The combustion of Alcohols investgation

PROBLEM

Investigate the heat energy in a range of alcohol’s used as fuels.

HYPOTHESIS

I predict that octanol will release the most heat energy. This is because there is more bond energy in that molecule than the other alcohols.

Within a molecule there are bond energies that are holding the atoms together. When the fuel combusts a chemical reaction takes place, this breaks the bonds, this requires energy, and makes new bonds this gives out energy. The energy differences between the two tell us how much energy was given out or taken in. We can show this on a graph.

ENERGY CHANGES DURING A REACTION

To find the bond energies in the molecule of the alcohol we have to look at the how much energy is in the separate bonds of the molecule. Below we have a table of bond energies.

Bond Bond Energy (kj/mol)

C – OH 402

C – H 435

C – C 347

H – O 464

C = O 805

If we draw out the structure of each molecule involved in the chemical reaction we can easily find out how much energy is in that molecule.

STRUCTURE OF THE MOLECULES INVOLVED

METHANOL

ETHANOL

PROPANOL

BUTANOL

PENTANOL

HEXANOL

HEPTANOL

OCTANOL

CARBON DIOXIDE

WATER

BALANCED EQUATIONS

If we work in the bond energies into these balanced equations, and we subtract the energy taken in by the breaking of the bonds from the energy given out by the formation of new bonds we will get the total energy released.

METHANOL

2CH3OH + 3O2(r) 2C02 + 4H20

ETHANOL

2C2H5OH + 6O2(r) 4C02 + 6H20

PROPANOL

2C3H5OH + 9O2(r) 6C02 + 8H20

BUTANOL

2C4H7OH + 12O2(r) 8C02 + 10H20

PENTANOL

2C5H9OH + 15O2(r) 10C02 + 12H20

HEXANOL

2C6H11OH + 18O2(r) 12C02 + 14H20

HEPTANOL

2C7H13OH + 21O2(r) 14C02 + 16H20

OCTANOL

2C8H15OH + 24O2(r) 16C02 + 18H20

We can see from the table above Octanol releases the most heat energy. This clearly shows that there is a correlation between the number of bonds in the molecules and how much total energy was released.

I also predict that the amount of heat energy released will increase with the number of carbon atoms in the alcohol. I can show this on a graph.

This graph shows us that the number of carbon atoms in the alcohol is directly proportional to the heat release. The line is a straight line and we can make a liner equation for this that links the number of carbon atoms to the heat released. This is,

Heat released = (-574 x No. of carbon atoms) x 2024

VARIABLES

My variables in my experiment will be the temperature of the water in the beaker and the mass of alcohol burnt. Things such as amount of water and how much the alcohol raises the water temperature must be controlled.

FAIR TEST

To keep this a fair test we have to bear certain aspects in mind. The beaker the water is contained in must be the same shape because if it is not the flame may have more surface area of where to heat the water. The alcohol must be weighed accurately with scales that weigh up to, least, one decimal point. During weighing the spirit lamp must be covered to avoid and evaporation of the alcohol. The alcohol has to be weighed accurately before and after the experiment. The alcohol has to be blown out immediately when the water temperature has been raised 30 degrees; it must be covered after the experiment to avoid evaporation. The thermometer must be swirled around the water before a reading can be taken, this insures that you are measuring the temperature of the whole water not just the bottom of the beaker. The shape of the spirit lamp must stay the same and so must the wick length. If all this is done we can ensure that we will get an accurate reading.

DIAGRAM

METHOD

1. Set up apparatus as diagram.

2. Weigh the spirit lamp + alcohol on scale note down weight.

3. Note down the water temperature.

4. Light the wick and let the alcohol heat up the water until it is raised by 30 degrees.

5. Blow out the flame and weigh the alcohol + spirit lamp immediately. Note down weight.

6. Work out the mass of alcohol burnt by subtracting the weight before from the weight after.

7. Work out how much heat energy was produced.

First find out how much heat energy the water gained by using this equation.

Specific heat capacity = 4.2 kj/kg/oC

Heat gained by water = mass (Kg) x temp. rise x Specific heat capacity

Once you have found that divide it by the mass of alcohol burnt and then multiply it by the mass of 1 mol of that alcohol.

8. Repeat steps 2 – 7 with the other alcohols and form a table of results.

ANALYSIS

Alcohol Alcohol used in, (g) Average

1st Experiment 2nd Experiment (g)

Methanaol 1.32 1.28 1.3

Ethanol 1.21 1.25 1.23

Propanol – – –

Butanol 0.79 0.72 0.805

Pentanol 0.69 0.73 0.71

Hexanol 0.69 0.70 0.695

Heptanol – – –

Octanol 0.66 0.64 0.65

Alcohol How much 1 mol weighs (g)

Methanaol 32

Ethanol 46

Propanol 60

Butanol 72

Pentanol 88

Hexanol 102

Heptanol 116

Octanol 130

Alcohol Energy released (kJmol)

Methanaol -310

Ethanol -471

Propanol –

Butanol -1237

Pentanol -1404

Hexanol -1849

Heptanol –

Octanol -2520

All the graphs are show as negative because it is an exothermic reaction, and exothermic reactions are shown as negative.

The line graph above shows that the heat released in combustion is directly proportional to the number of carbon atoms in the molecule.

The same is true when I did the experiment but the heat released is a considerable amount less than the theoretical heat released. The line of best fit is straight line and we can make a liner equation for this that links the number of carbon atoms to the heat released. This is,

Heat released = (-160.77 x No. of carbon atoms) x 262.8

Theoretical heat released

Experimental heat released

When we compare the two lines of best fit on the same graph we can see that the theoretical heat released is more than the experimental heat released. We can also see that the difference is increasing at a constant rate. This is because there is more energy in the fuel and because there was more energy in the fuel more energy was lost to the surroundings.

The results I have do confirm my initial predictions. We can see from the graph above that my prediction and my result do have the same trend.

EVALUATION

The results I got from my experiment were accurate enough to give me results I can relie on. I could have got more accurate results by modifying my plan for my experiment. When I was carrying out my experiment I saw four main things that could be improved to make the results more accurate.

1. Heat which never enters the water, because of draughts, for example.

2. Heat loss from the top and sides of the beaker.

3. Heat which is not conducted by the beaker.

4. Incomplete combustion – there is a restricted supply of oxygen, the alcohol was burning with an orange flame rather than blue. Some of the alcohol did not burn completely, giving carbon and carbon monoxide rather than carbon dioxide. A carbon deposit (soot) on the bottom of the beaker indicated this.

Improvements can be made to this by insulating the sides of the beaker, using a different material for the beaker, using a lid a providing a draught screen as shown below.

Alternatively we can remove all faults in planning by using an advanced technique such as a bomb calorimeter. This is the most accurate way of measuring bond energies and this will be as accurate as we can get in our results.