A Level Chemistry - How to Solve: Lattice Enthalpy and Born-Haber Cycles

How to Solve: Lattice Enthalpy and Born-Haber Cycles

Complete Guide For GCSE/A-Level Chemistry (Edexcel, AQA, OCR)

Introduction

"Mastering lattice enthalpy and Born-Haber cycles lets you predict whether a compound will form—and it’s worth up to 6 marks in your A-Level exam. Miss this, and you’ll lose easy marks on enthalpy calculations, ionic bonding, and even solubility trends."

WHAT YOU NEED TO KNOW FIRST

Before starting, you must understand:
1. Enthalpy changes (ΔH) – Exothermic (–ΔH) vs. endothermic (+ΔH).
2. Ionic bonding – How ions form and attract in a lattice.
3. Hess’s Law – The total enthalpy change is the same, regardless of the route taken.

KEY TERMS & FORMULAS

Key Terms

Formulas

1. Born-Haber Cycle Equation (for a 1:1 ionic compound, e.g., NaCl) ΔH_f° = ΔH_at (metal) + ΔH_at (non-metal) + ΔH_IE₁ (metal) + ΔH_EA₁ (non-metal) + ΔHₗₐₜₜ
2. ΔH_f° = Standard enthalpy of formation (given)
3. ΔH_at = Atomisation enthalpy (given)
4. ΔH_IE₁ = First ionisation energy (given)
5. ΔH_EA₁ = First electron affinity (given)
6. ΔHₗₐₜₜ = Lattice enthalpy (what we solve for)

MEMORISE THIS – You’ll rearrange it to find ΔHₗₐₜₜ.

1. For compounds with multiple ions (e.g., MgCl₂)
2. ΔH_IE₂ (second ionisation energy) is added.
3. 2 × ΔH_EA₁ (if two Cl⁻ ions form).

STEP-BY-STEP METHOD

Follow these steps for every Born-Haber cycle question:

1. Write the formation equation (1 mole of compound from elements in standard states). Example: Na(s) + ½Cl₂(g) → NaCl(s)

2. Draw the Born-Haber cycle as a diagram (or list steps in order):

3. Start with elements in standard states (Na(s) + ½Cl₂(g)).
4. Atomise the metal (Na(s) → Na(g)).
5. Atomise the non-metal (½Cl₂(g) → Cl(g)).
6. Ionise the metal (Na(g) → Na⁺(g) + e⁻).
7. Add electron to non-metal (Cl(g) + e⁻ → Cl⁻(g)).

8. Form the lattice (Na⁺(g) + Cl⁻(g) → NaCl(s)).

9. Label each step with its enthalpy change (given in the question or data booklet).

10. Apply Hess’s Law:

11. ΔH_f° = Sum of all steps in the cycle.

12. Rearrange to solve for ΔHₗₐₜₜ.

13. Check signs:

14. Exothermic steps (–ΔH): Lattice formation, electron affinity (usually).

15. Endothermic steps (+ΔH): Atomisation, ionisation.

16. Calculate and box your final answer.

WORKED EXAMPLES

Example 1 – Basic (NaCl)

Question: Calculate the lattice enthalpy of NaCl using the following data: - ΔH_f° (NaCl) = –411 kJ/mol - ΔH_at (Na) = +108 kJ/mol - ΔH_at (Cl) = +122 kJ/mol - ΔH_IE₁ (Na) = +496 kJ/mol - ΔH_EA₁ (Cl) = –349 kJ/mol

Solution:
1. Formation equation: Na(s) + ½Cl₂(g) → NaCl(s) ΔH_f° = –411 kJ/mol

1. Born-Haber steps:
2. Na(s) → Na(g) ΔH_at (Na) = +108 kJ/mol
3. ½Cl₂(g) → Cl(g) ΔH_at (Cl) = +122 kJ/mol
4. Na(g) → Na⁺(g) + e⁻ ΔH_IE₁ (Na) = +496 kJ/mol
5. Cl(g) + e⁻ → Cl⁻(g) ΔH_EA₁ (Cl) = –349 kJ/mol

6. Na⁺(g) + Cl⁻(g) → NaCl(s) ΔHₗₐₜₜ = ?

7. Apply Hess’s Law: ΔH_f° = ΔH_at (Na) + ΔH_at (Cl) + ΔH_IE₁ (Na) + ΔH_EA₁ (Cl) + ΔHₗₐₜₜ –411 = +108 + +122 + +496 + (–349) + ΔHₗₐₜₜ

8. Solve for ΔHₗₐₜₜ: –411 = +377 + ΔHₗₐₜₜ ΔHₗₐₜₜ = –411 – 377 = –788 kJ/mol

What we did and why: We used the Born-Haber cycle to break down the formation of NaCl into steps, then rearranged to find the lattice enthalpy. The negative sign confirms it’s exothermic (energy released when ions form a lattice).

