Gibbs Free Energy Calculator
ChemistryCalculate the Gibbs free energy change ΔG from enthalpy ΔH, entropy ΔS, and temperature T using ΔG = ΔH − TΔS to check for reaction spontaneity.
Reviewed by the thecalcu.com team · Last updated July 29, 2026
Gibbs Free Energy (ΔG°)
What is a Gibbs Free Energy?
The Gibbs Free Energy Calculator computes the standard Gibbs free energy change (ΔG°) from the enthalpy change (ΔH°), entropy change (ΔS°), and temperature (T) using the fundamental thermodynamic equation ΔG° = ΔH° − TΔS°. It also derives the equilibrium constant Kc from ΔG° = −RT ln(Kc) and classifies the reaction as spontaneous, non-spontaneous, or at equilibrium.
Gibbs free energy is the master thermodynamic criterion for chemical spontaneity. While enthalpy and entropy each capture part of the thermodynamic picture, energy release vs. disorder, ΔG combines both into a single number that directly answers whether a reaction will proceed. A negative ΔG at a given temperature means the reaction is spontaneous at those conditions; a positive ΔG means it is not. At ΔG = 0, the system is at equilibrium.
The temperature in the formula ΔG = ΔH − TΔS acts as a "weight" on the entropy term. This is why some reactions are spontaneous only above certain temperatures (entropy-driven at high T) and others only below certain temperatures (enthalpy-driven at low T). This temperature crossover behaviour is calculated directly by setting ΔG = 0 and solving for T = ΔH/ΔS.
Together with the Entropy Calculator (which extracts ΔS° from ΔH° and ΔG°) and the Equilibrium Constant Calculator (which connects Kc to ΔG°), this tool completes the core chemical thermodynamics toolkit.
Why Use a Gibbs Free Energy Calculator?
The most common error in ΔG calculations is a unit mismatch: ΔH° is typically tabulated in kJ/mol while ΔS° is tabulated in J/mol·K. Using both in the formula without converting ΔS° to kJ/mol·K causes a 1,000-fold error in the TΔS° term. This calculator handles the conversion automatically.
For students working through NCERT thermodynamics or JEE preparation, the ability to compute ΔG° and immediately see its equilibrium constant and spontaneity classification provides a complete answer to the standard exam question format: "calculate ΔG° and determine whether the reaction is spontaneous."
For process development, computing ΔG° at multiple temperatures identifies the crossover point, the temperature below or above which a reaction becomes non-spontaneous, critical for defining the operating temperature window in reactor design.
Who Should Use This Calculator?
Class 11–12 students covering Chemical Thermodynamics in NCERT. ΔG = ΔH − TΔS and the spontaneity criteria are core examination topics in CBSE, JEE Main, and JEE Advanced.
Undergraduate physical chemistry students at BSc and B.Pharm level computing ΔG° as part of thermochemistry coursework, equilibrium analysis, and electrode potential calculations (ΔG° = −nFE° for electrochemical cells).
Pharmaceutical scientists and formulation chemists evaluating thermodynamic stability of drug crystalline forms, solubility drivers, and protein-ligand binding energetics (ΔG_binding = −RT ln Kd).
Chemical engineers and process chemists determining feasible operating temperature ranges for reactions based on the ΔG crossover temperature, and computing equilibrium conversion at planned reactor temperatures.
Biochemists studying metabolic pathway thermodynamics, where ΔG° (and ΔG under cellular conditions) determines the direction of metabolic flux and coupled reactions.
What Insights Does the Gibbs Free Energy Calculator Give You?
ΔG° (kJ/mol) is the primary output, the standard Gibbs free energy change. A negative value means spontaneous under standard conditions; positive means non-spontaneous. The magnitude tells you how far from equilibrium the standard mixture is: a ΔG° of −100 kJ/mol means a far stronger thermodynamic drive toward products than a ΔG° of −5 kJ/mol.
Equilibrium Constant (Kc) is derived from ΔG° = −RT ln(Kc), converting the thermodynamic driving force directly into the equilibrium position. This cross-links ΔG° to measurable experimental quantities (equilibrium concentrations). Compare this Kc with values computed from concentration measurements using the Equilibrium Constant Calculator as a consistency check.
Spontaneity is the qualitative verdict: Spontaneous (ΔG < 0), Non-spontaneous (ΔG > 0), or At equilibrium (ΔG = 0). For the four possible combinations of signs of ΔH and ΔS, this output shows which temperature regime makes the reaction spontaneous.
How to use this Gibbs Free Energy calculator
- Find the standard enthalpy change ΔH° from a thermochemical table (NIST, NCERT appendix, or CRC Handbook). Enter it in kJ/mol in the Enthalpy Change (ΔH°) field, negative for exothermic.
- Find the standard entropy change ΔS° from the same source (or calculate it as the difference of standard molar entropies). Enter it in J/mol·K in the Entropy Change (ΔS°) field, note the units are J/mol·K, not kJ/mol·K.
- Enter the temperature in Kelvin in the Temperature field. For standard conditions, use 298 K; for a different temperature, note that ΔH° and ΔS° are assumed temperature-independent in this calculation.
- Read ΔG° (kJ/mol) and the Spontaneity verdict.
- Use the Kc output to connect to equilibrium analysis, compare to Kc values measured at the same temperature.
Show formula & methodology ↓Show less ↑
Formula & Methodology
Gibbs–Helmholtz equation:ΔG° = ΔH° − T × ΔS° (ΔH° in kJ/mol, T in K, ΔS° converted from J/mol·K to kJ/mol·K by ÷ 1000)Equilibrium constant from ΔG°:ΔG° = −RT ln(Kc) Kc = exp(−ΔG° × 1000 / (R × T)) [R = 8.314 J/(mol·K)]Crossover temperature (ΔG = 0):T_cross = ΔH° / ΔS° (ΔH° in J/mol, ΔS° in J/mol·K)Worked example, Haber process for ammonia: N₂(g) + 3 H₂(g) → 2 NH₃(g): ΔH° = −92.4 kJ/mol, ΔS° = −198.3 J/mol·K At T = 298 K:ΔG° = −92.4 − (298 × −0.1983) = −92.4 + 59.1 = −33.3 kJ/mol (spontaneous at 298 K) Kc = exp(33,300 / (8.314 × 298)) = exp(13.44) = 6.8 × 10⁵At T = 773 K (industrial operating temperature):ΔG° = −92.4 − (773 × −0.1983) = −92.4 + 153.3 = +60.9 kJ/mol (non-spontaneous at 773 K under standard conditions) Crossover: T = 92,400 / 198.3 = 466 KAbove 466 K, the Haber process is thermodynamically non-spontaneous under standard conditions, explaining why it is run at high pressure (to shift equilibrium) and uses a catalyst (to achieve acceptable rate at the lower temperatures that are thermodynamically favourable).
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