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Enthalpy of combustion calculations

Results

Calculations

Part 1

Mass of propan-1-ol burnt = 12.472 - 11.198 = 1.274 g

C₃H₇OH = 60.1

\ Number of moles of propan-1-ol burnt = 1.274/60.1 = 0.0212 mol

Enthalpy of combustion of propan-1-ol (from data book) = 2021.0 kJ mol^-1

\ Energy released on complete combustion = 0.0212 × 2021 = 42.8 kJ

Temperature rise = 27.9 - 18.7 = 9.2 °C

\ Heat capacity of calorimeter = 42.8/9.2 = 4.7 kJ K^-1 (Remember that the K and °C are the same size)

Part 2

Temperature rise = 35.05 - 25.25 = 9.8 °C

\ Energy released = 4.7 × 9.8 = 46 kJ

Mass of octan-1-ol burnt = 13.741 - 12.401 = 1.340 g

C₈H₁₇OH = 130.2

\ Number of moles of octan-1-ol burnt = 1.340/130.2 = 0.0103

\ DH_c (octan-1-ol) = 46/0.0103 = - 4430 kJ mol^-1

Don't forget to include the correct units, the sign (this reaction is clearly exothermic, hence the minus sign) and the correct number of significant figures.

Errors

The accepted value for the enthalpy of combustion of octan-1-ol from the data book is - 5294 kJ mol^-1. Our value is less exothermic than this. We need to consider the errors and the effect that they would have on the final answer.

Heat loss can never be completely avoided. It is unlikely that the heat loss will be exactly the same in both heating phases. We did not measure the time of each combustion phase. They were not largely different, but if one took longer than the other more heat would have been lost. In the case of the octan-1-ol the calorimeter water was a bit hotter to start with, and was raised by a slightly higher temperature range leading to increased heat loss. This heat loss would not be measured giving a less exothermic final answer, as we have.

Incomplete combustion is the major problem that will reduce the value obtained. It is difficult to get enough air into the apparatus to ensure complete combustion. This is particularly the case with the larger alcohols, as you will have seen on the video. You may find that your burner goes out periodically and you will generally observe a rather yellow flame, indicating partial combustion. As this means we get less energy out of the alcohol than is possible, this leads to a less exothermic value. Although this affects both heating phases, incomplete combustion is more significant with the larger octan-1-ol molecule - compare the yellowness of the flames on the video.

Another error is loss of alcohol by evaporation. It is helpful to use a cap on the burner once combustion is finished. Otherwise alcohol will be lost by evaporation (you will smell it - it gets quite sickly!) before you weigh it. This will suggest that a larger amount of alcohol was needed to achieve the measured temperature rise. The effect should be more significant with the lower boiling point propan-1-ol. This will lead to a higher heat capacity for the apparatus being calculated. This, in turn, will give you an answer for the octan-1-ol that is more exothermic than it should be. However, the loss of octan-1-ol by evaporation suggests that more moles were burnt than actually were. This would make the final answer less exothermic. Do you think this overall error is significant?

There will be a small weighing error due to the accuracy of the balance, but this will be random (leading to a bigger or smaller final answer) and very small in comparison with the other errors.

Table

As you can see there is a fairly consistent energy difference between adjacent alcohols. This leads us to use bond energies to make rough estimates of the enthalpy changes for reactions.