Lattice enthalpy

Why does table salt survive as a hard, high-melting solid instead of crumbling back into sodium metal and chlorine gas? The answer is a single number: the energy released when gaseous $\text{Na}^+$ and $\text{Cl}^-$ ions snap together into a crystal lattice. That number is the lattice enthalpy, and it is the single biggest factor deciding whether an ionic compound forms at all, and how strongly it holds together once it does. Lattice enthalpy cannot be measured directly with a thermometer — you cannot isolate a beaker of gaseous ions and watch them condense. Chemists instead extract it indirectly, using Hess's law dressed up as a closed loop of measurable steps called the Born–Haber cycle. This article builds that cycle, uses it to calculate the lattice enthalpy of $\text{NaCl}$ , and explains why some ionic solids (like $\text{MgO}$ ) bind far more strongly than others.

The core insight

Lattice enthalpy is the energy change when gaseous ions pack into an ionic solid (or the reverse), it is found indirectly via the Born–Haber cycle (Hess's law), and it grows larger with higher ionic charge and smaller ionic radius.

The core idea

Definition and sign convention. Lattice enthalpy ( $\Delta_{lattice}H$ ) can be defined two equivalent ways, and problems can quote either, so always check which one is meant:

\text{M}^+(g) + \text{X}^-(g) \rightarrow \text{MX}(s) \qquad \Delta_{lattice}H_{\text{formation}} \ (\text{exothermic, negative})

\text{MX}(s) \rightarrow \text{M}^+(g) + \text{X}^-(g) \qquad \Delta_{lattice}H_{\text{dissociation}} \ (\text{endothermic, positive})

where: the two values have the same magnitude and opposite sign; ISC textbooks commonly quote the dissociation (positive) convention. Formation is exothermic because bringing free ions together into an ordered lattice releases a large amount of Coulombic (electrostatic) energy — this release is what makes ionic solids so stable.

Getting at it indirectly: the Born–Haber cycle. Because gaseous ions cannot be isolated and weighed calorimetrically, lattice enthalpy is obtained by building a closed thermochemical cycle out of quantities that can be measured directly, then applying Hess's law (total enthalpy change is path-independent — the sum of enthalpy changes around a closed loop is zero). The steps needed to build $\text{MX}(s)$ from its elements via the gaseous-ion route are:

Sublimation of the solid metal to gaseous atoms, $\Delta_{sub}H$ (endothermic).
Ionisation of the gaseous metal atom to the gaseous cation, $\Delta_i H$ (endothermic).
Bond dissociation of the diatomic non-metal to gaseous atoms, $\Delta_{diss}H$ (endothermic; halve it if only one mole of atoms is needed).
Electron gain by the gaseous non-metal atom to form the anion, $\Delta_{eg}H$ (usually exothermic).
Lattice formation, ions condensing into the solid, $\Delta_{lattice}H_{\text{formation}}$ (exothermic).

These five steps, added together, must equal the directly measurable enthalpy of formation of the compound from its elements, $\Delta_f H$ :

\Delta_f H = \Delta_{sub}H + \Delta_i H + \tfrac{1}{2}\Delta_{diss}H + \Delta_{eg}H + \Delta_{lattice}H_{\text{formation}}

where: every term on the right is an enthalpy step of the cycle for a compound $\text{MX}(s)$ built from $\text{M}(s)$ and $\tfrac{1}{2}\text{X}_2(g)$ ; the $\tfrac{1}{2}$ appears because bond dissociation enthalpies are quoted per mole of $\text{X}_2$ , but only one mole of X atoms is needed per mole of MX. Since $\Delta_f H$ and the four other terms are all experimentally accessible, the equation can be rearranged to solve for whichever single term is unknown — almost always $\Delta_{lattice}H$ , since that is the one term nobody can measure directly.

What controls the size of lattice enthalpy. Two structural factors decide how strongly ions attract each other in the lattice: the charge on the ions and the distance between ion centres (essentially the sum of ionic radii). Coulombic attraction between two point charges scales with the product of the charges, $q_1 q_2$ , and inversely with the separation between them — so lattice enthalpy increases (becomes more exothermic in magnitude) as charges rise and as ions get smaller. Small, highly charged ions therefore always give the strongest lattices.

Figure

Bar chart comparing the lattice enthalpy magnitude of sodium chloride and magnesium oxide

Derivation

Derivation AUsing the Born–Haber cycle to extract the lattice enthalpy of sodium chloride from five directly measurable enthalpy steps

The goal of this derivation is to isolate the one unmeasurable quantity in the cycle, $\Delta_{lattice}H_{\text{formation}}(\text{NaCl})$ , by writing Hess's law for the closed loop connecting solid sodium and chlorine gas to solid $\text{NaCl}$ via two different routes, then substituting the four other steps (sublimation, ionisation, bond dissociation, electron gain) along with the measured enthalpy of formation, and solving the resulting linear equation for the single unknown term. Use $\Delta_{sub}H(\text{Na}) = +108\ \text{kJ mol}^{-1}$ , $\Delta_i H(\text{Na}) = +496\ \text{kJ mol}^{-1}$ , $\Delta_{diss}H(\text{Cl}_2) = +242\ \text{kJ mol}^{-1}$ , $\Delta_{eg}H(\text{Cl}) = -349\ \text{kJ mol}^{-1}$ , $\Delta_f H(\text{NaCl}) = -411\ \text{kJ mol}^{-1}$ .

