Energy changes in chemical reactionsBond energy calculations

The changes in energy that occur during a chemical reaction can be seen by examining the changes in chemical bonding. This can be used to classify reactions as exothermic or endothermic.

Part ofChemistry (Single Science)Chemical reactions and energy

Bond energy calculations

You can calculate the energy change in a reaction using average bond energies. Bond energy is the amount of energy needed to break one of a particular bond.

To calculate bond energy

  1. Add together the bond energies for all the bonds in the – this is the ‘energy in’.
  2. Add together the bond energies for all the bonds in the – this is the ‘energy out’.
  3. Calculate the energy change = energy in – energy out.

Worked example – an exothermic reaction

Hydrogen and chlorine react to form hydrogen chloride gas:

H−H + Cl−Cl → 2 × (H−Cl)

BondBond energy (kJ/mol)
H−H436
Cl−Cl243
H−Cl432
BondH−H
Bond energy (kJ/mol)436
BondCl−Cl
Bond energy (kJ/mol)243
BondH−Cl
Bond energy (kJ/mol)432
  1. Energy in = 436 + 243 = 679 kJ/mol (this is the energy absorbed when the bonds of the reactants break).
  2. Energy out = 2 × 432 = 864 kJ/mol (this is the energy released when the bonds of the products form).
  3. Energy change = in – out = 679 – 864 = –185 kJ/mol

The energy change is negative, due to the fact that the energy released by the bonds formed is greater than the energy absorbed by the bonds broken. This means that energy is released to the surroundings in an reaction.

Worked example – calculating bond energy [Higher tier only]

Hydrogen reacts with iodine to form hydrogen iodide.

H−H + I−I → 2 × (H–I)

BondBond Energy (kJ/mol)
H−I?
H−H436
I−I151
BondH−I
Bond Energy (kJ/mol)?
BondH−H
Bond Energy (kJ/mol)436
BondI−I
Bond Energy (kJ/mol)151

The energy change for this reaction is –3 kJ/mol. Calculate the bond energy of the H–I bond.

  1. Energy in = 436 + 151 = 587 kJ/mol
  2. Energy out = 2 × (bond energy of H–I) = 2(H–I)
  3. Energy change = in – out = 587 – 2(H–I) = –3 kJ/mol
  4. Rearrange to: 2(H–I) = 587 + 3 = 590 kJ/mol
  5. Therefore: (H–I) = 590 ÷ 2 = 295 kJ/mol