ΔHfθ (O2) = 0 kJmol-1
ΔHfθ (CO2) = -393.51 kJmol-1
ΔHfθ (H2O) = -285.83 kJmol-1
The value of A is -277.69, so -A will be 277.69.
The value of B is -787.02 – 857.49 = -1644.51.
C is calculated as 277.69 – 1644.51 = -1366.82
So ΔHcθ for ethanol is –1366.82 kJmol-1
The enthalpy cycle for propanol cannot be constructed, as the ΔHfθ (C3H7 OH) is not found in any data book. Without knowing this we cannot fully work out the value needed.
As it is not possible to use Hess’s Law to work out ΔHcθ for propanol we can use bond4 enthalpies instead.
Bonds Breaking Bonds Forming
C–C x 2 = 2(347) = 694 kJmol-1 O –H x 8 = 8(464 kJmol-1) = 3712
C–H x 7 = 7(413) =2891 kJmol-1 C=O x 6 = 6(805 kJmol-1) = 4830
C–O x 1 = 358 = 358 kJmol-1
O–H x 1 = 464 = 464 kJmol-1
O=O x 4.5 = 4.5(498) = 2241 kJmol-1
Total = 6648 kJmol-1 Total = 8542 kJmol-1
To work out the enthalpy change, take away the total of bonds forming from the total bonds breaking, leaving an energy value -1894 kJmol-1.
The problem with bond enthalpies is that they are average values calculated from many different molecules. The actual value of bond enthalpies depends on which molecule the bond is in.5
From these two values we can see that propanol has a higher enthalpy change of combustion than ethanol; meaning that more energy is released per mole of propanol that is combusted.
HYPOTHESIS
I predict that propanol will be the better fuel and that it will release the greatest amount of energy per mole. I will use the energy released per gram to compare the two alcohols. I predict this because I believe that as more carbon atoms are added the greater the amount of energy it will release. This is because the potential energy of a fuel increase as the carbon atoms increase, thus when released the energy value per gram will be greater. I can use the enthalpy change of combustion value of methanol and butanol in relation to my prediction. If I am correct methanol should be less than ethanol and butanol should be greater than the three.
ΔHcθ methanol is -730 kJmol-1 (data book value6)
ΔHcθ ethanol is -1366.82 kJmol-1 (calculated from enthalpy formations)
ΔHcθ propanol is -1894 kJmol-1 (calculated from bond enthalpies)
ΔHcθ butanol is -2670 kJmol-1 (data book value)
These values are coherent with what I previously suggested.
METHOD
The best way to work out enthalpy changes is to use a method called calorimetry. It is a technique that you can use to determine the amount of energy in a fuel per gram per fuel. Calorimetry measures the amount of energy that is needed to heat a known mass of water. The energy transferred is then used to calculate how much energy is released per mole of alcohol.
Scientific knowledge tells me that 4.2 J of energy are needed to raise the temperature of 1g of water by 1 K. This can be used to find the specific heat capacity of water, which is 4.2 Jg-1K-1.
Energy transfer is measured in joules and is calculated by the following equation:
Energy μ = m x c x ΔT joules
Where m is the mass of water (kg), c is the specific heat capacity of water and ΔT is the change in temperature.
The change for one mole can be worked out by (Energy μ / γ) x Mr of alcohol.
Where γ is the loss of mass of alcohol and Mr is the molecular mass. Mr of ethanol is 46 and the Mr of propanol is 60.
Apparatus
The apparatus that I used to take my measurements were: a copper calorimeter, a clamp and retort stand, a heatproof mat, a 10ml measuring cylinder, a spirit burner, a temperature probe and computer, an electronic balance, ethanol, propan-1-ol, a lid for the calorimeter and aluminium foil for insulation.
The reason why I used a copper calorimeter is because copper has a high thermal conduction value; this basically means that it is a good conductor of heat so a lot of the heat that the copper receives will be passed on to the water, which I am then able to measure. I decided to use a temperature probe instead of a regular thermometer because the probe is more accurate and can measure to point one of a degree. Also it reduces any human errors on reading the temperature. The electronic balance gives an accurate mass figure to 2 d.p. I used the fact that 1 ml of H2O weighs 1g, so while measuring out the different masses of water I can keep results accurate. Propanol can show structural isomerism so by using propan-1-ol the alcohol group is on the end carbon making it a straight chain alcohol, like ethanol (ethanol cannot show isomerism). I do not know whether the position of the alcohol group or branching within the molecule effects the enthalpy change. I need to use alcohols with as similar structures as possible, with the only difference being the number of carbon atoms within the molecule. The masses of water I decided to use for the experiment were 50g and 100g. I conducted two tests with mass 50g and two with 100g for each alcohol, thus obtaining four results for each. I used two different masses to determine whether the accuracy of my experiment increases with a greater or smaller amount of water.
On a trial run of the experiment it was found that too much heat was lost to the surroundings. For this experiment I can reduce the heat loss by putting a lid on the calorimeter with a small hole for the probe, by putting the calorimeter as close to the wick as possible, and by enclosing the set apparatus with aluminium foil, which would reflect the heat back. However space would be needed for the air to get in.
Measurements to be taken:
Mass of Water
Change in temperature = End temperature – initial temperature
Initial mass of alcohol = mass of burner and alcohol – mass of empty burner
Final mass of alcohol = mass of burner and alcohol at end of experiment – mass of empty burner.
