Chemistry Clarity · chemistryclarity.com

Bond Enthalpies

How can bond strengths be used to estimate the enthalpy change of a reaction?

IntermediateBonding & Molecular StructureLast reviewed 5 October 2026
Download lesson (PDF)

What is it?

The bond enthalpy (bond energy) is the energy needed to break one mole of a particular bond in gaseous molecules, forming gaseous atoms. For hydrogen:

HX2(g)→2 H(g)ΔH=+436 kJ/mol\begin{aligned} &\ce{H2(g) -> 2H(g)} \\[4pt] &\Delta H = +436\ \text{kJ/mol} \end{aligned}
  • Breaking a bond always needs energy (endothermic, positive).
  • Forming a bond always releases the same amount of energy (exothermic, negative).
BondBond enthalpy (kJ/mol)BondBond enthalpy (kJ/mol)
H–H436C–C348
Cl–Cl242C=C614
H–Cl431C≡C839
C–H413O=O498
O–H463C=O (in CO₂)799
N–H391N≡N945

Except for diatomic molecules such as H–H, these are average values: a C–H bond in methane is not exactly as strong as one in ethanol, so tables give an average over many compounds.

Key idea

A reaction breaks the old bonds and makes new ones. The enthalpy change is the energy taken in to break bonds minus the energy given out when new bonds form:

ΔH≈∑(bonds broken)−∑(bonds formed)\begin{aligned} &\Delta H \approx \textstyle\sum(\text{bonds broken}) \\[4pt] &\qquad - \textstyle\sum(\text{bonds formed}) \end{aligned}

If the new bonds are stronger than the old ones, more energy is released than taken in, and the reaction is exothermic.

Why does it matter?

  • Explaining energy changes. Fuels release energy because the bonds in COX2\ce{CO2} and HX2O\ce{H2O} are stronger than those in the fuel and oxygen.
  • Quick estimates. You can estimate ΔH for a reaction even when no enthalpies of formation are available.
  • Bond strength and reactivity. The very strong N≡N bond (945 kJ/mol) explains why nitrogen gas is so unreactive, and why making ammonia needs a catalyst.

How does it work?

1. Bond order, length and strength

The bond order is the number of shared pairs: 1 for single, 2 for double, 3 for triple. As bond order rises, the bond gets shorter and stronger:

BondOrderLength (pm)Enthalpy (kJ/mol)
C–C1154348
C=C2134614
C≡C3120839

A double bond is not twice as strong as a single bond, because the π bond is weaker than the σ bond.

2. Estimating ΔH: the steps

  1. Draw the structures and list every bond in the reactants and the products, using the coefficients.
  2. Add up the bond enthalpies of the bonds broken (reactants).
  3. Add up the bond enthalpies of the bonds formed (products).
  4. ΔH≈\Delta H \approx broken − formed.

3. Why it is only an estimate

The answer is approximate because the bond enthalpies are averages, and because they apply to gases: if a reactant or product is a liquid or solid, the energy of changing state is left out. Enthalpies of formation (Hess’s law) give more accurate values.

Think of it like this

Think of bonds as money. Breaking bonds is spending: you pay energy to pull atoms apart. Making bonds is earning: you are paid energy back. If you earn more from the new bonds than you spent on the old ones, you finish with a profit: the reaction gives out energy (exothermic).

More precisely

Bond enthalpies are measured for the gas phase, so the estimate works best when every species is a gas. For a diatomic molecule the bond enthalpy is an exact value (the bond dissociation enthalpy); for bonds in larger molecules it depends on the rest of the molecule. The value for C=O in CO₂ (799 kJ/mol) differs from the average C=O in aldehydes and ketones (about 745 kJ/mol), which is why the table says “in CO₂”. Some tables give 495 kJ/mol for O=O or 941 kJ/mol for N≡N; small differences like these change the estimate by a few kJ/mol.

Visualise it

Energy diagram for H2 plus Cl2 forming 2 HCl. From the reactants, an upward arrow of plus 678 kilojoules per mole breaks the bonds to give separate atoms 2H plus 2Cl. A larger downward arrow of minus 862 kilojoules per mole forms the new bonds. The products end 184 kilojoules per mole below the reactants: delta H is minus 184 kilojoules per mole.
Breaking bonds takes energy in; forming bonds gives more out. The difference is ΔH.

Worked example

Worked example: Hydrogen and chlorine

Question: Estimate ΔH for HX2(g)+ClX2(g)→2 HCl(g)\ce{H2(g) + Cl2(g) -> 2HCl(g)}.

