# Chem 2 - Third Law of Thermodynamics and Standard Molar Entropy V

Lumen Learning
2 de Jun de 2016
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### Chem 2 - Third Law of Thermodynamics and Standard Molar Entropy V

• 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)