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The Born-Haber cycle determines the lattice enthalpy of a salt, which cannot be measured directly: it decomposes the formation from the elements into measurable partial steps and applies Hess's law.
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Formula
ΔH_lattice in kJ/mol · ΔH°f in kJ/mol · ΔH_Sub in kJ/mol · I in kJ/mol · D in kJ/mol · EA in kJ/mol
| Symbol | Meaning | Unit |
|---|---|---|
| Lattice enthalpy (lattice formation from the gaseous ions) | kJ/mol | |
| ΔH°f | Standard enthalpy of formation of the salt | kJ/mol |
| Sublimation enthalpy of the metal | kJ/mol | |
| Ionization energy of the metal atom | kJ/mol | |
| Dissociation energy of the non-metal molecule | kJ/mol | |
| Electron affinity of the non-metal atom (usually negative) | kJ/mol |
\Delta H_{\text{Gitter}} = \Delta H_f^0 - \Delta H_{\text{Sub}} - I - \tfrac{1}{2} D - E_{\text{EA}}Original schematic by Quanta
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Max Born and Fritz Haber developed the cycle in 1919. The basis is Hess's law: the enthalpy of formation of the salt from the elements equals the enthalpy sum of the detour via gaseous atoms and ions. For NaCl: sublimation of sodium, ionization, half dissociation of Cl2, electron affinity of chlorine, finally the lattice formation. The smaller the ions and the higher their charge, the more strongly negative the lattice enthalpy (MgO around versus NaCl at −788 kJ/mol).
Holds exactly because enthalpy is a state function; it requires consistent signs and complete partial steps of the cycle at standard conditions.
The salt formation is written as a detour via gaseous atoms and ions; by Hess both paths have equal enthalpy.
Standard enthalpy of formation
The complete cycle in one line.
Electron affinity
Historically, quantities that were hard to measure were determined this way.
Calculate the lattice enthalpy of NaCl from the Born-Haber data.
°.
Forming the lattice from the gaseous ions is strongly exothermic.
KCl: ΔH°, , , , . Lattice enthalpy?
; smaller in magnitude than for NaCl because the K+ ion is larger.
Getting the sign of the electron affinity wrong.
For Cl, EA = −349 kJ/mol (exothermic); on rearranging it becomes +349.
Using the full dissociation energy of Cl2.
Only one Cl atom is needed per NaCl, hence ½D.
Confusing lattice enthalpy with the magnitude of the lattice energy.
Forming the lattice from gaseous ions is negative; tables often list the positive value for separating it.
Forgetting or double-counting partial steps.
Draw the full cycle: sublimation, ionization, dissociation, electron affinity, lattice.
Formula cluster
Applies Hess's law to ionic crystals.
NaCl: ΔH°, , , , (all kJ/mol). .
Explaining the stability of ionic crystals, comparing melting points and hardness, estimating salt solubility, ruling out hypothetical compounds such as NaCl2
Here is how to work through a typical Lattice Energy (Born-Haber Cycle) (ΔH_Gitter = ΔH°f − ΔH_Sub − I − ½D − EA) task step by step:
Task
Calculate the lattice enthalpy of NaCl from the Born-Haber data.
Solution path
°.
Forming the lattice from the gaseous ions is strongly exothermic.
Task
KCl: ΔH°, , , , . Lattice enthalpy?
Solution path
; smaller in magnitude than for NaCl because the K+ ion is larger.
The Born-Haber cycle is a thermodynamic trick to determine the lattice enthalpy of an ionic crystal, which cannot be measured directly. It relies on Hess's law: because enthalpy is a state function, the total energy depends only on the initial and final states, not on the path. You compare two paths from the elements to the solid salt: the direct path via the standard enthalpy of formation and a detour via gaseous atoms and ions. The detour consists of measurable partial steps: sublimation of the metal, ionization, dissociation of the non-metal, electron affinity and finally the sought lattice formation. Setting both paths equal lets you calculate the lattice enthalpy as the only unknown.
For NaCl the detour runs through five partial steps. First the sublimation of solid sodium to gaseous Na atoms, . Second the ionization of the Na atoms to Na+ ions, ionization energy . Third the dissociation of the chlorine molecule; since only one Cl atom is needed per formula unit, half the bond energy enters, . Fourth the uptake of an electron by the Cl atom, electron affinity , exothermic. Fifth the union of the gaseous ions into the solid crystal lattice, the sought lattice enthalpy. The sum of all five steps must equal the directly measured standard enthalpy of formation ΔH°. From this the lattice enthalpy follows.
The lattice enthalpy is defined as the energy released when isolated gaseous ions assemble into one mole of solid ionic crystal, or conversely the energy needed to separate them. There is no experiment that carries out just this one step: you cannot simply produce one mole of Na+ and one mole of Cl− gaseous ions and let them condense in a controlled way to measure the heat. Therefore one takes the detour via the Born-Haber cycle, in which all the other partial steps are individually measurable. The lattice enthalpy then follows by calculation as the difference. This approach is a classic example of how Hess's law makes inaccessible quantities calculable indirectly.
The lattice energy is essentially determined by the Coulomb attraction between the ions and therefore depends on two factors: the charge of the ions and their distance, that is their radii. The higher the charges, the stronger the attraction; that is why magnesium oxide MgO with doubly charged ions has a very large lattice energy in magnitude of about , while sodium chloride with singly charged ions is only around . The smaller the ions, the shorter the distance and the stronger the attraction; that is why the lattice energy of NaCl is larger in magnitude than that of KCl, because the potassium ion is larger. These relationships explain trends in melting points, hardness and solubility of salts.
In the Born-Haber cycle all partial steps must be added with the correct sign, otherwise the result is wrong. The electron affinity of chlorine is exothermic, the Cl atom releases energy when it takes up an electron, so . If you accidentally insert a positive value, the calculated lattice enthalpy shifts by almost , a gross error. When rearranging the cycle equation for the lattice enthalpy, the sign of the subtracted terms also flips, so that −349 becomes +349 there. This is exactly where most calculation errors happen. It helps to draw the cycle completely as a diagram and label each arrow with magnitude and sign before forming the sum.
Quanta exam set
Curated exam set for "Lattice Energy (Born-Haber Cycle)":
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Which formula describes Lattice Energy (Born-Haber Cycle)?
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How do you rearrange Lattice Energy (Born-Haber Cycle) for Standard enthalpy of formation?
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Which common mistake happens with Lattice Energy (Born-Haber Cycle)?
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Create a curated FSRS exam set for ΔH_Gitter = ΔH°f − ΔH_Sub − I − ½D − EA: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.
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