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Bond Enthalpies (AQA A Level Chemistry): Revision Note

Exam code: 7405

Written by: Stewart Hird

Reviewed by: Caroline Carroll

Updated on 10 July 2026

Bond Enthalpies

The amount of energy required to break one mole of a specific covalent bond in the gas phase is called the bond enthalpy

In symbols, the type of bond broken is written in brackets after E
- e.g., E (H-H) is the bond enthalpy of a mole of single bonds between two hydrogen atoms

Mean bond enthalpy

Bond enthalpies 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 enthalpy is known as the mean bond enthalpy and is defined as:
- the enthalpy (change) to break 1 mol of bonds (in the gaseous state), averaged over a range of compounds/molecules
Mean bond enthalpies are averaged over a range of different compounds, so the value for a specific bond in a specific molecule will differ from the mean
Bond enthalpy calculations assume all substances are in the gaseous phase
- If any reactant or product is a liquid (e.g., water, a liquid fuel), the calculation doesn't account for the enthalpy of vaporisation/condensation, making the result inaccurate
Since bond enthalpies cannot be determined directly, enthalpy cycles are used to calculate the mean bond enthalpy

Diagram comparing O–H bond energies in water and methanol, explaining more energy is needed to break the O–H bond in water than in methanol. — **Bond enthalpies are affected by other atoms in the molecule, so average bond enthalpies are listed in data tables**

Examiner Tips and Tricks

You may see the terms bond energy and average bond energy used in textbooks, but the term used in AQA is bond enthalpy, which you must use in exams.

Calculating enthalpy change from mean bond enthalpies

Bond enthalpies are used to find the ΔH_r^ꝋ of a reaction when this cannot be done experimentally
The formula is:

$Δ {H^{θ}}_{r} =$ $Σ$ Enthalpy change for bonds broken - $Σ$ enthalpy change for bonds formed

Values from mean bond enthalpy calculations differ from those determined using Hess's law
Hess's law calculations use actual enthalpy values specific to the compounds involved, while bond enthalpy calculations use averaged values across many different compounds — so the two methods give different (though related) answers

Examiner Tips and Tricks

Always use ΔH = Σ(bonds broken) – Σ(bonds formed). A common error is subtracting the wrong way round — [products] – [reactants] instead of [reactants] – [products].

When counting bonds in organic molecules, draw out the full structural formula. Students frequently miscount C–C bonds — e.g., butan-1-ol has three C–C bonds, not four.

Always write out the separate totals for bonds broken and bonds formed before doing the final subtraction. If you make a maths error but show clear intermediate values, you can still gain consequential marks.

Related topics

Worked Example

Calculating the enthalpy change in the Haber process

Calculate the change in enthalpy of reaction for the Haber process, producing ammonia from hydrogen and nitrogen:

N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g)

The relevant bond enthalpies are given in the table below:

Bond	Average Bond Enthalpy / kJ mol^-1
N $\equiv$ N	945
H-H	436
N-H	391

Answer

Step 1: Use the equation to work out the bonds broken and formed and set out the calculation as a balance sheet as shown below:

N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g)

Bonds broken / kJ mol^-1	Bonds formed / kJ mol^-1
1 $\times$ N $\equiv$ N = 1 $\times$ 945 =945 3 $\times$ H-H= 3 $\times$ 436 = 1308	6 $\times$ N-H = 6 $\times$ 391
Total = 2253	Total = 2346

Bonds broken / kJ mol^-1

Bonds formed / kJ mol^-1

1 $\times$ N $\equiv$ N = 1 $\times$ 945 =945

3 $\times$ H-H= 3 $\times$ 436 = 1308

6 $\times$ N-H = 6 $\times$ 391

Total = 2253

Total = 2346

Note! Values for bonds broken are positive (endothermic), and values for bonds formed are negative (exothermic)

Step 2: Calculate the standard enthalpy of reaction

ΔH = Σ(bonds broken) – Σ(bonds formed)

= (2253 kJ mol^-1) - (2346 kJ mol^-1)

= -93 kJ mol^-1

Worked Example

Calculating the enthalpy of combustion using mean bond enthalpies

The complete combustion of ethyne, C₂H₂, is shown in the equation below:

2H-C $\equiv$ C-H + 5O=O $\to$ 2H-O-H + 4O=C=O

Using the average bond enthalpies given in the table, what is the enthalpy of combustion of ethyne?

Bond	Mean bond enthalpy / kJ mol^-1
C-H	414
C $\equiv$ C	839
O=O	498
C=O	804
O-H	463
C-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 therefore be 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 / kJ mol^-1	Bonds formed / kJ mol^-1
1 $\times$ C $\equiv$ C = 1 $\times$ 839 =839 2 $\times$ C-H= 2 $\times$ 414 = 828 2½ $\times$ O=O = 2½ $\times$ 498 = 1245	2 $\times$ O-H = 2 $\times$ 463 = 926 4 $\times$ C=O = 4 $\times$ 804 = 3216
Total = 2912	Total = 4142

Bonds broken / kJ mol^-1

Bonds formed / kJ mol^-1

1 $\times$ C $\equiv$ C = 1 $\times$ 839 =839

2 $\times$ C-H= 2 $\times$ 414 = 828

2½ $\times$ O=O = 2½ $\times$ 498 = 1245

2 $\times$ O-H = 2 $\times$ 463 = 926

4 $\times$ C=O = 4 $\times$ 804 = 3216

Total = 2912

Total = 4142

ΔH = Σ(bonds broken) Σ(bonds formed)

= (2912 kJ mol^-1) - (4142 kJ mol^-1)

= -1230 kJ mol^-1

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Bond Enthalpies (AQA A Level Chemistry): Revision Note

Bond Enthalpies

Mean bond enthalpy

Examiner Tips and Tricks

Calculating enthalpy change from mean bond enthalpies

Examiner Tips and Tricks

Worked Example

Worked Example

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Author: Stewart Hird

Reviewer: Caroline Carroll