Breaking bonds absorbs energy; forming bonds releases it
In a chemical reaction, the bonds in the reactants break, the atoms rearrange, and the bonds of the products form. The potential energy curve from U2.2 shows that potential energy drops when two atoms bond, so:
- breaking a chemical bond absorbs energy;
- forming a chemical bond releases energy.
The energy needed to break 1 mol of a particular covalent bond in gaseous molecules is its bond enthalpy键焓, which is always positive. The larger the bond enthalpy, the stronger the bond. Between the same two atoms, a higher bond order means a shorter, stronger bond: C–C 348, C=C 614, C≡C 839 kJ/mol.
Estimating ΔH from bond enthalpies
Imagine the reaction in two steps: first break every bond in the reactants to get gaseous atoms (absorbing Σbroken), then let the atoms form the bonds of the products (releasing Σformed). So:
- If forming bonds releases more energy than breaking bonds absorbs, ΔH < 0 and the reaction is exothermic.
- If breaking bonds absorbs more energy than forming bonds releases, ΔH > 0 and the reaction is endothermic.
In practice, count only the bonds that change; bonds present both before and after cancel out.
Choose a reaction to see the two stages on the energy diagram, up to gaseous atoms and down to products, and compare with ΔH from the table of enthalpies of formation:
Why it’s only an estimate
Table values are average bond enthalpies: the same kind of bond (C–H, for example) is slightly stronger or weaker in different molecules, and the table gives an average over many molecules. Bond enthalpies also apply to gases; when liquids or solids take part, intermolecular forces matter too (for example, the heat released when water vapor condenses). So ΔH from bond enthalpies differs from ΔH from the table of enthalpies of formation by a few to a few tens of kJ/mol.
Working the other way, if ΔH for a reaction is known, you can treat one bond as the unknown and solve for its bond enthalpy. Click a bond to try it, and see how far the result is from the average bond enthalpy table:
Click a bond to make it the unknown x; the other bonds use average bond enthalpies.
Example 1: Count only the bonds that change
Estimate ΔH for C₂H₂(g) + 2H₂(g) → C₂H₆(g) from average bond enthalpies (kJ/mol): C≡C 839, H–H 436, C–C 348, C–H 413.
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Identify which bonds change. Draw the structures: C₂H₂ is H–C≡C–H and C₂H₆ is H₃C–CH₃. The 2 C–H bonds already in C₂H₂ are still there in C₂H₆, so leave them out:
- Broken: 1 C≡C, 2 H–H
- Formed: 1 C–C, 4 new C–H
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Energy absorbed breaking bonds. In kJ:
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Energy released forming bonds.
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Estimate ΔH. Broken minus formed:
Negative, so exothermic.
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Compare with the accurate value. The table of enthalpies of formation gives −311.4 kJ/mol, about 20 kJ/mol different: average bond enthalpies are averages over many molecules and don’t exactly match the actual bonds in these two, so they only give an estimate.
Example 2: From ΔH back to a bond enthalpy
For H₂(g) + Br₂(g) → 2HBr(g), ΔH = −103.7 kJ/mol. The bond enthalpy of H–H is 436 kJ/mol and that of Br–Br is 193 kJ/mol. Estimate the bond enthalpy of H–Br.
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List the bonds broken and formed. 1 H–H and 1 Br–Br are broken; 2 H–Br are formed. Let x be the H–Br bond enthalpy.
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Set up the equation. ΔH = energy absorbed breaking − energy released forming:
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Rearrange and solve.
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Substitute back.
This matches −103.7 (the difference is rounding).
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Connect to bond strength. ΔH < 0, so forming 2 mol of H–Br releases more energy (about 733 kJ) than breaking the H–H and Br–Br bonds absorbs (629 kJ); H–Br (366) is much stronger than Br–Br (193).
- H₂(g) → 2H(g)
An H–H bond breaks.
- 2Cl(g) → Cl₂(g)
A Cl–Cl bond forms.
- CH₄(g) → C(g) + 4H(g)
Four C–H bonds break.
- H(g) + Cl(g) → HCl(g)
An H–Cl bond forms.
- N₂(g) → 2N(g)
The N≡N triple bond breaks, which takes a lot of energy (941 kJ/mol).
- 2O(g) → O₂(g)
An O=O bond forms.
- H₂O(l) → H₂O(g)
Intermolecular forces are overcome; no covalent bonds break, but energy is still absorbed.
- Na⁺(g) + Cl⁻(g) → NaCl(s)
The ions attract each other and form a crystal, releasing energy.
Common mistakes
Mistake 1: “Breaking chemical bonds releases energy.” Breaking bonds always absorbs energy; only forming bonds releases energy. Burning a fuel releases heat because forming the bonds in the products releases more energy than breaking the bonds in the reactants absorbs.
