Born-Haber Cycle Calculator

What is a Born-Haber cycle?

A Born-Haber cycle is Hess’s law applied to an ionic solid. The solid can be made from its elements in one step, whose enthalpy change is the enthalpy of formation, or in stages: turn each element into gaseous atoms, turn the atoms into gaseous ions, then let the ions come together as a lattice. Both routes start and finish in the same place, so their enthalpy changes are equal: ΔfH = ΔatH(metal) + ΔatH(non-metal) + ΣIE + ΣEA + ΔlattH. Max Born and Fritz Haber set it out in 1919, and Kasimir Fajans did so independently the same year.

Every term except the lattice enthalpy can be measured, so the cycle is how lattice enthalpies are found: ΔlattH = ΔfH − (ΔatH + ΣIE + ΣEA). Run the other way, with a lattice enthalpy supplied, it gives the enthalpy of formation, which is how the cycle shows why a compound such as NaCl₂ is never made.

Using the Born-Haber cycle calculator

Choose a salt and the calculator loads its data, finds the lattice enthalpy and draws the cycle. The drawing is to scale, so every arrow’s length is its enthalpy change: arrows pointing up take in energy, arrows pointing down give it out, and the heavier arrow is the value being calculated. The working under the drawing adds the steps one at a time, so each number can be checked against your own.

Every value can be edited, for instance to match the data an exam question gives. “Your own values” also sets the metal and non-metal symbols and the charges on the ions, for any salt, including ones that do not exist. The Find menu swaps the unknown: give it a lattice enthalpy and it returns the enthalpy of formation. On a narrow screen the levels are numbered, and the list under the drawing names them.

Lattice enthalpy: formation or dissociation?

Lattice enthalpy is defined two opposite ways, and its sign depends on which. The lattice formation enthalpy is the enthalpy change when one mole of the solid forms from its gaseous ions, so it is always negative. The lattice dissociation enthalpy is the change when one mole of the solid breaks up into gaseous ions: the same size, and always positive. For sodium chloride they are −787.1 kJ/mol and +787.1 kJ/mol.

OCR, Edexcel and Cambridge International use the formation definition, AQA accepts either, and the IB uses dissociation, which is why its data booklet lists positive values. The calculator’s convention menu switches between the two: the lattice arrow turns round, the working is rewritten and the size stays the same.

Worked example: sodium chloride

The calculator opens on sodium chloride with NIST data. Making the gaseous ions takes four steps, and the lattice enthalpy is what is left:

• Atomise sodium, Na(s) → Na(g): +107.5 kJ/mol.
• Atomise chlorine, ½Cl₂(g) → Cl(g): +121.3 kJ/mol, half the bond enthalpy of Cl₂.
• Ionise sodium, Na(g) → Na⁺(g) + e⁻: +495.8 kJ/mol, which is NIST’s 5.13908 eV converted at 96.485 kJ/mol per eV.
• Give chlorine the electron, Cl(g) + e⁻ → Cl⁻(g): −348.6 kJ/mol.
• Elements to gaseous ions: 107.5 + 121.3 + 495.8 − 348.6 = +376.0 kJ/mol.
• Hess’s law, with ΔfH = −411.1 kJ/mol: ΔlattH = −411.1 − 376.0 = −787.1 kJ/mol.

So the ions give out 787.1 kJ/mol as they form the solid, more than twice the 376.0 kJ/mol it cost to make them, and the difference is the enthalpy of formation. That surplus is what makes the salt stable. It also explains why sodium stops at Na⁺. Its second electron would have to come from a full inner shell, as the electron configuration calculator shows, so sodium’s second ionisation energy is 4562.4 kJ/mol, more than nine times the first, and no lattice repays that.

Where the data come from

The salts share one table built from NIST reference data and rounded to 0.1 kJ/mol. Enthalpies of formation, of the salts and of the gaseous atoms, are those of the NIST Chemistry WebBook: the CODATA value where it lists one and the NIST-JANAF value otherwise. Ionisation energies come from the NIST Atomic Spectra Database and electron affinities from laser photodetachment measurements, both converted from electronvolts. Enthalpies of formation are themselves measured by calorimetry, the kind of experiment the calorimetry calculator works through.

The exception is oxygen’s second electron affinity, which MgO, CaO and Na₂O need. A free O²⁻ ion is not stable, so the step cannot be measured and is itself worked out from cycles like this one, and books disagree: the table uses +744 kJ/mol from Huheey’s Inorganic Chemistry, while Chemguide gives +844 kJ/mol. The lattice enthalpy moves one for one, so with Chemguide’s value magnesium oxide comes out at −3889.3 kJ/mol instead of −3789.3 kJ/mol.

What this model leaves out

• Temperature corrections. Ionisation energies and electron affinities are energies at absolute zero, while the other terms are enthalpies at 298 K. Strictly, each electron removed adds about 5/2 RT, or 6.2 kJ/mol, and each electron added takes the same away, so in a neutral salt they cancel and the simple sum stands.
• Lattice energy against lattice enthalpy. The cycle gives an enthalpy. The lattice energy, an internal energy, is smaller in size by nRT for n moles of gaseous ions, which is 5.0 kJ/mol for sodium chloride at 298 K.
• Covalent character. The cycle measures the real solid, whatever its bonding. A purely ionic model, built from the charges, the ion sizes and the crystal structure, agrees with it closely for the alkali metal halides and falls short for salts such as silver chloride, whose bonding is partly covalent.
• Entropy and equilibrium. Enthalpy alone does not decide whether a salt forms. That takes ΔG = ΔH − TΔS, which the Gibbs free energy calculator works out, and how far a reversible reaction goes is a matter for its equilibrium constant, which the ICE table calculator turns into concentrations.
• Dissolving. Whether a salt dissolves depends on the hydration enthalpies of its ions as well as its lattice enthalpy, and those belong to a different cycle.

Common mistakes

• Using the whole bond enthalpy. NaCl needs one chlorine atom, half a Cl₂ molecule, so atomising chlorine is +121.3 kJ/mol, not the full bond enthalpy of 2 × 121.3 = 242.6 kJ/mol.
• Forgetting to double. MgCl₂ holds two chloride ions, so chlorine’s atomisation and electron affinity both count twice: 2 × (−348.6) = −697.2 kJ/mol. In Na₂O it is sodium’s steps that double.
• Leaving out an ionisation energy. Making Mg²⁺ takes the first and the second: 737.7 + 1450.7 = 2188.4 kJ/mol.
• Giving the second electron affinity a minus sign. Adding an electron to O⁻ takes in energy, because the ion repels it, so the value is positive.
• Treating bromine as a gas. Bromine is a liquid at 298 K, so its atomisation enthalpy, +111.9 kJ/mol, includes evaporating the liquid as well as breaking the bond.
• Mixing the conventions. A lattice dissociation enthalpy from one book, put into a cycle written for the formation definition, flips the sign of the answer.