Bond Enthalpy Calculations (DP IB Chemistry): Revision Note
Bond enthalpy calculations
When bonds are broken or made enthalpy changes take place
A chemical bond is a force of attraction between two atoms
Breaking the bond requires the input of energy it is therefore an endothermic process
The energy change required to break the bond depends on the atoms that form the bond
The energy required to break a particular bond is called the bond dissociation enthalpy
This is usually just shortened to bond enthalpy or bond energy
Bond formation is the opposite of bond breaking and so energy is released when bonds are formed
It is therefore an exothermic process
Diagram to show bond breaking and bond making

The amount of energy released when a particular bond is formed has the same magnitude as the energy taken in when the bond is broken but has the opposite sign
Overall enthalpy changes
If more energy is released when new bonds are formed than energy is required to break bonds, the reaction is exothermic
The products are more stable than the reactants
If more energy is required to break bonds than energy is released when new bonds are formed, the reaction is endothermic
The products are less stable than the reactants
The relationship between bond breaking and bond making can be shown graphically like this:
Diagram to show the energy profiles of both exothermic and endothermic reactions

Bond enthalpy profiles
Average bond energy
Bond energies are affected by other atoms in the molecule (the environment)
Therefore, an average of a number of the same type of bond but in different environments is calculated
This bond energy is known as the average bond energy and is defined as
'The energy needed to break one mole of bonds in a gaseous molecule averaged over similar compounds

The average bond enthalpy of C-H is found by taking the bond dissociation enthalpy for the whole molecule and dividing it by the number of C-H bonds
The first C-H bond is easier to break than the second as the remaining hydrogens are pulled more closely to the carbon
However, since it is impossible to measure the energy of each C-H bond an average is taken
This value is also compared with a range of similar compounds to obtain an accepted value for the average bond enthalpy
Bond enthalpy calculations
Bond energies are used to find the ΔHꝋr of a reaction when this cannot be done experimentally
The process is a step-by-step summation of the bond enthalpies of the all the molecules present finishing with this formula:
ΔHꝊr = enthalpy change for bonds broken + enthalpy change for bonds formed
Worked Example
Calculate the enthalpy of reaction for the Haber process reaction.
The relevant bond energies are given in the table below:
Bond | Average Bond Energy (kJ mol-1) |
---|---|
N | 945 |
H | 436 |
N | 391 |
Answer:
Step 1: The chemical equation for the Haber process is:
N2 (g) + 3H2 (g) ⇌ 2NH3 (g)
N≡N 3 H-H 6 N-H
Step 2: Set out the calculation as a balance sheet as shown below:
Bonds broken:
1 x NN= 1 x 945 = 945kJ mol-1
3 x HH = 3 x 436 = 1308kJ mol-1
Total = +2253 kJ mol-1
Bonds formed:
6 N-H = 6 x 391 kJ mol-1
Total = -2346 kJ mol-1
Careful: Values for bonds broken are positive (endothermic) and values for bonds formed are negative (exothermic)
Step 3: Calculate the standard enthalpy of reaction
ΔHrꝋ = enthalpy change for bonds broken + enthalpy change for bonds formed
ΔHrꝋ = (+2253 kJ mol-1) + (-2346 kJ mol-1)
ΔHrꝋ = -93 kJ mol-1
Worked Example
The complete combustion of ethyne, C2H2, is shown in the equation below:
2C2H2 (g) + 5 O2 (g) → 2H2O (g) + 4CO2 (g)
Using the average bond enthalpies given in the table, what is the enthalpy of combustion of ethyne?
Bond | Average Bond Energy (kJ mol-1) |
---|---|
C | 414 |
C | 839 |
O | 498 |
C | 804 |
O | 463 |
O | 358 |
Answer:
Step 1: The enthalpy of combustion is the enthalpy change when one mole of a substance reacts in excess oxygen to produce water and carbon dioxide
The chemical reaction should be therefore simplified such that only one mole of ethyne reacts in excess oxygen:
H-C≡C-H + 2 ½ O=O → H-O-H + 2O=C=O
Step 2: Set out the calculation as a balance sheet as shown below:
Bonds broken:
2 x C-H = 2 x 414= 828 kJ mol-1
1 x C≡C= 1 x 839= 839 kJ mol-1
2½ O=O = 2½ x 498 = 1245kJ mol-1
Total = +2912 kJ mol-1
Bonds formed:
2 x OH= 2 x 463 = 926 kJ mol-1
4 x CO = 4 x 804 = 3216 kJ mol-1
Total= -4142 kJ mol-1
Step 3: Calculate the standard enthalpy of reaction
ΔHꝋr = enthalpy change for bonds broken + enthalpy change for bonds formed
ΔHꝋr = (+2912 kJ mol-1) + (- 4142 kJ mol-1)
ΔHꝋr = -1230 kJ mol-1
Examiner Tips and Tricks
The key to success in bond enthalpy calculations is to be very careful when accounting for every bond present. Always draw out the full displayed structures of the molecules so you don't miss any of the bonds.
Watch out for coefficients in the balanced equations as students often miss those, forget to multiply them by the bond enthalpies and get the answer wrong!
It is super important to show your steps because bond enthalpy calculations often carry 3 marks, 2 of which could be for workings if you get the final answer wrong
Unlock more revision notes. It’s free!
By signing up you agree to our Terms and Privacy Policy.
Already have an account? Log in
Did this page help you?