Bond Enthalpy

AP Chemistry· AP Chemistry CED — Thermodynamics· 14 min read

1. What Is Bond Enthalpy?★★☆☆☆⏱ 3 min

Bond enthalpy (also called bond energy or average bond dissociation enthalpy) is defined as the enthalpy change required to homolytically break one mole of a specific covalent bond in the gaseous state, averaged across many different compounds containing that bond. Common notation is $E (e x t X - Y)$ or $e x t b o n d e n t ha l p y$ , with standard units of kilojoules per mole ( $e x t k J m o l^{- 1}$ ).

Because bond enthalpies are averaged across different chemical environments, they are approximate values, not exact for any single molecule. This topic is a core skill in AP Chemistry Unit 6, appearing regularly in both multiple-choice and free-response sections.

📘 Definition

Average Bond Enthalpy

$E (e x t X - Y)$ , $e x t b o n d e n t ha l p y$

Enthalpy change required to break one mole of a given covalent bond in the gaseous state, averaged across all compounds containing that bond.

Example:

$E (e x t C - H) = 414 e x t k J m o l^{- 1}$ is the average value across all organic molecules

2. Calculating Reaction Enthalpy from Bond Enthalpies★★★☆☆⏱ 5 min

To calculate the net enthalpy change of a reaction, we first account for all energy required to break all reactant bonds into gaseous atoms, then subtract the energy released when those atoms form all product bonds. Breaking bonds contributes positive energy, while forming bonds contributes negative energy, leading to the core formula:

\Delta H_{\text{rxn}} = \sum E(\text{bonds broken}) - \sum E(\text{bonds formed})

This method only works for reactions where all reactants and products are in the gaseous state. Bond enthalpies do not account for extra enthalpy changes from phase changes of solids or liquids, so those reactions require Hess’s law or enthalpy of formation instead. Accurate counting of bonds is the most critical step for getting the correct result; drawing full Lewis structures for all species avoids miscounting.

📐 Worked Example

Calculate the enthalpy change for the complete combustion of one mole of gaseous methane, using the following bond enthalpy values: $E (C-H) = 414 kJ mol^{- 1}$ , $E (O=O) = 498 kJ mol^{- 1}$ , $E (C=O) = 799 kJ mol^{- 1}$ , $E (O-H) = 463 kJ mol^{- 1}$ .

1
Write the balanced gas-phase reaction:
$\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(g)$
2
Count and sum bonds broken in reactants: 4 C-H bonds (1 mol CH₄) + 2 O=O bonds (2 mol O₂):
$\sum E(\text{broken}) = (4 \times 414) + (2 \times 498) = 1656 + 996 = 2652 \text{ kJ}$
3
Count and sum bonds formed in products: 2 C=O bonds (1 mol CO₂) + 4 O-H bonds (2 mol H₂O):
$\sum E(\text{formed}) = (2 \times 799) + (4 \times 463) = 1598 + 1852 = 3450 \text{ kJ}$
4
Apply the formula to get the final enthalpy change:
$\Delta H_{\text{rxn}} = 2652 - 3450 = -798 \text{ kJ mol}^{-1}$

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

Bond enthalpy directly measures bond strength: the higher the bond enthalpy, the more energy required to break the bond, so the stronger the bond. For bonds between the same pair of atoms, bond enthalpy is inversely related to bond length: shorter bonds have higher bond enthalpy and are stronger. This trend follows bond order: as bond order increases (single → double → triple), bond length decreases and bond enthalpy increases. AP questions regularly ask to rank bonds by strength, or justify a given trend using bond enthalpy arguments, often in FRQ sections.

📐 Worked Example

Rank the following carbon-oxygen bonds from strongest to weakest, then explain the relationship between bond order, bond length, and bond enthalpy for this series: C-O in methanol ( $E = 358 kJ mol^{- 1}$ ), C=O in formaldehyde ( $E = 745 kJ mol^{- 1}$ ), C≡O in carbon monoxide ( $E = 1072 kJ mol^{- 1}$ ).

