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 …show more content…
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