For N_2 + 3H_2 → 2NH_3, bonds broken total 2252\ kJ and bonds formed total 2334\ kJ. Calculate H. [2 points]

United StatesChemistrySyllabus dot point

How can the enthalpy of a reaction be estimated from the energies of the bonds broken and formed?

Topic 6.7 Bond Enthalpies: estimate the enthalpy change of a reaction from average bond enthalpies, using the rule that breaking bonds absorbs energy and forming bonds releases it.

A focused answer to AP Chemistry Topic 6.7, covering average bond enthalpies, the principle that breaking bonds is endothermic and forming bonds is exothermic, and estimating the enthalpy of reaction as bonds broken minus bonds formed, with full worked examples.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-04

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Quick answer

A bond enthalpy is the energy required to break one mole of a particular bond in the gas phase; it is always positive, because breaking a bond requires energy. Forming the same bond releases the same energy. To estimate the enthalpy of a reaction, add up the energy to break all the bonds in the reactants (energy absorbed, positive) and the energy released forming all the bonds in the products (energy released, negative), then take bonds broken minus bonds formed: $\Delta H = \sum(\text{bonds broken}) - \sum(\text{bonds formed})$ . If the products have stronger bonds than the reactants, more energy is released than absorbed and the reaction is exothermic. Because tabulated values are averages over many molecules, this method gives an estimate, not an exact enthalpy.

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What this topic is asking
What a bond enthalpy is
The bond-energy rule
Why the method is an estimate
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What this topic is asking

The College Board (Topic 6.7) wants you to estimate the enthalpy of a reaction from average bond enthalpies, using the rule that breaking bonds absorbs energy and forming bonds releases energy. This gives an approximate $\Delta H$ from a table of bond energies, with no calorimetry needed.

What a bond enthalpy is

So the H-H bond enthalpy ( $436\ \text{kJ mol}^{-1}$ ) is the energy to break one mole of hydrogen molecules into atoms. A larger bond enthalpy means a stronger bond. Because a given bond type (say C-H) has slightly different strength in different molecules, the table value is an average, which is why the method is approximate.

The bond-energy rule

The sign of the result tells you the type of reaction. If the products' bonds are stronger (release more energy than was needed to break the reactants' bonds), $\Delta H$ is negative and the reaction is exothermic. If the reactants' bonds are stronger, $\Delta H$ is positive and the reaction is endothermic. The whole calculation rests on the difference between the bonds in and the bonds out.

Why the method is an estimate

Because bond enthalpies are averages, the calculated $\Delta H$ is an approximation, typically within a few percent of the true value for gas-phase reactions. It does not account for the exact molecular environment of each bond, and it strictly applies to gases (where there are no intermolecular forces to consider). For an exact enthalpy you use standard enthalpies of formation (Topic 6.8) or Hess's law (Topic 6.9). The bond-energy method is most useful when only bonds, not formation data, are available.

Estimating an enthalpy from bond enthalpies

For $\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(g)$ , use C-H $= 414$ , O=O $= 498$ , C=O $= 799$ , O-H $= 463\ \text{kJ mol}^{-1}$ .

step 1 Bonds broken in the reactants

$\text{CH}_4$ has four C-H ( $4 \times 414 = 1656$ ); two $\text{O}_2$ have two O=O ( $2 \times 498 = 996$ ). Total broken $= 1656 + 996 = 2652\ \text{kJ}$ .

step 2 Bonds formed in the products

$\text{CO}_2$ has two C=O ( $2 \times 799 = 1598$ ); two $\text{H}_2\text{O}$ have four O-H ( $4 \times 463 = 1852$ ). Total formed $= 1598 + 1852 = 3450\ \text{kJ}$ .

step 3 Apply the formula

$\Delta H = 2652 - 3450 = -798\ \text{kJ mol}^{-1}$ .

step 4 Interpret

The negative value confirms combustion is exothermic, because the product bonds (in $\text{CO}_2$ and water) are stronger overall than the reactant bonds.

Try this

Q1. For $\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$ , bonds broken total $2252\ \text{kJ}$ and bonds formed total $2334\ \text{kJ}$ . Calculate $\Delta H$ . [2 points]

Cue. $\Delta H = 2252 - 2334 = -82\ \text{kJ mol}^{-1}$ (exothermic).

Q2. Explain why the bond-enthalpy method gives only an estimate of $\Delta H$ . [2 points]

Cue. Tabulated bond enthalpies are averages over many molecules, so they do not capture the exact strength of each bond in a specific compound.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)4 marksSection II (long FRQ, part). For

\text{H}_2(g) + \text{Cl}_2(g) \rightarrow 2\text{HCl}(g)

, the average bond enthalpies are:

\text{H-H} = 436

\text{Cl-Cl} = 242

\text{H-Cl} = 431\ \text{kJ mol}^{-1}

. (a) Calculate the total energy needed to break the reactant bonds. (b) Calculate the total energy released forming the product bonds. (c) Calculate the enthalpy of reaction. (d) Justify the sign of your answer in terms of bond strengths.

Show worked answer →

A 4-point quantitative FRQ on bond enthalpies.

(a) Bonds broken (1 point): one H-H and one Cl-Cl: $436 + 242 = 678\ \text{kJ}$ absorbed.
(b) Bonds formed (1 point): two H-Cl: $2 \times 431 = 862\ \text{kJ}$ released.
(c) Enthalpy of reaction (1 point): $\Delta H = \text{(bonds broken)} - \text{(bonds formed)} = 678 - 862 = -184\ \text{kJ mol}^{-1}$ .
(d) Justify (1 point): the product H-Cl bonds are stronger (release more energy) than the reactant bonds absorbed in breaking, so more energy is released than absorbed and the reaction is exothermic.

Markers reward the bonds-broken sum, the bonds-formed sum, the enthalpy as broken minus formed, and the stronger-product-bonds reasoning.

AP 2021 (style)1 marksSection I (multiple choice). Using bond enthalpies, the enthalpy of a reaction equals (A) bonds formed minus bonds broken (B) bonds broken minus bonds formed (C) bonds broken plus bonds formed (D) the average of the two. Justify your choice.

Show worked answer →

A 1-point conceptual MCQ. The answer is (B).

Breaking bonds absorbs energy (positive) and forming bonds releases energy (negative), so $\Delta H = \sum(\text{bonds broken}) - \sum(\text{bonds formed})$ . The trap is reversing the order, which would give the wrong sign.

Related dot points

Sources & how we know this

AP Chemistry Course and Exam Description — College Board (2020)