Bond enthalpy — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: Definition of average bond enthalpy, calculation of reaction enthalpy via the bond enthalpy method, sign convention rules, bond strength comparisons, and prediction of reaction thermicity from bond changes for AP Chemistry CED Unit 6.

You should already know: Enthalpy change (ΔH) definition and sign conventions for endo/exothermic reactions. Balancing chemical equations for gas-phase reactions. The difference between average and exact molecular properties.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Bond enthalpy?

Bond enthalpy (also referred to as 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. The most common notation is $E(\text{X-Y})$ or $\Delta H_{\text{bond}}(\text{X-Y})$, with standard units of kilojoules per mole (${\text{kJ mol}}^{-1}$). Because bond enthalpies are averaged across different chemical environments, they are approximate values, not exact for any single molecule. Per the AP Chemistry CED, bond enthalpy is a core skill in Unit 6 (Thermodynamics) accounting for approximately 12% of the unit’s exam weight, and it appears regularly in both multiple-choice (MCQ) and free-response (FRQ) sections. A key foundational rule of bond enthalpy is that breaking any covalent bond is always endothermic (requires energy input, so ΔH for bond breaking is positive), while forming any covalent bond is always exothermic (releases energy, so ΔH for bond formation is negative). This core rule underpins all calculations using bond enthalpy.

2. Calculating Reaction Enthalpy from Bond Enthalpies

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 takes energy (positive contribution), while forming bonds releases energy (negative contribution), 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: $E(\text{C-H})=414{\text{ kJ mol}}^{-1}$, $E(\text{O=O})=498{\text{ kJ mol}}^{-1}$, $E(\text{C=O})=799{\text{ kJ mol}}^{-1}$, $E(\text{O-H})=463{\text{ kJ mol}}^{-1}$.

1. Write the balanced gas-phase reaction: ${\text{CH}}_4(g)+2{\text{O}}_2(g)\to {\text{CO}}_2(g)+2{\text{H}}_2\text{O}(g)$
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: $\Delta H_{\text{rxn}}=2652-3450=-798{\text{ kJ mol}}^{-1}$

Exam tip: Always draw full Lewis structures before counting bonds — the most common error is miscounting the number of double bonds (e.g., two C=O bonds in CO₂, not one), which leads to an incorrect result.

3. Bond Enthalpy, Strength, and Bond Length Trends

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 that require qualitative reasoning as well as calculation.

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 ($E=358{\text{ kJ mol}}^{-1}$), C=O in formaldehyde ($E=745{\text{ kJ mol}}^{-1}$), C≡O in carbon monoxide ($E=1072{\text{ kJ mol}}^{-1}$).

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.

Exam tip: The inverse bond length-bond enthalpy trend only applies to bonds between the same pair of elements. You cannot use this trend to compare, for example, H-F and I-I, because the bonded atoms have different sizes and electronegativities.

4. Predicting Reaction Thermicity from Bond Enthalpy

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 ${\text{C}}_2{\text{H}}_4(g)+{\text{H}}_2(g)\to {\text{C}}_2{\text{H}}_6(g)$. 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: $E(\text{C-C})>\frac{E(\text{C=C})}{2}$, so $E(\text{C=C})<2E(\text{C-C})$. 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.

Exam tip: When justifying the sign of ΔH on FRQ, you must explicitly compare total energy of bonds broken vs. bonds formed. You will not earn full credit for only stating that "bond breaking is endothermic and bond forming is exothermic" without the comparison.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Counting one C=O bond in CO₂ instead of two. Why: Students count bonds per molecule instead of bonds within each molecule. Correct move: Always draw full Lewis structures for all reactants and products before counting bonds of each type.
• Wrong move: Reversing the formula to $\Delta H=\sum E(\text{formed})-\sum E(\text{broken})$. Why: Students confuse the sign convention for bond formation. Correct move: Memorize the mnemonic "Break in, Make out": ΔH = total energy in (breaking) minus total energy out (making).
• Wrong move: Using bond enthalpy to calculate ΔH for a reaction with liquid water or solid reactants. Why: Students forget bond enthalpies only apply to gas-phase bonds and ignore phase change enthalpy contributions. Correct move: Only use the bond enthalpy method if all species are explicitly stated to be gaseous; use Hess's law or enthalpy of formation for reactions with condensed phases.
• Wrong move: Claiming shorter bonds are always stronger than longer bonds, regardless of the bonded atoms. Why: Students overgeneralize the inverse bond length-bond enthalpy trend. Correct move: Only use the bond length trend for bonds between the same pair of elements; compare given bond enthalpy values directly for bonds between different atoms.
• Wrong move: Forgetting to multiply bond counts by the stoichiometric coefficient of each molecule. Why: Students count bonds per molecule but forget to scale to the reaction as written. Correct move: After counting bonds per molecule, multiply by the molecule's stoichiometric coefficient to get the total moles of bonds.
• Wrong move: Treating bond enthalpy values as exact for a specific molecule. Why: Students forget bond enthalpies are averaged across many compounds. Correct move: Always note that a ΔH calculated from bond enthalpy is an approximation when asked for a justification on FRQ.

