Lattice Energy
Energy to separate ionic solid into gaseous ions
Born-Haber Cycle
ΔHf = ΔHsub + ΔHion + ΔHdiss + ΔHEA + U
- U = lattice energy
- ΔHf = enthalpy of formation
- ΔHsub = sublimation enthalpy
- ΔHion = ionization energy
- ΔHdiss = bond dissociation
- ΔHEA = electron affinity
Coulomb Estimate
U ∝ (Q₊ × Q₋) / r
Higher charges and smaller ionic radii lead to greater lattice energy.
Example: NaCl
Using Born-Haber cycle with experimental data:
ΔHf(NaCl) = -411 kJ/mol
ΔHsub(Na) = +108 kJ/mol
IE(Na) = +496 kJ/mol
½ ΔHdiss(Cl₂) = +122 kJ/mol
EA(Cl) = -349 kJ/mol
U = ΔHf - (ΔHsub + IE + ½ΔHdiss + EA)
U = -411 - (108 + 496 + 122 - 349) ≈ -788 kJ/mol
Answer: U ≈ -788 kJ/mol (energy released forming lattice)
Notes
- Higher lattice energy = more stable ionic compound.
- MgO has much higher U than NaCl due to +2/-2 charges.
- Sign convention varies; some define U as energy required (positive).