Chem 2 - The Second Law of Thermodynamics: Predicting Entropy Changes Qualitatively III

1. The Second Law of
Thermodynamics:
Predicting Entropy Changes
Qualitatively, plus Examples
(Pt 3)
By Shawn P. Shields, Ph.D.
This work is licensed by Shawn P. Shields-Maxwell under a Creative Commons
Attribution-NonCommercial-ShareAlike 4.0 International License.

2. Recall: The Second Law of Thermodynamics
The total entropy change for any
spontaneous process is positive
Positional disorder- The distribution of
molecules (particles) in space (related to
volume)
Thermal disorder- The distribution of
energy states among the particles
(related to temperature)

3. Predicting the Sign of S for Changes in Volume
Entropy increases with increasing
volume for the system.
Particles have more possible
positions in the larger volume.

4. Example 1: Predict the Sign of S
1 mol H2O (g) 1 mol H2O (g)
(1 atm, 25C) (0.01 atm, 25C)
No change in temperature, so no effect
on S.
Analyze the gas pressures…

5. Example 1: Predict the Sign of S
1 mol H2O (g) 1 mol H2O (g)
(1 atm, 25C) (0.01 atm, 25C)
Analyze the gas pressures…
PV = nRT
P V = constant

6. Example 1: Predict the Sign of S
1 mol H2O (g) 1 mol H2O (g)
(1 atm, 25C) (0.01 atm, 25C)
If the pressure is lower for the final state, the
volume must have increased (expansion).
Smaller to larger volume means positive S

7. Example 2: Predict the Sign of S
Solid KCl is dissolved in water

8. Example 2: Predict the Sign of S
Solid KCl is dissolved in water
𝐊𝐂𝐥 𝐬 → 𝐊+
𝐚𝐪 + 𝐂𝐥−
(𝐚𝐪)
The entropy increases (+S)
More positions are available
to the ions.
K+
CI

9. Example 3: Comparing S for Two Systems
Prepare two solutions with one mole
NaCl dissolved to the following
concentrations (at 25C):
5.0 M NaCl (aq) versus 1.0 M NaCl (aq)
Which solution has the higher entropy?

10. Example 3: Comparing S for Two Systems
Two solutions- one mole NaCl dissolved (at
25C): 5.0 M NaCl (aq) vs 1.0 M NaCl (aq)
Both solutions have the same number of Cl-
and Na+ ions…
The lower concentration must be in a larger
volume (diluted).
The more dilute solution has higher entropy.

11. Example 4: Sign of S for Chemical Reactions
Chemical reactions can also increase
or decrease entropy.
𝟒 𝐀𝐥 𝐬 + 𝟑𝐎 𝟐 𝐠 → 𝟐 𝐀𝐥 𝟐 𝐎 𝟑(𝐬)
Did entropy increase or decrease in
the reaction?

12. Example 4: Sign of S for Chemical Reactions
Analyze the phases in the reactants and
products:
𝟒 𝐀𝐥 𝐬 + 𝟑𝐎 𝟐 𝐠 → 𝟐 𝐀𝐥 𝟐 𝐎 𝟑(𝐬)
3 moles of gas on the reactant side, but zero
moles of gas in the products.
Gases have higher entropy than solids (and
liquids).
Entropy decreased ( S)

13. Next up…
Entropy and Heat (Pt 4)