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
| ΔH | ΔS | ΔG = ΔH − TΔS | Example |
|---|---|---|---|
| − | + | Negative at all T | Combustion of fuels |
| + | − | Positive at all T | Forming ozone from O₂ |
| − | − | Negative at low T | Water freezing, NH₃ synthesis |
| + | + | Negative at high T | Ice 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
| Field | What to enter |
|---|---|
| Solve for | Choose one: Gibbs energy change ΔG, Enthalpy change ΔH, Entropy change ΔS, Temperature T |
| Enthalpy change ΔH | Units: kJ/mol, J/mol, kcal/mol, cal/mol, eV (per particle) |
| Entropy change ΔS | Units: J/(mol·K), kJ/(mol·K), cal/(mol·K) |
| Temperature T | Units: K, °C, °F |
| Gibbs energy change ΔG | Units: 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))
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/molWorked 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.
Frequently asked questions
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.