Gibbs free energy is the single quantity that answers whether a change can happen on its own at constant temperature and pressure. It combines the two things pulling in opposite directions — the energy released or absorbed, and the change in how spread out that energy becomes — into one number whose sign carries the verdict. The equation is ΔG = ΔH − TΔS, and this calculator solves it for whichever of the four quantities you are missing.
It also does the thing most Gibbs calculators skip. Because temperature multiplies the entropy term and not the enthalpy term, there is generally one temperature at which ΔG passes through zero and the answer reverses. Arb Digital's calculator reports that crossover temperature alongside the main result, tells you which of the four sign regimes your reaction is in, and optionally converts ΔG° into an equilibrium constant. Everything below is educational explanation of the arithmetic, not laboratory guidance.
What This Gibbs Free Energy Calculator Does
Pick which variable you want and enter the other three. Solve for ΔG to get the standard result. Solve for T to find the temperature at which a reaction reaches a particular ΔG. Solve for ΔS when you have measured ΔH and ΔG and want the entropy change by difference. Solve for ΔH the same way.
Units are handled explicitly rather than assumed, because the unit mismatch between enthalpy and entropy is the most reliable source of wrong answers in this calculation. Enthalpy is tabulated in kilojoules per mole and entropy in joules per mole per kelvin, a factor of 1,000 apart. The tool converts everything to joules internally and converts back for display. Temperature can be entered in kelvin, Celsius or Fahrenheit and is always converted to kelvin before use, since the equation is only valid on an absolute scale.
The results panel breaks the answer into its two competing halves — the ΔH term and the −TΔS term — with a bar showing which one dominates. That breakdown is more informative than the total, because it tells you immediately whether the reaction is enthalpy-driven or entropy-driven, and therefore what will happen when the temperature changes.
How to Use It
- Choose what to solve for. The field you select is ignored as an input and filled in with the answer.
- Enter ΔH with its units. Tabulated standard enthalpies of formation are combined as products minus reactants to give a reaction ΔH°.
- Enter ΔS with its units, and check you have not confused J with kJ. Absolute entropies are tabulated per substance and combined the same way, products minus reactants.
- Set the temperature. Use 298.15 K if you are working with standard tabulated data, since that is the temperature the tables are referenced to.
- Read the regime line underneath the results. It states which of the four sign combinations you are in and what the crossover temperature means for that combination.
The Formula / How It's Calculated
The defining relation is G = H − TS, and for a change at constant temperature and pressure that becomes ΔG = ΔH − TΔS. Rearranged, the same equation gives ΔH = ΔG + TΔS, ΔS = (ΔH − ΔG) ÷ T, and T = (ΔH − ΔG) ÷ ΔS.
The crossover temperature is the special case where ΔG = 0, so Tcrossover = ΔH ÷ ΔS. At that temperature the two terms exactly cancel and the system is at equilibrium under standard conditions. Above it the entropy term wins; below it the enthalpy term does.
The equilibrium constant follows from ΔG° = −RT ln K, so K = e−ΔG°/RT, with R = 8.3145 J/(mol·K).
Work the default values through. The ammonia synthesis, N2 + 3H2 → 2NH3, has ΔH° = −92.22 kJ/mol and ΔS° = −198.75 J/(mol·K). Converting entropy to kJ gives −0.19875 kJ/(mol·K). At 298.15 K the entropy term is TΔS = 298.15 × (−0.19875) = −59.26 kJ/mol, and subtracting it gives ΔG° = −92.22 − (−59.26) = −32.96 kJ/mol. Negative, so the reaction is favourable at room temperature.
The crossover sits at T = −92.22 ÷ −0.19875 = 464.0 K, which is 190.8 °C. Above that temperature ΔG° turns positive and the reaction becomes unfavourable under standard conditions. The equilibrium constant at 298.15 K is e32,960 ÷ (8.3145 × 298.15) = e13.30, roughly 6 × 105. Values of this kind for thousands of substances are compiled in the NIST Chemistry WebBook, Standard Reference Database 69.
The Four Sign Regimes, and Why Only Two Have a Crossover
Everything about the temperature behaviour of ΔG falls out of the signs of ΔH and ΔS. There are four combinations and they behave completely differently.
