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Bond enthalpiesEdexcel International A Level Chemistry: Revision notes

Section 1

Bond enthalpy and mean bond enthalpy

Bond enthalpy is the enthalpy change when 1 mol of a covalent bond is broken in the gas phase. Breaking bonds is endothermic (ΔH positive); making bonds is exothermic (ΔH negative).

The same type of bond has slightly different values in different molecules, so data tables give a mean bond enthalpy: the average enthalpy change for breaking 1 mol of a bond, averaged over a range of compounds containing that bond.

Typical values (kJ mol⁻¹): C–H 413, C–C 347, C=C 612, O–H 464, O=O 498, H–H 436, N≡N 945.

Key termsbond enthalpymean bond enthalpy

Section 2

Calculating ΔH from bond enthalpies

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

  1. Draw displayed formulae and count every bond that changes.
  2. Add the bond enthalpies of all bonds broken (energy in).
  3. Add the bond enthalpies of all bonds formed (energy out).
  4. Subtract: broken – formed.

Worked example: CH₄ + 2O₂ → CO₂ + 2H₂O (all gases) Broken: 4(413) + 2(498) = 2648. Formed: 2(805) + 4(464) = 3466. ΔH = 2648 – 3466 = –818 kJ mol⁻¹.

Common mistake

Subtracting the wrong way round. For bond enthalpies it is broken minus formed, the opposite of formation data.

Section 3

Limitations of the method

Results are only estimates:

  • Mean values are averages, not exact for the bonds in the particular molecules.
  • The data apply only to gaseous substances; if a reactant or product is a liquid or solid, extra enthalpy changes (e.g. of vaporisation) are needed.
  • The calculated value usually differs from the experimental value, which is more reliable and is found from formation or combustion data using Hess's law.

The sign and approximate size are usually correct.

Exam tip

In an evaluation question, state that the bond enthalpies are mean values and that the species must be gases.

Section 4

Calculating a mean bond enthalpy

Use an enthalpy cycle to find the enthalpy change for converting the molecule into gaseous atoms, then divide by the number of bonds.

Worked example: mean C–H in CH₄ CH₄(g) → C(g) + 4H(g) ΔH = –ΔfH(CH₄) + ΔatH(C) + 4 × ΔatH(H) = +75 + 717 + 4(218) = +1664 kJ mol⁻¹ Four C–H bonds, so mean C–H bond enthalpy = 1664 ÷ 4 = 416 kJ mol⁻¹.

Section 5

Bond enthalpy and reactivity

Bond enthalpy data show which bond breaks first and how easily.

  • The bond with the lowest bond enthalpy breaks first.
  • A high bond enthalpy means a lot of energy is needed to break the bond, so the reaction is slow at room temperature.
  • A low bond enthalpy means the bond breaks easily, so the reaction is faster.

Examples: nitrogen is unreactive at room temperature as N≡N has a very high bond enthalpy (945 kJ mol⁻¹). The C–I bond (228) is weaker than C–Br (290) and C–Cl (346), so iodoalkanes react faster than bromoalkanes and chloroalkanes.

Must Know

  • Breaking bonds endothermic; making bonds exothermic
  • ΔH = Σ bonds broken – Σ bonds formed
  • Mean bond enthalpies are averages and apply only to gases, so results are estimates
  • Mean bond enthalpy from ΔH of atomisation ÷ number of bonds
  • Lowest bond enthalpy breaks first; high bond enthalpy gives slow reaction at room temperature

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Bond enthalpies

  1. A gas hob burns methane in air: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g). An engineer estimates the enthalpy change using mean bond enthalpies (kJ mol⁻¹): C–H 413, O=O 498, C=O (in CO₂) 805, O–H 464.
    The experimental standard enthalpy change of combustion of methane is –890 kJ mol⁻¹. Give two reasons why the value calculated from mean bond enthalpies is different.2 marks
  2. Chemists use bond enthalpy data to predict which bonds break most easily and so how quickly a compound reacts at room temperature. Mean bond enthalpies (kJ mol⁻¹): C–Cl 346, C–Br 290, C–I 228, N≡N 945, O=O 498.
    The compound BrCH₂CH₂Cl contains a C–Br bond and a C–Cl bond. State which bond breaks first and explain why. Predict, with a reason, whether iodoethane reacts faster or slower than bromoethane.2 marks
  3. In the Haber process nitrogen and hydrogen react to form ammonia: N₂(g) + 3H₂(g) → 2NH₃(g). A student uses mean bond enthalpies (kJ mol⁻¹): N≡N 945, H–H 436, N–H 391.
    Define the term mean bond enthalpy and explain why bond enthalpy data tables quote mean values.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).