Born-Haber Cycles and Lattice Enthalpy Revision Notes | A Level AQA Chemistry
A Born-Haber cycle uses Hess's law to calculate lattice enthalpy indirectly from measurable enthalpy changes. Free AQA notes and 15 questions answered.
Updated 5 Sept 2026 · Revise Right Team
Born-Haber Cycles and Lattice Enthalpy is part of Thermodynamics (A-level only) in A Level AQA Chemistry.
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What this lesson covers
A Born-Haber cycle uses Hess's law to calculate lattice enthalpy indirectly from measurable enthalpy changes. Free AQA notes and 15 questions answered.
Construct Born-Haber cycles using standard enthalpy changes to calculate lattice enthalpy
Use Hess's Law to calculate any one unknown enthalpy change from a Born-Haber cycle
Compare experimental lattice enthalpies from Born-Haber cycles with theoretical values from the perfect ionic model
Evaluate evidence for covalent character in ionic compounds using discrepancies between experimental and theoretical lattice enthalpies
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Use this lesson to stay inside Thermodynamics (A-level only) while connecting the detail back to the wider Physical chemistry topic.
Lattice Enthalpy: The enthalpy change when one mole of an ionic compound is formed from its constituent gaseous ions under standard conditions. Always exothermic (negative) under the formation convention used at A-Level.
Born-Haber Cycle: An application of Hess's Law that constructs a closed enthalpy cycle linking the standard enthalpy of formation of an ionic compound to a series of measurable enthalpy changes, allowing lattice enthalpy to be determined indirectly.
Enthalpy of Atomisation: The enthalpy change when one mole of gaseous atoms is formed from an element in its standard state. Always endothermic (positive).
Perfect Ionic Model: A theoretical model that treats ions as hard, non-polarising spheres with purely electrostatic interactions. Used to calculate a theoretical lattice enthalpy for comparison with the experimental Born-Haber value.
Revision notes
Read the core explanations from Born-Haber Cycles and Lattice Enthalpy before testing yourself from memory.
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.
Enthalpy Change
Symbol
Sign
What It Represents
Standard enthalpy of formation
$\Delta_f H^\circ$
Usually negative
Elements in standard states $\rightarrow$ ionic compound (1 mol)
Enthalpy of atomisation (metal)
$\Delta_{at} H^\circ$
Always positive (+)
Metal in standard state $\rightarrow$ 1 mol gaseous metal atoms
Enthalpy of atomisation (non-metal)
$\Delta_{at} H^\circ$
Always positive (+)
Non-metal in standard state $\rightarrow$ 1 mol gaseous non-metal atoms
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$^{-1}$.
$\Delta_f H^\circ = \Delta_{at} H^\circ(\text{Na}) + \Delta_{at} H^\circ(\text{Cl}) + IE_1 + EA_1 + \Delta_{latt} H^\circ$ — The direct route (formation) equals the sum of all indirect steps. This is the core Hess's Law statement for a Born-Haber cycle.
$\Delta_{latt} H^\circ = \Delta_f H^\circ - \Delta_{at} H^\circ(\text{Na}) - \Delta_{at} H^\circ(\text{Cl}) - IE_1 - EA_1$ — Isolate the unknown by subtracting all known indirect steps from the formation enthalpy.
$\Delta_{at} H^\circ(\text{Na}) + \Delta_{at} H^\circ(\text{Cl}) + IE_1 + EA_1 = (+107) + (+122) + (+496) + (-349) = +376$ kJ mol$^{-1}$ — Add carefully, keeping track of signs. The electron affinity is negative, so it reduces the total.
$\Delta_{latt} H^\circ = (-411) - (+376) = -787$ 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.
$\Delta_{latt} H^\circ = -787$ 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.
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.
$\Delta_f H^\circ = \Delta_{at} H^\circ(\text{K}) + \Delta_{at} H^\circ(\text{Cl}) + IE_1(\text{K}) + EA_1(\text{Cl}) + \Delta_{latt} H^\circ$ — Start from the standard Hess's Law relationship for a 1:1 ionic compound.
$\Delta_{at} H^\circ(\text{K}) = \Delta_f H^\circ - \Delta_{at} H^\circ(\text{Cl}) - IE_1(\text{K}) - EA_1(\text{Cl}) - \Delta_{latt} H^\circ$ — Isolate $\Delta_{at} H^\circ$(K) by subtracting all other known terms.
$\Delta_{at} H^\circ(\text{K}) = (-437) - (+122) - (+419) - (-349) - (-711)$ — Subtracting a negative value is equivalent to adding a positive value. Take care with each sign.
$\Delta_{at} H^\circ(\text{K}) = +82$ 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.
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.
Sources and content standards
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-09-05.
