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Comparing the enthalpy changes of combustion ofdifferent alcohols.

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Introduction

Comparing the enthalpy changes of combustion of different alcohols. What are alcohols? Alcohol is the common family name for the hydrocarbon group alkanols. At least one of the hydrogen groups is replaced by an -OH group. They are all organic compounds. The general formulas for the alcohols are: CnH(2n+1)OH Where n represents a number. The first and simplest member of the alkanols family is methanol. Its molecular formula is CH3OH. You can now see that each member of the alcohol family has a different number of carbons in its structure. They increase by one carbon atom and two hydrogen atoms each time. The question that this piece of coursework is going to answer is "Is there a relationship between the number of carbon atoms in a chain and the enthalpy change of combustion?" Aim The aim of this experiment is to prove that the longer the hydrocarbon chain the higher the amount of energy transferred to the water, therefore fewer moles of fuel will be used to achieve a temperature rise of 15�C. The enthalpy change of combustion is the amount of energy transferred when one mole of a substance burns completely in oxygen (always -ve). To work out the enthalpy change of combustion of the alcohols the energy output must be measured. The easiest way to do this accurately is to use the thermal energy of combustion to raise the temperature of a substance with a known specific heat capacity, (the amount of energy required to increase the temperature of 1g of a liquid by 1�C).We will be using water. Prediction I predict that the longer the hydrocarbon chain the more energy will be transferred to the water, therefore fewer moles of fuel will be used to achieve the same temperature rise. So as the number of carbon atoms increases the enthalpy of combustion will become more negative. Alcohols to be used to compare * Methanol = CH3OH * Pentan-1-ol = C5H11OH * Propan-1-ol = C3H7OH * Propan-2-ol = C3H7OH * Butan-1-ol ...read more.

Middle

266.71g 263.26g 168.03g Change in mass of spirit burner (g) 1.20g 1.19g 1.18g Starting temperature of water (�C) 21�C 20�C 23�C Final temperature of water (�C) 36�C 35�C 38�C Propan-2-ol Replicate one Replicate two Replicate three Weight of spirit burner before (g) 238.43g 237.23g 236.08g Weight of spirit burner after (g) 237.23g 236.08g 234.87g Change in mass of spirit burner (g) 1.20g 1.15g 1.21g Starting temperature of water (�C) 22�C 21�C 20�C Final temperature of water (�C) 37�C 36�C 35�C The mass of the alcohols both before, after and the change in mass is to 2 decimal places. This is because the digital balance weighs to this degree of accuracy. The measurement taken with the thermometer is to one whole number because this is the degree of accuracy that the thermometer has. Calculations. Now that I have my results, I will work out the enthalpy change of combustion (?Hc) for each alcohol. I will work out the ?Hc for each replicate for each alcohol and then work out the average ?Hc. Here is an example of how I will work out the ?Hc. I will use replicate one of Methanol. 1. It takes 4.2J of energy to heat 1g of water by 1�C. so the amount of energy taken in by the water is 200 x 15 x 4.2 = 12600J 2. Working out the number of moles of methanol burnt using Mass/Molar mass Molar mass of Methanol (CH3OH) = 32 3.15/32 = 0.0984375 moles 3. If 0.0984375 moles of Methanol release 12600J of energy, 1 mole would release 12.60kJ/0.0984375 = 128kJmol-1 Therefore the ?Hc of methanol is -128KJmol-1 (Enthalpy of combustion is ALWAYS negative) I have calculated the enthalpy change of combustion (?Hc) using the same method for each of the replicates for each of the fuels. The chart below shows the enthalpy change of combustion for each alcohol and the average enthalpy change of combustion overall. ...read more.

Conclusion

I would need to find a way of introducing an 'oxygen feed' into the experiment so that there will be enough oxygen for complete combustion and also clean off any soot that had accumulated on the copper calorimeter. The reason that the incomplete combustion would have such a large effect on the overall results would be that the products formed from incomplete combustion release less energy than the products that are formed from complete combustion. The equations for the complete combustion of the alcohols are, Methanol CH3OH(l) + 11/2O2 O(g) � CO2(g) + 2H2O(l) Propan-1-ol and Propan-2-ol C5H11OH(l) + 8O2(g) � 5CO2(g) + 6H2O(l) Butan-1-ol and Butan-2-ol C4H9OH(l) + 61/2O2(g) � 4CO2(g) + 5H2O(l) Pentan-1-ol C5H11OH(l) + 8O2(g) � 5CO2(g) + 6H2O(l). Even if I could weigh out all of the soot and work out the number of moles of carbon, there would be no way of getting all of the soot off the can. The best way would be to minimize the overall distance between the calorimeter and the flame and have a calorimeter with a larger surface area to absorb more heat from the flame. In order to reproduce the experiment and obtain a higher degree of accuracy of results that are closer to the actual values in the data book, there are some subtle changes that will need to take place. 1. The calorimeter we used consisted of a basic copper can. In reality a calorimeter looks like this I would use this piece of equipment. I could not use this in the experiment that I carried out because the school does not have access to this equipment. 2. The calorimeter would have a lid on it to stop any heat escaping that way. 3. As already said an oxygen feed would be set up to ensure that complete combustion occurs. 4. The distance between flame and calorimeter would therefore be reduced. 5. Surface area has been increased. ...read more.

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