Case 1
- Enthalpy change ΔH
- -100000J/mol
- Absolute temperature T
- 298.15K
- Entropy change ΔS
- 50J/(mol·K)
-114,907.50J/mol
Open with these valuesGibbs Free Energy Calculator
-114,907.50J/mol
Result: -114,907.50 J/molΔG = ΔH − T × ΔS decides whether a reaction runs on its own: negative means spontaneous, zero means equilibrium, positive means it needs a push. The temperature must be absolute — 25 °C is 298.15 K, and entering 25 instead is the single most common mistake here.
The numbers at a glance
Held fixed: Enthalpy change ΔH -100,000.00 J/mol, Absolute temperature T 298.15 K.
| Entropy change ΔS (J/(mol·K)) | Result (J/mol) |
|---|---|
| 0.00 | -100,000.00 |
| 20.00 | -105,963.00 |
| 40.00 | -111,926.00 |
| 50.00Your value | -114,907.50 |
| 60.00 | -117,889.00 |
| 80.00 | -123,852.00 |
| 100.00 | -129,815.00 |
Worked examples
Case 2
- Enthalpy change ΔH
- 100000J/mol
- Absolute temperature T
- 1000K
- Entropy change ΔS
- 200J/(mol·K)
-100,000.00J/mol
Open with these valuesCase 3
- Enthalpy change ΔH
- 50000J/mol
- Absolute temperature T
- 298.15K
- Entropy change ΔS
- -100J/(mol·K)
79,815.00J/mol
Open with these valuesHow it's calculated
ΔG = ΔH − T × ΔS
- StepEnter the enthalpy change in J/mol; negative means exothermic.
- StepEnter the absolute temperature in kelvin — add 273.15 to a Celsius value.
- StepEnter the entropy change in J/(mol·K); negative means the disorder falls.
- ResultRead ΔG in J/mol; divide by 1000 for kJ/mol.
Reference table
| ΔH, T, ΔS | What it means | ΔG in J/mol |
|---|---|---|
| -100000, 298.15, 50 | exothermic and spontaneous | -114907.50 |
| 100000, 1000, 200 | endothermic, spontaneous when hot | -100000.00 |
| -80000, 200, -50 | spontaneous while it stays cold | -70000.00 |
| -50000, 0, 100 | at absolute zero only ΔH is left | -50000.00 |
| 1.5, 2, 0.25 | unit check of the formula | 1.00 |
| 50000, 298.15, -100 | never spontaneous at any T | 79815.00 |
Questions
How do I calculate the change in Gibbs free energy?
Subtract the temperature times the entropy change from the enthalpy change: ΔG = ΔH − T × ΔS. With ΔH = −100000 J/mol, T = 298.15 K and ΔS = 50 J/(mol·K) that is −114907.5 J/mol.
Must the temperature be in kelvin?
Yes, and this is the one field in the calculator where Celsius is genuinely wrong. T multiplies ΔS, so it needs the absolute scale: 25 °C is 298.15 K, and typing 25 changes the answer by more than 13000 J/mol.
What does a negative or positive ΔG mean?
Negative means the reaction is spontaneous and releases free energy; positive means it will not proceed without energy put in. At exactly zero the reaction sits at equilibrium with no net change either way.
Why does temperature decide whether a reaction runs?
Because the entropy term T × ΔS grows with temperature while ΔH does not. An endothermic reaction with rising entropy only turns spontaneous once T × ΔS outgrows ΔH, and a reaction with falling entropy can stop being spontaneous as things heat up.
J/mol or kJ/mol?
This calculator works in joules throughout: ΔH in J/mol, ΔS in J/(mol·K), ΔG in J/mol. Tables often quote ΔH in kJ/mol next to ΔS in J/(mol·K), so multiply such an ΔH by 1000 before entering it.
Sources and last check
Information, not professional advice.