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Standard Enthalpy of Combustion

Standard Enthalpy of Combustion: ΔHθc,298

Standard enthalpy of combustion is defined as the enthalpy change when 1 mole of a compound is completely burnt in oxygen gas at 298K and 1 bar pressure.


C8H18(l) + 12½O2(g) → 8CO2(g) + 9H2O(l) : ΔHθc,298 = -5512kJ.mol-1

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) : ΔHθc,298 = -2816kJ.mol-1

Many enthalpy changes are very difficult to measure directly, particularly ΔHθf values, which are often impossible to measure directly, e.g. the standard heat of formation of ethanol, ΔHθf,298 [CH3CH2OH(l)] is given by the equation:

2C(graphite) + 3H2(g) + ½O2(g) → CH3CH2OH(l) : ΔHθf,298 = -278kJ.mol-1

There is very little chance indeed of graphite combining with hydrogen gas and oxygen gas at 298K and 1 bar pressure to give liquid ethanol. So how do we know that the enthalpy change is -278kJ.mol-1?

Graphite, hydrogen and ethanol all readily burn in oxygen to give combustion products. Using the law of conservation of energy we can set up a simple cycle.

In general terms

For ethanol

If you examine the two Hess Cycles above you should see the following relationship: ΔH clockwise = ΔH anticlockwise.

Click on the expressions to complete the equations listed below.

ΔH1 = ?

ΔH2 = +

ΔH3 =

Click on the appropriate option.  ΔHθf,298 [CH3CH2OH(l)]   ΔHθc,298 [C(graphite)]   2ΔHθc,298 [C(graphite)]   ΔHθc,298 [H2(g)]   2ΔHθc,298 [H2(g)]   3ΔHθc,298 [H2(g)]   ΔHθc,298 [CH3CH2OH(l)]

If you wish to use the standard data without answering the questions or carrying out the practical, press , or

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