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" (ΔGsys, ΔHsys, ΔSsys)
➞ if ΔG < 0, the process is spontaneous (at constant T and P)
➞ if ΔG > 0, the process is not spontaneous (at constant T and P)
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?
Br2(l) → Br2(g)
ΔH° = 31.0 kJ/mol
ΔS° = 93.0 J/K.mol
Also, what's the normal boiling point of Br2(l)?
➞ 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 Br2 coexist at 1 atm.
Therefore, the normal boiling point (bp) is also 333 K.
Entropy Changes (ΔS) in Chemical Reactions
TIP - Focus on the amount of gas molecules or atoms.
Gases - highest level of disorder, so highest amount of entropy (ΔS)
Solids - lowest level of disorder, so lowest amount of entropy (ΔS)
ex: Does entropy increase or decrease in the following reactions?
a. N2(g) + 3H2(g) → 2NH3(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. 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(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:
Al2O3(s) + 3H2(g) → 2Al(s) + 3H2O(g)
➞ 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 H2O is a more complex molecule than H2, and thus has a higher standard entropy value (S°).
Why? Because H2O'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) + O2(g) → 2 CO2(g)
➞ 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 (ΔGf°)
ΔGf° = 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, Gf°
➞ value of ΔGf° = 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?
C2H4(g) + H2O(l) → C2H5OH(l)
➞ we need to find Gf° 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).