combustion equation

Study of the combustion equation of the complete combustion of a hydrocarbon applied to engine pollution control.

More: forum on combustion

We start from the generic formula of complete combustion of alkanes:

CnH(2n+2) + (3n+1)/2*(O2+3.76N2) –> nCO2 + (n+1)H2O+(3n+1)/2*3.76N2

1) Volume study of the complete combustion equation:

Considering the exhaust gases in CNPT.
1 25 mole gas = L
Let's talk about the combustion of one mole of CnH fuel (2n + 2)

The previous equation therefore gives us the exhaust:
25n L CO2
25 (n + 1) L H2O
25 (3n 1 +) / 2 3.76 * L N2

A total of 25n 25 + (n + 1) + 25 (3n 1 +) / * 2 3.76 25 = (7.64n 2.88 +) = n + 191 72 L gas.

Note: For n = 0 the 72 L correspond to the mole of H2O and to the 1.88 moles of N2 resulting from the combustion of pure hydrogen.

For a given alkane so we have respectively:

25n / (191n 72 +)% of CO2
25 (n + 1) / (191n 72 +)% of H2O
(25(3n+1)/2*3.76)/(191n+72) % de N2

A division by 25 simplify formulas.

This is valid in the case of complete combustion (no creation of CO or particles) and ideal (no creation of Nox)

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2) Mass study of the complete combustion equation:

Let us study the mass rejections of the complete equation.

[CO2]=12+2*16=44 g/mol
[H2O] = 2 1 * + = 16 18 g / mol
[N2] = 2 14 * = 28g / mol

The calculation on the N2 is useless in the case of an ideal combustion (no creation of Nox) since this element does not intervene, it is an inert gas.

therefore the respective masses would be:
for CO2: 44n
for H2O: 18 (n + 1)

Application to gasoline (pure octane). n = 8
[C8H18] = 8 12 * + = 18 1 114 * g / mol.
The mass of CO2 released per mole of octane consumed is: 44 * 8 = 352 g.
The mass of H2O released per mole of octane consumed is: 18 (8 + 1) = 162 g.
The ratio of gasoline consumption to CO2 emissions is 352/114 = 3.09

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As the unit of volume is more common when talking about fuel, it is preferable to convert this ratio to grams of CO2 per liter of gasoline consumed.

Knowing that the density of gasoline is 0.74 kg / l and that 1 gram of gasoline burned rejects 3.09 grams of CO2, it comes to: 0.74 * 3.09 = 2.28 kg of CO2 per liter of gasoline burned.

These 2.28 kg occupy a volume of 2280/44 * 25 = 1295 L of CO2 released per liter of gasoline consumed.

Likewise for H2O: The ratio of gasoline consumption to CO2 emissions is 162/114 = 1.42
hence: 0.74 * 1.42 = 1.05 kg of H2O per liter of gasoline burnt.


A vehicle consuming 1 L of gasoline will therefore reject a little more than a kilo of water and 2.3 kg of CO2.

The water will condense quite quickly directly or in the form of a cloud and will fall back in liquid form quite quickly (because we must not forget that water vapor is a very good greenhouse gas, much more "powerful" than water. CO2), this is not the case with CO2 which has a lifespan of around 100 years.

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For other fuels, just replace the n with the fuel used. For example, diesel is made up of alkanes having an n varying between 12 and 22. It would also be interesting to calculate the CO2 emissions compared to the energy supplied by a given fuel. This may be the subject of another page.

Anyway, an article will follow with the study of incomplete combustion (creation of CO) and non-ideal (creation of Nox)

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