Learning objective

Calculate lattice enthalpy from Born-Haber data.

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Flashcards

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Topic

Thermodynamics (A-level only)

Subtopic

Born-Haber cycles (A-level only)

AQA A Level ChemistryPhysical chemistry

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Short explanation

In the subtopic Born-Haber cycles (A-level only), this AQA A-Level Chemistry 7405 learning objective focuses on calculate lattice enthalpy from Born-Haber data. It belongs to Thermodynamics (A-level only), so revision should stay anchored to this exact subtopic rather than drifting into a generic GCSE-level chemistry summary. Approved keywords to use include lattice enthalpy. Lattice enthalpy. means the energy change when one mole of an ionic compound is formed from its gaseous ions Avoid the mistake of students often forget to include the sign of the lattice enthalpy when calculating it from Born-Haber data, leading to incorrect conclusions about the stability of the ionic compound; instead, to fix this, remember that lattice enthalpy is exothermic and should be treated as a negative value in calculations. Use the formula: Lattice Enthalpy = ΔHf (formation enthalpy) - (ΔHsub + ΔHionization + ΔHelectron + ΔHbond). Substitute the values correctly, ensuring to apply the negative sign for lattice enthalpy. For example, if ΔHf = -400 kJ/mol, ΔHsub = +100 kJ/mol, ΔHionization = +200 kJ/mol, ΔHelectron = -100 kJ/mol, and ΔHbond = +50 kJ/mol, the calculation would be: Lattice Enthalpy = -400 - (100 + 200 - 100 + 50) = -400 - 250 = -650 kJ/mol. Thus, the lattice enthalpy is -650 kJ/mol For exam answers, use the Born-Haber cycle to calculate lattice enthalpy by summing the enthalpy changes for each step

Key concepts

lattice enthalpyBorn-Haber cycle

Why it matters

This objective helps connect Born-Haber cycles (A-level only) to exam-style questions, flashcards, and revision notes for Thermodynamics (A-level only).

Common mistakes

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  • Incorrect Lattice Enthalpy Calculation: To fix this, remember that lattice enthalpy is exothermic and should be treated as a negative value in calculations. Use the formula: Lattice Enthalpy = ΔHf (formation enthalpy) - (ΔHsub + ΔHionization + ΔHelectron + ΔHbond). Substitute the values correctly, ensuring to apply the negative sign for lattice enthalpy. For example, if ΔHf = -400 kJ/mol, ΔHsub = +100 kJ/mol, ΔHionization = +200 kJ/mol, ΔHelectron = -100 kJ/mol, and ΔHbond = +50 kJ/mol, the calculation would be: Lattice Enthalpy = -400 - (100 + 200 - 100 + 50) = -400 - 250 = -650 kJ/mol. Thus, the lattice enthalpy is -650 kJ/mol.

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Calculate lattice enthalpy from Born-Haber data. | Thermodynamics… | ExamCompanion