What extra steps are in the Born-Haber cycle for magnesium oxide?
The Born-Haber cycle for magnesium oxide has two extra steps compared with NaCl: the second ionisation energy of magnesium and the second electron affinity of oxygen. Both are needed because Mg²⁺ and O²⁻ are the ions present in the lattice.
Why is the second electron affinity always endothermic?
The second electron affinity is always endothermic because an electron is being forced onto an already negatively charged ion, and electrostatic repulsion makes this energetically unfavourable. It must be included in Born-Haber cycles for compounds such as magnesium oxide, which has O²⁻ ions.
How do you calculate lattice enthalpy from a Born-Haber cycle?
To calculate lattice enthalpy, use Hess's law: the enthalpy of formation equals the sum of all the indirect steps, so lattice enthalpy = enthalpy of formation − (atomisation enthalpies + first ionisation energy + first electron affinity). For NaCl the indirect steps sum to +376 kJ mol⁻¹, so lattice enthalpy = −411 − 376 = −787 kJ mol⁻¹.
What is a Born-Haber cycle?
A Born-Haber cycle is an enthalpy cycle that uses Hess's law to calculate lattice enthalpy indirectly from measurable enthalpy changes, such as enthalpies of formation and atomisation, ionisation energy and electron affinity. It is needed because gaseous ions cannot be put into a calorimeter to measure lattice enthalpy directly.
Why do experimental and theoretical lattice enthalpies differ?
Experimental and theoretical lattice enthalpies differ when bonding is not purely ionic. The theoretical value uses the perfect ionic model, so a more negative experimental Born-Haber value shows covalent character caused by the cation polarising the anion, as seen in AgCl and AgI.
What is the perfect ionic model?
The perfect ionic model treats ions as perfectly spherical, non-polarising point charges with purely electrostatic interactions between them. It uses ionic charges and ionic radii to calculate a theoretical lattice enthalpy, which agrees closely with the Born-Haber value only when a compound has no covalent character.
Why can't lattice enthalpy be measured directly?
Lattice enthalpy cannot be measured directly because gaseous ions cannot be put into a calorimeter to snap together into a crystal; steps such as ionising a metal atom happen in the gas phase under conditions that make direct calorimetry impossible. A Born-Haber cycle uses Hess's Law to calculate it indirectly.
What is the lattice enthalpy of NaCl from a Born-Haber cycle?
The lattice enthalpy of NaCl from a Born-Haber cycle is −787 kJ mol⁻¹. The atomisation of Na (+107) and Cl (+122), the first ionisation energy of Na (+496) and the first electron affinity of Cl (−349) sum to +376, and −411 − 376 = −787 kJ mol⁻¹.
Why is the atomisation enthalpy of oxygen half its bond enthalpy?
The atomisation enthalpy of oxygen is half the bond dissociation enthalpy of O₂ because it refers to forming one mole of gaseous oxygen atoms. Oxygen exists as a diatomic molecule, so only half a mole of O₂ is needed to produce one mole of O atoms.
How do you find an unknown enthalpy change from a Born-Haber cycle?
An unknown in a Born-Haber cycle is found by writing the Hess's Law equation, substituting known values with their signs and rearranging. For KCl, the atomisation of potassium = (−437) − (+122) − (+419) − (−349) − (−711) = +82 kJ mol⁻¹, which is positive as atomisation should be.
What is polarisation in ionic compounds?
Polarisation in ionic compounds is the distortion of the electron cloud of an anion by the electric field of a neighbouring cation. Greater polarisation leads to greater sharing of electron density between the ions, introducing covalent character into what would otherwise be a purely ionic bond.
What are Fajans' rules?
Fajans' rules summarise the conditions that favour polarisation: a small, highly charged cation with high charge density polarises a large, highly polarisable anion most effectively. A larger anion has a more diffuse electron cloud that is more easily distorted, so anion size is a primary factor and anion charge is secondary.
Why do silver halides show covalent character?
AgCl and AgI show covalent character because their experimental lattice enthalpies are considerably more negative than the theoretical values from the perfect ionic model. The discrepancy grows from AgCl to AgI because the iodide ion is larger and more polarisable than the chloride ion, while NaCl and NaI give close values.
Why does nitrogen have a positive first electron affinity?
Nitrogen has a positive first electron affinity, an exception to the usual negative value, because of its half-filled 2p subshell. Most first electron affinities are negative, while second electron affinities are always positive.
Why is lattice enthalpy of formation always negative?
Lattice enthalpy of formation is always negative because energy is released when the lattice forms from gaseous ions. The more negative the value, the stronger the ionic lattice. If a Born-Haber calculation gives a positive lattice enthalpy under this convention, the signs in the rearrangement should be checked.