Enthalpy Changes in Combusting of Different Alcohols.
Enthalpy Changes in Combusting of Different Alcohols
Introduction:
Different fuels produce different amounts of energy when they are burnt in oxygen. I am going to investigate the different energy contents of alcohols. I am going to use just alcohols because when I find out the energy per mole produced by different alcohols I can compare them. I am going to use Methanol, Ethanol, Propanol, Butanol and Pentanol.
When fuels are burnt in oxygen, water and carbon dioxide are formed. The combustion also produces heat. This is because when the alcohol burns energy from the heat is used to break the bonds and when they form carbon dioxide and water they are making bonds and energy is released. This energy becomes heat. I am going to investigate how much energy is produced by each alcohol I burn and compare the results when I have finished.
Prediction:
For my prediction I am going to calculate how much energy each alcohol should produce. I am going to do this by using a secondary source of information. This is a book called 'AS Chemistry' published by 'Collins' by "Nicholls and Radcliff." I will uses this by taking the values given in this book of how many kJ/mol are required to break/make certain bonds. The information that I am going to use is:
To break a C?H (Carbon to Hydrogen) bond it requires 413 Kjmol-1 of energy.
To break a C?O (Carbon to Oxygen) bond it requires 360 Kjmol-1 of energy.
To break an O?H (Oxygen to Hydrogen) bond it requires 463 Kjmol-1 of energy.
To break a C?C (carbon to Carbon) bond it requires 347 Kjmol-1 of energy.
To break an O=O (double oxygen) bond it requires 498 Kjmol-1 of energy.
To make a C=O (Carbon to Oxygen double) bond it produces -743Kjmol-1 of energy.
To make an O?H (oxygen to Hydrogen) bond it produces - 463 kjmol-1 of energy.
I am now going to use this information to calculate the total energy required to break the bonds on the left hand side of each equation and then how much energy is produced when it makes the bonds on the right hand side and therefore predict how much energy per mole is going to be produced at the end.
Methanol
Methanol + Oxygen = Carbon dioxide + water
2CH3OH + 3O2 ? 2CO2 + 4H2O
On the Left-Hand Side of the equation:
2 x ?3 x C?H (413 kjmol-1)
2 x ?1 x C?O (360 kjmol-1)
2 x ?1 x O?H (463 kjmol-1)
3 x O=O (498 kjmol-1)
Total energy required to break the bonds:
(6 x 413) + (2 x 360) + (2 x 463) + (3 x 498) =
2478 + 720 + 926 + 1494 = 5618 kjmol-1
On the Right hand side of the equation:
2 x? 2 x C=O (- 743 kjmol-1)
4 x ?2 x O?H (- 463 kjmol-1)
Total energy produced to make the bonds:
(4 x - 743) + (8 x - 463) =
(- 2972) + (- 3704) = - 6676 kjmol-1
Therefore the expected enthalpy of combustion of Methanol is=
(-6676 + 5618) = ? 1058 kjmol -1 for 2 moles because the balanced equation had 2 moles of Methanol. Therefore, the prediction is
-1058 / 2 = - 529 kjmol-1
Ethanol
Ethanol + Oxygen = Carbon dioxide + water
C2H5OH + 3O2 = 2CO2 + 3H2O
On the left-hand side of the equation:
5 x C?H (413 kjmol-1)
x C?C (347 kjmol-1)
x C?O (360 kjmol-1)
x O?H (463 kjmol-1)
3 x ?1 x O=O (498 kjmol-1)
Total energy required to break the bonds:
(5 x 413) + (1 x 347) + (1 x 360) + (1 x 463) + (3 x 498) =
2065 + 347 + 360 + 463 + 1494 = 4729 kjmol-1
On the right-hand side of the equation:
2 x ?2 x C=O (- 743 kjmol-1)
3 x ?2 x O?H (- 463 kjmol-1)
Total energy produced by making the bonds:
(4 x - 743) + (6 x - 463) =
- 2972 + - 2778 = - 5750 Kjmol-1
Therefore the expected enthalpy of combustion of Ethanol is:
(- 5750) + 4729 = - 1021 kjmol-1
Propanol
Propanol + Oxygen = Carbon dioxide + water
C3H7OH + 41/2O2 = 3CO2 + 4H2O
= 2C3H7OH + 9O2 = 6CO2 + 8H20
