Exothermic reaction Essay Example for Free

Exothermic response EssayThe object of this investigation is to determine the enthalpy change for the reception CaCO3 (s) i CaO (s) + CO2 (g) by an indirect method based on Hess Law. Hesss law states that the enthalpy change for any chemical reaction is independent of the route taken provided that the initial and final conditions are identical. So the temperature change during these reactions below sens be measured and the enthalpy changes ? H1 and ? H2 envisiond. For Example Using Hess law with the calculated values for ? H1 and ? H2 it is possible to calculate a value for ? H3. Results Table.Temperature change during reaction The results for the temperatures are to the nearest whole number as it is unreal to measure to a point of a ? C with this type of thermometer and the masses rounded up to 2 decimal places for greater accuracy. Calculations Its possible to use the formula E = mc ? T, where E = energy transferred, m = mass of HCl, c= specific ignite capacity of HCl an d ? T = temperature change. This formula can be apply for calculating the energy transferred in the following reactions ? ?H1, CaCO3 (s) + HCl and ? H2, CaO (s) + HCl.Seeing as the molar mass of CaCO3 = coulomb. 00 ?H1 = 420 x (1 x 0. 0250) = 16. 80 kJmol-1 I will non include the last result in my average for ? H1, which is 16. 80 kJmol-1. This is because its way off the other results and would significantly run into my average results, its an anomaly. AVERAGE for the ? H1 for the reaction between CaCO3 + HCl (- 25. 09 kJmol-1) + (- 24. 90 kJmol-1) 2 ?H1 = 25. 00 kJmol-1 This value for ? H1 is negative because heat is lost to the surroundings. Its an exothermic reaction. Calculations for ? H2 for the reactions between CaO (s) + HCl 1.I will not include the 102. 86 kJmol-1 result in my average for ? H2. This is because its way off the other results and would significantly affect my average results, its an anomaly. AVERAGE for the ? H2 for the reaction between CaO + HCl (-128. 05 kJmol-1) + (- 111. 43 kJmol-1) 2 ?H2 = 119. 74 kJmol-1 This value for ? H1 is negative because heat is lost to the surroundings. Its an exothermic reaction. Using Hess cycle I will use the values that I concur calculated for ?H1 and ? H2 to work out the value for ? H3. ?H3= ? H1 ? ?H2 = (- 25. 00 kJmol-1) (- 111. 43 kJmol-1)= 86. 43 kJmol-1 This value is positive because heat is engrossed from the surroundings. Its an endothermic reaction. I eat up been told the actual value for ? H3, which is 178. 00. So I will calculate the luck by which my value is out by the actual value. 178. 00 ? 86. 43 = 91. 57 (91. 57 ? 178. 00) x 100 = 51% Evaluation Errors in procedure When the CaO and CaCO3 were put into the cup there was a delay out front the lid was put on. This could have caused heat to escape out of the cup and the temperature change would not have been as great compared to if there was no delay.