Chemistry » Thermochemistry » Enthalpy

# Enthalpy of Combustion

## Enthalpy of Combustion

Standard enthalpy of combustion$$\left(\text{Δ}{H}_{C}^{\text{°}}\right)$$ is the enthalpy change when 1 mole of a substance burns (combines vigorously with oxygen) under standard state conditions; it is sometimes called “heat of combustion.” For example, the enthalpy of combustion of ethanol, −1366.8 kJ/mol, is the amount of heat produced when one mole of ethanol undergoes complete combustion at 25 °C and 1 atmosphere pressure, yielding products also at 25 °C and 1 atm.

$${\text{C}}_{2}{\text{H}}_{5}\text{OH}\left(l\right)+3{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}2{\text{CO}}_{2}+3{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{298}^{°}=\text{−1366.8 kJ}$$

Enthalpies of combustion for many substances have been measured; a few of these are listed in the table below. Many readily available substances with large enthalpies of combustion are used as fuels, including hydrogen, carbon (as coal or charcoal), and hydrocarbons (compounds containing only hydrogen and carbon), such as methane, propane, and the major components of gasoline.

Standard Molar Enthalpies of Combustion
SubstanceCombustion ReactionEnthalpy of Combustion, $$\text{Δ}{H}_{c}^{°}$$$$\left(\frac{\text{kJ}}{\text{mol}}\phantom{\rule{0.2em}{0ex}}\text{at}\phantom{\rule{0.2em}{0ex}}25\phantom{\rule{0.2em}{0ex}}\text{°C}\right)$$
carbon$$\text{C}\left(s\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}\left(g\right)$$−393.5
hydrogen$${\text{H}}_{2}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\cfrac{1}{2}\phantom{\rule{0.1em}{0ex}}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O}\left(l\right)$$−285.8
magnesium$$\text{Mg}\left(s\right)+\phantom{\rule{0.1em}{0ex}}\cfrac{1}{2}\phantom{\rule{0.1em}{0ex}}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{MgO}\left(s\right)$$−601.6
sulfur$$\text{S}\left(s\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{SO}}_{2}\left(g\right)$$−296.8
carbon monoxide$$\text{CO}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\cfrac{1}{2}\phantom{\rule{0.1em}{0ex}}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}\left(g\right)$$−283.0
methane$${\text{CH}}_{4}\left(g\right)+2{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}\left(g\right)+2{\text{H}}_{2}\text{O}\left(l\right)$$−890.8
acetylene$${\text{C}}_{2}{\text{H}}_{2}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\cfrac{5}{2}\phantom{\rule{0.1em}{0ex}}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}2{\text{CO}}_{2}\left(g\right)+{\text{H}}_{2}\text{O}\left(l\right)$$−1301.1
ethanol$${\text{C}}_{2}{\text{H}}_{5}\text{OH}\left(l\right)+3{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{2CO}}_{2}\left(g\right)+3{\text{H}}_{2}\text{O}\left(l\right)$$−1366.8
methanol$${\text{CH}}_{3}\text{OH}\left(l\right)+\phantom{\rule{0.1em}{0ex}}\cfrac{3}{2}\phantom{\rule{0.1em}{0ex}}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}\left(g\right)+2{\text{H}}_{2}\text{O}\left(l\right)$$−726.1
isooctane$${\text{C}}_{8}{\text{H}}_{18}\left(l\right)+\phantom{\rule{0.1em}{0ex}}\cfrac{25}{2}\phantom{\rule{0.1em}{0ex}}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}8{\text{CO}}_{2}\left(g\right)+9{\text{H}}_{2}\text{O}\left(l\right)$$−5461

## Example: Using Enthalpy of Combustion

As the figure below suggests, the combustion of gasoline is a highly exothermic process. Let us determine the approximate amount of heat produced by burning 1.00 L of gasoline, assuming the enthalpy of combustion of gasoline is the same as that of isooctane, a common component of gasoline. The density of isooctane is 0.692 g/mL.

The combustion of gasoline is very exothermic. (credit: modification of work by “AlexEagle”/Flickr)

SolutionStarting with a known amount (1.00 L of isooctane), we can perform conversions between units until we arrive at the desired amount of heat or energy. The enthalpy of combustion of isooctane provides one of the necessary conversions. The table above gives this value as −5460 kJ per 1 mole of isooctane (C8H18).

Using these data,

$$\require{cancel}1.00\phantom{\rule{0.2em}{0ex}}\cancel{\text{L}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{8}{\text{H}}_{18}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\cfrac{1000\phantom{\rule{0.2em}{0ex}}\cancel{\text{mL}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{8}{\text{H}}_{18}}}{1\phantom{\rule{0.2em}{0ex}}\cancel{\text{L}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{8}{\text{H}}_{18}}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\cfrac{0.692\phantom{\rule{0.2em}{0ex}}\cancel{\text{g}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{8}{\text{H}}_{18}}}{1\phantom{\rule{0.2em}{0ex}}\cancel{\text{mL}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{8}{\text{H}}_{18}}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\cfrac{1\phantom{\rule{0.2em}{0ex}}\cancel{\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{8}{\text{H}}_{18}}}{114\phantom{\rule{0.2em}{0ex}}\cancel{\text{g}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{8}{\text{H}}_{18}}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\cfrac{-5460\phantom{\rule{0.2em}{0ex}}\text{kJ}}{1\phantom{\rule{0.2em}{0ex}}\cancel{\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{8}{\text{H}}_{18}}}\phantom{\rule{0.2em}{0ex}}=-3.31\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\text{kJ}$$

The combustion of 1.00 L of isooctane produces 33,100 kJ of heat. (This amount of energy is enough to melt 99.2 kg, or about 218 lbs, of ice.)

Note: If you do this calculation one step at a time, you would find:

$$\begin{array}{l}\\ 1.00\phantom{\rule{0.2em}{0ex}}\text{L}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{8}{\text{H}}_{18}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}1.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{mL}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{8}{\text{H}}_{18}\\ 1.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{mL}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{8}{\text{H}}_{18}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}692\phantom{\rule{0.2em}{0ex}}\text{g}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{8}{\text{H}}_{18}\\ 692\phantom{\rule{0.2em}{0ex}}\text{g}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{8}{\text{H}}_{18}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}6.07\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{8}{\text{H}}_{18}\\ 692\phantom{\rule{0.2em}{0ex}}\text{g}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{8}{\text{H}}_{18}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}-3.31\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\text{kJ}\end{array}$$