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The aim of this experiment is to compare the amount of energy released when a number of alcohols are burned.

Extracts from this document...

Introduction

Planning Aim The aim of this experiment is to compare the amount of energy released when a number of alcohols are burned. Introduction An alcohol is a series of organic, homologous compounds with the general formula CnH2n+1OH. Alcohols react with the oxygen in the air to form water and carbon dioxide. In this coursework we are going to carry out the experiment on a maximum of eight alcohols. On the next page I have drawn the structure of each of these alcohols, as the structure will determine the formula of the alcohol. This is because the formula includes the number of carbon atoms and hydrogen atoms in one molecule of the alcohol, so by drawing the structure of each alcohol we can clearly see the number of carbon and hydrogen atoms in its molecule. This will give its formula. The formula of each alcohol will be needed when working out the equation for the burning of the alcohols. What affects the amount of heat released by the alcohol, in this coursework? Volume of water - if there is more water, more energy will be required to raise the temperature of it by a fixed temperature. And if there is a small volume of water, less energy will be required to raise the temperature by the same value. This is because when there is more water, there are more water particles that have to be provided with energy so that they move around quicker. But when there is less water there are less water particles, so not as much energy has to be inputted to make the particles move around. Distance between burner and can - when there is a larger distance between the two, the alcohol will have to release more energy to raise the temperature of water by a certain temperature because there will be more heat loss in the air between the burner and can. ...read more.

Middle

0.40 23 33 10 250 178.57 178.17 0.40 Average 0.40 Hexan-1-ol 22 32 10 250 182.31 181.93 0.38 23 33 10 250 181.93 181.54 0.39 Average 0.39 Heptan-1-ol 23 33 10 250 185.83 185.43 0.40 22 32 10 250 185.43 184.97 0.46 Average 0.43 Octan-1-ol 23 33 10 250 182.92 182.41 0.51 24 34 10 250 182.41 182.02 0.39 Average 0.45 Analysis By looking at the results I obtained from the investigation, it can be generally seen that as the total number of bonds in the alcohol increased, so did the H. Relating back to the prediction I made in the Planning section of the coursework, I can see that my results agreed with it. I predicted that as the number of carbon atoms in one molecule of the alcohol increases, so will the negative H. This was true because when there was one more carbon added to the alcohol, there were also two more hydrogen atoms added. This increased the overall RMM of the alcohol, which subsequently caused the value for 'number of moles x 1000', decrease. As this value reduced, the value for enthalpy change increased because the number I was dividing by got smaller so the final answer got larger. This is also true for the formula where only the 'mass of alcohol reacted' is the denominator, because as the RMM gets larger the numerator increases numerically. When the numerator is higher the denominator divides into it more times, so the final value is also higher. This confirms my prediction. There is another explanation behind the reason of why the enthalpy change increased as the number of carbon atoms in one molecule of the alcohol, increased. When there are more bonds that make up the alcohol, more energy is required to break these bonds up, so that the atoms are free. Subsequently, when there are more free atoms about, more energy will be released when the new products are formed. ...read more.

Conclusion

One way we could do this is to repeat the experiment again. A third set of findings could be found and this would improve the results because when there are more findings the results become more reliable and accurate. Then the experiment could again be repeated at the points where there were disagreements between the first and second findings, such as with Methanol and Octan-1-ol. This would hopefully wipe out the discrepancies because repeating the experiment would highlight the findings that were anomalous, and the average would become more accurate. Another way of extending this experiment is to carry out the experiment on more alcohols. This would give more points on the graph and improve the line of best fit, as there would be more points that it has to go through. We could carry out this same experiment, but with a 30oC temperature rise of water each time. This would show whether the trend of H is the same. It is likely that the line of best fit would be even lower than it is with this experiment, because when there is a higher temperature rise, there will be more time for extra heat loss. When extra heat is lost, more alcohol than is needed will be burnt, so the 'mass of alcohol reacted' will increase, decreasing the final value. If this is true then the line of best fit will be lower. So by doing this experiment, we will be able to compare the trends. Overall, by looking at the way I carried out the experiment, I think the investigation was relatively successful as I only obtained two anomalous results on my graph. And although we obtained some anomalies, these were bound to be obtained, as there were possibilities of such inaccuracies and errors occurring. But the trend that I initially predicted did come out to be true. And as I have identified, there are other ways that I could have carried out the experiment, in order to improve the reliability and obtain more accurate results. I have also explained ways in which this experiment can be extended for further study. ...read more.

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