Comparing the enthalpy changes of combustion of different alcohols.

Bond Enthalpy = 5590.5 – 8144 = -2553.5 KJ mol -1 

Butan-1-ol

CH3 (CH2)2CH2OH (l) + 6O2 (g)                4CO2 (g) + 5H2O (l)

Bond Enthalpy = 6336 – 10550 = -4214KJ mol-1

Pentan-1ol

CH3 (CH2)3CH2OH (l) + 71/2O2  (g)            5CO2 (g) + 6H2O (l)

Bond Enthalpy = 8733.5 – 12956 = -4222.5 KJ mol-1 

Hexan-1-ol

CH3 (CH2)4CH2OH (l) + 8O2 (g)               6CO2 (g) + 7H2O (l)

Bond Enthalpy = 9808 – 5554 = 4254 KJ mol-1 

Apparatus

1 x goggles

1 x thermometer

1 x 100cm3 measuring cylinder  

1 x copper can

1 x balance

1 x Bunsen burner

1 x Retort stand and clamp

2 x heat proof mats-draught shielding

6 x spirit burners containing

1. Methanol
2. Ethanol
3. Propan-1-ol
4. Butan-1-ol
5. Pentan-1-ol
6. Hexan-1-ol  

Diagram

Method  

1. Put 100cm3 of cold water in a copper calorimeter and record its temperature.
2. Place the copper can in the clamp on a retort stand over the spirit burner containing the relevant alcohol. Make sure that the copper can is 5cm away from the wick of the spirit burner.
3. Weigh the spirit burner.
4. Replace the burner under the calorimeter and light the wick.
5. Use the thermometer to stir the water all of the time it is being heated. Keep heating until the temperature of the water has risen by 200C.
6. Extinguish the burner and keep stirring the water and record the highest temperature reached.
7. Weigh the burner to see what mass of fuel has been burned.  

Fair Test 

In this experiment there are certain variables that need to be controlled. These are,

Mass of water- I will control this by using the same volume of water each time by using a measuring cylinder and measuring exactly to 100cm3.  

Amount of wick on the burner-I will keep this constant at 1cm so that the height of the flame will be the same.

Height of can above the flame-the height of the copper can above the flame will be 5cm for each experiment.  

Time of burning-This would depend on how long the water takes to make the thermometer a difference of 200C.

Type of alcohol-The position of the OH group would affect this. The bond energies between two atoms would vary according to where in the molecule the pair of atoms is. I have decided to use propan-1-ol, butan-1-ol, pentan-1-ol and hexan-1-ol because this means that the alcohols that I compare will be all straight chain alcohols and they will all have their OH group joined onto the end carbon. I will need to use alcohols with the only difference being the number of carbon atoms in the molecule.

Number of times experiment is repeated-I will repeat the experiment three times for each alcohol to improve the accuracy of my results.  

Same copper can- I will use the same copper can each time so that this will improve the conductivity of heat from the flame to the water.

Purity of the alcohol-all of the alcohols will be 99% pure.  

Risk Assessment

There are some risks with my experiment, which I will need to identify and prevent from becoming hazardous. These are:

1. It will be vital that I keep the lid on the spirit burner, because the alcohols ignite above 130C.
2. The vapours of the alcohols produce narcotic effects if inhaled therefore victims should be removed to fresh air.
3. If swallowed the mouth must be washed out with water.
4. If it is spilt on eyes, flood with running tap for 10 minutes, then seek medical attention.
5. If spilt on clothing or skin remove clothing to prevent fire risk and wash the affected area.
6. If spilt shut off all sources of ignition and wash with mineral absorbent.  

Results

*The results are shown in the table over the page.

I worked out the total enthalpy changes of the different alcohols by firstly using the equation from the beginning, which gave me 8400J as the enthalpy transferred to one mole of all of the fuel. I then used this to work out the total enthalpy change of combustion.  This was done using a certain number of steps, which are shown below in working out the enthalpy change for methanol.  

1. Write the formula of the fuel-Methanol = CH3OH 

2.Work out the mass of 1 mole of the fuel using the equation, mass/molar mass = 1.10/32 = 0.03moles 

3.Work out the energy transferred by using one mole of the fuel = 1mole/0.03 x 8400J = 280 000J

4.The enthalpy change of combustion is worked out by doing, 280 000/1000 = -280KJ mol-1

The same steps were conducted for each of the alcohols and the answers are shown in the table below.

I used these results to draw a graph to show these results against the number of carbon atoms to emphasis the trend that my results imply. Then I drew another graph to show these results against the bond enthalpies I worked out at the beginning to show the differences in enthalpy change between theoretical and experimental results.

