Combustion of Alcohols

July 31, 2017 Chemistry

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

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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. f 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. THIS IS THE BOMB Reference: Letts, GCSE Chemistry. Bob McDuell; Understanding Chemistry for Advanced level. Lister & Renshaw; Nuffield Advanced science Chemistry.

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