**Second Law of Thermodynamics**

__2nd Law of Thermodynamics__ = the entropy (S) of the universe is constantly increasing. In any spontaneous process, the entropy of the universe increases.

The change in entropy of the universe is given the symbol, ΔS_{univ}

➞ if ΔS_{univ} > 0 , the process is spontaneous.

➞ if ΔS_{univ} < 0 , the __reverse__ process is spontaneous.

➞ if ΔS_{univ} = 0 , the process has no tendency to occur (it's at equilibrium).

➞ Sometimes ΔS_{sys} and ΔS_{surr} have opposite signs, and their magnitudes determine whether or not the overall process is spontaneous ( ΔS_{surr} > 0 ).

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**Gibbs Free Energy, G**

__Gibbs Free Energy, G__ = a thermodynamic function that not only is related to spontaneity, but also deals with the temperature dependence of spontaneity:

**ΔG = ΔH - TΔS**

➞ if ΔG < 0 , the process is spontaneous.

➞ if ΔG > 0 , the process is __not__ spontaneous.

➞ To calculate ΔG, you must compare and consider the ΔH and ΔS values, and you must consider the __Kelvin__ temperature (T).

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*ex:* At what temperatures is the following process spontaneous at 1atm?

** Br _{2}(l) ➞ Br_{2}(g)**

ΔH° = 31.0 kJ/mol, and

ΔS° = 93.0 J/K^{.}mol

Well, ΔG must be negative in **ΔG° = ΔH° - TΔS°** (° symbol = 25°C, 1atm conditions)...

... So we can find the __threshold temperature (T)__ by setting __ΔG° = 0__

*NOTE* - When T = 333K, the free-energy change (ΔG) = 0 and the

__liquid__and

__gas__phases of Br

_{2}

__coexist__at 1atm.

-- so, the normal boiling point (bp) = **333K**

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**What is Entropy?**

**Entropy Changes in Chemical Reactions**

-- It's most important to focus on the amount of * gas* molecules.

*ex:* Does entropy increase or decrease in the following reactions?

**i. N _{2}(g) + 3H_{2}(g) **➞

**2NH**

_{3}(g)- 4 gas molecules "goes to" 2 gas molecules, and __fewer__ gas molecules means __fewer__ possible configurations.

- fewer gas molecules = less disorder = **entropy decreases, ΔS < 0**

**ii. 4NH _{3}(g) + 5O_{2}(g) ** ➞

**4NO(g) + 6H**

_{2}O(g)- **entropy increases, ΔS > 0**, because 9 gas molecules becomes 10 gas molecules.

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**Standard Entropy Values**

__Standard Entropy Values (S°)__ are listed in a lengthy table or appendix in the back of your textbook. You can find one here as well: Table of Standard Entropies.

➞ notice that indeed S° increases for a substance in going from solid to a liquid to a gas.

➞ S° has units of " J / K^{.}mol "

➞ To * calculate ΔS°* for a reaction:

ΔS°_{rxn} = (sum of all entropies for products) - (sum of all entropies of reactants) , or...

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*ex:* Calculate ΔS° for this reaction:

** Al _{2}O_{3}(s) + 3H_{2}(g) **➞

**2Al(s) + 3H**

_{2}O(g)- First, we will need a Table of **Standard Entropy Values (S°)** from the back of our textbook, or use the link I gave you earlier in this article (above).

- This gives us...

** Question:** In the sample problem above, if 3 gas molecules (reactants) are converted to 3 gas molecules (products), how come the ΔS° value is large instead of nearly 0 J/K ?...

* Answer:* The large ΔS° value is because

*is*

**H**_{2}O__more complex__than

*, and thus has a higher standard entropy value (S°).*

**H**_{2}➞ * Why?* Because H

_{2}O's molecular structure, being more sophisticated, has a greater number of vibrations/rotations/etc. which increase its randomness and disorder (entropy).

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In my next post covering **SECTION 16 - Spontaneity, Entropy, and Free Energy**,

We'll talk extensively about Spontaneous Reactions and the Gibbs Free Energy Change, ΔG°