Calculate the Gibbs free energy change ΔG = ΔH − TΔS from the enthalpy change, entropy change and temperature. See whether the process is spontaneous, at which temperature it switches (ΔG = 0), the equilibrium constant, and how ΔG changes with temperature.
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Enter your values to see the result.
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How to use the Gibbs Free Energy Calculator
Enter ΔH (negative for exothermic) and pick its unit.
Enter ΔS and its unit; most tables list entropy in J/(mol·K) while ΔH is in kJ/mol.
Enter the temperature in K, °C or °F.
Read ΔG, the spontaneity verdict and the crossover temperature.
Check the chart and table to see how ΔG changes at other temperatures.
The four ΔH / ΔS cases
ΔH
ΔS
ΔG = ΔH − TΔS
Spontaneous?
Example
− (exothermic)
+
Always negative
At all temperatures
Combustion of fuels
+ (endothermic)
−
Always positive
Never
6 CO₂ + 6 H₂O → glucose + 6 O₂ without energy input
−
−
Negative at low T
Below T = ΔH/ΔS
Water freezing, ammonia synthesis
+
+
Negative at high T
Above T = ΔH/ΔS
Ice melting, CaCO₃ decomposition
Formulas used
Quantity
Formula
Notes
Gibbs free energy change
ΔG = ΔH − TΔS
T in kelvin, ΔH and ΔS in matching energy units
Crossover temperature
T = ΔH ÷ ΔS
Only meaningful when ΔH and ΔS have the same sign
Equilibrium constant
K = e^(−ΔG°/RT)
R = 8.314462618 J/(mol·K)
Kelvin from Celsius
T(K) = T(°C) + 273.15
298.15 K = 25 °C (standard temperature for ΔG°)
Example: for ammonia synthesis, N₂ + 3 H₂ → 2 NH₃, ΔH° ≈ −92.2 kJ/mol and ΔS° ≈ −198.7 J/(mol·K). At 298.15 K, ΔG° = −92,200 − 298.15 × (−198.7) ≈ −32.96 kJ/mol (spontaneous), and the reaction stops being spontaneous above about 464 K.
Frequently asked questions
What is the Gibbs free energy equation?
ΔG = ΔH − TΔS, where ΔH is the enthalpy change, T the absolute temperature in kelvin and ΔS the entropy change. If ΔG is negative the process is spontaneous at that temperature, if it is positive it is not, and if it is zero the system is at equilibrium.
Why do I get a wrong answer when mixing kJ and J?
Enthalpy is usually tabulated in kJ/mol and entropy in J/(mol·K). Multiplying T × ΔS gives J/mol, so you must convert one of them before subtracting. This calculator converts everything to joules automatically; forgetting the factor of 1000 is the most common mistake.
How do you find the temperature at which a reaction becomes spontaneous?
Set ΔG = 0 and solve: T = ΔH ÷ ΔS (in consistent units). For CaCO₃ → CaO + CO₂ with ΔH ≈ +178.3 kJ/mol and ΔS ≈ +160.5 J/(mol·K), T ≈ 178,300 ÷ 160.5 ≈ 1,111 K, so the decomposition is spontaneous above about 838 °C.
When is a reaction spontaneous at all temperatures?
When ΔH is negative and ΔS is positive, both terms make ΔG negative, so it is spontaneous at every temperature. When ΔH is positive and ΔS negative it is never spontaneous. When both have the same sign, the temperature decides.
How is ΔG related to the equilibrium constant?
For standard values, ΔG° = −RT ln K, so K = e^(−ΔG°/RT) with R = 8.314 J/(mol·K). A negative ΔG° gives K > 1 (products favoured) and a positive ΔG° gives K < 1.
Does spontaneous mean fast?
No. Spontaneity is about thermodynamics, not speed. The conversion of diamond to graphite has a negative ΔG at room temperature but is so slow it is never observed. Reaction rate depends on activation energy, which ΔG does not tell you.
Are ΔH and ΔS constant with temperature?
Not exactly, but they change slowly, so treating them as constant is the usual approximation and gives good estimates over moderate temperature ranges. For precise work at very different temperatures, use values measured at that temperature.
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