Bond Enthalpy Calculations with Worked Examples

Bond enthalpies let you estimate the energy change of a reaction without doing an experiment: add up the energy needed to break the bonds in the reactants, subtract the energy released by forming the bonds in the products. This article sets out the method once and then works through six examples of increasing difficulty, including a reverse calculation and an explanation of why the answers differ from measured values.

The method

Breaking bonds absorbs energy (endothermic, +). Making bonds releases energy (exothermic, −). So:

ΔH = Σ(bond enthalpies of bonds broken) − Σ(bond enthalpies of bonds formed)

or, more briefly, ΔH = broken − made.

• If more energy is released making bonds than is used breaking them, ΔH is negative (exothermic).
• If breaking takes more than making gives back, ΔH is positive (endothermic).

Steps

1. Write a balanced equation.
2. Draw displayed formulas so you can see every bond (see single, double and triple bonds).
3. List and count the bonds broken in the reactants.
4. List and count the bonds formed in the products.
5. Multiply by the bond enthalpies and add up each side.
6. Calculate broken − made. Include the sign and units.

Shortcut: bonds that appear unchanged on both sides (for example, C–H bonds that survive a reaction) can be left out, because they cancel. It’s safest, though, to count everything until you’re confident.

Data used in these examples

Average bond enthalpies in kJ mol⁻¹ (values differ slightly between data books; always use the ones given in your question):

Bond	kJ mol⁻¹	Bond	kJ mol⁻¹
H–H	436	C=O	805 (in CO₂)
Cl–Cl	243	C–O	358
H–Cl	432	O–H	463
C–H	413	O=O	498
C–C	347	N≡N	945
C=C	614	N–H	391
C–Cl	346	Br–Br	193
C–Br	290	H–Br	366

Example 1: hydrogen and chlorine

H₂(g) + Cl₂(g) → 2HCl(g)

• Broken: 1 × H–H + 1 × Cl–Cl = 436 + 243 = 679 kJ
• Made: 2 × H–Cl = 2 × 432 = 864 kJ
• ΔH = 679 − 864 = −185 kJ mol⁻¹

Exothermic: the two H–Cl bonds formed are stronger in total than the H–H and Cl–Cl bonds broken. This reaction can be explosive in sunlight, which splits Cl₂ into atoms to start a chain reaction.

Example 2: making ammonia

N₂(g) + 3H₂(g) → 2NH₃(g)

• Broken: 1 × N≡N + 3 × H–H = 945 + 3(436) = 945 + 1,308 = 2,253 kJ
• Made: 2 NH₃ each contain 3 N–H, so 6 × N–H = 6 × 391 = 2,346 kJ
• ΔH = 2,253 − 2,346 = −93 kJ mol⁻¹ (per mole of N₂, i.e. for 2 mol NH₃)

The measured value is −92 kJ for this equation, so the estimate is excellent here. Note how much of the “broken” total comes from the N≡N triple bond — that’s why the reaction needs a catalyst and high temperature to get going, even though it’s exothermic.

Example 3: combustion of methane

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

Displayed formulas: methane has 4 C–H bonds; each O₂ has one O=O; CO₂ is O=C=O (two C=O); each water has two O–H.

• Broken: 4 × C–H + 2 × O=O = 4(413) + 2(498) = 1,652 + 996 = 2,648 kJ
• Made: 2 × C=O + 4 × O–H = 2(805) + 4(463) = 1,610 + 1,852 = 3,462 kJ
• ΔH = 2,648 − 3,462 = −814 kJ mol⁻¹

The measured enthalpy of combustion of methane is −890 kJ mol⁻¹. Why the difference? Partly because bond enthalpies are averages, and partly because the standard enthalpy of combustion forms liquid water. Bond enthalpies only apply to gases, so our calculation gives water as a gas. Condensing 2 mol of steam releases about 2 × 44 = 88 kJ, bringing our estimate to about −902 kJ mol⁻¹, much closer to the measured value.

Example 4: adding bromine to ethene

CH₂=CH₂(g) + Br₂(g) → CH₂Br–CH₂Br(g)

Here we can use the shortcut. The four C–H bonds are unchanged, so leave them out.

