8.9-8.11 - Bond Enthalpies

Energy Changes When Bonds Break And Form

A common trap is to say that exothermic reactions happen because "breaking bonds releases energy". That is backwards. A covalent bond is an attraction between nuclei and shared electrons, so energy is needed to separate the bonded atoms.

In a reaction, two energy changes are happening in the reacting system:

The overall enthalpy change depends on the balance between those two totals. If bond formation releases more energy than bond breaking requires, the reaction is exothermic and $\Delta H$ is negative. If bond breaking requires more energy than bond formation releases, the reaction is endothermic and $\Delta H$ is positive.

Problem opener: hydrogen and chlorine react to form hydrogen chloride:

The reaction is exothermic even though the H-H and Cl-Cl bonds must be broken first. This is an energy-accounting route, not a claim that every reactant bond breaks simultaneously before any product forms. The reaction is exothermic because forming two H-Cl bonds releases more energy than was needed to break one H-H bond and one Cl-Cl bond.

Bond Enthalpy and Mean Values

Bond enthalpy puts a number on the energy needed to separate bonded atoms in the gas phase.

This averaging is needed because the exact strength of a bond depends on its molecular environment. For example, an O-H bond in water is not in exactly the same electronic environment as an O-H bond in an alcohol or a carboxylic acid.

That distinction matters in calculations:

Common-error contrast: do not treat a mean bond enthalpy as a measured enthalpy change for one named reaction. It is an average model. It can still predict whether a reaction is roughly exothermic or endothermic, but it will not usually match a precise calorimetry or Hess's law value.

Calculating Reaction Enthalpy

Mean bond enthalpy calculations use a simple energy route: break the reactant bonds into gaseous atoms, then make the product bonds from those atoms.

[DIAGRAM: asset_name: Bond Breaking And Forming; asset_slug: edexcel_a_level_chemistry_l045_bond_breaking_and_forming; recommended_method: deterministic_chart; description: Conceptual bond enthalpy route showing reactant bonds being broken to separated atoms, then product bonds being formed.]

The calculation is:

Unchanged bond types can cancel in a mean-value calculation. For $\ce{CH3OH(g) + HCl(g) -> CH3Cl(g) + H2O(g)}$, three C-H bonds and one O-H bond occur on both sides; after cancellation, break C-O and H-Cl and form C-Cl plus one additional O-H. Counting every bond gives the same answer.

The values in the table are treated as positive numbers. The subtraction gives the sign of the reaction enthalpy.

Worked example: calculate $\Delta H$ for forming steam from hydrogen and oxygen.

Given mean bond enthalpies:

Count the bonds using the balanced equation, not just the formula of one molecule.

Now subtract:

$\Delta H = 1370 - 1856$
$\Delta H = -486 \mathrm{kJ\,mol^{-1}}$