Energy from Combustion

How does burning a stable fuel yield energy?

Three questions

  1. What does stable mean?
  2. In combustion: what vanishes; what appears?
  3. Can generated energy be simply predicted?

Take home messages

  1. Combustion produce energy by burning stable (usually) molecule to more stable end products. (Note: O2 binding energy is zero.)
  2. In combustion: reaction conserves number of each atom species; compute energy per standand number of molecules (mol) and not weight.
  3. On weight basis, methane produces the most energy.

Conserve number of atoms of each element

Consider methane CH4 burning in oxygen O2.

Is this correct? CH4 + O2 ⇒ CO2 + H2O. No!
Problem: LHS: 4 H & 2 O while RHS: 2 H & 3 O.
How about?
   CH4 + 2 O2 ⇒ CO2 + 2 H2O? Yes!

Balancing the equation: number of each atom species is conserved in reaction.

Note: every molecule in reaction is  stable. And yet we get energy. How?

First we need to define binding energy of a molecule.

Binding energy of a molecule.

Molecular binding energy is difference between energy of molecule and its elements in their most stable state.

Chemists call this Enthalpy of Formation; the complete definition prescribes the measurement. Measurements are hard; in 30 years numbers have changed 1/2 percent.

Set binding energy zero for most stable oxygen state O2.
Binding energies of other stable molecules are negative. (really?)

Units: kJ/mol. What is a mol? Standard number of molecules (∼ 6 1023). Unit eases balancing chemical equations; imagine using kJ/kg. (∼20 mols in Dasani [335 ml].)

Sample Combustion Calculation

Binding energies (kJ/mol) of some molecules
Molecule CH4C2H6C3H8 COCO2 NONO2 H2O
B.E. -75-85-104 -111-394 +91+34 -242
            both unstable  
Combustion Energy from Methane (0.016 kg/mol)
CH4+2 O2 CO2+ 2 H2O
-75+0 = -394+2(-242)+"heat"
or   heat = 394+2(242)-75
  = 803 kJ/mol
  = 50 MJ/kg.
Note: (i) mucho energy/unit-weight; (ii) no pollution (w/o N2 ); (iii) efficient combustion; (iv) tranportable.

Can we get more energy?

Ethylene C2H4: , 0.0284 kg/mol
C2H4 +3 O2 2 CO2+2 H2O
52+0= 2(-394)+2(-242)+"heat"
 heat = 2(394+242)=52
  = 1344 kJ/mol
  = 47.3 MJ/kg.
While ethylene is not bound, its greater weight per mole counterbalances that to give about same energy

Heat comes from burning molecule into more stable end products.
Light atoms have the potential to be produce more energy.

What about other "alkanes"?

ethane Ethane: -84.7 kJ/mol; 0.0301 kg/mol
C2H6 + 3.5 O2 ⇒ 2 CO2 + 3 H2O
½ [-84.7 -2(-393.5) -3(-241.8)]/0.0301
= 23.7 MJ/kg
propane Propane: -103.9 kJ/mol, 0.0302 kg/mol
C3H8 + 5 O2 ⇒ 3 CO2 + 4 H 2O
(1/3)[-103.9 -3(-393.5) -4(-241.8)]/0.0441
= 15.4 MJ/kg
butane Butane: -124.7 kJ/mol; 0.0581 kg/mol
C4H10 + 6.5 O2 ⇒ 4 CO2 + 5 H2O
¼[-124.7 -4(-393.5) -5(-241.8)]/0.0581
= 11.4 MJ/kg


To cite this page:
Getting Energy from Combustion
<http://www.physics.ohio-state.edu/~wilkins.5/energy/Resources/Lectures/combenergy-land.html>
[Sunday, 17-Dec-2017 18:19:39 EST]
Edited by: wilkins@mps.ohio-state.edu on Tuesday, 04-Oct-2011 08:59:57 EDT