An Investigation into the Burning of Fuels

The experiments were done again but changing the alcohol being burned to ethanol, propanol and then butanol last because it created a lot of soot. Goggles were worn throughout the experiments.

Preliminary experiments:

I did some preliminary experiments to decide on the distance from the spirit burner wick and the conical flask, the volume of water in the conical flask and the time taken to burn the fuel.

Preliminary results:

I have decided to keep the distance from the wick and the conical flask at 10cm because at 15cm away the temperature rise was too small and at 5cm the temperature rise was too great so the readings may have been innacurate. I have decided to use 100ml of water and burn the alcohol for 2 minutes because any longer created a large temperature rise.

Theory:

For a combustion reaction to take place a certain amount of energy (activation energy) is needed to break the bonds. Once the activation energy is achieved the bonds break an d then new bonds form which provides energy.

The total energy released from Methanol can be calculated as follows:

                H

        H        C        O        H

                H

CH3 OH + 1.5 O2                 CO2 + 2 H2O

Bonds needed to be broken are:

3 C – H                3 x 410 = 1230kJ

1 C – O                1 x 360 = 360 kJ

1 O – H                1 x 460 = 460 kJ

1.5 O = O                1.5 x 496 = 744 kJ

(Where 410, 360, 460 and 496 are the bond energy levels)

Total energy level needed to react 1 mole of CH3 OH and 1.5 mole of O2 = 2794kJ

The energy released can be calculated as follows:

2 C = O                2 x 740 = 1480

4 H – O                4 x 460 = 1840

(Where the bond energies are 740 and 460)

Total energy released in forming 1 mole of CH3 OH and 1.5 mole of O2 = 3320kJ

Total energy released = 3320 – 2794 = 526kJ/mole

The energy released in other alcohols (per mole) is shown below with the equation:

Ethanol = 1012kJ

C2H5OH + 3O2                 2CO2 + 3H2O

Propanol = 1498Kj

C3H7OH + 4.5O2                   3CO2 + 4H2O

Butanol =  1984kJ

C4H9OH + 6O2                 4CO2 + 5H2O

Therefore the energy released per gram is:

Methanol

Mr = 12 + (3x1) + 16 + 1 = 32

Since 1 mole of methanol is 32g

∴32g of methanol releases 526kJ of energy

∴1g releases 526/32 = 16.4kJ/g

Ethanol

22.0kJ/g

 

Propanol

25.0kJ/g

Butanol

26.8kJ/g

Prediction:

I predict that Methanol will produce the least amount of energy because there are fewer bonds to be broken and formed therefore producing less energy.

I predict that Methanol will produce 16.4kJ/g of energy as found in the theory. I predict that Ethanol will produce 22.0kJ/g, Propanol will produce 25.0kJ/g and Butanol will produce 26.8kJ/g.

Analysis:

From my results I have found that Methanol releases the least energy per gram, then ethanol, then porpanol, then butanol. The energy released does not really increase by regular amounts. The increase from propanol to butanol was very small, this is because butanol did not burn efficiently because it produced a lot of soot.

Evaluation:

My procedure did give me sufficient results but a lot of energy was lost because of draughts in the room making the flame wobble about. This could have been overcome by using a chimney around the flame. The butanol result was an anomalous result because it lost a lot of energy producing soot. The quality of my results were generally quite accurate but some of the temperature rises were different to the repeats, that is why I did three repeats; so that any anomalous results would not be so obvious.