Enthalpy Essay, Research Paper
I am going to investigate the difference in enthalpy of combustion for a number of alcohols, the enthalpy of combustion being the ‘enthalpy change when one mole of any substance is completely burnt in oxygen under the stated conditions’. I will be attempting to find how the number of carbon atoms the alcohol contains effects the enthalpy change that occurs during the combustion of the alcohol. Method
I plan to measure the enthalpy change by burning the alcohol, using a spirit burner, I will then use the heat produced during the combustion of the alcohol to heat 100ml of water that will be situated in a copper calorimeter directly above the burning alcohol. The calorimeter is made of copper as copper has a high thermal conduction value, this basically means that it is a good conductor of heat so a lot of the heat the copper receives will be passed on to the water which I am then able to measure.
During the experiment I will be taking a number of measurements, I will firstly take the initial temperature of the water and initial mass of the alcohol I will then burn the alcohol until an increase in temperature of 20oc has occurred in the water I will then reweigh the alcohol.
The measurements
* Mass of alcohol burned (g)
* Temperature increase (oc) will tell me what mass of alcohol is used during combustion to cause the temperature increase of 20oc in the water, I can then work out the energy released per mole and compare these values and see which has the highest enthalpy of combustion. I will need to repeat my experiment a number of times and take an average so I am sure of an accurate result.
The set up of the apparatus that I plan to use is shown below
The set up of the apparatus as you can see is very simple, the calorimeter, which contains the 100ml of water, is held directly above the spirit burner by a retort stand and clamp. The calorimeter has a mercury thermometer in it, which are very accurate, this will be used to measure the water temperature. I have decided that the calorimeter should be held 1cm above the top of the flame produced by the burning alcohols as so to keep the experiment fair, this being as apposed to having it at a random height. I have also decided that the size of the wick from which the alcohol burns from should be constant on all the spirit burners, so to keep the experiment as fair as possible so I will adjust them accordingly so they are all the same length. I have decided that the length should be should be one cm, I will do this so that all the alcohols burn from the same surface area, this will mean that I will also have to use wicks of the same thickness. The experiments will be taking place in a laboratory so this means that the environment each experiment takes place in should be pretty constant i.e. room temperature etc, this will also help improve my results. Prediction
I am expecting that the alcohols with a greater number of carbon atoms within the molecule to have a higher enthalpy of combustion than the ones with less.
For any reaction to take place bonds must be broken and made, bond breaking requires energy while bond making releases energy. Bonds between different atoms require or release different amounts of energy when broken or made because they are different in strength. By looking at the equation for the reaction and more importantly looking at the bonds that are being broken and made, it is possible to work out an estimation for the amount of energy that will be released in the reaction. The estimation is worked out by applying the average bond enthalpies, an example for doing this is shown below for methanol
Methanol (CH3OH)
The balanced equation for the combustion of methanol is CH3OH(l) + 1.5 O2(g) CO2(g) + 2H2O(l)
Below is the type and number of bonds within each mole of reactants and products, they are shown with the amount of energy, measured in kilo joules per mole (DH/KJ mol-1) released or required for the particular bond
Methanol Oxygen Carbon Dioxide Water
CH3OH 1.5 O2 CO2 2H2O
3 C__H 413 1.5O=O 497 2C=O 740 4O__H 463
C__O 360
O__H 463
2062 745.5 1480 1852
The total energy required to break The total energy released in the forming the bonds in the reactants is of the bonds in the products is 2807.5 DH/KJ mol-1 3332 DH/KJ mol-1 the difference between the reactants and products is
-524.5 DH/KJ mol-1 From above, there are 2807.5 kJ mol-1 of energy absorbed initially by the reaction when the bonds are broken. Then 3332KJ mol-1 of energy is released by the reaction when the new bond are formed, overall this leaves a difference of 524.5 kJ mol-1 between the reactants and products, this energy is released by the reaction in the form of heat energy.
