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To determine the enthalpy change of a reaction

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To determine the enthalpy change of a reaction Analysis Results These are the results I obtained from reacting CaCo3 with Hcl and CaO with Hcl, these 2 experiments were done after one another. The temperature changes of each reaction will then be used to determine the enthalpy change. Reaction with CaCO3 CaO Mass used (g) 2.48 1.41 Temperature of acid (Hcl) initially (�c) 24 24 Temperature of solution after mixing (�c) 26 36 Temperature change during the reaction (�c) 2 12 Looking at these results it suggests that CaO was a more reactive reaction and a more exothermic reaction than CaCO3 as more energy was released than the reaction with CaCO3. However, we can see that both reactions were exothermic as when the reaction finished both temperatures of the solution after mixing the reactant with the Hcl rise. We can see that these reactions were not endothermic as the temperatures of the solution would have lowered as energy would have been absorbed. Calcium carbonate, CaCO3 decomposes with heat. CaCO3 (s) --> CaO(s) + CO2 (g) Calcium oxide reacted with Hydrochloric acid: CaO(s) + 2Hcl(aq) --> CaCl(aq) + H2O Both calcium oxide and calcium carbonate react readily with 2 mol dm-3 hydrochloric acid (Hcl). ...read more.


Therefore following the calculation we made before the enthalpy change of ?H3 CaO and CO2 will be: (-17.63 kj mol-1) - (-103.64 kj mol-1) = 86.01 kj mol-1 Therefore this shows that the enthalpy change of ?H3 is: CaCO3(s) CaO(aq) + CO2(g) is 86.01 kj mol-1. This equation is a endothermic reaction as the reaction absorbs energy. We can tell this by the positive ?H. In this reaction the temperature often falls. All answers are given to 2 decimal places The correct theoretical value for ?H3 should be 178 kj mol-1 As quoted from http://www.chemguide.co.uk/inorganic/group2/thermstab.html As we can see from the results I obtained and the theoretical value, it shows that my answer is quite far off to the theoretical value of calcium carbonate. We will look at reasons why this may be later on. Level of accuracy in apparatus Equipment errors = error / how much u used/measured x 100 * 250cm3 measuring cylinder- this was correct to every 0.5 cm3 +/-, therefore the percentage error will be worked out by: 0.5 / 50 x 100 =1.00% (exp 1+2) * 0-100�c thermometer 1�c +/- % error = 1/24 x 100 = 4.17% (exp. ...read more.


* Surface area on the calcium oxides and calcium carbonate: different sizes of calcium oxide/carbonate would have affected the rate of the reaction, so to improve this factor I would suggest to either use powdered calcium oxide or relatively large pieces of CaO which are very similar to each other size wise. * Thermometer: this went up in graduations of 1�c. This means that the results we collected could have varied by +/- 0.5cm3. To improve the accuracy and reliability of the experiment, a digital thermometer to 2dp would have been the better choice to ensure a reliable and accurate reading for the experiment. * Measuring balance: These were accurate to 2 decimal places and the balance error was to the nearest +/- 0.005. I have made calculations above to show the percentage error and although the percentage error is not relatively large, to increase the experiment, we should use balances accurate to at least 3 decimal places. Therefore it is obvious that these significant errors may have made my results less accurate and the improvements which have been suggested e.g. use a digital thermometer may increase the reliability of the experiment. As we have several variances of results which were used to obtain the correct enthalpy change. The variance in calculations has caused the ending results for ?H3 to be inaccurate. ?? ?? ?? ?? 1 ...read more.

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