**Gibbs Free Energy, G**

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

**ΔG = ΔH - TΔS**

__Note__: all functions are from the point of view of the "system" (ΔG_{sys}, ΔH_{sys}, ΔS_{sys})

➞ * if ΔG < 0*, the process is spontaneous (at constant T and P)

➞ * if ΔG > 0*, the process is

*spontaneous (at constant T and P)*

__not__----------

__To calculate ΔG__, you must compare and consider the ΔH and ΔS values, and you must consider the __kelvin__ temperature (T).

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

** Br _{2}(l) → Br_{2}(g)** ΔH° = 31.0 kJ/mol

ΔS° = 93.0 J/K

^{.}mol

Also, what's the normal boiling point of Br_{2}(l)?

_________**answer:**

➞ ΔG must be negative (-) for the process to be spontaneous

➞ We can find the __threshold temperature (T)__ by setting * ΔG° = 0*...

So, the **threshold temperature is 333 K**.

Above 333 K, the forward reaction is spontaneous because at higher temperatures, the "TΔS" term is dominant in:

**ΔG° = ΔH° - TΔS°**

When T = 333 K, the free energy change (ΔG) = 0 and the __liquid__ and __gas__ phases of Br_{2} __ coexist__ at 1 atm.

Therefore, the normal boiling point (bp) is also 333 K.

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**Entropy Changes (ΔS) in Chemical Reactions**

TIP - Focus on the amount of **gas** molecules or atoms.

__Gases__ - highest level of * dis*order, so highest amount of entropy (ΔS)

__Solids__ - lowest level of * dis*order, so lowest amount of entropy (ΔS)

----------

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

**a. 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, so __entropy decreases__, ΔS < 0

**b. 4NH _{3}(g) + 5O_{2}(g) → 4NO(g) + 6H_{2}O(g)**

➞ 9 gas molecules "goes to" 10 gas molecules, and more gas molecules means more possible configurations.

➞ more gas molecules = more disorder, so __entropy increases__, ΔS > 0

==========

**Standard Entropy Values (S°)**

__Standard Entropy Values (S°)__ are listed in a lengthy table or appendix in the back of your textbook.

Or you can find them here: List of Entropy Values

- Notice that indeed S° increases for a substance in going from s **→** l **→** g

- S° has units of " J/K^{.}mol "

- To __calculate__ ΔS° for a reaction:

I've boxed up the 2nd version of the entropy equation above for a reason. It's easier to remember.

So use that one!

----------

*ex:* Calculate ΔS° for this reaction:

** Al _{2}O_{3}(s) + 3H_{2}(g) → 2Al(s) + 3H_{2}O(g)**

_________

**answer:**➞ ΔS°_{rxn} = [sum of product entropy values] - [sum of reactant entropy values]

➞ don't forget to take # of moles of each substance into account

----------

* QUESTION* - In the above reaction, 3 gas molecules "goes to" 3 gas molecules.

So how come the ΔS° value is large (179 J/K) instead of nearly 0 J/K ?...

➞ * ANSWER* - the large ΔS° value is because

**H**is a

_{2}O__more complex__molecule than

**H**, and thus has a

_{2}__higher__standard entropy value (S°).

**Why?** Because H_{2}O's molecular structure, being more sophisticated, has a greater number of vibrations/rotations/etc. which increases its randomness and disorder (entropy).

==========

**Free Energy (G) and Chemical Reactions**

**The standard free energy change (ΔG°)** is the change in free energy that occurs when the __reactants__ in their standard states are converted to __products__ in their standard states.

Knowing ΔG° is useful because the more negative its value is, the more likely it is that a reaction will occur (*or* "shift right" to reach equilibrium).

----------

**There are 3 Ways to Calculate ΔG°**

1. Using equation: ΔG° = ΔH° - TΔS° (at constant T, P)

➞ where both ΔH° and ΔS° are calculated using a Table of Thermodynamic Values.

**2. Using procedures similar to Hess' Law (see "Section 6")**

*ex:* From the data provided below, calculate ΔG° for this reaction:

** 2 CO(g) + O _{2}(g) → 2 CO_{2}(g)**

Data:

_________**answer:**

➞ the only way to get " 2 CO(g) " as a reactant is to flip the first reaction. So...

3. Using __standard free energies of formation__ (ΔG_{f}°)

**ΔG _{f}° ** = change in free energy that accompanies the formation of 1 mole of substance from its elements, with all of the reactants and products in their standard states.

You can find them here: List of Free Energy Values, G_{f}°

➞ value of ΔG_{f}° = 0 for elements in their standard states.

➞ To __calculate__ ΔG° for a reaction:

Look familiar? Use the boxed up version.

It's easier to remember...

----------

*ex:* Is the following reaction

__spontaneous__under standard conditions?

**C _{2}H_{4}(g) + H_{2}O(l) → C_{2}H_{5}OH(l)**

_________

**answer:**➞ we need to find G_{f}° values from the back of your textbook, or via the link above...

==========

Next up in **SECTION 16 - Spontaneity, Entropy, and Free Energy**, we'll continue with this discussion...

And we'll talk about the mathematical relationship between free energy (ΔG) and the equilibrium constant (K).