# Gibbs Free Energy Calculator

URL: https://speedycalculator.com/calculators/chemistry/gibbs-energy/
Last updated: 2026-10-30

> The Gibbs free energy calculator solves ΔG = ΔH − TΔS for any one unknown – Gibbs energy, enthalpy, entropy or temperature – and adds the equilibrium constant, the crossover temperature and whether the reaction is spontaneous, enthalpy-driven or entropy-driven.

## Inputs

- Solve for
- Enthalpy change ΔH (kJ/mol, J/mol, kcal/mol, cal/mol, eV (per particle))
- Entropy change ΔS (J/(mol·K), kJ/(mol·K), cal/(mol·K))
- Temperature T (K, °C, °F)
- Gibbs energy change ΔG (kJ/mol, J/mol, kcal/mol, cal/mol, eV (per particle))

## Results

- Gibbs energy change ΔG
- Enthalpy change ΔH
- Entropy change ΔS
- Temperature T
- Note
- Entropy term T·ΔS
- Spontaneity
- Character
- Crossover temperature ΔH/ΔS (ΔG = 0)
- Equilibrium constant K = e^(−ΔG/RT) (if ΔG = ΔG°)
- log₁₀ K

## Formula

```
ΔG = ΔH − T·ΔS · K = e^(−ΔG/RT) · T(ΔG = 0) = ΔH / ΔS · T in kelvin, ΔS in kJ/(mol·K) when ΔH is in kJ/mol
```

## Explanation

Gibbs free energy combines the two driving forces of chemistry into one number. Reactions tend to release heat (negative ΔH) and to increase disorder (positive ΔS); at constant temperature and pressure, the change in Gibbs energy ΔG = ΔH − TΔS tells you which way the balance falls. A negative ΔG means the process is spontaneous – it can proceed without outside work – while a positive ΔG means it needs a continuous energy input, and ΔG = 0 means the system is at equilibrium.

Because temperature multiplies the entropy term, many reactions switch direction at a particular temperature. Limestone (CaCO₃) does not decompose at room temperature, but above roughly 850 °C the entropy gain from releasing CO₂ gas wins and lime kilns can operate. Ice melting is the opposite case of the same idea: below 0 °C the enthalpy term wins and water freezes, above it the entropy term wins.

The standard Gibbs energy change also fixes the equilibrium constant through ΔG° = −RT ln K. A ΔG° of −5.7 kJ/mol at 25 °C corresponds to K ≈ 10, and every further −5.7 kJ/mol multiplies K by about ten.

### How to use the Gibbs Free Energy Calculator

- **Solve for**
- **Enthalpy change ΔH**
- **Entropy change ΔS**
- **Temperature T**
- **Gibbs energy change ΔG**

Results update as you type, every field has its own unit menu, and everything is calculated in your browser.

### How it works

Pick the quantity to solve for and enter the other three. The calculator converts ΔS from J/(mol·K) to kJ/(mol·K) and temperature to kelvin before applying ΔG = ΔH − TΔS (or its rearrangements), then reports K = exp(−ΔG/RT), the crossover temperature ΔH/ΔS and a plain-language verdict.

### Worked example

Thermal decomposition of limestone, CaCO₃(s) → CaO(s) + CO₂(g). From standard data, ΔH° = −635.1 − 393.5 − (−1206.9) = +178.3 kJ/mol and ΔS° = 38.1 + 213.8 − 92.9 = +159.0 J/(mol·K). At 298.15 K: ΔG° = 178.3 − 298.15 × 0.1590 = **+130.9 kJ/mol** – not spontaneous. The crossover temperature is 178,300 / 159.0 = 1,121 K (848 °C); above it the entropy term wins and the decomposition becomes spontaneous.

### The four sign combinations

ΔHΔSΔG = ΔH − TΔSExample−+Negative at all TCombustion of fuels+−Positive at all TForming ozone from O₂−−Negative at low TWater freezing, NH₃ synthesis++Negative at high TIce melting, CaCO₃ decomposition

### Tips

- Use kelvin for every temperature in thermodynamic formulas – the unit menus convert for you.
- Watch the units: ΔH is usually in kJ/mol and ΔS in J/(mol·K).
- Tabulated values are for standard conditions (1 bar, usually 25 °C).
- Everything is calculated in your browser.

### Common mistakes to avoid

- Mixing kJ and J between ΔH and ΔS.
- Assuming ΔH and ΔS do not change with temperature over large ranges.
- Using Celsius in ln(T₂/T₁).