  1. Bonds broken: one H–H and one Cl–Cl:

    436 kJ/mol+242 kJ/mol=678 kJ/mol\begin{aligned} &436\ \text{kJ/mol} \\[4pt] &\quad + 242\ \text{kJ/mol} \\[4pt] &= 678\ \text{kJ/mol} \end{aligned}

  2. Bonds formed: two H–Cl:

    2×431 kJ/mol=862 kJ/mol\begin{aligned} &2 \times 431\ \text{kJ/mol} \\[4pt] &= 862\ \text{kJ/mol} \end{aligned}

  3. Enthalpy change:

    ΔH≈678 kJ/mol−862 kJ/mol=−184 kJ/mol\begin{aligned} &\Delta H \approx 678\ \text{kJ/mol} \\[4pt] &\quad - 862\ \text{kJ/mol} \\[4pt] &= -184\ \text{kJ/mol} \end{aligned}

    Exothermic: the H–Cl bonds are stronger in total than the bonds broken.

Worked example: Burning methane

Question: Estimate ΔH for CHX4(g)+2 OX2(g)→COX2(g)+2 HX2O(g)\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g)}.

  1. Broken: 4 C–H and 2 O=O:

    4×413 kJ/mol+2×498 kJ/mol=2648 kJ/mol\begin{aligned} &4 \times 413\ \text{kJ/mol} \\[4pt] &\quad + 2 \times 498\ \text{kJ/mol} \\[4pt] &= 2648\ \text{kJ/mol} \end{aligned}

  2. Formed: 2 C=O (in COX2\ce{CO2}) and 4 O–H (in 2 HX2O\ce{H2O}):

    2×799 kJ/mol+4×463 kJ/mol=3450 kJ/mol\begin{aligned} &2 \times 799\ \text{kJ/mol} \\[4pt] &\quad + 4 \times 463\ \text{kJ/mol} \\[4pt] &= 3450\ \text{kJ/mol} \end{aligned}

  3. Enthalpy change:

    ΔH≈2648 kJ/mol−3450 kJ/mol=−802 kJ/mol\begin{aligned} &\Delta H \approx 2648\ \text{kJ/mol} \\[4pt] &\quad - 3450\ \text{kJ/mol} \\[4pt] &= -802\ \text{kJ/mol} \end{aligned}

    This agrees closely with the measured value for gaseous water.

Worked example: Making ammonia

Question: Estimate ΔH for NX2(g)+3 HX2(g)→2 NHX3(g)\ce{N2(g) + 3H2(g) -> 2NH3(g)} and compare it with the measured value of −92 kJ/mol.

  1. Broken: 1 N≡N and 3 H–H: 945 kJ/mol+3×436 kJ/mol=2253 kJ/mol945\ \text{kJ/mol} + 3 \times 436\ \text{kJ/mol} = 2253\ \text{kJ/mol}

  2. Formed: 2 NHX3\ce{NH3} contain 6 N–H: 6×391 kJ/mol=2346 kJ/mol6 \times 391\ \text{kJ/mol} = 2346\ \text{kJ/mol}

  3. Enthalpy change:

    ΔH≈2253 kJ/mol−2346 kJ/mol=−93 kJ/mol\begin{aligned} &\Delta H \approx 2253\ \text{kJ/mol} \\[4pt] &\quad - 2346\ \text{kJ/mol} \\[4pt] &= -93\ \text{kJ/mol} \end{aligned}

  4. Within about 1 kJ/mol of the measured −92 kJ/mol. The difference comes from using an average N–H value.

Common mistake

Common mistake: Subtracting the wrong way round

It is broken − formed (reactants − products). This is the opposite order to the enthalpy-of-formation formula (products − reactants), because bond enthalpies are energies taken in. Check: forming strong bonds should make ΔH negative.

Common mistake: Forgetting the coefficients or missing bonds

In CHX4+2 OX2\ce{CH4 + 2O2} there are two O=O bonds to break, and two water molecules contain four O–H bonds. Draw every molecule and count every bond.

Common mistake: Using a single-bond value for a double bond

C=O, C=C and N≡N have their own values. Using C–O (358 kJ/mol) for the C=O in CO₂ gives a badly wrong answer.

Notation note

  • Bond enthalpy is sometimes written DD or EE, e.g. D(H–H)=436D(\text{H–H}) = 436 kJ/mol.
  • 1 pm (picometre) = 10−1210^{-12} m.

Remember this

Remember this

  • Bond enthalpy: energy to break 1 mol of a bond in the gas phase (always positive).
  • Breaking bonds is endothermic; forming bonds is exothermic.
  • ΔH≈∑(broken)−∑(formed)\Delta H \approx \sum(\text{broken}) - \sum(\text{formed}).
  • Higher bond order: shorter and stronger (C–C 348, C=C 614, C≡C 839 kJ/mol).
  • Results are estimates: values are averages and apply to gases.

Test yourself

Check your understanding before moving on.

