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Enthalpy of Combustion

  • Always associated with burning in oxygen.

  • Process is exothermic.

  • The products of complete combustion need to be known, so carbon monoxide and sulphur trioxide do not count! Only carbon dioxide and sulphur dioxide.If you are combustion sulphur or carbon!

  • Need to be precise about the reactants if there are different forms of the same element in existence.


S(rhombic, solid) + O2(g) DHq = -296.9 kJ mol-1

S(monoclinic, solid) + O2(g) DHq = -297.2kJ mol-1

The most important of these are organic compounds, hydrocarbons (methane), and carbohydrates (glucose). The products are always, carbon dioxide and water.


 CH4(g) + 2O2(g) CO2(g) + 2H2O(g)        DHq = -882 kJ mol-1

Sucrose gives more energy per mole.

C12H22O11(s) + 12O2(g) 12CO2(g) + 11H2O(g)        DHq = -5644 kJ mol-1

In the real world the water is either produced as liquid or gas depending on the end conditions. This will give a slightly different result.

Often the enthalpy of formation and combustion can appear to be the same, especially if the formed compound is an oxide.


            C(graphite, solid) + O2(g)    CO2(g)        DHq = -393 kJ mol-1

As well as being the enthalpy of combustion of carbon in its standard state (graphite), this is also the enthalpy of formation of carbon dioxide (see definition earlier).

Using diamond however would result in an incorrect value for the formation of CO2, as diamond is less thermodynamically stable than graphite.

An enthalpy of formation must also be 1 mole of the substance, so the following would not be an enthalpy of formation.

            2Li(s) + O2(g) Li2O(s)                  DHq = -596 kJ mol-1