Calculating enthalpy change from bonds, from Hess's law, and — at Higher Level — from formation/combustion data, lattice enthalpy and Born–Haber cycles.
This HL lesson contains the full SL content — bond enthalpies (R1.2.1) and Hess's law (R1.2.2) — and then the Additional Higher Level material: standard enthalpies of formation and combustion (R1.2.3), lattice enthalpy and Born–Haber cycles (R1.2.4), and enthalpies of solution and hydration (R1.2.5). AHL sections are marked in purple.
Breaking a chemical bond always absorbs energy — it is endothermic (positive). Forming a bond always releases energy — it is exothermic (negative). A reaction's enthalpy change is the balance between the two:
ΔH = Σ(bonds broken) − Σ(bonds formed)
The values are average bond enthalpies: the mean energy to break one mole of a covalent bond in the gaseous state, averaged over many molecules. So the method gives an estimate — the specific bond differs from the average, and every species must be gaseous for it to apply.
Pick a gas-phase reaction. The calculator breaks every reactant bond (energy in, +) and forms every product bond (energy out, −), then takes the difference. Values in kJ mol⁻¹ (IB data booklet).
Hess's law: the enthalpy change of a reaction is independent of the route — it depends only on the initial and final states. This follows from energy conservation. If a reaction is hard to measure directly, we use an alternative route whose steps we know, and add them (reversing a step reverses the sign of its ΔH).
Two special enthalpy changes let us build Hess cycles from tabulated data:
The standard enthalpy of formation, ΔHf°, is the enthalpy change when one mole of a compound forms from its elements in their standard states (100 kPa, a stated temperature, usually 298 K). By definition, an element in its standard state has ΔHf° = 0. The standard enthalpy of combustion, ΔHc°, is the enthalpy change when one mole of a substance burns completely in oxygen under standard conditions.
From formation data, a reaction's enthalpy is products minus reactants:
ΔHr° = Σ ΔHf°(products) − Σ ΔHf°(reactants)
(From combustion data the cycle flips the other way: ΔHr° = Σ ΔHc°(reactants) − Σ ΔHc°(products), because the elements/compounds are joined to a common combustion product.)
For an ionic compound we cannot measure the enthalpy of forming the lattice from gaseous ions directly, so we use a Born–Haber cycle — a Hess cycle built from steps we can measure. In the IB the lattice enthalpy, ΔHlat°, is defined as the energy to break one mole of the solid ionic lattice into its gaseous ions. That is endothermic, so IB data-booklet lattice enthalpies are positive. A larger value means stronger ionic bonding (higher ionic charge and/or smaller ions).
The other steps for NaCl are the enthalpy of atomisation of Na and of Cl (endothermic), the first ionisation energy of Na (endothermic), and the first electron affinity of Cl (exothermic). Around the cycle, the direct formation of NaCl from its elements (ΔHf°) must equal the indirect route:
When an ionic solid dissolves, two things happen: the lattice is pulled apart into gaseous ions (endothermic, = +ΔHlat°), then those ions are surrounded by water — hydrated — releasing energy (exothermic, the enthalpy of hydration ΔHhyd°, one value per ion). The enthalpy of solution is their sum:
ΔHsol° = ΔHlat° + Σ ΔHhyd°
Because lattice breaking (+) and hydration (−) nearly cancel, enthalpies of solution are usually small — sometimes slightly endothermic, sometimes slightly exothermic. Hydration is more exothermic for ions with higher charge and smaller radius (stronger ion–dipole attraction to water).
Seven questions (five SL, two HL) — instant feedback, nothing saved.
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