Prediction
The pattern of results that I am looking for is that the longer the chains of alcohols become the more energy is given off. Therefore methanol will give off the least energy and pentanol the most. I think this is what will happen because the longer the chains are the more atoms are made and broken in the full reaction. As the alcohol increases by one carbon atom, in the combustion equation there is 2 extra C-H bonds, one extra C-C bond and 1 and a half extra O=O bonds that are made. These all need energy and take in energy from their surroundings. However, there are also 2 extra C=O bonds and 2 extra O-H bonds made each time that there is an extra carbon atom. Because these bonds are high-energy bonds of 805 and 464 they give off a lot more energy than the other bonds take in. This explains why each time you add a CH2 to the alcohol the amount of energy increases. By adding up how much energy is taken in and given off using the bond energies I can predict how much the energy difference is between each alcohol. 1920kJ/mol is taken in by the bonds been made and 2538kJ/mol is given off by all the bonds been made. The difference between these is 618kJ/mol and therefore I can predict that as you add a carbon atom and extra 618kJ/mol is given off.
I also think this is the pattern I will get because I have looked up the values in a data book and have found the same thing:
Methanol: -726.0kJ/mol
Ethanol: -1367.3kJ/mol
Propanol: -2021.0kJ/mol
Butanol: -2675.6kJ/mol
Pentanol: -3328.7kJ/mol
Finally, I can do bond calculations to see a specific pattern. The energies for each bond used are shown below in kJ/mol:
C-H: 413
C-C: 347
C=O: 805
C-O:358
O=O:498
O-H: 464
I then did the calculations for each alcohol using these bond energies. An example is for Methanol:
CH3OH +11/2O2 → CO2 + 2H2O
2061 + 747 1610 + 1856
2028 - 3466
=
Preliminary Work
In my preliminary work I burnt butanol to see what sort of number I would get and to ensure the equation would work.
The final answer was –695.8KJ/mol. I now know that I will find my results are around this number and that most of them will be smaller than this.
Initially I was planning to burn each fuel for a length of 2 minutes and look at the temperature change but I will not do this because the concept of time in the equation is irrelevant. Instead, I will now keep the temperature change constant and ignore the time.
In my preliminary experiment I found out a lot of things, which I need to do make my experiment better. I started by using a tripod to put the calorimeter on but I now know that this would not work because there would be too much heat lost because the tripod it too high off the top of the burner. In my preliminary experiment I placed the thermometer in the calorimeter but I realised this would not give me accurate reading because it would be measuring the temperature of the metal and not the water. I also realised that I should use 50cm3 of water and that the wick of the burner should be about 2cm below the bottom of the calorimeter to ensure as little as possible heat is lost.
Equipment
The diagram below shows how some equipment can be set up to measure the heat produced when fuels burn. The heat produced warms the water in the metal can. If the rise in temperature of the water is measured the heat produced and be calculated. It is possible to find the mass of the alcohols burnt and then from this I can calculate the heat produced when one mole of fuel burns.
When carrying out my experiment I will use the following equipment and it will be arranged as shown.
- Clamp stand
- 2 x clamps
- Calorimeter
- Thermometer
- Burner
- 5 alcohols
- Scales
Method
- The equipment will be set up as shown above.
- The mass of the fuel inside of the burner will be weighed using the electronic scales.
- The burner will then be put in place and the calorimeter will be set up 2cm above the wick of the burner.
- The fuel will be burnt until the temperature of the water reaches 50 C.
- When it reaches 50 C the burner will be turned off and weighed again.
- The difference of the two masses will be recorded.
- The whole test will be repeated for each alcohol three times and all the differences in mass will be recorded
- Finally the results will be used to work out the KJ/mol released.
Variables
There are many things that I will have to do to ensure that my experiment is a fair test. I will only change one factor and that will be the fuel that is been burnt.
- For each test the temperature change will be 30 C. This means the water will rise from a temperature of 20 C to 50 C in each test.
- I will repeat each test three times so I can obtain three sets of results for each alcohol and then take an average. This also makes it possible to see any anomalies
- The calorimeter will always be 2cm above the top of the wick of the burner for each test so that the same amount of heat is transferred in each test.
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In each test I will fill the calorimeter with 50cm3 of water.
- The thermometer in each test will be suspended in the calorimeter using a clamp on the clamp stand.
- After each test I will clean down the bottom of the calorimeter of any soot that has built up as this could affect the conductivity of the calorimeter in the next test.
