Gibbs Free Energy Calculator (ΔG = ΔH − TΔS)
Work out ΔG = ΔH − TΔS and whether a reaction is spontaneous. Solve for ΔG, ΔH, ΔS or the temperature, find the crossover temperature where spontaneity flips, convert between ΔG° and K, and get the actual ΔG from the reaction quotient Q.
How to Use
- Choose what to solve for: ΔG, ΔH, ΔS or the temperature, or switch to K from ΔG°, ΔG° from K, or ΔG from Q.
- Enter ΔH in kJ/mol, J/mol or kcal/mol and ΔS in J/(mol·K), kJ/(mol·K) or cal/(mol·K). The units are converted for you, so the classic kJ-against-J mistake cannot happen.
- Enter the temperature in °C, K or °F; it is converted to kelvin, and anything at or below absolute zero is rejected.
- Read ΔG, whether the reaction is spontaneous, the crossover temperature T = ΔH ÷ ΔS and K. The graph shows ΔG against temperature with the spontaneous region shaded, and the table marks which of the four sign cases applies.
- For K or Q, enter ΔG° and the temperature (and Q); Show Work gives each step of ΔG° = −RT ln K or ΔG = ΔG° + RT ln Q.
Worked Example
Ammonia synthesis at 25 °C. For N₂ + 3H₂ → 2NH₃, standard textbook (NBS) data at 298 K give ΔH° = 2 × (−46.11) = −92.22 kJ/mol and ΔS° = 2(192.45) − 191.61 − 3(130.684) = −198.76 J/(mol·K). ΔG° = −92.22 − 298.15 × (−0.19876) = −32.96 kJ/mol: spontaneous. Both terms are negative, so it is spontaneous only below T = −92,220 ÷ −198.76 = 464.0 K (190.8 °C).
From ΔG° to K. At 298.15 K, RT = 2.479 kJ/mol, so ln K = 32.96 ÷ 2.479 = 13.30 and K = 5.95 × 10⁵.
The common mistake: mixing kJ and J. ΔH is usually given in kJ/mol and ΔS in J/(mol·K). Typing −92.22 − 298.15 × (−198.76) gives +59,168, a huge positive ΔG that says the reaction is impossible. Convert ΔS to −0.19876 kJ/(mol·K) first and the answer is −32.96 kJ/mol. Using 25 instead of 298.15 for T is the other classic slip: it gives −87.25 kJ/mol.
Show Work
Formulas
Gibbs, Helmholtz and Free Energy
Josiah Willard Gibbs, professor of mathematical physics at Yale, published “On the Equilibrium of Heterogeneous Substances” in parts between 1875 and 1878 in the Transactions of the Connecticut Academy. In it he defined the function now called the Gibbs energy, G = H − TS, and showed that at constant temperature and pressure a system moves towards lower G. Hermann von Helmholtz reached the matching idea at constant volume, the Helmholtz energy, in 1882.
The relation ΔG° = −RT ln K joined thermodynamics to the equilibrium constants that Guldberg and Waage had measured, and Gilbert N. Lewis and Merle Randall’s 1923 textbook Thermodynamics and the Free Energy of Chemical Substances made tables of free energies a working tool for chemists. The ΔH and ΔS values in the presets are standard 298 K textbook data (NBS tables: NH₃ −46.11 kJ/mol and 192.45 J/(mol·K); ice 6.01 kJ/mol to melt; water 40.65 kJ/mol to boil at 100 °C); values differ slightly between sources, so the crossover for ice comes out at 273.2 K rather than exactly 273.15 K.
About This Tool
This calculator solves ΔG = ΔH − TΔS for any one of its four quantities, tells you whether the reaction is spontaneous, and finds the crossover temperature T = ΔH ÷ ΔS where that changes. The graph shows ΔG against temperature with the spontaneous region shaded, and the sign-case table highlights which of the four cases your reaction is. It also converts ΔG° to K and back, and works out the actual ΔG for any reaction quotient Q.
