Determinaing Enthalpy Change of Potassium

2150 WordsDec 5, 20109 Pages
Determining the enthalpy change for the thermal decomposition of potassium hydrogen carbonate into potassium carbonate. Controlled Variables: 1. Volume of HCl ± 0.5 cm3 (± 2%) 2. Concentration of HCl, 3. Same mass of K2CO3 and KHCO3 within specified ranges of 2.5 – 3.0g and 3.25 – 3.75g respectively 4. Same calorimeter used i.e. polystyrene cup is used in this experiment 5. Same thermometer will be used ± 0.10K 6. Same source of K2CO3, KHCO3 and HCl Raw Data Results: The raw data table shows, the temperatures at initial point, after and the change in temperature of the reaction between K2CO3(s) with HCl (aq) The change in temperature is calculated as: After temperature – initial temperature= change in temperature…show more content…
That is 30 cm3. Hence the uncertainty in measuring 30.000 cm3 will equal (0.5/30) X 100 = 1.7 % The uncertainty for measuring the temperature of K2CO3 has already been calculated above, that is (0.9 %). Therefore the total uncertainty can be calculated by summing up all the uncertainties that has been calculated. That is 1.7 + 0.9 = 2.6 % Therefore -0.282/0.0193 = - 14.619 KJ mol -1 ± 6.6 % 2.6 + 4.0 = ± 6.6 % The energy change for KHCO3 = (30/1000) X 4.18 X (+ 7.6) = 0.953 KJ ± 2.9 % / 0.0354 moles ± 3 % Therefore 0.953/ 0.0354 = 26.875 KJ mol -1 ± 5.9 % The uncertainty is calculated by summing up ± 1.7 % (as shown above) and of ± 1.2 % as the uncertainty for measuring temperature. Therefore 1.7 + 1.2 = 2.9 % Therefore the total uncertainty of the enthalpy change is 2.9 + 3.0 = ± 5.9 % By using Hess’s law the enthalpy change for the thermal decomposition of potassium hydrogen carbonate into potassium carbonate can be found. KCl (aq) + CO2(g) + H2O(l) = + 26.875 KJ/mol KHCO3 (s) + HCl(aq) 2KCl (aq) + CO2 (g) + H2O(l) = - 14.619 KJ/mol K2CO3(s) + 2HCl (aq) K2CO3 + CO2 + H2O 2KHCO3 2 HCl + CO2 + H2O Therefore according to Hess’s Law: H1= 2 X H2 - H3 Therefore : H1 = (2

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