In this case, ? H is negative, which means that more energy is released than is taken in. a & b = the reactants E = energy in system c & d = the products AE = activation energy Propanol & Butanol Propanol From these calculations, I can draw the conclusion that ethanol will not be an efficient alternative for petrol, as petrol produces much more energy per gram than ethanol. Also based on these calculations, I can conclude that the larger the alcohol molecule, the more energy is given out. However, in practice, a proportion of this energy will be lost to the surroundings.
In order to verify these theoretical results, I need to make some actual measurements. My experiment is set out in the method. Prediction I predict that as the size of the alcohol molecule increases, so will the amount of energy per mole released. I also predict that ethanol will not release as much energy per mole than petrol. Method Apparatus list: Measuring cylinder Copper calorimeter Crucible & glass wool wick [spirit burner] Heat proof mat Clamp & stand Goggles Alcohols – ethanol, propanol and butanol. I will set up the apparatus as shown, using the above equipment.
From my preparation experiments, I know that to achieve the best results, I need to use 50cc’s [50ml or 50g] of water, and set the calorimeter at 10cm above the top of the spirit burner, which used a 3. 5cm high wick. Before I start the experiment, I will need to weigh the spirit burner with alcohol, so that after I have heated the water 10i?? C, I can weigh it again, and find the mass of alcohol burnt. I can then put this information into the formula: Mass of water x ? t of water x specific heat capacity of water = energy given out This formula tells me how much energy (in joules) I get from X grams of fuel.
I can already substitute in a few pieces of information, to give: 50 x 10i?? C x 4. 2 = energy given out 50 x 10i?? C x 4. 2 = 2100J Thus, for every X grams of alcohol I burn, I should get 2100J of energy, although how much of this will be lost to the surroundings, I do not know. I will repeat this experiment three times for each alcohol, in order to ensure that I don’t just get an anomalous result. Results Once I had collected my data, I put it into the following tables, and worked out how much energy one gram and how much one mole of the alcohol produced. I already know that X grams produces 2100J.
If I divide 2100 by X, I get the amount of energy one gram produces. 2 1 mass of fuel burnt gives J of energy 1. 27g 2100 1g 1653 46g 76038 1 mole 76kJ expt. 2 mass of fuel burnt gives J of energy1. 2g 2100 1g 1750 46g 80500 1 mole 80. 5kJ expt. 3 mass of fuel burnt gives J of energy 1 mole 71kJ average mass of fuel burnt gives J of energy 1 1649 46 75854 1 mole 75kJ propanol mass of fuel burnt gives J of energy expt. Analysis or results As can be seen from this graph, there is a discrepancy between the theoretical results, and what I actually recorded. This is because not all the energy went into the water, a lot was dissipated into the surroundings, i. e. the calorimeter, the heatproof mat, the air, etc.
However, we can still see a definite positive correlation between the size of the molecule and the amount of energy released as it is burnt. As the size increases, so does the amount of energy released. This is exactly what I predicted. Knowing that there is a positive correlation between the size of the molecule and the energy released, I can calculate the efficiency of my practical results. This is done by putting them into the following formula:
Theoretical Expt. Ethanol Propanol ButanoAs the size of the molecules increases, efficiency should decrease, as there will be more carbon to be oxidised, and therefore less energy as not all the carbon is turned into carbon dioxide (some will be turned into carbon monoxide; some into soot [just carbon]). However, this does not appear to be the case, as the efficiency was greatest for propanol, and least for ethanol. Evaluation I am satisfied with my results, I received no anomalous results, and I repeated my experiment three times, ensuring a good level of accuracy. I achieved a reasonable level of efficiency, although a higher level would have been much better.
The results back each other up; there is a definite positive correlation between the size of the alcohol molecule and the energy released per mole, suggesting a very firm conclusion. This supports my original prediction. Also, even theoretically, none of the alcohols produced as much energy as petrol. So I don’t think ethanol would be a viable alternative for petrol, although, some larger alcohols maybe. But, these molecules require more energy to get them started. There is also a trend in the balanced formulas for the complete combustion of the alcohols.
With each extra carbon atom in alcohol molecule, there are two extra hydrogen atoms, and 1. 5 extra oxygen molecules are needed to completely combust the alcohol. An extra carbon dioxide and an extra water molecule are also produced. This can be turned into a mathematical formula, thus: This sort of formula is called an homologous series. Alcohols are a homologous series, and other such series include alkanes, and alkenes. What could I do to make this experiment better?
To make this experiment work better, and achieve a higher efficiency, I could do several things. First, I could conduct the experiment inside a sealed container, as this would stop air heated by the combustion of the alcohol escaping. The container could also be silvered (this would stop heat being radiated away), as well as being filled with oxygen, which would ensure that as much of the alcohol was oxidised as possible. I could also use a much cleaner copper calorimeter, as this would ensure a good transfer of energy to the water.