The value above for the energy released by the alcohol is only an approximation for the combustion of methanol this is because firstly the bond enthalpies vary slightly from one molecule to another and so the values used are only an average. The values given for the bond enthalpies also assume that the reactants and products are in a gaseous state but as you can see from the equation they are clearly not, with the water and the alcohol’s both being in a liquid state.
The alcohols that I plan to use in my investigation are methanol, ethanol, propanol and butanol. The estimation for the enthalpy of combustion, using the bond enthalpies are worked out below for each of the alcohols. Methanol CH3HO(l) + 1.5O2 (g) CO2(g) + 2H2O(l)
3 C__H 413 1.5O=O 497 2C=O 740 4O__H 463
C__O 369
O__H 436
2062 745.5 1480 1852 2807.5 DH/KJ mol-1 3332 DH/KJ mol-1
-524.5 DH/KJ mol-1
Ethanol C2H5HO(l) + 3O2 (g) 2CO2(g) + 3H2O(l)
C__C 346 3O=O 497 4C=O 740 6O__H 463
5 C__H 413
C__O 369
O__H 436
3234 1491 2960 2778 4723 DH/KJ mol-1 5738 DH/KJ mol-1
-1013 DH/KJ mol-1
Propanol C3H7HO(l) + 4.5O2 (g) 3CO2(g) + 4H2O(l)
2 C__C 346 4.5O=O 497 6C=O 740 8O__H 463
7 C__H 413
C__O 369
O__H 436
4406 2236.5 4440 3704 6642.5 DH/KJ mol-1 8144 DH/KJ mol-1
-1501.5 DH/KJ mol-1
Butanol C4H9HO(l) + 6O2 (g) 4CO2(g) + 5H2O(l)
3 C__C 346 6O=O 497 8C=O 740 10O__H 463
9 C__H 413
C__O 369
O__H 436
5578 2982 5920 4630 8560 DH/KJ mol-1 10550 DH/KJ mol-1
-1990 DH/KJ mol-1
The bond enthalpies worked out above clearly show an increase in the overall energy that is released as the alcohols increased in size. The table below shows the Calculated enthalpy of combustion, using bond enthalpy’s for the stated alcohols. The difference in the enthalpies of combustion from alcohol to alcohol, as they get larger is constant, this is no surprise though as the only difference as the alcohols get larger is an increase in size of one carbon and two hydrogen’s each time. Whether this difference is as constant in practice is another thing.
AlcoholCalculated enthalpy of combustion (DHc)Difference In Enthalpy of combustion (DHc)
Methanol524.5 DH/KJ mol-1
Ethanol1013 DH/KJ mol-1488.5 DH/KJ mol-1
Propanol1501.5 DH/KJ mol-1488.5 DH/KJ mol-1
Butanol1990 DH/KJ mol-1488.5 DH/KJ mol-1 Methanol is the smallest alcohol, it releases 524.5 kJ mol-1, the next largest, Ethanol is one carbon and two hydrogen’s larger, It release 1013 kJ mol-1 this is a difference of 488.5 kJ mol-1 this trend continues as the alcohols get larger. This is effectively because the only difference between the alcohol’s is a increase in size by one carbon and two hydrogen’s each time.
For my investigation I am going to use propan-1-ol and butan-1-ol as representatives of propanol and butanol. propanol and butanol are large enough molecules to form isomers etc, I have decided to use propan-1-ol and butan-1-ol because they have the closest structural arrangement to the other alcohols that I am going to be testing. Using butan-1-ol and propan-1-ol means all the alcohols that I am comparing have their OH group joined onto an end carbon and they are all straight chain alcohol’s. I need to keep as many of the factors in my experiments as I can the same, only changing what I have to, the variables, so that I get as accurate results as possible showing the correct pattern. I don’t know how or whether the positions of the OH group on the alcohol, and whether branching within the molecule effects the enthalpy change. I need to use alcohols with as similar structure as possible, with the only difference being the number of carbon atoms within the molecule as this is what I am investigating.
Risk Assessment
There are obvious risks with the experiment that I am going to do in the alcohol’s that I am will be using
* The alcohol’s are obviously flammable so need to be handled with care, avoiding spillages and kept in a suitable container, they should only used out of the way of other naked flames.
