Bond Enthalpies and Estimating Reaction Energy
The bond enthalpy is the energy needed to break 1 mol of a particular bond in the gas phase, and it can be used to estimate the enthalpy change of a reaction as ΔH ≈ Σ(bonds broken) − Σ(bonds formed). In brief: for H2 + Cl2 → 2HCl, breaking one H–H (436 kJ) and one Cl–Cl (243 kJ) bond and forming two H–Cl bonds (2 × 431 kJ) gives ΔH ≈ 679 − 862 = −183 kJ.
When data on the enthalpy of formation are not available, average bond enthalpies provide a quick estimate of the heat of a reaction. They also explain why some reactions release energy and others do not. This article gives a table of average values, explains the method, shows four worked examples and compares the estimates with the true values.
What is bond enthalpy?
Breaking a bond always requires energy, so bond enthalpies are positive. Forming the same bond releases the same amount of energy. The value for a diatomic molecule such as H2 is exact: H2(g) → 2H(g), ΔH = +436 kJ/mol. In a polyatomic molecule the energy needed to break each C–H bond of methane is slightly different, so the tables list an average over many compounds. A higher bond enthalpy means a stronger bond. For the same two atoms, multiple bonds are stronger than single bonds: C–C 346, C=C 614 and C≡C 839 kJ/mol.
Average bond enthalpies
| Bond | Energy | Bond | Energy | Bond | Energy |
|---|---|---|---|---|---|
| H–H | 436 | C–C | 346 | O=O | 498 |
| C–H | 413 | C=C | 614 | N≡N | 945 |
| O–H | 463 | C≡C | 839 | H–Cl | 431 |
| N–H | 391 | C–O | 358 | Cl–Cl | 243 |
| H–Br | 366 | C=O (in CO2) | 799 | C–Cl | 339 |
The C=O bond in aldehydes and ketones has an average of about 745 kJ/mol, and the C=O bond in carbon dioxide has an even higher value of 799 kJ/mol. Check which value your table uses, and apply it consistently.
How do you estimate the enthalpy of a reaction?
- Write the balanced equation and draw the structures of all the molecules so that you can see all the bonds.
- List all the bonds that are broken in the reactants and add their enthalpies.
- List all the bonds that are formed in the products and add their enthalpies.
- Calculate ΔH ≈ Σ(broken) − Σ(formed). Note that this is the reverse of the order used with the enthalpies of formation.
The method assumes that all reactants and products are gases. If a liquid or solid is involved, the enthalpy of vaporization or sublimation must be added.
Worked examples
Example 1: hydrogen and chlorine
H2 + Cl2 → 2HCl. Broken: 436 + 243 = 679 kJ. Formed: 2 × 431 = 862 kJ. ΔH ≈ 679 − 862 = −183 kJ.
Example 2: combustion of methane
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
- Broken: 4 C–H (4 × 413 = 1652) + 2 O=O (2 × 498 = 996) = 2648 kJ.
- Formed: 2 C=O (2 × 799 = 1598) + 4 O–H (4 × 463 = 1852) = 3450 kJ.
- ΔH ≈ 2648 − 3450 = −802 kJ.
The product is water vapor, not liquid water. Condensing 2 mol of water releases another 2 × 44 = 88 kJ, which gives about −890 kJ for liquid water.
Example 3: ammonia synthesis
N2 + 3H2 → 2NH3. Broken: N≡N 945 + 3 × H–H (3 × 436 = 1308) = 2253 kJ. Formed: 6 N–H = 6 × 391 = 2346 kJ. ΔH ≈ 2253 − 2346 = −93 kJ. The N≡N bond is so strong that a lot of energy is needed to start the reaction, even though the reaction is exothermic overall.
Example 4: hydrogenation of ethene
C2H4 + H2 → C2H6. Broken: C=C 614 + H–H 436 = 1050 kJ. Formed: C–C 346 + 2 × C–H (826) = 1172 kJ. ΔH ≈ 1050 − 1172 = −122 kJ.
How accurate are the estimates?
| Reaction | Estimate from bond enthalpies (kJ) | Value from formation data (kJ) |
|---|---|---|
| H2 + Cl2 → 2HCl | −183 | −184.6 |
| N2 + 3H2 → 2NH3 | −93 | −91.8 |
| CH4 + 2O2 → CO2 + 2H2O(g) | −802 | −802 |
| C2H4 + H2 → C2H6 | −122 | −137 |
The estimates are typically within 5–10 % of the true values, and can be less accurate for reactions with strained or conjugated molecules.
Why are the results only estimates?
- Averages: the actual bond energy depends on the molecule and its environment.
- Gas phase only: intermolecular forces in liquids and solids are not included.
- Resonance and strain: the delocalization in benzene or ring strain in cyclopropane make the bonds different from the average values.
- Rounding of data: tables from different sources have slightly different values.
Where do bond enthalpies come from?
Bond energies were first derived from thermochemical data on the combustion and formation of compounds. In the 1930s, Linus Pauling used them to show that the bond between two different atoms is stronger than the average of the two homonuclear bonds, and he related the difference to the electronegativity difference, which led to the Pauling electronegativity scale. Modern values come from spectroscopy, calorimetry and quantum-chemical calculations.
Which mistakes are common?
- Reversing the sign: ΔH = broken − formed, not formed − broken.
- Missing bonds: draw all the structures, so that all the bonds are counted.
- Counting bonds inconsistently: bonds that stay unchanged cancel out, but a bond counted on one side and skipped on the other gives a wrong result.
- Using liquid water in the estimate: bond enthalpies refer to the gas phase.
- Mixing different tables: use data from one source.