Gibbs Free Energy Calculator

Gibbs free energy decides whether a reaction goes

The Gibbs free energy change of a reaction is ΔG = ΔH − TΔS: the enthalpy change minus the absolute temperature times the entropy change. A negative ΔG means the reaction can go forward on its own at that temperature, a positive ΔG means the reverse reaction is the one that can, and zero means equilibrium. For the ammonia synthesis loaded above, N₂ + 3H₂ → 2NH₃ at 298.15 K, ΔG = −91.8 − 298.15 × (−0.1981) = −32.7 kJ/mol, so it is spontaneous.

Enter any three of the four values and choose Solve for this on the fourth. ΔH goes in kJ/mol and ΔS in J/(mol·K), the units data tables use, and the calculator divides ΔS by 1000 before combining them, which the working shows. To find the temperature at which a reaction becomes spontaneous, set ΔG to 0 and solve for T. The readouts say whether the reaction is spontaneous at the temperature entered, which of the four sign cases it falls in, the temperatures at which it is spontaneous, and the equilibrium constant a standard ΔG° implies.

ΔG weighs two tendencies against each other. ΔH is the heat the reaction gives out or takes in, and giving out heat favours a reaction. ΔS measures how much more spread out matter and energy become, and spreading out favours it too, by an amount the temperature multiplies. That is why the temperature matters at all: it sets how much the entropy term counts against the enthalpy term. IUPAC calls the quantity the Gibbs energy, G = H − TS, after Josiah Willard Gibbs, who defined it in 1876. UK A-level courses say feasible where this page says spontaneous: a reaction is feasible when ΔG is zero or negative.

The four sign cases

Because only the entropy term changes with temperature, the signs of ΔH and ΔS settle most of the question before any arithmetic. Where they favour the same direction, no temperature can change the answer. Where they pull against each other, one temperature, T = ΔH ÷ ΔS, divides the range where the reaction is spontaneous from the range where it is not.

Carbon burning to carbon monoxide gives out heat and turns one molecule of gas into two, so both terms favour it. Oxygen turning into ozone takes in heat and packs three molecules into two, so both oppose it, and ozone forms only where energy is put in from outside, as ultraviolet light does in the upper atmosphere. Freezing, condensing and reactions that give out heat while turning more molecules of gas into fewer are the third case; melting, boiling and decompositions that take in heat and release gas are the fourth.

Worked example: ammonia synthesis

For N₂ + 3H₂ → 2NH₃, the tables give ΔH° = 2 × (−45.9) = −91.8 kJ/mol and ΔS° = 2 × 192.8 − 191.6 − 3 × 130.7 = −198.1 J/(mol·K). Is it spontaneous at 25 °C, and up to what temperature?

• Convert ΔS first: −198.1 J/(mol·K) ÷ 1000 = −0.1981 kJ/(mol·K).
• At 298.15 K, TΔS = 298.15 × (−0.1981) = −59.06 kJ/mol, so ΔG° = −91.8 − (−59.06) = −32.74 kJ/mol. Spontaneous, and the readout gives K = 5.43 × 10⁵.
• The crossover is T = ΔH ÷ ΔS = −91.8 ÷ (−0.1981) = 463.4 K, about 190 °C. Below it ΔG° is negative; above it the entropy term wins, and ammonia tends to break back down into nitrogen and hydrogen.

Industrial plants still run at 400 to 500 °C, well above the crossover, because the reaction is too slow at lower temperatures even over a catalyst, and they use pressures of about 150 to 250 atm to push the equilibrium back towards ammonia (OpenStax Chemistry 2e, section 13.3). At 450 °C the calculator gives ΔG° = +51.5 kJ/mol and K = 1.92 × 10⁻⁴. The NIST-JANAF tables, which follow ΔH and ΔS as they change with temperature, give about +60 kJ/mol there and a K near 5 × 10⁻⁵, about four times smaller: the size of error to expect from 298 K values used more than 400 K away.