• Broken: 1 × C=C + 1 × Br–Br = 614 + 193 = 807 kJ
• Made: 1 × C–C + 2 × C–Br = 347 + 2(290) = 347 + 580 = 927 kJ
• ΔH = 807 − 927 = −120 kJ mol⁻¹

The key idea: a C=C double bond (614) is replaced by a C–C single bond (347), so only about 267 kJ is spent breaking the π bond, but two new C–Br bonds are made. That’s why addition to alkenes is typically exothermic (see sigma and pi bonds).

Example 5: combustion of ethanol

C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(g)

Assume ethanol is a gas for the bond enthalpy calculation. Ethanol’s displayed formula: H₃C–CH₂–O–H.

Bonds in ethanol: 5 × C–H, 1 × C–C, 1 × C–O, 1 × O–H.

• Broken:
  • 5 × C–H = 5 × 413 = 2,065
  • 1 × C–C = 347
  • 1 × C–O = 358
  • 1 × O–H = 463
  • 3 × O=O = 3 × 498 = 1,494
  • Total = 4,727 kJ
• Made:
  • 4 × C=O (in 2 CO₂) = 4 × 805 = 3,220
  • 6 × O–H (in 3 H₂O) = 6 × 463 = 2,778
  • Total = 5,998 kJ
• ΔH = 4,727 − 5,998 = −1,271 kJ mol⁻¹

The measured standard enthalpy of combustion is −1,367 kJ mol⁻¹. Again, the gap comes from average bond enthalpies and states: real ethanol starts as a liquid (vaporising it costs about 42 kJ mol⁻¹) and the water ends as a liquid (condensing 3 mol releases about 132 kJ). Correcting for both gives about −1,271 + 42 − 132 ≈ −1,361 kJ mol⁻¹ — very close to the measured value.

Example 6: finding an unknown bond enthalpy

H₂(g) + Br₂(g) → 2HBr(g) ΔH = −103 kJ mol⁻¹

Given H–H = 436 and Br–Br = 193, calculate the H–Br bond enthalpy.

• Broken: 436 + 193 = 629 kJ
• Made: 2 × x
• ΔH = broken − made: −103 = 629 − 2x
• 2x = 629 + 103 = 732
• x = 366 kJ mol⁻¹

This matches the table value. Reverse calculations like this are common in exams: set up the same equation and solve for the unknown.

Why bond enthalpy answers are approximate

Reason	Effect
Average values: the actual bond enthalpy in a specific molecule differs from the average across many molecules	Small errors in each term add up
Gas phase only: bond enthalpies assume all species are gases	Liquids and solids involve extra energy changes (evaporation, condensation)
Neighbouring atoms affect bond strength	E.g. C=O in CO₂ (805) differs from C=O in aldehydes and ketones (about 745)

For accurate values, chemists use enthalpies of formation and Hess’s law, which are based on measurements of the specific compounds. Bond enthalpy calculations are best for estimates and for understanding why a reaction releases or absorbs energy.

Common mistakes

• Doing “made − broken”. It’s broken − made. (If you prefer, write bonds broken as + and bonds made as −, then add.)
• Miscounting bonds. Draw every displayed formula. Remember CO₂ has two C=O bonds, and each water has two O–H bonds.
• Forgetting coefficients. 3O₂ means three O=O bonds.
• Treating C=O as two C–O bonds. Use the C=O value.
• Missing the sign or units. Always write, for example, −814 kJ mol⁻¹.

Practice

Use the data table to estimate ΔH for:

1. CH₄(g) + Cl₂(g) → CH₃Cl(g) + HCl(g)
2. C₂H₆(g) → C₂H₄(g) + H₂(g)

Answers:

1. Broken: C–H + Cl–Cl = 413 + 243 = 656. Made: C–Cl + H–Cl = 346 + 432 = 778. ΔH = −122 kJ mol⁻¹.
2. Broken: C–C + 6 C–H = 347 + 2,478 = 2,825. Made: C=C + 4 C–H + H–H = 614 + 1,652 + 436 = 2,702. ΔH = +123 kJ mol⁻¹ (endothermic — this is why cracking needs high temperatures).

Key takeaways

• ΔH = bonds broken − bonds made.
• Breaking absorbs energy; making releases it.
• Draw displayed formulas and count carefully, including coefficients.
• Results are estimates: averages and gas-phase assumptions cause differences from measured values.

For the ideas behind these numbers, read bond length and bond strength and exothermic vs endothermic reactions.