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Analysis of the Standard Enthalpy of Combustion for Alcohols

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Introduction

Investigate an Aspect of Organic Chemistry Investigation of the standard enthalpy change of combustion for alcohols Sarah van der Post HL Chemistry IB Internal 24 – 6 - 2012 Design Aim: To investigate the standard enthalpy change of combustion for 5 consecutive alcohols in the alcohol homologous series, methanol, ethanol, propan-1-ol, butan-1-ol and pentan-1-ol, by using a calorimetric method to calculate the heat gained by the 100cm3 water in the experiment, and thus the heat lost by the alcohol lamp at standard temperature and pressure (298 K and 101.3 kPa). Background Knowledge: Alcohols are organic compounds containing Oxygen, Hydrogen and Carbon. The alcohols are a homologous series containing the functional –OH group. As we move down the homologous series of alcohols, the number of Carbon atoms increase. Each alcohol molecule differs by –CH2; a single Carbon atom and two Hydrogen atoms. Combustion is the oxidation of carbon compounds by oxygen in air to form CO2 and H2O. Combustion produces heat as well as carbon dioxide and water. The enthalpy change of combustion is the enthalpy change that occurs when 1 mole of a fuel is burned completely in oxygen. When alcohol undergoes complete combustion it produces carbon dioxide and water as products, and energy is released. The standard enthalpy of combustion of an alcohol (âH°comb) is the enthalpy change when one mole of an alcohol completely reacts with oxygen under standard thermodynamic conditions (temperature of 25°C and pressure of 101.3 kPa). The standard enthalpy change of combustion of alcohols varies depending on their molecular size. The greater the number of carbons, the higher the standard enthalpy of combustion, as there is the presence of more bonds. ...read more.

Middle

1. Room temperature. Room temperature, and thus the surroundings of the experiment was 19.0 °C ± 0.1 °C 1. Results when 100cm3 of water is heated with a spirit burner of different alcohols. Initial and final temperature of 100 cm3 ± 0.08 cm3 heated distilled water and the initial and final mass of spirit burner for methanol/ethanol/ propan-1-ol/ butan-1-ol/ pentan-1-ol after heating water for 180 seconds methanol ethanol propan-1-ol butan-1-ol pentan-1-ol Temperature (°C ± 0.1 °C) trial 1 initial 18.3 17.2 18.3 17.6 18.3 final 36.0 53.4 44.0 42.8 46.6 trial 2 initial 18.0 17.2 18.3 18.1 19.6 final 35.1 51.9 41.7 39.1 51.2 Trial 3 initial 18.3 17.1 18.4 17.9 19.7 final 35.7 52.6 42.5 42.2 53.0 Mass (g ± 0.01 g) Trial 1 initial 203.98 236.36 288.17 220.31 239.34 final 202.95 234.80 287.36 219.61 238.60 Trial 2 initial 202.95 234.80 287.36 219.61 238.60 final 202.01 233.18 286.52 218.91 237.84 Trial 3 initial 201.24 233.18 286.52 218.91 237.84 final 200.15 231.96 285.73 218.13 237.06 1. Graph of temperature of 100cm3 ± 0.08 cm3 water when heated for 180 seconds by propan-1-ol. Trial 3 Trial 2 Trial 1 30 seconds before the lamp was lit by a match, to ensure the temperature was constant initially. Qualitative Data: 1. The alcohols had a strong characteristic odour and were all a colourless liquid. 2. The water in the conical flask began to boil, and increased in temperature 3. The flask also increased in temperature 4. As the number of carbons in the alcohol increased the amount of soot left on the bottom of the flask increased, and the rate it formed increased as well. ...read more.

Conclusion

+ 0.5O2 (g) → CO2 (g) + 2H2O (l) m(CH3OH) = 1.02 g ± 0.074% M(CH3OH) = 12.01+4×1.01+ 16=32.05 n=mM n(CH3OH) =1.02 g ± 0.074%32.05 n(CH3OH) = 0.0318 mol± 0.074% 1. Calculating the mass of water Absolute Uncertainty → Percentage uncertainty 0.08 g100 g×100=0.0008 % Mass of Water (m) = Volume x Density = (100 cm3 ± 0.08 cm3) x 1g/cm3 = 100 g ± 0.08 g = 100 g ± 0.0008 % 1. Calculating the heat gained by water. q = mcâT q =100 g ±0.0008 %×4.18 J°C-1g-1×(17.4 °C ±1.724%) q = 7273.2 J ±1.7248 % 1. Calculating the heat lost by alcohol Assuming that there is no heat loss to surroundings and that all energy is transferred from the alcohol to the water. heat gained by water = - heat lost by alcohol heat lost by alcohol = - 7273.2 J ±1.7248 % 1. Calculating the standard enthalpy change of combustion Percentage Uncertainty → absolute uncertainty 1.7988 %100 %×228.71698 kJmol-1=4 kJmol-1 âH°comb =qnâH°comb =-7273.2 J±1.7248 %0.0318 mol± 0.074% =- 228716.98 Jmol-1 ±1.7988 % =- 228.71698 kJmol-1 ±1.7988 % =- 229 kJmol-1 ±4kJmol-1 Similar procedures are followed to determine the standard enthalpy of combustion of the rest of the 5 alcohols. The overall results are as shown in a table below: Alcohol Number of Carbons Molecular Mass Change in enthalpy of combustion Methanol 1 32.02 - 229 kJmol-1 ±4 kJmol-1 Ethanol 2 46.08 - 465 kJmol-1 ±8 kJmol-1 Propan-1-ol 3 60.11 - 756 kJmol-1 ±10 kJmol-1 Butan-1-ol 4 74.14 - 997 kJmol-1 ±20 kJmol-1 Pentan –1- ol 5 88.17 - 1348 kJmol-1 ±20 kJmol-1 The above table of results has been graphed in the graph below to explain the trends in Standard Enthalpy of Combustion of alcohols. ...read more.

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