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To investigate the enthalpy change of combustion for various alcohols.

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Chemistry Coursework P l a n n i n g Aim: To investigate the enthalpy change of combustion for various alcohols Alcohols will be burnt to heat up water. The aim is to find out how much energy is produced when burning the following alcohols : ethanol, propanol and butanol. Alcohols react with oxygen in the air to form carbon dioxide, water and energy is liberated (exothermic reaction), because the reactants' energy is higher than that of the products'. The alcohols form a series of compounds which are related to each other. This is called a homologous series and has the formula : Cn H 2n+1 OH In this investigation the alcohols used were: ethanol, propanol and butanol. Ethanol : C2H5OH Propanol : C3H7OH Butanol : C4H9OH Combustion Reactions: Ethanol : C2H5OH + 3O2 2CO2 + 3H2O + heat Propanol : C3H7OH + 4.5O2 3CO2 + 4H2O + heat Butanol: C4H9OH + 6O2 4CO2 + 5H2O + heat Heat (joules) = T x mass (g) x specific heat capacity Heat energy = rise in temperature x mass of water (g) x 4.18 The specific heat capacity is the number of joules required to raise the temperature of 1g of water by 1oC (1g of H2O raised by 1oC = 4.18 joules) The combustion of the alcohols is not exactly as shown in the equation because some carbon is deposited on the calorimeter base ie. C2H5OH + O2 2C + 3H2O Additionally, some carbon monoxide is produced ie. C2H5OH + 2O2 2CO + 3H2O Therefore the equation provided for the complete combustion is an over simplification. Prediction: When bonds are broken, energy is absorbed by the system (endothermic process); and when bonds are formed, energy is released to the external environment (exothermic process). When alcohols are burnt, the reaction is always overall exothermic. Therefore more energy is being released than is being consumed. Thus, a qualitative prediction is that each alcohol will burn exothermally and therefore the temperature of the water will rise. ...read more.


* The calorimeter was weighed again, making sure it is still the same weight. If the weight was more/ less, the new weight was recorded, taking it into account when working out the mass of water. * Experiment was then repeated 5 times for each alcohol. R e s u l t s Using the relationship: heat = temperature rise x mass water x 4.18 No. of moles = mass/ RFM ETHANOL: Experiment Temperature rise (oC) Mass of water(g) Heat (J) Number of moles ?Hc (kJmol-1) 1 30.0 54.2 6796.6800 0.01300 522.8 2 26.5 54.2 6003.7340 0.00978 613.9 3 26.3 54.2 5958.4228 0.01150 518.1 4 26.0 54.2 5890.4560 0.01130 521.3 5 27.0 54.2 6117.0120 0.01196 511.5 Method of calculating results obtained: 1.) 30.0 x 54.2 x 4.18 = 6796.68 joules energy produced by 0.6g ethanol no. of moles of ethanol = mass/RFM = 0.6 / 46 = 0.013 As 0.013 moles of ethanol liberate 6796.68 joules energy Therefore 1 mole of ethanol liberates 6796.68 / 0.013 = 522821.5 joules or 522.8 kJmol-1 2.) 26.5 x 54.2 x 4.18 = 6003.734 joules energy produced by 0.45g ethanol no. of moles of ethanol = mass/RFM =0.45 / 46 = 0.00978 As 0.00978 moles of ethanol liberate 6003.734 joules energy Therefore 1 mole of ethanol liberates 6003.734 / 0.00978 = 613878.7 joules or 613.9 kJmol-1 simply from comparing this result to the other results, I can see that it is anomalous. However, I would need to draw a graph to compare how abnormal the result is. However, comparing the result with the literary values, this result may be the closest to the true values (least amount of heat lost to the atmosphere). 3.) 26.3 x 54.2 x 4.18 = 5958.4228 joules energy produced by 0.53g ethanol no. of moles of ethanol = mass/RFM = 0.53 / 46 = 0.0115 As 0.0115 moles of ethanol liberate 5958.4228 joules energy Therefore 1 mole of ethanol liberates 5958.4228 / 0.0115 = 518123.7 joules or 518.1 kJmol-1 4.) ...read more.


Sufficiency of evidence: The experiment was repeated five times for each alcohol. This gave only five results to plot on a graph. If there had been more anomalous results, I would not have had enough data to compare them with so as to determine the anomalous results. Improvement: Obvious improvements would be to make the combustion more efficient so that no carbon was deposited and no carbon monoxide was produced. This could be achieved by designing a burner rather than using a wick. As with a gas oven, the fuel needs to be in gaseous form for the most efficient combustion. One reason why the results obtained were inaccurate is that some of the heat liberated was used to vaporise the liquid prior to its combustion. A suitably designed burner could get around this - presumably using several fine holes rather than one big hole to get better mixing of the vapour with the oxygen. Another modification might be to use oxygen enriched air or even pure oxygen to ensure the complete combustion of the alcohol. Perhaps a cylindrical reflective metal sheet could be placed around the burner and calorimeter to prevent the heat being lost to the surroundings e.g. a sheet of aluminium. Further work to provide additional evidence or extend enquiry: To further my enquiry I would investigate in the same way the combustion of methanol (CH3OH) and pentanol (C5H11OH). This would give me further evidence to conclude that: the larger the alcohol molecule, the more bonds will be broken and formed, and therefore the more heat will be produced. For this enquiry, I would predict that methanol would produce less heat than all of the alcohols, and that pentanol would produce the most heat. In conclusion, the experiment was worthwhile as it showed the relationship between different sized alcohol molecules in a combustion reaction. However, the evidence is not accurate enough to show any other information, other than the fact that the larger the alcohol molecule the more heat is produced during combustion. Tania Lapa ...read more.

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