Change in mass of alcohol = initial mass – final mass of alcohol
Procedure
I placed the empty calorimeter on the balance to obtain its mass. Then Filled up the measuring cylinder, to the 10ml mark, with water and added it to the calorimeter. I did this 10 times, checking the weight on the balance, and making sure that it increased by 100g (approximately). Next I weighed the spirit burner with the lid, then added the alcohol and weighed it again. Both weights were recorded.
I set up the computer and the temperature probe and clamped the calorimeter with the clamp arm and placed the lid on it. After I placed the probe inside the lid and then recorded the starting temperature of the experiment.
I placed the spirit burner on the heatproof mat and lowered the clamp to about 3cm above the burner, to minimise heat loss.
Finally I used a sheet of foil to envelope the set up leaving enough space for air to get in. then I removed the lid of the burner and lit the wick.
When the temperature had risen by a significant amount (+/-50oC) I replaced the lid back on the burner, extinguishing the flame.
I waited for the temperature to reach its highest point then recorded this figure, which was the final temperature.
The spirit burner had to be weighed again in order to calculate the loss of mass of alcohol. I made sure that when not lit the lid was always on the burner as there could be a loss in mass due to evaporation.
I repeated this procedure again with 100g of water, and then I did it twice using 50g of water. Instead of filling the calorimeter 10 times, this time I did it 5.
This method needs to be carried out for both alcohols.
RESULTS TABLE
ANALYSIS
Here is an example of the way in which to work out ΔHcθ ethanol.
Using my results from the first column in my table and the equations for energy transferred and enthalpy change that I stated before, I can calculate the energy transferred through combusting 1 mole of ethanol.
Energy μ = 100 x 4.2 x 45.3
= 19026 J
ΔHcθ = (19026/2.26) x 46 Jmol-1
= 387254 Jmol-1
= 387.25 kJmol-1
Value for ΔHcθ(C2 H5 OH) = -387.25 kJmol-1
From my results the average can now be calculated for each alcohol.
For ethanol the average is calculated as the total of the enthalpy changes divided by four, which would give –1509.6/4 = -377.4 kJmol-1.
For propanol the average is calculated as the total of the enthalpy changes divided by four, which would give –2398.68/4 = -599.67 kJmol-1.
Now that I have obtained these averages I can calculate the enthalpy change of combustion for 1g of each fuel.
Ethanol = (19026/2.26 + 18354/2.22 + 11655/1.42 + 11172/1.41) / 4
= 8204.33 Jg-1
= 8.20 kJg-1
Propanol = (21168/1.94 + 19404/2.13 + 9954/1.03 + 9366/0.91) / 4
= 9994.50 Jg-1
= 10.0 kJg-1
According to my results propanol can be said to be the better fuel as it gives off the greater amount of energy per gram. As an average for the enthalpy change of combustion propanol was also the greater result. I can relate these results to my predictions.
The value of ΔHcθ for ethanol that I calculated using an enthalpy cycle was -1366.82 kJmol-1.
The value of ΔHcθ for propanol that I calculated using bond enthalpies was -1894 kJmol-1.
Calculate values of enthalpy change of combustion gives the ΔHcθ for ethanol as -1370 kJmol-1 and for propanol as –2010 kJmol-1. This shows how inaccurate working out ΔHcθ using bond enthalpies for propanol is.
These results show an immense difference with those that I acquired from my experiment. This means that the energy from the fuel was not efficiently transferred to heat up the water. In other words the amount of heat lost to the surroundings was still too great despite my attempts to reduce the loss.
Putting all this to one side I can still say that my prediction was correct, that propanol would release the greater amount of energy per mole.
EVALUATION
The bad point about my procedure was the weak attempt to reduce the heat loss to the surroundings. Although the foil acts as a good insulator it could not prevent the vast amount of heat that was lost.
The good point about my procedure was that I used electronic devices when available for measuring masses and temperature. The electronic balance gave an accurate figure to 2 d.p., while the computer gave us a reading for temperature to 0.1 degree.
There were not any significant anomalous results, however the results for the enthalpy changes are not the same. For ethanol the range of the results is 22.77 kJmol-1, and for propanol the range is 105.12 kJmol-1. The fact that the range for the enthalpy changes of propanol is so large could be because of the different amount of heat lost. The highest result for ΔHcθ of propanol was 654.71 kJmol-1, the fact that this was so high compared to the other results may be because it was done on a separate day.
My experiment was fairly accurate however inaccuracies could have occurred from the lid of the calorimeter. The first day a plastic lid was used and some of it melted. The second day foil was used instead. This would change how much heat was lost. Inaccuracies could also have occurred by how well insulated the experiment was.
The method that was more accurate was by using Hess’s Law. Even though this proved to be more accurate it is a specific value and for each experiment that is done conditions change and so would the results
One way of improving my results without redoing another experiment would be by measuring the energy transferred to the calorimeter. The energy form the fuel is used to heat up the copper calorimeter, which then transfers the heat to the water. Not all the heat goes to water. The specific heat capacity of copper is 0.385 Jg-1K-1, this value can be used to measure the energy transfer with the known mass of the calorimeter. Unfortunately this was not measured and therefore can not be calculated, but it is worth considering.
A way of improving the experiment would be to use a method that does not use specific heat capacities. There are two ways to do this. One way would be to use a flame calorimeter and the other a bomb calorimeter.
REFERENCES
1 –
2 –
3 – Atkins, P.W., Physical Chemistry. UK: Oxford University Press, 1993
4 & 5 – Ratcliff, Brian, ed., et al. Chemistry 1. UK: Cambridge University Press,
2001
6 – Coulson, E.H., and Ingle, Richard, eds. Nuffield: Advanced Science. UK: Longman Group Ltd, 1994
