion.

Students also lose marks on diatomic elements. If the non-metal exists as Cl₂, O₂, or similar in its standard state, you must account for the bond.

For one oxygen atom:

1/2O₂(g) → O(g)

For two chlorine atoms:

Cl₂(g) → 2Cl(g)

This is why stoichiometry matters so much. A Born–Haber cycle is really a thermodynamic accounting exercise. Every atom and every electron must balance.

Common errors:

- confusing atomisation with bond dissociation
- forgetting 1/2 for O₂
- missing second ionisation energy for Group 2 metals
- using 2 × first electron affinity instead of first + second for oxide
- giving lattice formation when the question asked for lattice dissociation
  > **Test Tip:** Before calculating, write the ions that must appear in the gaseous stage. If your final gaseous ions are wrong, every later step will also be wrong.

### Calculating unknown values other than lattice enthalpy

The exam can hide the unknown anywhere in the cycle. You may be asked for atomisation enthalpy, second electron affinity, or enthalpy of formation. The method stays the same: write the full Hess relationship, then rearrange.

Suppose for MgO you are given:

- ΔHf = −602 kJ mol⁻¹
- atomisation of O = +249
- IE1 of Mg = +738
- IE2 of Mg = +1450
- EA1 of O = −141
- lattice formation = −3890
- atomisation of Mg = unknown

The Hess relationship is:

ΔHf = atomisation of Mg + atomisation of O + IE1 + IE2 + EA1 + EA2 + lattice formation

If the unknown is atomisation of Mg, rearrange so it is isolated. The logic is always the same: move every known value to the other side with the correct sign.

This is why understanding the cycle matters more than memorising one formula. The same cycle can produce several different questions.

A useful check is whether your answer is chemically sensible:

- atomisation enthalpy should usually be positive
- ionisation energies are positive
- first electron affinity is often negative
- second electron affinity for O is positive
- lattice formation is negative
- lattice dissociation is positive

If your answer breaks one of these usual patterns, check your algebra and arrow directions first.

### Experimental and theoretical lattice enthalpy: what the difference means

AQA does not only test calculations. You must also explain what lattice enthalpy values tell you about bonding.

> **Key Term: perfect ionic model**
> A model that assumes ions are perfectly spherical, bonding is purely ionic, and charge is evenly distributed.

Theoretical lattice enthalpy is calculated from this perfect ionic model. Experimental lattice enthalpy is obtained from a Born–Haber cycle using real data. Because the theoretical model assumes fully ionic bonding, any major difference between the two values shows that the real compound is not perfectly ionic.

If the **experimental** value is **more exothermic** than the **theoretical** value, the bonding in the real lattice is stronger than the ionic model predicts. This is evidence for **covalent character**.

Why? Because the cation can polarise the anion. In simple terms, the positive ion distorts the electron cloud of the negative ion. Once the electron cloud is distorted, the bonding is no longer purely electrostatic. There is some sharing of electron density, so the compound has partial covalent character.

> **Key Term: covalent character**
> Partial sharing of electron density in a bond that is not purely ionic.

This effect is greatest when:

- the cation is small and highly charged, so it has high charge density
- the anion is large, so its electron cloud is easier to distort

That is why magnesium halides show more covalent character than sodium halides.

Consider these patterns:

- sodium chloride: experimental and theoretical values are quite close
- magnesium iodide: experimental value is much more negative than theoretical

The reasoning is not just “magnesium compounds are more covalent.” You need to explain the mechanism:

- Mg²⁺ is smaller and more highly charged than Na⁺
- I⁻ is large and easily polarised
- so Mg²⁺ distorts I⁻ strongly
- this increases covalent character
- the real lattice is stronger than the perfect ionic model predicts
- so the experimental lattice enthalpy is more exothermic

A common weak answer is: “There is covalent character.” A stronger answer links the difference in values to polarisation and stronger bonding.

Another common mistake is the reverse conclusion. If the values are close together, that means the perfect ionic model works well, so there is **little** covalent character.

> **Test Tip:** For explanation questions, use this chain: experimental more negative than theoretical → bonding stronger than perfect ionic model predicts → anion polarised by cation → some covalent character.

## Key Point Checklist

_This article has covered the following key knowledge points:_

- A Born–Haber cycle applies Hess's law to the formation of an ionic solid
- The direct route is the standard enthalpy of formation from elements in their standard states
- The indirect route uses atomisation, ionisation energies, electron affinities, and lattice formation
- Enthalpy of lattice formation is negative, while lattice dissociation enthalpy is positive
- Lattice formation and lattice dissociation have the same magnitude but opposite signs
- Group 2 compounds need successive ionisation energies because 2+ ions are formed
- Oxide ions require both first and second electron affinity terms
- For diatomic elements such as O₂ and Cl₂, stoichiometry and bond breaking must be handled carefully
- You must multiply enthalpy terms by the number of atoms or ions required by the formula
- Unknown values other than lattice enthalpy can be calculated by rearranging the Hess relationship
- Theoretical lattice enthalpy uses a perfect ionic model, while experimental lattice enthalpy comes from a Born–Haber cycle
- If the experimental value is more exothermic than the theoretical value, this suggests covalent character caused by polarisation

## Key Terms and Concepts

- Born–Haber cycle
- Hess's law
- standard enthalpy of formation
- enthalpy of lattice formation
- lattice dissociation enthalpy
- first electron affinity
- second electron affinity
- perfect ionic model
- covalent character


Thermodynamics - Born-Haber cycles