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Measuring the Enthalpy of Combustion for Alcohols

Autor:   •  October 14, 2018  •  2,031 Words (9 Pages)  •  942 Views

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Average mass of ethanol combusted

[pic 50]

Average mass of butan-1-ol combusted

[pic 51]

Average mass of pentan-1-ol combusted

[pic 52]

Mols of alcohol combusted (mol)

[pic 53]

% uncertainty mol = % uncertainty , since there is no uncertainty associated to the molar mass.[pic 54]

Mols of ethanol combusted

[pic 55]

Mols of butan-1-ol combusted

[pic 56]

Mols of pentan-1-ol combusted

[pic 57]

Enthalpy of combustion of alcohols ()[pic 58]

[pic 59]

% uncertainty on = % uncertainty of Q + % uncertainty of mol[pic 60]

Enthalpy of combustion of ethanol

[pic 61]

Enthalpy of combustion of butan-1-ol

[pic 62]

Enthalpy of combustion of pentan-1-ol

[pic 63]

Average % uncertainty on the enthalpy of combustion = [pic 64]

Number of carbon atoms

Table 4: Number of carbon atoms in the alcohols

Alcohol

Condensed Formula

Number of carbon atoms

Ethanol

CH3CH2OH

2

Butan-1-ol

CH3(CH2)2CH2OH

4

Pentan-1-ol

CH3(CH2)3CH2OH

5

A relation between the number of carbon atoms in the alcohol and the enthalpies of combustion can be plotted. For the trend to be shown more clearly, the absolute value of the enthalpies will be plotted.

Graph 1: Number of Carbon Atoms vs. Enthalpies of Combustion (experimentally determined)

[pic 65]

Percentage Error

In order to calculate the percentage error I gathered reference values for the enthalpies of combustion of the three alcohols[1].

Ethanol: [pic 66]

Butan-1-ol: [pic 67]

Pentan-1-ol: [pic 68]

error = * 100 [pic 69][pic 70]

→ using this formula we can determine the percentage error of the experimentally determined values.

Percentage error for ethanol

error = [pic 71][pic 72]

Percentage error for butan-1-ol

error = [pic 73][pic 74]

Percentage error for pentan-1-ol

error = [pic 75][pic 76]

Average % error

[pic 77]

A second graph may be used to graph the reference trend and the experimental one together. Also in this case the absolute values will be used.

Graph 2: Experimental Trend vs. Reference Trend

[pic 78]

In this graph the pink line represents the trend set by the reference values. The black line, instead, represents the trend set by the experimental values.

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Conclusion

With this experiment I was able to calculate the enthalpies of combustion of ethanol, butan-1-ol and propan-1-ol. The corresponding results are: , and . I used the three alcohols to heat up a beaker of 100 ± 1 of water. I therefore calculated the energy absorbed by the water (, took its negative, and assumed it to by the energy released by the combustion of the alcohol. Diving by the number of mols of alcohol consumed gave me the change in enthalpy. [pic 79][pic 80][pic 81][pic 82][pic 83][pic 84][pic 85]

As we have observed previously from the strong positive correlation shown in the graph, the absolute value of the change in enthalpy increases together with the number of carbon atoms present in the alcohol molecule. When a reaction occurs, energy is first absorbed to break the bonds, and eventually released in the formation of the new bonds. Therefore the first part of the reaction is endothermic whilst the second is exothermic. Even though the number of carbons in the reactants increases, the change in enthalpy progressively decreases. This means that as the chain grows the energy absorbed increases because of the longer chain, but even more energy is then released in the successive formation of the bonds. These are the corresponding equations:

→ Ethanol: C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

→ Butan-1-ol: C4H9OH(l) + 6O2(g) → 4CO2(g) + 5H2O(l)

→ Pentan-1-ol: C5H11OH(l) + 7.5 O2(g) → 5CO2(g) + 6H2O(l)[pic 86]

From butan-1-ol to pentan-1-ol, the number of carbon atoms in the chain increases by one. As the chain increases also the oxygen need does. Therefore one more C-C, two more C-H bonds and 1.5 more O=O bonds are broken, but two more C=O bonds and two more O-H bonds have to be formed.

Looking at the table you can see that the energy needed to form the bonds increases more than the one needed to break them as the number of carbons

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