Energy changes and bonding

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I am doing some work on enthalpy changes of combustion of different alcohols, but I am not sure how to explain how the structure of the alcohol affects this and why each successive alcohol releases more energy.

Jenkin says
You need to think about the bonds broken and those made during the combustion reactions. Let’s start with ethanol, C₂H₅OH. The equation for the reaction is
C₂H₅OH + 3¹/₂ O₂ -> 2CO₂ + 3H₂O
The bonds broken are
in the ethanol, 5 x C-H
1 x C-C
1 x C-O
and 1 x O-H
and in the oxygen, 3 ¹/₂ O=O
A specific quantity of energy is needed to break each of these bonds.
 
Now, in the products, the bonds formed are
in the carbon dioxide, 4 x C=O
and in the water, 6 x O-H
 
A specific quantity of energy will be released when each of these bonds is formed. The enthalpy change of combustion is the difference between the energy released when bonds are formed and the energy absorbed when bonds are broken.
 
The next member of the alcohol series is propan-1-ol. The equation for its combustion is
 
C₃H₇OH + 5O₂ -> 3CO₂ + 4H₂O
 
Comparing this reaction with that of methanol, we can see that the extra bonds broken are
 
in the propanol, 2 x C-H and 1x C-C
and in the oxygen, 1 ¹/₂ x O=O
 
The extra bonds formed are 2 x C=O and 2 x O-H
 
Next, butan-1-ol.
C₄H₉OH + 6¹/₂ O₂ -> 4CO₂ + 5H₂O
 
Extra bonds broken are once again 2 x C-H, 1 x C-C and 1¹/₂ x O=O, and extra bonds formed are 2 x C=O and 2 x O-H.
 
So we can see that as we move up the series form one alcohol to the next, a specific extra quantity of energy will be released.
 
I haven’t included methanol in this discussion since it has no C-C bond. What about isomers such as propan-2-ol? Well, the bonds broken and formed are the same as in propan-1-ol, but the slightly different environments of some of the bonds in the propan-2-ol molecule will mean the bond-breaking energies will differ slightly.

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updated: 14 December 2006