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How do we find the lattice enthalpy of an ionic solid, and what makes it large?

Lattice enthalpy and its determination by Born-Haber cycles, the enthalpy changes involved (formation, atomisation, ionisation, electron affinity), enthalpies of solution and hydration, and the effect of ionic charge and radius.

An OCR H432 module 5 answer on lattice enthalpy: the Born-Haber cycle and its enthalpy terms, calculating lattice enthalpy by Hess's law, enthalpies of solution and hydration, and the effect of ionic charge and radius on their magnitudes.

Generated by Claude Opus 4.813 min answer

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  1. What this topic is asking
  2. Lattice enthalpy
  3. The Born-Haber cycle
  4. Charge and radius
  5. Enthalpies of solution and hydration
  6. Examples in context
  7. Try this

What this topic is asking

OCR specification point 5.2.1 wants you to define lattice enthalpy, construct Born-Haber cycles and use them with Hess's law to calculate lattice enthalpy, define and use enthalpies of solution and hydration, and explain how ionic charge and radius affect the magnitude of lattice and hydration enthalpies. This extends the Hess's law of Module 3 to ionic compounds.

Lattice enthalpy

The Born-Haber cycle

By Hess's law, the formation enthalpy equals the sum of all the steps to the gaseous ions plus the lattice enthalpy. Rearranging gives the lattice enthalpy.

Charge and radius

Enthalpies of solution and hydration

Examples in context

Example 1. Why some salts dissolve and others do not. Whether a salt dissolves depends on the balance in ΔsolH=∑ΔhydH−ΔLEH\Delta_{sol}H = \sum\Delta_{hyd}H - \Delta_{LE}H: if hydration releases nearly as much energy as the lattice took to break, dissolving is energetically accessible, which is why charge and radius govern solubility trends.

Example 2. The stability of ionic ceramics. Magnesium oxide is used as a refractory (high-temperature) material precisely because its very exothermic lattice enthalpy reflects strong, hard-to-break ionic bonding, a direct consequence of the high charges and small ions.

Try this

Q1. State what is meant by the lattice enthalpy of an ionic compound. [2 marks]

  • Cue. The enthalpy change when one mole of the ionic solid is formed from its gaseous ions (an exothermic process).

Q2. State and explain whether the lattice enthalpy of NaF or NaCl is more exothermic. [2 marks]

  • Cue. NaF, because the fluoride ion is smaller than the chloride ion, so the ions are closer and the electrostatic attraction is stronger.

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20205 marksUse the following data to calculate the lattice enthalpy of sodium chloride by a Born-Haber cycle: ΔfH=−411\Delta_f H = -411, atomisation of Na =+107= +107, first ionisation energy of Na =+496= +496, atomisation of 12Cl2=+122\tfrac{1}{2}\text{Cl}_2 = +122, electron affinity of Cl =−349= -349 (all kJ mol−1\text{kJ mol}^{-1}).
Show worked answer →

By Hess's law, the formation enthalpy equals the sum of the steps to gaseous ions plus the lattice enthalpy (the formation of the lattice from gaseous ions) (1).

ΔfH=ΔatH(Na)+IE1(Na)+ΔatH(12Cl2)+EA(Cl)+ΔLEH\Delta_f H = \Delta_{at}H(\text{Na}) + IE_1(\text{Na}) + \Delta_{at}H(\tfrac{1}{2}\text{Cl}_2) + EA(\text{Cl}) + \Delta_{LE}H (1).

Rearranging: ΔLEH=ΔfH−[ΔatH(Na)+IE1+ΔatH(Cl)+EA]\Delta_{LE}H = \Delta_f H - [\Delta_{at}H(\text{Na}) + IE_1 + \Delta_{at}H(\text{Cl}) + EA] (1).

ΔLEH=−411−[107+496+122+(−349)]=−411−376=−787 kJ mol−1\Delta_{LE}H = -411 - [107 + 496 + 122 + (-349)] = -411 - 376 = -787\ \text{kJ mol}^{-1} (1)(1).

Markers reward the Born-Haber relationship, the rearrangement, the substitution and the value of about −787 kJ mol−1-787\ \text{kJ mol}^{-1}.

OCR 20223 marksExplain why the lattice enthalpy of magnesium oxide is much more exothermic than that of sodium chloride.
Show worked answer →

Magnesium oxide contains Mg2+\text{Mg}^{2+} and O2−\text{O}^{2-} ions, which carry double the charge of the Na+\text{Na}^+ and Cl−\text{Cl}^- ions in sodium chloride (1). The ions in MgO are also smaller (1), so the ions are closer together and the electrostatic attraction between them is much stronger, giving a far more exothermic lattice enthalpy (1).

Markers reward the higher ionic charges, the smaller ionic radii, and the link to stronger electrostatic attraction.

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