Bond EnergiesCourse level: High School / HonorsCourse level: AP Chemistry

Bond Enthalpies and Estimating Enthalpy Change

What you'll be able to do: Estimate a reaction enthalpy from average bond enthalpies and explain why the result is only an estimate.

Best after: Enthalpy of Reaction and Thermochemical Equations, Standard Enthalpies of Formation

Introduction

Bond enthalpies let you predict whether a reaction releases energy using nothing but a Lewis structure and a table of averages. The trade-off is accuracy, and knowing why the answer is approximate is part of the skill.

These are recommended, not required. You can start this lesson at any time.

Learning objectives

  • Explain why bond breaking is endothermic and bond making exothermic
  • Estimate ΔH from average bond enthalpies
  • Count bonds correctly from structural formulas
  • Explain the limitations of the bond enthalpy method

Breaking costs, making pays

A bond exists because the bonded arrangement is lower in energy. Pulling the atoms apart requires that energy back, so bond dissociation is always endothermic and the tabulated values are always positive. Forming the same bond releases the same magnitude.

Separating bonded gaseous atoms raises their energy; a positive bond-dissociation enthalpy is the cost of that specified process. Recombining those same atoms to restore the same bond has the opposite sign. Burning a fuel releases energy from the net replacement of bonds, not from breaking the fuel's bonds alone. Melting ice also should not be described as breaking its covalent O–H bonds: that phase change chiefly alters intermolecular interactions.

There is no such thing as a negative bond enthalpy in a table. The sign appears from which side of the calculation the bond is on.

The estimating relationship

Every reactant bond that changes must be broken and every product bond must be formed. Adding the costs and subtracting the payback gives the net enthalpy change.

Imagine taking gaseous reactants to separated atoms, then assembling gaseous products. The first leg adds bond-breaking energies, the second subtracts the listed dissociation energies for product bonds. That Hess cycle explains ΔH ≈ ΣD(broken) − ΣD(formed). It is not the real reaction mechanism and does not imply every bond actually breaks simultaneously. Each term must include both the number of such bonds per molecule and the molecule's equation coefficient.

ΔH = S(bonds broken) - S(bonds formed)

Break bonds before forming the new ones. Read this visual alongside the section; the text alternative describes its scientific content and assumptions.

Counting bonds properly

Draw or visualise the structures. Methane has four C-H bonds, not one. O₂ is a double bond, N₂ is a triple bond, and CO₂ contains two C=O bonds. Miscounting the structures, rather than the arithmetic, is the usual source of error.

For H₂ + Cl₂ → 2 HCl, the ledger contains one H–H and one Cl–Cl broken, two H–Cl formed. With 436, 242 and 431 kJ/mol, ΔH ≈ 436 + 242 − 2(431) = −184 kJ. For methane combustion to gaseous water, count four C–H and two O=O broken; two C=O and four O–H formed. A coefficient of two waters means four O–H bonds, not two. Triple and double bonds use their own tabulated D values rather than three or two copies of a single-bond value.

CH₄: 4 x C-H; CO₂: 2 x C=O; N₂: 1 x N N triple

Why the answer is approximate

A C-H bond in methane is not identical to a C-H bond in ethanol, so tables list an average across many compounds. The method also assumes every species is gaseous, because condensing to a liquid releases additional energy that bond enthalpies do not capture.

Average bond values vary with the surrounding molecule, so cancelling “unchanged C–H bonds” is an approximation when their environments change. For ethene hydrogenation, the useful net ledger replaces C=C and H–H with C–C and two C–H bonds; it estimates the gas-phase heat. To estimate a liquid product, add the appropriate condensation or other phase correction. Formation enthalpies for the exact molecules and states generally provide the better thermochemical result when available. A difference between two methods can arise from both averaging and phase choice; do not assign all discrepancy to one cause without checking.

If a question quotes water as a liquid, expect a bond enthalpy estimate to disagree noticeably with the formation-data answer.

When to use which method

Use formation enthalpies when tabulated values exist and states matter, and use bond enthalpies for gas-phase reactions or when you only need to know whether a reaction is exothermic and roughly by how much.

