Gibbs Free Energy Calculator
Calculate the Gibbs free energy change (ΔG) from enthalpy (ΔH), entropy (ΔS), and temperature, and instantly see whether the reaction is spontaneous under those conditions.
Find ΔG and Check if the Reaction Is Spontaneous
Gibbs Free Energy Formula
Where:
- \(\Delta G\) — Gibbs free energy change (J/mol or kJ/mol)
- \(\Delta H\) — enthalpy change of the reaction (J/mol)
- \(T\) — absolute temperature (kelvin)
- \(\Delta S\) — entropy change of the reaction (J/(mol·K))
The formula combines the two driving forces of a chemical reaction — the release or absorption of heat (\(\Delta H\)) and the change in disorder (\(\Delta S\)) — into a single value that tells you whether the reaction can proceed on its own at the given temperature.
Gibbs Free Energy – Calculation Example
Example: the Haber reaction \(N_2 + 3H_2 \rightarrow 2NH_3\) has \(\Delta H = -92.2\) kJ/mol and \(\Delta S = -198.75\) J/(mol·K). Is ammonia formation spontaneous at 25 °C?
\[ T = 25 + 273.15 = 298.15 \text{ K} \]
\[ \Delta G = -92{,}200 - 298.15 \times (-198.75) = -92{,}200 + 59{,}257.31 = -32{,}942.69 \text{ J/mol} \]
Result: \(\Delta G \approx -32.94\) kJ/mol (about \(-7.87\) kcal/mol). Since \(\Delta G < 0\), the reaction is spontaneous at 25 °C — but the calculator also warns that spontaneity flips near 191 °C (464 K), so above that temperature ammonia formation stops being favorable.
How to Use the Gibbs Free Energy Calculator
- Enter the Enthalpy Change (ΔH) of the reaction and select its unit (kJ/mol, J/mol, or kcal/mol). Use a negative value for exothermic reactions and a positive value for endothermic ones.
- Enter the Entropy Change (ΔS) and choose the unit (J/(mol·K), kJ/(mol·K), or cal/(mol·K)). Positive means disorder increases.
- Enter the Temperature and pick °C, K, or °F. The calculator converts everything to consistent units automatically.
- Read the Gibbs Free Energy Change in both kJ/mol and kcal/mol, plus the Predicted Spontaneity of the reaction under those exact conditions.
What Is Gibbs Free Energy?
Gibbs free energy, named after Josiah Willard Gibbs, is a thermodynamic quantity that measures the maximum amount of useful work a reaction can perform at constant temperature and pressure. Its change during a reaction, \(\Delta G\), is the single most convenient indicator of whether a process is thermodynamically favorable: a reaction can proceed spontaneously only if it lowers the Gibbs free energy of the system.
The sign of \(\Delta G\) carries all the meaning. If \(\Delta G 0\), the forward reaction is non-spontaneous (endergonic), and energy must be supplied to drive it; the reverse reaction is then the spontaneous one. If \(\Delta G = 0\), the system is at equilibrium and no net reaction occurs in either direction.
Temperature plays a decisive role because the entropy term is multiplied by \(T\). When \(\Delta H\) and \(\Delta S\) have opposite signs, the outcome is the same at every temperature. When they have the same sign, temperature decides: an endothermic reaction with a positive entropy change (like melting ice) becomes spontaneous only above a certain temperature, while an exothermic reaction with negative entropy change (like the Haber example above) is spontaneous only below a crossover temperature given by \(T = \Delta H / \Delta S\).
One important caveat: spontaneity is about thermodynamics, not speed. A reaction can have a strongly negative \(\Delta G\) and still be extremely slow — diamond converting to graphite is the classic example. For rates and activation barriers, see the Arrhenius Equation calculator.
Sign Combinations at a Glance
| ΔH | ΔS | ΔG = ΔH − TΔS | Outcome |
|---|---|---|---|
| − | + | always negative | Spontaneous at all temperatures |
| + | − | always positive | Never spontaneous |
| − | − | negative at low T | Spontaneous below the crossover temperature |
| + | + | negative at high T | Spontaneous above the crossover temperature |
Assumptions and Practical Tips
- Standard vs. actual values: tabulated \(\Delta H\) and \(\Delta S\) values usually refer to standard conditions (298.15 K, 1 bar). The calculator applies them at any temperature you enter, which assumes both quantities are roughly temperature-independent — a good approximation over moderate ranges.
- Constant conditions: the formula is strictly valid at constant temperature and pressure; that covers most bench-top chemistry and biological processes.
