23.1 A2 Level

Lattice energy and Born-Haber cycles

8 flashcards to master this topic

Definition Flip

Define the term 'enthalpy change of atomisation' (ΔHₐt).

Answer Flip

The enthalpy change of atomisation is the enthalpy change when one mole of gaseous atoms is formed from its element in its standard state under standard conditions.

Example: the atomisation of sodium, Na(s) → Na(g), is endothermic.
Definition Flip

Define 'lattice energy' (ΔHlatt) for an ionic solid.

Answer Flip

Lattice energy is the enthalpy change when one mole of a solid ionic compound is formed from its gaseous ions under standard conditions. It is always exothermic (negative value) because energy is released when ions come together to form a stable lattice.

Key Concept Flip

What is the purpose of a Born-Haber cycle?

Answer Flip

A Born-Haber cycle is an application of Hess's Law to calculate the lattice energy of an ionic compound. It links the enthalpy change of formation of an ionic solid with other enthalpy changes (

Example: ionisation energy, electron affinity, atomisation).
Key Concept Flip

Outline the steps required to construct a Born-Haber cycle for NaCl.

Answer Flip

1. Start with elements in their standard states: Na(s) + 1/2Cl₂(g). 2. Atomisation: Na(g) + Cl(g). 3. Ionisation: Na⁺(g) + Cl(g) + e⁻. 4. Electron affinity: Na⁺(g) + Cl⁻(g). 5. Lattice formation: NaCl(s).

Calculation Flip

Given the following enthalpy changes, calculate the lattice energy of MgO: ΔHformation = -602 kJ/mol, ΔHat (Mg) = +148 kJ/mol, IE₁ (Mg) = +738 kJ/mol, IE₂ (Mg) = +1451 kJ/mol, ΔHat (O) = +249 kJ/mol, EA₁ (O) = -141 kJ/mol, EA₂ (O) = +798 kJ/mol.

Answer Flip

Using the Born-Haber cycle: ΔHformation = ΔHat(Mg) + IE₁ + IE₂ + ΔHat(O) + EA₁ + EA₂ + ΔHlatt. Therefore, ΔHlatt = ΔHformation - [ΔHat(Mg) + IE₁ + IE₂ + ΔHat(O) + EA₁ + EA₂] = -602 - [148 + 738 + 1451 + 249 - 141 + 798] = -3845 kJ/mol

Key Concept Flip

How does ionic charge affect the magnitude of lattice energy?

Answer Flip

Lattice energy is directly proportional to the product of the ionic charges. Higher ionic charges lead to stronger electrostatic attractions and thus a more negative (larger magnitude) lattice energy.

Example: MgO (charges +2 and -2) has a much larger lattice energy than NaCl (charges +1 and -1).
Key Concept Flip

How does ionic radius affect the magnitude of lattice energy?

Answer Flip

Lattice energy is inversely proportional to the sum of the ionic radii. Larger ionic radii lead to weaker electrostatic attractions (ions are further apart) and thus a less negative (smaller magnitude) lattice energy.

Example: the lattice energy of NaCl is less negative than that of LiCl because Na⁺ has a larger radius than Li⁺.
Key Concept Flip

Explain why the second electron affinity of oxygen is endothermic.

Answer Flip

The second electron affinity of oxygen is endothermic because a negatively charged O⁻ ion is forced to accept another negatively charged electron. This requires energy to overcome the electrostatic repulsion between the existing negative charge and the incoming electron.

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22.2 Mass spectrometry 23.2 Enthalpies of solution and hydration