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# To find how much energy is produced when burning different alcohols.

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Introduction

Heat of combustion Aim I Aim to find how much energy is produced when burning different alcohols. To do this I will heat a beaker of water by burning alcohols. I will burn five different alcohols, methanol, ethanol, propanol, butanol and pentanol. I will also use these results to predict the heat of combustion of hexonal. Prediction I predict that all the experiments will be exothermic. This is because there are more bonds formed that there are made and bond forming is an exothermic process. I think that the more carbon atoms each alcohol has the higher energy per mole the more heat given out Theory The reaction that is involved in burning alcohols is exothermic because heat is given out. The formulae of the alcohols that I will be using are... � Methanol C1H3 OH � Ethanol C2 H5 OH � Propanol C3 H7 OH � Butanol C 4H9 OH � Pentanol C5 H 11OH Amount of bond energy C-H 413 C-O 358 O-H 464 O=O 498 C=O 806 C-C 347 Type of alcohol Energy required Methanol 2100 Ethanol 3270 Propanol 3740 Butanol 4560 Pentanol 5380 Methanol Balancing equations 2 C1H30H +3 O2 -> 2 CO2 + 4 H2O Diagram of compound C-C = 0 O-H = 1 C-H = 3 O=O = 3 C-O = 1 C=O = Apparatus Alcohols, methanol, ethanol, propanol, butanol, pentanol. ...read more.

Middle

Energy evolved in (kj) Methanol 21 000 21 Ethanol 21 000 21 Propanol 21 000 21 Butanol 21 000 21 Pentanol 21 000 21 To find out how much energy is produced per gram we use the formula... Energy per gram of fuel = Energy evolved x Mass of fuel burnt Energy per gram of fuel = 21kj x ? Below is a table showing how much energy is produced per gram when burning the alcohols in question... Type of alcohol Energy per gram (kj) Methanol 9.63 Ethanol 5.36 Propanol 4.16 Butanol 2.76 Pentanol 2.19 As you can see the energy per gram decreases as the length of the molecule increases. This is because more fuel is burnt so there is more of it to be filled with the energy. This is shown in graph 1. To find out how much energy is produced per mole you have to use this formula.. Energy per mole = Energy per gram x Formula mass Here is a list comprising all the formula masses of the chosen alcohols... � Methanol 32g � Ethanol 46g � Propanol 60g � Butanol 74g � Pentanol 88g Below is a table showing the energy produced per mole... Type of alcohol Energy per mole(kj/mol) Methanol 308.16 Ethanol 245.56 Propanol 249.60 Butanol 206.46 Pentanol 192.72 Again the results decrease as the molecule length increases suggesting that during the experiment more energy was last in the longer alcohols, this is shown in graph 2. ...read more.

Conclusion

This would give a better graph reading and a wider range of results to support a firm conclusion. On the other hand, if I had started below room temperature, so that the amount of energy gained, from room temperature might equal the energy lost at temperatures higher than room temperature. Next time reducing heat lost would be my main priority. Improving insulation techniques would be a valuable asset in obtaining the most reliable data I could. Another error is that of incomplete combustion. Complete combustion occurs if there are lots of oxygen atoms available when the fuel burns, then you get carbon dioxide (carbons atoms bond with two oxygen atoms). If there is a limited supply of oxygen then you get carbon monoxide (each carbon atom can only bond with one oxygen atom). This is when incomplete combustion has occurred. This is so because the carbon monoxide could react some more to make carbon dioxide. If the oxygen supply is very limited then you get some atoms of carbon released before they can bond with any oxygen atoms. This is what we call soot. Since heat is given out when bonds form, less energy is given out by incomplete combustion. So this is why it affects the outcome of the experiment. To overcome this problem, I would have to make sure a sufficient supply of oxygen was involved in the reaction. This could also be improved by making a better draught proof Chrissie Cassley 10E2 02/12/01 ...read more.

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