On the left-hand side of the equation:
2 x ?7 x C-H (413 kjmol-1)
2 x ?2 x C-C (347 kjmol-1)
2 x ?1 x C-O (360 kjmol-1)
2 x ?1 x O-H (463 kjmol-1)
9 x O=O (498 kjmol-1)
Total energy required to break the bonds:
(14 x 413) + (4 x 347) + (2 x 360) + (2 x 463) + (9 x 498) =
5782 + 1388 + 720 + 926 + 4482 = 13298 kjmol-1
On the right-hand side of the equation:
6 x ?2 x C=O (- 743 kjmol-1)
8 x ?2 x O-H (- 463 kjmol-1)
Total energy produced by making bonds:
(12 x -743) + (16 x -463) =
-8916 + -7408 = -16324 kjmol-1
Therefore the expected enthalpy of combustion for Propanol is:
(-16324) + 13298 = - 3026 kjmol-1 for 2 moles because the balanced equation had 2 moles of Propanol. Therefore, the prediction is
- 3026 / 2 = - 1513 kjmol-1
Butanol
Butanol + Oxygen = Carbon dioxide + water
C4H9OH + 6O2 = 4CO2 + 5H2O
On the left-hand side of the equation:
9 x C-H (413 kjmol-1)
3 x C-C (347 kjmol-1)
x C-O (360 kjmol-1)
x O-H (463 kjmol-1)
6 x O=O (498 kjmol-1)
Total energy required to break the bonds:
(9 x 413) + (3 x 347) + (1 x 360) + (1 x 463) + (6 x 498) =
3717 + 1041 + 360 + 463 + 2988 = 8569 kjmol-1
On the right-hand side of the equation:
4 x ?2 x C=O (- 743 kjmol-1)
5 x ?2 x O-H (- 463 kjmol-1)
Total ...
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9 x C-H (413 kjmol-1)
3 x C-C (347 kjmol-1)
x C-O (360 kjmol-1)
x O-H (463 kjmol-1)
6 x O=O (498 kjmol-1)
Total energy required to break the bonds:
(9 x 413) + (3 x 347) + (1 x 360) + (1 x 463) + (6 x 498) =
3717 + 1041 + 360 + 463 + 2988 = 8569 kjmol-1
On the right-hand side of the equation:
4 x ?2 x C=O (- 743 kjmol-1)
5 x ?2 x O-H (- 463 kjmol-1)
Total energy produced by making bonds:
(8 x - 743) + (10 x - 463) =
- 5944 + - 4630 = -10574 kjmol-1
Therefore the expected enthalpy of combustion for Butanol is:
(-10754) + 8569 = - 2005 kjmol-1
Pentanol
Pentanol + Oxygen = Carbon dioxide + water
C5H11OH + 71/2 O2 = 5CO2 + 6H2O
= 2C5H110H + 15O2 = 10CO2 + 12H2O
On the left-hand side of the equation:
2 x ?11 x C-H (413 kjmol-1)
2 x ?4 x C-C (347 kjmol-1)
2 x ?1 x C-O (360 kjmol-1)
2 x ?1 x O-H (463 kjmol-1)
5 x O=O (498 kjmol-1)
Total energy required to break the bonds:
(22 x 413) + (8 x 347) + (2 x 360) + (2 x 463) + (15 x 498) =
9086 + 2776 + 720 + 926 + 7470 = 20978 kjmol-1
On the right-hand side of the equation:
0 x ?2 x C=O (- 743 kjmol-1)
2 x ?2 x O-H (- 463 kjmol-1)
Total energy produced by making bonds:
(20 x -743) + (24 x -463) =
( -14860) + (-11112) = -25972 kjmol-1
Therefore the expected enthalpy of combustion for Pentanol is:
(-25972) + 20978 = - 4994 kjmol-1, for 2 moles because the balanced equation had 2 moles of Propanol. Therefore, the prediction is:
4984 / 2 = -2497 kjmol-1
This is therefore a table of all my calculations together:
Fuel
Expected Enthalpy of Combustion
Methanol
-529 kjmol-1
Ethanol
-1021 kjmol-1
Propanol
-1513 kjmol-1
Butanol
-2005 kjmol-1
Pentanol
-2447 kjmol-1
Although my calculations are correct I do not expect my results to come out like this because I will not be able to get the experiment totally accurate. Mainly because some of the heat will have been lost to the surrounding air and my method of finding the amount of energy produced may not be completely accurate. The measurements that I take may also be not exact due to human error.
The amount of Carbons in the fuel changes its amount of energy that it will produce. This is because the more carbons there are per molecule of fuel, the larger the molecule will be. Therefore there will be more products; so on the other side of the equation there will be more moles of CO2 and H2O which will have been formed. This means that there will be more bonds formed and so the amount of energy given out will be greater.
