Here are first two pages with part I and part II and calculations from another c

ID: 522329 • Letter: H

Question

Here are first two pages with part I and part II and calculations from another chegg person:

3. AH (neutralization) of 1.00 M Huso. and 1.00 M NaoH Data Initial temperature of NaoH solution in calorimeter (50 mL of 1.00 M NaOH) Initial temperature of H2So. solution in graduated cylinder (50 mL of 1.00 M H2SOe) Final temperature of mixture in calorimeter (50 mL 1.00 M NaOH 50 mL 1.00 M HSO40 Calculations a) Temperature change of NaoH solution in calorimeter ATC b) Temperature change of added H2Soesolution ATG -Ta )calculate the heat gained by the NaOH solution. d) Calculate the heat gained by the H:sousolution. e) Calculate AHradsea AHC-AHR (f Calculate the moles of added H2SO. mol H2Sos Vo(L) M (mol/L) g) calculate Hmacso per mole of reactants by dividing the results of part III (e) by the number of moles of Hiso. calculated in part III (f. Express the results in units ofk/mol. Be sure to keep the proper sign of the reaction enthalpy.) Average Trial 1 Trial 2 22.2. 29,7 oc Trial 2 Trial 1 kJ/mol kJ/mol kJ/mol

Explanation / Answer

We will assume that the density (d) of the solutions is d = 1.00 g/mL and the specific heat of both the solutions is the same as the specific heat of water, i.e, CS = 4.18 J/g.C.

Calculations

Trial 1

Trial 2

a) Temperature change of NaOH solution in Calorimeter I (TC = Tfinal - Tinitial) (C)

28.8 – 22.2 = 6.6

28.7 – 22.1 = 6.6

b) Temperature change of added H2SO4 solution (TG = Tfinal – Tinitial) (C)

28.8 – 21.9 = 6.9

28.7 – 21.8 = 6.9

c) Calculate the heat gained by the NaOH solution.

HC = VC*d*CS*TC

(50 mL)*(1.00 g/mL)*(4.18 J/g.C)*(6.6C) = 1379.4 J

d) Calculate the heat gained by the H2SO4 solution.

HG = VG*d*CS*TG

(50 mL)*(1.00 g/mL)*(4.18 J/gC)*(6.9C) = 1442.1 J

e) Calculate Hreaction.

Hreaction = -HC - HG

-(1379.4 J) – (-1442.1 J) = 63 J

f) Calculate the moles of added H2SO4.

VG(L)*M (mol/L)

(50 mL)*(1 L/1000 mL)*(1.00 mol/L) = 0.05 mole

e) Hreaction per mole

(63 J)/(0.05 mole) = 1260 J/mol = (1260 J/mol)*(1 kJ/1000 J) = 1.26 kJ/mol

Average

1.26 kJ/mol