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General Chemistry II · Chemical Thermodynamics

Gibbs Free Energy

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On this page 9 sections
  1. In 30 seconds
  2. Why this matters
  3. The college version
  4. Eli explains
  5. Worked example
  6. Key takeaway
  7. Quick check
  8. Study tools
  9. Sources & references

In 30 seconds

Gibbs free energy (G) is the thermodynamic function that packages the Second Law into a system-only quantity. Its change is ΔG = ΔH − TΔS, and at constant temperature and pressure, ΔG < 0 means the process is spontaneous, ΔG > 0 means it is nonspontaneous (the reverse is spontaneous), and ΔG = 0 means the system is at equilibrium. Because ΔG = −T·ΔS_universe, checking ΔG is exactly equivalent to checking the Second Law — but far more convenient, since ΔH and ΔS refer only to the system.

Why this matters

Gibbs free energy is the single most used spontaneity test in chemistry and biology. It tells chemists whether a reaction will run, biochemists whether a metabolic step is favorable, and engineers how much useful (non-PV) work a process can deliver at constant T and P. ATP hydrolysis, combustion, and battery discharge are all quantified through ΔG.

The college version

Core Concept

Gibbs free energy (G) is the thermodynamic function that packages the Second Law into a system-only quantity. Its change is ΔG = ΔH − TΔS, and at constant temperature and pressure, ΔG < 0 means the process is spontaneous, ΔG > 0 means it is nonspontaneous (the reverse is spontaneous), and ΔG = 0 means the system is at equilibrium. Because ΔG = −T·ΔS_universe, checking ΔG is exactly equivalent to checking the Second Law — but far more convenient, since ΔH and ΔS refer only to the system.

Key Ideas

  • One number, two drivers. ΔG balances the enthalpy tendency (ΔH) against the entropy tendency (TΔS).
  • TΔS scales with temperature. The higher T, the more the entropy term matters (and its units are energy, because K × J/(mol·K) = J/mol).
  • Sign convention is everything. Negative ⇒ spontaneous/forward-favored; positive ⇒ reverse-favored; zero ⇒ equilibrium.
  • Units discipline. ΔH in kJ/mol and ΔS in J/(mol·K) must be reconciled (convert ΔS to kJ/(mol·K) or ΔH to J/mol) before combining.
  • Kelvin only. T must be absolute temperature.

Equations and Variables

SymbolMeaningCommon units
ΔGGibbs free-energy changekJ/mol
ΔHEnthalpy changekJ/mol
ΔSEntropy changeJ/(mol·K)
TAbsolute temperatureK

ΔG = ΔH − TΔS

How It Works

  1. Obtain ΔH and ΔS for the process (from tables or measurement) at the temperature of interest.
  2. Put ΔS and ΔH in compatible units (e.g., convert ΔS to kJ/(mol·K)).
  3. Multiply T × ΔS.
  4. Subtract: ΔG = ΔH − TΔS.
  5. Read the sign: negative → spontaneous; positive → nonspontaneous; zero → equilibrium.

Worked Example

Calculate ΔG° at 25 °C for H₂O(l) → H₂O(g). Is vaporization spontaneous at this temperature?

Data: ΔH° = +44.0 kJ/mol, ΔS° = +118.9 J/(mol·K) = +0.1189 kJ/(mol·K).

  • T = 25 + 273 = 298 K
  • TΔS° = 298 K × 0.1189 kJ/(mol·K) = 35.4 kJ/mol
  • ΔG° = ΔH° − TΔS° = +44.0 − 35.4 = +8.6 kJ/mol

ΔG° > 0, so liquid water does not spontaneously vaporize at 25 °C (the reverse, condensation, is favored). At 100 °C (373 K), TΔS° = 373 × 0.1189 = 44.4 kJ/mol, giving ΔG° ≈ 44.0 − 44.4 ≈ 0 — equilibrium, the boiling point.

How it works

  1. Obtain ΔH and ΔS for the process (from tables or measurement) at the temperature of interest.
  2. Put ΔS and ΔH in compatible units (e.g., convert ΔS to kJ/(mol·K)).
  3. Multiply T × ΔS.
  4. Subtract: ΔG = ΔH − TΔS.
  5. Read the sign: negative → spontaneous; positive → nonspontaneous; zero → equilibrium.

