Gibbs Free Energy Calculator

Find the Gibbs free energy change from enthalpy, entropy and temperature.

Solves for any one of ΔG, ΔH, ΔS or T, with four energy units and three temperature scales.

Updated October 11, 2026

CreatorFrank Zhao

Loading calculator…

What the calculator does

Gibbs free energy answers one blunt question about a chemical reaction: on its own, at a fixed temperature and pressure, will it run by itself or does it need a push? This calculator takes the enthalpy change, the entropy change and the temperature, and returns the Gibbs free energy change from the equation below.

Gibbs free energy equation

\Delta G = \Delta H - T\,\Delta S

Solve in any direction

Fill in three values and the fourth appears automatically. Type into the result box instead and the calculator re-picks which value to solve for — that reverse direction is how you find the temperature at which a reaction changes its mind.

Units stay consistent

Enthalpy, entropy and Gibbs energy share the same four-unit menu (J, kJ, cal, kcal), and temperature accepts Celsius, Fahrenheit or kelvin. A mixed set — $\mathrm{kJ}$ for $\Delta H$ and $\mathrm{J/K}$ for $\Delta S$ — is fine; everything is converted before the multiplication.

The sign is the verdict

A negative $\Delta G$ means the reaction is spontaneous (exergonic) under those conditions; a positive one means it needs energy input (endergonic); exactly zero means the system sits at equilibrium.

Worked example in four steps

The best-known use of this equation is the industrial ammonia synthesis, $\mathrm{N_2 + 3\,H_2 \rightarrow 2\,NH_3}$ . Here is the full path at room temperature.

Gather the thermochemical data

Per mole of reaction as written, standard values give $\Delta H^\circ = -91.88\ \mathrm{kJ}$ (twice the enthalpy of formation of ammonia gas, −45.94 kJ/mol) and an entropy change of $\Delta S^\circ = -198.11\ \mathrm{J/K}$ from the standard molar entropies (NH₃, N₂, H₂ in the NIST Chemistry WebBook).

Pick the temperature

Room temperature is $25\,^\circ\mathrm{C}$ ; type it with the Celsius unit selected, or enter $298.15$ in kelvin — the result is identical either way.

Enter ΔH and ΔS, read ΔG

Use any units you like — say −91.88 kJ and −198.11 J/K — and the Gibbs free energy box fills itself in:

\Delta G

=

\Delta H - T\,\Delta S

=

-91.88\ \mathrm{kJ} - 298.15\,\mathrm{K} \times (-0.19811\ \mathrm{kJ/K})

=

-32.81\ \mathrm{kJ}

Read the verdict

$\Delta G \approx -32.8\ \mathrm{kJ}$ is negative, so ammonia formation is thermodynamically spontaneous at room temperature. Notice both terms pulled the same way here — the reaction releases heat and becomes more ordered, and the ordering penalty is what you are watching for in the next section.

Finding the tipping temperature

The same ammonia numbers hide a trade-off. Both $\Delta H$ and $\Delta S$ are negative, so the heat term favors the reaction while the entropy term fights it — and the entropy term grows with temperature. Run the reverse solve: tell the calculator you want $T$ to be the unknown by entering $\Delta H$ , $\Delta S$ and $\Delta G = 0$ , and it returns the tipping point:

Temperature at which ΔG changes sign

T = \frac{\Delta H}{\Delta S}

T

=

\frac{-91880\ \mathrm{J}}{-198.11\ \mathrm{J\,K^{-1}}}

=

463.8\ \mathrm{K}\ (\approx 191\,^\circ\mathrm{C})

Below 464 K

The entropy term is smaller than the heat term, so $\Delta G < 0$ and the forward reaction forms ammonia on its own.

Above 464 K

The entropy term wins, so $\Delta G > 0$ and the reverse reaction — decomposing ammonia — becomes the spontaneous direction.

This is exactly why industrial ammonia plants operate at elevated temperatures despite the thermodynamic penalty — they are trading spontaneity for speed, a compromise the Gibbs equation makes visible. You can re-run the forward solve at 700 K in the calculator and watch $\Delta G$ come out positive.

Working backwards from a measurement

The reverse direction is not just a convenience. Suppose a reaction releases $52.0\ \mathrm{kJ}$ but at 298.15 K is still not spontaneous, with $\Delta G = +8.6\ \mathrm{kJ}$ . What entropy change does that imply? Enter $\Delta H$ , $\Delta G$ and $T$ , and the calculator gives

\Delta S = \frac{\Delta H - \Delta G}{T}

=

\frac{52.0 - 8.6}{298.15}\ \mathrm{kJ/K}

=

145.6\ \mathrm{J/K}

A large positive entropy change means the reaction becomes much more disordered, which is the kind of reaction that turns spontaneous once you heat it. The same calculation from the calculator shows this one crosses zero near 357 K. If you need the entropy change itself from stoichiometry rather than from $\Delta G$ , the entropy change calculator handles the product-minus-reactant bookkeeping.

