ChemistryEnergetics

Born-Haber Cycle

The Born-Haber cycle is a thermochemical application of Hess's Law used to calculate the lattice energy of ionic crystalline solids. It relates the standard enthalpy of formation of an ionic compound to the energy required to atomize and ionize the constituent elements.

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Δ H_{f} = Δ H_{a t} + I E + E A + Δ H_{L E}

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Core idea

Overview

When to use: Use this cycle when direct experimental measurement of lattice enthalpy is not feasible. It is applicable for calculating any missing energetic component of an ionic compound's formation when the other thermodynamic values are known.

Why it matters: This cycle allows scientists to evaluate the strength of ionic bonds and the stability of crystals. Discrepancies between theoretical lattice energy and values derived from the cycle often reveal the degree of covalent character in a bond.

Remember it

Memory Aid

Phrase: Friends Always Invite Every Legend

Visual Analogy: Climb a ladder to break atoms (Atomisation) and strip electrons (IE), then slide down as electrons join (EA) and the lattice snaps together.

Exam Tip: Always draw the cycle arrows. Formation is the direct path; ensure signs (+/-) match the arrow direction relative to the standard definitions.

Why it makes sense

Intuition

Imagine a closed energy cycle, like a multi-stage journey, where the total energy change for forming an ionic compound from its elements is the sum of the energy changes for each intermediate step of atomization

Symbols

Variables

$Δ$ $H_{f}$ = Enthalpy of Formation, $Δ$ $H_{a t}$ (M) = Atomization (Metal), $Δ$ $H_{a t}$ (X) = Atomization (Non-metal), IE = Ionization Energy, EA = Electron Affinity

Δ H_{f}

Enthalpy of Formation

kJ/mol

Δ H_{a t} (M)

Atomization (Metal)

kJ/mol

Δ H_{a t} (X)

Atomization (Non-metal)

kJ/mol

Ionization Energy

kJ/mol

Electron Affinity

kJ/mol

Δ H_{L E}

Lattice Enthalpy

kJ/mol

Walkthrough

Derivation

Understanding the Born-Haber Cycle

Applies Hess’s Law to calculate lattice enthalpy by breaking ionic solid formation into gaseous steps.

Cycle steps are theoretical and use standard enthalpy values.

Use Hess’s Law Around the Cycle:

Formation enthalpy equals the sum of intermediate steps plus lattice enthalpy (with correct signs).

Δ_{f} H^{⊖} = \sum (atomisation, IE, EA) + Δ_{latt} H^{⊖}

Note: Exact steps depend on the ionic compound (number of ionisations/electron affinities).

Result

Δ_{f} H^{⊖} = \sum (atomisation, IE, EA) + Δ_{latt} H^{⊖}

Source: OCR A-Level Chemistry A — Energetics (Born–Haber cycles)

Where it shows up

Real-World Context

Explaining why NaCl is stable.

Avoid these traps

Common Mistakes

Sign errors (endo vs exo).
Forgetting atomization of diatomic elements.
Wrong electron affinity values.

Study smarter

Tips

Ensure stoichiometry is correct: if the formula is MX₂, ensure you double the EA and use appropriate atomization values.
Lattice enthalpy and enthalpy of formation are almost always negative (exothermic).
Ionization energy is always positive (endothermic), while electron affinity is usually negative for the first electron.
Check that all values use consistent units, typically kJ/mol.

Common questions

Frequently Asked Questions

Applies Hess’s Law to calculate lattice enthalpy by breaking ionic solid formation into gaseous steps.

Use this cycle when direct experimental measurement of lattice enthalpy is not feasible. It is applicable for calculating any missing energetic component of an ionic compound's formation when the other thermodynamic values are known.

This cycle allows scientists to evaluate the strength of ionic bonds and the stability of crystals. Discrepancies between theoretical lattice energy and values derived from the cycle often reveal the degree of covalent character in a bond.

Sign errors (endo vs exo). Forgetting atomization of diatomic elements. Wrong electron affinity values.

Explaining why NaCl is stable.

Ensure stoichiometry is correct: if the formula is MX₂, ensure you double the EA and use appropriate atomization values. Lattice enthalpy and enthalpy of formation are almost always negative (exothermic). Ionization energy is always positive (endothermic), while electron affinity is usually negative for the first electron. Check that all values use consistent units, typically kJ/mol.