AP Chemistry · Unit 6: Thermochemistry
6.7 Bond Enthalpies
What is bond enthalpy?
Bond enthalpy is the average energy required to break one mole of a specific bond in a gaseous molecule, always a positive value since breaking bonds requires energy input (endothermic). Forming bonds releases energy (exothermic), so bond formation has a negative enthalpy of the same magnitude. Stronger, shorter bonds (like triple bonds) have higher bond enthalpies than weaker, longer bonds (like single bonds).
Estimating ΔH with bond enthalpies
Since reactions involve breaking bonds in reactants and forming bonds in products, ΔH°rxn ≈ Σ(bonds broken) − Σ(bonds formed), using average bond enthalpies. This gives an *estimate*, not exact ΔH, because tabulated values are averages across many molecules — actual bond strength depends on the specific molecular environment. This method only works for reactions in the gas phase.
Trends and limitations
Bond enthalpy increases with bond order (triple > double > single) and generally decreases as bond length increases. Because bond enthalpies are averaged over many compounds, calculations using them give approximate ΔH values that may differ from experimentally measured enthalpies (e.g., from calorimetry or Hess's law using ΔH°f data), which are more accurate.
Key terms
- Bond enthalpy
- Average energy required to break one mole of a specific bond in gaseous molecules; always positive.
- Formula for ΔH using bond enthalpies
- ΔH°rxn ≈ Σ(bonds broken) − Σ(bonds formed)
- Why breaking bonds requires energy
- Energy must be added to overcome the attractive forces holding atoms together (endothermic).
- Bond order vs. bond enthalpy
- Higher bond order (triple > double > single) means stronger, shorter bonds and higher bond enthalpy.
- Limitation of bond-enthalpy calculations
- They give only approximate ΔH values since bond enthalpies are averages across many compounds, not exact for one specific molecule.
Try an AP-style question
Data
Bond enthalpies (kJ/mol): H–H 436, Cl–Cl 243, H–Cl 431.
Estimate ΔH for H₂ + Cl₂ → 2HCl.
- −248 kJ
- −183 kJ
- +248 kJ
- +183 kJ
Show the answer
B. (436 + 243) − 2(431).
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