- ΔH negative, ΔS positive. Both terms push ΔG negative. The reaction is favourable at every temperature, and there is no crossover. Combustion reactions mostly live here.
- ΔH positive, ΔS negative. Both terms push ΔG positive. Unfavourable at every temperature. The reverse reaction is the one in the first category.
- ΔH negative, ΔS negative. Enthalpy favours it, entropy opposes it, and the entropy term grows with temperature. Favourable below the crossover, unfavourable above it. Ammonia synthesis, freezing and most condensations are here.
- ΔH positive, ΔS positive. The mirror image: unfavourable at low temperature, favourable above the crossover. Melting, boiling and thermal decompositions are here.
The crossover temperature only exists for the last two. In the first two, ΔH ÷ ΔS is negative, which is not a physically reachable temperature, and the calculator says so rather than printing a meaningless number. This is also why the reported crossover matters more than the ΔG at one temperature: a reaction that is marginally favourable at 298 K and has a crossover at 310 K is a very different proposition from one whose crossover is at 2,000 K.
What Spontaneity Does Not Mean
This is the section worth reading twice, because "spontaneous" is a technical term that misleads almost everyone who meets it in plain English.
It says nothing about rate. A negative ΔG means the change is thermodynamically downhill. It does not say the system will get there this century. The conversion of diamond to graphite has a negative ΔG at room temperature and pressure; the activation barrier is so high that diamonds are stable on any human timescale. Petrol and air have a strongly negative ΔG together and sit quietly in a tank until something supplies an ignition source. Rate is kinetics, governed by activation energy and mechanism, and thermodynamics is silent on it.
It says nothing about needing no input. "Spontaneous" does not mean unaided. Every reaction needs to get started, and many favourable reactions are run with heating simply to make them fast enough to be useful.
ΔG° and ΔG are different quantities. The tabulated standard value applies at one bar and, for solutions, one molar for every species. Real mixtures are nowhere near that, and the actual driving force is ΔG = ΔG° + RT ln Q, where Q is the reaction quotient. A reaction with a positive ΔG° can proceed perfectly well if the products are removed as they form, which is exactly what the Haber process does industrially and what coupled reactions do in biochemistry.
The sign belongs to a direction, not to a substance. Reverse the reaction as written and every sign flips. A ΔG of −33 kJ/mol for making ammonia is a ΔG of +33 kJ/mol for decomposing it. Always know which direction the number was quoted for.
Zero ΔG is not nothing happening. At equilibrium the forward and reverse processes both continue at equal rates. ΔG = 0 describes a balance, not a stop.
Why the Ammonia Crossover Explains an Industrial Compromise
The worked example above is the classic illustration of why thermodynamics alone never decides a process. Ammonia synthesis has a negative ΔG° at room temperature and a crossover near 464 K, which says the yield is best cold. The reaction is also extremely slow cold, because the nitrogen triple bond is hard to break, so the equilibrium position that thermodynamics promises is unreachable in any practical time.
The industrial answer is to accept a worse equilibrium in exchange for a workable rate: run hot enough for the kinetics, use an iron catalyst to lower the barrier further, and then claw back the yield thermodynamics has cost you by raising the pressure, which favours the side with fewer gas molecules. None of that changes ΔH or ΔS. It changes which compromise you accept, and the crossover temperature is what tells you how much you are giving away by going hot.
The general lesson carries over to any temperature-dependent process. When ΔH and ΔS have the same sign, there is a tension between yield and rate, and the crossover is the number that quantifies it.
Getting ΔH and ΔS From Tables Without Introducing Errors
Reaction values are assembled from tabulated per-substance values, and there are two traps in doing so. For enthalpy you use standard enthalpies of formation, ΔHf°, combined as the sum over products minus the sum over reactants, each multiplied by its stoichiometric coefficient — so the coefficients from a balanced equation matter, and our chemical equation balancer is the place to settle those first. An element in its standard state has ΔHf° of exactly zero by definition.
