Heat of combustion is the heat released when one mole of a substance is completely burnt in excess oxygen, for example C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l). It is measured from the temperature rise of water using Q = mcθ.
The heat of combustion is the heat released when one mole of a substance is completely burnt in excess oxygen. Combustion is strongly exothermic, which is why fuels release energy, and the Form 5 Thermochemistry chapter uses the alcohols as the standard fuels to study. The exam tests the balanced equation, the definition, the calculation from a temperature rise, the trend down the series, and the reasons an experimental value falls short of the theoretical one.
The balanced equation
Complete combustion turns every carbon atom into carbon dioxide and every hydrogen atom into water. For ethanol:
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)
Other fuels follow the same pattern, for example methanol and propan-1-ol:
2CH3OH(l) + 3O2(g) → 2CO2(g) + 4H2O(l)
2C3H7OH(l) + 9O2(g) → 6CO2(g) + 8H2O(l)
The value quoted as the heat of combustion is always per one mole of the fuel burnt, so the equation is written for one mole of fuel where possible.
Conditions required
- The fuel is burnt completely in excess (plenty of) oxygen, so that carbon dioxide and water are the only products.
- The heat released is directed at a known mass of water in a metal container (for example a copper can), whose temperature rise is measured.
- The apparatus is shielded from draughts to reduce heat loss.
Observations
- The temperature of the water rises, showing the reaction is exothermic.
- The flame should be a clean blue flame; a yellow, sooty flame signals incomplete combustion, which lowers the heat delivered.
- The mass of fuel burnt is found by weighing the burner before and after.
How the value is calculated
The heat gained by the water is Q = mcθ, where m is the mass of water, c is its specific heat capacity (4.2 J per g per degree C) and θ is the temperature rise. Dividing this heat by the number of moles of fuel burnt gives the heat of combustion per mole. The step that most often loses marks is converting the mass of fuel to moles using its molar mass, so keep the moles of fuel and the heat released clearly linked.
The trend down the alcohol series
As you go from methanol to ethanol to propanol, each fuel has one more -CH2− unit, so more carbon and hydrogen atoms are burnt and more energy is released per mole. The heat of combustion therefore increases steadily down the homologous series. Plotting heat of combustion against the number of carbon atoms gives a rising line, a graph the exam likes to set.
Why experiment falls below theory
The measured heat of combustion is usually lower than the theoretical value for three reasons: heat is lost to the surroundings and the apparatus rather than all reaching the water; the fuel may burn incompletely, releasing less heat; and some fuel vapour may escape without burning. Naming these losses, and suggesting an insulated or shielded set-up to reduce them, is a common evaluation question.
Common mistakes to avoid
Define the quantity per one mole of fuel completely burnt. Use the mass of the water, not the fuel, in Q = mcθ, then divide by moles of fuel. Give complete-combustion products (carbon dioxide and water), and remember state symbols. Explain the low experimental value by heat loss and incomplete combustion, not by “the thermometer being wrong”.
How it appears in the SPM exam
In Paper 2 (4541/2) you may write the combustion equation, define heat of combustion, calculate it from data with Q = mcθ, interpret the trend down the series, and evaluate sources of heat loss. In Paper 3 (4541/3), heating water with a burning fuel and recording the temperature rise is a set practical. Keep one mole of fuel and the water’s temperature rise at the centre, and every heat-of-combustion question stays manageable.
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