Independent Variable: [Alcohol]
I shall change the fuel that I use for each experiment. I will use 5 different alcohols, these are, Methanol, Ethanol, Propanol, Butanol and Pentanol. I will heat a boiling tube of 25cm3 of water above each fuel, one at a time. I am going to measure the temperature of the water, before and after the experiment in ?C and the mass of the fuel before and after the experiment. I will use these figures in my calculations to find out how much energy was produced by the fuel in kjmol-1.
Dependant Variable:
My dependant variables are measuring the temperature of the water before and after the experiment and the mass of the fuel before and after the experiment. I have decided that it does not matter how long the experiment takes and therefore I do not need to time my experiment. The temperature difference between the beginning of the experiment and the end will be reflected in how long it took to increase the difference in mass that I will find between the beginning of the experiment and the end. If I leave the experiment longer and allow the temperature difference to increase, the mass difference will also increase so therefore it does not matter.
To find out how much energy produced per mole by each alcohol, I will go through a number of stages, these are:
- Calculate the temperature rise of the water. By finding the difference between the beginning temperature of the water and taking it away from the final temperature of the water. This will be in ?C.
- Calculate the mass of fuel used. By weighing the fuel at the beginning and at the end of the experiment and subtracting the beginning one from the end to find the difference. I will measure this in grams. I will weigh the fuel while it is in the burner both times so that I can ignore the weight of the burner because it will not change.
- Calculate the number of moles of fuel used.
To do this I must use this equation:
Mass = Number of Moles used
Molar Mass
Molar Mass of each alcohol using the Periodic Table:
Methanol = CH3OH (12 + 3 + 16 +1) = 32
Ethanol = C2H5OH (24 + 5 + 16 + 1) = 46
Propanol = C3H7OH (36 + 7 + 16 + 1) = 60
Butanol = C4H9OH (48 + 9 + 16 + 1) = 74
Pentanol = C5H11OH (60 +11 + 16 + 1) = 88
- Calculate the energy produced.
To do this I will use this equation:
Energy Used Mass Specific The temperature
& Produced = of x Heat x rise of the
To Heat the Water Water (g) Capacity Water (?C)
of water
The Specific Heat Capacity is the amount of energy required to raise one gram of a substance by 1?C. For water this is 4200 J /Kg /?C. This means that because I am using grams for the mass of the water, I must use grams for the S.H.C therefore, using 4200 ?1000 (1kg = 1000g), making 4.2 J /g /?C. This answer will be in joules.
- And finally, calculate the energy produced per mole:
To do this I will use the equation=
Energy energy produced to
Produced = heat the water (q.4 ) (J)
Per mole ????????????
Number of moles of fuel used
This will be measured in kjmol-1. This will then be compared to the figures that I predicted should happen for each different alcohol.
Control Variable:
For my experiment, I must keep a number of things controlled. They will be:
- The amount of water used in the experiment, which will be 25cm3.
- The distance between the test-tube containing the water and the burning flame from the fuel. It must be the same in each experiment, so that I can make sure that the heat that may be lost will be the same in each experiment, therefore making it a fair test. I will measure this before I begin, with a ruler and will try to make it the same for each experiment.
- I will not need to keep the time controlled. This is because it has no relevance to the information that I need for the calculations.
- The most important thing that I must do is to not let the water reach 100?C, this is because if the water reaches its boiling point it will not get any higher and if it cannot get any higher the temperature rise will be inaccurate for the amount of fuel that has been used up. The fuel burner will still be using up fuel but this will not be reflected in the temperature rise of the water if it goes above the boiling point of water (100?C). To stop the temperature of the water reaching 100?C, I will monitor the temperature of the water on the thermometer and when it reaches approximately 90?C I will stop the experiment.
- Throughout the experiments I will not change the volume of water to make the fuel burn longer but keep it consistent all the way through.
- I decided to try out two different ways of doing my experiment when I did the first one it was not very good at producing accurate results. This was to surround the experiment on three sides by heatproof mats; this was to keep the heat in to help prevent heat loss to the atmosphere during the experiment. I did this and have got my results, which I will use as my preliminary experiment. I instead though decided to use tin foil to surround my experiment and have a hole at the top. I decided this because the shiny surface reflects heat back in so I should not lose as much heat although it can be conducted through the metal.
Equipment List:
- A fuel burner for each fuel containing each of, Methanol, Ethanol, Propanol, Butanol and Pentanol.
- A boiling tube.
- A measuring cylinder to measure out 25cm3 of distilled water for each experiment.
- A thermometer
- 2 clamps, a clamp stand and a ruler.
