Entropy & Gibbs Free Energy Calculator

Calculate ΔS, ΔG, Boltzmann entropy and more — six thermodynamic laws in one tool.

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Entropy sits at the heart of chemistry, physics and engineering — it governs whether reactions happen spontaneously, how much useful work an engine can extract, and why mixed substances resist separation. This calculator puts six foundational thermodynamic laws in one place, each rearranged so you can solve for any unknown rather than only the textbook output variable.

What this calculator covers

Six modes are available from the drop-down:

  1. Gibbs free energy — ΔG = ΔH − TΔS. Solve for ΔG, ΔH, T or ΔS and read off whether the process is spontaneous.
  2. Boltzmann entropy — S = k_B · ln W. Enter the number of microstates W (or moles with a per-mole degeneracy) to get the absolute entropy in joules per kelvin.
  3. Clausius inequality — ΔS = Q_rev / T. The classical definition: entropy gained equals reversible heat divided by absolute temperature.
  4. Entropy of mixing — ΔS_mix = −R Σ nᵢ ln xᵢ. Works for two- or three-component ideal systems; just enter the moles of each component.
  5. Isothermal expansion — ΔS = nR · ln(V₂/V₁). Rearrangeable for V₁, V₂ or n.
  6. Van ‘t Hoff — derives ΔG° = −RT ln K first, then extracts ΔS° from the Gibbs relation using your supplied ΔH°.

Every mode shows the substituted formula as a copyable working block so you can paste it straight into a lab report or homework solution.

How it works

All six calculations run entirely client-side using the exact IUPAC-recommended constants:

  • R = 8.314 J/(mol·K) — universal gas constant
  • k_B = 1.380649 × 10⁻²³ J/K — Boltzmann constant (2019 SI exact value)
  • N_A = 6.02214076 × 10²³ mol⁻¹ — Avogadro constant

No data leaves your browser. Results update on every keystroke.

Worked example — Gibbs free energy of water formation

The formation of liquid water from hydrogen and oxygen at 298.15 K has approximate standard values of ΔH° = −285,830 J/mol and ΔS° = −163.2 J/(mol·K).

Substituting into ΔG = ΔH − TΔS:

ΔG = −285830 − (298.15 × −163.2) = −285830 + 48659 = −237171 J/mol

The large negative ΔG confirms the reaction is strongly spontaneous under standard conditions. The negative ΔS reflects that a gas (H₂, O₂) is consumed to form a liquid — fewer positional microstates, lower entropy.

Formula reference

ModeFormulaKey constants
GibbsΔG = ΔH − TΔS
BoltzmannS = k_B · ln Wk_B = 1.381 × 10⁻²³ J/K
ClausiusΔS = Q_rev / T
MixingΔS_mix = −R Σ nᵢ ln xᵢR = 8.314 J/(mol·K)
IsothermalΔS = nR · ln(V₂/V₁)R = 8.314 J/(mol·K)
Van ‘t HoffΔG° = −RT ln K then ΔS° = (ΔH° − ΔG°)/TR = 8.314 J/(mol·K)

All temperatures must be in kelvin (K). To convert from Celsius: T(K) = T(°C) + 273.15. Energies are in joules; divide by 1000 to convert to kJ.

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