Joules against kilojoules

Tables give enthalpies in kJ/mol and entropies in J/(mol·K), so ΔS has to be divided by 1000 before TΔS is taken away from ΔH. Leave that out and the ammonia example gives −91.8 − 298.15 × (−198.1) = +58,972: the entropy term is a thousand times too big, and even the sign of ΔG comes out wrong. That one is easy to spot. For a reaction with a small ΔS the wrong answer can look ordinary: diamond turning into graphite, with ΔH = −1.89 kJ/mol and ΔS = +3.36 J/(mol·K), comes out at −1003.7 kJ/mol instead of −2.9. The calculator takes ΔS in joules and converts it, and it warns when ΔS is under 1 J/(mol·K), which is what a value already converted to kJ/(mol·K) looks like.

ΔG° and the equilibrium constant

When ΔH and ΔS are standard values, as in a data table, the ΔG they give is the standard ΔG°, with every gas at 1 bar and every dissolved species at 1 mol/L, and it fixes the equilibrium constant: ΔG° = −RT ln K, or K = e^(−ΔG°/RT) with R = 8.314 J/(mol·K). A negative ΔG° gives K above 1 and a positive one K below 1, and at 298.15 K every 5.71 kJ/mol is a factor of ten in K.

A positive ΔG° does not mean nothing happens. Water evaporating at 25 °C has ΔG° = 44.01 − 298.15 × 0.1188 = +8.59 kJ/mol, yet puddles dry. The K that goes with it is 0.0313, and because liquid water counts as 1 in K, that is the pressure of water vapour at equilibrium in bar: 3.13 kPa, close to the 3.17 kPa measured at 25 °C. Evaporation outpaces condensation until the air above the water holds that much vapour.

Real mixtures are rarely at standard conditions, and ΔG = ΔG° + RT ln Q corrects for that, Q being the reaction quotient built from the actual pressures and concentrations. A cell potential gives ΔG° as well, through ΔG° = −nFE°, which the standard reduction potentials page works through for the Daniell cell.

Common mistakes

• Mixing joules and kilojoules. ΔS from a table is in J/(mol·K) and ΔH in kJ/mol. Divide ΔS by 1000 before multiplying by T, or the entropy term swamps everything else.
• Using Celsius as if it were kelvin. Typing 25 where 298.15 K is meant gives ΔG = −86.8 kJ/mol for ammonia synthesis instead of −32.7, a number that looks just as believable. The unit menu beside T is the guard.
• Reading ΔG° as ΔG. ΔG° describes every species at its standard state. Whether a particular mixture reacts depends on ΔG = ΔG° + RT ln Q, which can have the opposite sign.
• Forgetting that the numbers belong to the equation as written. Halving N₂ + 3H₂ → 2NH₃ halves ΔH, ΔS and ΔG and takes the square root of K; reversing it flips every sign and turns K into 1/K.
• Taking spontaneous to mean fast. Diamond turning into graphite has ΔG° = −2.9 kJ/mol at 25 °C and does not happen on any human timescale. How fast a reaction goes depends on its activation energy, which the Arrhenius equation calculator puts numbers on and the collision theory simulator shows.
• Trusting 298 K values far from 298 K. ΔH and ΔS drift with temperature, and the further a question sits from 25 °C the more the answer is an estimate.

What this calculator does not cover

It holds ΔH and ΔS fixed as the temperature changes, the approximation textbooks make. Water shows its size: its 25 °C values put the boiling point at 370.5 K, but under 1 bar water boils at 99.6 °C, which is 372.8 K. By then the enthalpy of vaporisation has fallen from 44.01 to 40.67 kJ/mol (OpenStax Chemistry 2e, section 10.3) and the entropy of vaporisation from 118.8 to 109.1 J/(mol·K), and their ratio has moved with them. The calculator warns when the temperature is more than 100 K from 298.15 K.

It does not look values up or add them for you. ΔH and ΔS for a reaction are products minus reactants from a table such as OpenStax Appendix G or the NIST Chemistry WebBook; in the lab, ΔH comes from heat measured in a calorimeter, which the calorimetry calculator works out. At a melting or boiling point ΔG = 0, so ΔS = ΔH ÷ T, and the latent heat calculator gives the heat that phase change takes. It also works only with the values entered, not with RT ln Q, and it says nothing about rate.

Converting units first? Use the temperature conversion table.