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# Combustion of Alcohols

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Introduction

Combustion of Alcohols In this investigation, I will be using a spirit burner to burn three different alcohols, to heat up a calorimeter containing water. My task is to investigate the heat produced and the combusting capabilities of each fuel. The fuels I will use are ethanol (C2H5OH), propanol (C3H7OH) and butanol (C4H9OH). The alcohol homologous series has the general formula CnH2n+1OH. I aim to discover how much energy is produced when burning these alcohols. When alcohols are burned, or combust, they react with oxygen in the air to form water and carbon dioxide, e.g.: CH3OH + 1.5O2 CO2 + 2H2O Methanol + Oxygen Carbon Dioxide + Water The energy produced from this is called the heat of combustion. The reaction is also exothermic. This tells us that the reactant energy is higher than the product energy, e.g. using methanol again: Ea = Activation Energy Energy Reactants CH3OH + 1.5O2 Products CO2 + 2H2O Energy is given out when forming bonds between the carbon dioxide and water molecules. The amount of heat energy produced can be calculated by using the following equation: Amount of heat produced (J) = 4.2 x Volume of Water x Temperature Rise The 4.2 in the equation is the specific heat capacity (SHC) of water. This means that 4.2J are needed to heat 1gram of water by 1�C. I am using water in this investigation as it is readily available, safe and has a reliable SHC. Method I have decided that I will: 1. ...read more.

Middle

In this investigation, I will use Ethanol. Breaking Bonds I predict that the larger the molecule, the more energy will be required to break its bonds, therefore the more heat produced. This is because large molecules have many bonds, but small molecules have few bonds. I can demonstrate this by working out the theoretical heat of combustion of the three alcohols by using the following tables and equation: ?H = Bonds Broken - Bonds Formed Type of Bond Energy required to break bond C-C 348 kJ/mol C-O 360 kJ/mol C-H 412 kJ/mol O-H 463 kJ/mol C=O 743 kJ/mol O=O 496 kJ/mol Combustion of Ethanol C2H5OH + 3O2 2CO2 + 3H2O Type of Bond Energy required to break bond(s) Type of Bond Formed Energy required to form bond(s) C-H x 5 2060 kJ C=O x 4 2972 kJ C-O 360 kJ O-H x 6 2778 kJ O-H 463 kJ C-C 348kJ O=O x 3 1488 kJ Energy Combustion of Propanol C3H7OH + 41/2O2 3CO2 + 4H2O Type of Bond Energy required to break bond(s) Type of Bond formed Energy required to form bond(s) C-C x 2 696kJ C=O x 6 4458kJ C-H x 7 2884kJ O-H x 8 3704kJ C-O 360 kJ O-H 463kJ O=O x 4.5 2232kJ Energy Combustion of Butanol C4H9OH + 6O2 4CO2 + 5H2O Type of Bond Energy required to break bond(s) Type of Bond formed Energy required to form bond(s) C-C x 3 1044kJ C=O x 8 5944kJ C-H x 9 3708kJ O-H x 10 4630kJ C-O 360kJ O-H 463kJ O=O x 6 2976kJ Energy Alcohol Energy required to break all bonds in the alcohol (kJ) ...read more.

Conclusion

46 Heat per mole = 21 = 420 kJ 0.05 The results from these tests can be tabulated for easier reference: Test Variable Changed Heat produced Heat per Mole 1 None 42 kJ 420kJ 2 Volume of water - 150cm3 instead of 200cm3 31.5 kJ 350 kJ 3 Desired temperature rise - 25�C instead of 50�C. 21 kJ 420 kJ If I had more time, I would have conducted more tests and changed different variables. I have noticed a couple of things: 1. By halving the desired temperature rise, I halved the heat produced; however, the heat per mole is not altered. 2. By taking a quarter of the volume of water off (200cm3 - 1/4 x 200cm3 = 150cm3), a quarter of the heat produced is taken off (42 kJ - 1/4 x 42 kJ = 31.5 kJ). However, a quarter of the heat per mole value is not removed from itself. Therefore, whatever you do to a variable, i.e. if you double it, halve it, square it, etc. you do to the same amount of heat produced. My results are lower than those found in the text books because the fuel burned with a yellow flame and carbon soot was deposited on the bottom of the calorimeter. This indicates that incomplete combustion had taken place. Therefore, the fuel was not burning fully, and as a consequence, less heat energy was produced. I can draw a table to show the percentage of heat energy given out to heat the water: Test My results (kJ/mol) Expected result (kJ/mol) Percentage of Energy given out to heat water 1. -420 -1371 30.6% 2. -350 -1371 25.5% 3. -420 -1371 30.6% ...read more.

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