Lattice Energy (Born-Haber Cycle)
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
\Delta H_{\text{Gitter}} = \Delta H_f^0 - \Delta H_{\text{Sub}} - I - \tfrac{1}{2} D - E_{\text{EA}}Variables & units – Lattice Energy (Born-Haber Cycle)
| Symbol | Meaning | Unit |
|---|---|---|
| ΔH_Gitter | Lattice enthalpy (lattice formation from the gaseous ions) | kJ/mol |
| ΔH°f | Standard enthalpy of formation of the salt | kJ/mol |
| ΔH_Sub | Sublimation enthalpy of the metal | kJ/mol |
| I | Ionization energy of the metal atom | kJ/mol |
| D | Dissociation energy of the non-metal molecule | kJ/mol |
| EA | Electron affinity of the non-metal atom (usually negative) | kJ/mol |
Derivation & background – Lattice Energy (Born-Haber Cycle)
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 −3800 kJ/mol versus NaCl at −788 kJ/mol).
Exam blueprint
Validity range
Holds exactly because enthalpy is a state function; it requires consistent signs and complete partial steps of the cycle at standard conditions.
Derivation steps
The salt formation is written as a detour via gaseous atoms and ions; by Hess both paths have equal enthalpy.
- 1Direct path: elements → salt (ΔH°f); detour: sublimation, ionization, ½ dissociation, electron affinity, lattice.
- 2ΔH°f = ΔH_Sub + I + ½D + EA + ΔH_lattice; solve for ΔH_lattice.
Rearrangements
Standard enthalpy of formation
The complete cycle in one line.
Electron affinity
Historically, quantities that were hard to measure were determined this way.
Task variant
Calculate the lattice enthalpy of NaCl from the Born-Haber data.
ΔH_lattice = ΔH°f − ΔH_Sub − I − ½D − EA = −411 − 108 − 496 − 122 − (−349) = −788 kJ/mol. Forming the lattice from the gaseous ions is strongly exothermic.
KCl: ΔH°f = −437, ΔH_Sub = +89, I = +419, ½D = +122, EA = −349 kJ/mol. Lattice enthalpy?
ΔH_lattice = −437 − 89 − 419 − 122 + 349 = −718 kJ/mol; smaller in magnitude than for NaCl because the K⁺ ion is larger.
Common mistakes
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.
Exam context
- Setting up the cycle for NaCl or MgO, calculating a missing quantity and comparing lattice enthalpies via ionic radius and charge.
These mistakes cost points in real exams. The set drills them until they stick.
Formula cluster
Chemical energetics
Applies Hess's law to ionic crystals.
Worked example
NaCl: ΔH°f = −411, ΔH_Sub(Na) = +108, I(Na) = +496, ½D(Cl₂) = +122, EA(Cl) = −349 (all kJ/mol). ΔH_lattice = −411 − 108 − 496 − 122 + 349 = −788 kJ/mol.
Applications
Explaining the stability of ionic crystals, comparing melting points and hardness, estimating salt solubility, ruling out hypothetical compounds such as NaCl₂
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How do you rearrange ΔH_Gitter = ΔH°f − ΔH_Sub − I − ½D − EA for Standard enthalpy of formation?
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Which common mistake happens with Lattice Energy (Born-Haber Cycle)?
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Frequently asked questions about Lattice Energy (Born-Haber Cycle)
What is the Born-Haber cycle?+
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.
Which partial steps belong to the Born-Haber cycle of NaCl?+
For NaCl the detour runs through five partial steps. First the sublimation of solid sodium to gaseous Na atoms, ΔH_Sub = +108 kJ/mol. Second the ionization of the Na atoms to Na⁺ ions, ionization energy I = +496 kJ/mol. Third the dissociation of the chlorine molecule; since only one Cl atom is needed per formula unit, half the bond energy enters, ½D = +122 kJ/mol. Fourth the uptake of an electron by the Cl atom, electron affinity EA = −349 kJ/mol, 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°f = −411 kJ/mol. From this the lattice enthalpy follows.
Why can the lattice energy not be measured directly?+
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.
What determines the magnitude of the lattice energy?+
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 −3800 kJ/mol, while sodium chloride with singly charged ions is only around −788 kJ/mol. 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.
Why is the sign of the electron affinity so important?+
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 EA = −349 kJ/mol. If you accidentally insert a positive value, the calculated lattice enthalpy shifts by almost 700 kJ/mol, 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.
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How do you calculate with Lattice Energy (Born-Haber Cycle)?
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:
- 1
Task
Calculate the lattice enthalpy of NaCl from the Born-Haber data.
Solution path
ΔH_lattice = ΔH°f − ΔH_Sub − I − ½D − EA = −411 − 108 − 496 − 122 − (−349) = −788 kJ/mol. Forming the lattice from the gaseous ions is strongly exothermic.
- 2
Task
KCl: ΔH°f = −437, ΔH_Sub = +89, I = +419, ½D = +122, EA = −349 kJ/mol. Lattice enthalpy?
Solution path
ΔH_lattice = −437 − 89 − 419 − 122 + 349 = −718 kJ/mol; smaller in magnitude than for NaCl because the K⁺ ion is larger.