• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Combustion of Alcohols

Extracts from this document...

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.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our GCSE Organic Chemistry section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Organic Chemistry essays

  1. Marked by a teacher

    Experiment to investigate the heat of combustion of alcohols.

    4 star(s)

    Therefore I predict that: * The longer the molecule of alcohol, the larger the amount kJ/mol is given out by burning the substance. To back this prediction, I have calculated some theoretical values to show how the amount of energy released increases as the number of carbon atoms rises.

  2. Investigating the Combustion of Alcohols

    0.82 + 0.94 = 0.87 g 3 This means that in three trials, the average mass of propan-1-ol burned was 0.87g. Energy transferred (J) = mass of water (g) � specific heat capacity � temperature change Q (J) = M/g � -4.2 J g-1 K-1 � ?T �C Q (J)

  1. Molar Heat of Combustion of Alcohols

    92.44 94.41 66.23 Final mass of Ethanol (g) 92.02 93.96 65.82 Difference in mass (g) 0.42 0.45 0.41 Mass of Water (g)* Specific Heat Capacity of Water (4.2)* Temp Change (OC) = Energy Supplied (j) Energy Supplied/ Amount of Alcohol Burnt= Energy Supplied for 1g Energy Supplied By One Gram* Weight of One Mole of Alcohol= Molar Heat

  2. "Could Sainsbury's add value to their business by using an alternative fuel for their ...

    emissions at the point of use and are extremely quiet * Electric vehicles produce no tailpipe carbon dioxide emissions * Electric vehicles are not subject to Vehicle Excise Duty * Lower Personal Benefit in Kind (BIK) tax liability * Enhanced capital allowance rate of 100% in the first year *

  1. To Investigate the Combustion of Fuels

    and divide that number by two (because there are two moles of methanol in the equation), to find the total energy released per mole of methanol. (6640 - 5588) / 2 = 526 KJ mol-1 Using the same method, I calculated the energy released per mole burnt for ethanol, propanol

  2. Combustion of Alcohols Investigation.

    This renders the hypothesis correct, as does the fact that as noticed, pentanol is indeed the most effective fuel of all the alcohols we investigated. The most energetic bond is the bond between oxygen and carbon, so the alcohol with the most carbon would logically have the biggest enthalpy.

  1. Comparing the enthalpy changes of combustion of different alcohols.

    3.15g 1.85g 1.52g Starting temperature of water (�C) 21�C 21�C 20�C Final temperature of water (�C) 36�C 36�C 35�C Pentan-1-ol Replicate one Replicate two Replicate three Weight of spirit burner before (g) 214.88g 214.04g 210.73g Weight of spirit burner after (g)

  2. GCSE Chemistry Revision Notes - everything!

    The heat operates at about 5 volts but with currents of up to about 100,000 amps. The heat generated by the huge current keeps the electrolyte molten. The two major costs of extracting aluminium are the costs of electricity and also the replacing of the anodes as they burn away.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work