# Determinaing Enthalpy Change of Potassium Essay B
• Words: 1793

• Pages: 7

Get Full Essay

Get Access

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. ))/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. KHCO3| The temperatures of each trial at ± 0. 10K| Trials| Initial ± 0. 5%| After ± 0. 7%| Change is ( + ) ± 1. 2%| 1| 21. 700| 14. 300| 7. 4| 2| 22. 000| 14. 200| 7. 8| The average can be calculated for the temperature by using the same equation as shown above. Therefore: Average of the temperature change of KHCO3 = (7. 4 + 7. 8)/2 = 7. 6 0K ± 1. 2% When using a thermometer that measures ± 0. K the uncertainty shown in the table of trial 1 for example, K2CO3 measures 21. 7000K will equal to (0. 1/21. 600) X 100 = ± 0. 5% for trial 2: (0. 1/14. 300) X 100 = ± 0. 4 % The overall uncertainty for K2CO3 for the temperature change is the summing up of all the uncertainties. 0. 5 + 0. 4 = ± 0. 9% ( this has been used above ) This is the same method for the calculation of the uncertainties for KHCO3, For trial 1: (0. 1/21. 000) X 100 = ± 0. 5 % For trial 2: (0. 1/ 21. 700) X 100 = ± 0. 7 % The overall for KHCO3 is 0. 5 + 0. 7 = ± 1. 2 %

As the mass was used from a range, for K2CO3 and KHCO3 from specified ranges of 2. 5 – 3. 0g and 3. 25 – 3. 75g respectively. Therefore: The raw data for the two trials for both the compounds has been tabled with the average worked out for each species. For K2CO3| For KHCO3| Trials| The mass that was used from specified range of 2. 5 – 3. 0g ± 0. 010g| Average| Trials| The mass that was used from specified range of 3. 25 – 3. 75g ± 0. 010g| Average| 1| ± 0. 4%2. 729± 0. 4%2. 613| ± 0. 8%| 1| ± 0. 3%| ± 0. 6%| | | 2. 671| | 3. 455| 3. 555| 2| | | 2| ± 0. 3%| | | | | | 3. 65| | The average was worked out as: = the mass of trials (1 + 2)/2 When using a balance that weighs ± 0. 010g the uncertainty shown in the table of trial 1 for K2CO3 weighs 2. 729g will equal to (0. 010/2. 729) X 100 = 0. 4% Hence for trial 2 = (0. 010/2. 613) X 100 = 0. 4% Therefore the overall uncertainty for the mass of K2CO3 is summing up all the individual uncertainties for both the trials. Hence 0. 4 + 0. 4 = 0. 8% For KHCO3: Trial 1: (0. 010/3. 455) X 100 = 0. 3% Trial 2: (0. 010/3. 665) X 100 = 0. 3% Therefore the overall uncertainty equals 0. 3 + 0. 3 = 0. 6%

The moles of each species are required to be calculated. The calculation is, Moles = mass/Mr Therefore the Mr of K2CO3 = (2 X K) + (1 X C) + (3 X 0) Mr = (2 X 39. 10) + 12. 01 + (3 X 16) = 138. 01 Therefore as the mass is 2. 671, then Moles = 2. 67/ 138. 21 = 0. 0193 M For KHCO3 Mr = 39. 10 + 12. 01 + (3 X 16) + 1. 01 = 100. 11 ± 4 % Moles = 3. 555/ 100. 11 = 0. 0354 ± 3 % Specific heat capacity of solution (Jg-1 0C-1) Temp. change (0C) Mass of solution being heated (g) The enthalpy change is required to be calculated. Therefore the enthalpy change can be calculated by the following equation:

Energy change = X X Since the solution has the same density as water (1g/cm3) and the same heat capacity as water (4. 18 Jg -1 0C-1) Therefore the energy change for for K2CO3 = (30/1000) X 4. 18 X -2. 25 = (-0. 282/1000) KJ ± 2. 6 % / ± 4 % 0. 0193 moles Due to using 30cm3 the uncertainty of the measuring cylinder is ± 0. 5 cm3. 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. + 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 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 + 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. the decomposition reaction was endothermic. This is where the reactants have less enthalpy than the products. This is an increase in energy taken from the surrounding, and the products are less energetically stable than the reactants.

This is because in endothermic reactions heat is absorbed from the surroundings, as the bonds in the reactants are stronger than that of the bonds in the products. So it requires energy due to the fact that the reactant will not spontaneously decompose by itself. So in this reaction, the heat transferred from the water to the reaction can be calculated by measuring the lowered temperature of the known mass. The enthalpy for a reaction depends on the difference between the enthalpy of the products and the enthalpy of the reactants. It is independent on the route of which the reaction might occur on.

Therefore the enthalpy change for a reaction is the sum of the individual enthalpy changes for each step. Therefore Hess’s is a very beneficial method for determining the enthalpy change for the decomposition, due to the reaction being difficult to measure directly. Hess’s law is illustrated with an aid of a cycle. (as shown above for this reaction). Error| Type of error| Quantity of error| Explanation for error| Improvements| Measurement in measuring cylinder| Systematic error| ± 0. 5 cm3| Equipment limitation, this is because the line where each of the reading might not be precise. i. . the manufacturer has not placed the line at the exact place on the measuring cylinder. | Utilise different measuring cylinders from different manufacturers, in order to be more precise in the measurements. Also must make multiple trials, in order for the accuracy to increase. | | Random error| | Unequal measurement of the meniscus, due to the fact that it is unclear where exactly the meniscus lies. | Utilise different measuring cylinders from different manufacturers, in order to be more precise in the measurements. Also must make multiple trials, in order for the accuracy to increase. Concentration| Systematic error. | | Manufacturer limitation, unsure whether the exact concentration was made precisely. | Use different manufacturers and increase the number of trials obtained. In order to increase accuracy. | Measurement in the scale. | Systematic error| ± 0. 010g| Equipment limitation. This can be because the measurement of the mass might be inaccurate. | Use multiple scales and increase the number of trials done. Also send back the electronic balance to the manufacturer to re-calibrate it. | Thermometer| Systematic error| ± 0. 10K| Equipment limitation.

This is because the actual reading might be designated at another place. i. e placed incorrectly. | Use multiple scales and increase the number of trials done. In order to increase the accuracy. | Source of K2CO3, KHCO3 and HCl| Systematic error| | Trust on the manufacturer, to give the right source of each of the compounds. | Use multiple manufacturers and increase the number of trials. | Powder not completely removed from dish. | Random error| | Not all the powder might have been placed into the reaction causing an incomplete reaction of the limiting reagent. Use diluted water onto the dish, to wash the powder into the calorimeter. In which the extra volume added must be included in the calculation. | Effervescence (fizzing)| Random error| | This may cause a loss of volume, due to the lid being placed on top of the polystyrene causing the fizzing to be sprayed on to the lid. | A larger cup of polystyrene is required to be used in order for the fizzing of the solution to not travel far away from the reaction. | As mentioned before the experimental value obtained is well below the accepted literature value this also could be due to several reasons. 1.

The end point being subjective. i. e. the method was unclear as to when the temperature should be checked. This could have been the potential heat loss during the stirring or when the potential heat loss after the reaction had completed, i. e. the reaction had stopped effervescence. The method could be improved by creating test trials. In which, multiple readings could be recorded, where the optimum i. e. the highest temperature change recorded. The time that it took for the highest could be used as a controlled variable, where the reaction would take place for only that period of time. So the results obtained more precisely. . 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. . 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.