Comparing the Enthalpy Changes of Combustion of Different Alcohols

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Comparing the enthalpy changes of combustion of different alcohols


What are alcohols?

Alcohol is the common family name for the hydrocarbon group alkanols.
They are part of a homologous series. At least one of the hydrogen groups in the molecule is replaced by an OH group.



They are all organic compounds.

The general formulas for the alcohols are:


Where n represents a number.

The first and simplest member of the alkanols family is methanol. Its molecular formula is CH3OH.

You can now see that each member of the alcohol family has a different number of carbons in its structure. They increase by one carbon atom and two
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So as the number of carbon atoms increases the enthalpy of combustion will become more negative. I think this because as you add a H-C-H bond you increase the energy required to break the bonds by 826KJ/Mol.
This is worked out by the average bond enthalpy of the C-H bond, which is 413kJmol-1 being multiplied by two because there are two extra C-H bonds in the molecule.

Using this I will predict the enthalpy change of combustion value for methanol. To do this I will use the equation

CH3OH(l) + 1½O2 O(g) † CO2(g) + 2H2O(l)

In order to make this easier to work out and for people to understand
I will multiply everything by 2 so that I don’t have any values in the equation. This gives the new equation of;

2CH3OH(l) + 3O2 O(g) † 2CO2(g) + 4H2O(l)

On the left hand side of this equation you have

2 x (3 x C-H(413))

2 x (1 x C-O(358))

2 x (1 x O-H(464))

3 x (1 x O=O(498))

In total the total energy required to break the bonds is

(6 x 413 kJmol-1) + (2 x 358 kJmol-1) + (2 x 464 kJmol-1) + (3 x 498 kJmol-1)

2478 kJmol-1 + 716 kJmol-1 + 928 kJmol-1 + 1494 kJmol-1 = 5616kJmol-1

On the right hand side of the equation

2 x (2 x C=O(-805 kJmol-1))

4 x (2 x O-H(-464 kJmol-1))

In total the total energy required to form the bonds is

(4 x -805 kJmol-1) + (8 x -464 kJmol-1)

-3220 kJmol-1 + -3712

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