1
Recall that higher bond enthalpy equals stronger bond, so order by bond enthalpy from highest to lowest: C≡O (1072 kJ mol⁻¹) > C=O (745 kJ mol⁻¹) > C-O (358 kJ mol⁻¹).
2
Identify the bond order for each bond: C≡O has bond order 3, C=O has bond order 2, C-O has bond order 1.
3
For bonds between the same two atoms (carbon and oxygen here), as bond order increases, bond length decreases and bond enthalpy increases, resulting in stronger bonds.

4. Predicting Reaction Thermicity from Bond Enthalpy★★★☆☆⏱ 3 min

AP FRQ frequently asks to justify whether a reaction is endothermic or exothermic using bond enthalpy arguments, without requiring a full numerical calculation. If the total energy required to break reactant bonds is greater than the total energy released when forming product bonds, ΔH is positive (endothermic). If the total energy released from forming product bonds exceeds the energy required to break reactant bonds, ΔH is negative (exothermic). This skill is often tested in contextual questions about polymerization, fuel combustion, or bond cleavage reactions.

📐 Worked Example

The hydrogenation of ethene follows the reaction $C_{2} H_{4} (g) + H_{2} (g) \to C_{2} H_{6} (g)$ . Given that the average bond enthalpy of a C-C single bond is greater than half the bond enthalpy of a C=C double bond, predict if the reaction is endothermic or exothermic and justify your answer.

1
Count bonds broken and formed: Bonds broken = 1 C=C (ethene) + 1 H-H (H₂); Bonds formed = 1 C-C (ethane) + 2 new C-H bonds.
2
Write the ΔH expression:
$\Delta H = [E(\text{C=C}) + E(\text{H-H})] - [E(\text{C-C}) + 2E(\text{C-H})]$
3
Apply the given condition: $E (C-C) > \frac{E ( C=C )}{2}$ , so $E (C=C) < 2 E (C-C)$ . The total energy of bonds broken is less than the total energy of bonds formed.
4
ΔH = (smaller value) - (larger value) = negative, so the reaction is exothermic.

5. AP-Style Practice Problems★★★★☆⏱ 6 min

✓ Quick check

Test your calculation skills with this multiple-choice question:

Which of the following correctly gives the enthalpy change for the reaction below, using the provided bond enthalpies? $H_{2} (g) + Cl_{2} (g) \to 2 HCl (g)$ Bond enthalpies: $E (H-H) = 436 kJ mol^{- 1}$ , $E (Cl-Cl) = 242 kJ mol^{- 1}$ , $E (H-Cl) = 431 kJ mol^{- 1}$ .
- -184 kJ mol⁻¹
- +184 kJ mol⁻¹
- -248 kJ mol⁻¹
- +248 kJ mol⁻¹

📐 Worked Example

Ethene undergoes polymerization to form polyethylene according to the reaction: $n C_{2} H_{4} (g) \to - (CH_{2} CH_{2})_{n} - (g)$ (end groups are negligible for large $n$ ). Use the bond enthalpies $E (C=C) = 614 kJ mol^{- 1}$ , $E (C-C) = 348 kJ mol^{- 1}$ to answer: (a) Calculate ΔH per mole of ethene monomer. (b) Justify the sign of ΔH. (c) If starting with liquid ethene, is the magnitude of ΔH larger or smaller? Explain.

1
(a) All C-H bonds are unchanged, so they do not contribute to ΔH. Per mole of monomer, 1 C=C bond is broken, and 1 full mole of new C-C bonds is formed:
$\Delta H = 614 - (2 \times 348) = -82 \text{ kJ mol}^{-1}$
2
(b) ΔH is negative because the energy required to break the C=C bond (614 kJ/mol) is less than the energy released forming the two new C-C bonds (696 kJ/mol). Net energy is released, so the reaction is exothermic.
3
(c) The magnitude of ΔH will be smaller. Converting liquid ethene to gaseous ethene is endothermic, so the positive energy input partially offsets the negative polymerization enthalpy, leading to a smaller net magnitude.

📐 Worked Example

The Haber process produces ammonia for fertilizer: $N_{2} (g) + 3 H_{2} (g) ⇌ 2 NH_{3} (g)$ . Bond enthalpies: $E (N\equivN) = 945 kJ mol^{- 1}$ , $E (H-H) = 436 kJ mol^{- 1}$ , $E (N-H) = 391 kJ mol^{- 1}$ . A plant claims the reaction releases usable energy. Calculate ΔH per mole of ammonia produced, and confirm if the claim is supported.

