Born-Haber Cycles and Lattice Enthalpy Revision Notes | A Level AQA Chemistry

1. Why Lattice Enthalpy Cannot Be Measured Directly

You cannot put gaseous sodium ions and gaseous chloride ions into a calorimeter and watch them snap together into a crystal. The individual steps — ionising a metal atom, adding an electron to a non-metal atom — happen in the gas phase under conditions that make direct calorimetry impossible.

This is where the Born-Haber cycle earns its place. It uses Hess's Law — the principle that enthalpy is a state function, so the total enthalpy change is independent of the route taken — to calculate lattice enthalpy indirectly from a set of measurable enthalpy changes.

The more negative the lattice enthalpy, the stronger the ionic lattice. Lattice enthalpy becomes more negative as ionic charge increases and as ionic radius decreases — both factors increase the electrostatic attraction between ions.

2. The Building Blocks: Enthalpy Changes in a Born-Haber Cycle

A Born-Haber cycle is assembled from several types of enthalpy change. You need to know the sign and direction of each one before you can draw the cycle correctly.

The second electron affinity is always endothermic because you are forcing an electron onto an already negatively charged ion — electrostatic repulsion makes this energetically unfavourable. This is a detail that catches many students out when constructing cycles for compounds like magnesium oxide.

3. Constructing a Born-Haber Cycle: Sodium Chloride

The cycle for NaCl is the standard starting point. Picture it as a vertical energy ladder: you start at the elements in their standard states, climb upward through endothermic steps, then descend in one large exothermic drop — the lattice enthalpy — to arrive at the ionic compound.

Hess's Law tells us that the direct route (formation of NaCl from elements) must equal the sum of all the indirect steps. This gives us the equation we use to calculate lattice enthalpy.

Worked example: Calculating the Lattice Enthalpy of NaCl

Use the following data to calculate the lattice enthalpy of sodium chloride. $\Delta_f H^\circ$(NaCl) = $-411$ kJ mol$^{-1}$; $\Delta_{at} H^\circ$(Na) = $+107$ kJ mol$^{-1}$; $\Delta_{at} H^\circ$(Cl) = $+122$ kJ mol$^{-1}$; $IE_1$(Na) = $+496$ kJ mol$^{-1}$; $EA_1$(Cl) = $-349$ kJ mol$^{-1}$.

1. $\Delta_f H^\circ = \Delta_{at} H^\circ(\text{Na}) + \Delta_{at} H^\circ(\text{Cl}) + IE_1 + EA_1 + \Delta_{latt} H^\circ$ — The direct route (formation) equals the sum of all indirect steps. This is the core Hess's Law statement for a Born-Haber cycle.
2. $\Delta_{latt} H^\circ = \Delta_f H^\circ - \Delta_{at} H^\circ(\text{Na}) - \Delta_{at} H^\circ(\text{Cl}) - IE_1 - EA_1$ — Isolate the unknown by subtracting all known indirect steps from the formation enthalpy.
3. $\Delta_{at} H^\circ(\text{Na}) + \Delta_{at} H^\circ(\text{Cl}) + IE_1 + EA_1 = (+107) + (+122) + (+496) + (-349) = +376$ kJ mol$^{-1}$ — Add carefully, keeping track of signs. The electron affinity is negative, so it reduces the total.
4. $\Delta_{latt} H^\circ = (-411) - (+376) = -787$ kJ mol$^{-1}$ — Subtracting a positive number makes the result more negative. The lattice enthalpy is exothermic, which is physically correct — energy is released when the lattice forms.
5. $\Delta_{latt} H^\circ = -787$ kJ mol$^{-1}$ (negative = exothermic = lattice formation releases energy) — Lattice enthalpy under the formation convention must always be negative. If you get a positive value, check your signs in the rearrangement.

Answer: $\Delta_{latt} H^\circ$(NaCl) = $-787$ kJ mol$^{-1}$

4. Extending the Cycle: Magnesium Oxide

Magnesium oxide introduces two additional steps compared to NaCl: the second ionisation energy of magnesium ($IE_2$) and the second electron affinity of oxygen ($EA_2$). Both must be included because $\text{Mg}^{2+}$ and $\text{O}^{2-}$ are the ions present in the lattice.

The enthalpy of atomisation of oxygen refers to forming one mole of gaseous oxygen atoms from $\text{O}_2(g)$. Because oxygen exists as a diatomic molecule, you need half a mole of $\text{O}_2$ to produce one mole of O atoms — so the atomisation enthalpy is half the bond dissociation enthalpy of $\text{O}_2$.

