Bond enthalpies — AQA A-Level Chemistry
Test yourself on Bond enthalpies with AQA A-Level practice questions.
A week of Premium from your first lesson · Then the free plan continues · No card, no charge
Bond enthalpies explained
A mean bond enthalpy is the enthalpy change when one mole of a stated covalent bond is broken in the gaseous state, averaged over a range of different compounds containing that bond.
Read the full explanation
The two halves both matter. Gaseous state, because breaking a bond in a liquid would also have to pay for the intermolecular forces. Averaged, because the O−H bond in water is not quite the same strength as the O−H bond in ethanol, so the tabulated value is a compromise that fits neither exactly. Bond breaking always takes energy in, so mean bond enthalpies are always positive and examiners refuse any definition that says energy released. The averaging is also why any ∆H calculated from them is only approximate.
Students should be able to: define the term mean bond enthalpy use mean bond enthalpies to calculate an approximate value of ∆H for reactions in the gaseous phase explain why values from mean bond enthalpy calculations differ from those determined using Hess’s law.
Mean bond enthalpy is the average enthalpy change to break one mole of a specified covalent bond in the gaseous state, averaged across many different compounds. To use these values, treat the reaction as pulling reactant molecules apart into atoms and reassembling them: ∆H = Σ(bonds broken) − Σ(bonds made). For example, calculating ∆H for the combustion of methane by counting all C-H, O=O, C=O, and O-H bonds. Count each bond type from displayed formulae and multiply by balancing numbers. The answer often disagrees with a Hess's law value because tabulated figures are averages, not exact values for specific molecules, and the method assumes all species are gaseous, ignoring intermolecular forces in liquids or solids.
Your focus
- Define mean bond enthalpy in a form that includes one mole of bonds, the gaseous state and the averaging over different compounds.
- Explain why every mean bond enthalpy is positive by reference to what happens to the bond.
- Justify why the O−H mean bond enthalpy does not exactly describe the O−H bond in any one named molecule.
Show all 7 objectives
- Write an equation, with state symbols, whose enthalpy change is the bond enthalpy of a stated diatomic bond.
- Define mean bond enthalpy as an average value for breaking one mole of a bond in the gaseous state.
- Calculate an approximate ∆H for a gas phase reaction from mean bond enthalpies, counting each bond type from displayed formulae.
- Explain two distinct reasons why a mean bond enthalpy result differs from a value obtained using Hess's law.
Bond enthalpies exam tips
Marking Points
- one mark for enthalpy change to break one mole of bonds
- one mark for specifying the gaseous state
- one mark for averaged over a range of different compounds or molecules
- one mark for recognising the value as positive, since breaking a bond requires energy
- One mark for defining mean bond enthalpy as the energy to break one mole of a covalent bond in the gas phase, averaged across compounds.
- One mark for the correct expression, total bonds broken minus total bonds made.
- One mark for counting every bond in each species and multiplying by the balancing numbers.
- One mark for explaining a difference from a Hess's law value as averaging over compounds, or as data applying only to the gaseous state.
Examiner Tips
- 💡Learn the definition as four chunks: enthalpy change, to break one mole of bonds, in the gaseous state, averaged over different compounds.
- 💡Heat energy change is accepted in place of enthalpy change, but enthalpy released is never accepted.
- 💡If the question names a specific molecule, still say averaged over a range of compounds; the average is the defining feature.
- 💡Draw the displayed formulae of everything in the equation and tick off bonds as you count them.
- 💡When asked why the value differs from one found by Hess's law, give both reasons: averaging, and the assumption that all species are gaseous.
Common Mistakes
- defining it as the energy released when a bond is formed, which examiners explicitly refuse
- omitting the gaseous state, so the definition would include the energy to vaporise the substance
- omitting the averaging, which is the point that makes it a mean bond enthalpy rather than a bond dissociation enthalpy
- writing per mole of molecules rather than per mole of bonds
- assuming the tabulated value applies exactly to the compound in the question
- Reversing the expression to bonds made minus bonds broken, which gives the right size and the wrong sign.
- Missing the unchanged bonds in a spectator part of the molecule, or double counting the C−H bonds in a large molecule.
- Explaining the discrepancy as experimental error rather than as a consequence of using averaged, gas phase data.