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AQA A-Level Chemistry 7405 · 3.1.4 Energetics

Part 4: Bond enthalpies & fuel comparisons

All four parts available · diagram placeholders included. Reviewed 1 October 2026.

1. Mean bond enthalpy

The mean bond enthalpy is the enthalpy change to break one mole of a specified covalent bond in gaseous molecules, averaged over a range of different compounds.

  • Breaking bonds is always endothermic (positive). Making bonds is always exothermic (negative).
  • A stronger bond has a larger bond enthalpy.
  • Breaking a bond never releases energy; energy is released only when new bonds form.

Why “mean”? The same type of bond has slightly different strengths in different molecules. Even within CH4, removing the four H atoms one by one takes four different amounts of energy, because each removal leaves a different fragment (CH3, then CH2…). The C−H bonds in intact methane are equivalent; it is the successive steps that differ.

Some values are not means. O=O occurs only in O2, so its value is a bond dissociation enthalpy specific to O2, rather than an average across different compounds.

Mean bond enthalpies used here (kJ mol⁻¹): C−H 413, C−C 347, C=C 612, H−H 436, O=O 498, C=O (in CO₂) 805, O−H 464, Cl−Cl 243, C−Cl 346, H−Cl 432.

2. ΔH from bond enthalpies

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

Diagram placeholder

Bond breaking and making as an enthalpy diagram

Labels to include:

  • Vertical axis: Enthalpy
  • Lower left: reactants (gaseous molecules)
  • Top: separate gaseous atoms
  • Upward arrow reactants → atoms: Σ bonds broken (+)
  • Downward arrow atoms → products: Σ bonds formed (−)
  • Right: products level; ΔH arrow reactants → products

Going up to atoms costs energy; coming down to products releases energy. If more is released than was put in, ΔH is negative.

3. Counting every bond

Draw out each molecule and multiply by its coefficient. Useful counts:

  • A straight-chain alkane CnH2n+2 has (n − 1) C−C bonds and (2n + 2) C−H bonds.
  • O2 has one O=O; CO2 has two C=O; H2O has two O−H.

4. Finding an unknown bond enthalpy

5. Why the answers differ from data-book values

Values from mean bond enthalpies are approximate because the actual bonds in a particular molecule are not exactly the average strength across many compounds. This is a limitation of the method, not experimental error. Answers using ΔfH or ΔcH data for the specific compounds are more accurate.

6. Gas-phase states and corrections

Bond enthalpies apply to gases. If a question's equation has liquid water, include the extra state change:

  • H2O(g) → H2O(l) is condensation, exothermic (about −44 kJ mol−1). Converting n mol of gaseous product water to liquid adds n × (−44), making ΔH more exothermic.
  • A reactant that is liquid must first be vaporised; add its (positive) enthalpy of vaporisation where provided.

Carbon in formation cycles

If you build a formation cycle from bond enthalpies, solid carbon cannot be “bond-broken” directly: you need the enthalpy of atomisation of carbon, C(s) → C(g), which the question will supply. (Full atomisation and Born–Haber cycles belong to 3.1.8 Thermodynamics.)

7. Homologous series

Each extra CH2 in an alkane or alcohol makes ΔcH more exothermic by a roughly constant amount (about −650 kJ mol−1 for alkanes): methane −890, ethane −1560, propane −2220, butane −2877.

Why: each extra CH2 means more bonds to break (one C−C, two C−H, 1½ O=O ≈ +1920) and more bonds to form (two C=O, two O−H ≈ −2538). The extra bond-making outweighs the extra bond-breaking, giving about −618 kJ mol−1 (with H2O(g)) per CH2.

8. Comparing fuels fairly

“Energy per mole” favours big molecules. Fairer comparisons:

energy per gram = |ΔcH| ÷ M (kJ g−1)
energy per dm3 = density (g dm−3) × |ΔcH| ÷ M (kJ dm−3)
CO2 per kJ = (mol CO2 in the equation) ÷ |ΔcH|

Quick checks

These are Finesse practice questions. The step-by-step answers are indicative worked solutions, not official AQA mark allocations.

Mean bond enthalpies used here (kJ mol⁻¹): C−H 413, C−C 347, C=C 612, H−H 436, O=O 498, C=O (in CO₂) 805, O−H 464, Cl−Cl 243, C−Cl 346, H−Cl 432.

Q1. Calculate ΔH for CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g).Show answer

Step 1: three C−H terms cancel (same bond type and same mean value on both sides). Broken: C−H 413 + Cl−Cl 243 = 656.

Step 2: formed: C−Cl 346 + H−Cl 432 = 778.

Step 3: ΔH = 656 − 778 = −122 kJ mol−1.

Q2. Calculate ΔH for C2H6(g) + 3½O2(g) → 2CO2(g) + 3H2O(g). Then give the value for liquid water (ΔHcond = −44 kJ mol−1).Show answer

Step 1: broken: 347 + 6(413) + 3.5(498) = 4568.

Step 2: formed: 4(805) + 6(464) = 6004.

Step 3: ΔH = 4568 − 6004 = −1436 kJ mol−1.

Step 4: liquid water: −1436 + 3(−44) = −1568 kJ mol−1.

Q3. H2(g) + ½O2(g) → H2O(g), ΔH = −242 kJ mol−1. Using H−H 436 and O=O 498, find the O−H bond enthalpy in water.Show answer

Step 1: broken = 436 + ½(498) = 685.

Step 2: −242 = 685 − 2x, so 2x = 927.

Step 3: x = 464 kJ mol−1 (463.5).

Q4. Give two reasons why ΔcH calculated from mean bond enthalpies differs from the data-book value.Show answer

Mean bond enthalpies are averages across many compounds, not the exact values in these molecules.

Bond enthalpies assume gaseous species, whereas the standard combustion has H2O(l) (and some fuels are liquids), so state changes are missing unless corrected.

Q5. Ethanol: ΔcH = −1367 kJ mol−1, M = 46.0, density 789 g dm−3. Find energy per gram, energy per dm3 and mol CO2 per kJ. Compare with octane in the table above.Show answer

Step 1: 1367 ÷ 46.0 = 29.7 kJ g−1.

Step 2: 789 × 29.72 = 2.34 × 104 kJ dm−3.

Step 3: C2H5OH forms 2CO2: 2 ÷ 1367 = 1.46 × 10−3 mol kJ−1.

Ethanol gives less energy per gram and per dm3 than octane, and almost the same CO2 per kJ. Other advantages or drawbacks require considering production route, renewability, cost, storage and other emissions.

Sources

Sources and examiner guidance (reviewed 1 October 2026)

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