3. A student gets the following data for the Heat of Neutralization Trial using
ID: 492507 • Letter: 3
Question
3. A student gets the following data for the Heat of Neutralization Trial using 1.0 M NaoH and 1.0 M HCl at the First Nations University of Canada Chemistry Laboratory. Trial 1 A. Mass of calorimeter (g) B. Mass of HCl and calorimeter (g) 73.2 Mass of calorimeter HCI NaoH (g) 94.2 01. Mass of HCI (g) (B-A) D2. Mass of NaOH (g) (C-B) Mass of Solution (g) (01 t D2) Initial Temperature HCl measured 35.8 (thermometer 1) Initial Temperature HCI corrected Initial Temperature NaoH (C) measured 30.8 (thermometer 2) nitial Temperature NaoH corrected E. Average Initial Temperature (C) corrected Final Temperature measured C) 38.0 (thermometer 1) F. Final Temperature (C) corrected G. Temperature change C) corrected (F-E) Using this data, calculate the Molar Enthalpy of Neutralization for this trial. The moal is 0.973 and bale is 1.7 for both thermometer 1 and thermometer 2 and the calculated heat capacity of the calorimeter was 32.9 J C 1 (5 marks)Explanation / Answer
3. Experiment : mixing 1 M NaOH + 1M HCl
mass of solution = 94.2 - 52.7 = 41.5 g
Temp. HCl crrected = 35.8 - 1.7 = 34.1 oC
Temp. NaOH corrected = 30.8 - 1.7 = 29.1 oC
Average initial temperature = 31.6 oC
final corected temp. = 38 - 1.7 = 36.3 oC
Temperature change = 36.3 - 31.6 = 4.7 oC
Molar enthalpy of neutralization = 41.5 x 4.184 x 4.7/1 M x 0.0205 L + 32.9/4.7 = 46.81 kJ/mol
2. Heat capacity of calorimeter
Heat lost by hot water
q(hot) = mCpdT
= (50.2 - 0.973) x 4.18 x (15.2 - 1.7) = 2777.88 J
Heat gained by hot water
q(cold) = mCpdT
= (51 - 0.973) x 4.18 x (8.7 - 1.7) = 1463.79 kJ
Heat gained by calorimeter = q(hot) - q(cold) = 1314.09 J
Heat capacity of calorimeter = 1314.09/(8.7 - 1.7) = 187.73 J/oC
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