Enthalpy of phase change: fusion, vaporisation, sublimation
When a solid melts or a liquid boils, the temperature stays constant until the whole sample has changed. The heat goes into pulling molecules apart, not into speeding them up.
- Enthalpy of fusion ΔfusH: heat to melt 1 mol of a solid at its melting point. Ice: +6.01 kJ mol⁻¹ at 0 °C.
- Enthalpy of vaporisation ΔvapH: heat to turn 1 mol of a liquid into vapour at its boiling point. Water: +40.79 kJ mol⁻¹ at 100 °C.
- Enthalpy of sublimation ΔsubH: solid straight to vapour (like naphthalene balls or dry ice). At the same temperature, ΔsubH = ΔfusH + ΔvapH (Hess's law).
All three are endothermic. The reverse changes (freezing, condensing, deposition) have the same size with a negative sign. A bigger ΔvapH means stronger forces between the molecules.
Enthalpy of combustion
The standard enthalpy of combustion ΔcH° is the enthalpy change when 1 mol of a substance burns completely in oxygen, all reactants and products being in their standard states. It is always negative (exothermic).
Examples: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), ΔcH° = −890.3 kJ mol⁻¹. H₂(g) + ½O₂(g) → H₂O(l), ΔcH° = −285.8 kJ mol⁻¹.
Fuels and foods are compared by the heat they give per gram (calorific value). Combustion values are measured in a bomb calorimeter and are used with Hess's law to find enthalpies of formation.
Enthalpy of atomisation and bond enthalpy
Enthalpy of atomisation ΔaH: the enthalpy change to break 1 mol of a substance completely into atoms in the gas phase. For H₂(g) → 2H(g), ΔaH = +435 kJ mol⁻¹. For a metal like Na(s) → Na(g), it equals the enthalpy of sublimation.
Bond dissociation enthalpy: the enthalpy change to break 1 mol of a particular covalent bond in gaseous molecules. For a diatomic molecule it equals the atomisation enthalpy (H–H: 435, Cl–Cl: 242 kJ mol⁻¹).
Mean bond enthalpy
In CH₄ the four C–H bonds need slightly different energies to break one after another. So we take the average: CH₄(g) → C(g) + 4H(g), ΔaH = 1665 kJ mol⁻¹, so the mean C–H bond enthalpy = 1665 ÷ 4 = 416 kJ mol⁻¹.
Reaction enthalpy from bond enthalpies (gas phase)
ΔrH° = Σ bond enthalpies of reactants (broken) − Σ bond enthalpies of products (formed). Bond breaking is endothermic, bond making is exothermic. The answer is approximate because mean values are used.
Lattice enthalpy and the Born–Haber cycle
Lattice enthalpy of an ionic compound is the enthalpy change when 1 mol of the solid breaks into its gaseous ions: NaCl(s) → Na⁺(g) + Cl⁻(g), ΔlatticeH° = +788 kJ mol⁻¹.
We cannot measure it directly, so we use a Born–Haber cycle (Hess's law in a loop): sublime Na (+108.4), break ½Cl₂ (+121), ionise Na (+496), add an electron to Cl (electron gain, −348.6) and compare with the enthalpy of formation of NaCl (−411.2). The lattice enthalpy is the missing step that closes the loop.
Bigger charges and smaller ions give larger lattice enthalpies (MgO is much larger than NaCl).
Enthalpy of solution, hydration and dilution
Enthalpy of solution ΔsolH: the enthalpy change when 1 mol of a substance dissolves in a large amount of solvent. Think of it in two steps: (1) break the lattice into gaseous ions (lattice enthalpy, +); (2) water molecules surround the ions (hydration enthalpy, −).
ΔsolH = ΔlatticeH + ΔhydH. For NaCl: 788 + (−784) = +4 kJ mol⁻¹, slightly endothermic. When hydration gives out more than the lattice needs, dissolving is exothermic (like anhydrous CaCl₂ or NaOH); when less, the solution cools (like NH₄NO₃, ΔsolH ≈ +25.7 kJ mol⁻¹).
Enthalpy of dilution is the heat change when more solvent is added to a solution; it depends on how much solvent is added. Some books also define enthalpies of neutralisation (strong acid + strong base ≈ −57 kJ per mol of water) and ionisation.
In the exam
Definitions (1 mark), bond-enthalpy numericals (2–3 marks) and Born–Haber / ΔsolH reasoning (3 marks) are common.
