Bond enthalpiesEdexcel A-Level Chemistry: Revision notes
Section 1
Bond enthalpy
Bond enthalpy is the enthalpy change when one mole of a covalent bond is broken in the gaseous state. Breaking bonds needs energy, so bond enthalpies are always positive (endothermic). Making a bond releases the same amount of energy.
A large bond enthalpy means a strong bond. Multiple bonds are stronger than single bonds between the same atoms (C=C > C–C), though a double bond is less than twice the single bond.
Bond enthalpies refer to the gaseous state, so the energy to separate molecules (for example in a liquid) is not included.
Section 2
Mean bond enthalpy
In a molecule such as methane every C–H bond is identical, but in different compounds (CH₃OH, C₂H₆, …) the same type of bond has slightly different strengths because it has different neighbouring atoms.
The mean bond enthalpy is the average enthalpy change to break one mole of a given type of bond in the gaseous state, averaged over a range of compounds containing that bond. Tables give mean values, so results are only estimates.
Section 3
Calculating an enthalpy change of reaction
Treat a reaction as breaking all (or only the changing) bonds in the reactants, then making the bonds in the products:
Multiply each bond enthalpy by the number of moles of that bond in the balanced equation. If bonds made release more energy than is needed to break the bonds, ΔH is negative (exothermic).
Worked example: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g), with C–H 413, O=O 498, C=O 805, O–H 464 kJ mol⁻¹.
Broken: 4(413) + 2(498) = 1652 + 996 = 2648 kJ. Made: 2(805) + 4(464) = 1610 + 1856 = 3466 kJ. ΔH = 2648 − 3466 = −818 kJ mol⁻¹.
Reversing the subtraction. Energy absorbed to break bonds comes first: ΔH = broken − made. Exothermic reactions must give a negative answer.
Section 4
Calculating a mean bond enthalpy
If ΔH for a reaction is known and all but one bond enthalpy are known, set up the same equation with the unknown bond as x and solve.
Worked example: ½N₂ + 3/2 H₂ → NH₃, ΔH = −46 kJ mol⁻¹, N≡N 945, H–H 436. Broken = ½(945) + 3/2(436) = 1126.5. Made = 3x. 1126.5 − 3x = −46, so x = 391 kJ mol⁻¹.
Watch the fractions: the equation must form one mole of the product, so half moles of N₂ and 3/2 moles of H₂ are used.
Make a table of bonds broken and bonds made with their numbers before writing the equation. It stops you missing bonds, especially in organic molecules.
Section 5
Limitations of the method
Calculated values often differ from experimental ones because:
- Mean values are averages, so they are not exact for the molecules in the reaction.
- Bond enthalpies apply to gaseous species. If a reactant or product is a liquid or solid, the energy to change state is missing from the calculation.
- The experimental ΔH may refer to different states (for example H₂O(l) rather than H₂O(g)).
Values from enthalpy cycles using formation or combustion data are more accurate, since they use measured data for the actual substances.
In an evaluate question, name two limitations and say which direction the error goes, e.g. the energy released on condensing water makes the experimental value more exothermic.
Must Know
- Bond enthalpy: break one mole of bonds in the gas phase, always positive
- Mean bond enthalpy: average over different compounds
- ΔH = Σ(bonds broken) − Σ(bonds made)
- Rearrange to find an unknown bond enthalpy
- Limitations: averages and gaseous state only
- Count the number of each bond from the balanced equation
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Bond enthalpies
- A student is estimating the enthalpy change for the reaction H₂(g) + Cl₂(g) → 2HCl(g) using mean bond enthalpies (kJ mol⁻¹): H–H 436, Cl–Cl 243, H–Cl 432.Use the bond enthalpy values to explain why this reaction is exothermic.2 marks
- Ethene reacts with hydrogen in the presence of a nickel catalyst: C₂H₄(g) + H₂(g) → C₂H₆(g). Mean bond enthalpies (kJ mol⁻¹): C=C 612, C–C 347, C–H 413, H–H 436. The experimental enthalpy change for the reaction is −137 kJ mol⁻¹.Calculate the enthalpy change for the reaction using the mean bond enthalpies.2 marks
- Ammonia has a standard enthalpy change of formation of −46 kJ mol⁻¹. Mean bond enthalpies (kJ mol⁻¹): N≡N 945, H–H 436. Hydrazine, H₂N–NH₂, which contains one N–N bond and four N–H bonds, decomposes: N₂H₄(g) → N₂(g) + 2H₂(g), ΔH = −95 kJ mol⁻¹.Calculate the mean bond enthalpy of the N–H bond from the standard enthalpy change of formation of ammonia, for ½N₂(g) + 3/2 H₂(g) → NH₃(g).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).