Bond energies: break first, then form

By the end of this lesson

By the end you can calculate a reaction's enthalpy change from bond enthalpies, using the "bonds broken minus bonds formed" route through gaseous atoms — and say why the answer is only approximate for a real compound.

The two-step imagination

Bond enthalpies are defined for gaseous species. Breaking a bond always costs energy; making a bond always releases energy.

So picture any reaction as two steps:

Break every bond in the reactants, giving separate gaseous atoms. This costs the sum of the reactant bond enthalpies.
Form every bond in the products from those atoms. This releases the sum of the product bond enthalpies.

The atoms in the middle are the same in both steps, so the cycle closes — that is Hess's law doing the work again.

$\Delta H_r = \sum \text{bonds broken} - \sum \text{bonds formed}$

Definition

Mean bond enthalpy

The average energy needed to break one mole of a particular covalent bond, with the species in the gaseous state, averaged over a range of compounds.

It is an average because the same bond (say C–H) is not quite identical in every molecule. That is why bond-enthalpy calculations give values close to, but not exactly, the measured enthalpy change.

Why "broken minus formed" and not the other way round

Breaking bonds costs energy, so it appears with a positive sign. Forming bonds releases energy, so it appears with a negative sign.

If the bonds you make are stronger than the bonds you break, more energy is released than taken in, and the reaction is exothermic: $\Delta H$ is negative.

That is the whole idea of an exothermic reaction in bond terms — stronger bonds on the product side.

Worked example: Worked example: hydrogen and chlorine

Calculate $\Delta H_r$ for

$\text{H}_2\text{(g)} + \text{Cl}_2\text{(g)} \rightarrow 2\text{HCl(g)}$

using bond enthalpies: H–H $= 436$ , Cl–Cl $= 242$ , H–Cl $= 431\ \text{kJ mol}^{-1}$ .

Have a go first?

Worked example: Worked example: substituting on methane

Calculate $\Delta H_r$ for

$\text{CH}_4\text{(g)} + \text{Cl}_2\text{(g)} \rightarrow \text{CH}_3\text{Cl(g)} + \text{HCl(g)}$

using C–H $= 412$ , Cl–Cl $= 242$ , C–Cl $= 338$ , H–Cl $= 431\ \text{kJ mol}^{-1}$ .

Have a go first?

Quick check

Multiple choiceQuestion 1 · Recall

Multiple choiceQuestion 2 · Recall

CalculateQuestion 3 · Apply

Answer the 3 questions above to continue.