### Inputs explained

FieldWhat to enterSolve forChoose one: Gibbs energy change ΔG, Enthalpy change ΔH, Entropy change ΔS, Temperature TEnthalpy change ΔHUnits: kJ/mol, J/mol, kcal/mol, cal/mol, eV (per particle)Entropy change ΔSUnits: J/(mol·K), kJ/(mol·K), cal/(mol·K)Temperature TUnits: K, °C, °FGibbs energy change ΔGUnits: kJ/mol, J/mol, kcal/mol, cal/mol, eV (per particle)

### What each result means

- **Gibbs energy change ΔG** = [G]
- **Enthalpy change ΔH** = [H]
- **Entropy change ΔS** = [S]
- **Temperature T** = [T]
- **Entropy term T·ΔS** = [T]×[S]÷1000
- **Equilibrium constant K = e^(−ΔG/RT) (if ΔG = ΔG°)** = exp(-[G]×1000÷(8.314462618×[T]))
- **log₁₀ K** = -[G]×1000÷(8.314462618×[T]×ln(10))

## Worked example

Inputs — Solve for: Gibbs energy change ΔG; Enthalpy change ΔH: 178.3 kJ/mol; Entropy change ΔS: 159 J/(mol·K); Temperature T: 298.15 K.

Results — **Gibbs energy change ΔG = 130.8942 kJ/mol**; Entropy term T·ΔS = 47.4059 kJ/mol; Spontaneity = Non-spontaneous (endergonic, ΔG > 0); Character = Entropy-driven: spontaneous only above the crossover temperature; Crossover temperature ΔH/ΔS (ΔG = 0) = 1,121.3836 K; Equilibrium constant K = e^(−ΔG/RT) (if ΔG = ΔG°) = 1.1704e-23; log₁₀ K = -22.9317.

## Key terms

- **Gibbs free energy (G)**: G = H − TS; its change at constant temperature and pressure is the maximum non-expansion work a process can do.
- **Exergonic / endergonic**: A process with negative / positive ΔG.
- **Crossover temperature**: The temperature ΔH/ΔS at which ΔG = 0 and a reaction changes direction.
- **Standard state**: The reference conditions for tabulated values: 1 bar for gases, 1 mol/L for solutes, the pure substance for liquids and solids.

## FAQ

**What does a negative ΔG mean?**
The process is spontaneous (exergonic) at that temperature and pressure: it can happen without outside work. It says nothing about how fast it happens.

**How do I calculate ΔG from ΔH and ΔS?**
Convert ΔS to kJ/(mol·K) and T to kelvin, then ΔG = ΔH − TΔS. For CaCO₃ → CaO + CO₂ at 298.15 K: 178.3 − 298.15 × 0.1590 = +130.9 kJ/mol.

**At what temperature does limestone decompose?**
Under standard conditions ΔG = 0 at T = ΔH/ΔS = 178,300/159.0 = 1,121 K, about 848 °C – close to the temperatures used in lime kilns.

**Why do I have to convert ΔS to kJ?**
ΔH is usually tabulated in kJ/mol but ΔS in J/(mol·K). Forgetting the factor of 1,000 is the most common mistake.

**What is the relationship between ΔG° and K?**
ΔG° = −RT ln K. At 25 °C, ΔG° = −11.4 kJ/mol gives K ≈ 100 and +11.4 kJ/mol gives K ≈ 0.01.

**Is ΔG the same as ΔG°?**
ΔG° is for all species in their standard states (1 bar, 1 M). The actual ΔG = ΔG° + RT ln Q depends on the current concentrations.

**Why is the equilibrium constant tiny for limestone at room temperature?**
ΔG° = +130.9 kJ/mol gives K = e^(−52.8) ≈ 1 × 10⁻²³, so the equilibrium CO₂ pressure over CaCO₃ at 25 °C is negligible.

**What formula does the Gibbs Free Energy Calculator use?**
ΔG = ΔH − T·ΔS · K = e^(−ΔG/RT) · T(ΔG = 0) = ΔH / ΔS · T in kelvin, ΔS in kJ/(mol·K) when ΔH is in kJ/mol.


## Sources

- [OpenStax – Chemistry 2e, 16.4 Free energy](https://openstax.org/books/chemistry-2e/pages/16-4-free-energy)
- [NIST Chemistry WebBook](https://webbook.nist.gov/chemistry/)
- [IUPAC Compendium of Chemical Terminology (Gold Book)](https://goldbook.iupac.org/)