AP Chemistry Gibbs Free Energy

Using ΔG = ΔH − TΔS, the crossover temperature, and the link to K, in four worked examples.

Gibbs free energy is the one number that settles whether a process is thermodynamically favored, and the exam asks for it in three forms: a calculation from ΔH and ΔS, a temperature at which the answer flips, and a conversion to or from the equilibrium constant. The Unit 9 review introduces all of them; this page works each through with numbers and flags the unit trap that costs the most points.

The Gibbs Free Energy Equation and What Its Sign Means

ΔG = ΔH − TΔS, with T in kelvins. A negative ΔG means the process is thermodynamically favored; a positive ΔG means it is not; zero means equilibrium. AP materials use the phrase “thermodynamically favored,” so use it in free-response answers. A favored process is not necessarily fast, because speed depends on the activation energy, covered in activation energy.

The signs of ΔH and ΔS decide how ΔG changes with temperature. Each graph shows ΔG against T for one combination, with the favored region (ΔG < 0) shaded green.

Worked Example 1: Calculating ΔG From ΔH and ΔS (the Haber Process)

For N₂(g) + 3 H₂(g) → 2 NH₃(g), ΔH° = −92.2 kJ and ΔS° = −198.7 J/K. Is the reaction thermodynamically favored at 298 K?

Match the units first: ΔS° = −198.7 J/K = −0.1987 kJ/K.
ΔG° = ΔH° − TΔS° = −92.2 kJ − (298 K)(−0.1987 kJ/K)
= −92.2 + 59.2 = −33.0 kJ.
ΔG° is negative, so the reaction is thermodynamically favored at 298 K.

Both ΔH and ΔS are negative here, so this is the “favored at low temperature” case: the favorable enthalpy wins until the −TΔS term grows large enough. That happens at T = ΔH/ΔS = (−92.2)/(−0.1987) = 464 K, above which the reaction is no longer favored at standard conditions.

Worked Example 2: Finding the Crossover Temperature for Melting Ice

Melting ice has ΔH = +6.01 kJ/mol and ΔS = +22.0 J/(mol·K). Find the temperature above which melting is thermodynamically favored.

Set ΔG = 0: T = ΔH/ΔS = 6010 J/mol ÷ 22.0 J/(mol·K) = 273 K, which is 0°C.
Check at 298 K: ΔG = 6010 − (298)(22.0) = −546 J/mol, negative, so melting is favored.
Check at 250 K: ΔG = 6010 − (250)(22.0) = +510 J/mol, positive, so melting is not favored (freezing is).

Both signs are positive, the “favored at high temperature” case. The calculation reproduces the freezing point from thermodynamic data, which is a good check that you set the problem up correctly. The energy change itself is the kind classified in endothermic and exothermic problems.

Worked Example 3: Converting ΔG° to the Equilibrium Constant K

Use ΔG° = −33.0 kJ for the Haber reaction at 298 K to find K.

ΔG° = −RT ln K, so ln K = −ΔG°/(RT).
Units: R = 8.314 J/(mol·K), so write ΔG° as −33,000 J.
ln K = 33,000 ÷ (8.314 × 298) = 33,000 ÷ 2478 = 13.3.
K = e^13.3 = 6 × 10⁵.

A negative ΔG° gives K > 1, as it must: products are favored at equilibrium at 298 K. The reverse works too, since a K less than 1 means a positive ΔG°. Setting up and using K is covered in equilibrium constants.

Worked Example 4: Matching Each Sign Combination to a Real Reaction

Predict the temperature behavior of each reaction from the signs of ΔH and ΔS.

2 H₂O₂(l) → 2 H₂O(l) + O₂(g): exothermic (ΔH < 0), and a gas is produced (ΔS > 0). Both terms favor products, so it is favored at all temperatures.
3 O₂(g) → 2 O₃(g): endothermic (ΔH > 0), and fewer gas molecules (ΔS < 0). Both terms oppose it, so it is never favored.
N₂O₄(g) → 2 NO₂(g): endothermic and more gas molecules (ΔH > 0, ΔS > 0), so it is favored only at high temperature.
Freezing of water: exothermic and more ordered (ΔH < 0, ΔS < 0), so it is favored only at low temperature.

The entropy signs come from counting gas molecules and changes of order, the reasoning in the Unit 9 review. Always check that the conclusion matches the graph for that sign pair.

Five Gibbs Free Energy Practice Questions With Answers

1. A reaction has ΔH = +40 kJ and ΔS = +100 J/K. Is it thermodynamically favored at 300 K, and at what temperature does it become favored?
   Show answer

   ΔG = 40 − (300)(0.100) = +10 kJ, so not favored at 300 K. Crossover T = 40 kJ ÷ 0.100 kJ/K = 400 K; it is favored above 400 K.
2. When is a reaction with ΔH < 0 and ΔS < 0 thermodynamically favored?
   Show answer

   At low temperatures, where the favorable ΔH outweighs the unfavorable −TΔS term.
3. A reaction has ΔG° = +5.0 kJ/mol at 298 K. Is K greater than or less than 1?
   Show answer

   Less than 1. A positive ΔG° means reactants are favored at equilibrium.
4. Calculate ΔG at 298 K for ΔH = −50 kJ and ΔS = −80 J/K.
   Show answer

   ΔG = −50 − (298)(−0.080) = −50 + 23.8 = −26 kJ, thermodynamically favored.
5. Why is ΔH alone not enough to decide whether a process is favored?
   Show answer

   Entropy matters too. An endothermic process (ΔH > 0) can still be favored if the entropy increase is large enough that TΔS exceeds ΔH.

Common Gibbs Free Energy Mistakes

• Mixing kJ and J. Convert ΔS to kJ/K, or ΔH to J, before combining them. This single slip changes the answer by 1000.
• Using Celsius for T. The temperature must be in kelvins.
• Reading “favored” as “fast.” ΔG says nothing about rate.
• Looking for a crossover temperature when the signs differ. It exists only when ΔH and ΔS have the same sign.
• Forgetting the minus sign in ΔG° = −RT ln K. A negative ΔG° must give K > 1; use that to catch an error.

Free energy also links to electrochemistry through ΔG° = −nFE°, developed in galvanic cells and cell potential. When you want to see how a heavy unit like this one moves an overall score, the AP Chem Score Calculator is a quick way to test it.

Related Resources

• AP Chem Score Calculator
• Unit 9 Review: Thermodynamics and Electrochemistry
• AP Chemistry Endothermic vs Exothermic
• AP Chemistry Equilibrium Constant Kc and Kp
• AP Chemistry Galvanic Cells and Cell Potential
• AP Chemistry Activation Energy
• AP Chemistry Study Guide

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