Methanol - Bond Energy Calculations

= C = O x 8 (803)

O - H x 10 (464)

= 11064kj

?E = 8760 - 11064kj

= -2304

= -3575kj/mol

Pentanol

Bond Energy Calculations:

Pentanol + Oxygen › Carbon dioxide + Water

2C5H11 OH + 15O2 › 10CO2 + 12H2O

= 2 moles of Pentanol

Structure of Reactants:

2x 15x

H H H H H

H - C - C - C - C - C - O - H O = O

H H H H H

= H - C x 22 (435)

C - O x 2 (358)

O - H x 2 (464)

O = O x 15 (497)

C - C x 8 (347)

= 21445kj

Structure of Products:

0x 12x

O = C = O H - O - H

= C = O x 20 (803)

O - H x 24 (464)

= 27196kj

?E = 21445 - 27196kj

= -5751/2

= -2875.5kj/mol

Note:

Due to the fact that all these bond energies are negative they are all exothermic reactions, and so take energy from their surroundings.

j

Preliminary - method:

Foremost, I will identify the variables as follows:-

The fuel to be burned; [fixed variable]

Volume/Mass of Fuel; [fixed variable]

Amount of water heated;

Beaker/Test tube, shape, size etc

Distance of burning fuel from the beaker/test tube;

Size of watch glass/crucible etc;

Room temp;

Heat shield;

Method of lighting the fuel.

I propose to carry out a number of short preliminary experiments in order to establish the quantity of different variables and to find the best amounts and techniques so as to give me the most accurate results in the subsequent full experiment.

Note:

The amount of fuel is to be fixed at 1 ml, although there is not an equivalent mass for each fuel this has been noted. This also applies for the fuel to be burned obviously that has to be changed.

First, I plan to experiment with the amount of water to be heated in order to gain the clearest temperature difference results. I plan to experiment with 3 different amounts of water; 25, 50 and 75 ml. because I am using 1ml of fuel I am unable to use a beaker other than one that is of 100ml volume due to the fact that a larger beaker would require more than 75ml water in order to submerge the bulb of the thermometer.

The results were as follows:-

Fuel - Methanol

25ml

50ml

75ml

2

2

2

Start Temp. 21?C

29?C

29?C

27?C

28?C

24?C

23?C

This clearly shows that the biggest change in temperature was using 25 ml of fuel, consequently I will use this amount as I believe it will give me the clearest results as well as submerging the bulb of the thermometer.

Although I am unable to test any other variables I have thought about them and noted my reasons for particular ones below:

Due to the way I will carry out my experiment (see below) I have decided to use a 10 cm diameter watch glass in which to place my fuel for burning, this is because there is a larger surface area for the flame to be in contact with the underside of the baker, however I have considered there will be more evaporation in proportion to a smaller watch glass.

I will also use a windshield to firstly direct my flame towards the beaker and to prevent the flame from being interfered with; also I hope to lose less heat to the surroundings. In addition the height of the windshield shall also be level with the bottom of the beaker. 6cm from the watch glass to beaker underside

In terms of the room temperature I cannot control this however it should always be between 19?C - 21?C and should not influence my results greatly

Preliminary variables identified:

Therefore, I will use a 100ml beaker with 25ml of water, the beaker will be 6cm above a 10cm diameter watch glass and I shall use 1 ml of fuel. I am unable to control the room temperature.

Once I had the results for the preliminary experiment I set forth designing the apparatus to us.

Option 1 Option 2

I decided to use option one because there is much less heat loss into the surroundings

Final Diagram:

Equipment:

. 100ml beaker

2. Water

3. Fuel

4. Watch Glass

5. Heat shield

6. Splint

Method:

As stated above I will use 1ml of fuel, either Methanol, Ethanol, Propanol, Butanol or Pentanol and I shall heat 25 ml of water on the above appropriate apparatus, chosen, dually as it shall give me optimum results, whilst also being safe. To calculate my results I will take starting and finishing temperatures, subtract the two and take an average temperature rise, subsequently I will be able to work out the energy produced, the energy that should have been produced the comparison of fuel efficiency.

To obtain optimum results I have devised the above set up of apparatus in order to exploit the efficiency and fuel transferral. This apparatus consists of a 100ml beaker, windshield and watch glass. Though the beaker has a small surface area it covers fully the flame and decreases the amount of heat lost due to conduction and convection through the beaker. In addition the windshield lessens the flickering of the flame, keeping it heating the beaker more consistently also heat is less likely to be lost between the flame and beaker. Lastly, the watch glass gives the flame a large surface area, so as to heat the water in a more concentrated fashion, it is also easier to observe the flame and light it safely. The experiment for each fuel shall be carried out three times so as to give me an average result.

Safety:

In terms of safety I shall wear goggles throughout out the experiment, in addition I will use a splint to light any fuel, so as to not endanger myself getting burnt. Whilst using a Bunsen burner I will use a safety flame. Moreover, I followed all safety advice from my teachers. Also my apparatus is chosen as it is safe, both water and fuel would be hard to spill and the fuel will also be lit safely. Lastly whilst carrying out all experiments in the laboratory I shall stand, enabling me to move quickly if needed. Overall the main safety hazards are getting burnt or by spilling alcohol over bare skin.

Prediction:

My prediction is based on the molar energy calculations; I have calculated the following.

Energy Released per 1 ml

Methanol: = -589.5 kj/32g =-18.42 x 0.793 = 14.60j

Ethanol: = -1161 kj/46g =-25.24 x 0.789 = 19.91j

Propanol: = -1732.5 kj/60g =-28.88 x 0.804 = 23.22j

Butanol: = -2304 kj/74g =-31.14 x 0.810 = 25.22j

Pentanol: = -2875.5 k/j80g =-32.68 x 0. 815 = 26.63j

This shows me that he higher the bond energies the more transferral of heat will take place, due to the increased number of breaking bonds. Also because I am measuring the fuel by ml there are different densities. Therefore effectively there is more Pentanol than Methanol. In short, due to the more carbon atoms the alcohol compound has a higher bond energy and subsequent heat transferral.

Task:

To investigate how the combustion of 1ml, of at least four alcohols raises the temperature of a particular amount of water.

Results:

Four tests for each fuel and the average temp rise

Fuel

Exp. 1

Exp. 2

Exp. 3

Exp. 4

Temp Rise

Methanol

Ethanol

Propanol

Butanol

Pentanol