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

Which of the following correctly gives the enthalpy change for the reaction below, using the provided bond enthalpies? $${\text{H}}_2(g)+{\text{Cl}}_2(g)\to 2\text{HCl}(g)$$ Bond enthalpies: $E(\text{H-H})=436{\text{ kJ mol}}^{-1}$, $E(\text{Cl-Cl})=242{\text{ kJ mol}}^{-1}$, $E(\text{H-Cl})=431{\text{ kJ mol}}^{-1}$. A) $-184{\text{ kJ mol}}^{-1}$ B) $+184{\text{ kJ mol}}^{-1}$ C) $-248{\text{ kJ mol}}^{-1}$ D) $+248{\text{ kJ mol}}^{-1}$

Worked Solution: First, sum the total bond enthalpy of bonds broken: 1 mol of H-H and 1 mol of Cl-Cl gives $\sum E(\text{broken})=436+242=678$ kJ. Next, sum the total bond enthalpy of bonds formed: 2 mol of H-Cl gives $\sum E(\text{formed})=2\times 431=862$ kJ. Applying the formula $\Delta H_{\text{rxn}}=\sum E(\text{broken})-\sum E(\text{formed})$ gives $678-862=-184$ kJ mol⁻¹. Reversing the formula gives the incorrect positive value (option B). The correct answer is A.

Question 2 (Free Response)

Ethene undergoes polymerization to form polyethylene according to the reaction: $n{\text{ C}}_2{\text{H}}_4(g)\to -({\text{CH}}_2{\text{CH}}_2{)}_n-(g)$ (for large $n$, end groups are negligible). Use the bond enthalpies $E(\text{C=C})=614{\text{ kJ mol}}^{-1}$, $E(\text{C-C})=348{\text{ kJ mol}}^{-1}$ to answer the questions below. (a) Calculate the total enthalpy change per mole of ethene monomer for this reaction. Show your working. (b) Justify the sign of the enthalpy change you calculated in terms of bonds broken and bonds formed. (c) If the reaction were carried out starting with liquid ethene instead of gaseous ethene, would the magnitude of the overall enthalpy change be larger or smaller than the value you calculated in (a)? Explain your answer.

Worked Solution: (a) All C-H bonds are conserved, so they do not contribute to ΔH. Per mole of ethene monomer, 1 C=C double bond is broken. Opening the C=C bond creates two reactive carbon sites that each form a new C-C bond with a neighboring monomer, so per mole of monomer, 1 full mole of new C-C bonds is formed (shared between two monomers). The calculation is: $$\Delta H=E(\text{bonds broken})-E(\text{bonds formed})=614-(2\times 348)=614-696=\boxed{-82{\text{ kJ mol}}^{-1}}$$ (b) The sign of ΔH is negative because the total energy required to break the C=C double bond (614 kJ per mole of monomer) is less than the total energy released when forming the two new C-C single bonds (696 kJ per mole of monomer). Net energy is released, so the reaction is exothermic with a negative ΔH. (c) The magnitude of the overall enthalpy change will be smaller. Converting liquid ethene to gaseous ethene (the phase where bond enthalpy applies) is an endothermic process that requires energy input. This positive energy contribution partially offsets the negative enthalpy of the gas-phase polymerization, resulting in a smaller net magnitude of ΔH for the overall reaction starting with liquid monomer.

Question 3 (Application / Real-World Style)

The Haber process produces ammonia for agricultural fertilizer, according to the gas-phase reaction: ${\text{N}}_2(g)+3{\text{H}}_2(g)\rightleftharpoons 2{\text{NH}}_3(g)$. Average bond enthalpies are: $E(\text{N≡N})=945{\text{ kJ mol}}^{-1}$, $E(\text{H-H})=436{\text{ kJ mol}}^{-1}$, $E(\text{N-H})=391{\text{ kJ mol}}^{-1}$. A fertilizer plant claims the reaction releases usable energy to offset operating costs. Calculate the enthalpy change per mole of ammonia produced, and state whether the claim is supported.

Worked Solution: First, calculate total bonds broken: 1 N≡N bond + 3 H-H bonds gives $\sum E(\text{broken})=945+(3\times 436)=2253$ kJ. Total bonds formed: 6 N-H bonds (for 2 moles of NH₃) gives $\sum E(\text{formed})=6\times 391=2346$ kJ. ΔH for the reaction as written (2 moles of NH₃) is $2253-2346=-93$ kJ. Per mole of ammonia, $\Delta H=\frac{-93}{2}=-46.5$ kJ mol⁻¹. The negative ΔH confirms the reaction is exothermic and releases energy, so the manufacturer's claim is supported.

7. Quick Reference Cheatsheet

8. What's Next

Bond enthalpy is the foundational skill for understanding reaction energy profiles, activation energy, and how covalent bond structure impacts reaction kinetics, which you will study next 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.

Enthalpy of formation Hess's law Bond order and Lewis structures Reaction energy profiles