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Results: 1. Gravimetric Determination of Ke Mass of Evap Dish (tare) 22.7998% Mass of Evap Dish (gross) 123. 7460g Calcu

Posted: Wed May 18, 2022 12:41 pm
by answerhappygod
Results 1 Gravimetric Determination Of Ke Mass Of Evap Dish Tare 22 7998 Mass Of Evap Dish Gross 123 7460g Calcu 1
Results 1 Gravimetric Determination Of Ke Mass Of Evap Dish Tare 22 7998 Mass Of Evap Dish Gross 123 7460g Calcu 1 (54.08 KiB) Viewed 64 times
Results 1 Gravimetric Determination Of Ke Mass Of Evap Dish Tare 22 7998 Mass Of Evap Dish Gross 123 7460g Calcu 2
Results 1 Gravimetric Determination Of Ke Mass Of Evap Dish Tare 22 7998 Mass Of Evap Dish Gross 123 7460g Calcu 2 (42.14 KiB) Viewed 64 times
Results 1 Gravimetric Determination Of Ke Mass Of Evap Dish Tare 22 7998 Mass Of Evap Dish Gross 123 7460g Calcu 3
Results 1 Gravimetric Determination Of Ke Mass Of Evap Dish Tare 22 7998 Mass Of Evap Dish Gross 123 7460g Calcu 3 (42.02 KiB) Viewed 64 times
Results 1 Gravimetric Determination Of Ke Mass Of Evap Dish Tare 22 7998 Mass Of Evap Dish Gross 123 7460g Calcu 4
Results 1 Gravimetric Determination Of Ke Mass Of Evap Dish Tare 22 7998 Mass Of Evap Dish Gross 123 7460g Calcu 4 (66.15 KiB) Viewed 64 times
Results 1 Gravimetric Determination Of Ke Mass Of Evap Dish Tare 22 7998 Mass Of Evap Dish Gross 123 7460g Calcu 5
Results 1 Gravimetric Determination Of Ke Mass Of Evap Dish Tare 22 7998 Mass Of Evap Dish Gross 123 7460g Calcu 5 (66.15 KiB) Viewed 64 times
Results 1 Gravimetric Determination Of Ke Mass Of Evap Dish Tare 22 7998 Mass Of Evap Dish Gross 123 7460g Calcu 6
Results 1 Gravimetric Determination Of Ke Mass Of Evap Dish Tare 22 7998 Mass Of Evap Dish Gross 123 7460g Calcu 6 (71.97 KiB) Viewed 64 times
Results: 1. Gravimetric Determination of Ke Mass of Evap Dish (tare) 22.7998% Mass of Evap Dish (gross) 123. 7460g Calculated Values record cach value until two are within 2 mg of one another Avg Final Mass (gross) Net mass of NH.NO, Moles of NH.NO, 23.402g 0.60229 7523x103 10. OBM 251.2 X .100005 Molarity of NH.NO, 2. Determination of Ksp using lons Selective Electrode (ISE) Ion Analyzed : Ammonium nitrate Nitrate (NO.; FW = 62.01 g/mole) Molarity of stock solution including formula 1.39253x10-3 Dilution of stock solution uL of stock solution diluted to mL total volume 500 L diluted to 100 mL total ISE Reading in mg/L for diluted stock solution 251.2 m/ 1.3925 ISE Reading in mg/L for diluted ammonium nitrate solution Calculated Molarity of diluted stock solution from dilutica factor Calculated Molarity of diluted stock solution from ISE valus Percent d'erence between the diluted stock solution marities Calculated Molarity of the ion in the diluted ammonium sitrate solution Calculated molarity of ammonium nitrate in the saturated solution
3. Determination of the Heat of Solution of Ammonium Nitrate Room Temperature Trial Number 29.7°C 1 2 Mass of NH.NO: 4.01609 Moles of NH.NO, 4.03049 5.03SX102 24.3°C 19°C 15.0178102 24.7°C 19.1 Initial Temperature (T.) T at time=1.00 min (TO) AT T-T 5.6c 5.3°C 9 water 4 cal 4 rxn ΔΗχολη Average AH soln
1. Defermination and comparison of thermodynamic values Gravimetric Values ISE Values Kap INH. INO K, [NH"][NO1 Percent Difference KG-KISEX 100% Ky from ref Gravimetric Calculated Values ISE Values 46° AG AG Percent Error Calc - Expt x 100% Calc AH ΔΗ AH Percent Error Calc - Expt x 100% Calc AS AS AS° Percent Error Calc - Expt x 100% Calc
Calculations Gravimetric analysis of Ksp for ammonium wiats. Page 114 1.1 Determine the mass of solid ammonium present in the section Convert this into moles (FW NHANO is 8096 g/mol). The concert of ammonium nitrate in the saturated solution must be the name of makes you calculate divided by the volume in liters of the solution you removed 500 104 ge d 4.1.2 The molarity of the ammonium nitrate must be the molarity of mind nitrate in the solution. The Ksp is then the product of these two Determine a value for the Ksp of ammonium nitrate You Determination of Ksp from ISE data ווד molarity) 4.2.1 Calculate the molarity of the diluted stock