Enthalpy Changes of Combustion of Different Alcohols

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Enthalpy Changes of Combustion of Different Alcohols

Aim:

The aim of this experiment is to find out how the enthalpy change

(total energy released when the alcohols are completely combusted in a

plentiful supply of air) for 5 different alcohols is affected by the

number of carbon atoms in the alcohol and other factors contributing

to the molecular structure.

Prediction:

I predict that as the amount of carbon atoms in the alcohol increases,

the higher the enthalpy of combustion will be. I have made this

prediction, using the values for the enthalpy change of combustion for

each alcohol, calculated using bond enthalpies and Hess' law.

Methanol's molecular formula is CH3OH. This is the basic structure for

all the alcohols, then to make the larger ones an extra carbon is

added to the existing carbon each time and the oxygen-hydrogen

molecule gets added to the atoms added to the new carbon atom

When methanol combusts in air, it reacts with oxygen molecules to from

water and carbon dioxide. The balanced equation fort this is:

[IMAGE] CH3OH (l) + 1.5O2 (g) CO2 (g) + 2H2O (l)

This means that the bonds broken are; 3 carbon- hydrogen, 1

carbon-oxygen, 1 oxygen-hydrogen and 1.5 oxygen- oxygen (double bond)

and the bonds broken are; 2 carbon- oxygen (double bond) and 4

oxygen-hydrogen.

Constructing a Hess' law cycle will show how these are linked

together:

[IMAGE] CH3OH (l) + 1.5O2 (g) CO2 (g) + 2H2O (l)

[IMAGE][IMAGE]

C (g) + 4H (g) + 4O (g)

N.B. bond enthalpies are for elements in their gaseous states

If a calculation for the amount of energy needed to break the bonds is

made and then the amount of energy given out from bond formation, the

resultant energy difference (negative because the reaction is

exothermic) is the enthalpy change of combustion.

Average bond enthalpies for elements in their gaseous states

(kJmol-1):

Carbon - Carbon (C-C) = +347

Carbon - Hydrogen (C-H) = +413

Oxygen - Hydrogen (O-H) = +464

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