Step 1. Write Hess's law for the two paths from elements to solid NaCl. Path A goes directly, $\text{Na}(s) + \tfrac{1}{2}\text{Cl}_2(g) \rightarrow \text{NaCl}(s)$ , with enthalpy change $\Delta_f H$ . Path B goes via gaseous atoms and then gaseous ions, summing the five individual steps. Because enthalpy is a state function, both paths must give the same total: $\Delta_f H = \Delta_{sub}H + \Delta_i H + \tfrac{1}{2}\Delta_{diss}H + \Delta_{eg}H + \Delta_{lattice}H_{\text{formation}}$ .

Step 2. Substitute the four known steps and the measured formation enthalpy, leaving only the lattice term unknown: $-411 = 108 + 496 + \tfrac{1}{2}(242) + (-349) + \Delta_{lattice}H_{\text{formation}}$ .

Step 3. Collect the known terms into a single number. $108 + 496 + 121 - 349 = 376\ \text{kJ mol}^{-1}$ , so the equation becomes $-411 = 376 + \Delta_{lattice}H_{\text{formation}}$ .

Step 4. Solve for the unknown by rearranging: $\Delta_{lattice}H_{\text{formation}} = -411 - 376 = -787\ \text{kJ mol}^{-1}$ . The negative sign confirms the expected exothermic release when gaseous ions condense into the solid lattice; equivalently, the lattice enthalpy of dissociation is $+787\ \text{kJ mol}^{-1}$ .

Worked example

Explain, without a full numerical calculation, why the lattice enthalpy of $\text{MgO}(s)$ (about $3795\ \text{kJ mol}^{-1}$ ) is far larger than that of $\text{NaCl}(s)$ (about $787\ \text{kJ mol}^{-1}$ ), found above.

Step 1. Compare ionic charges. NaCl is built of $\text{Na}^+$ and $\text{Cl}^-$ , each singly charged, so the charge product $q_1 q_2 = 1 \times 1 = 1$ . MgO is built of $\text{Mg}^{2+}$ and $\text{O}^{2-}$ , each doubly charged, so $q_1 q_2 = 2 \times 2 = 4$ .

Step 2. Apply the Coulombic scaling. Since attraction between ions is proportional to $q_1 q_2$ , the charge effect alone makes MgO's ions attract roughly four times as strongly per unit distance as NaCl's ions.

Step 3. Compare ionic sizes. $\text{Mg}^{2+}$ and $\text{O}^{2-}$ are also smaller than $\text{Na}^+$ and $\text{Cl}^-$ respectively, so the interionic distance in MgO is shorter — and attraction increases further as distance decreases.

Step 4. Combine both effects. Higher charge and shorter distance both push in the same direction, compounding to give MgO a lattice enthalpy roughly five times that of NaCl. This is also why MgO melts at about 2852°C versus NaCl's 801°C — a far stronger lattice needs far more thermal energy to break apart.

Common traps

Mixing up the sign convention — formation (ions to solid) is exothermic and negative, dissociation (solid to ions) is endothermic and positive with the same magnitude; always check which one a question quotes or asks for.
Forgetting to halve the bond dissociation enthalpy of a diatomic non-metal like $\text{Cl}_2$ when only one mole of Cl atoms is needed per mole of NaCl.
Assuming lattice enthalpy depends on charge alone, or radius alone — both factors act together and can partly offset each other when comparing different compounds.
Expecting to recall numerical enthalpy values from memory — the syllabus expects applying Hess's law with given data, not memorising specific figures.

Ask the tutor: Want to work through a Born–Haber cycle for a different compound, or check your own lattice enthalpy calculation step by step?

Why can't we just measure lattice enthalpy directly in a calorimeter?

Because the process it describes, gaseous ions condensing into (or a solid dissociating into) free gaseous ions, cannot be set up and observed on its own in the lab. You cannot isolate a container of gaseous $\text{Na}^+$ and $\text{Cl}^-$ ions and watch them combine. Instead, chemists measure the surrounding steps — sublimation, ionisation, bond dissociation, electron gain, and overall formation enthalpy — each of which can be measured directly, and use Hess's law to back out the one missing term.

Does a bigger lattice enthalpy always mean a higher melting point?

Generally yes, for compounds of similar structure, because both properties measure how strongly the ions are held in the crystal — more energy is needed to pull ions apart (melting) or fully separate them (lattice dissociation) when the lattice is more strongly bound. This is exactly why MgO, with its far larger lattice enthalpy, melts at a much higher temperature than NaCl. It is a useful qualitative check, though it is not an exact numerical equivalence since melting only loosens the lattice rather than fully separating every ion.

If ionic charge quadruples the attraction going from 1+/1- to 2+/2-, why isn't MgO's lattice enthalpy exactly four times NaCl's?

The charge product effect alone predicts roughly a fourfold increase, and that is indeed most of the difference. But lattice enthalpy also depends on interionic distance, and $\text{Mg}^{2+}$ and $\text{O}^{2-}$ are both smaller than $\text{Na}^+$ and $\text{Cl}^-$ , shortening the interionic distance and adding extra attraction on top of the charge effect. The two factors compound, which is why the real ratio (about 3795 to 787, roughly fivefold) comes out a bit higher than the charge-only prediction of four.

Lattice enthalpy

The core idea

Derivation

Worked example

Common traps

Before this

Where this leads