*I will be using a naked flame, so I will need keep my experiments out of the way of other experiments, people and flammable objects
* My experiment also produces heat and so the apparatus will heat up so they will need to be handled with care during and after the experiment has taken place.
Overall though my experiment is pretty safe, as long as it is carried out sensibly taking heed of the general laboratory rules.Results
The results I have attained from my experiments are shown below in the table, I repeated the experiment six times for each alcohol, this was decided mainly by the time that I had, but I also thought it was a suitable number to test. The experiment went pretty much to plan although I did have to make changes to the experiment to increase the accuracy of the results that I got, the details of this are explained later. Below in the table are the results that I got from my experiments ALCOHOLINITIAL TEMPERATURE (oc)FINALTEMPERATURE(oc)+/-(oc)INITIALWEIGHT (g)FINALWEIGHT (g)+/-(g)
Methanol214322213.18212.061.12
214322216.06211.080.98
204121211.08210.051.03
204020208.49209.581.09
214322205.70204.770.93
21 4120204.77203.761.01
Ethanol224321289.49288.720.77
204020288.72288.010.71
234320287.89287.200.69
204121287.1286.340.76
204020286.34285.610.73
204121285.5284.740.75
Propan-1-ol224220287.45286.930.52
224220287.01286.510.5
224220286.46286.060.4
214120286.04285.540.5
214221285.50284.930.57
214120284.9284.390.51
Butan-1-ol203919280.08279.580.5
193920278.34277.750.59
193819277.47277.000.47
183820276.57276.070.5
194021276.03275.500.53
173720275.56275.090.47 I have drawn a graph, below, that shows the temperature increase per gram for each alcohol. The temperature increase per gram although can not be used to compare for the enthalpy of combustion, as that is a measure of the enthalpy change per mole it highlights a problem that I found with my results for butan-1-ol. I started to notice that the results that I got from the butan-1-ol experiments were not showing the decrease in mass of alcohol needed to causes the temperature rise almost straight away and could not understand why this was happening. I checked the apparatus and then I noticed that the bottom of the calorimeter had a black sooty type substance on it, which I knew must have been affecting my results for butan-1-ol. I came to the conclusion that the black substance was carbon from the alcohol that had not been completely combusted in the reaction and was being deposited on the calorimeter, like soot in a chimney.
Although this may have also happened on the smaller alcohols it will not have been as severe as the activation energy for the combustion of the alcohols as they get smaller decreases. This can be explained by looking back to when I examined the bond energies, the smaller alcohol’s needed less energy to break the bonds in the reactants and so less energy is needed to initiate the reaction. The results that I got from my experiments for butan-1-ol will have been affected in two ways
*Firstly carbon is a better insulator than copper, it has a lower thermal conductivity value and so it will stop as much heat getting through to heat the water as there should be *Secondly the carbon has come from the spirit burner and for my results I am assuming that all the alcohol that leaves the spirit burner is combusting and releasing heat to contribute to the heating of the water which it is quiet clearly not. This will mean I am assuming that more alcohol is needed than really is to cause the increase in temperature. Apart from the butan-1-ol my results are as expected showing a clear increase in the energy released per unit mass of alcohol. So that it is possible for me to see which alcohol has the highest enthalpy of combustion I need to find which alcohol releases the most energy per mole. From the measurements I have taken I have the temperature increase per gram.
If you raise the temperature of an object you increase the energy of the particles it is made from, to do this you need to supply energy. The energy needed to raise the temperature is proportional to the mass of the substance and the temperature rise
Energy µ mass x temp increase The constant of proportionality depends on the substance you are heating, it is called the specific heat capacity.
