Born–Haber Cycle

How the cycle works

Hess's law says that the total enthalpy change between two states of a system depends only on the states, not on the path. Born and Haber turned that abstract principle into a calculation tool. Take any ionic solid, draw two paths from its elements (in their standard states) to the crystal, and equate them.

The "direct" path is one step: combine the elements, measure ΔH_f in a calorimeter. The "long" path passes through gas-phase atoms, then gas-phase ions, then the crystal — five steps, all individually measurable. Setting the two paths equal exposes the lattice energy as the only unknown.

            ΔH_f (measured)
   Na(s) + ½Cl₂(g) ──────────────────→ NaCl(s)
       │                                  ↑
       │  ΔH_atom(Na)              U      │  lattice energy
       ↓                                  │
     Na(g) + ½Cl₂(g) ──→ Na(g) + Cl(g)    │
       │ IE(Na)         ↑                 │
       ↓                │ ½ΔH_diss(Cl₂)   │
     Na⁺(g) + e⁻ + Cl(g) ───────────→ Na⁺(g) + Cl⁻(g)
                          EA(Cl)

Reading clockwise: solid sodium evaporates to atoms, those atoms lose an electron, half a chlorine molecule dissociates, the chlorine atom gains the electron, and finally the gas-phase ions snap into the lattice. Sum of arrows up + sum of arrows down = ΔH_f.

Worked example — lattice energy of NaCl

Plug in the standard tabulated values (kJ/mol, 298 K):

Step                                    ΔH (kJ/mol)
1. Na(s) → Na(g)             ΔH_atom        +108
2. Na(g) → Na⁺(g) + e⁻       IE₁(Na)        +496
3. ½Cl₂(g) → Cl(g)           ½ΔH_diss(Cl₂)  +122
4. Cl(g) + e⁻ → Cl⁻(g)       EA(Cl)         −349
5. Na⁺(g) + Cl⁻(g) → NaCl(s) U                ?
─────────────────────────────────────────────────
   Total:                    ΔH_f(NaCl)      −411

Hess's law:
  ΔH_f = ΔH_atom + IE₁ + ½ΔH_diss + EA + U
  −411 = 108 + 496 + 122 − 349 + U
  −411 = 377 + U
  U = −788 kJ/mol

The off-by-one from the more commonly quoted −787 reflects rounding. The Born–Landé prediction we computed under "Lattice Energy" was −762 kJ/mol — the cycle gives the experimental value the model is being judged against.

Tabulated cycle inputs

Quantity	NaCl	MgO	CaCl₂	AgCl
ΔH_atom (metal)	+108	+148	+178	+285
IE₁	+496	+738	+590	+731
IE₂	—	+1451	+1145	—
½ΔH_diss(non-metal)	+122	+249 (O atom)	+244 (2 Cl)	+122
EA₁	−349	−141	−349 (×2)	−349
EA₂	—	+744	—	—
ΔH_f (measured)	−411	−602	−796	−127
U (cycle output)	−787	−3795	−2258	−905

All values in kJ/mol. Note three subtleties: the second ionization of magnesium is roughly twice the first (electron removed from a deeper shell); the second electron affinity of oxygen is endothermic (you must do work to add a second electron to an already-anionic O⁻); and silver chloride's enthalpy of formation is small in magnitude despite a large lattice energy because silver is hard to atomize.

Energy-level diagram (NaCl)

Energy (kJ/mol)
   +800 ┤            ●  Na⁺(g) + Cl(g) + e⁻
        │           ╱│
   +600 ┤          ╱ │ EA = −349
        │   IE +496  ●  Na⁺(g) + Cl⁻(g)
   +400 ┤        ╱   │
        │       ●    │
        │      ╱     │ U = −787
   +200 ┤     ╱      │
        │    ● Na(g) + Cl(g)
        │   ╱½diss=+122
      0 ┤  ●─Na(g) + ½Cl₂(g)
        │ ╱ atom=+108
        │●─Na(s) + ½Cl₂(g)              ●  NaCl(s) (final state, ΔH_f = −411)
   −200 ┤ ╲                            ╱
        │  ╲                          ╱
   −400 ┤   ╲────────────────────────●

  Two paths from elements to NaCl(s); both end at the same point.

