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

Entropy and feasibility

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

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. Formation route = atomise both + ionise + electron affinity + lattice formation.
  2. −411 = 107 + 496 + 122 + (−349) + ΔH(latt)
  3. ΔH(latt) = −411 − 376 = −787 kJ mol⁻¹

3.1.8.1Dissolving: lattice vs hydration

3.1.8.2ΔG worked example

Worked example — decomposing CaCO₃. ΔH = +178 kJ mol⁻¹, ΔS = +160 J K⁻¹ mol⁻¹.
  1. Convert: ΔS = 0.160 kJ K⁻¹ mol⁻¹.
  2. Feasible when ΔG ≤ 0: T ≥ ΔH/ΔS = 178 ÷ 0.160 ≈ 1110 K — 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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