- A heatproof mat.
- Goggles and lab coat for safety.
Safety:
I must be careful with the fuel burners so that I do not spill fuel anywhere, because it highly flammable and is a fire hazard. I must wear safety goggles so that I do not get anything in my eyes. I must also use tongs to remove the boiling tube from the clamp after it has been heated.
Method:
The set up of apparatus looks like this:
Method:
- Set up a heatproof mat, clamp stand, and 2 clamps and measure out 25cm3 of H2O in a boiling tube.
- Tightly secure the boiling tube containing the H2O to the clamp stand, directly above the place where the fuel burner will be.
- Secure a thermometer to a clamp above the H2O and place in the water making sure that it is in the centre and that it does not touch the edges. This is because I want to record the temperature of all the water and because it may different at the top to the bottom If I take the temperature in the centre it should be accurate.
- Measure the distance between the bottom of the boiling tube containing the water and the clamp stand. In cm.
- Measure the temperature of the water in ?C.
- Weigh the first fuel burner which contains CH3OH and record the mass in grams, (g).
- Place the fuel burner under the boiling tube and light the wick.
- Place the tin foil wrap around the experiment, surrounding the fuel burner and the bottom of the boiling tube.
- Wait until the temperature of the water reaches approximately 90?C and extinguish the flame on the fuel burner. Record the temperature when the experiment was stopped; then weigh and record the mass of the fuel burner containing the fuel.
- Then repeat all of the steps above for all the other fuels, Ethanol, Propanol, Butanol, and Pentanol. Making sure that the distance measured between the clamp stand and the bottom of the boiling tube is the same each time, this is to keep the experiment fair by making sure that the heat has to travel the same distance for each time.
- Repeat all of this 3 times so that you can average out all the results for each fuel, to make it a more accurate way of finding the results.
I will record all of my results in a table like this:
Fuel
Temp.
Before (?C)
Temp.
After (?C)
Mass
Before (g)
Mass
After (g)
Distance
(cm)
Methanol
Ethanol
Propanol
Butanol
Pentanol
The last column labelled 'distance' is the distance between the boiling tube and the clamp stand and I have added this column so that if the distance does vary at all between each fuel, it will explain if any of my results are inaccurate. Although it could vary between each fuel, I will try to keep it the same each time so that it is a fair experiment.
Results:
The results that I have found, I have put into a table. The first sets of results were of a preliminary experiment where I used 3 heatproof mats to surround the experiment to contain the heat. Because this did not work very well, with the rest of my results I used tin foil to surround the experiment so that as much heat as possible would be reflected back in and not lost to the atmosphere. From this results table, I will calculate the energy produced for each fuel that I have done and then average the ones that have been done when the experiment was surrounded by tin foil.
Fuel
Mass before (g)
Mass After (g)
Temp. Before
(?C)
Temp. After (?C)
Distance (cm)
Methanol
221.11
218.314
8
92
0.5
Ethanol
206.77
205.02
24.5
95
0.4
Propanol
230.06
228.94
8.5
78.5
0.5
Butanol
230.58
227.36
8
91
0.5
Pentanol
219.59
216.65
9
91
0.5
Methanol
228.08
226.65
20
91
0.4
Ethanol
246.4
244.4
20
92
0.5
Propanol
216.4
214.8
20
93
0.3
Butanol
219.1
218.3
20
90
0.5
Pentanol
225
223.6
9
89
0.5
Methanol
220.3
218.4
20
90
0.4
Ethanol
236
233.5
21
90
0.6
Propanol
227.2
225.9
21
90
0.5
Butanol
216.9
215.4
21
90
0.4
Pentanol
237.7
235.3
21
90
0.5
Methanol
217.7
215.5
22
91
0.5
Ethanol
234.5
232.3
22
92
0.6
Propanol
223.6
222.3
22
90
0.6
Butanol
237.7
236.8
21
90
0.6
Pentanol
235.3
234
22
89
0.5
st Methanol:
. Temperature rise= (91-20)= 71?C
2. Mass of fuel used = (228.08 - 226.65) = 1.43 g
3. Molar mass of CH3OH = 32
No. of moles = mass .
