Gibbs Free Energy: Predict Thermodynamic Favorability
Gibbs free energy combines enthalpy, entropy, and temperature into one value that helps predict whether a process is thermodynamically favorable as written.
The two common relationships are Delta G = Delta H - T Delta S and Delta G standard = -RT ln K.
This makes Gibbs free energy a bridge between heat, disorder, temperature, and equilibrium. A reaction can release heat but still be unfavorable if the entropy term is strongly negative at the temperature of interest. Another reaction can absorb heat but become favorable at high temperature if the entropy increase is large enough.
How to use this calculator
- Choose the thermodynamic relationship you need.
- Enter enthalpy, entropy, temperature, Delta G, or K as required.
- Use kelvin for absolute temperature, or enter Celsius and let the tool convert.
- Review the sign of Delta G and the calculated equilibrium constant.
Interpretation
Negative Delta G is thermodynamically favorable as written, positive Delta G is non-favorable as written, and Delta G near zero indicates equilibrium under the stated conditions.
Standard versus nonstandard conditions
Delta G standard refers to a defined standard-state reference. Real reaction mixtures may not be at standard conditions, so the actual Delta G can differ depending on concentrations, pressures, and reaction quotient. The equation with K describes the standard free-energy difference and equilibrium position, not necessarily the instantaneous driving force in every mixture.
Thermodynamics is not speed
A negative Delta G does not mean a reaction happens quickly. Diamond converting to graphite is thermodynamically favorable under ordinary conditions, but it is extremely slow because the kinetic barrier is large. Catalysts, enzymes, heat, and reaction pathways affect rate; Gibbs free energy describes favorability and equilibrium direction.
Worked example
If Delta H is -100 kJ/mol and Delta S is -200 J/mol K at 298.15 K, then Delta G = -100 - 298.15 x (-0.200) = about -40.37 kJ/mol. The negative value indicates the process is favorable as written at that temperature. The corresponding equilibrium constant is large, meaning products are favored under standard-state assumptions.
