- Level: International Baccalaureate
- Subject: Chemistry
- Word count: 2168
Determination of potassium hydrogen carbonate into potassium carbonate
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Introduction
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 K2CO3 The temperatures of each trial at � 0.10K Trials Initial � 0.5% After � 0.4% Change � 0.9% 1 21.600 23.700 -2.100 2 22.200 24.600 -2.400 Therefore the Average can be calculated for the temperature: The equation shows, Average = change in temperature of trials (1 + 2)/2 Therefore the Average = (-2.1 +( - 2.4))/2 = -2.250K � 0.9% The table below shows the raw data of the temperatures at their initial point, after and the change in temperature of the reaction between KHCO3 with HCl. ...read more.
Middle
Therefore according to Hess's Law: H1= 2 X H2 - H3 Therefore : H1 = (2 X 26.875) - (- 14.619) = 68.369 KJ / mol � 12.5 % The total uncertainty that is shown above is from the total of the enthalpy changes of K2CO3 and KHCO3. 6.6 + 5.9 = � 12.5 % Conclusion and Evaluation: As shown above, of the calculation of the enthalpy change for the thermal decomposition of potassium hydrogen carbonate into potassium carbonate is calculated as + 68.396 KJ / mol � 12 %. (� 8.207 KJ/mol). I.e. 12/100 X 68.396 = � 8.207 KJ mol -1. Therefore, 68.396 - 8.207 = 60.188 KJ/ mol. However the Literature value given is + 92 KJ/mol. This shows that the value obtained from the experiment is below the literature values. That is 23.604 JK/ mole of a difference. The enthalpy change was of a decomposition reaction. This means that a chemical reaction, where the molecules of a compound (in this case KHCO3 ) breakdown to form simpler molecules of two or more new substances. I.e K2CO3, CO2 and H2O. By the enthalpy change of not just the literature value but also the experimental value. It can be seen that the change was positive. I.E. ...read more.
Conclusion
So the results obtained more precisely. 2. The stirring technique was not uniform. This is a random error, which created problems as it could have caused the reaction to incomplete at incorrect time. So the method could be improved by defining the time of stirring and the stirring technique. 3. The conditions of temperature and pressure might be slightly different to than of the standard enthalpy conditions would need to be. Which is 298 K and 1 atm. However this error will be minor. By controlling the laboratory environment can control the standard enthalpy conditions. This can be by using a water bath for the temperature at 298 K. 4. In calculation the density of the solution was assumed to be 1g cm -3 and specific heat capacity was taken as 4.18 Jg -1 0C-1. This is only for pure water and the values would differ here, introducing an error in calculation. 5. During the experiment the calorimeter was washed, but not sure all of the previous solution was removed. Therefore, use different calorimeters of polystyrene. 6. The concentration of HCl was supplied yet the error of the concentration was unknown, therefore if the concentration was much lower than 1 mol dm-3 this would have resulted in a lower enthalpy value. ?? ?? ?? ?? ...read more.
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