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The bond dissociation enthalpy (often just "bond enthalpy") is the enthalpy change required to break one mole of a specified bond in the gas phase. It is always positive (endothermic) because breaking bonds requires input of energy.
Example: H2(g) → 2H(g), ΔH = +436 kJ mol⁻¹ - the bond enthalpy of H–H is +436 kJ mol⁻¹.
The strength of, say, a C–H bond is not the same in every molecule. In CH4 the average C–H bond enthalpy is +413 kJ mol⁻¹, but:
The mean bond enthalpy is an average value taken over a representative set of molecules containing that bond. Tables quote one figure that you apply universally, accepting a modest loss of accuracy.
| Bond | Mean bond enthalpy / kJ mol⁻¹ |
|---|---|
| H–H | +436 |
| C–H | +413 |
| C–C | +347 |
| C=C | +612 |
| C≡C | +838 |
| O=O | +498 |
| C–O | +358 |
| C=O | +805 |
| O–H | +464 |
| N≡N | +945 |
| N–H | +391 |
| Cl–Cl | +243 |
| H–Cl | +432 |
These are typical OCR data-booklet values - always use the values provided in the exam paper.
For a gas-phase reaction:
ΔrH° = Σ(bonds broken in reactants) − Σ(bonds formed in products)
Bonds broken takes energy in (+), bonds formed releases energy out (−). The difference is the net enthalpy change. Note the sign convention: because bond enthalpies are tabulated as positive values (for breaking), we use reactant bonds minus product bonds, not the other way round.
Q: Using mean bond enthalpies, estimate ΔrH° for 2H2(g) + O2(g) → 2H2O(g).
A:
Bonds broken (reactants):
Bonds formed (products):
ΔrH° = 1370 − 1856 = −486 kJ mol⁻¹
The literature value for 2H2O(g) is approximately −484 kJ mol⁻¹, agreement to ~0.4%.
Important: The reaction is for gaseous water. If the question specified H2O(l) the calorimetric value would be ~−572 kJ mol⁻¹ (including condensation). Mean bond enthalpies always give gas-phase values.
Q: Use mean bond enthalpies to estimate ΔrH° for C2H4(g) + H2(g) → C2H6(g).
A:
Bonds broken:
Bonds formed:
ΔrH° = 2700 − 2825 = −125 kJ mol⁻¹
The Hess-cycle value (Lesson 4) was −137 kJ mol⁻¹ - a discrepancy of 12 kJ, about 9%. This is typical: mean bond enthalpies give approximate answers.
Shortcut: The four C–H bonds in ethene and four of the C–H bonds in ethane are essentially unchanged, so you can ignore them. The net change is "break C=C and H–H, form C–C and 2 × new C–H":
ΔH = [612 + 436] − [347 + 2 × 413] = 1048 − 1173 = −125 kJ mol⁻¹ ✓
Q: Estimate ΔH for H2(g) + Cl2(g) → 2HCl(g).
A:
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