Bond Enthalpies

CIE A-Level Chemistry· 5.3(b)· 25 min read

1. Definition and Key Properties of Bond Enthalpy★★☆☆☆⏱ 7 min

📘 Definition

Average Bond Enthalpy

$\overset{ˉ}{E} (X - Y)$

The average energy required to break one mole of a particular covalent bond in the gaseous state, averaged over a range of different compounds containing the bond.

Example:

The average C-H bond enthalpy is 413 kJ mol⁻¹, averaged over compounds like methane, ethane and propane.

Breaking any covalent bond requires an input of energy (endothermic), so all bond enthalpy values are positive by definition. When a bond forms, the same amount of energy is released (exothermic), so the enthalpy change for bond formation is equal to $- \overset{ˉ}{E} (X - Y)$ .

📐 Worked Example

Explain why the measured O-H bond enthalpy in a single water molecule differs from the average O-H bond enthalpy given in data tables.

1
By definition, average bond enthalpy is calculated by averaging the bond enthalpy of the O-H bond across many different compounds, not just water.
2
Compounds containing O-H bonds include water, hydrogen peroxide, alcohols and carboxylic acids, each with slightly different actual O-H bond energies.
3
The measured value in pure water is for the specific O-H bonds in H₂O only, so it differs from the averaged value.

2. Calculating Enthalpy Change of Reaction★★★☆☆⏱ 10 min

To calculate the overall enthalpy change of a reaction, we add the total energy required to break all bonds in the reactants, then subtract the total energy released when all new bonds are formed in the products. The formula is:

\Delta H = \sum (\text{Bond enthalpies of bonds broken}) - \sum (\text{Bond enthalpies of bonds formed})

📐 Worked Example

Calculate the enthalpy change for the reaction $C H_{4} (g) + C l_{2} (g) \to C H_{3} C l (g) + H C l (g)$ using the following bond enthalpies (kJ mol⁻¹): C-H = 413, Cl-Cl = 242, C-Cl = 346, H-Cl = 431.

1
Step 1: Count bonds broken in reactants: 1 C-H bond, 1 Cl-Cl bond
2
Step 2: Calculate total energy for bond breaking: $413 + 242 = 655$ kJ mol⁻¹
3
Step 3: Count bonds formed in products: 1 C-Cl bond, 1 H-Cl bond
4
Step 4: Calculate total energy from bond formation: $346 + 431 = 777$ kJ mol⁻¹
5
Step 5: Apply the formula: $Δ H = 655 - 777 = - 122$ kJ mol⁻¹

3. Limitations of Average Bond Enthalpy Calculations★★★☆☆⏱ 5 min

Average bond enthalpy calculations always give approximate results, for two key reasons: (1) values are averaged over many different compounds, so do not match the exact bond energy in a specific molecule; (2) bond enthalpy calculations only account for covalent bond breaking and forming, and do not account for intermolecular forces in solids or liquids.

📐 Worked Example

The enthalpy of combustion of methane calculated using bond enthalpies is -890 kJ mol⁻¹, while the experimental value is -898 kJ mol⁻¹. Explain the difference.

1
Step 1: Bond enthalpy calculations assume all products and reactants are gaseous. In experimental combustion of methane, water is produced as a liquid, not a gas.
2
Step 2: Condensation of gaseous water to liquid water is exothermic, so it adds extra energy released to the overall reaction.
3
Step 3: The 8 kJ mol⁻¹ difference comes from this phase change effect, plus small differences from the average bond enthalpy approximation.

4. Bond Enthalpy, Bond Length and Bond Strength★★☆☆☆⏱ 3 min

There is a clear relationship between these three properties: shorter bonds have higher bond enthalpy and are stronger. This is because more bonding electrons between two nuclei pull the nuclei closer together, creating a stronger attraction that requires more energy to break.

Bond Type	Bond Length (pm)	Bond Enthalpy (kJ mol⁻¹)
C-C single	154	347
C=C double	134	612
C≡C triple	120	838
C-O single	143	360
C=O double	120	749

📐 Worked Example

Explain why the N≡N bond has a much higher bond enthalpy than the N-N single bond.

1
Step 1: A N≡N triple bond has three shared pairs of bonding electrons between the two nitrogen atoms, compared to one shared pair in a N-N single bond.
2
Step 2: More bonding electrons create a stronger electrostatic attraction between the two nitrogen nuclei, pulling them closer together to form a shorter bond.
3
Step 3: More energy is required to overcome this stronger attraction and break three covalent bonds compared to one, so N≡N has a much higher bond enthalpy (944 kJ mol⁻¹ vs 163 kJ mol⁻¹ for N-N).

5. Common Pitfalls

Wrong move:

Swapping the formula order, calculating $Δ H = \sum bonds formed - \sum bonds broken$

Why:

This gives the opposite sign for the enthalpy change, reversing whether the reaction is exothermic or endothermic.

Correct move:

Always remember the formula: $Δ H =$ total energy in (breaking bonds) $-$ total energy out (forming bonds).

Wrong move:

Miscounting the number of bonds, for example forgetting the O=O bond in combustion reactions, or the N-N bond in hydrazine.

Why:

All bonds in reactants must be broken and all bonds in products formed to get the correct total. Even one missing bond gives an incorrect result.

Correct move:

Always draw the full displayed structural formula of every molecule before counting bonds of each type.

Wrong move:

Using bond enthalpies for reactions with liquid or solid reactants/products without correcting for phase changes.

Why:

Bond enthalpies only account for covalent bond energy changes, not the enthalpy change of changing between solid/liquid and gas.

Correct move:

Add the enthalpy of vaporization (for liquid → gas) or condensation (gas → liquid) to your final result to correct for phase changes.

Wrong move:

Stating that bond enthalpy can be negative because forming bonds is exothermic.

Why:

Bond enthalpy is explicitly defined as the energy required to break bonds, not form them.

Correct move:

Remember all bond enthalpy values are positive. Negative enthalpy change only applies to the overall reaction when more energy is released forming bonds than is used breaking them.

6. Quick Reference Cheatsheet

Concept	Key Fact / Formula
Average Bond Enthalpy	Avg energy to break 1 mol of gaseous covalent bond, averaged over many compounds
Enthalpy Change Formula	$Δ H = \sum (bonds broken) - \sum (bonds formed)$
Bond Enthalpy Sign	All bond enthalpy values are positive (bond breaking is endothermic)
Bond Trend	Shorter bond = higher bond enthalpy = stronger bond
Multiple Bond Trend	Triple > double > single in bond enthalpy; shorter bond length for multiple bonds
Limitations	Only for gaseous species; results are always approximate

7. Frequently Asked

Why are all bond enthalpy values positive?

Bond enthalpy is defined as the energy required to break bonds, which is always an endothermic process that absorbs energy, so values are always positive.

Why do bond enthalpy calculations give approximate values?

Values are averaged across many different compounds, so do not reflect the exact bond energy in a specific molecule. They also only account for covalent bonds, not intermolecular forces in solids/liquids.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · 2
Calculate enthalpy change of reaction
2022 · 1
Bond enthalpy multiple choice
2021 · 2
Explain bond enthalpy limitations

What's Next

Bond enthalpy is a foundational concept that connects the microscopic structure of chemical bonds to the macroscopic energetics of whole reactions. You will use this understanding later when exploring activation energy in kinetics, and when comparing the energy output of different fuels. The approximation of average bond enthalpies also highlights why more accurate methods like Hess's law using enthalpy of formation or combustion are preferred for many calculations. Next, you will deepen your understanding of enthalpy calculations, before moving on to entropy and Gibbs free energy, which allow us to predict reaction spontaneity.