1
Calculate total bonds broken for the full reaction:
$\sum E(\text{broken}) = 945 + (3 \times 436) = 2253 \text{ kJ}$
2
Calculate total bonds formed for 2 moles of NH₃:
$\sum E(\text{formed}) = 6 \times 391 = 2346 \text{ kJ}$
3
Scale ΔH to per mole of ammonia:
$\Delta H_{\text{rxn}} = 2253 - 2346 = -93 \text{ kJ} \implies \Delta H_{\text{per mole}} = \frac{-93}{2} = -46.5 \text{ kJ mol}^{-1}$
4
The negative ΔH confirms the reaction is exothermic and releases energy, so the plant's claim is supported.

6. Common Pitfalls

Wrong move:

Counting one C=O bond in CO₂ instead of two

Why:

Students miscount multiple bonds within a single molecule

Correct move:

Always draw full Lewis structures for all species before counting bonds

Wrong move:

Reversing the formula to ΔH = Σ E(formed) - Σ E(broken)

Why:

Confusion over the sign convention for bond formation

Correct move:

Use the mnemonic 'Break in, Make out': ΔH = energy in (breaking) minus energy out (making)

Wrong move:

Using bond enthalpy for reactions with solid/liquid reactants/products

Why:

Forgetting bond enthalpies only apply to gas-phase bonds and ignore phase change enthalpy

Correct move:

Only use bond enthalpy method if all species are gaseous; use Hess's law/enthalpy of formation otherwise

Wrong move:

Claiming shorter bonds are always stronger than longer bonds regardless of atoms

Why:

Overgeneralizing the inverse bond length-bond enthalpy trend

Correct move:

Only use the trend for bonds between the same two elements; compare given bond enthalpies directly for different atoms

Wrong move:

Forgetting to multiply bond counts by stoichiometric coefficients

Why:

Counting bonds per molecule but not scaling to the reaction as written

Correct move:

Multiply bonds per molecule by the molecule's stoichiometric coefficient to get total moles of bonds

Wrong move:

Treating bond enthalpy values as exact for a specific molecule

Why:

Forgetting bond enthalpies are averaged across many different compounds

Correct move:

Always note that ΔH calculated from bond enthalpy is an approximation for FRQ justifications

7. Quick Reference Cheatsheet

Category	Formula / Rule	Notes
Reaction Enthalpy	$Δ H_{rxn} = \sum E (bonds broken) - \sum E (bonds formed)$	Only for all-gas-phase reactions; result is approximate
ΔH for Bond Breaking	$Δ H = + E (bond)$	Always endothermic, positive sign
ΔH for Bond Forming	$Δ H = - E (bond)$	Always exothermic, negative sign
Bond Enthalpy vs Strength	Higher $E (bond)$ = stronger bond	Applies to all bonds
Bond Enthalpy vs Length	Higher $E (bond)$ = shorter bond length	Only valid for bonds between the same two elements
Bond Enthalpy vs Bond Order	Higher bond order = higher $E (bond)$	Only for bonds between the same two elements
ΔH per mole of species	$Δ H_{per mole} = \frac{Δ H _{rxn}}{stoichiometric coefficient}$	Used when asked for enthalpy per specific species

What's Next

Bond enthalpy is a foundational skill for understanding reaction energy profiles, activation energy, and how covalent bond structure impacts reaction kinetics, which you will explore further in AP Chemistry Unit 5 (Kinetics) and Unit 6 (Thermodynamics). Without mastering bond enthalpy calculations and sign conventions, you will not be able to correctly calculate activation energy from bond energies or justify why some reactions have higher energy barriers than others. Beyond kinetics and thermodynamics, bond enthalpy is used to compare the stability of different allotropes (e.g., diamond vs graphite) and calculate the energy content of different fuels, a common context for AP FRQ questions. Building on this topic, you will next apply enthalpy calculation rules to other methods that work for non-gas phase reactions.