MadSci Network: Chemistry
Query:

Re: What is the fuel value (kJ/g) of ethanol (ethyl alcohol)?

Date: Sun Oct 15 17:20:39 2000
Posted By: Paul Henderson, Undergraduate, Small Physics, Fortismere
Area of science: Chemistry
ID: 970036739.Ch
Message:

Word Up Andy!

The short answer to this is Yes. You can find out a fuel value by looking at 
percentage compositions, but not of elements. You have to count the number of 
bonds in a compound and what kind they are, eg: C-H bonds. Ethanol is a very 
good example of how to do this.
Here's a little step-by-step:
1. find out what the molecule looks like, ethanol "looks" like this:
       H   H 
       |   |
   H - C - C - O - H
       |   |
       H   H

2. Count the different bonds:
C-H bonds = 5
C-C bonds = 1
C-O bonds = 1
O-H bonds = 1

3. Then you find out what it's going to react to form, when ethanol (or any 
hydrocarbon for that matter) combusts in excess oxygen, water and carbon dioxide 
are made. So we see what these molecules look like:
Oxygen: O=O
O=O bonds = 1

Water: H-O-H
O-H bonds = 2
Carbon Dioxide: O=C=O
C=O bonds = 2

4. Make a balanced equation to find the ratios the chemicals react with:

C2H5OH + 2O2 ---> 3H20 + 2CO2

this shows us that we have twice the number of O=O bonds on the left hand side. 
And 3 times the number of O=H bonds and twice the number od C=O bonds on the 
right hand side.

Now more counting. On the left hand side we have a total of:
C-H bonds = 5
C-C bonds = 1
C-O bonds = 1
O-H bonds = 1
+
O=O bonds = 2

And on the right hand side:
O-H bonds = 6
+
C=O bonds = 4

5. Still with me? Good, now dig out that big fat Chemistry text book, or find 
one somehow, and find out what the standard Bond Enthalpies (energy of one mole 
of this kind of bond - kJ/mol) are for C-H, C-C, C-O, O-H, O=O and C=O bonds.
Generously, I've done this for you:
C-H = 413 kJ/mol
C-C = 347 kJ/mol
C-O = 358 kJ/mol
O-H = 464 kJ/mol
O=O = 497 kJ/mol
C=O = 740 kJ/mol

then sum the enthalpies on the left hand side:
reactants = 5(C-H) + 1(C-C) + 1(C-O) + 1(O-H) + 2(O=O)
          = 5(413) + 347 + 358 + 464 + 2(497)
          = 4228 kJ/mol

and the enthalpies on the right hand side:
products = 6(O-H) + 4(C=O)
         = 6(464) + 4(740)
         = 5744 kJ/mol

6. subtract products from reactants:
energy difference = 4228 - 5744
                  = -1516 kJ/mol  (the negative sign means energy is given out)

7. You want this in kJ/g ? you need to find out relative molecular masses here 
(big fat periodic table nessecary)again I've taken the liberty of finding this 
out for you:
add up all the relative molecular masses of the elements in ethanol:
Carbons = 2,  RMM = 12
Hydrogens = 6,  RMM = 1
Oxygens = 1,  RMM = 16

add 'em up:
relative molecilar mass of ethanol = 2(12) + 6(1) + 1(16)
                                   = 46 g/mol

now divide the energy given out by the relative molecular mass:
kJ/mol = -1516/46
       = 32.96 kJ/g (to 4 s.f)

8. There you go! It's a little less than that for unleaded petrol.  Hope this 
makes sense / isn't too long.

A future fuel for our cars? Maybe... Or we'll all have mini fusion reactors in 
our autos.

Paul Henderson BSc (Bronze Swimming Certificate)

oh and I don't know how to do subscripts on this so please forgive 
C2H50Hs and stuff



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