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To determine the enthalpy change of a reaction.

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To determine the enthalpy change of a reaction Aim: The object of this practical exercise is to determine the enthalpy change for this reaction by an indirect method based on Hess's law - Calcium carbonate, CaCO3 decomposes with heat. CaCO3(s) CaO(s) + CO2 (g) An enthalpy change ?H is the heat energy exchange with the surrounding at constant pressure. Analysis I performed an experiment of observing the temperature change before and after each of the reactions between CaCO3(s) and HCl and CaO(s) with hydrochloric acid. Equipment and chemicals were provided. My results from my experiment are presented in the table below: Mass of CaCO3(s) + weighing bottle 12.06g Mass of empty weighing bottle 9.59g Mass of CaCO3(s) used 2.47g Temperature of HCl acid initially 24.5 0c Temperature of solution after mixing 25.0 0c Temperature change during reaction -1 0c Mass of CaO(s) + weighing bottle 11.18g Mass of empty weighing bottle 9.59g Mass of CaO(s) used 1.59g Temperature of acid initially 24.5 0c Temperature of solution after mixing 34.8 0c Temperature change during reaction -10.3 0c My results table shows clearly that calcium oxide reacted more than calcium carbonate as their was a higher and faster temperature rise being10.3 0c for calcium oxide while reacting with hydrochloric acid, then calcium carbonate of only having a temperature rise of 1 0c. ...read more.


m= 50g c= 4.2 Jg-1k-1 ?T= (34.8-24.50c) = 10.30c E= m x c x ?T Therefore: 50g x 4.2 Jg-1k-1 x 10.30c E= 2,163.0J This means that 0.028mols of CaO react to form 2.163.0J. I need to work out how much energy 1 mole of CaO is. 0.028mol: 2,163.0J I will divide both sides by 0.028mols 1mol:77250.0 I will convert the answer into KJ by dividing the answer by 1000 So: 77250.0/1000 = 77.25kJmol-1 ?H2 = -77.25kJmol-1 Using Hess's Law to find ?H3 I need to work out the standard enthalpy change for the thermal decomposition of CaCO3 to CaO using Hess's law. Hess's law states that the 'total enthalpy change for a chemical reaction is independent of the route by which the reaction takes place , provided initial and final conditions are the same' - quote from 'Chemistry 1 ocr text book'. ?H3 CaCO3(s) CaO (s) + CO2 (g) + 2HCl (aq) ?H1 CaCl2 (aq) + CO2 (g) + H2O (l) Route 1= Route 2 This means that ?H3 = ?H1 - ?H2 I calculated the values for ?H1 which is - 5.25 kJmol-1 and ?H2 which is -77.25kJmol-1 Therefore: ?H3 = - 5.25 kJmol-1 - ( -77.25kJmol-1) ...read more.


If more time was available, the temperature of the room could have been measured so it stays constant at all times. * While measuring the 2 mol dm-3 concentration of hydrochloric acid in a measuring cylinder, there may have been a parallax error as I may have not observed carefully the divisions on the cylinder at my eye level to make sure it was exactly 50cm3 of HCl. * The entire amount of CaCO3 (s) and CaO(s) used to react with the HCl may have not been entirely complete as there were still deposits of the reactants left at the bottom of the beaker. * Also there was a parallax error when measuring the temperature from the thermometer. As my eye level did not meet the level of temperature formed on the thermometer. I may have read the graduations to 1 0c incorrectly resulting in inaccuracy and reduction of reliability of readings which would effect my calculations of enthalpy change of the reaction slightly. * Also when weighing out the calcium oxide, which had to be in the range 1.3 to 1.5g, was done inaccurately as I weighed 1.59g. A little more over the exceeding amount. This may have resulted in different results that expected and there may have been excess of CaO(s) during the reaction with hydrochloric acid. ...read more.

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