- Concentrations matter: this calculator uses the simple \(\Delta G = \Delta H - T\Delta S\) form. If reactant and product concentrations differ from standard values, the actual \(\Delta G\) will shift (the full expression involves the reaction quotient \(Q\)).
- Check your units: the classic mistake is mixing kJ with J. If you enter \(\Delta H\) in kJ/mol but \(\Delta S\) in J/(mol·K), let the unit selectors handle the conversion — the calculator always converts internally before applying the formula.
- Per mole: values must refer to the reaction as written (per mole of reaction). If you double the stoichiometric coefficients, \(\Delta H\) and \(\Delta S\) double too.
Once you know a reaction is thermodynamically favorable, you can plan the rest of your stoichiometry: use the Limiting Reactant calculator to find how much product the available reagents can make, and the Percent Yield calculator to compare your actual output with the theoretical maximum. For solutions chemistry, the Molarity calculator is a natural next step.
Frequently Asked Questions
How do I calculate Gibbs free energy by hand without a calculator?
Convert everything to consistent units first:
ΔH in J/mol, ΔS in J/(mol·K), and T in kelvin (add 273.15 if your temperature is in °C). Then multiply T by ΔS and subtract the result from ΔH:
ΔG = ΔH − T·ΔS. For example, with ΔH = −92,200 J/mol, ΔS = −198.75 J/(mol·K), and T = 298 K, you get ΔG = −92,200 − 298 × (−198.75) ≈ −32,940 J/mol. Finally, divide by 1000 to express it as −32.94 kJ/mol.
What is a good or typical value of ΔG?
There is no universal "good" value — the sign matters more than the magnitude. A negative ΔG means the reaction is thermodynamically favorable; a positive one means it is unfavorable as written. Magnitudes vary hugely: ΔG around −5 to −50 kJ/mol is common for everyday reactions, while ATP hydrolysis in cells is about −30.5 kJ/mol under standard conditions. Very large negative values (hundreds of kJ/mol) indicate strongly driven reactions such as combustion.
What's the difference between ΔG and ΔG° (standard Gibbs free energy)?
ΔG° refers to standard conditions: 298.15 K, 1 bar pressure, and all solutes at 1 M concentration. It is what tabulated values and this simple formula give you. The actual ΔG at non-standard concentrations or pressures is ΔG = ΔG° + RT·ln Q, where Q is the reaction quotient. So a reaction with a positive ΔG° can still proceed spontaneously if the concentrations make Q small enough, as often happens in coupled biological reactions.
How is ΔG related to the equilibrium constant K?
They are linked by ΔG° = −RT·ln K, where R is the gas constant (8.314 J/(mol·K)) and T is the temperature in kelvin. A strongly negative ΔG° corresponds to K much greater than 1 (products favored), a positive ΔG° to K less than 1 (reactants favored), and ΔG° = 0 to K = 1. You can rearrange this to extract K from a calculated ΔG°, which is a common exam task in chemistry courses.
Why must temperature be in kelvin, and how do I convert °C or °F?
Kelvin is the absolute temperature scale, and the formula ΔG = ΔH − T·ΔS only works with absolute values because T acts as a multiplier of the entropy term. To convert, use T(K) = T(°C) + 273.15, or T(K) = (T(°F) − 32) × 5/9 + 273.15. Entropy units also contain K (J/(mol·K)), so mixing kelvin with a ΔS expressed per kelvin keeps the units consistent. The calculator handles all these conversions automatically if you pick the right unit from the dropdown.
What are the most common mistakes when computing ΔG?
The classic error is mixing kJ and J: if ΔH is −92.2 kJ/mol but ΔS is −198.75 J/(mol·K), entering both as plain numbers gives a wildly wrong answer — multiply kJ by 1000 first. Other frequent mistakes include using °C instead of K, forgetting that entropy values in tables are often in J/(mol·K) while enthalpies are in kJ/mol, and scaling values inconsistently when the balanced equation is doubled or halved (both ΔH and ΔS scale with stoichiometry).
Does a spontaneous reaction always happen quickly?
No. Spontaneity only tells you the direction a reaction is thermodynamically favored — it says nothing about speed. Diamond converting to graphite has a negative ΔG yet takes millions of years because the activation barrier is enormous. Combustion of wood is also spontaneous at room temperature in terms of ΔG, but it needs a spark to overcome the kinetic barrier. For reaction rates, use the Arrhenius equation rather than Gibbs free energy.