Data Tables: Part A: Neutralization of 1.00 M HCI with 1.00 M NaOH Trial 1 Trial

Question

Data Tables: Part A: Neutralization of 1.00 M HCI with 1.00 M NaOH Trial 1 Trial 2 Total volume after mixing 7-4 Part B: Neutralization of 0.500 M HCI with 0.500 M NaOH Trial 2 Trial 1 Total volume after mixing 28, 2°C Calculations HC (0.500 M and 0.500 M NaOH) Trial 2 Part A Part B (1.00 M HCl and 1.00 M NaOH) Trial 1 Trial 2 Trial 1 AT at mixing Mass of calorimeter, em 359 334 33.46 134.5613 Mass of calorimeter, full 133.599-1-133.41 Mass of resulting solution ! qsolution Greaction AHneutralizaton Moles of HCI reacted Hneutralization (molar) Average molar Hneutral Overall average molar AHneutralization Difference between two average molar Percent difference between two average molar AHneutrelization Enthalpy of Neutralization

Explanation / Answer

Data

Part A : Neutralization of 1.0 M HCl with 1.0 M NaOH

Trial 1,

qreaction = 100 x 4.18 x 7 = 2926 J

qsolution = 2929 J

qneutralization = -2926 J

moles HCl = 1 M x 0.050 L = 0.050 mol

dHneutralization = -2926 J/0.050 mol x 1000 = -58.52 kJ/mol

Part B : Neutralization of 0.5 M HCl with 0.5 M NaOH

Trial 1,

qreaction = 100 x 4.18 x 3.4 = 1421.2 J

qsolution = -1421.2 J

qneutralization = -1421.2 J

moles HCl = 0.5 M x 0.050 L = 0.025 mol

dHneutralization = -1421.2 J/0.025 mol x 1000 = -56.85 kJ/mol

average dHneutralization = -57.685 kJ/mol

difference between the two = 1.67 kJ/mol

1. These are exothermic reaction as dHneutralization are -ve vales. The final temperature of the solution is greater than the initial temperature of the solutions. The average value matches with this.

2. moles HCl = molariy x volume

                     = 0.250 M x 0.050 L

                     = 0.0125 mol

heat evolved here = 0.0125 mol x -57.685 kJ/mol

                              = -0.721

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