Flashcards

10 cards

  1. Question
    What is bond enthalpy?
    Answer

    The energy needed to break one mole of a bond in gaseous molecules, forming gaseous atoms.

  2. Question
    Is breaking a bond endothermic or exothermic? And forming one?
    Answer

    Breaking: endothermic (energy in). Forming: exothermic (energy out).

  3. Question
    Give the formula for estimating ΔH from bond enthalpies.
    Answer

    ΔH ≈ Σ(bonds broken) − Σ(bonds formed)

  4. Question
    How do bond length and strength change with bond order?
    Answer

    Higher order: shorter and stronger. C–C 154 pm, 348 kJ/mol; C=C 134 pm, 614 kJ/mol; C≡C 120 pm, 839 kJ/mol.

  5. Question
    Why is a C=C bond less than twice as strong as C–C?
    Answer

    The second bond is a π bond, which is weaker than a σ bond.

  6. Question
    Why are bond-enthalpy calculations only estimates?
    Answer

    Most values are averages over many compounds, and they apply only to gases.

  7. Question
    Estimate ΔH for H₂ + Cl₂ → 2HCl (H–H 436, Cl–Cl 242, H–Cl 431 kJ/mol).
    Answer

    (436 + 242) − 2 × 431 = −184 kJ/mol

  8. Question
    Why is nitrogen gas so unreactive?
    Answer

    The N≡N triple bond is very strong (945 kJ/mol).

  9. Question
    When is a reaction exothermic, in terms of bonds?
    Answer

    When the bonds formed are stronger in total than the bonds broken.

  10. Question
    How many O–H bonds are formed in CH₄ + 2O₂ → CO₂ + 2H₂O?
    Answer

    Four (two in each water molecule).

Quiz

7 questions

  1. Question 1EasyWhich statement is correct?
    Show answer

    Answer: Breaking bonds requires energy

    Energy must be supplied to pull atoms apart, so bond breaking is endothermic; forming bonds releases the same energy. Bond enthalpies are positive.

  2. Question 2EasyWhich bond is the shortest and strongest?
    Show answer

    Answer: C≡C

    Higher bond order means a shorter, stronger bond: C≡C (120 pm, 839 kJ/mol) beats C=C and C–C.

  3. Question 3MediumEstimate ΔH for H₂ + Br₂ → 2HBr (all gases). H–H 436, Br–Br 193, H–Br 366 kJ/mol.
    Show answer

    Answer: −103 kJ/mol

    Broken: 436 + 193 = 629 kJ/mol. Formed: 2 × 366 = 732 kJ/mol. ΔH ≈ 629 − 732 = −103 kJ/mol. −263 forgets the factor of 2.

  4. Question 4MediumEstimate ΔH for 2H₂ + O₂ → 2H₂O (gases). H–H 436, O=O 498, O–H 463 kJ/mol.
    Show answer

    Answer: −482 kJ/mol

    Broken: 2 × 436 + 498 = 1370 kJ/mol. Formed: 4 O–H = 4 × 463 = 1852 kJ/mol. ΔH ≈ 1370 − 1852 = −482 kJ/mol.

  5. Question 5MediumWhy is the ΔH from bond enthalpies only an estimate?
    Show answer

    Answer: Most bond enthalpies are averages, and apply to gases only

    A C–H bond is slightly different in every molecule, so tables give averages; changes of state are also left out.

  6. Question 6HardFor H₂ + I₂ → 2HI, ΔH = −9 kJ/mol. With H–H 436 and I–I 151 kJ/mol, what is the H–I bond enthalpy?
    Show answer

    Answer: 298 kJ/mol

    −9 = (436 + 151) − 2x, so 2x = 596 and x = 298 kJ/mol. 596 kJ/mol is the energy for two H–I bonds.

  7. Question 7HardIn terms of bonds, why is burning methane exothermic?
    Show answer

    Answer: The bonds in CO₂ and H₂O are stronger in total than those in CH₄ and O₂

    Bonds broken total 2648 kJ/mol; bonds formed total 3450 kJ/mol. More energy is released than absorbed: ΔH ≈ −802 kJ/mol.

Notes and downloads

  • Worksheet

    Bond Enthalpies Worksheet

    8 questions on bond enthalpy, bond order and length, estimating ΔH from bonds broken and formed, and finding an unknown bond enthalpy. Answer key included.

    IntermediateFree

References

  1. Brown, T. L.; LeMay, H. E., Jr.; Bursten, B. E.; Murphy, C. J.; Woodward, P. M.; Stoltzfus, M. W. Chemistry: The Central Science, 15th ed.; Pearson, 2022.

Practise this topic with flashcards and a quiz at chemistryclarity.com/chemistry/bond-enthalpies/

Spotted a mistake? Let us know and we'll fix it.