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Gibbs Free Energy Calculator (ΔG = ΔH − TΔS)
In short: Gibbs free energy calculator: solve ΔG = ΔH − TΔS for ΔG, ΔH, T or ΔS, find the crossover temperature, convert between ΔG° and K, and correct ΔG for non-standard conditions. It is free, works in your web browser and needs no download.
Enter any three of ΔG, ΔH, T and ΔS to find the fourth. ΔS can be in J or kJ, and T in °C or K. The result shows whether the process is spontaneous, which of the four sign cases applies and the crossover temperature. Other tabs convert ΔG° to K (shown in scientific notation, even for huge values) and apply ΔG = ΔG° + RT ln Q.
Fill in three values and leave the one you want to find blank.
ΔG° = −RT ln K. Fill in either ΔG° or K.
Gibbs free energy formulas
ΔG = ΔH − TΔS crossover temperature (ΔG = 0): T = ΔH / ΔS
ΔG° = −RT ln K = −2.303RT log K K = e^(−ΔG°/RT) ΔG = ΔG° + RT ln Q
R = 8.314 J mol⁻¹ K⁻¹ and T is in kelvin. ΔH and ΔG are usually in kJ/mol while ΔS is in J/(mol·K), so divide ΔS by 1000 before combining them. ΔG < 0 means the process is spontaneous at that temperature. It says nothing about how fast it goes.
Spontaneity: the four sign cases
| ΔH | ΔS | ΔG = ΔH − TΔS | Spontaneous? | Example |
|---|---|---|---|---|
| − (exothermic) | + | Always negative | At all temperatures | 2H₂O₂ → 2H₂O + O₂ |
| + (endothermic) | − | Always positive | Never (reverse is spontaneous) | 3O₂ → 2O₃ |
| − | − | Negative at low T | Below T = ΔH/ΔS | N₂ + 3H₂ → 2NH₃; water freezing |
| + | + | Negative at high T | Above T = ΔH/ΔS | CaCO₃ → CaO + CO₂; ice melting |
ΔG° and K at 298.15 K
| ΔG° (kJ/mol) | K |
|---|---|
| −100 | 3.3 × 10¹⁷ |
| −50 | 5.8 × 10⁸ |
| −20 | 3.2 × 10³ |
| −10 | 56.5 |
| −5 | 7.52 |
| 0 | 1 |
| +5 | 0.133 |
| +10 | 0.0177 |
| +20 | 3.1 × 10⁻⁴ |
| +50 | 1.7 × 10⁻⁹ |
| +100 | 3.0 × 10⁻¹⁸ |
Every 5.71 kJ/mol of ΔG° changes K by a factor of 10 at 25 °C.
Worked examples
Haber process at 25 °C
N₂ + 3H₂ → 2NH₃: ΔH° = −92.2 kJ/mol, ΔS° = −198.7 J/(mol·K).
ΔG° = −92.2 − 298.15 × (−0.1987) = −92.2 + 59.2
ΔG° = −33.0 kJ/mol (spontaneous)
Above what temperature is NH₃ formation non-spontaneous?
Both ΔH and ΔS are negative, so it is spontaneous only below T = ΔH/ΔS.
T = −92 200 / −198.7
T = 464 K (191 °C)
Decomposition of limestone (NEET style)
CaCO₃ → CaO + CO₂: ΔH° = +178.3 kJ/mol, ΔS° = +160.5 J/(mol·K).
T = 178 300 / 160.5
Spontaneous above 1111 K (838 °C)
K from ΔG°
ΔG° = −33.0 kJ/mol at 298.15 K.
ln K = 33 000 / (8.314 × 298.15) = 13.31
K = 6.0 × 10⁵
ΔG° from Ka (JEE style)
Acetic acid: Ka = 1.8 × 10⁻⁵ at 25 °C.
ΔG° = −8.314 × 298.15 × ln(1.8 × 10⁻⁵) = −2479 × (−10.93)
ΔG° = +27.1 kJ/mol
Non-standard conditions
ΔG° = −33.0 kJ/mol and Q = 100 at 298.15 K.
ΔG = −33.0 + 8.314 × 298.15 × ln 100 / 1000 = −33.0 + 11.4
ΔG = −21.6 kJ/mol (still forward)
Units and common mistakes
- J vs kJ. ΔS tables are in J/(mol·K). Convert to kJ (÷1000) before using ΔG = ΔH − TΔS with ΔH in kJ. This is the most common error.
- Kelvin only. 25 °C is 298.15 K. Using 25 instead of 298 makes TΔS twelve times too small.
- ΔG vs ΔG°. ΔG° is for standard states (1 bar, 1 M). The real driving force under other conditions is ΔG = ΔG° + RT ln Q.
- Spontaneous is not fast. Diamond turning into graphite has ΔG° < 0 but is extremely slow. Rate depends on activation energy.
- ΔH and ΔS change a little with T. T = ΔH/ΔS assumes they are constant, which is a good approximation over moderate ranges.
- Use the same R in both places. With ΔG° in J/mol use R = 8.314 J mol⁻¹ K⁻¹; for kJ/mol use 0.008314.
Drawing energy diagrams or reactions? Sketch reactions, mechanisms and lab set-ups in MolDraw, a free online chemical structure editor.
Open MolDraw editorFrequently asked questions
How do I calculate Gibbs free energy?
Use ΔG = ΔH − TΔS with T in kelvin and ΔS converted to kJ/(mol·K). For the Haber process at 298.15 K: ΔG° = −92.2 − 298.15 × (−0.1987) = −33.0 kJ/mol.
What does a negative ΔG mean?
The process is spontaneous (thermodynamically favourable) in the forward direction under those conditions. A positive ΔG means the reverse direction is favourable. ΔG = 0 means equilibrium.
How do I find the temperature at which a reaction becomes spontaneous?
Set ΔG = 0: T = ΔH/ΔS. This crossover only exists when ΔH and ΔS have the same sign. For CaCO₃ decomposition, T = 178 300 / 160.5 = 1111 K.
How are ΔG° and the equilibrium constant related?
ΔG° = −RT ln K. A negative ΔG° gives K greater than 1 (products favoured); a positive ΔG° gives K less than 1. At 25 °C, each 5.71 kJ/mol changes K tenfold.
What is the difference between ΔG and ΔG°?
ΔG° is the free energy change with everything in its standard state. ΔG applies to the actual mixture: ΔG = ΔG° + RT ln Q. As a reaction proceeds, Q approaches K and ΔG approaches zero.
Why do I need to divide ΔS by 1000?
Entropy is usually tabulated in J/(mol·K) while enthalpy is in kJ/mol. Dividing ΔS by 1000 puts both terms in kJ/mol so they can be subtracted.
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