Bond enthalpiesAQA A-Level Chemistry: Revision notes
Section 1
Bond breaking and bond making
Chemical reactions involve breaking bonds in the reactants and making new bonds in the products.
- Breaking bonds is endothermic: energy must be supplied to overcome the attraction between the bonded atoms.
- Making bonds is exothermic: energy is released.
The overall enthalpy change depends on the balance between the two. If more energy is released making bonds than is absorbed breaking bonds, the reaction is exothermic and ΔH is negative.
Saying that bond breaking releases energy. Breaking bonds always needs energy.
Section 2
Mean bond enthalpy
The mean bond enthalpy is the enthalpy change when one mole of a given type of bond is broken in gaseous molecules, averaged over a range of different compounds containing that bond. It is in kJ mol⁻¹ and is always positive.
The value is a mean because a bond such as C–H has a slightly different strength in methane, ethane or ethanol, so tables use an average. Bond enthalpies are given in the question, and need not be recalled.
A double bond (C=C) is stronger than a single bond (C–C) but less than twice as strong, and a triple bond is stronger still.
Always include the idea of averaging over different compounds and the gaseous state in a definition.
Section 3
Calculating ΔH from mean bond enthalpies
ΔH = Σ(bonds broken) − Σ(bonds made)
- Draw out the structures so that every bond can be counted.
- Add up the bond enthalpies for all bonds broken (reactants).
- Add up the bond enthalpies for all bonds made (products).
- Subtract: bonds broken minus bonds made.
Worked example: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g).
Broken: 4 × C–H (413) + 2 × O=O (496) = 2644 kJ. Made: 2 × C=O (805) + 4 × O–H (463) = 3462 kJ. ΔH = 2644 − 3462 = −818 kJ mol⁻¹.
The answer is only approximate, because mean values are used.
Forgetting to multiply by the number of moles in the balanced equation, for example using three N–H bonds instead of six for 2NH₃.
Section 4
Why only gaseous reactions
Mean bond enthalpies refer to bonds broken in the gas phase, so the method is only valid when all reactants and products are gases. If a substance is a liquid or solid, extra energy is needed to vaporise it, and this is not included in the bond enthalpy data.
For example, using mean bond enthalpies for the combustion of methane to give H₂O(g) is valid, but combustion giving H₂O(l) would need the enthalpy of vaporisation of water as well.
Section 5
Why bond enthalpy values differ from Hess's law
A value from mean bond enthalpies is usually different from one calculated using Hess's law and enthalpies of formation or combustion, because:
- mean bond enthalpies are averages over many compounds, so are not exact for the bonds in the particular molecules
- Hess's law uses experimental data for the actual substances, so is more accurate
- bond enthalpy calculations assume the reactants and products are gases
For methane combustion, the bond enthalpy value is −818 kJ mol⁻¹, but Hess's law with enthalpies of formation gives −802.3 kJ mol⁻¹.
Must know
- Breaking bonds is endothermic; making bonds is exothermic
- ΔH = Σ(bonds broken) − Σ(bonds made)
- Mean bond enthalpy: one mole of a bond broken in gaseous molecules, averaged over different compounds
- Only valid for gases, and approximate
- Differs from Hess's law because of averaging
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Bond enthalpies
- Hydrogen reacts with chlorine in the gas phase: H₂(g) + Cl₂(g) → 2HCl(g). The mean bond enthalpies, in kJ mol⁻¹, are: H–H 436, Cl–Cl 243 and H–Cl 432.Explain, in terms of bonds, why this reaction is exothermic.2 marks
- Ethene reacts with hydrogen in the gas phase in the presence of a nickel catalyst: C₂H₄(g) + H₂(g) → C₂H₆(g). The mean bond enthalpies, in kJ mol⁻¹, are: C=C 612, C–C 347, C–H 413 and H–H 436.Define the term mean bond enthalpy.2 marks
- Methane burns completely in oxygen with all species in the gas phase: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g). The mean bond enthalpies, in kJ mol⁻¹, are: C–H 413, O=O 496, C=O 805 and O–H 463. The standard enthalpies of formation, in kJ mol⁻¹, are: CH₄(g) −74.8, CO₂(g) −393.5 and H₂O(g) −241.8.Use the mean bond enthalpies to calculate the enthalpy change for the combustion of methane.3 marks
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).