S16E3 - Spontaneous Reactions and the Gibbs Free Energy Equation

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 not spontaneous (at constant T and P)

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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)?
_________
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 Br₂ 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 disorder, so highest amount of entropy (ΔS)

Solids - lowest level of disorder, so lowest amount of entropy (ΔS)

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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

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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!

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ex:  Calculate ΔS° for this reaction:

        Al₂O₃(s)  +  3H₂(g)  →  2Al(s)  +  3H₂O(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

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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).

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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).

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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)

   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...

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ex:  Is the following reaction spontaneous under standard conditions?

        C₂H₄(g)  +  H₂O(l)  →  C₂H₅OH(l)
_________
answer:

➞  we need to find G_f° values from the back of your textbook, or via the link above...

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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).