Andrew Trummer AP Chemistry Thermodynamics Enthalpy of Reaction and Hess s Law Objective The purpose of the lab was to verify Hess s Law using the three acid base reactions Data Part 1 Data Table Determination of the Heat Capacity of the Calorimeter Initial Temperature Celsius 50 0 mL H20 room temperature 50 0 mL H20 heated 24 4 degrees Celsius 64 4 degrees Celsius Mixing Data Time sec Temperature Celsius Time sec Temperature Celsius 20 42 5 120 41 5 40 42 140 41 4 60 41 8 160 41 3 80 41 7 180 41 2 100 41 6 Tmix C 42 358 C qcal J 4 9563 J Tavg Ccal J C 7 5324 J C C 41 7 C Part 2 Data Table Determination of Heats of Reaction Reaction 1 HCl aq NaOH aq NaCl aq H2O l Initial Temperature Celsius 50 0 ml 2 0 M HCl 50mL 2 0 M NaOH 22 6 degrees Celsius 21 6 degrees Celsius Mixing Data Time sec Temperature Celsius Time sec Temperature Celsius 20 35 1 120 34 7 40 35 140 34 6 60 34 9 160 34 5 80 34 8 180 34 5 100 34 7 Tmix C 35 139 C Qcal H kJ mol 5 712 kJ mol J 5712 025 J Reaction 2 NH4Cl aq NaOH aq NH3 aq NaCl aq H2O l Initial Temperature Celsius 50 0 mL 2 0 M NH4Cl 50 0 mL 2 0 M NaOH 20 6 degrees Celsius 21 6 degrees Celsius Mixing Data Time sec Temperature Celsius Time sec Temperature Celsius 20 23 120 23 1 40 23 140 23 1 60 23 160 23 1 80 23 180 23 1 100 23 1 T mix C 22 972 C q cal J 791 02 J H kJ mol 0 7910 kJ mol Reaction 3 NH3 aq HCl aq NH4Cl Initial Temperature Celsius 50 0 mL 2 0 M NH3 50 0 mL 2 0 M HCl 21 0 degrees Celsius 20 9 degrees Celsius Mixing Data Time sec Temperature Celsius Time sec Temperature Celsius 20 32 7 120 23 6 40 32 8 140 32 5 60 32 7 160 32 4 80 32 6 180 32 4 100 32 6 T mix C 32 822 C q cal H kJ mol 5 0219 kJ mol J 5021 9462 J Calculations Part 1 Calculate the Heat Capacity of the Calorimeter 1 2 y 0 0069x 42 358 y 0 0069 0 42 358 y 42 358 C 3 What I did to find the average temperature is that I took all the temperatures and added them together Then I divided them by the how many temperatures there were In this case there are 9 temperatures so I divide by 9 42 5 42 41 8 41 7 41 6 41 5 41 4 41 3 41 2 375 9 41 7 4 Qwater grams of water x specific heat of water x Tmix Tavg qwater x x 4 18 J g C x 42 358 41 70 qwater 4 9563 J qwater qcal 4 9563 J qcal 4 9563 J qcal 5 Ccal Ccal 7 5324 J C Part 2 Calculate the Enthalpy of Reaction 1 HCl aq NaOH aq NaCl aq H20 l NH4Cl aq NaOH aq NH3 aq NaCl aq H2O l NH3 aq HCl aq NH4Cl aq 2 qrxn heat absorbed by solution heat absorbed by calorimeter qrxn grams of solution x specific heat x Tsolution Ccal x Tsolution Reaction 1 qrxn 103 g of solution x 4 18 J g C x 35 139 22 1 7 5324 x 35 139 22 1 qrxn 5613 81 98 215 qrxn 5712 025 J Reaction 2 qrxn 103 g of solution x 4 18 J g C x 22 97 21 1 7 5324 x 22 97 21 1 Qrxn 791 02 Reaction 3 qrxn 103 g of solution x 4 18 J g C x 32 822 20 95 7 5324 x 32 82220 95 qrxn 5021 9462 J 3 Reaction 1 H H 5712 025 J x 1kJ 1000 J 5 0712 kJ mol Reaction 2 H H 791 02 J x 1 kJ 1000 J 0 7910 kJ mol Reaction 3 H H 5021 9462 J x 1kJ 1000 J 5 0219kj mol Part 3 Verify Hess s Law 1 HCl aq NaOH aq NaCl aq H20 l H Cl Na OH Na Cl H2O l H OH H2O l NH4Cl aq NaOH aq NH3 aq NaCl aq H2O l NH4 Cl Na OH NH3 Na Cl H2O l NH4 OH NH3 H2O l NH3 aq HCl aq NH4Cl aq NH3 H Cl NH4 ClNH3 H NH4 2 H OH H2O l NH4 OH NH3 H2O l NH3 H2O l NH4 OH H 5 712 kJ mol H 0 791 kJ mol H 0 791 kJ mol NH3 H NH4 3 3rd Reaction Difference in percentage H 4 921 kJ mol 4 921 5 0219 0 9799 x 100 97 99 100 97 99 2 01 There is a 2 01 difference between the H calculated through calorimetry and the H calculated through Hess s Law Post Lab Questions 1 Calorimetry means the measure of heat flow Basically it is the heat that is measured between both reactants and products 2 Graphical analysis improves the accuracy of data by showing how close the data is to the trend line From there you can prove that the data is either really close to the trend line or really far from the trend line 3 The meaning of the negative sign in front of the equation shows the heat that was absorbed by the calorimeter So it is showing that the number is the same value of the reaction but has the opposite sign 4 The lab results do not support Hess s Law The values are too small to be the actual heats 5 The procedure could be modified to make it more accurate by improving the instruments that were used during the lab 6 1 H OH H2O l H H 0 kJ mol H 515 83 kJ mol OH H 230 kJ mol Water H 285 83kj mol 2 NH4 OH NH3 H2O l H 728 62 kJ mol NH4 H 132 5 kJ mol OH H 230 kJ mol NH3 H 80 29 kJ mol Water H 285 83kj mol 3 NH3 H NH4 NH3 H 80 29 kJ mol H 212 79 kJ mol H H 0 kJ mol NH4 H 132 5 kJ mol These values do support Hess s Law All these values do support Hess s Law because they are the values that that are assigned to each of the chemicals And when you subtract the 2nd equation by the 1st you will get the third equation Conclusion The purpose of the lab was to use three acid base reactions to verify Hess s Law We had to calorimetry in order to find what we need to find in the end of the experiment We had to use several key equations like q C x m x T to figure out the q s …