1.2.1—Bond energies
- Syllabus
- First assessment 2025
- Objective
- 1.2.1
- Level
- HL
ΔH≈Σ(bondsbroken)−Σ(bondsformed)
Breaking bonds absorbs energy; forming bonds releases energy. Count every bond with its stoichiometric multiplicity before applying the signed sum.
For H₂ + Cl₂ → 2HCl, break one H–H and one Cl–Cl bond, then form two H–Cl bonds. Average bond enthalpies give an estimate because the tabulated value averages that bond across different gaseous molecules.
Worked example — bond enthalpies: for CX2HX4(g)+HBr(g)CX2HX5Br(g), the local course book gives C−H=414, C=C=614, H−Br=366, C−C=346 and C−Br=285kJmol−1. ΔH=[4(414)+614+366]−[5(414)+346+285]=2636−2701=−65kJmol−1. The negative estimate means the bonds formed release more energy than the bonds broken absorb; it remains approximate because the values are gaseous averages.
Representative question
Calculate the enthalpy change for the reaction, ΔH. Use section 12 of the data booklet.
Alternative 1:
«bonds broken»
121O=O+4C−H+Cl−Cl/121×498+4×414+242/2645 «kJ mol −1 » «bonds formed»
C=O+2C−O+2H−O+2H−Cl/804+2×358+2×463+2×431/3308 « kJ mol−1 » ΔH= «2645-3308=»-663 «kJ mol-1» (correct calculation: reactants-products) OR
Alternative 2:
«Bonds broken» [12(414)+2(346)+242+1.5(498)] =6649 V
«Bonds formed» [8(414) + 804+2(358)+2(346)+2(431)+2(463)]=7312.
ΔH= «6649-7312=» -663 « kJmol−1 » (correct calculation: reactants-products)
Marking guidance:
Award [3] for correct final answer.
Award [2 max] for +663 « kJmol−1 ».
Accept breaking and remaking all the other bonds.
Retrieve the route: count bond breaking/forming, manipulate Hess equations, define standard formation/combustion values, apply product–reactant sums, and track every Born–Haber energy term.
Check equation direction, coefficients, signs, standard states, lattice enthalpy convention, and one-versus-two-electron steps.