4.2.1Additional lessons
Bond Enthalpy Calculations — IB Chemistry HL Revision Notes & Topical Questions
5 min read 5 parts Lesson 73 of 172
Part 1 of 5The Concept
Definition
- 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 Breaking bonds is endothermic, so the bond enthalpy value is positive.
Making bonds
- Making bonds Forming bonds releases the same amount of energy, so it contributes a negative term.
Formula
- Formula ΔH = Σ(bonds broken) − Σ(bonds formed), using values for every bond in the molecules.
Phase
- Phase All species must be gases, otherwise the energy of changes of state must be included.
| Bond | Average enthalpy / kJ mol⁻¹ |
|---|---|
| C–H | 414 |
| O=O | 498 |
| C=O (in CO₂) | 804 |
| O–H | 463 |
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
- Average values Bond enthalpies are averages from many compounds, so they differ slightly from the true value in a particular molecule.
Gas phase only
- Gas phase only Reactions involving liquids or solids give errors because intermolecular forces are ignored.
Estimates
- Estimates Results usually differ from values obtained by Hess's law or calorimetry by a few percent.
Hess alternative
- 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.
- Bonds broken4 C–H and 2 O=O = 4(414) + 2(498) = 1656 + 996 = 2652 kJ.
- Bonds formed2 C=O and 4 O–H = 2(804) + 4(463) = 1608 + 1852 = 3460 kJ.
- ΔH = 2652 − 3460.
- 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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