Cornell Note: Born-Haber Cycle

Cues / Keywords

Hess's Law
Lattice energy
Endothermic steps
Sublimation, IE, EA

Main Notes

The Born-Haber cycle applies Hess's Law to calculate lattice energy indirectly since it cannot be measured directly.

Steps for MX (e.g., NaCl):

Step	Process	Enthalpy Sign	Example (NaCl)
1. Sublimation	M(s) → M(g)	+ve (endothermic)	+108 kJ/mol
2. Bond dissociation	½ $X_{2}$ (g) → X(g)	+ve (endothermic)	+121 kJ/mol
3. Ionisation energy	M(g) → $M^{+}$ (g) + $e^{-}$	+ve (endothermic)	+496 kJ/mol
4. Electron affinity	X(g) + $e^{-}$ → $X^{-}$ (g)	−ve (exothermic)	−349 kJ/mol
5. Lattice formation	$M^{+}$ (g) + $X^{-}$ (g) → MX(s)	−ve (exothermic)	−787 kJ/mol
Net	M(s) + ½ $X_{2}$ (g) → MX(s)	−ve	−411 kJ/mol

Equation (Hess's Law): $\Delta H_f = \Delta H_{sub} + \frac{1}{2}\Delta H_{diss} + IE + EA + U$

Rearranging for lattice energy: $U = \Delta H_f - \Delta H_{sub} - \frac{1}{2}\Delta H_{diss} - IE - EA$

NaCl Calculation: $U = -411 - 108 - 121 - 496 - (-349) = -787 \text{ kJ/mol}$

Factors affecting lattice energy: $U \propto \frac{q^+ \times q^-}{r^+ + r^-}$ Higher charge → higher |U|. Smaller ions → higher |U|.

Summary / Recall

Born-Haber: 5 steps to get U. Only lattice formation is exothermic (besides EA). Higher charges and smaller ions give larger (more negative) lattice energy. Use: U = $\Delta Hf$ − sum of all other steps.