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7.3 Properties and combustion of hydrocarbons

How boiling point, viscosity and flammability change with the chain length of hydrocarbons, how this affects their use as fuels, and the complete combustion of hydrocarbons with balanced equations.

AQA Edexcel OCR 1 min read

Properties depend on chain length

Key terms

  • Viscosity: how thick a liquid is; how easily it flows.
  • Flammability: how easily a substance catches fire.

As the number of carbon atoms (chain length) in a hydrocarbon increases:

  • boiling point increases: bigger molecules have stronger intermolecular forces, so more energy is needed to separate them
  • viscosity increases: long chains tangle and flow less easily
  • flammability decreases: long-chain hydrocarbons are harder to ignite
Bar chart of alkane boiling points rising from −162 °C for methane to 126 °C for octane, with boiling point and viscosity increasing and flammability decreasing as chains get longer, and the complete combustion equation for methane
Bar chart of alkane boiling points rising from −162 °C for methane to 126 °C for octane, with boiling point and viscosity increasing and flammability decreasing as chains get longer, and the complete combustion equation for methane

So short-chain hydrocarbons make the best fuels: they are runny, evaporate easily and catch fire easily.

Short and long chains
PropertyShort chains (e.g. petrol)Long chains (e.g. fuel oil)
Boiling pointlowhigh
Viscosityrunnythick
Flammabilityvery flammablehard to ignite

Complete combustion

Key terms

  • Complete combustion: burning a fuel in plenty of oxygen, so that the carbon and hydrogen are fully oxidised to carbon dioxide and water.

hydrocarbon + oxygen → carbon dioxide + water

The reaction is exothermic: it releases a lot of energy, which is why hydrocarbons are used as fuels. The carbon and hydrogen are both oxidised.

Examples:

  • methane: CH4 + 2O2 → CO2 + 2H2O
  • propane: C3H8 + 5O2 → 3CO2 + 4H2O

Balancing combustion equations

  1. Balance the carbon atoms (gives the number of CO2).
  2. Balance the hydrogen atoms (gives the number of H2O).
  3. Balance the oxygen last. If you need half an O2, double everything.

Example: 2C2H6 + 7O2 → 4CO2 + 6H2O

Test yourself on this lesson

5 exam questions · 9 marks · with mark schemes and model answers

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