Lattice Enthalpy

Lattice enthalpy is a measure of the strength of an ionic lattice — the energy change associated with forming one mole of an ionic solid from its constituent gaseous ions under standard conditions. When gaseous ions come together to form a crystalline lattice, electrostatic attractions between oppositely charged ions release a large amount of energy. The more negative the lattice enthalpy (defined as the enthalpy of formation of the lattice), the more stable and tightly held together the ionic compound is. For example, the lattice enthalpy of sodium chloride (NaCl) is approximately −788 kJ mol⁻¹, meaning 788 kJ of energy is released per mole of NaCl lattice formed from Na⁺(g) and Cl⁻(g) ions. Lattice enthalpy cannot be measured directly by experiment, because it is impossible to bring gaseous ions together cleanly without other processes occurring. Instead, chemists use an indirect method called the Born–Haber cycle, which applies Hess's Law to a series of measurable enthalpy changes — such as ionisation energies, electron affinities, atomisation enthalpies, and the overall enthalpy of formation — to calculate the lattice enthalpy indirectly. This elegant approach was developed by Max Born and Fritz Haber in 1919. The magnitude of lattice enthalpy depends on two key factors: the charges of the ions and the ionic radii. Higher ionic charges create stronger electrostatic attractions, giving a more negative (larger magnitude) lattice enthalpy. Smaller ionic radii allow the ions to pack closer together, also increasing attraction. These trends are captured by a simplified version of Coulomb's law: the lattice energy is proportional to the product of the ionic charges divided by the sum of the ionic radii. Understanding lattice enthalpy helps chemists explain why some ionic compounds are very stable solids with high melting points while others dissolve readily in water or have lower thermal stability.

Key takeaways
  • Define the Target: Enthalpy of Formation
  • Step 1 – Atomisation of Sodium
  • Step 2 – Atomisation of Chlorine
  • Step 3 – First Ionisation Energy of Sodium
  • Step 4 – First Electron Affinity of Chlorine
  • Step 5 – Applying Hess's Law to Find Lattice Enthalpy
  • Trends in Lattice Enthalpy

Define the Target: Enthalpy of Formation

The overall enthalpy of formation (ΔHf°) of NaCl(s) from its elements in standard states is the energy change for: Na(s) + ½Cl₂(g) → NaCl(s). This is experimentally measured as −411 kJ mol⁻¹. The Born–Haber cycle breaks this single arrow into a series of steps whose enthalpies are all individually measurable.

Step 1 – Atomisation of Sodium

Solid sodium must first be converted to gaseous atoms: Na(s) → Na(g). This endothermic process is the standard enthalpy of atomisation (ΔHat°) of sodium = +107 kJ mol⁻¹. Energy must be supplied to overcome metallic bonding and vaporise the metal.

Step 2 – Atomisation of Chlorine

Chlorine gas exists as diatomic Cl₂ molecules, so half a mole must be dissociated into atoms: ½Cl₂(g) → Cl(g). The bond dissociation enthalpy for Cl₂ is +242 kJ mol⁻¹, so for ½ mol: ΔH = +121 kJ mol⁻¹. This is endothermic because covalent Cl–Cl bonds must be broken.

Step 3 – First Ionisation Energy of Sodium

The gaseous sodium atom loses one electron to form Na⁺(g): Na(g) → Na⁺(g) + e⁻. The first ionisation energy of Na = +496 kJ mol⁻¹. This is always endothermic; energy is needed to remove an electron against the nuclear attraction.

Step 4 – First Electron Affinity of Chlorine

The gaseous chlorine atom gains one electron to form Cl⁻(g): Cl(g) + e⁻ → Cl⁻(g). The first electron affinity of Cl = −349 kJ mol⁻¹. This is exothermic; chlorine's high nuclear charge strongly attracts an incoming electron into its 3p subshell.

Step 5 – Applying Hess's Law to Find Lattice Enthalpy

By Hess's Law, the sum of all enthalpy changes around the cycle must equal zero (energy is conserved). Therefore: ΔHf° = ΔHat°(Na) + ΔHat°(½Cl₂) + IE₁(Na) + EA₁(Cl) + ΔHlattice. Rearranging: ΔHlattice = ΔHf° − [ΔHat°(Na) + ΔHat°(½Cl₂) + IE₁(Na) + EA₁(Cl)]. Substituting: ΔHlattice = −411 − (107 + 121 + 496 − 349) = −411 − 375 = −786 kJ mol⁻¹. This matches the experimentally estimated value of approximately −788 kJ mol⁻¹.