Track the changed bonds, not every unchanged bond

Average bond enthalpy estimates can be simplified by canceling identical bonds present on both sides. For hydrogenation of a C=C, the net bond changes are breaking one C=C and one H-H, then forming one C-C and two C-H.

This is an estimate for gas-phase bonds; it does not include phase changes or exact molecule-specific bond energies.

ΔH ≈ bonds broken minus bonds formed.

Key ideas

Definition
Bond enthalpy

Energy required to break one mole of a specified bond in the gas phase, always positive.

Rule
Broken minus formed

ΔH is approximately the sum of bonds broken minus the sum of bonds formed.

Key concept
Average values

Table entries are averaged over many molecules, so results are estimates.

Assumption
Gas phase only

The method assumes all reactants and products are gases.

Key concept
Bond strength trend

For the same pair of atoms, triple bonds are stronger and shorter than double, which are stronger than single.

Equation
Enthalpy from bond enthalpies

ΔH = Σ D(bonds broken) - Σ D(bonds formed)

  • D = average bond enthalpy in kJ/mol, always a positive number

Worked examples

Worked example 1

Estimate ΔH for CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(g) using D(C-H) = 413, D(O=O) = 498, D(C=O) = 799, D(O-H) = 467 kJ/mol.

Try it first: List every bond on each side with its count before touching a number.

    0 of 4 steps revealed.

    Worked example 2

    Estimate ΔH for H₂(g) + Cl₂(g) → 2 HCl(g) using D(H-H) = 436, D(Cl-Cl) = 242, D(H-Cl) = 431 kJ/mol.

      0 of 3 steps revealed.

      Worked example 3

      A reaction has bonds broken totalling 1200 kJ and bonds formed totalling 1150 kJ. Is it exothermic or endothermic, and what does that say about relative bond strengths?

        0 of 3 steps revealed.

        Worked example 4

        Estimate ΔH for C₂H₄ + H₂ → C₂H₆ using D(C=C)=614, D(H-H)=436, D(C-C)=347, D(C-H)=413 kJ/mol.

        Try it first: The four original C-H bonds cancel from both sides.

          0 of 3 steps revealed.

          Common mistakes

          Common mistakes and better checks
          MistakeWhy it's wrongCheck instead
          Writing bonds formed minus bonds broken.It reverses the sign of the whole estimate.Broken first, because breaking is the cost you pay upfront.
          Treating bond formation as requiring energy.Forming a bond moves the system to a lower energy state, releasing energy.Breaking absorbs, forming releases.
          Counting one C-H bond in methane instead of four.Every bond present must be accounted for.Sketch the structure and count explicitly.
          Expecting the bond enthalpy answer to match the formation-data answer exactly.Table values are averages and the method assumes gases.Report it as an estimate and mention the phase assumption.

          Scroll horizontally if needed to read all columns.

          After the lesson · Question bank

          Independent practice

          Work through new problems on your own. The lesson's quick checks and worked examples stay above if you need to review a step.

          27 topic questions across 3 difficulty levels.

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          What you should now know

          Bond breaking always absorbs energy and bond making always releases it, so all tabulated bond enthalpies are positive. For a gas-phase reaction, ΔH is approximately the total energy of bonds broken minus the total energy of bonds formed. The values are averages taken across many molecules, so results are estimates, typically within a few tens of kilojoules. The method also assumes all species are gases, which is why it fails when liquids or solids are involved.

          • Bond breaking absorbs energy, bond forming releases it
          • Tabulated bond enthalpies are always positive
          • ΔH is approximately bonds broken minus bonds formed
          • Count every bond from the structure, not the formula alone
          • Results are estimates and assume all species are gases

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          Sources and further reading

          • Chemistry 2e, Section 7.5: Strengths of Ionic and Covalent Bonds
            Paul Flowers, Klaus Theopold, Richard Langley, William R. Robinson · OpenStax, Rice University · Chapter 7.5
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            The current online Chemistry 2e edition lists CC BY-NC-SA 4.0. The source version used for this lesson and its applicable rights remain unverified.

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            Recorded attribution / license claim (applicability unverified): Chemistry 2e, OpenStax, Rice University, licensed CC BY 4.0.

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