Topic 9.3 Notes – Gibbs Free Energy and Thermodynamic Favorability

1. Gibbs Free Energy and What ΔG° Means

Gibbs free energy (G) combines heat flow (enthalpy, ΔH°) and disorder (entropy, ΔS°) into one value that predicts thermodynamic favorability.

When everything is in its standard state

• Pure solids or liquids
• 1.0 M solutions
• Gases at 1.0 atm (or 1 bar)

the change is written as ΔG° and usually reported in kJ/mol.

Here’s how to read the sign:

• ΔG° < 0 → thermodynamically favored (historically called spontaneous)
• ΔG° > 0 → thermodynamically unfavored
• ΔG° = 0 → system is at equilibrium

“Thermodynamically favored” means the process lowers free energy under those conditions. It does not mean:

• Fast
• Explosive
• No activation energy

Rusting iron is favored. It just takes time.

2. The Equation That Connects Enthalpy and Entropy

The relationship tying this together is:

$$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$

• $\Delta H^\circ$ in kJ/mol
• $\Delta S^\circ$ in kJ/mol}\\cdot\\text{K (or convert from J!)
• $T$ in Kelvin

Temperature controls how powerful the entropy term is. As $T$ increases, the size of $T\Delta S^\circ$ increases.

Sign logic becomes mechanical once you trust the equation:

• ΔH° < 0 (exothermic) helps make ΔG° negative
• ΔS° > 0 also helps, because subtracting a positive makes ΔG° smaller
• High temperature magnifies the entropy effect

Quick example:

A reaction has
$\Delta H^\circ = 40 \text{ kJ/mol}$
$\Delta S^\circ = 0.150 \text{ kJ/mol}\\cdot\\text{K}$

At 300 K:

$$\Delta G^\circ = 40 - (300)(0.150) = 40 - 45 = -5 \text{ kJ/mol}$$

Even though it’s endothermic, it’s favored at this temperature because entropy wins.

Always convert °C to K. That mistake shows up constantly on quizzes.

3. All Possible ΔH° and ΔS° Combinations

There are only four sign combinations. Once you know them, you can predict temperature effects instantly.

Two cases require no math:

• ΔH° < 0 and ΔS° > 0 → always favored
• ΔH° > 0 and ΔS° < 0 → never favored

The other two depend on temperature.

Freezing of water
$\ce{H2O(l) -> H2O(s)}$
- ΔH° < 0
- ΔS° < 0
Favored only at low temperature.

Dissolving sodium nitrate
$\ce{NaNO3(s) -> Na+(aq) + NO3-(aq)}$
- ΔH° > 0 (solution gets cold)
- ΔS° > 0
Favored at higher temperature because disorder increases.

These are classic AP examples.

4. Calculating ΔG° from ΔGf° Values

Sometimes you aren’t given ΔH° and ΔS°. Instead, you’ll use standard Gibbs free energies of formation.

$$\Delta G^\circ_{\text{reaction}} = \sum \Delta G_f^\circ(\text{products}) - \sum \Delta G_f^\circ(\text{reactants})$$

Rules:

• Multiply each value by its coefficient.
• Subtract reactants from products.
• Elements in their standard states have $\Delta G_f^\circ = 0$.
  Example: $\ce{O2(g)}$, $\ce{N2(g)}$, metals in solid form.

Example:

$$\ce{2SO2(g) + O2(g) -> 2SO3(g)}$$

If given formation values, you would multiply each by its coefficient, sum products, subtract reactants, and check the sign.

This setup is identical in structure to how you calculate ΔH° from ΔHf°.

5. Enthalpy-Driven vs Entropy-Driven Processes

Sometimes one term clearly dominates.

Enthalpy-driven

• Large negative ΔH°
• Even if ΔS° is negative, heat release makes ΔG° negative
• Example: freezing below 0°C

Entropy-driven

• ΔH° may be positive
• Large positive ΔS° makes ΔG° negative
• Example: dissolution of some ionic solids

On free-response questions, they often want you to explicitly state which factor drives the favorability and why temperature matters.