For entropy you use absolute entropies, S°, not entropies of formation, and this is the trap: absolute entropies are never zero for an element at 298 K, because the third law sets zero at absolute zero rather than at 25 °C. Forgetting this and treating elemental entropies as zero the way you treat elemental formation enthalpies produces a ΔS that is wrong by a large margin. The LibreTexts treatment of Gibbs free energy works through the combination step with examples.
The second trap is temperature. Tabulated ΔH° and ΔS° are 298.15 K values, and using them at 800 K assumes both are temperature-independent. That approximation is usually acceptable over a few hundred kelvin because the heat capacity difference between products and reactants is small, but it degrades, and it degrades fastest across a phase change. Treat a crossover temperature far from 298 K as an estimate rather than a measurement. If you are also tracking a first-order rate process at the same time, our half-life calculator handles the kinetics side, and the scientific notation converter is useful for equilibrium constants that come out with large exponents.
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Browse Free Tools Contact Arb DigitalCommon Mistakes to Avoid
- Mixing kJ and J — enthalpy is tabulated in kJ/mol and entropy in J/(mol·K), a factor of 1,000 apart, and forgetting to reconcile them makes the entropy term vanish or explode.
- Using Celsius in the equation — the T in TΔS is absolute, so 25 °C is 298.15 K, and using 25 makes the entropy term twelve times too small.
- Treating elemental absolute entropies as zero — that rule applies to enthalpies of formation, not to S°, which is non-zero for every substance above absolute zero.
- Reading a negative ΔG as a promise of speed — thermodynamics gives direction, kinetics gives rate, and the two are entirely independent.
- Quoting a crossover temperature as exact — it assumes ΔH and ΔS do not change with temperature, which is an approximation that weakens the further you get from 298 K.
Related Free Tools From Arb Digital
Settle the stoichiometry first with the chemical equation balancer, work out formula masses with the molar mass calculator, convert quantities with the moles to grams calculator, and switch between energy units with the energy converter. For temperature conversions outside this page, the temperature converter covers kelvin, Celsius and Fahrenheit.
Frequently Asked Questions
It means the change is thermodynamically favoured at that temperature and pressure — the system can move in that direction without external work being done on it. It says nothing about how fast the change will occur. Diamond converting to graphite has a negative ΔG at room temperature and takes geological time.
Use ΔG = ΔH − TΔS with T in kelvin. Convert the entropy to the same energy unit as the enthalpy first, since ΔH is normally tabulated in kJ/mol and ΔS in J/(mol·K). For ΔH = −92.22 kJ/mol and ΔS = −198.75 J/(mol·K) at 298.15 K, ΔG comes to −32.96 kJ/mol.
It is the temperature at which ΔG passes through zero, found by dividing ΔH by ΔS. Below it one term dominates and above it the other does, so the reaction reverses its favourability. It only exists as a real temperature when ΔH and ΔS share the same sign.
Because the T in TΔS multiplies an absolute quantity and the equation is only valid on an absolute scale. Using 25 instead of 298.15 shrinks the entropy term by a factor of twelve, and using a negative Celsius value flips its sign, which changes the conclusion entirely.
ΔG° is the value under standard conditions — one bar for gases and one molar for solutes. ΔG is the value under the conditions you actually have, and the two are linked by ΔG = ΔG° + RT ln Q, where Q is the reaction quotient. A reaction with a positive ΔG° can still proceed if the products are continuously removed.
Yes, from ΔG° = −RT ln K, so K equals e raised to the power of −ΔG° divided by RT, with R as 8.3145 J/(mol·K). This only applies to the standard-state value. A ΔG° of −32.96 kJ/mol at 298.15 K corresponds to a K of roughly 6 × 10⁵.
No. It means the system is at equilibrium, with the forward and reverse processes proceeding at equal rates so there is no net change. Both directions continue at the molecular level. A phase change at its transition temperature is the everyday example.
It is an approximation. Tabulated ΔH° and ΔS° values are referenced to 298.15 K, and treating them as constant assumes the heat capacity difference between products and reactants is negligible. That usually holds over a few hundred kelvin but breaks down across a phase change, so a distant crossover temperature is an estimate.
This tool is provided for educational and study use. It performs the arithmetic of a thermodynamic relationship only, and nothing on this page is laboratory, process, handling or chemical safety guidance.
