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Course homeModule II-13 · Recommended before: Before thermodynamics and spontaneity
Entropy, Gibbs energy and equilibrium
Thermodynamics combines signed energy changes. Entropy must be in compatible energy units and temperature must be in kelvin.
Learn the method
ΔG = ΔH − TΔS; ΔG° = −RT ln K
- Convert ΔS into kJ/K if ΔH is in kJ.
- Calculate TΔS.
- Compute ΔG = ΔH − TΔS.
- Interpret a negative ΔG as spontaneous under specified conditions.
- For standard equilibrium, use ΔG° = −RT ln K.
Common mistake: Using ΔS in J/K when ΔH is in kJ gives a thousand-fold unit error.
Worked example
Problem: At 300 K, ΔH = 20 kJ and ΔS = 0.100 kJ/K. Find ΔG.
Step-by-step solution: ΔG = 20 − (300)(0.100) = −10 kJ; favorable under the stated conditions.
Check the units and whether the answer is reasonable before continuing.
More applications of this mathematics
Try each extension before expanding its solution.
Free energy and equilibrium
At 298 K, if K = 10, ΔG° = −RT ln K ≈ −(8.314)(298)(2.303) = −5.71 kJ/mol.
Temperature threshold
With ΔH=+20 kJ and ΔS=+0.050 kJ/K, ΔG changes sign near T=ΔH/ΔS=400 K (assuming ΔH and ΔS constant).
Reaction entropy from tables
Compute ΔS°rxn=ΣνS°products−ΣνS°reactants with stoichiometric coefficients; if table units are J/(mol·K), convert before combining with ΔH in kJ.
Independent practice — five questions
Suggested mastery threshold: four out of five. Incorrect answers display a worked explanation after submission.
Supporting reference
OpenStax Chemistry 2e / NIST — reference for this subject. This lesson’s wording, examples and questions are original.
IntoChemistry original instructional content, 2026. Formative feedback is not an official course grade. Sources are linked for independent reading. Accessibility testing with your institution and assistive technology is still recommended.