Energy = specific heat capacity x mass x temp increase
(j) (J/g/ oc) (g) (oc)
The results that I am going to use are
AlcoholInitial TemperatureFinal Temperature+/-Mass of Alcohol Burnt
Methanol27oc49oc22oc1.12g First I need to find the heat exchanged to the water Specific heat Capacity of water,4.2
Mass of water heated,100g
Heat energy exchanged = Mass x Specific Heat x Temperature
To the water (g) Capacity Rise
(J) (J/g/ oc) (oc)
= 100 x 4.2 x 22
= 9240J
9240J is the heat energy taken in during the experiment, this needs to be converted to heat taken in per mole of alcohol burned 1.12g Methanol 9240J
1 g ” 9240J/1.12
32g ” 9240J/1.12 x 32
26400Jmole-1
So there is 264000 J released from 32g which is one mole, to change to KJ simply divide by 1000
Which gives
264kJmole-1
There is also heat absorbed by the calorimeter that I can also work out
Specific heat capacity of copper, 0.385
Mass of calorimeter 61.55g
Heat energy exchanged = Mass x Specific Heat x Temperature
To the copper (g) Capacity Rise
(J) (J/g/ oc) (oc) = 66.55 x 0.385 x 22
=521.3J
Again this is the heat taken in during the experiment and needs to be converted to heat taken in per mole of alcohol burned 1.12g Methanol 521.3J
1g ” 521.3J/1.12g
32g ” 521.3J/1.12g x32g
14894.3 Jmole-1
This can also be changed to kJm-1 by dividing by 1000 giving
14.9kJm-1 Adding the two values worked out above for the energy absorbed by the water and the calorimeter gives the total gives the total energy that I have measured to having been released by the combustion of the alcohol.
The total energy that I measured for the combustion of methanol example above is
264kJmole-1 + 14.9kJmole-1 = 278.9 kJmole-1 Below I have put in a table the average enthalpy of combustion for my results AlcoholAverage measured enthalpy of combustion
Methanol278.4 kJmole-1
Ethanol539.2 kJmole-1
Propon-1-ol995.6 kJmole-1
Butan-1-ol1214.6 kJmole-1 My results show a clear increase in the enthalpy of combustion as the alcohols get larger, Butan-1-ol, the largest alcohol that I have tested shows the highest enthalpy of combustion and methanol, the smallest in size has the smallest value for the enthalpy of combustion. This is as I had expected, as the enthalpy of combustion that I estimated earlier using average bond enthalpy’s had predicted. Below is a graph, which compares my results to the bond enthalpy values that I had worked out earlier.
The changing of my results into the form kjmole-1 has eliminated the error that I highlighted before. My results, which had shown propan-1-ol, releasing more energy than butan-1-ol, now tell a different story. This is because although they may have been affected it has not been enough to alter them completely, as changing them to form kjmole-1 has meant that they now follow the pattern that they should. Butan-1-ol now releases more energy than propan-1-ol. This change has occurred because one mole of butan-1-ol is heavier than one mole of propan-1-ol so although they release about the same energy per gram butan-1-ol releases more energy per mole
The graph above shows that my results show the same pattern for the enthalpy of combustion as the average bond enthalpy estimation worked out, but they do not show the same total amount of energy being released per mole. Remembering that the bond enthalpies are only estimations I need to compare my results to other more reliable results to see how accurate the results that I have obtained are.
It is actually possible using something called a bomb calorimeter to measure the exact enthalpy of combustion. This means that I can compare these values against my results and it will be possible to work out how exact the results are. The bomb calorimeter apparatus’s are specially designed to avoid heat loss by completely surrounding the ‘bomb’ with water. Heat losses can be eliminated altogether if the thermo chemical investigation is coupled with an electrical calibration. First of all, the chemical reaction is carried out in the calorimeter and the temperature is plotted against time before, during and after the reaction. The experiment is now repeated, but this time an electrical heating coil replaces the reactants. The current in the coil is carefully adjusted so as to give a temperature/time curve identical to that obtained in the chemical reaction. By recording the current during the time of this electrical calibration, it is possible to calculate the electrical energy supplied with great accuracy. This electrical energy is exactly the same as the energy change in the reaction. As it includes both the heat absorbed by the system and the energy lost in the system and the heat lost from the system it eliminates the need for heat loss correction. The values for the enthalpy of combustion given by the bomb calorimeter are