CHEM20018 · Chemistry: Reactions And Synthesis
Inorganic Thermodynamics: Structures & Lattice Enthalpy
This section turns ionic solids into a thermodynamics problem. You start from close-packed structures (CCP/FCC and HCP) and the standard ionic structure types (NaCl, CsCl, zinc blende, wurtzite, CaF₂), then quantify their stability with lattice enthalpy. The central tool is the Born–Haber cycle, a Hess's-law loop that links a solid's formation enthalpy to atomisation, ionisation energy, electron affinity and lattice enthalpy — so any one unknown leg can be solved. Calculated lattice enthalpies come from the Born–Landé / Born–Mayer equations (needing a Madelung constant) or Kapustinskii (which does not), and the same ideas explain trends in thermal stability and solubility of ionic compounds.
What this chapter covers
- 01Close packing CCP (= FCC) and HCP; coordination number 12; octahedral and tetrahedral holes
- 02Standard structure types: NaCl, CsCl, zinc blende, wurtzite, CaF₂ (fluorite)
- 03Born–Haber cycle: a Hess loop relating ΔHf°, atomisation, IE, EA and lattice enthalpy
- 04Born–Landé / Born–Mayer lattice enthalpy (Madelung constant A, charge product, interionic distance d, repulsion term)
- 05Kapustinskii equation: estimates ΔHL without a Madelung constant, using number of ions and thermochemical radii
- 06ΔHL ∝ |z₊z₋| and ∝ 1/d (higher charge / smaller ions → larger lattice enthalpy)
- 07Thermal stability of carbonates and sulfates: large cation stabilises large anion (lattice-mismatch argument)
- 08Solubility of ionic compounds as a balance of lattice enthalpy vs hydration enthalpy
Born–Haber cycle to extract an electron affinity
- 2 marks — correct cycle with every leg placed and signedWrite the cycle for K(s) + ½Cl₂(g) → KCl(s). The direct route is ΔHf°. The stepwise route is: atomise K (+89), ionise K (+419), atomise ½Cl₂ (+122), add the electron to Cl (electron affinity, EA, unknown), then form the lattice (ΔHL = −701).
- 1 mark — Hess equation set upApply Hess's law (the two routes have equal enthalpy): ΔHf° = ΔHatom(K) + IE(K) + ½D(Cl₂) + EA(Cl) + ΔHL.
- 1 mark — correct substitutionSubstitute: −437 = 89 + 419 + 122 + EA + (−701).
- 2 marks — arithmetic and final value with the correct (negative) signSum the knowns: 89 + 419 + 122 − 701 = −71. So −437 = −71 + EA → EA = −437 + 71 = −366 kJ mol⁻¹.
Key terms
- Lattice enthalpy
- The enthalpy change associated with forming one mole of an ionic solid from its gaseous ions (or, by convention, the reverse); a large negative value indicates a stable lattice.
- Born–Haber cycle
- A Hess's-law thermodynamic cycle linking the formation enthalpy of an ionic solid to atomisation, ionisation energy, electron affinity and lattice enthalpy.
- Madelung constant
- A purely geometric factor (A) for a given ionic structure type that sums the Coulombic interactions of an ion with all others in the lattice.
- Kapustinskii equation
- An empirical estimate of lattice enthalpy that avoids the Madelung constant, using the number of ions in the formula unit, the charge product and thermochemical radii.
- Close packing
- The most efficient arrangement of equal spheres (CCP/FCC or HCP), giving each sphere 12 nearest neighbours and creating octahedral and tetrahedral holes for counter-ions.
Inorganic Thermodynamics: Structures & Lattice Enthalpy FAQ
When do I use Born–Mayer versus Kapustinskii?
Use Born–Mayer (or Born–Landé) when you know the structure type and so can look up its Madelung constant. Use Kapustinskii when you do not have a Madelung constant — for example for a polyatomic-ion salt such as a sulfate — since it only needs the number of ions in the formula unit and thermochemical radii.
Why does a bigger cation give a more thermally stable carbonate or sulfate?
Decomposition makes a small oxide from the salt. A small cation gains a lot of lattice enthalpy by forming that small oxide, providing a strong driving force to decompose. A large cation matches the large carbonate/sulfate anion well and gains relatively little by forming the small oxide, so its salt is more thermally stable and decomposes at a higher temperature.
What controls whether an ionic compound is soluble?
Solubility is a balance: breaking the lattice costs the lattice enthalpy, while hydrating the separated ions releases hydration enthalpy. When hydration enthalpy roughly compensates for lattice enthalpy (and entropy is favourable) the salt dissolves; a very large lattice enthalpy that hydration cannot offset gives low solubility.
Exam move
Drill the Born–Haber cycle as a sign-bookkeeping exercise until you can build it from a blank page and solve for any leg. Memorise the two proportionalities ΔHL ∝ |z₊z₋| and ∝ 1/d — they answer most qualitative trend questions in one line. For the equations, know which to reach for (Born–Mayer if you have a Madelung constant, Kapustinskii if you do not) rather than the constants themselves, which are supplied. Practise converting ionic radii into an interionic distance d in pm before substituting. Tie the structure types to coordination numbers and hole-filling so you can justify why a given salt adopts NaCl vs CsCl vs fluorite. Finish with thermal-stability and solubility reasoning, both of which reduce to lattice-enthalpy-versus-(hydration or anion-size) arguments.
