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Lesson 4 · 25 min

Bond enthalpies and fuels

Breaking bonds costs, forming bonds pays: average bond enthalpies estimate gas-phase reaction enthalpies — until a molecule like benzene breaks the estimate. Why do fuels release heat? The answer hides in the weak O=O bond.

In this lesson

  • Define bond enthalpy and see why breaking bonds costs energy while forming them releases it
  • Estimate ΔH of gas-phase reactions from average bond enthalpies
  • Explain why bond estimates differ from the formation-enthalpy route (averaging, physical state, resonance)
  • Compare fuels per gram, per mole of O₂ and per MJ of CO₂ emitted

The Space Shuttle’s main engines burned liquid hydrogen, a car burns petrol, your cells “burn” glucose. Per gram, hydrogen releases nine times as much heat as glucose; yet per mole of oxygen consumed, petrol, natural gas, alcohol and glucose all release almost the same, between 400 and 480 kJ. Where does the heat of burning come from? The answer lies in chemical bonds — though not in any “energy stored in the fuel’s bonds”.

Predict first

Estimate ΔH for CHX4(g)+2 OX2(g)⟶COX2(g)+2 HX2O(g)\ce{CH4(g) + 2O2(g)} \longrightarrow \ce{CO2(g) + 2H2O(g)} from average bond enthalpies (kJ/mol): C–H 413, O=O 495, C=O 799, O–H 463. Which is closest?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

Bond enthalpy: the price of breaking one bond

The bond enthalpy D(X−Y)D(\mathrm{X{-}Y}) is the enthalpy change for breaking one mole of X–Y bonds in the gas phase into two gaseous fragments — for example HX2(g)→2 H(g)\ce{H2(g)} \mathrel{\ce{->}} \ce{2H(g)}, D=436D = 436 kJ/mol. It is always positive: pulling apart two atoms that attract each other always costs energy. Higher bond order and smaller atoms make stronger bonds: C–C 348, C=C 614, C≡C 839 kJ/mol.

The same kind of bond is not equally strong in every molecule — the C–H bonds of methane average 416 kJ/mol, while in other molecules C–H ranges from about 380 to over 440. Tables therefore list average bond enthalpies over many molecules. To estimate a reaction enthalpy with them, imagine the reaction in two steps: break every bond in the reactants to make gaseous atoms, then assemble those atoms into the bonds of the products.

ΔrH≈∑D(bonds broken)−∑D(bonds formed)\Delta_\mathrm{r}H \approx \sum D(\text{bonds broken}) - \sum D(\text{bonds formed})
Valid only when reactants and products are all gases.
DD
average bond enthalpy / (kJ/mol), always positive
Breaking costs, forming pays · checked against formation data

Bond-enthalpy ledger

NX2(g)+3 HX2(g)⟶2 NHX3(g)\ce{N2(g)} + 3\,\ce{H2(g)} \longrightarrow 2\,\ce{NH3(g)}

Bonds broken (absorbed)

  • 3 × H–H × 436 +1308
  • 1 × N≡N × 941 +941
  • Σ+2249 kJ

Bonds formed (released)

  • 6 × N–H × 391 −2346
  • Σ−2346 kJ
Bond estimate ΔH ≈ Σbroken − Σformed
−97 kJ
ΔrH° from formation data
−91.9 kJ
Error of the estimate
−5 kJ (−6 %)
reactants gaseous atoms +2249 products (estimate) −97 products (measured) −91.905001000150020002500Enthalpy relative to reactants / kJ

Molecule by molecule: sum of average bonds vs actual atomisation enthalpy

MoleculeBondsΣ average DActual ΔatHDifference
NX2(g)\ce{N2(g)}N≡N941945.4−4.4
HX2(g)\ce{H2(g)}H–H436436.0+0.0
NHX3(g)\ce{NH3(g)}3 N–H11731172.6+0.4
Computed: average bond enthalpies from Brown et al., Chemistry: The Central Science, Table 8.4 (the same table as the bonding unit); the “measured” route uses 298 K formation enthalpies, with atomisation enthalpy = Σ ΔfH°(gaseous atoms) − ΔfH°(molecule) (CODATA). The two routes differ only because of the averaging: the same bond is not equally strong in different molecules. Benzene is counted as one Kekulé structure. Illustrative: the horizontal placement in the level diagram.
Compare them one by one: ammonia synthesis and hydrogen + chlorine come within a few per cent; hydrogenating benzene misses by about 170 kJ — look at benzene’s row in the table: its real atomisation enthalpy exceeds “three single + three double bonds” by 161 kJ.

Why estimates miss

Bond estimates differ from the formation-enthalpy route for three reasons:

  1. Averaging: the table values average over many molecules; in a particular molecule a bond can be stronger or weaker, typically giving errors of ±5–10 %.
  2. Physical state: bond enthalpies hold for gases only. If the product is liquid water you must add its enthalpy of condensation (−44 kJ per mole) — the attraction between molecules in a liquid sits in no covalent bond at all.
  3. Delocalisation and strain: benzene’s π electrons spread over the whole ring, making it about 150–160 kJ/mol more stable than three isolated double bonds (the resonance energy); small rings such as cyclopropane are strained and their bonds weaker than average.

