∆Hcombustion = 2809 – 3332 = -523kJ per mole
Ethanol + oxygen = carbon dioxide + water
C2H5OH + 3O2 = 2CO2 + 3H2O
H H + O=O = O=C=O + H-O-H
H-C-C-O-H + O=O = O=C=O + H-O-H
H H + O=O = + H-O-H
Reactants-bonds present Products-bonds present
5XC-H=5X413=2065kJ 4XO=C=4X740=2960kJ
1XC-C=1X346=346kJ 6XO-H=6X463=2778kJ
1XC-O=1X360=360kJ
1XO-H=1X463=463kJ
3XO=O=3X498=1494
TOTAL ENERGY IN=4728kJ total energy out=5738kJ
∆Hcombustion = 4728 – 5738 = -1010kJ per mole
Propanol + oxygen = carbon dioxide + water
C3H7OH + 4.5 O2 = 3CO2 + 4H2O
H H H O=O O=C=O H-O-H
H-C-C-C-O-H + O=O = O=C=O + H-O-H
H H H O=O O=C=O H-O-H
0.5(O=O) H-O-H
Reactants-bonds present Products-bonds present
7XC-H=7X413=2891kJ 6XO=C=6X740=4440kJ
2XC-C=2X346=692kJ 8XO-H=8X463=3704kJ
1XC-O=1X360=360kJ
1XO-H=1X463=463kJ
4.5xO=O=4.5x498=2241
TOTAL ENERGY IN=6647kJ TOTAL ENERGY OUT=8144kJ
∆Hcombustion = 6647 – 8144 = -1497kJ per mole
Butanol + oxygen = carbon dioxide + water
C4H9OH + 6O2 = 4CO2 + 5H2O
H H H H O=O O=C=O H-O-H
H-C-C-C-C-O-H+ O=O = O=C=O + H-O-H
H H H H O=O O=C=O H-O-H
O=O O=C=O H-O-H
O=O H-O-H
O=O
Reactants-bonds present products-bonds present
9XC-H=9X413=3717kJ 8XO=C=8X740=5920kJ
3XC-C=3X346=1038kJ 10XO-H=10X463=4630kJ
1XC-O=1X360=360kJ
1XO-H=1X463=463kJ
6XO-O=6X498=2988kJ
TOTAL ENERGY IN=8566kJ TOTAL ENERGY OUT=10550kJ
∆Hcombustion = 8566 – 10550 = -1984kJ per mole
Prediction
The theoretical values show the amount of energy that should be given off by each fuel. The results show that the alcohol with the most components in it, the greater amount of energy is given off. This is because it takes more energy to break all the bonds with the bigger alcohols between atoms but also a greater amount is given off when the new bonds are formed, and so, the greater the size of the alcohol, the greater the amount of energy given off when the new bonds are formed1.
I predict that the longer the hydrocarbons in the chain of the alcohols, the more heat it will produce. However, I also think the actual values will be less than the theoretical values, as in the experiment, energy will be lost into the atmosphere, transferred to the atoms in the copper beaker, causing heat energy, etc.
The products of the reactions will be carbon dioxide and water as well as heat energy to heat the water, which will crack the hydrocarbons into simpler/ smaller molecules and then will combust later, making this experiment an exothermic reaction.
I have plotted a graph on the next page of the theoretical values and my predicted values.
Preliminary work
I took the temperature of the water every minute for 5 minutes and decided that after three minutes the temperature increase started to slow after 3 or 4 minutes, so decided, in my experiment, to record the temperature and weight after 3 minutes. I also noticed that the size of the flame changed, even if the wicks were about the same size. I noticed that I would have to make the wick a lot smaller with the alcohols with longer hydrocarbons. I decided that the distance between the spirit burner and the bottom of the copper beaker needed to be smaller, so less energy would escape into the atmosphere instead of going to the water. This is especially noticeable with the preliminary work for ethanol, as the temperature difference is very different to the rest of the substances.
Results
To calculate the energy released by each alcohol, the following calculation must be used:
Energy transferred = mass of water x specific heat capacity x temperature
to _ water (J) (50g) of water (4.2J) increase (see table)
To calculate the heat combustion of each alcohol, the following calculation must be used:
∆Hcombustion = energy transferred to water x molar mass of fuel
mass of fuel used
However, this will give a big numbered answer in Joules. Therefore, to convert the answer into an easier, smaller number, in kiloJoules, the answer obtained by the equation above must be divided by 1000.
