Unit 9 · Topic 9.3

9.3 Gibbs Free Energy and Thermodynamic Favorability

A process is thermodynamically favored when its standard Gibbs free energy change, ΔG°, is negative. You can find ΔG° from ΔG° = ΔH° − TΔS° or from tables of free energies of formation, and the signs of ΔH° and ΔS° tell you whether a process is favored at all temperatures, no temperatures, or only high or low ones.

Key terms

Gibbs free energy (ΔG°)
thermodynamically favored
ΔG° = ΔH° − TΔS°
standard free energy of formation
temperature dependence

What ΔG° tells you

The Gibbs free energy change combines the two things that drive change: releasing energy (negative ΔH) and spreading out (positive ΔS). The ° means all substances are in their standard states: pure solids and liquids, 1 M solutions, and gases at 1 atm (or 1 bar).

If ΔG° < 0, the process is thermodynamically favored: under standard conditions, products are favored. If ΔG° > 0, it is thermodynamically unfavored. You may see the old word 'spontaneous' for ΔG° < 0, but 'thermodynamically favored' is preferred, because a favored process isn't necessarily fast or sudden (topic 9.4).

Calculating ΔG°

From enthalpy and entropy: ΔG° = ΔH° − TΔS°, with T in kelvins. Watch your units. ΔH° is usually in kJ, but ΔS° is usually in J/K. Convert ΔS° to kJ/K (divide by 1000) before combining.

From tables: ΔG°rxn = ΣnΔG°f(products) − ΣnΔG°f(reactants). Just like ΔH°f, the ΔG°f of an element in its standard state is zero. This method gives ΔG° at the temperature of the table (usually 25 °C) only.

The four sign combinations

Because −TΔS° grows with temperature, the entropy term matters more at high temperature. The signs of ΔH° and ΔS° tell you when ΔG° is negative.

ΔH°	ΔS°	Favored (ΔG° < 0) at	Example
Negative	Positive	All temperatures	2H₂O₂(l) → 2H₂O(l) + O₂(g)
Positive	Negative	No temperature	2H₂O(l) + O₂(g) → 2H₂O₂(l)
Positive	Positive	High temperatures only	Melting ice; decomposing CaCO₃
Negative	Negative	Low temperatures only	Freezing water; N₂ + 3H₂ → 2NH₃

The crossover temperature

In the two mixed cases, the switch happens where ΔG° = 0, so T = ΔH° / ΔS° (with matching units). Above this temperature, the entropy term wins; below it, the enthalpy term wins.

For ice melting, ΔH° = +6.01 kJ/mol and ΔS° = +22.0 J/(mol·K), so T = 6010 ÷ 22.0 ≈ 273 K, which is 0 °C. Above 0 °C, melting is favored; below it, freezing is.

This is why both enthalpy and entropy matter: freezing water releases energy but decreases entropy, and dissolving sodium nitrate absorbs energy but increases entropy. Neither one alone decides.

Two assumptions to keep in mind

When you use ΔG° = ΔH° − TΔS° at temperatures other than 25 °C, you're assuming ΔH° and ΔS° don't change much with temperature. That's a reasonable estimate, and it's what AP questions expect.

ΔG° describes standard conditions. Under other conditions, whether a reaction runs forward depends on Q compared with K (topic 7.7), and the sign of ΔG° tells you whether K is bigger or smaller than 1 (topic 9.5).

