Energy changes
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How do the structures of methanol, cyclohexane, propan-1-ol and propan-2-ol affect the enthalpy changes of combustion? Why are some enthalpy changes higher than others?
Igloo writes
To answer this question you first need to write an equation for the combustion process and then construct the structural formulae for all the reactants and products.
Let’s take cyclohexane for example:
C6H12 + 9O2 -> 6CO2 + 6H2O
The bonds of the reactants will all be broken:
6 C-C bonds
12 C-H bonds
9 O=O bonds
This will require a great deal of energy which can be calculated if you know the values for the three bond energies concerned.
When the products are formed, many bonds are formed:
12 C=O bonds
12 O-H bonds
Much energy is evolved during this bond-forming process.
Given that more energy is evolved in creating 12 C=O and 12 O-H bonds than is required to break 6 C-C, 12 C-H and 9 O=O bonds, this extra energy is given out to the surroundings, and we say that the reaction is exothermic.
If a similar exercise is carried out on the other compounds in your list, you will realize that since the same products are formed in each case, the energy evolved has everything to do with the structure of the organic compound being burnt.
Since the difference between the energies needed to break bonds and that evolved in forming the bonds of the products is bound to differ from one compound to the next it follows that the enthalpy changes of combustion will be higher for some compounds than others. I hope you can see that compounds with relatively weak bonds (low enthalpy changes of atomization) will have relatively high enthalpies of combustion.
Incidentally, since propan-1-ol and propan-2-ol are isomers and contain the same bonds (2 C-C, 7 C-H, 1 C-O and 1 O-H) they might be expected to share the same enthalpy changes of combustion. However these energies are slightly different. Why might this be, do you think?
Risk assessment
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updated: 20 January 2006
