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Measuring the Enthalpy Change of Combustion of Different Fuels

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Introduction

Measuring the Enthalpy Change of Combustion of Different Fuels Kristen Schroeder Table of Contents 1. Apparatus and Materials 2. Equipment 3. Procedure 4. Risk Assessment 5. Explanation and sources 6. Data 7. Data Analysis, Amount of Heat Absorbed by the Water 8. Calculating Enthalpy Changes 9. Conclusion 10. Evaluation 11. Procedure Assessment Materials and Apparatus Chemicals Chemicals Empirical Formulas Molecular Structure Methanol CH3OH H O H - C - H H Ethanol C2H5OH or CH3CH2OH H H H - C - C - O - H H H Propan-1-ol C3H7OH or CH3CH2CH2OH H H H H - C - C - C - O - H H H H Butan-1-ol C4H9OH or CH3CH2OCH2CH3 H H O H H H - C - C - C - C - H H H H H Apparatus Equipment * 2 bench mats * 10� C - 50� C thermometer * access to a balance * 1 draught shield * 1 small aluminium can * 4 spirit burners containing o methanol o ethanol o propan-1-ol o butan-1-ol * 100 ml of water (for each experiment) * 1 Bunsen burner * matches * tweezers * pipette * extra amounts of the chemicals * 100 cm3 measuring cylinder * splints (small and slim) pieces of wood Procedure 1. Take a spirit burner that is for methanol. If it is not full to the top, use tweezers to open the spirit burner and then use a pipette to place more methanol into the burner. 2. Place the spirit burner with the cap on a balance and record the mass given in grams. 3. Measure 100 cm3 of water in the 100cm3 measuring cylinder. 4. Pour the water carefully from the cylinder into the aluminium can. 5. Use the thermometer to record the water temperature. In the future experiment, if the water won't go to the same temperature just raise the temperature of the water to the same amount of degrees each time. ...read more.

Middle

3. Methanol (4.2 Jg-1K-1) x (100 grams) x (20.0� C) energy transferred= 8400 J 8400 J = 8.40 kJ mass of fuel used = 0.83 grams molar mass of methanol = 32.00 grams 0.83 grams/ 32.00 grams = 0.0259375 grams = 0.025 grams ?Hc = 8.40 kJ/0.025 grams = 336.000 kJ mol-1 Therefore, the enthalpy change for combustion in Methanol 3 is -336.000 kJ mol-1. 1. Ethanol (4.2 Jg-1K-1) x (100 grams) x (20.0� C) energy transferred= 8400 J 8400 J = 8.40 kJ mass of fuel used = 0.60 grams molar mass of ethanol = 46.00 grams 0.60 grams/ 46.00 grams = 0.0130434783 grams = 0.013 grams ?Hc = 8.40 kJ/0.013 grams = 646.1538462 kJ mol-1 = 646.154 kJ mol-1 Therefore the enthalpy change for combustion in Ethanol 1 is -646.154 kJ mol-1. 2. Ethanol (4.2 Jg-1K-1) x (100 grams) x (20.5� C) energy transferred= 8610 J 8610 J = 8.61 kJ mass of fuel used = 0.60 grams molar mass of ethanol = 46.00 grams 0.60 grams/ 46.00 grams = 0.0130434783 grams = 0.013 grams ?Hc = 8.61 kJ/0.013 grams = 662.3076923 kJ mol-1 = 662.307 kJ mol-1 Therefore, the enthalpy change for combustion in Ethanol 2 is -662.307 kJ mol-1. 3. Ethanol (4.2 Jg-1K-1) x (100 grams) x (20.5� C) energy transferred= 8610 J 8610 J = 8.61 kJ mass of fuel used = 0.58 grams molar mass of ethanol = 46.00 grams 0.58 grams/ 46.00 grams = 0.0126086957 grams = 0.013 grams ?Hc = 8.61 kJ/0.02 grams = 662.3076923 kJ mol-1 = 662.307 kJ mol-1 Therefore, the enthalpy change for combustion in Ethanol 3 is -662.307 kJ mol-1. 1. Propan-1-ol (4.2 Jg-1K-1) x (100 grams) x (20.5� C) energy transferred= 8610 J 8610 J = 8.61 kJ mass of fuel used = 0.55 grams molar mass of propan-1-ol = 60.00 grams 0.55 grams/ 60.00 grams = 0.009166666 grams = 0.009 grams ?Hc = 8.61 kJ/0.009 grams = 956.66666 kJ mol-1 = 956.667 kJ mol-1 Therefore, the enthalpy change for combustion in Propan-1-ol 1 is -956.667 kJ mol-1. ...read more.

Conclusion

Heat can also stay in the can itself and escape along the outside of the can into the environment. Incomplete combustion is also a main source of error because the flame was yellow (rather than blue showing complete combustion). The spirit burner wasn't getting enough oxygen to achieve complete combustion and some of the fuel was not converted into the purest form of carbon dioxide and water (both in the form of gases). This affects my result because obviously not all the fuel that was used in the experiment was completely converted, which if they had been, it would produce more heat Overall, I think my results were as precise as they could be in the conditions that the equipment allowed. My results seem to be pretty close together, and therefore I think they are precise, but they are not accurate because they are not spot-on due to the conditions that I was able to use. Procedure Assessment To make the procedure more accurate and to improve my results, the best solution would be to use alternative apparatus, such as using a balance that goes more than two decimal places, you could also use a different form of the draught shielding and metal calorimeter, or you could use a more accurate thermometer with more increments on it. A different form of apparatus looks like this: Using a different balance with more decimal places would improve my results by making them more accurate because it would give a more specific measurement. Using the particular different form of apparatus would improve my results because it would keep the draft away from the water because it is surrounded by glass. It also keeps heat from escaping the water as well as evaporation because there is a tube allowing the hot air or exhaust to escape through the top. Also, the coil through the water allows the heat to go throughout the whole of the water and spread throughout it. Lastly, using a different thermometer with more increment would be more specific in its measurement because it will be able to go further than one half increments. ...read more.

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