Molar mass
No. of moles = 1.43 / 32 = 0.0446875 moles of fuel used
4. Energy used & produced to = mass of water x S.H.C x Temp rise
heat the water
Energy = 25 x 4.2 x 71
= 7455 joules
5. Energy produced per mole = energy produced to heat the water
No. of moles of fuel used
Energy produced per mole = 7455 / 0.0446875
= 166825.17 / 1000
= 166.825 kjmol-1
I repeated this calculation for the 2nd and 3rd Methanol and then with all the other alcohols. This is a table of my average results, which I can then compare with the results that I predicted:
Fuel
Energy Produced
Methanol
31.999 kjmol-1
Ethanol
52.819 kjmol-1
Propanol
317.121 kjmol-1
Butanol
544.332 kjmol-1
Pentanol
401.29 kjmol-1
Predicted Results:
Fuel
Energy Produced
Methanol
529
Ethanol
021
Propanol
513
Butanol
2005
Pentanol
2447
Comparing the results from my actual experiments and averaged out to those of my preliminary experiment. It can be seen that the second method that I used of using tin foil instead of 3 heatproof mats was a lot more accurate. The heat was being contained a lot better, whereas with the heatproof mats there was a lot more heat lost to the atmosphere, one because it only covered 3 sides and had a large amount of open space where the heat could escape and two, the tin foil was shiny and could help reflect the heat back into the experiment whereas the heatproof mats could not.
Analysis and Conclusion:
This experiment showed us that as you increase the amount of Carbons in the alcohol, it's energy content increases. It shows us that when an alcohol has one extra carbon it's energy produced per mole increases by exactly 492 kjmol-1
The amount of energy produced per mole increases because the more carbons there are per molecule of fuel, the larger the molecules will be. Therefore there will be more products; so on the other side of the equation there will be more moles of CO2 and H2O, which will have been formed. This means that there will be more bonds formed and so the amount of energy given out will be greater.
This links to my original prediction that this would happen but when put in to practice, my results have not come out anyway near the results that I predicted but this is due to experimental error. Although my experiment did not show exactly these results, it still followed the trend of increasing apart from Pentanol, where one of the results was a long way out and that could have been for any reason. I am still confident that what I predicted is correct and that there is a link between the energy produced per mole and the number of Carbon atoms in that alcohol.
Evaluation:
The results that obtained were really different to those that I had predicted. On my graphs this is shown very much, I have many anomalous points and hardly any of the points are on the line of best fit. My results only have a vague positive correlation but the actual results I calculated show a very accurate correlation - proved by the formula:
492n + 37 = enthalpy of combustion. ('n' is the number of Carbons in the fuel.)
With the calculated results there is a special pattern. If you find the differences between the different enthalpy of combustions and relate them to the amount of Carbons there is a formula that relates the amount of Carbons to the enthalpy of combustion. To do this I found that there is exactly 492 kjmol-1 between each one as a carbon is added. This means that the formula had something to do with 492. I found that if you times the number of carbons by 492 and then add 37 you get the expected enthalpy of combustion.
There are many reasons for the inaccuracy of my results:
* Due to human error. Either in reading measurements, not being careful enough and allowing heat to escape, also leaving the lid off of the fuel burner so that the fuel can be evaporated out.
* The energy that is lost in the light produced by the flame.
* The energy lost in heating up the wick until the flame starts to burn properly.
* The heat being lost to heating up the glass in the boiling tube.
* The heat lost to the surroundings, mainly from the tin foil having nothing over the top so heat could escape that way. Also there is the time between when you put the foil on and when you are pulling it off, where heat could be lost.
* The thermometer only takes the temperature of the water around it whereas it may not be totally accurate because the water further down may be warmer and the water above may be cooler.
* The distance between the bottom of the boiling tube and the clamp stand varied by a few millimetres which may have made a difference in the distance the heat has to travel to heat the water.
* The scales that measured the mass of the fuel in the burner were not digital and so there could have been some human error in reading them and some inaccuracy.
* The time between, when we put the flame out and checked the temperature of the water and the mass of the fuel might have been slightly altered each time, which meant that maybe the temperature had continued to increase by the time we measured it.
There weren't any major problems, and the experiment was done safely. There were two incidents that we could prevent in the future where the water got to boiling point quickly and wasn't stopped in time apart from that everything went relatively well.
If we were to do this experiment again, I think that we should use more tin foil so that we can cover also the top of the experiment because heat was being lost there. I think that we could be more careful in reading off measurements from rulers or scales. I also think that we should have definitely made sure that the distance between the clamp stand and the bottom of the boiling tube were exactly the same each time that we did it instead of being roughly the same.
Next time we could measure a different variable such as seeing how long each fuel takes to heat a boiling tube of water to 100?C. This would be able to show us how quickly each fuel can produce enough energy to bring the water to boiling point. We could also use the five different fuels and have exactly the same amount of mass in each and burn them, and time which one burns up the fastest. This would also show us how quickly each fuel breaks and makes bonds.
