Coaching Centre
Bond energies and enthalpy
Explaining enthalpy changes as bond breaking and bond forming, the law of conservation of energy, average bond energies, estimating ΔH from bond energies, and why estimates differ from measured values.
Here is a statement that sounds wrong but isn't: burning a fuel releases energy, yet no bond ever releases energy when it breaks. Breaking a bond always costs energy. The energy from a flame comes entirely from the new bonds that form afterwards. In Year 10 you used this idea to add up bond energies; now you'll see why it works, how far to trust it, and where it fails.
Breaking and forming bonds
A covalent bond holds two atoms together because both nuclei attract the shared electrons. Pulling the atoms apart means doing work against that attraction, so:
- Breaking a bond absorbs energy (endothermic).
- Forming a bond releases energy (exothermic), exactly the same amount as breaking that bond would absorb.
Any reaction can be imagined as two steps: break every bond in the reactants to make separate gaseous atoms, then join the atoms up into products. By the law of conservation of energy, the overall ΔH must equal the energy taken in by the first step minus the energy given out by the second:
$$\Delta H \approx \sum E(\text{bonds broken}) - \sum E(\text{bonds formed})$$
If the new bonds are stronger than the old ones, more energy comes out than went in: exothermic. If the bonds broken are stronger, the reaction is endothermic. Real reactions don't actually take this route (atoms never all fly apart), but because ΔH depends only on the start and finish, any imaginary route gives the same answer. That idea is Hess's law, which is the subject of 4F.
Bond energies
The bond energy (bond enthalpy) is the energy needed to break one mole of a particular bond in gaseous molecules, forming gaseous atoms. For a diatomic molecule such as H₂ or Cl₂ it can be measured exactly. For a bond such as C–H, which occurs in thousands of different molecules with slightly different strengths, tables give an average bond energy.
| Bond | kJ mol⁻¹ | Bond | kJ mol⁻¹ | Bond | kJ mol⁻¹ |
|---|---|---|---|---|---|
| H–H | 436 | C–C | 346 | O=O | 498 |
| C–H | 413 | C=C | 614 | C=O (in CO₂) | 805 |
| O–H | 463 | C≡C | 839 | C–O | 358 |
| N–H | 391 | N≡N | 945 | Cl–Cl | 242 |
| H–Cl | 431 | C–Cl | 339 | Br–Br | 193 |
Different data books list slightly different averages (C–C is given as 346 or 348, for example), so in an exam always use the values provided. Two patterns are worth noticing: multiple bonds are stronger than single bonds between the same atoms (but a double bond is less than twice a single), and bonds to hydrogen are generally strong.
Estimating ΔH from bond energies
The reliable method is to draw out every molecule with all its bonds, then count.
Use bond energies to estimate ΔH for CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g).
For H₂(g) + Br₂(g) → 2HBr(g), ΔH = −103 kJ mol⁻¹. Given H–H = 436 and Br–Br = 193 kJ mol⁻¹, calculate the H–Br bond energy.
A shortcut that avoids errors: bonds that appear unchanged on both sides (such as the C–H bonds in ethene + Br₂ → C₂H₄Br₂) can be left out of both sums, because breaking and re-forming them cancels.
- For H₂ + Cl₂, press Break bonds then Form bonds. The estimate is −184 kJ against data of −184.6 kJ. Why is this one so accurate?
- For Ethene + H₂ the estimate is −122 kJ but the data say −136.9 kJ. Which bonds use average values here?
- For the Haber process, compare the size of the two arrows with the size of ΔH. Why does a small percentage error in a big number matter?
- For Methane burns, read the note: why would the estimate be much further from the data if the water were liquid?
Why estimates differ from measured values
- Bond energies are averages. The C–H bond in methane is not exactly the same strength as one in ethanol. Using one value for all of them introduces a few percent error.
- Bond energies are for gases only. If a reactant or product is a liquid or solid, the energy of the intermolecular forces (for example the 44 kJ mol⁻¹ released when steam condenses) is not included. Combustion data for liquid water therefore disagree badly with bond-energy estimates.
- ΔH is a small difference between two large numbers. A 1% error in 2700 kJ is 27 kJ, which can be 20% of a small ΔH.
That's why measured values (from calorimetry, 4B and 4C) or values calculated from enthalpies of formation (4G) are preferred whenever they exist. Bond energies are best for estimates, for reactions nobody has measured, and for explaining why a reaction is exo- or endothermic.
Test yourself
Use bond energies (N≡N 945, H–H 436, N–H 391 kJ mol⁻¹) to estimate ΔH for N₂(g) + 3H₂(g) → 2NH₃(g).
Broken: 945 + 3(436) = 2253 kJ. Formed: 6(391) = 2346 kJ. ΔH ≈ 2253 − 2346 = −93 kJ mol⁻¹.
A reaction is exothermic. In terms of bonds, this means
ΔH = broken − formed is negative when the energy released in forming bonds exceeds that absorbed in breaking them.
The bond-energy estimate for CH₄ + 2O₂ → CO₂ + 2H₂O is −814 kJ mol⁻¹, but the heat of combustion is −890 kJ mol⁻¹. The main reason for the difference is that
Condensing 2 mol of steam releases about 88 kJ more, which bond energies cannot account for. (O=O is a diatomic, so its value is exact.)
For Cl₂(g) + CH₄(g) → CH₃Cl(g) + HCl(g), which bonds need to be included in the calculation if unchanged bonds are cancelled?
Three C–H bonds survive unchanged and cancel. Both options 1 and 2 give the same ΔH ≈ (413 + 242) − (339 + 431) = −115 kJ mol⁻¹; option 2 is the shortcut.
Label the bond-energy cycle.
Does each process absorb or release energy?
Breaking a bond energy and forming a bond energy. ΔH is approximately the sum of the bonds minus the sum of the bonds formed. Values for bonds such as C–H are , and all bond energies apply only to substances in the state.
Ethene reacts with chlorine: C₂H₄(g) + Cl₂(g) → C₂H₄Cl₂(g).
(a) Use bond energies (C=C 614, C–C 346, C–Cl 339, Cl–Cl 242 kJ mol⁻¹) to estimate ΔH. (3 marks)
(b) The measured ΔH is −182 kJ mol⁻¹. Explain why the two values differ. (2 marks)
(a) The four C–H bonds are present on both sides, so they cancel. Bonds broken: C=C + Cl–Cl = 614 + 242 = 856 kJ. Bonds formed: C–C + 2 C–Cl = 346 + 2(339) = 1024 kJ. ΔH ≈ 856 − 1024 = −168 kJ mol⁻¹.
(b) The C=C, C–C and C–Cl values are average bond energies measured over many different compounds. The actual strengths of these bonds in 1,2-dichloroethane differ slightly from the averages, so the estimate is about 14 kJ (8%) less exothermic than the measured value. (All species are gases, so state is not the cause here.)