Example 2 – Medium (MgO)

Question: Calculate the lattice enthalpy of MgO using: - ΔH_f° (MgO) = –602 kJ/mol - ΔH_at (Mg) = +148 kJ/mol - ΔH_at (O) = +249 kJ/mol - ΔH_IE₁ (Mg) = +738 kJ/mol - ΔH_IE₂ (Mg) = +1451 kJ/mol - ΔH_EA₁ (O) = –141 kJ/mol - ΔH_EA₂ (O) = +798 kJ/mol

Solution:
1. Formation equation: Mg(s) + ½O₂(g) → MgO(s) ΔH_f° = –602 kJ/mol

1. Born-Haber steps:
2. Mg(s) → Mg(g) ΔH_at (Mg) = +148 kJ/mol
3. ½O₂(g) → O(g) ΔH_at (O) = +249 kJ/mol
4. Mg(g) → Mg⁺(g) + e⁻ ΔH_IE₁ (Mg) = +738 kJ/mol
5. Mg⁺(g) → Mg²⁺(g) + e⁻ ΔH_IE₂ (Mg) = +1451 kJ/mol
6. O(g) + e⁻ → O⁻(g) ΔH_EA₁ (O) = –141 kJ/mol
7. O⁻(g) + e⁻ → O²⁻(g) ΔH_EA₂ (O) = +798 kJ/mol

8. Mg²⁺(g) + O²⁻(g) → MgO(s) ΔHₗₐₜₜ = ?

9. Apply Hess’s Law: ΔH_f° = ΔH_at (Mg) + ΔH_at (O) + ΔH_IE₁ (Mg) + ΔH_IE₂ (Mg) + ΔH_EA₁ (O) + ΔH_EA₂ (O) + ΔHₗₐₜₜ –602 = +148 + +249 + +738 + +1451 + (–141) + +798 + ΔHₗₐₜₜ

10. Solve for ΔHₗₐₜₜ: –602 = +3243 + ΔHₗₐₜₜ ΔHₗₐₜₜ = –602 – 3243 = –3845 kJ/mol

What we did and why: MgO has a 2+ ion, so we included two ionisation energies for Mg and two electron affinities for O. The large negative lattice enthalpy shows strong ionic bonding.

Example 3 – Exam-Style (CaF₂)

Question: Calculate the lattice enthalpy of CaF₂ using: - ΔH_f° (CaF₂) = –1220 kJ/mol - ΔH_at (Ca) = +178 kJ/mol - ΔH_at (F) = +79 kJ/mol - ΔH_IE₁ (Ca) = +590 kJ/mol - ΔH_IE₂ (Ca) = +1145 kJ/mol - ΔH_EA₁ (F) = –328 kJ/mol

Solution:
1. Formation equation: Ca(s) + F₂(g) → CaF₂(s) ΔH_f° = –1220 kJ/mol

1. Born-Haber steps:
2. Ca(s) → Ca(g) ΔH_at (Ca) = +178 kJ/mol
3. F₂(g) → 2F(g) ΔH_at (F) = +79 kJ/mol × 2 = +158 kJ/mol
4. Ca(g) → Ca⁺(g) + e⁻ ΔH_IE₁ (Ca) = +590 kJ/mol
5. Ca⁺(g) → Ca²⁺(g) + e⁻ ΔH_IE₂ (Ca) = +1145 kJ/mol
6. F(g) + e⁻ → F⁻(g) ΔH_EA₁ (F) = –328 kJ/mol × 2 = –656 kJ/mol

7. Ca²⁺(g) + 2F⁻(g) → CaF₂(s) ΔHₗₐₜₜ = ?

8. Apply Hess’s Law: ΔH_f° = ΔH_at (Ca) + ΔH_at (F) + ΔH_IE₁ (Ca) + ΔH_IE₂ (Ca) + 2 × ΔH_EA₁ (F) + ΔHₗₐₜₜ –1220 = +178 + +158 + +590 + +1145 + (–656) + ΔHₗₐₜₜ

9. Solve for ΔHₗₐₜₜ: –1220 = +1415 + ΔHₗₐₜₜ ΔHₗₐₜₜ = –1220 – 1415 = –2635 kJ/mol

What we did and why: CaF₂ has two F⁻ ions, so we doubled the atomisation and electron affinity values. The large lattice enthalpy reflects strong ionic bonds in CaF₂.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP

"Right, listen up—this is your last-minute lifesaver for lattice enthalpy and Born-Haber cycles. Here’s what you must remember:

1. Born-Haber cycles break down the formation of an ionic compound into steps: atomisation, ionisation, electron affinity, and lattice formation.
2. Lattice enthalpy is always exothermic (–ΔH)—it’s the energy released when gaseous ions form a solid.
3. For 2+ or 2– ions (e.g., MgO, CaF₂), double the atomisation and electron affinity values.
4. Write the full equation first, then rearrange to find ΔHₗₐₜₜ.
5. Check signs: Lattice formation = –, ionisation = +, electron affinity = usually –.

If you see MgO or CaF₂ in the exam, don’t panic—just remember to include all ionisation steps and double the non-metal values. Now go smash that question!"