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4.2.1Additional lessons

Bond Enthalpy Calculations — IB Chemistry HL Revision Notes & Topical Questions

5 min read 5 parts Lesson 73 of 172

📚 IB Chemistry HL Hub / Topical Revision
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Part 1 of 5The Concept

4.2.1 Bond Enthalpy Calculations
Definition
  • Bond enthalpy is the energy needed to break one mole of a particular bond in gaseous molecules, averaged over many compounds.
Breaking bonds
  • Breaking bonds is endothermic, so the bond enthalpy value is positive.
Making bonds
  • Forming bonds releases the same amount of energy, so it contributes a negative term.
Formula
  • ΔH = Σ(bonds broken) − Σ(bonds formed), using values for every bond in the molecules.
Phase
  • All species must be gases, otherwise the energy of changes of state must be included.
BondAverage enthalpy / kJ mol⁻¹
C–H414
O=O498
C=O (in CO₂)804
O–H463

Part 2 of 5Calculation Steps

1
Draw the structures of all molecules so that every bond is visible, including multiple bonds.
2
List and count the bonds broken in the reactants and multiply by their bond enthalpies.
3
List and count the bonds formed in the products and multiply by their bond enthalpies.
4
Subtract the total energy of bonds formed from the total energy of bonds broken.
5
Report the result with units of kJ mol⁻¹ and interpret the sign as exothermic or endothermic.

Part 3 of 5Limitations

Limitations
Average values
  • Bond enthalpies are averages from many compounds, so they differ slightly from the true value in a particular molecule.
Gas phase only
  • Reactions involving liquids or solids give errors because intermolecular forces are ignored.
Estimates
  • Results usually differ from values obtained by Hess's law or calorimetry by a few percent.
Hess alternative
  • For more accurate answers, use enthalpies of formation or combustion where data are available.

Part 4 of 5Worked Example

Question Estimate ΔH for CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g) using the bond enthalpies above.

  1. Bonds broken4 C–H and 2 O=O = 4(414) + 2(498) = 1656 + 996 = 2652 kJ.
  2. Bonds formed2 C=O and 4 O–H = 2(804) + 4(463) = 1608 + 1852 = 3460 kJ.
  3. ΔH = 2652 − 3460.
  4. ΔH = −808 kJ mol⁻¹.

Answer ΔH = −808 kJ mol⁻¹ (exothermic)

Part 5 of 5Exam Checklist & Tips

4.2.1 – Key Points for the Exam

  • Bond enthalpy The energy required to break one mole of a bond in the gas phase, averaged over compounds.
  • Bonds broken The bonds in the reactants that must be split.
  • Bonds formed The new bonds in the products.
  • Average value A mean over different molecules containing the same bond.

Common Mistakes

  • Reversing the formula and getting formed minus broken.
  • Forgetting multiple bonds such as O=O and C=O.
  • Using the bond enthalpy for one mole of molecules instead of one mole of bonds.
  • Using the method for reactions involving liquid water without correction.

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