🟣 This is Higher Level (HL) content.
📘 What is a Born-Haber Cycle?
A Born-Haber cycle is a thermochemical cycle based on Hess's Law that breaks down the formation of an ionic compound into individual, measurable steps. Its primary purpose is to calculate lattice enthalpy indirectly, since this value cannot be measured experimentally.
Key Enthalpy Terms
| Term | Symbol | Definition | Sign |
|---|---|---|---|
| Enthalpy of formation | ΔHf⦵ | Enthalpy change when 1 mol of compound forms from elements in standard states | Usually – |
| Enthalpy of atomisation | ΔHat⦵ | Enthalpy change to produce 1 mol of gaseous atoms from element in standard state | Always + |
| 1st ionisation energy | IE₁ | Energy to remove 1 mol of electrons from 1 mol of gaseous atoms | Always + |
| 2nd ionisation energy | IE₂ | Energy to remove a second electron from M⁺(g) | Always + |
| 1st electron affinity | EA₁ | Energy change when 1 mol of gaseous atoms gains 1 electron | Usually – |
| 2nd electron affinity | EA₂ | Energy change when X⁻(g) gains another electron | Always + (endothermic) |
| Lattice enthalpy (IB def.) | ΔHlatt⦵ | Energy to separate 1 mol of ionic solid into gaseous ions at infinite distance | Always + |
Born-Haber Cycle for NaCl
Calculating Lattice Enthalpy
By Hess's Law, the enthalpy of formation equals the sum of all other steps:
\( \Delta H_f^\ominus = \Delta H_{at}(M) + \Delta H_{at}(X) + IE + EA - \Delta H_{latt}^\ominus \)
Rearranging for lattice enthalpy:
\( \Delta H_{latt}^\ominus = -\Delta H_f^\ominus + [\Delta H_{at}(M) + \Delta H_{at}(X) + IE + EA] \)
Worked Example: Lattice Enthalpy of NaCl
Given data:
- ΔHat(Na) = +107 kJ mol⁻¹
- ΔHat(Cl) = +121 kJ mol⁻¹
- IE₁(Na) = +496 kJ mol⁻¹
- EA₁(Cl) = –349 kJ mol⁻¹
- ΔHf(NaCl) = –411 kJ mol⁻¹
Calculation:
ΔHlatt = –(–411) + [107 + 121 + 496 + (–349)]
= 411 + 375
ΔHlatt = +786 kJ mol⁻¹
MgCl₂: Key Differences
When constructing a Born-Haber cycle for MgCl₂, watch the stoichiometry:
- Double the atomisation of Cl: need 2 mol Cl(g), so multiply ΔHat(Cl) by 2
- Two ionisation energies: Mg forms Mg²⁺, so include both IE₁ and IE₂
- Double the electron affinity: forming 2 mol Cl⁻(g), so multiply EA₁(Cl) by 2
Theoretical vs Experimental Lattice Enthalpy
| Compound | Theoretical | Experimental | Discrepancy |
|---|---|---|---|
| NaCl | +766 | +786 | Small (mostly ionic) |
| NaBr | +732 | +742 | Small (mostly ionic) |
| AgCl | +770 | +905 | Large (significant covalent character) |
| AgI | +736 | +876 | Large (significant covalent character) |
Why the discrepancy? Theoretical values assume a perfect ionic model with 100% ionic bonding. When the experimental value is significantly larger than theoretical, this indicates covalent character due to polarisation: a small, highly charged cation (e.g. Ag⁺) distorts the electron cloud of a large anion (e.g. I⁻).
⚠️ Common Mistakes
- Forgetting to multiply atomisation/EA for MX₂ or M₂X compounds
- Missing the 2nd ionisation energy for 2+ cations
- Not including state symbols on every step
- Confusing the sign: IB defines lattice enthalpy as endothermic (+)
⚠️ Exam Tip
If asked to draw a Born-Haber cycle, always: (1) start from elements in their standard states, (2) go upward through atomisation and ionisation steps, (3) come down through electron affinity and lattice formation, (4) show the electrons explicitly at each step.