About the Gibbs Free Energy Calculator
This Gibbs free energy calculator applies ΔG = ΔH − TΔS to tell you whether a reaction or phase change is thermodynamically favourable at a chosen temperature. Enter the enthalpy change in kJ/mol, the entropy change in J/(mol·K) and the temperature in kelvin, Celsius or Fahrenheit, and it returns ΔG, a plain-English spontaneity verdict, the TΔS term and the equilibrium constant K = e^(−ΔG/RT).
It is aimed at general and physical chemistry students, and at anyone checking whether heating or cooling will make a process run. The calculator also finds the crossover temperature where ΔG changes sign (T = ΔH ÷ ΔS) and plots ΔG against temperature, so you can see at a glance whether a reaction is favoured at low temperature, high temperature, always or never.
The calculation assumes ΔH and ΔS stay constant over the temperature range, which is a good approximation for moderate ranges but not exact. Use standard values (ΔH°, ΔS°) to get the standard free energy ΔG° and the thermodynamic equilibrium constant. Note the unit trap: ΔS is usually tabulated in J/(mol·K), so it is divided by 1,000 before combining with ΔH in kJ/mol.
With the default inputs, the gibbs free energy δg is -32.96 kJ/mol. Change any value above to recalculate instantly.
How to use the gibbs free energy calculator
- 1Enter ΔH in kJ/mol (negative for exothermic reactions).
- 2Enter ΔS in J/(mol·K), keeping its sign.
- 3Enter the temperature and choose K, °C or °F.
- 4Read ΔG and the spontaneity verdict.
- 5Use the crossover temperature and chart to see where the reaction switches direction.
Formula and method
The Gibbs free energy change combines the enthalpy change (heat released or absorbed) with the entropy change scaled by absolute temperature. A negative ΔG means the process can proceed spontaneously at constant temperature and pressure; a positive ΔG means it cannot without an input of work, and the reverse process is favoured. ΔS is converted from J/(mol·K) to kJ/(mol·K) before it is multiplied by T.
Setting ΔG = 0 gives the crossover temperature ΔH ÷ ΔS, which only exists when ΔH and ΔS have the same sign. With standard values, ΔG° = −RT ln K links free energy to the equilibrium constant, using R = 8.314462618 J/(mol·K). The model treats ΔH and ΔS as temperature-independent.
- ΔG
- Gibbs free energy change (kJ/mol)
- ΔH
- Enthalpy change (kJ/mol)
- ΔS
- Entropy change (J/(mol·K), divided by 1000 in the formula)
- T
- Absolute temperature (K)
- R
- Gas constant, 8.314462618 J/(mol·K)
- K
- Thermodynamic equilibrium constant
Worked examples
Haber process (N₂ + 3H₂ → 2NH₃) at 25 °C
TΔS = 298.15 × (−0.1987) = −59.24 kJ/mol, so ΔG = −92.2 − (−59.24) = −32.96 kJ/mol: spontaneous. Because both ΔH and ΔS are negative, it stops being favourable above 92,200 ÷ 198.7 ≈ 464 K, yet ammonia plants run at roughly 400–500 °C for speed, accepting a lower equilibrium yield and offsetting it with high pressure, an iron catalyst and recycling of unreacted gas.
Limestone decomposition (CaCO₃ → CaO + CO₂) at room temperature
ΔG = 178.3 − 298.15 × 0.1606 = +130.42 kJ/mol, so limestone does not decompose at room temperature. Both terms are positive, so it becomes spontaneous above 178,300 ÷ 160.6 ≈ 1,110 K — the reason lime kilns run at around 900 °C.
Ice melting at 10 °C
At 283.15 K, TΔS = 6.229 kJ/mol, slightly more than ΔH = 6.01 kJ/mol, so ΔG ≈ −0.22 kJ/mol and ice melts. The crossover of 6,010 ÷ 22.0 ≈ 273.2 K is the normal melting point.
Frequently asked questions
What does a negative ΔG mean?+
A negative Gibbs free energy change means the process is thermodynamically spontaneous at that temperature and pressure: it can proceed without outside work. It says nothing about speed — a spontaneous reaction can still be extremely slow if its activation energy is high.
Why do I divide ΔS by 1000?+
Entropy changes are normally tabulated in J/(mol·K) while enthalpy changes are in kJ/mol. Multiplying T by ΔS gives J/mol, so it must be divided by 1,000 to match ΔH in kJ/mol before subtracting.
How do I find the temperature at which a reaction becomes spontaneous?+
Set ΔG = 0 and solve: T = ΔH ÷ ΔS (with ΔS in kJ/(mol·K)). This crossover exists only when ΔH and ΔS share a sign. If both are positive the reaction is spontaneous above T; if both are negative it is spontaneous below T.
How is ΔG related to the equilibrium constant?+
For standard free energy, ΔG° = −RT ln K. A ΔG° of −5.7 kJ/mol at 298 K corresponds to K ≈ 10; each further −5.7 kJ/mol multiplies K by roughly another factor of ten.
What is the difference between ΔG and ΔG°?+
ΔG° is the free energy change with all species in their standard states (1 bar, 1 M). The actual ΔG at other concentrations is ΔG° + RT ln Q, where Q is the reaction quotient. This calculator gives ΔG° when you enter standard ΔH° and ΔS° values.