Combustion of Alcohols Investigation.

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Combustion of Alcohols Investigation

PLAN

Aim: To find out how much energy is required to burn enough of the following alcohols: methanol, ethanol, propanol and pentanol, in order to heat a beaker of water by 40°.

        The combustion of an alcohol is an exothermic reaction, meaning that more energy is given off during the process than is being taken in. by knowing the individual bond energies for each bond, we can calculate what the energy values obtained from our investigation should aspire to to be accurate. Each bond (eg. C-H, where the ‘-‘ signifies a single bond, or O=O, where the ‘=’ denotes a double bond), have different energies. They are as follows:

        MOLECULE DIAGRAMS

 

        

             

The following are the reactions which occur when each alcohol combusts.

METHANOL: CH3OH + 1½ O2                 CO2 + 2H2O

ETHANOL: CH3CH2OH + 3O2         2CO2 + 3H2O

PROPANOL: CH3CH2CH2OH + 4½ O2           3CO2  + 4H2O

PENTANOL: CH3CH2CH2CH2CH2OH + 7½ O2          5CO2 + 6H2O

Given this knowledge we can calculate how much energy should be given out per mole of alcohol that is combusted. To do this we add up the bond energies for each bond that is either broken or made during the combustion process. We then simply subtract the total energy given out from the total energy used up. We should in theory obtain a negative number from this, as energy is lost in the reaction.

CALCULATIONS:

METHANOL:  (C-H) x3 = 435 x 3

Energy input         (O-H) x1 = 464

         (C-O) x1 = 358

         (O=O) x1.5 = 497 x 1.5

         total = 2872.5

Energy output   (C=O) x2 = 803 x 2

         (H-O) x4 = 464 x 4

         total = 3462

        DIFFERENCE = -589.5

ETHANOL:        (C-H) x5 = 435 x 5

Energy input        (O-H) x1 = 464

        (C-O) x1 = 358

        (C-C) x1 = 347

        (O=O) x3 = 497 x 3

        total = 4835

Energy output   (C=O) x4 = 803 x 4

        (H-O) x6 = 464 x 6

        total = 5996

        DIFFERENCE = -1161

PROPANOL:        (C-H) x7 = 435 x 7

Energy input        (O-H) x1 = 464

        (C-O) x1 = 358

        (C-C) x2 = 347 x 2

        (O=O) x4.5 = 497 x 4.5

        total = 6797.5

Energy output   (C=O) x6 = 803 x 6

        (H-O) x8 = 464 x 8

        total = 8530

        DIFFERENCE = -1732.5

PENTANOL:   (C-H) x11 = 435 x 11

Energy input        (O-H) x1 = 464

        (C-O) x1 = 358

        (C-C) x4 = 347 x 4

        (O=O) x7.5 = 497 x 7.5

        total = 10722.5

Energy output        (C=O) x10 = 803 x 10

        (H-O) x12 = 464 x 12

        total = 13598

        DIFFERENCE = -2875.5

Using these calculations for the differences, or enthalpy change, we can construct a graph of predicted results, that we can use to compare with the experimental results.

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From this graph I can construct a hypothesis.

HYPOTHESIS:

        The further you go down the alcohol series (i.e. starting with methanol, followed by ethanol etc.) the more heat is given off during combustion, with both weight and mass as constant. This means that of the four alcohols we investigate, pentanol would be the most efficient as a fuel.

        The following is the method for a trial run experiment we will undertake in order to familiarise ourselves with the process of this experiment, and so we can find any problems we need to improve upon for the final ...

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