Comparing NaCl (−788 kJ mol⁻¹) with MgO (−3791 kJ mol⁻¹) illustrates the charge effect: Mg²⁺ and O²⁻ each carry double the charge of Na⁺/Cl⁻, and the four-fold increase in charge product (2×2 vs 1×1) leads to a roughly five-fold larger lattice enthalpy magnitude. Within the same charge group, lattice enthalpies decrease as ionic radius increases — for example, from LiF (−1037 kJ mol⁻¹) to CsF (−744 kJ mol⁻¹) — because larger ions mean greater interionic distance and weaker electrostatic attraction.

The Key Equation

ΔHlattice=ΔHf∘−[ΔHat(metal)+ΔHat(non-metal)+IE1+EA1]\Delta H_{\text{lattice}} = \Delta H_f^\circ - \left[\Delta H_{\text{at}}(\text{metal}) + \Delta H_{\text{at}}(\text{non-metal}) + IE_1 + EA_1\right]
  • \Delta H_{\text{lattice}} — Lattice enthalpy (kJ mol⁻¹): Enthalpy change when 1 mol of ionic lattice forms from gaseous ions; negative (exothermic) for stable lattices.
  • \Delta H_f^\circ — Standard enthalpy of formation (kJ mol⁻¹): Enthalpy change when 1 mol of compound forms from its elements in their standard states at 298 K.
  • \Delta H_{\text{at}} — Enthalpy of atomisation (kJ mol⁻¹): Enthalpy needed to produce 1 mol of gaseous atoms from an element in its standard state; always endothermic.
  • IE_1 — First ionisation energy (kJ mol⁻¹): Energy required to remove 1 mol of electrons from 1 mol of gaseous atoms in their ground state; always positive.
  • EA_1 — First electron affinity (kJ mol⁻¹): Enthalpy change when 1 mol of gaseous atoms each gain one electron; typically negative (exothermic) for non-metals.

Worked Example

Problem. Use the following data to calculate the lattice enthalpy of potassium fluoride, KF(s): • Standard enthalpy of formation of KF(s): −568 kJ mol⁻¹ • Enthalpy of atomisation of K(s): +89 kJ mol⁻¹ • Enthalpy of atomisation of ½F₂(g): +79 kJ mol⁻¹ • First ionisation energy of K(g): +419 kJ mol⁻¹ • First electron affinity of F(g): −328 kJ mol⁻¹

ΔHlattice(KF)=−827 kJ mol−1\Delta H_{\text{lattice}}(\text{KF}) = -827 \ \text{kJ mol}^{-1}

Answer. The lattice enthalpy of KF(s) = −827 kJ mol⁻¹

Figure
Diagram
Born–Haber Cycle for NaClEnthalpy / kJ mol⁻¹Na⁺(g) + e⁻ + Cl(g)EA₁(Cl)−349Na⁺(g) + Cl⁻(g)ΔHlattice−788NaCl(s)IE₁(Na)+496Na(g) + ½Cl₂(g)ΔHat(½Cl₂)+121ΔHat(Na)+107Na(s) + ½Cl₂(g)(Elements, standard state)ΔHf°−411Key (all values in kJ mol⁻¹)■Atomisation■Ionisation■Electron affinity■Formation■Lattice enthalpy
Origins · 1 minFritz Haber and how this was discovered
FH
Fritz Haber
1868–1934 · German

Fritz Haber, together with Max Born, developed the thermodynamic cycle in 1919 that bears their names — the Born–Haber cycle — which uses Hess's Law to calculate lattice enthalpies of ionic compounds indirectly from measurable thermodynamic quantities. Haber is perhaps most famous for the Haber–Bosch process for synthesising ammonia from nitrogen and hydrogen gas, which enabled large-scale production of fertilisers and transformed global agriculture. He was awarded the Nobel Prize in Chemistry in 1918. Haber was so devoted to science that he reportedly continued working in his laboratory on the morning of his own wedding to Clara Immerwahr, the first woman to receive a chemistry doctorate from a German university.

  • Ionic Bonding and Ion Formation — Lattice enthalpy arises directly from the electrostatic forces in ionic bonds; understanding how ions form (electron transfer, oxidation states) is a prerequisite for interpreting Born–Haber cycles.
  • Hess's Law and Enthalpy Cycles — The Born–Haber cycle is an application of Hess's Law. Mastery of Hess's Law — that enthalpy change is independent of pathway — is essential for setting up and solving Born–Haber calculations.
  • Ionisation Energies and Electron Affinities — Ionisation energies and electron affinities are key data inputs into every Born–Haber cycle. Trends in these values across periods and groups explain trends in lattice enthalpies.
  • Enthalpy of Solution and Hydration — Lattice enthalpy competes with hydration enthalpy to determine whether an ionic compound dissolves in water. The relative magnitudes of these two quantities explain solubility trends.