All chemical reactions involve breaking old bonds and forming new ones. Bond breaking is endothermic as it requires energy, while bond formation is exothermic as it releases energy. The heat absorbed or released in a reaction comes from the breaking and forming of chemical bonds. The greater the bond energy, the stronger the bond. Bond energies can be used to calculate the enthalpy change (ΔH) of a reaction by adding the bond energies of reactants and subtracting the bond energies of products.

1. BOND ENERGY

2. BOND ENERGY
All chemical reactions involve the breaking of old
bonds followed by the making of new bonds.
The heat absorbed or released in a reaction comes
from the chemical bonds being broken or made
respectively.

3. BOND ENERGY
Bond breaking is endothermic.
C-H + energy  C H
Bond formation is exothermic.
H H  H-H + energy
The greater the bond energy is, the stronger the bond

4. BOND ENERGY
Table of average bond energies:

5. BOND ENERGY
The bond energies on the table are averages, because
they may differ depending on adjacent bonds.
i.e. the C-H bond requires 435kJ to break, but
breaking all 4 bonds of CH4 requires 1652kJ of
energy, this the average is 413kJ/mol
435kJ/mol H
H C H 1652kJ/mol = 413kJ/mol
4 bonds
H

6. BOND ENERGY
Use bond energies to find the ΔH° of a reaction:
Ex 1. H2(g) + Br2(g)  2HBr(g)
Bond energies 432kJ/mol 193kJ/mol 2 x -363kJ/mol
There are 2 Bond formation
moles of H-Br is exothermic
Add the bond energies together:
ΔH° = 432kJ/mol + 193kJ/mol + 2 x (-363kJ/mol)
ΔH° = -101kJ/mol
.: ΔH° is -101kJ/mol

7. BOND ENERGY
Use bond energies to find the ΔH° of a reaction:
Ex 1. H2(g) + Br2(g)  2HBr(g)
432kJ/mol 193kJ/mol 363kJ/mol
Can be solved using this formula as well
ΔH° = Σ bond energy of reactants –Σ bond energy of products
ΔH° = (432kJ/mol + 193 kJ/mol) – (2 x
363kJ/mol)
ΔH° = -101kJ/mol
.: ΔH° is -101kJ/mol

8. BOND ENERGY
For more complex molecules…
Ex 2. Calculate the enthalpy of combustion
for ethanol using average bond energies.
C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(g)
Draw the structures to determine the types and number of bonds
347

9. BOND ENERGY
Ex 2. Calculate the enthalpy of combustion for ethanol
using average bond energies.
347
+ (1 x C-C)
+ (1 x 347)
+ 347
+ 347 kJ/mol = 4731 kJ/mol

10. BOND ENERGY
Ex 2. Calculate the enthalpy of combustion for ethanol
using average bond energies.
= 4731 kJ/mol – 5782 kJ/mol
= -1051 kJ/mol
.: ΔH° is -1.05x103kJ
The molar enthalpy of combustion of ethanol based on bond
energies is -1051 kJ/mol. The accepted value is -1368 kJ/mol
(using Hess’ Law).
There difference is due to the use of average bond energies.

11. BOND ENERGY
Ex 3. Calculate the enthalpy of combustion for methoxy
methane (CH3OCH3) using average bond energies.
CH3OCH3(l) + 3O2(g)  2CO2(g) + 3H2O(g)

12. BOND ENERGY
Ex 3. Calculate the enthalpy of combustion for methoxy
methane (CH3OCH3) using average bond energies.

13. BOND ENERGY
Ex 3. Calculate the enthalpy of combustion for methoxy
methane (CH3OCH3) using average bond energies.
.: ΔH° is -1.09x103kJ

14. BOND ENERGY
Bond dissociation energy (D) is also a
measure of bond strength in a chemical bond.
It is the change in enthalpy of a homolysis
reaction at abolute zero (0 kelvin) where a
molecule is broken down into two free radicals.
ΔH = __kJ/mol
It is not the same as average bond energy.