solution by accounting for boon. then the calculated molarity would be (0.7500ml/5.00mL.)"stock solution for example you diluted 750 L of the stock solution to a total volume 5.00 L 42.2 Calculate the molarity of ammonium or nitrate ion by converting the most M(FW for NH4 is 18.05: FW for NO; is 62.01 g/mol) for the diluted stated solution. Since the ISE value is in mg/l. then converting the number of mg of the ion to moles of the ion will be the ion molarity in the solution 4.2.3 Calculate the molarity of the ammonium or nitrate ion in the diluted stock solo that you measured. From the dilution determine the actual molarity expected this solution as well as the percent error for this value. (this is the difference between the calculated and observed molarity divided by the calculated molarity he 100%) 424 Calculate the molarity of the ammonium or nitrate ion in the dilued ammedin nitrate solution. In addition determine the molarity of the ammonium nitrate saturated solution by accounting for dilution (here multiplying by 200 to account for the 0.500mL to 100.0 mL) 1. TE Determination of heat of solution from calorimeter data 13.1 You should have a value for T. and Trfor each of the calorimetry runs. From this determine AT which is TT.. We will assume that the heat lost by the ammonium nitrate solution is the same as if it was lost by water and that the mass of the water is the same as the volume (density equal to one). Calculate the heat absorbed by the reaction using the s of water to be 4.184 J/g-C, the density of water as 1.00 g/ml and the ms of the calorimeter as being 10JCC.
CHM 200 Lab 11 Thermodynamics Rev205-19 Page 8 of 14 You should have a heat term for the water as well as the calorimeter as follows (note both of these q values should be negative): 4 water - m s AT = (mass of water used) x 4,184 1g-°C x (Tf - To) 4 calmsAT - 10/°C x (Tr-T) Determine the values for 4 water and 4 cal and enter this information into the table provided. 4.3.2 The heat lost by the water and the calorimeter plus the heat absorbed by the reaction is equal to zero: 4 total = 0 - 4 rx + 4 water + 4 calorimeter This equation can be solved for the heat given off by the reaction: 4 rxn--9 water +- 9 calorimeter Calculate the value for a reaction and enter it into the table provided. The enthalpy change for the reaction is the heat given off divided by the number of moles of limiting reagent. Divide the 4 rxn by the moles of material used and enter this value as the AHsoln in the table (note that the 4 rxn and the AHsoln will both be positive). 4.3.3 You should have two values for the enthalpy solution from the two trials so determine the average value and record this in the table provided. 4.4 Determination and comparison of thermodynamic values 4.4.1 The relevant equations are: AGAH-TAS and AG =-RTINK. You should also know how to calculate AH, AG and AS values from data as described in Appendix 3 of the Chang text 4.4.2 The data for ammonium nitrate as well as for the aqueous ammonium ion and nitrate ion are given below. Substance ΔΗ, Ammonium Nitrate (NH4NO/..) -365.6 kJ/mol Ammonium ion (NH) -132.8 kJ/mol -79.5 kJ/mol 112.8 J/mol K Nitrate ion(NO) -206.6 kJ/mol -110.5 kJ/mol 146.4J/mol-K AG -184.0 kJ/mol AS 151.1 J/mol-K From this data determine the standard values of AH, AG and AS for ammonium nitrate solid becoming ammonium nitrate solution. 4.4.3 From the molarity of the ammonium and nitrate ion determined from the gravimetric analysis calculate the value of the Ksp for ammonium nitrate. Repeat this calculation using the ISE data. 4.4.4 From the relationship between the standard Gibbs Free Energy and the equilibrium constant calculate a value for the standard Gibbs Free Energy using each of the Ksp values in 4.4.3. 4.4.5 From the calculated AG values and the average AH value calculate the AS values for this reaction using both values of AG (gravimetric and ISE). 4.4.6 Determine the percent error between the calculated values from 4.4.2 and the average heat of solution, both of AG° (gravimetric and ISE) values and both of AS° (gravimetric and ISE), values.