R1.2 Energy cycles in reactionsIB Chemistry SL: Revision notes

Breaking a covalent bond always absorbs energy (endothermic) because the attraction between the bonded atoms and the shared electron pair must be overcome. Forming a bond releases the same amount of energy (exothermic).

A reaction is exothermic overall when more energy is released forming the new bonds than is absorbed breaking the old ones, and endothermic when the reverse is true.

The bond enthalpy is the energy needed to break one mole of a given covalent bond in gaseous molecules. For bonds found in many compounds (C–H, O–H, C–C) the data booklet gives an average bond enthalpy, the mean over a range of compounds. For diatomic molecules such as H₂, Cl₂ and HCl the value is exact for that molecule.

ΔH = Σ(bonds broken) − Σ(bonds formed)

Picture every structure and count the bonds. Example, C₂H₄ + H₂ → C₂H₆: broken C=C (614) + H–H (436) = 1050; formed C–C (346) + 2 C–H (828) = 1174; ΔH = −124 kJ mol⁻¹. You can count only the bonds that change, or break every bond and form every bond: the answer is the same.

Values from bond enthalpies are approximate because:

• average values are not exact for the bonds in a particular molecule;
• bond enthalpies apply only to gases, so if a reactant or product is a liquid or solid, the energy of the change of state is missing.

When a calculated value differs from experiment, check the physical states first: liquid bromine and solid iodine give much bigger discrepancies than gaseous chlorine.

Hess's law: the enthalpy change for a reaction is independent of the pathway between the initial and final states. It follows from conservation of energy.

So if a reaction cannot be measured directly, you can find its ΔH by going round an energy cycle through steps whose enthalpy changes are known.

Treat the given equations like algebra so that they add up to the target equation:

• reverse an equation → change the sign of ΔH;
• multiply an equation → multiply ΔH by the same factor;
• add the equations and cancel species that appear on both sides.

Example: S + O₂ → SO₂ (−297 kJ) and 2SO₂ + O₂ → 2SO₃ (−198 kJ). For S + 1½O₂ → SO₃: −297 + ½(−198) = −396 kJ.