Getting trustworthy numbers

Keep the entropy unit in mind

The entropy box shares the energy unit menu, so choosing kilojoules there means kilojoules per kelvin. A value like $-0.19811$ kJ/K and $-198.11$ J/K are the same number — pick whichever matches your source so you avoid shifting the decimal by 1000.

Signs matter more than magnitudes

The most common mistake is dropping the minus sign on an exothermic $\Delta H$ or a decreasing entropy. A reaction with $\Delta H = -100\ \mathrm{kJ}$ and $\Delta S = -0.20\ \mathrm{kJ/K}$ flips from spontaneous to non-spontaneous right at 500 K; get one sign wrong and you predict the opposite behavior at every temperature.

Sanity-check the scale

For a reaction written per mole, sensible magnitudes are tens to hundreds of kJ for $\Delta H$ and tens to hundreds of J/K for $\Delta S$ . If the calculator flags your value as far above any laboratory-scale reaction, the likely culprit is a unit mix-up or an extra zero, not an unusual chemistry.

Spontaneous does not mean fast

A strongly negative $\Delta G$ says nothing about how long the reaction takes — diamond converting to graphite is spontaneous, and mercifully slow. For the speed side of the story, use the Arrhenius equation calculator to work with activation energy and rate constants.

Common questions

Why does ΔG equal zero at equilibrium?

At equilibrium there is no net driving force left — the forward and reverse tendencies exactly cancel, so the maximum useful work the reaction can deliver is zero. That is also why setting $\Delta G = 0$ and solving for $T$ gives the equilibrium (tipping) temperature for reactions whose $\Delta H$ and $\Delta S$ barely change with temperature.

Can the entropy change of a reaction be negative?

Yes. $\Delta S$ describes the change, not the absolute entropy, so any reaction that consumes gas molecules or builds a more ordered product — ammonia synthesis being the classic — has a negative $\Delta S$ . Absolute entropies themselves are always positive; only differences can carry a minus sign.

My temperature came out negative. Is that an error?

If you solved for $T$ and got a value at or below 0 K, the calculator shows it in red rather than hiding it, because the arithmetic is valid but the state is physically impossible — no reaction runs below absolute zero. Check the signs of $\Delta H$ , $\Delta G$ and $\Delta S$ ; this combination usually means two of the three disagree about the direction of the reaction.

Does a positive ΔG mean the reaction can never happen?

Not at that temperature, and not under those conditions — but heating can change the verdict whenever $\Delta S$ is positive, since the $-T\,\Delta S$ term grows with temperature. Coupling the reaction to a strongly spontaneous one is the other standard trick biology uses to drive otherwise unfavorable chemistry.

Limitations & Disclaimers

The equation assumes constant temperature and pressure with only pressure–volume work. Reactions coupled to electrical work, non-expansion work, or strongly non-ideal mixtures need a fuller treatment than $\Delta G = \Delta H - T\,\Delta S$ .
Using tabulated $\Delta H^\circ$ and $\Delta S^\circ$ values at a different temperature treats both as roughly constant over the gap. For wide temperature swings, heat-capacity corrections become noticeable and the single-point numbers here will drift from the true value.
Spontaneity is a statement about direction and extent, never about rate. A negative $\Delta G$ tells you the reaction wants to happen; it says nothing about whether it will happen on any timescale you care about without a catalyst or sufficient temperature.
Results are for education and quick estimation. For process design, safety reviews or published thermochemistry, work from vetted reference data and have the full calculation reviewed by a qualified chemist or chemical engineer.

Related Calculators

BIP39 Passphrase Generator

Generate and validate BIP39 mnemonics (seed phrases) from entropy with multi-language wordlists. Runs locally in your browser.

Try Now

Password Strength Analyser

Estimate password strength with a brute-force cracking model. See crack time, entropy, character set size, and a simple score — all locally in your browser.

Try Now

Annealing Temperature Calculator

Estimate PCR primer annealing temperature (Ta) from primer and target melting temperatures (Tm). Supports °C / °F / K.

Try Now

Crickets Chirping Thermometer

Estimate the temperature outside by counting cricket chirps with this crickets chirping thermometer calculator based on Dolbear's law.

Try Now

Chemical Equation Balancer

Balance a chemical reaction and get the smallest whole-number coefficients. Supports bracketed groups such as Fe(OH)3, user-supplied coefficients, and unicode subscripts.

Try Now

Net Ionic Equation Calculator

Reduce a molecular equation to its net ionic form. Balances the reaction, splits soluble compounds into ions, and names the spectator ions that cancel out.

Try Now