Trial 3: Mass of solution = 50.051 g + 4.264 g = 54.315 g qsurroundings 54.315 g 4.184 J/goC -935.2 J = Step 3: Calculate the moles of NaNO3 (-4.1°C). = Molar mass of NaNO3 = 85 g/mol Trial 1: Moles = 4.288 g / 85 g/mol = 0.0504 mol Trial 2: Moles 4.288 g / 85 g/mol = 0.0504 mol = Trial 3: Moles = 4.264 g / 85 g/mol = 0.0502 mol Step 4: Calculate the enthalpy change (AH) for each trial AH=qsurroundings/moles of NaNO3 Trial 1: AH-967.2 J/0.0504 mol = -19190 J/mol = =-19.19 kJ/mol Trial 2: AH = -961.9 J/0.0504 mol -19081 J/mol -19.08 kJ/mol Trial 3: AH = -935.2 J/0.0502 mol = -18631 J/mol = -18.63 kJ/mol Step 5: Calculate the average enthalpy change Average AH (-19.19 kJ/mol - 19.08 kJ/mol - 18.63 = kJ/mol)/3=-18.97 kJ/mol Therefore, the average enthalpy change for the dissolution of NaNO3 is approximately -18.97 kJ/mol. This negative value indicates that the dissolution process is exothermic.
Trial 3: Mass of solution = 50.051 g + 4.264 g = 54.315 g qsurroundings 54.315 g 4.184 J/goC -935.2 J = Step 3: Calculate the moles of NaNO3 (-4.1°C). = Molar mass of NaNO3 = 85 g/mol Trial 1: Moles = 4.288 g / 85 g/mol = 0.0504 mol Trial 2: Moles 4.288 g / 85 g/mol = 0.0504 mol = Trial 3: Moles = 4.264 g / 85 g/mol = 0.0502 mol Step 4: Calculate the enthalpy change (AH) for each trial AH=qsurroundings/moles of NaNO3 Trial 1: AH-967.2 J/0.0504 mol = -19190 J/mol = =-19.19 kJ/mol Trial 2: AH = -961.9 J/0.0504 mol -19081 J/mol -19.08 kJ/mol Trial 3: AH = -935.2 J/0.0502 mol = -18631 J/mol = -18.63 kJ/mol Step 5: Calculate the average enthalpy change Average AH (-19.19 kJ/mol - 19.08 kJ/mol - 18.63 = kJ/mol)/3=-18.97 kJ/mol Therefore, the average enthalpy change for the dissolution of NaNO3 is approximately -18.97 kJ/mol. This negative value indicates that the dissolution process is exothermic.
Chemistry: Matter and Change
1st Edition
ISBN:9780078746376
Author:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Publisher:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Chapter14: Mixtures And Solutions
Section: Chapter Questions
Problem 106A: Which solute has the greatest effect on the boiling pointof 1.00 kg of water: 50.0 g of strontium...
Related questions
Question
For the experiment of Investigating Entropy
NaOH
- Trial 1: ΔT=7.8∘C, qsurroundings=1.633 kJ
- Trial 2: ΔT=7.8∘C, qsurroundings=1.632 kJ
- Trial 3: ΔT=9.1∘C, qsurroundings=1.906 kJ
The average enthalpy change (ΔH) for the dissolution process is:
ΔH=−33.71 kJ/mol
For NaOH3 is on the paper (see below please)
a) Was each dissolution process endothermic or exothermic? Explain how you determined this.
b) Based solely on your observations in lab, was the dissolution of each solid spontaneous or nonspontaneous? How could you tell?
c) Based solely on your calculations for ΔS, did the entropy increase or decrease during each dissolution process? How do you know? Is this consistent with what you expect?
d) Based on your results of this experiment, what makes a process spontaneous? And how many trials did you do? Do you think you have enough trials to have data that is
reliable and reproducible? Why or why not?
Transcribed Image Text:Trial 3:
Mass of solution = 50.051 g + 4.264 g = 54.315 g
qsurroundings 54.315 g 4.184 J/goC
-935.2 J
=
Step 3: Calculate the moles of NaNO3
(-4.1°C).
=
Molar mass of NaNO3
=
85 g/mol
Trial 1: Moles = 4.288 g / 85 g/mol = 0.0504 mol
Trial 2: Moles 4.288 g / 85 g/mol = 0.0504 mol
=
Trial 3: Moles = 4.264 g / 85 g/mol = 0.0502 mol
Step 4: Calculate the enthalpy change (AH) for each
trial
AH=qsurroundings/moles of NaNO3
Trial 1: AH-967.2 J/0.0504 mol = -19190 J/mol
=
=-19.19 kJ/mol Trial 2: AH = -961.9 J/0.0504 mol
-19081 J/mol -19.08 kJ/mol Trial 3: AH = -935.2
J/0.0502 mol = -18631 J/mol = -18.63 kJ/mol
Step 5: Calculate the average enthalpy change
Average AH (-19.19 kJ/mol - 19.08 kJ/mol - 18.63
=
kJ/mol)/3=-18.97 kJ/mol
Therefore, the average enthalpy change for the
dissolution of NaNO3 is approximately -18.97 kJ/mol.
This negative value indicates that the dissolution
process
is exothermic.
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