Hess\'s Law Partner(s) REPORT Data and Calculations 1. AH (neutralization & dilu
ID: 522024 • Letter: H
Question
Hess's Law Partner(s) REPORT Data and Calculations 1. AH (neutralization & dilution) of 10 M H2Sou and 1.00 M NaoH Data Trial 2 Trial 1 Initial temperature of 10 M H2SO4 (5.0 mL in a graduated cylinder) 2 ,3 oc 21,7 oc Initial temperature of calorimeter (50.0 mL of 1.00 M NaOH 45.0 mL water) Final temperature of the mixed solutions in calorimeter (50 mL 1.00 M NaoH 45.0 mL water 5.0 mL 10 M H soa) 32 3 oc 32.2 oc Trial 1 Trial 2 Calculations a) Temperature change of NaoH solution in calorimeter l b) Temperature change of added H2SO. solution ATG Tamal -Tiniti) c) Calculate the heat gained by the NaOH solution. Assume that he product of density heat capacity d C) is 4.10 J.mL I.K 1. AHC Vcd C, TC d) Calculate the heat gained by the H2SO4solution. Assume that the product of density heat capacity d C) is 4.10 J.ml I.K AH Vod C, TG (e) Calculate AHn AH reaction -AHC-AHG Calculate the moles of added H2SO4 mol H2SO4 VG (L) M (mol/L) mol mol Calculate AH per mole of reactants by dividing the results of part I (e) by the number of moles of H2SO4 calculated n part (f. Express the final resul keep the proper sign of the calculated reaction enthalpy. kJ/mol kJ/mol Average kJ/molExplanation / Answer
Calculation (Ist page)
For Trial 1
(a) Temperature Change NaOH dT = 32.8 - 22.5 = 10.3 oC
(b) Temperature Change H2SO4 dT = 32.8 - 24.8 = 8 oC
(c) dHc NaOH = 50 x 4.10 x 10.3 = 2111.5 J
(d) dHo H2SO4 = 5 x 4.10 x 8 = 164 J
(e) dHrxn = 1.95 kJ
(f) moles H2SO4 = 10 M x 0.005 ml = 0.05 mol
(g) dHrxn/mol H2SO4 = 1.95/0.05 = 39 kJ/mol
IInd page
(a) Temperature change water dT = 28.1 - 22.2 = 5.9 oC
(b) Temperature change H2SO4 dT = 28.1 - 22.9 = 5.2 oC
(c) dHc of water = 90 x 4.10 x 5.9 = 2177.1 J
(d) dHo for H2SO4 = 10 x 4.10 x 5.2 = 213.2 J
(e) dHrxn = 1.964 kJ
(f) moles of H2SO4 = 10 M x 0.010 L = 0.1 mol
(g) dHrxn/mol H2SO4 = 19.64 kJ/mol