Analysis

My results show that the heat produced by combustion is related to the number of carbon atoms in the alcohol. The more carbon atoms there are and the longer the chain is, the higher the enthalpy change of combustion. This is due to increased presence of hydrogen atoms, which means that there are more bonds to break. Therefore the more bonds that are broken, the more heat that is produced and so the higher the enthalpy of combustion. Methanol is the shortest alcohol that I have tested with the least carbon atoms therefore this has the lower energy of combustion, whereas Hexanol is the longest alcohol that I have tested with more carbon atoms; therefore this has the highest energy of combustion.  I expected these results, because of my predicted enthalpies which I made earlier on using the bond energies. My graph re-enforces my results as it shows a clear increase in gradient as the number of carbon atoms increases. These results occurred, because the more carbon atoms there are the more energy is required to break and then make the new bonds. Therefore the more bonds that are broken the higher the overall enthalpy change would be. My second graph shows that the enthalpy changes of combustion for the estimated bond enthalpies of the alcohols are higher than my results. This is because the data that I used to work out the enthalpies of combustion will have been collected under controlled standard conditions (298Kelvin and 1 atmospheric pressure), using specialized equipment; therefore the overall energy transferred is likely to be higher and more reliable.    

 

The precision of each of the measurements that I have made is indicated by the % error. This is shown below,

I have identified that there will have been energy loss from the spirit burner as there was not adequate draught shielding. All that was available to me was heatproof mats, which were fairly efficient, but did not restrict all of the heat loss to the surroundings; therefore this will have caused the energy transfer to the copper can to become less and consequently will have affected the final energy transfer calculation. The volume of the water may not have been completely accurate, which will have also had an effect on the overall energy transfer calculation. This is because if the measurement were over 100 cm3 then I would have expected the percentage error to be higher, whereas if it had been slightly below 100 cm3 then the percentage error will have been lower. Also there may have been incomplete combustion and evaporation therefore this will have led to the percentage error becoming higher, because the products will have been carbon monoxide and carbon which can be combusted further to make up the energy loss. I think that the lack of appropriate and reliable draught shielding had the biggest effect on my results overall, because if there had been access to specialized equipment then there will have likely to have been less energy loss to the surroundings; therefore my results will have been more accurate.  

I can see how accurate my answers are compared to Data Book answers by working out the percentage difference for each of the alcohols and then comparing which is the most reliable answer. This procedure is shown below,

(i) Methanol

My Answer = 280 KJ mol-1

Data Book Answer = 726 KJ mol-1

% Difference = 280/726 x 100 = 0.38 %

(ii) Ethanol

My Answer = 420 KJ mol-1

Data Book Answer = 1367 KJ mol-1

% Difference = 420/1367 x 100 = 30.72 %

(iii) Propan-1-ol

My Answer = 646 KJ mol-1

Data Book Answer = 2021 KJ mol-1

% Difference = 646/2021 x 100 = 31.96 %

(iv) Butan-1-ol 

My Answer = 933.33 KJ mol-1  

      Data Book Answer = 2676 KJ mol-1

% Difference = 933.33/2676 x 100 = 34.87 %

(v) Pentan-1-ol 

My Answer = 1400 KJ mol-1

Data Book Answer = 3331 KJ mol-1

% Difference = 1400/3331 x 100 = 42.02 %

(vi) Hexan-1-ol

My Answer = 2800 KJ mol-1

Data Book Answer = 3984 KJ mol-1

% Difference = 2800/3984 x 100 = 70.28 %

 

The most accurate answer was for Methanol. This might be because there are less bonds to break and make therefore there is less chance for any excess energy to escape, whereas in Hexanol there is a bigger percentage difference because there are more bonds to break and make therefore my answer is likely to be smaller than the Data book answer because there is more energy needed and therefore more excess energy loss.

 

Evaluation

By looking my experiment I can see that the accuracy of my results was fairly good because there provided a good trend to demonstrate that my prediction was correct. However the percentage errors and percentage differences are quite high. I think that these may have occurred due to procedural errors and not the precision of my calculations. I have identified some of these procedural errors, which are the measuring of the alcohol, the reading of the thermometer, (as I only could read to the nearest degree of accuracy) measuring of the water to be heated and insufficient draught shielding.

If I were to do this experiment again I would make a few alterations. To prevent less heat escaping I would shorten the distance between the burner and the copper can. However this would not prevent all of the heat form escaping therefore I would propose that I would insulate the can and the spirit burner to ensure that the water does not lose any heat and that all of the heat for the burner goes to the water in the can and doesn’t escape. As there may have been incomplete combustion I would overcome this next time by ensuring that the wick is long enough and that there is a good air supply as\oxygen is a vital component in combustion. The measuring of the water temperature in the can was a problem therefore if I had al longer thermometer that read to the nearest 0.1 0C then this would make my readings more accurate. Despite the fact that stirring the water will have made a difference, it will have been done in an enclosed beaker, which will have increased the chances of evaporation and consequently reducing the temperature of the water. Also the time taken to reweigh the fuel after burning may have affected the results because there may have been evaporation as the fuel cooled down therefore this will have decreased the mass.

Another way in which I could improve the reliability of my results would be to user a bomb calorimeter, because this would prevent any of the problems occurring which will have affected the results from before. By using this type of equipment I would expect the results to be closer to the predicted enthalpies of combustion. Below is a diagram of a bomb calorimeter.

 

There were also other areas of this experiment that I would have liked to have investigated but time was limited. If it was possible I will have investigated into whether the position of the –OH group within the molecule affects the enthalpy of combustion and also if branching in the molecule has any effect on the enthalpy of combustion.  

References: Salters Advanced Chemistry – Chemical Ideas (Second Edition) and Chemical Storylines (Second Edition), Chemistry in Context (Third Edition) Graham Hill and John Holman