Born–Haber vs related thermo cycles

	Born–Haber	Hess cycle (general)	Bond-energy cycle	Solvation cycle
Domain	Ionic solids	Any reaction	Covalent gas reactions	Dissolution
Unknown solved for	Lattice energy U	Any missing step	Bond enthalpy	Hydration enthalpy
Path through	Gas-phase ions	Any intermediate set	Atomized gas	Gas-phase ions, then aq
Inputs needed	ΔH_atom, IE, EA, ΔH_diss, ΔH_f	Any subset of ΔH steps	BDEs, ΔH_f	Lattice U, ΔH_solv
Compounds	Pure ionic	Any	Covalent molecules	Salts in water
Year	1919	1840 (Hess)	varies	1920s (Bernal–Fowler)
Typical accuracy	±0.5%	±0.5%	±2% (BDEs are less precise)	±2%

Variants and applications

• Multivalent oxides. For Al₂O₃ you ionize aluminium three times (5137 kJ/mol cumulative), atomize two oxygens, and add electrons twice each. The cycle has more steps but the same logic. Reproduces experimental U ≈ −15,916 kJ/mol.
• Predicting "impossible" compounds. Use the cycle in reverse: estimate U from Born–Landé for a hypothetical compound (NaCl₂, with Na²⁺), then check whether ΔH_f comes out positive (unstable) or negative (stable). NaCl₂ fails because IE₂(Na) = 4562 kJ/mol can't be paid back.
• Solubility analysis. Combine the Born–Haber cycle with the dissolution step ΔH_solv = −U − ΔH_hydration to get ΔH_dissolution. This is why high-lattice-energy salts (MgO, BaSO₄) are insoluble while low-lattice-energy salts dissolve.
• Polarization corrections. When the predicted ΔH_f from a "purely ionic" cycle disagrees badly with the measured value (silver halides, copper halides), the discrepancy is attributed to covalent character and is quantified by Fajans's rules.
• Mixed-cation cycles. For double salts (KMgCl₃, alums) the cycle uses additional atomization and ionization steps for each cation. Provided every elementary step is tabulated, the closure remains exact.

Common pitfalls

• Sign errors on electron affinity. By convention EA is reported as the energy released when an electron attaches, but tables sometimes list it as the energy required (positive value for the same process). Mixing conventions throws the cycle by twice the EA — typically a 700 kJ/mol error.
• Forgetting the second electron affinity is endothermic. O(g) + e⁻ → O⁻(g) is exothermic (−141 kJ/mol). But O⁻(g) + e⁻ → O²⁻(g) requires energy (+744 kJ/mol) because pushing a second electron onto an already-negative ion needs work. Neglecting this for oxide cycles gives wildly wrong U.
• Using ΔH_f at the wrong temperature. Standard tabulated ΔH_f are at 298 K. Lattice energies derived from them are 298 K values too; comparing to a 0 K Born–Landé prediction introduces small thermal-correction errors.
• Treating ½ΔH_diss as a bond energy. The atomization step requires breaking one Cl–Cl bond per two Cl atoms — so half the bond-dissociation enthalpy of Cl₂. Plugging in the full bond energy doubles the dissociation contribution.
• Skipping the atomization step. If your metal is liquid or gaseous in standard state, ΔH_atom may be small or zero. But for solids — most metals — you must include the sublimation/atomization step or you'll undercount by 100–300 kJ/mol.
• Confusing IE₁ + IE₂ with the cumulative ionization. For Mg²⁺ you need both ionizations summed (738 + 1451 = 2189 kJ/mol), not just one. Using IE₁ alone underestimates the cost of forming the dication by half.