AP Chemistry 9.2: Gibbs Free Energy, Temperature, and Coupled Reactions

Unit 6 focused heavily on:

• ΔH
• You learned:
• ΔH < 0 → exothermic
• ΔH > 0 → endothermic

But whether a reaction releases heat does not by itself determine whether the reaction is thermodynamically favored.

Both enthalpy and entropy matter.

• This leads to Gibbs free energy.

Gibbs free energy is represented by:

• G
• The change in Gibbs free energy is:
• ΔG
• Under standard conditions:
• ΔG°

The key equation is:

• ΔG° = ΔH° − TΔS°
• where:
• ΔG° = standard Gibbs free-energy change
• ΔH° = standard enthalpy change
• T = Kelvin temperature
• ΔS° = standard entropy change

This is a very common source of mistakes.

• ΔH is often given in:
• kJ/mol
• while ΔS is often:
• J/(mol·K)
• Before using:
• ΔG = ΔH − TΔS
• make sure ΔH and TΔS use matching energy units.
• For example:
• 150 J/(mol·K) = 0.150 kJ/(mol·K)

The forward process is:

• thermodynamically favored

The forward process is:

• thermodynamically unfavored
• The reverse process is favored.

The system is at:

• equilibrium

Start with:

• ΔG = ΔH − TΔS
• Whether ΔG is negative depends on the signs of ΔH and ΔS.

ΔH < 0

• ΔS > 0
• Both terms favor negative ΔG.
• Therefore the reaction is:
• favored at all temperatures

ΔH > 0

• ΔS < 0
• Both terms work against the reaction.
• Therefore the reaction is:
• unfavored at all temperatures

ΔH < 0

• ΔS < 0
• The enthalpy term favors the reaction.
• The entropy term works against it because:
• −T(negative ΔS) = positive contribution
• Therefore the reaction tends to be:
• favored at low temperature
• and
• unfavored at high temperature

ΔH > 0

• ΔS > 0
• The enthalpy term works against the reaction.
• The entropy term favors it because:
• −T(positive ΔS) = negative contribution
• Therefore the reaction tends to be:
• unfavored at low temperature
• and
• favored at high temperature

This table is worth knowing.

At the temperature where favorability changes:

• ΔG = 0
• So:
• 0 = ΔH − TΔS
• Rearrange:
• T = ΔH / ΔS
• Use matching units.
• The resulting temperature should normally be in Kelvin.

Standard free energy of formation is represented by:

• ΔG°f
• It is the free-energy change associated with forming 1 mol of a substance from its elements in their standard states.
• For an element in its standard state:
• ΔG°f = 0
• Examples include substances such as:
• O₂(g)
• N₂(g)
• H₂(g)
• when those are the element's standard forms.

Use:

• ΔG°rxn = ΣnΔG°f(products) − ΣnΔG°f(reactants)
• This works like the ΔH°f equation from Unit 6.

Always include coefficients.

Sometimes a thermodynamically unfavored reaction can occur by being connected to a strongly favored reaction.

• This is called:
• reaction coupling
• Suppose:
• Reaction 1:
• ΔG° = +20 kJ
• Reaction 2:
• ΔG° = −50 kJ
• If they are coupled properly:
• ΔG°overall = +20 + (−50)
• = −30 kJ
• The overall process is thermodynamically favored.

Free energy behaves like enthalpy in this way.

If reactions are added:

• add ΔG values

If a reaction is reversed:

• change the sign of ΔG

If the equation is multiplied:

• multiply ΔG by the same factor