Methanol
-
Energy transferred = 50g x 4.2g x 30°c = 6300J
-
Energy transferred = 50g x 4.2g x 30°c = 6300J
-
Energy transferred = 50g x 4.2g x 32°c = 6720J
AVERAGE ENERGY = (6300 + 6300 + 6720) = 6440J
3
∆Hcombustion = 6440 x 32/1.27 = 162268/1000 = 162kJ
Ethanol
-
E = 50 x 4.2 x 27 = 5670J
-
E = 50 x 4.2 x 28 = 5880J
-
E = 50 x 4.2 x 21 = 4410J
AVERAGE ENERGY = 5320J
∆Hcombustion = 5320 x 46/1.27 = 192693/1000 = 193kJ
Propanol
-
E = 50 x 4.2 x 27 = 5670J
-
E = 50 x 4.2 x 29 = 6090J
-
E = 50 x 4.2 x 15 = 3100J
AVERAGE ENERGY = 4970J
∆Hcombustion = 4970 x 60/1.27 = 234803/1000 = 235kJ
Butanol
-
E = 50 x 4.2 x 23 = 4830J
-
E = 50 x 4.2 x 23 = 4830J
-
E = 50 x 4.2 x 27 = 5670J
AVERAGE ENERGY = 5110J
∆Hcombustion = 5110 x 74/1.5 = 252093/1000 = 252kJ
I have plotted a graph on the next page, showing my results from the experiment, compared to my predicted results and the theoretical results. However, the answers above (for ∆Hcombustion) will actually be recorded as negative numbers, as they are exothermic reactions, so energy is given out.
(Methanol= -162kJ; Ethanol= -193kJ; Propanol= -235kJ; Butanol= -252kJ).
Conclusion
My prediction was correct, as the more complex alcohols gave out more energy. I found that the more hydrocarbons in chain molecules in each alcohol, the more heat will be produced. This is shown by butanol, as it gives out the highest amount of energy, and also has the highest amount of carbon molecules in the hydrocarbon chain out of the four alcohols.
The experiment results were at a much lower rate than the theoretical values, as I expected. However, my predicted regression line was still a lot higher than my results, as I expected more of the total energy to transfer to the water. The experiment results were a lot lower than the theoretical values because the energy would have been lost into the surrounding air, by convection, as heat and light energy, so not all the energy was used to heat the water. Also, it was very hard to determine the wick length needed to ensure a similar flame size for each fuel. Obviously, the flame size would not have been exactly the same, so this will have affected the results. However, I did learn how to make the flame sizes more similar from the preliminary work, along with other things, such as the distance of the clamp from the work surface.
As the number of carbon atoms increase, this is because with larger molecules of alcohol, energy is released as heat through a faster rate. This means that the energy transfer is even more inefficient, because it means a quicker and larger convection current in the air so more heat loss and also the copper beaker is not a perfect conductor and so can only conduct a certain amount of energy. Therefore, as the molecular size of the alcohol increases then so does the rate of energy production.
I noticed that the results are just about in a straight line and almost fit on the regression lines. This tells me that the molecular size of the alcohol is proportional to the rates of combustion.
Evaluation
The experiment did not allow me to obtain very accurate results, as energy was lost to the surroundings, even though I tried to decrease the loss slightly after my preliminary work. However, I could have also done more, such as set up some sort of draft excluder. I could have made a draft excluder by using 3 boards in a triangle and a board on top around the apparatus, although energy would still have been lost before I put the guard up after lighting the wick. I could move the beaker even closer to the flame, to lose less energy, but this could cause incomplete combustions, since there would be less oxygen to react with, and so the energy transferred would be different than expected and the carbon that forms on the bottom of the beaker can cause inefficient heat transfer2.
Stirring the water may have helped with the heat transfer, as the heat would transfer to more water molecules, rather than relying on convection of heat through the water while it is stationary. However, the stirring may increase the rate of evaporation. Another cause of inaccuracy can be the time it takes to put out the flame, and the time it takes to weigh the spirit burner, as some of the alcohol is likely to have continued evaporating whist being transported to the scales.
A digital thermometer would have been better to get a more accurate reading, as most digital thermometers give a reading to 2 decimal places. Something else that may have affected the results was that the copper beaker was heated for one trial, and then the beaker would still be warm for the next experiment. A better way may have been to have numerous identical beakers, to be used for the experiments as long as they all start at the same temperature, i.e. room temperature.
The measurement of the water was not very accurate, as it was very easy to be a millilitre or two over or under. Digital scales may have been better to measure the water, or even a measuring cylinder with a smaller diameter, as this would make it easier to read the scale. The inaccuracy makes the results less reliable, as different volumes of water means that different amounts of molecules will have to be heated.
A more detailed trend would have been obtained by continuing the experiment with more alcohols that have larger molecules (more carbon atoms), such as pentanol and hexanol etc. 3. It would have been easier to draw a regression line with more points to link/ see a trend between. This would help to estimate the heat of combustion of alcohols with different molar masses. Also, more trials would have been better, but I did not have a sufficient amount of time to keep repeating the experiments.
Plus, it was hard to determine if the end of the wick was actually in the alcohol; the alcohol may not actually have been burned. This would have altered the results, as it would change the overall outcome if used in the result table.
There were not any anomalous results on my graph. However, the end temperature, and the temperature difference, in the third trial for propanol was irregular, as it is over ten degrees cooler than the other results for propanol. This result may be unusual because the flame size was smaller than the other trials. Also, I had another trial for ethanol, but the results were extremely unusual, which I think was mainly caused by the flame size. I decided not to include these results in the experiment, as it may have considerably affected the calculations.
I would also like to mention that there is some inaccuracy, since most of my calculations were in whole numbers; my highest was to two d.p.
Further work relating to this experiment could include seeing how long it takes for each alcohol to heat water to a certain temperature/ by a certain amount. For this, I would predict that the alcohols with the most carbon atoms would heat the water in the least amount of time.