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Is decomposing limestone favored?
For CaCO₃(s) → CaO(s) + CO₂(g), ΔH° = +178.3 kJ/mol and ΔS° = +160.7 J/(mol·K). (a) Is the reaction favored at 298 K? (b) Above what temperature does it become favored? Assume ΔH° and ΔS° don't change with temperature.
Show the solutionHide the solution
1. Step 1: (a) Convert ΔS° to kJ: 160.7 J/(mol·K) = 0.1607 kJ/(mol·K).
2. Step 2: ΔG° = 178.3 − (298)(0.1607) = 178.3 − 47.9 = +130.4 kJ/mol. Positive, so not favored at 298 K.
3. Step 3: (b) Both ΔH° and ΔS° are positive, so it becomes favored at high temperature. Set ΔG° = 0: T = ΔH° / ΔS° = 178.3 kJ/mol ÷ 0.1607 kJ/(mol·K) = 1110 K (about 837 °C).
4. Step 4: Above about 1110 K, the TΔS° term is larger than ΔH°, so ΔG° is negative. This is why limestone is heated strongly in kilns to make lime (CaO).
Answer: (a) ΔG° = +130.4 kJ/mol, not favored; (b) favored above about 1110 K
Example 2
ΔG° from free energies of formation
Given ΔG°f(NO₂, g) = +51.3 kJ/mol and ΔG°f(N₂O₄, g) = +97.9 kJ/mol, find ΔG° for 2NO₂(g) → N₂O₄(g) at 25 °C. Is it favored?
Show the solutionHide the solution
1. Step 1: ΔG° = ΣnΔG°f(products) − ΣnΔG°f(reactants) = 97.9 − 2(51.3).
2. Step 2: ΔG° = 97.9 − 102.6 = −4.7 kJ/mol.
3. Step 3: Negative, so the reaction is thermodynamically favored under standard conditions, though only slightly (it's close to zero; see topic 9.5).
Answer: ΔG° = −4.7 kJ/mol; favored
Example 3Calculator allowed
Trap: forgetting to convert units
For N₂(g) + 3H₂(g) → 2NH₃(g), ΔH° = −92.2 kJ/mol and ΔS° = −198.7 J/(mol·K). A student calculates ΔG° at 298 K as −92.2 − (298)(−198.7) = +59,000 and concludes it isn't favored. Find the correct ΔG°.
Show the solutionHide the solution
1. Step 1: The student mixed kJ and J. Convert ΔS°: −198.7 J/(mol·K) = −0.1987 kJ/(mol·K).
2. Step 2: ΔG° = −92.2 − (298)(−0.1987) = −92.2 + 59.2 = −33.0 kJ/mol.
3. Step 3: Negative, so the reaction is favored at 298 K. Both ΔH° and ΔS° are negative, so it is favored only below T = 92.2 ÷ 0.1987 ≈ 464 K.
Answer: ΔG° = −33.0 kJ/mol (favored at 298 K)

Common mistakes

Combining ΔH° in kJ with ΔS° in J without converting.
Using °C instead of K for T.
Thinking 'favored' means 'fast'. ΔG° says nothing about rate.
Assuming an endothermic process can never be favored. With a large enough positive ΔS°, it can, at high enough temperature.

On the exam

Expect to calculate ΔG° and state whether a process is thermodynamically favored, often with a follow-up asking at what temperature that changes. Show the unit conversion.
For sign-only questions, the table of four cases is enough; when ΔH° < 0 and ΔS° > 0 (or the reverse), you don't need to calculate anything.

Connected topics

Videos

How to Tell If a Process Is Thermodynamically-Favored - AP Chem Unit 9, Topic 3a
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Check yourself

4 questions on 9.3 Gibbs Free Energy and Thermodynamic Favorability. Pick an answer to see if you got it, and why.

Question 1 of 4

A reaction has ΔH° > 0 and ΔS° > 0. Assuming both stay constant, under which conditions is the reaction thermodynamically favored?

Question 2 of 4Calculator allowed

At 25 °C, ΔG°f for NO₂(g) is +51.3 kJ/mol and ΔG°f for N₂O₄(g) is +97.9 kJ/mol. What is ΔG° for 2NO₂(g) → N₂O₄(g) at 25 °C, and is the reaction thermodynamically favored?

Question 3 of 4

For H₂O(l) → H₂O(s), ΔH° = −6.01 kJ/mol and ΔS° = −22.0 J/(mol·K). Which of the following best explains why water freezes on its own only below about 273 K?

Question 4 of 4Calculator allowed

For CaCO₃(s) → CaO(s) + CO₂(g), ΔH° = +178.3 kJ/mol and ΔS° = +160.6 J/(mol·K). Assuming these values don't change with temperature, what is ΔG° at 1,200 K?

0 of 4 answered