Try it at home: feel an enthalpy change
(1) Put a drop of water on the back of one hand and a drop of hand sanitiser (alcohol) on the other. Blow gently. The alcohol side feels colder: it evaporates faster, taking ΔvapH from your skin. (2) Dissolve a spoon of urea fertiliser or a cold-pack powder in a glass of water and touch the glass: it cools, so ΔsolH > 0. Now open free play and compare the bars for 'Vaporisation' and 'Solution of NH₄NO₃'.
Key formulas and definitions
- ΔsubH = ΔfusH + ΔvapH (same temperature)
- Heat for n mol: q = n × ΔH (e.g. melting: q = n ΔfusH)
- Mean bond enthalpy = ΔaH ÷ number of such bonds (e.g. C–H = 1665 ÷ 4)
- ΔrH ≈ Σ(bond enthalpies broken) − Σ(bond enthalpies formed) (gas phase)
- ΔsolH = ΔlatticeH + ΔhydH
- Born–Haber: ΔfH = ΔsubH + ½ΔbondH + IE + ΔegH − ΔlatticeH
- Water: ΔfusH = 6.01, ΔvapH = 40.79 kJ mol⁻¹; ΔcH(CH₄) = −890.3 kJ mol⁻¹
Worked examples
1. How much heat is needed to melt 36 g of ice at 0 °C? (ΔfusH = 6.01 kJ mol⁻¹)
Moles = 36 ÷ 18 = 2 mol. q = 2 × 6.01 = 12.02 kJ.
2. Find the heat needed to boil away 90 g of water at 100 °C (ΔvapH = 40.79 kJ mol⁻¹).
n = 90 ÷ 18 = 5 mol. q = 5 × 40.79 = 203.95 ≈ 204 kJ.
3. At 0 °C, ΔfusH of ice = 6.01 kJ mol⁻¹ and ΔvapH of water = 45.07 kJ mol⁻¹. Find ΔsubH of ice at 0 °C.
ΔsubH = ΔfusH + ΔvapH = 6.01 + 45.07 = 51.08 kJ mol⁻¹.
4. How much heat is released when 8 g of methane burns? (ΔcH = −890.3 kJ mol⁻¹)
Moles of CH₄ = 8 ÷ 16 = 0.5 mol. Heat released = 0.5 × 890.3 = 445.2 kJ (ΔH = −445.2 kJ).
5. ΔaH of CH₄ is 1665 kJ mol⁻¹. Find the mean C–H bond enthalpy.
CH₄ has 4 C–H bonds. Mean C–H = 1665 ÷ 4 = 416.25 ≈ 416 kJ mol⁻¹.
6. Use bond enthalpies (H–H 435, Cl–Cl 242, H–Cl 431 kJ mol⁻¹) to find ΔrH for H₂(g) + Cl₂(g) → 2HCl(g).
Broken: 435 + 242 = 677 kJ. Formed: 2 × 431 = 862 kJ. ΔrH = 677 − 862 = −185 kJ.
7. Find ΔrH for CH₄ + Cl₂ → CH₃Cl + HCl. Bond enthalpies: C–H 414, Cl–Cl 242, C–Cl 330, H–Cl 431 kJ mol⁻¹.
Only one C–H and one Cl–Cl break; one C–Cl and one H–Cl form. Broken = 414 + 242 = 656. Formed = 330 + 431 = 761. ΔrH = 656 − 761 = −105 kJ mol⁻¹.
8. Born–Haber for NaCl: ΔsubH(Na) = +108.4, ½ΔbondH(Cl₂) = +121, IE(Na) = +496, ΔegH(Cl) = −348.6, ΔfH(NaCl) = −411.2 kJ mol⁻¹. Find the lattice enthalpy.
ΔfH = ΔsubH + ½ΔbondH + IE + ΔegH − ΔlatticeH. So ΔlatticeH = 108.4 + 121 + 496 − 348.6 + 411.2 = +788 kJ mol⁻¹.
9. Lattice enthalpy of NaCl = +788 kJ mol⁻¹ and its hydration enthalpy = −784 kJ mol⁻¹. Find ΔsolH. Will the water warm or cool?
ΔsolH = 788 + (−784) = +4 kJ mol⁻¹. Positive, so the water cools very slightly.
Common mistakes
- Writing 'bonds formed − bonds broken'. The correct order is broken − formed (reactants − products for bond enthalpies).
- Forgetting to count how many bonds break: 2HCl has 2 H–Cl bonds, CH₄ has 4 C–H bonds.
- Giving combustion enthalpy a positive sign. Burning always releases heat: ΔcH < 0.
- Adding ΔfusH and ΔvapH measured at different temperatures to get ΔsubH. They must be at the same temperature.