The exact approach uses atomisation enthalpies: the formation enthalpies of gaseous atoms give the total bond energy of any molecule, ΔatH=∑ΔfH∘(atoms,g)−ΔfH∘(molecule,g)\Delta_\mathrm{at}H = \sum \Delta_\mathrm{f}H^\circ(\text{atoms}, \mathrm{g}) - \Delta_\mathrm{f}H^\circ(\text{molecule}, \mathrm{g}). Using these instead of averages reproduces the formation-enthalpy route exactly.

Worked example

Burning methane: the books, line by line

Estimate ΔH for CHX4(g)+2 OX2(g)⟶COX2(g)+2 HX2O(l)\ce{CH4(g) + 2O2(g)} \longrightarrow \ce{CO2(g) + 2H2O(l)} (note: liquid water) from average bond enthalpies and compare with the exact −890.6 kJ.

Fuels: where the heat comes from

Fuels can be compared with different rulers:

  • Per gram: what rockets and aircraft care about. Hydrogen 142 kJ/g, natural gas 55.5, petrol (octane) 47.9, ethanol 29.7, glucose 15.6 kJ/g.
  • Per mole of O₂: fuels made of C, H and O all land near 400–480 kJ (Thornton’s rule). Burning releases heat mainly because the O=O bond in oxygen (498 kJ/mol) is unusually weak per oxygen atom; when it is traded for the strong C=O bonds of CO₂ and O–H bonds of water, the difference comes out as heat. The fuel’s own bonds do not “store” especially much energy.
  • CO₂ per MJ: the more hydrogen a fuel contains, the less CO₂ it emits for the same heat — natural gas about 49 g/MJ, petrol 64 g/MJ, coal (as pure carbon) 112 g/MJ.

Also distinguish the higher heating value (liquid water formed) from the lower heating value (water vapour formed): most burners exhaust vapour, so the lower value is what they can actually use.

Every combustion enthalpy computed from formation data

Comparing fuels

What rockets and aircraft care about: heat per gram carried. Hydrogen leads by far — it is the lightest.

kJ/g

Selected fuel: Methane CHX4(g)\ce{CH4(g)}

CHX4(g)+2 OX2(g)⟶COX2(g)+2 HX2O(l)\ce{CH4(g)} + 2\,\ce{O2(g)} \longrightarrow \ce{CO2(g)} + 2\,\ce{H2O(l)}

−ΔcH°
890.6 kJ/mol
Molar mass
16.04 g/mol
Heat per gram
55.51 kJ/g
Heat per mole of O₂
445 kJ
CO₂ emitted per MJ
49 g

Methane: 55.5 kJ per gram, 445 kJ per mole of O₂, 49 g CO₂ per MJ.

Computed: combustion equations are balanced automatically from the formula (C → CO₂(g), H → H₂O, N → N₂) and −ΔcH° comes from 298 K formation enthalpies; the higher heating value uses liquid water, the lower heating value water vapour. Illustrative: bar colours only group the fuels, and every row carries its name and value. Real fuels (petrol, coal, natural gas) are mixtures with values close to, but not identical with, these pure substances.
Switch the yardstick: per gram, hydrogen is far ahead; per mole of O₂, everything but hydrogen and ammonia lines up; per MJ of CO₂, coal comes last. Then switch the water to vapour and see how much each fuel loses.

Concept check

Which statement about breaking a chemical bond is correct?

Try the calculation

From average bond enthalpies (C=C 614, H–H 436, C–C 348, C–H 413 kJ/mol), estimate ΔH (kJ) for hydrogenating ethene, CX2HX4(g)+HX2(g)⟶CX2HX6(g)\ce{C2H4(g) + H2(g)} \longrightarrow \ce{C2H6(g)}.

kJ

Key takeaways

  • Bond enthalpy D is always positive: breaking bonds absorbs energy, forming them releases it.
  • For gas reactions ΔrH ≈ ΣD(broken) − ΣD(formed); averaging costs ±5–10 %.
  • Liquids need their phase-change enthalpy added; delocalised molecules such as benzene are far more stable than the bond sum (resonance energy about 150–160 kJ/mol).
  • Combustion heat comes mainly from trading the weak O=O bond for strong C=O and O–H bonds: C/H/O fuels release about 400–480 kJ per mole of O₂.
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Data and references

  1. T. L. Brown et al., Chemistry: The Central Science, 14th ed., Pearson (2018), Table 8.4 (average bond enthalpies).
  2. J. D. Cox, D. D. Wagman, V. A. Medvedev, CODATA Key Values for Thermodynamics (1989) — enthalpies of formation of gaseous atoms.
  3. K. Schmidt-Rohr, “Why combustions are always exothermic, yielding about 418 kJ per mole of O₂”, J. Chem. Educ. 92, 2094 (2015).
  4. W. M. Thornton, “The relation of oxygen to the heat of combustion of organic compounds”, Phil. Mag. 33, 196 (1917).
  5. W. M. Haynes (ed.), CRC Handbook of Chemistry and Physics, 97th ed. (2016), standard thermodynamic properties (combustion enthalpies computed from them).