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Born-Haber cycles

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A Born-Haber cycle is an enthalpy cycle for making an ionic compound from its elements. It splits the reaction into steps you can measure (atomising the elements, ionising the metal, adding electrons to the non-metal) plus the one you can't: forming the lattice from gaseous ions. Because of Hess's law the steps must add up to the enthalpy of formation, so the cycle lets you calculate the lattice enthalpy.

What is a Born-Haber cycle?

Lattice enthalpy measures how strongly the ions in an ionic solid are held together. You can't measure it directly, because you can't make a mole of separate gaseous ions and let them crash into a crystal. But you can measure the enthalpy change of forming the compound from its elements, and every other step on the way round. Hess's law says the total enthalpy change is the same whichever route you take, so:

\Delta_f H = \text{(sum of the other steps)} + \Delta_{LE} H

Here\Delta_f His the enthalpy of formation and\Delta_{LE} His the lattice enthalpy of formation. Rearranging gives the lattice enthalpy, which is what most Born-Haber questions ask for.

The enthalpy changes in a Born-Haber cycle

You need to be able to define each of these and write an equation for it. Each is for one mole, with everything in the state shown.

Enthalpy changeWhat happensExampleSign
Enthalpy of formation1 mol of compound forms from its elements in their standard statesCs(s) + ½Cl₂(g) → CsCl(s)Usually −
Enthalpy of atomisationThe element in its standard state becomes 1 mol of gaseous atomsCs(s) → Cs(g)+
Bond enthalpy1 mol of a covalent bond breaks, in the gas phaseCl₂(g) → 2Cl(g)+
First ionisation energy1 mol of electrons is removed from 1 mol of gaseous atomsCs(g) → Cs⁺(g) + e⁻+
Second ionisation energy1 mol of electrons is removed from 1 mol of gaseous 1+ ionsMg⁺(g) → Mg²⁺(g) + e⁻+
First electron affinity1 mol of gaseous atoms each gain an electronCl(g) + e⁻ → Cl⁻(g)−
Second electron affinity1 mol of gaseous 1− ions each gain an electronO⁻(g) + e⁻ → O²⁻(g)+
Lattice enthalpy of formation1 mol of ionic solid forms from its gaseous ionsCs⁺(g) + Cl⁻(g) → CsCl(s)−

For chlorine, the enthalpy of atomisation makes one mole of Cl atoms, so it is half the Cl-Cl bond enthalpy. The second electron affinity is positive because you are pushing an electron onto an ion that is already negative, and it repels it.

How to draw a Born-Haber cycle

  1. Draw a line for the elements in their standard states, at zero.
  2. Atomise the metal: an arrow up to a new line, with the metal now as gaseous atoms.
  3. Atomise the non-metal: another arrow up.
  4. Ionise the metal: an arrow up for each ionisation energy. Write the electrons in: e⁻.
  5. Add the electrons to the non-metal: an arrow down for a first electron affinity (up for a second).
  6. Form the lattice: a long arrow down from the gaseous ions to the solid compound.
  7. Finally, the enthalpy of formation arrow goes from the elements straight to the compound.

Arrows pointing up are endothermic (positive) and arrows pointing down are exothermic (negative). Here is the cycle for caesium chloride, with values in kJ mol⁻¹:

+77+121+376−349lattice−443Cs(s) + ½Cl₂Cs + ½Cl₂Cs + ClCs⁺ + e⁻ + ClCs⁺ + Cl⁻CsCl(s)
Born-Haber cycle for CsCl in kJ mol⁻¹ (not to scale). Every species above the elements line is a gas: (g) is left off to save space

How to calculate lattice enthalpy

Example 1: caesium chloride

Use these values to calculate the lattice enthalpy of formation of caesium chloride.

Enthalpy changekJ mol⁻¹
Enthalpy of formation of CsCl(s)−443
Enthalpy of atomisation of caesium+77
Enthalpy of atomisation of chlorine+121
First ionisation energy of caesium+376
First electron affinity of chlorine−349

Going round the cycle from the elements to the gaseous ions:

77 + 121 + 376 + (-349) = +225\ \text{kJ mol}^{-1}

Then, by Hess's law, the enthalpy of formation equals this plus the lattice enthalpy:

\Delta_{LE} H = -443 - 225 = -668\ \text{kJ mol}^{-1}

So the lattice enthalpy of formation of CsCl is −668 kJ mol⁻¹, and its lattice dissociation enthalpy is +668 kJ mol⁻¹.

Example 2: magnesium chloride

In MgCl₂ the magnesium loses two electrons and there are two chlorine atoms, so the cycle has two ionisation energies and two lots of the chlorine values.

