1. The Third Law of
Thermodynamics and
Standard Molar Entropy (S)
(Pt 5)
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. Standard Molar Entropy and the Third Law
of Thermodynamics
Recall that the entropy of a
substance is related to the number
of microstates.
We discussed thermal and
positional entropy.

3. Molecular Motion and
Standard Molar Entropy
Molecular motion contributes to the
entropy of a given substance.
Molecular motions can include
Translations
Vibrations
Rotations

4. Molecular Motion and
Standard Molar Entropy
The more possible molecular
motions, the higher the entropy.
The higher the temperature, the
more molecular motions.
So what would happen if we
reduced the temperature to 0 K?

5. The Third Law of Thermodynamics and
Standard Molar Entropy (S)
The entropy of a substance at 0 K
is zero.
S = 0 at 0 K
This is a perfectly ordered crystal
with only one microstate.

6. The Third Law and
Standard Molar Entropy (S)
The Third Law means that
absolute entropies for a
given substance can be
calculated.

7. Standard Molar Entropy (S)
The Standard Molar Entropy (S)
is the entropy of 1 mol of a
substance under standard
conditions (at 25 C or 298 K)
These values are tabulated in a
table with 𝑆0
values.

8. Examples of Standard Molar Entropy
Values (S) in a Table
An example is given below.
Here is the column that provides S values (in J/mol K).

9. The Second Law of Thermodynamics
The total entropy change of the
universe for any spontaneous
process is positive
+ S
∆𝐒 = 𝐒 𝐟 − 𝐒𝐢
Where Si and Sf are the entropy of the initial and
final states, respectively.

10. Calculating ∆S0
Using Standard
Molar Entropies (S)
We can calculate ∆S0
from these tabulated
standard molar entropies.
For the reaction
aA + bB  cC + dD
∆𝐒 𝟎
= 𝐜 𝐒 𝟎
𝐂 + 𝐝 𝐒 𝟎
𝐃 − 𝐚 𝐒 𝟎
𝐀 + 𝐛 𝐒 𝟎
𝐁

11. Calculating ∆S0
Using Standard
Molar Entropies (S)
Another way to write the equation:
For the reaction aA + bB  cC + dD
∆𝐒 𝟎
= 𝐜 𝐒 𝟎
𝐂 + 𝐝 𝐒 𝟎
𝐃 − 𝐚 𝐒 𝟎
𝐀 + 𝐛 𝐒 𝟎
𝐁
∆𝐒 𝟎
= 𝐧 𝐩 𝐒 𝟎
𝐩𝐫𝐨𝐝𝐮𝐜𝐭𝐬 − 𝐧 𝐫 𝐒 𝟎
𝐫𝐞𝐚𝐜𝐭𝐚𝐧𝐭𝐬

12. Next up…
Gibbs Free Energy and
Spontaneous Reactions (Pt 6)