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In order to calculate the enthalpy change of Calcium Carbonate to Calcium Oxide, which is exceedingly hard to control and measure the energy change we must use Hess' law

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Results CaCO Test 1 Test 2 Average Mass of CaCO Used :- 2.48g 2.49g 2.485g Temperature of HCl Initially :- 20�C 20�C 20�C Temperature of HCl after mixing with CaCO: - 22�C 22�C 22�C Total Temperature Change :- 2�C 2�C 2�C CaO Test 1 Test 2 Average Mass of CaO Used :- 1.40g 1.40g 1.40g Temperature of HCl Initially :- 19�C 20�C 19.5�C Temperature of HCl after mixing with CaO: - 38�C 36�C 37�C Total Temperature Change :- 18�C 16�C 17�C In order to calculate the enthalpy change of Calcium Carbonate to Calcium Oxide, which is exceedingly hard to control and measure the energy change we must use Hess' law which states:- "The total enthalpy change for a chemical reaction is independent of the route by which the reaction takes lace, provided initial and final conditions are the same" So therefore on adding Hydrochloric acid, which reacts readily with both, Calcium Carbonate and Calcium Oxide they both form Calcium Chloride crating a enthalpy energy triangle. ?H3 CaCO (s) CaO(s) + CO (g) ?H1 ?H2 CaCl (aq) Therefore:- ?H1 - ?H2 = ?H3 But we cannot enter our results directly into Hess' equation, we must convert them into Enthalpy energies. ...read more.


= 126136.6937 Jmol = 126 kJmol to 3sf With my result and the result produced by the data book I can calculate my percentage error. The information produced in the data book is under Standard Conditions. These conditons are:- * A pressure of 100 kilopascals * A temperature of 298K * The reactants must be in their natural physical state * All solutions must have a concetration of 1 mol dm With these accurate results from the data book I can calculate my percentage error:- Data book Information on "The Standard Enthalpy Change of Formation" ?H1 - Calcium Carbonate = -1207 kJmol ?H2 - Calcium Oxide = -635 kJmol ?H2 - Cabon Dioxide = -394 kJmol So on putting our Standard Enthalpy results into Hess's equation we get a Standard Enthalpy Change for Calcium Carbonate to Calcium Oxide:- ?H1 - ?H2 = ?H3 ==> -1207 - ((-635) + (-394)) = 178 kJmol Percentage error x 100 Standard Result - Our result Standard Result x 100 ==>178 - 126 178 =29.21% = 29% to 2 sf Evaluation Our experiment, when put into practice was not very reliable this was due to the factor of reactants surface area. ...read more.


There was also a problem with the chemicals used, the calcium oxide, calcium carbonate and hydrochloric acid may have been contaminated with other chemicals. It might not all have reacted the way it should have. I could nullify this by using a brand new sample of chemicals and have no interactions with other experiments or uses. This would guarantee reliable results. My method also had faults: The method did not specify a accurate measurement for Calcium carbonate and Calcium oxide, this would portray inaccurate and unreliable results, as an increase in the amount of reactants could increase the enthalpy value and could decrease the enthalpy value for a lower amount of reactants. Conclusion From my results I can tell that both reactions were exothermic, but that the reaction with calcium oxide was much more exothermic than the reaction with calcium carbonate. They were different because the bonds that were made in the calcium oxide reaction required less energy to be made than in the calcium carbonate reaction. My Hess' Law cycle can be labelled correctly:- ?H3 CaCO (s) -178 kJmol CaO(s) + CO (g) ?H1 ?H2 - 1207 kJmol -1029 kJmol CaCl (aq) If I were to do the experiment again I would enforce my changes mentioned to achieve a closer result to the Standard enthalpy value. Also I would further investigate enthalpy changes of alcohol's. ...read more.

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