StepkJ mol⁻¹
Enthalpy of formation of MgCl₂(s)−642
Atomisation of magnesium+147
First ionisation energy of magnesium+738
Second ionisation energy of magnesium+1451
Atomisation of chlorine, × 22 × (+121) = +242
First electron affinity of chlorine, × 22 × (−349) = −698
147 + 738 + 1451 + 242 - 698 = +1880
\Delta_{LE} H = -642 - 1880 = -2522\ \text{kJ mol}^{-1}

The lattice enthalpy of formation of MgCl₂ is −2522 kJ mol⁻¹, nearly four times that of CsCl. The values in both examples are rounded from standard data tables; different data books can differ by a few kJ mol⁻¹, so always use the values in the question.

Check your understanding.

practice
Pick an answer, then check it.
In the Born-Haber cycle for sodium oxide, Na₂O, which of these steps is exothermic?

What affects the size of the lattice enthalpy?

A more exothermic lattice enthalpy of formation means stronger attraction between the ions. Two things increase it:

  • Higher charges on the ions. Mg²⁺ attracts Cl⁻ more strongly than Cs⁺ does.
  • Smaller ions. Smaller ions can get closer together, so the attraction between them is stronger. Mg²⁺ is also much smaller than Cs⁺.

Both effects together explain why MgCl₂ has a far more exothermic lattice enthalpy than CsCl.

Born-Haber values and the perfect ionic model

A lattice enthalpy can also be calculated in theory, by treating the compound as perfect spheres of charge held together only by electrostatic attraction: the perfect ionic model. Comparing that value with the Born-Haber value, which comes from real experimental data, tells you about the bonding:

  • The two values are close: the bonding is almost purely ionic.
  • The Born-Haber value is more exothermic: the compound has some covalent character. A small, highly charged cation pulls the electron cloud of a large anion towards it (it polarises the anion), adding extra attraction that the perfect ionic model leaves out.

Lattice enthalpy and dissolving

Lattice enthalpies also decide the enthalpy change when an ionic solid dissolves in water. AQA's specification says cycles are used to calculate enthalpies of solution from lattice enthalpies and enthalpies of hydration: the lattice is broken into gaseous ions, then the ions are hydrated. Our enthalpy of solution guide works through that cycle, including caesium chloride with the lattice enthalpy from Example 1.

Common mistakes

  • Forgetting to double. In MgCl₂ there are two chlorine atoms, so use 2 × atomisation and 2 × electron affinity.
  • Using the bond enthalpy instead of atomisation. For ½Cl₂ to Cl, use half the bond enthalpy.
  • Missing the second ionisation energy for a 2+ ion.
  • Sign errors. Write every value with its sign, and remember the first electron affinity is negative but the second is positive.
  • Mixing up formation and dissociation. They are the same size but opposite signs.
  • Writing the wrong states. Ionisation energies and electron affinities are always for gaseous atoms and ions.

Practice questions

  1. Write an equation for the second ionisation energy of calcium.
  2. Write an equation for the lattice enthalpy of formation of magnesium chloride.
  3. The lattice enthalpy of formation of CsCl is −668 kJ mol⁻¹. Use this and the other values in Example 1 to calculate the first electron affinity of chlorine.
  4. For oxygen, why is the second electron affinity positive when the first is negative?

Answers: 1) Ca⁺(g) → Ca²⁺(g) + e⁻. 2) Mg²⁺(g) + 2Cl⁻(g) → MgCl₂(s). 3) −443 = 77 + 121 + 376 + EA + (−668), so EA = −443 − 77 − 121 − 376 + 668 = −349 kJ mol⁻¹. 4) The electron is being added to an O⁻ ion, which is already negative and repels it, so energy must be put in.

Practise with a short timed paper

Born-Haber cycles and entropy are in the second year of A-level Chemistry. The daily A-level Chemistry: Year 2 paper is a short timed paper of 8 questions, with a new one every day.

Common questions

What is a Born-Haber cycle?
An enthalpy cycle that breaks the formation of an ionic compound into steps (atomisation, ionisation, electron affinity and lattice formation), so the lattice enthalpy can be calculated using Hess's law.
How do you calculate lattice enthalpy from a Born-Haber cycle?
Add up the enthalpy changes from the elements to the gaseous ions, then subtract that total from the enthalpy of formation. The answer is the lattice enthalpy of formation.
Why can't lattice enthalpy be measured directly?
It is not possible to make a mole of separate gaseous ions and measure the energy released as they form a solid, so it is found indirectly with a Born-Haber cycle.
Is lattice enthalpy positive or negative?
The lattice enthalpy of formation is negative (exothermic). The lattice dissociation enthalpy is the same size but positive.
Why is the second electron affinity positive?
The electron is added to a negative ion, which repels it, so energy has to be supplied.
What does a Born-Haber cycle show about covalent character?
If the Born-Haber lattice enthalpy is more exothermic than the value from the perfect ionic model, the compound has some covalent character.

Revision cards

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