Mistake 2: “ΔH = Σformed − Σbroken.” That’s backward. Breaking absorbs (positive) and forming releases (negative), so ΔH ≈ Σbroken − Σformed.
Mistake 3: “ΔH from bond enthalpies matches the experimental value exactly.” Average bond enthalpies are averages over many molecules and apply only to gases, so the ΔH they give is an estimate.
Quick check
Use bond enthalpies (kJ/mol: H–H 436, Cl–Cl 242, H–Cl 431) to estimate ΔH for H₂(g) + Cl₂(g) → 2HCl(g).
(436 + 242) − 2 × 431 = 678 − 862 = −184 kJ/mol. C forgets that 2 H–Cl bonds form.
A gas-phase reaction is exothermic. Which statement is correct?
ΔH ≈ Σbroken − Σformed < 0, so forming bonds releases more. The comparison is of energy, not numbers of bonds.
Use bond enthalpies (kJ/mol: H–H 436, F–F 155, I–I 151, H–F 567, H–I 299) to compare H₂(g) + F₂(g) → 2HF(g) with H₂(g) + I₂(g) → 2HI(g). The first is far more exothermic. What is the main reason?
Both reactions break 1 H–H and 1 X–X bond and form 2 H–X bonds. Estimates: 436 + 155 − 2 × 567 = −543 kJ/mol and 436 + 151 − 2 × 299 = −11 kJ/mol. Nearly all of the difference comes from the strength of the H–X bonds.
For N₂(g) + 3H₂(g) → 2NH₃(g), ΔH = −92.2 kJ/mol; N≡N is 941 kJ/mol and H–H is 436 kJ/mol. Estimate the N–H bond enthalpy.
941 + 3 × 436 − 6x = −92.2, so 6x = 2341.2 and x ≈ 390 kJ/mol. A uses the wrong sign for ΔH; B counts only 3 N–H bonds.
Breaking the bond in 1 mol of which molecule takes the most energy: N₂, O₂, or F₂?
N≡N 941, O=O 495, F–F 155 kJ/mol. A higher bond order means a stronger bond.
Bond enthalpies give about −485 kJ/mol for 2H₂(g) + O₂(g) → 2H₂O(l), while the table of enthalpies of formation gives −571.6 kJ/mol. What mainly explains the difference?
Bond enthalpies give ΔH for forming H₂O(g). Condensing 2 mol of water vapor releases about 2 × 44.0 = 88 kJ more (at 25 °C).
AP-style free response
Methane burns with all substances gaseous: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g). Average bond enthalpies (kJ/mol): C–H 413, O=O 495, C=O 799, O–H 463.
(a) List the bonds broken and formed in the reaction, with how many of each.
(b) Estimate ΔH for the reaction from the average bond enthalpies.
(c) Explain in terms of bond breaking and bond forming why the reaction is exothermic.
(d) ΔH for this reaction from the table of enthalpies of formation is −802.3 kJ/mol. Using average values for the other bonds, calculate the enthalpy of the C=O bond in CO₂.
(e) For CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), ΔH = −890.3 kJ/mol, which is more negative than the value in (d). Explain why.
Model answer and scoring
(a) Broken: 4 C–H and 2 O=O; formed: 2 C=O and 4 O–H.
(b) Broken: 4 × 413 + 2 × 495 = 2642 kJ; formed: 2 × 799 + 4 × 463 = 3450 kJ; ΔH ≈ 2642 − 3450 = −808 kJ/mol.
(c) Forming the C=O and O–H bonds releases more energy (3450 kJ) than breaking the C–H and O=O bonds absorbs (2642 kJ); the difference is released as heat.
(d) 2642 − (2x + 4 × 463) = −802.3, so 2x = 2642 − 1852 + 802.3 = 1592.3 and x ≈ 796 kJ/mol.
(e) To form liquid water, 2 mol of water vapor must also condense, and condensation releases heat (about 44.0 kJ per mole): −802.3 − 88.0 = −890.3 kJ/mol.
Scoring: (a) kinds and numbers of bonds; (b) Σbroken − Σformed; (c) forming releases more energy than breaking absorbs; (d) setup and solution; (e) condensation releases heat.
Summary
- Breaking bonds absorbs energy; forming bonds releases energy. ΔH ≈ Σbroken − Σformed, counting only the bonds that change.
- If forming bonds releases more energy than breaking bonds absorbs, the reaction is exothermic; otherwise it is endothermic.
- With a known ΔH, you can treat one bond as the unknown and solve for its bond enthalpy.
- Average bond enthalpies are averages over many molecules and apply only to gases, so the ΔH they give is an estimate.
AP tip: To explain whether a reaction is endothermic or exothermic, compare “energy absorbed to break the bonds in the reactants” with “energy released when the bonds in the products form”; never say that breaking bonds releases energy. For bond enthalpy calculations, draw or write the structures first and count each kind of bond broken and formed.