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The Combustion of Alchohols

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Introduction

The Combustion of Alchohols Chemistry Coursework Introduction In order to carry out this investigation, I first have to define the terms of the objectives so that an understanding is present and a clear process to acquire the answer can be devised. What are alcohols? Alcohols are a specific group of substances that are hydroxyl derivatives of hydrocarbons but more explicitly they are types of fuel. A fuel is a substance (e.g. methane) that readily undergoes combustion (the combustion of a substance involves its reaction with oxygen and the release of energy) and gives out large amounts of energy. The combustion of fuels is usually an exothermic reaction. The distinction between alcohols and other fuels, such as methane, is that alcohols are a homologous series of compounds that contain the -OH group of atoms as the functional group. The functional group in a molecule is the group of atoms within the structure that determines the characteristic reactions of that substance. All the earlier, smaller alcohols have a neutral pH, and are colourless liquids that do not conduct electricity. The general chemical structure of an alcohol is as follows: CnH(2n+1)OH By far the most well known alcohol is ethanol, which often goes by the common name of alcohol itself. Its chemical formula is C2H5OH. It is produced in industry by reacting ethene and steam together. They are first compressed to 60 atmospheres and passed over a catalyst (immobilised phosphoric(v) acid) at 300�C. 300�C, 60 atmospheres ethene + steam -------� ethanol phosphoric acid C2H4(g) + H2O(g) --� C2H5OH(g) It can also be produced via fermentation because ethanol and CO2 are the natural waste products of yeasts when they ferment sugar. However, ethanol is toxic to yeast so once the ethanol concentration has reached around 14% or the sugar runs out then the multiplying yeast die and the fermentation ends. The Babylonians and Egyptians discovered this fermentation and found the product of crushed grapes (also containing sugar) ...read more.

Middle

Put the thermometer in the can and wait a minute, then record the initial temperature of the water. Light the wooden splint from the bunsen burner and then after taking the lid off the burner, use the splint to ignite the spirit burner. From the moment you have ignited the burner, watch the temperature of the water with the thermometer. Make sure the thermometer always remains in the water. As soon as the temperature of the water has increased by 20�C (from the initial recorded temperature) then extinguish the burner immediately using its lid. Re-record the mass of the spirit burner using the scales. Carry out the combustion of the same fuel another two times (the repetition of the test in order to dampen the effect of anomalous results). Then after recording the results carry out the whole test for the other four fuels - ethanol, propanol, butanol & pentanol. Results: Below are the results from the main experiment for all five alcohols. Fuel Mass at Start (g) Mass at End (g) Mass of Fuel Burnt (g) (1) 1 173.99 173.09 0.90 Methanol 2 173.09 172.20 0.89 CH3OH 3 145.09 144.27 0.82 (2) 1 106.73 106.24 0.49 Ethanol 2 106.23 105.83 0.40 C2H5OH 3 105.80 105.40 0.40 (3) 1 146.06 145.61 0.45 Propanol 2 145.59 145.08 0.51 C3H7OH 3 145.08 144.56 0.52 (4) 1 250.73 250.46 0.27 Butanol 2 250.45 250.16 0.29 C4H9OH 3 250.18 249.85 0.33 (5) 1 88.74 88.46 0.28 Pentanol 2 88.46 88.17 0.29 C5H11OH 3 88.17 87.89 0.28 Now I have to find the average of these results because each alcohol has been tested thrice. Fuel Average Mass of Fuel Burnt (g) 1 Methanol 0.87 2 Ethanol 0.43 3 Propanol 0.49 4 Butanol 0.30 5 Pentanol 0.28 On the next page is a graphical representation of these results in the form of a bar chart. I have used a bar chart instead of a scatter graph because on one axis will be the number of carbon atoms in the alcohol molecule. ...read more.

Conclusion

6932 - 5616.9 = 1315.1 kj mol-1 However, this is the energy released by two moles of methanol as you can see from the equation so: 1315.1 ? 2 = 657.55 kj mol-1 Ethanol (5 x 413) + 347 + 358 + 464 + (3 x 498.3) --� (4 x 805) + (6 x 464) = 4728.9 kj mol-1 --� 6004 kj mol-1 6004 - 4728.9 = 1275.1 kj mol-1 Propanol (14 x 413) + (4 x 347) + (2 x 358) + (2 x 464) + (9 x 498.3) --� (12 x 805) + (16 x 464) = 13298.7 kj mol-1 --� 17084 kj mol-1 17084 - 13298.7 = 3785.3 kj mol-1 3785.3 ? 2 = 1892.65 kj mol-1 Butanol (9 x 413) + (3 x 347) + n358 + 464 + (6 x 498.3) --� (8 x 805) + (10 x 464) = 8569.8 kj mol-1 --� 11080 kj mol-1 11080 - 8569.8 = 2510.2 kj mol-1 Pentanol (22 x 413) + (8 x 347) + (2 x 358) + (2 x 464) + (15 x 498.3) --� (20 x 805) + (24 x 464) = 20980.5 kj mol-1 --� 27236 kj mol-1 27236 - 20980.5 = 6255.5 kj mol-1 6255.5 ? 2 = 3127.75 kj mol-1 These values can also be plotted onto a graph so that I can see the trend between each. These results also form a straight line like the values obtained from the website. However, I want to compare these results with the ones researched and with my own ones to see how different they are. Combining all three plots onto one scatter graph can easily do this. I have done this on the following page. As you can see, and as I expected, my results are plotted well below the other two lines. This shows that the energies I calculated to have been released by the combustion were lower than the theoretical/actual values. This is obviously because of the loss of energy in the experiment. ?? ?? ?? ?? Investigation into the Combustion of Specific Alcohols - Page 21 - ...read more.

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