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Comparing the enthalpy changes of combustion of different alcohols

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Introduction

Oliver White - Comparing the enthalpy changes of combustion of different alcohols. Theory When chemical reactions occur, bonds between atoms are broken and new bonds are made. The amount of energy needed to break a particular bond is called its bond enthalpy(H). We cannot measure the enthalpy of a substance but we can measure the change in enthalpy(?H) when a reaction occurs. ?H = Hproducts - Hreactants Hproducts is the final enthalpy of the system, measured in joules. In a chemical reaction, Hproducts is the enthalpy of the products. Hreactants is the initial enthalpy of the system, measured in joules. In a chemical reaction, Hreactants is the enthalpy of the reactants. Enthalpy change in a chemical reaction gives the gives the quantity of energy transferred to or from the surroundings, when the reaction is carried out openly. In an exothermic reaction, the enthalpy of the reacting system decreases. ?H is negative. In an endothermic reaction, the enthalpy of the reacting systems increases. ?H is positive. When chemists talk about enthalpy changes they often refer to the system meaning the reactants and the products of the reaction they are interested in. The system can lose or gain enthalpy depending on the surroundings meaning the rest of the world: the test tube, the air etc. I will be investigating the enthalpy change of combustion of different alcohols. The standard enthalpy change of combustion (?Hc?) is the enthalpy change, when one mole of a substance burns completely in oxygen under standard conditions. ?Hc? is always negative. Fig 1 Enthalpy change diagram showing the combustion of methanol. Plan I am investigating the enthalpy change of combustion of several alcohols so that I can find how and why enthalpy change is affected by the molecular structure of the alcohol. The alcohols I will be using are methanol, ethanol, propan-1-ol, butan-1-ol and pentan-1-ol. These will give me clear results as they range from short-chain hydrocarbons to long-chain hydrocarbons, so patterns should be fairly clear in the results and my conclusion. ...read more.

Middle

To finish my calculations, I will be using the figures of energy released and will work out the number of moles of fuel that has been burnt by working out the difference between the mass of fuel before the water heated and the mass of fuel afterwards. Then I will work out the energy given out by 1 mole of the substance, leaving me and answer in J/mol-1, meaning I will have to finish by dividing the final calculation by 1000 to get an answer KJ. 1) Mass of water x 4.2 x temperature rise = X 2) Mass = Y Mr 3) Heat given out by fuel (X) Number of moles (Y) 4) Final Answer 1000 I will use methanol again as an example to show my working out and then put all my final readings into a table, to be able to compare. 1) 100 x 4.2 x 17 = 7140 j 2) 0.65 32 [Mr = CH3OH = C bonds (1 x 12) + O bonds (1 x 16) + H bonds (4 x 1)] = 0.0203 (4 d.p) 3) 7140 0.0203 =351724.1379 J/mol-1 4) 351724.1379 1000 = - 351.72 KJ/mol-1 (2 d.p) [answer is negative because it is an exothermic reaction] Now I can say that the enthalpy change of combustion of one mole of ethanol is equal to 351.72 KJ. Below is a table for the results of the rest of the calculations of the other alcohols. Alcohol Mass Burnt (g) Energy Released (J) Molar Mass (g) ?Hc/KJ mol-1 Methanol 0.65 7140 32 -351.72 Ethanol 0.65 6720 46 -435.46 Propan-1-ol 0.41 7140 60 -1050 Propan-2-ol 0.47 7140 60 -915.38 Butan-1-ol 0.35 6720 74 -1429.79 Table 2: Final Results My results show fairly clearly that the enthalpies of combustion increase. I think this is because of the atoms and bonds present in each alcohol. As you go down the homologous series of alcohols, another carbon atom is added and two hydrogen atoms to each alcohol, meaning two more H-C bonds and one C-C bond to break each time. ...read more.

Conclusion

The last percentage error is when weighing the spirit burners. This percentage error won't be as high, as the balance is quite accurate. Nevertheless I could only read it to an accuracy of +/-0.005, giving me a percentage error of 0.0047%. This is very low because the mass of the object is more than double all of the other masses I have been working out. Percentage error = 0.01 x 100 214.581 (rather than 100 for the measuring cylinder) =0.0047% The reading is 214.581 because rather than working this out for all 10 times I used the scales, I added all the weights up including before and after and then divided by 10 to get an average. There are other points to look at which barely affect the experiment, but must still be pointed out, for example did I constantly stir the water the same for each alcohol or even how long I left for the temperature to rise, after extinguishing the heat. I believe I did do these fairly correctly as I did with how quickly I extinguished the flame itself after burning but there may still be errors involved which accumulate to the overall errors when doing this experiment. I my results were useful in identifying which alcohol has the highest enthalpy change in combustion, and the pattern involved. I believe that they were as accurate as I could do them with the apparatus available and the time given to me to perform this experiment. If I had the opportunity to redo the experiment with the chance to use the improvements I have suggested, I would get better results but they would still be far from the expected results under standard conditions, which is simply due to human error and the fact that I am not able to perform this experiment under standard conditions. Even though the improvements I stated would improve my final readings and results, overall my experiment succeeded in finding out which alcohol produces the highest enthalpy change of combustion out of the 5 I used and the trend involved as you go down the homologous series. ...read more.

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