Topic 6 · page 13 of 15

Bond enthalpies and enthalpy changes

Breaking bonds takes in energy and making bonds gives it out. Add up the bonds broken, subtract the bonds made, and you have an estimate of ΔH for a reaction of gases.

14 min read6.9

Read first: Enthalpy changes of combustion, neutralisation and atomisation

Bond enthalpy and mean bond enthalpy 6.9

Every reaction breaks some bonds in the reactants and makes new bonds in the products. Breaking a bond takes in energy, so it is endothermic. Making a bond gives out energy, so it is exothermic. ΔH is the balance of the two.

For a bond in a diatomic molecule, such as H–H or Cl–Cl, there is only one bond environment, so this is an exact value. The C–H bond in methane is not quite the same as the C–H bond in ethanol, so a single table value for C–H has to be an average.

All bond enthalpies are positive because breaking bonds always takes in energy. The reverse, making a bond, has the same size with a negative sign.

Bond Mean bond enthalpy / kJ mol⁻¹
H–H 436
C–H 413
O–H 464
N–H 391
H–Cl 432
C–C 347
C=C 612
O=O 498
C=O (in CO₂) 805
Cl–Cl 243
F–F 158
N≡N 945

In the exam you are given a table like this in the question or the data booklet. The values on these pages are used in the worked examples; there is no need to learn them.

Using bond enthalpies to find ΔH 6.9

MethodCalculating ΔH from bond enthalpies

  1. Write the balanced equation and make sure all species are gases.
  2. Draw the displayed formula of every molecule, so every bond is visible.
  3. List the bonds broken in the reactants, with the number of each, taken from the equation. Multiply by the bond enthalpy and add: Σ(bonds broken).
  4. List the bonds made in the products in the same way: Σ(bonds made).
  5. ΔH = Σ(bonds broken) − Σ(bonds made)

The route is: break every bond in the reactants to give separate atoms (energy in), then make every bond in the products from those atoms (energy out).

Energy level diagram for hydrogen plus chlorine making hydrogen chloride, drawn to scale. Reactants on a middle level. An upward arrow of plus 679 kilojoules breaks the H-H and Cl-Cl bonds to give separate gaseous atoms on a high level. A downward arrow of minus 864 kilojoules makes two H-Cl bonds, ending on the product level, slightly below the reactants. The overall delta H is minus 185.
Break the bonds (up), make the new bonds (down). The net step from reactants to products is ΔH.

Worked exampleHydrogen with chlorine

Calculate ΔH for HX2(g)+ClX2(g)→2 HCl(g)\ce{H2(g) + Cl2(g) -> 2HCl(g)} using the table.

  1. Bonds broken: one H–H and one Cl–Cl: 436 + 243 = +679 kJ.
  2. Bonds made: two H–Cl (the equation makes 2 HCl): 2 × 432 = 864 kJ.
  3. ΔH = 679 − 864 = −185 kJ mol⁻¹.

Worked exampleBurning ethanol vapour

Calculate ΔH for CX2HX5OH(g)+3 OX2(g)→2 COX2(g)+3 HX2O(g)\ce{C2H5OH(g) + 3O2(g) -> 2CO2(g) + 3H2O(g)}. Use the table and C–O = 358 kJ mol⁻¹.

Ethanol is CH₃–CH₂–OH. It has 5 C–H, 1 C–C, 1 C–O and 1 O–H. Each CO₂ has two C=O. Each H₂O has two O–H.

Bonds Energy / kJ
Broken 5 × C–H = 5 × 413 2065
1 × C–C 347
1 × C–O 358
1 × O–H 464
3 × O=O = 3 × 498 1494
Total +4728
Made 4 × C=O = 4 × 805 (2 CO₂) 3220
6 × O–H = 6 × 464 (3 H₂O) 2784
Total −6004

ΔH = 4728 − 6004 = −1276 kJ mol⁻¹.

Check yourself

  1. Define bond enthalpy.
  2. Calculate ΔH for 2 HX2(g)+OX2(g)→2 HX2O(g)\ce{2H2(g) + O2(g) -> 2H2O(g)} using H–H 436, O=O 498 and O–H 464 kJ mol⁻¹.
  3. Calculate ΔH for CHX4(g)+ClX2(g)→CHX3Cl(g)+HCl(g)\ce{CH4(g) + Cl2(g) -> CH3Cl(g) + HCl(g)} using C–H 413, Cl–Cl 243, C–Cl 346 and H–Cl 432 kJ mol⁻¹.
  4. Calculate ΔH for NX2(g)+3 HX2(g)→2 NHX3(g)\ce{N2(g) + 3H2(g) -> 2NH3(g)} using N≡N 945, H–H 436 and N–H 391 kJ mol⁻¹. The value from Hess's Law is −92 kJ mol⁻¹; comment.
  5. Explain why the same bond, such as C–H, has a mean bond enthalpy, but H–H has just a bond enthalpy.
  6. State why a bond enthalpy calculation needs every reactant and product to be a gas.

Answers cover these until you have tried the questions

  1. The enthalpy change when one mole of a particular covalent bond is broken, in gaseous molecules.
  2. Broken: 2 × H–H + 1 × O=O = 2 × 436 + 498 = 872 + 498 = 1370. Made: 4 × O–H = 4 × 464 = 1856. ΔH = 1370 − 1856 = −486 kJ mol⁻¹.
  3. Broken: C–H + Cl–Cl = 413 + 243 = 656. Made: C–Cl + H–Cl = 346 + 432 = 778. ΔH = 656 − 778 = −122 kJ mol⁻¹.
  4. Broken: 945 + 3 × 436 = 945 + 1308 = 2253. Made: 6 × N–H = 6 × 391 = 2346. ΔH = 2253 − 2346 = −93 kJ mol⁻¹. This is close to −92: the bond enthalpy calculation is an estimate because the values are mean values.
  5. H–H occurs in only one environment (the H₂ molecule), so its enthalpy is exact. C–H appears in many different molecules, with slightly different strengths each time, so the table gives the average over a range of compounds.
  6. Bond enthalpies are defined for bonds broken in gaseous molecules, and the method takes every species to and from separate gaseous atoms. If a reactant or product is a liquid or solid, the energy to evaporate or melt it, or the energy given out when a gas condenses, is not in the calculation, so the answer is not the enthalpy change for the equation as written.