A-level chemistry · Chapter 11

Thermodynamics

Born–Haber cycles, entropy, Gibbs free energy

Lattice enthalpy of formation — the enthalpy change when one mole of a solid ionic compound is formed from its gaseous ions. Always exothermic (large and negative).

Born–Haber cycles

A Born–Haber cycle compares two routes to the same ionic solid. Route 1 is direct formation from the elements. Route 2 atomises both elements, ionises the metal, adds the electron affinity of the non-metal, then forms the lattice. By Hess's law the two routes are equal, so any missing term can be calculated.

Check it on NaCl: ΔH°f −411 = (+107 atomisation Na) + (+496 IE₁) + (+122 atomisation ½Cl₂) + (−349 EA) + lattice, which gives a lattice enthalpy of about −787 kJ mol⁻¹.

Comparing real lattices with the perfect ionic model is revealing. Agreement is good for NaCl, but large differences (e.g. AgI) mean covalent character — a small, highly charged cation polarises the anion.

Entropy and feasibility

ΔG = ΔH − TΔS  ·  feasible when ΔG ≤ 0  ·  switch-over at T = ΔH/ΔS

Entropy S rises with disorder — solid < liquid < gas — and with more moles of gas. An endothermic reaction can still be feasible if TΔS outweighs ΔH; that is exactly why ice melts above 0 °C.

Watch the units trap: ΔS is usually in J K⁻¹ mol⁻¹, so divide by 1000 before mixing it with ΔH in kJ.

3.1.8.1The NaCl Born–Haber cycle, fully worked

Worked example. Find the lattice enthalpy of formation of NaCl (kJ mol⁻¹): ΔH°f(NaCl) = −411; atomisation Na +107; first ionisation Na +496; atomisation ½Cl₂ +122; electron affinity Cl −349.
  1. The formation route equals atomise both elements + ionise + electron affinity + lattice formation.
  2. Substitute the numbers: −411 = 107 + 496 + 122 + (−349) + ΔH(latt).
  3. Rearranging gives ΔH(latt) = −411 − 376 = −787 kJ mol⁻¹.
The Born–Haber cycle for NaCl — the cycle closes
The Born–Haber cycle for NaCl — the cycle closes: +107 +122 +496 −349 −787 = −411 kJ mol⁻¹.Diagram: ChemLab original

3.1.8.1Dissolving: lattice vs hydration

ΔH(solution) = −ΔH(lattice formation) + Σ ΔH(hydration). For NaCl that is +787 + (−406 −378) = ≈ +3 kJ mol⁻¹ — slightly endothermic, so dissolving is driven by the entropy increase instead.

Dissolving NaCl
Dissolving NaCl: lattice dissociation up (+787), hydration down (−784) — the small difference (+3 kJ mol⁻¹) is ΔH of solution.Diagram: ChemLab original

Hydration enthalpy is more exothermic for smaller, more highly charged ions, because the ion–dipole attraction to water is stronger.

3.1.8.2ΔG worked example

Worked example — decomposing CaCO₃. ΔH = +178 kJ mol⁻¹, ΔS = +160 J K⁻¹ mol⁻¹.
  1. Convert the entropy first: ΔS = 0.160 kJ K⁻¹ mol⁻¹.
  2. The reaction is feasible when ΔG ≤ 0, so T ≥ ΔH/ΔS = 178 ÷ 0.160 ≈ 1110 K — which is why limekilns run hot.
Exam tip. Two classic traps: (1) J vs kJ in ΔS; (2) electron affinity questions — the FIRST is exothermic, the SECOND endothermic (forcing an electron onto an anion), and lattice enthalpy values quoted as "dissociation" flip the sign of everything.
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