5. Using the Cycle to Find Other Enthalpy Changes

The Born-Haber cycle is not limited to finding lattice enthalpy. If lattice enthalpy is known (from a theoretical calculation), you can rearrange the Hess's Law equation to find any other unknown — for example, an electron affinity that is difficult to measure directly.

The method is identical: write the Hess's Law equation, substitute all known values with their correct signs, and rearrange algebraically for the unknown. The key discipline is keeping track of signs throughout — a single sign error will give a completely wrong answer.

Worked example: Finding an Unknown Enthalpy of Atomisation from a Born-Haber Cycle

For potassium chloride, the following data are available: $\Delta_f H^\circ$(KCl) = $-437$ kJ mol$^{-1}$; $IE_1$(K) = $+419$ kJ mol$^{-1}$; $\Delta_{at} H^\circ$(Cl) = $+122$ kJ mol$^{-1}$; $EA_1$(Cl) = $-349$ kJ mol$^{-1}$; $\Delta_{latt} H^\circ$(KCl) = $-711$ kJ mol$^{-1}$. Calculate the enthalpy of atomisation of potassium.

1. $\Delta_f H^\circ = \Delta_{at} H^\circ(\text{K}) + \Delta_{at} H^\circ(\text{Cl}) + IE_1(\text{K}) + EA_1(\text{Cl}) + \Delta_{latt} H^\circ$ — Start from the standard Hess's Law relationship for a 1:1 ionic compound.
2. $\Delta_{at} H^\circ(\text{K}) = \Delta_f H^\circ - \Delta_{at} H^\circ(\text{Cl}) - IE_1(\text{K}) - EA_1(\text{Cl}) - \Delta_{latt} H^\circ$ — Isolate $\Delta_{at} H^\circ$(K) by subtracting all other known terms.
3. $\Delta_{at} H^\circ(\text{K}) = (-437) - (+122) - (+419) - (-349) - (-711)$ — Subtracting a negative value is equivalent to adding a positive value. Take care with each sign.
4. $= -437 - 122 - 419 + 349 + 711 = +82$ kJ mol$^{-1}$ — Working left to right: $-437 - 122 = -559$; $-559 - 419 = -978$; $-978 + 349 = -629$; $-629 + 711 = +82$ kJ mol$^{-1}$.
5. $\Delta_{at} H^\circ(\text{K}) = +82$ kJ mol$^{-1}$ (positive = endothermic) — Atomisation is always endothermic — energy is required to break metallic bonds and produce gaseous atoms. A positive answer is physically correct.

Answer: $\Delta_{at} H^\circ$(K) = $+82$ kJ mol$^{-1}$

6. Theoretical Lattice Enthalpies and the Perfect Ionic Model

Alongside the experimental Born-Haber value, chemists can calculate a theoretical lattice enthalpy using the perfect ionic model. This model treats ions as perfectly spherical, non-polarising point charges with purely electrostatic (Coulombic) interactions between them.

The theoretical calculation uses ionic charges and ionic radii to predict how strongly the ions attract one another. If the compound were truly ionic — with no covalent character whatsoever — the theoretical and experimental values would agree closely.

When the two values differ significantly, this is evidence that the bonding is not purely ionic. A larger experimental lattice enthalpy (more negative than the theoretical value) indicates that additional bonding interactions are present — specifically, covalent character arising from polarisation of the anion by the cation.

Fajans' rules summarise the conditions that favour polarisation: a small, highly charged cation (high charge density) polarises a large, highly polarisable anion most effectively. For the anion, large size and high polarisability are the primary factors — a larger anion has a more diffuse electron cloud that is more easily distorted, while higher charge on the anion increases polarisability but is secondary to size. This is why compounds such as aluminium chloride show a much larger discrepancy between experimental and theoretical lattice enthalpies than sodium chloride does.

Notice that for NaCl and NaI the two values are close, indicating predominantly ionic bonding. For AgCl and AgI the experimental values are considerably more negative than the theoretical predictions — the discrepancy grows from AgCl to AgI as the iodide ion is larger and more polarisable than the chloride ion.

7. Exam Technique and Common Mistakes

Born-Haber cycle questions are highly structured, which means marks are predictable — and so are the errors. The most reliable way to gain full marks is to write the Hess's Law equation first, substitute values with explicit signs, and rearrange algebraically before reaching for a calculator.