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  • Level: GCSE
  • Subject: Science
  • Document length: 971 words

An experiment for comparing the enthalpy change of combustion of different alcohols.

Extracts from this essay...

Introduction

An experiment for comparing the enthalpy change of combustion of different alcohols Aim To find out the enthalpy change of combustion for a number of alcohols. Apparatus Thermometer Measuring cylinder Electrical balance Stand Burners of different alcohols (methanol ethanol propan-1-ol butan-1-ol) Water Copper can Clamp Boss Heatproof mal Matches Method 1.Measure 200cm of water (at room temperature) into the measuring cylinder; record the temperature of the water. 2.Set up the apparatus as shown below in fig 1: Thermometer Stand Copper can Thermometer Boss & clamp Copper Can Stand Alcohol burner Heatproof Fig 1. The apparatus to measure enthalpy change 3.Pour the water from the measuring cylinder into the copper can, which is on the stand and clamped in position, and put a thermometer into the water to measure the temperature change. 4.Weigh the alcohols that are to be investigated, and record the weight. 5.Place the alcohol burner under the copper can, which is on the stand. 6.Ignite the burner with a match to heat the copper can, which contains water.

Middle

The steps described in the method are detailed and clear and progress from setting up the experiment to obtaining results. Result tables This is the result table of the weight of the alcohols used for rising up 20C of water (from 21C to 41C) Methanol Ethanol Propan-1-ol Butan-1-ol Beginning weight of the alcohols 232.39g 231.41g 248.30g 247.46g Finishing weight of the alcohols 231.16g 230.51g 247.49g 246.77g Alcohol used 1.23g 0.90g 0.81g 0.69g This is the result table of the changing in temperature Methanol Ethanol Propan-1-ol Butan-1-ol Beginning temperature 20ºC 21ºC 21ºC 22ºC Finishing temperature 40ºC 41ºC 41ºC 42ºC Temperature change 20ºC 20ºC 20ºC 20ºC Calculations The beginning temperature is 21C 4.17 X 200 X 20 = 16680J=16.68KJ Methanol: Hc = 16.68 X 32 / 1.23 = 433.95kJ/mol-1 The Hc of methanol is 433.95KJ/mol-1 Ethanol: Hc = 16.68 X 46 / 0.90 = 852.53KJ/mol-1 The Hc of ethanol is 852.53KJ/mol-1 Propan-1-ol: Hc =16.68 X 60 / 0.81 = 1235.56KJ/mol-1 The Hc of propan-1-ol is 1235.56KJ/mol-1 Butan-1-ol: Hc = 16.68 X 74/0.69 = 1788.87KJ/mol-1 The Hc of butan-1-ol is 1788.87/KJmol-1 Procedural errors The official values of the energy produce by methanol,

Conclusion

+/- 0.005 / 230.51 x100 % = 0.0021 % Mass of the propan-1-ol burner (reading): Before the combustion +/- 0.005 / 248.30 x100 % = 0.0020 % After the combustion +/- 0.005 / 247.49 x100 % = 0.0020 % Mass of the butan-1-ol burner (reading): Before the combustion +/- 0.005 / 247.46 x100 % = 0.0020 % After the combustion +/- 0.005 / 246.77 x100 % = 0.0020 % Measuring cylinder: +/- 0.5 / 200 X 100 % = 0.25 % Temperature: +/- 0.5 / 20.0 X 100 % = 2.5 % Conclusion The generalized equation for the combustion of alcohols is: Alcohol + Oxygen Carbon dioxide + Water The longer chain molecules such as butan-1-ol and propan-1-ol can produce more energy than shorter chain molecules like methanol and ethanol, because the energy is contained in the bonds, Therefore an alcohol with the greatest number of carbons will have greatest amount of energy, and will produce the increase in temperature for the least amount of fuel used. Reference Heinemann, Salters Advanced Chemistry, Chemical idea (Second Edition), Central Team George, Burton, John Holman, John Lazonby, Gwen Pilling and David Waddington

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