Common confusions

  • "ΔG < 0 means the reaction is fast." — Wrong. ΔG says nothing about rate; many ΔG < 0 reactions are kinetically slow.
  • "Use °C in ΔG = ΔH − TΔS." — Wrong. T must be in Kelvin.
  • "A negative ΔH always makes ΔG negative." — Wrong. If ΔS is negative and T is large, the −TΔS term can outweigh ΔH and make ΔG positive.
  • "ΔG = 0 means the reaction stopped." — Wrong. It means dynamic equilibrium: forward and reverse rates are equal.
  • "ΔG and ΔH have the same units as ΔS." — Wrong. ΔH and ΔG are kJ/mol; ΔS is J/(mol·K) — convert before combining.

Quick review

  • ΔG = ΔH − TΔS; T in Kelvin.
  • ΔG < 0 spontaneous, > 0 nonspontaneous, = 0 equilibrium.
  • ΔG = −T·ΔS_universe.
  • Entropy term scales with temperature.
  • Convert ΔS to kJ/(mol·K) before combining with ΔH.
Eli, the EliExplains learning guide

Eli explains

The same idea, in plain words

Explain it like I’m 10

Picture a seesaw with two kids. On one side sits "heat" (ΔH) — a reaction likes to give off heat, which pushes toward happening. On the other side sits "spread-out" (ΔS) — a reaction likes to spread things out, which also pushes toward happening. Temperature is the weight of the "spread-out" kid: crank up the heat and the entropy kid gets heavier and can tip the seesaw by himself. If the seesaw tips "forward," the reaction happens on its own; that's ΔG being negative. (The limit: ΔG is really about available energy to do work, not a literal seesaw, but "two competing pushes, one of which gets stronger with temperature" is the correct picture.)

Worked example

Worked Example

Calculate ΔG° at 25 °C for H₂O(l) → H₂O(g). Is vaporization spontaneous at this temperature?

Data: ΔH° = +44.0 kJ/mol, ΔS° = +118.9 J/(mol·K) = +0.1189 kJ/(mol·K).

  • T = 25 + 273 = 298 K
  • TΔS° = 298 K × 0.1189 kJ/(mol·K) = 35.4 kJ/mol
  • ΔG° = ΔH° − TΔS° = +44.0 − 35.4 = +8.6 kJ/mol

ΔG° > 0, so liquid water does not spontaneously vaporize at 25 °C (the reverse, condensation, is favored). At 100 °C (373 K), TΔS° = 373 × 0.1189 = 44.4 kJ/mol, giving ΔG° ≈ 44.0 − 44.4 ≈ 0 — equilibrium, the boiling point.

Key takeaways

  • ### High-Yield Facts
  • ΔG = ΔH − TΔS (T in Kelvin).
  • ΔG < 0 ⇒ spontaneous; ΔG > 0 ⇒ nonspontaneous; ΔG = 0 ⇒ equilibrium.
  • ΔG = −T·ΔS_universe (so ΔG < 0 ⇔ ΔS_universe > 0).
  • TΔS has energy units; a large T amplifies the entropy contribution.
  • Convert ΔS (J/mol·K) to kJ/mol·K before subtracting from ΔH (kJ/mol).
  • At constant T and P, ΔG is the maximum non-expansion work available.

Quick check

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Question 1 of 1

When a reaction is at equilibrium, which of the following is true?

Choose an answer, then check it.

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Study tools & related lessonsYou’ll learn to · Related

You’ll learn to

  • Define Gibbs free energy and explain what ΔG = ΔH − TΔS means.
  • Use the sign of ΔG to classify a process as spontaneous, nonspontaneous, or at equilibrium.
  • Calculate ΔG from given ΔH, ΔS, and temperature (in Kelvin).
  • Explain how ΔG combines the enthalpy and entropy criteria of spontaneity.

Sources & references

  1. OpenStax, *Chemistry 2e*, Ch. 16.4 "Free Energy." https://openstax.org/books/chemistry-2e/pages/16-4-free-energy
  2. OpenStax, *Chemistry 2e*, Ch. 16.2 "Entropy." https://openstax.org/books/chemistry-2e/pages/16-2-entropy
  3. NIST Chemistry WebBook. https://webbook.nist.gov/chemistry/

This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.

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