Explain why the standard enthalpy of combustion of propane (C₃H₈) is more negative than that of methane (CH₄).

IB DP Chemistry Standard Level (2023 syllabus) — R1.3 Energy from fuels · Explain · 4 marks · View as Markdown

Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).

Standard enthalpies of combustion: methane = −890 kJ mol⁻¹; propane = −2220 kJ mol⁻¹.

Model answer (4 marks)

Propane contains more C–H and C–C bonds than methane, so its combustion breaks more bonds. The reaction forms the same CO₂ and H₂O products, but because the fuel has more bonds, more bonds are formed in the products. The energy released in forming these bonds outweighs the energy absorbed in breaking the fuel bonds by a larger amount for propane, giving a more negative enthalpy of combustion.

Examiner tips

Use the word ‘more negative’ to show exothermicity; mention bond breaking/formation; quantify that propane has more bonds; link to larger energy release.
Keep answer concise – 4 marks can be earned with 3–4 short points.

Common mistakes

Confusing the sign of ΔH (writing positive instead of negative); not mentioning bond breaking/formation; ignoring that both fuels form the same products; over‑expanding with unrelated details.

Mark scheme (4 marks)

Propane has more carbon–hydrogen (and carbon–carbon) bonds than methane / propane has a greater number of covalent bonds.
Combustion involves breaking bonds in the fuel and forming bonds in the products (CO₂ and H₂O).
More bonds are formed in the products when propane combusts, releasing more energy overall.
The enthalpy of combustion is more negative (more exothermic) for propane because the total energy released forming bonds exceeds the total energy absorbed breaking bonds by a greater amount than for methane.

Key terms in this question

standard enthalpy of combustion