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# To work out the enthalpy change of combustion of alcohols

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Introduction

Enthalpy of combustion planning Aim: To work out the enthalpy change of combustion of alcohols To work out the enthalpy change of combustion of the alcohols, the energy output must be measured. This simplest way of doing this accurately is to use the thermal energy of combustion to raise the temperature of a substance with a known specific heat capacity. The rise in temperature of this substance can then used to work out the thermal energy used to do this work. Basic plan Apparatus Metal calorimeter Thermometer measures to 0.1 oc Clamp stand Alcohol with spirit burner Heat proof mat Scales that measure to two decimal places Wooden splint 100cm3 measuring cylinder that measures 1cm3 For safety To begin with prepare a table remove all items of stationary from the surface and place them in bags, leaving only the paper and pen that are to be used to record the results. Remove bags from under the desks and place them in the designated positions. With the exception of the scales collect all the apparatus above, and place them on the prepared table. To ensure the weights recorded on the scales are reliable. Switch the scales on, wait until the balance has settled, once the balance has settled press the tare button to ensure that the balance reads 0.00 and the balance registers the entire weight of the spirit burner alcohol. Weigh the alcohol in the burner to two decimal places with burner lid on. This weight should be recorded straight away in the table below, to ensure that a simple mistake in the recall of a figure weight doesn't corrupt the reliability of the final findings. The weighed alcohol should be placed with the equipment on the prepared desk. Next 100cm3 should be carefully measured out ensuring the bottom of the meniscus rests on the 100cm3 line. This water should then be carefully poured into the calorimeter, (no water should be present before hand and if the calorimeter has been previously used it should be dried with a cloth as it could affect the results) ...read more.

Middle

alcohol/g Estimated average mass of water heated/g Methanol 10 0.4 99.67 Ethanol 10 0.38 99.84 Propan-1-ol 10 0.34 100.00 Butan-1-ol 10 0.24 100.00 From the table containing the average changes that I observed in the combustion alcohols, I can clearly see a trend in the change in mass of fuel, burnt in order to raise the temperature of approximately 100cm3 of water by 10oc As the length of the chain increases, (the table is ordered by number of carbons in a chain increasing by one each row) mass of alcohol used to give out the energy required to raise the temperature 10 oc decreases. As the temperature change is kept constant the amount of energy that each alcohol is required to give out is constant. This suggests that energy is contained in a smaller volume. Which means that there are more covalent bonds being broken per cm 3 of liquid. Manifestly there most be an increase in the density of the alcohol as the length of the chain increases. Calculation of the enthalpy of combustion Assumptions made to make this calculation 1cm 3 of water has a mass of 1g The average mass of water used is exactly half way between the start and end mass The specific heat capacity of water is 4.2J/g/K Methanol q = cm?T q represents the quantity of energy used to obtain the rise in temperature of water c represents the specific heat capacity, (the amount of energy needed to raise 1 gram of a substance's temperature by 1 oK m represents mass of the substance being heated ?T represents the temperature change observed in the substance being heated q=4.2J/g/K x 99.67gx10 oK Note the intervals in the Kelvin scale are the same as the intervals of the Celsius scale therefore a change of 10 oc is the same as change of 10 oK q= 4186.14J The average heat energy absorbed by water from the combustion of methanol was 4190J To the enthalpy of combustion is energy given out (thermal) ...read more.

Conclusion

that measures to 1cm-3 is 0.005 The average value of water recorded was 100.00cm3 0.005 x 100 = 5.00x 10-3% 100 The uncertainty in reading a balance that meat measure to 2 decimal places it 0.005 the average mass recorded is the average (start mass + average end mass)/2 192.17+191.93 =192.05 2 0.005 x 100=2.60 x 10-3% 1920.05 The percentage my result is out is 1300 x 100=48.6% 2676 48.6-100= -51.4% from the actual enthalpy change of combustion of butan-1-ol My total percentage uncertainty is 0.51/% As I have established that the percentage uncertainty hasn't had a considerable effect on my experiment the only way I can think of improving my procedure is eradicate some of the procedural errors. The easiest error to remove, is the incomplete combustion of the alcohol, The experiment simply has to take place in an oxygen rich atmosphere to ensure that there is oxygen way in excess of quantity needed for the complete combustion of the alcohols. Also if I could redesign the procedure to remove the space between the burner and the calorimeter, and introduced oxygen feed to ensure enough oxygen is supplied I believe that it is quite possible to cut my percentage out from 2/3 to 1/3. The reason why incomplete combustion could have a drastic effect on the worked enthalpy of combustion is the fact that the products release less energy then the products of complete combustion. My investigation is based on the premises that complete combustion has occurred, and even if I did manage to collect all of the soot weigh it and work out how many moles of carbon is present, there is no way of getting it all. Therefore there is no way of working out the ratio of complete combustion to incomplete. Hence unless I get rid of the incomplete combustion there is no way gaining an accurate result. Reducing the distance between the burner, to a minimal will maximise the efficiency of the heat transfer from flame to the water ...read more.

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