ΔH and ΔS are treated as constant with temperature, which is a good approximation over a few hundred kelvin but not exact. Everything runs in your browser; nothing is sent anywhere.
Related tools: Chemical Equilibrium (ICE Table) Calculator, Reaction Rate & Arrhenius Calculator, and Specific Heat Calculator.
Frequently Asked Questions
How do you calculate Gibbs free energy?
ΔG = ΔH − TΔS, with T in kelvin and ΔH and ΔS in matching energy units. For melting ice, ΔH = 6.01 kJ/mol and ΔS = 22.0 J/(mol·K) = 0.0220 kJ/(mol·K). At 25 °C (298.15 K), ΔG = 6.01 − 298.15 × 0.0220 = −0.549 kJ/mol, negative, so ice melts.
What is the crossover temperature?
The temperature where ΔG = 0, T = ΔH ÷ ΔS, at which spontaneity flips. For decomposing limestone, CaCO₃ → CaO + CO₂, ΔH = 178.3 kJ/mol and ΔS = 160.6 J/(mol·K), so T = 178,300 ÷ 160.6 = 1,110 K, about 837 °C; lime kilns run hotter than that. For ammonia synthesis it is 464.0 K, above which the reaction is no longer spontaneous.
When is a reaction spontaneous at all temperatures?
When ΔH is negative and ΔS is positive: both terms make ΔG negative. If ΔH is positive and ΔS negative it is never spontaneous. If both are negative it is spontaneous only below T = ΔH ÷ ΔS, and if both are positive only above it, as for melting ice (above 273 K) and boiling water (above about 373 K).
How are ΔG° and the equilibrium constant related?
ΔG° = −RT ln K, so K = e^(−ΔG°/RT). For ammonia synthesis at 25 °C, ΔG° = −32.96 kJ/mol gives K = e^(13.30) = 5.95 × 10⁵. At 25 °C every 5.708 kJ/mol of ΔG° changes K by a factor of 10, and K = 1.0 × 10⁻⁵ means ΔG° = +28.54 kJ/mol.
What is the difference between ΔG and ΔG°?
ΔG° is for standard conditions (1 bar, 1 M); the actual ΔG depends on the mixture through ΔG = ΔG° + RT ln Q. For ammonia synthesis at 25 °C with Q = 1.0 × 10⁶, ΔG = −32.96 + 2.479 × ln(10⁶) = +1.288 kJ/mol: Q is already past K = 5.95 × 10⁵, so the reaction runs in reverse despite its negative ΔG°.
How do I use the Gibbs Free Energy Calculator (ΔG = ΔH − TΔS)?
Simply type your numbers and read the result, which refreshes the instant you change something. There is nothing to submit and nothing to wait for.
Do I need to install or sign up for anything?
Not at all — it runs in the browser with nothing to install and no account. After it loads once, it even works without an internet connection.
Is my information private?
Yes. Everything happens in your browser. Nothing you type is sent to a server or saved anywhere.
Common Use Cases
Ammonia synthesis
ΔG = −32.96 kJ/mol at 25 °C, but +61.45 kJ/mol at 500 °C, which is why industry needs high pressure as well as heat for speed.
Melting and boiling
With ΔH 6.01 kJ/mol and ΔS 22.0 J/(mol·K) ice has its crossover at 273.2 K; water vapour, 40.65 kJ/mol and 108.9 J/(mol·K), at 373.3 K.
Lime kilns
Limestone decomposition has ΔG = +130.4 kJ/mol at 25 °C and −26.17 kJ/mol at 1,000 °C, crossing zero at 1,110 K.
Biochemistry
ATP hydrolysis, with the textbook ΔG°′ of −30.5 kJ/mol, has K = 1.37 × 10⁵ at 37 °C.
Equilibrium constants
K = 1.0 × 10⁻⁵ at 25 °C corresponds to ΔG° = +28.54 kJ/mol; every 5.708 kJ/mol is a factor of 10 in K.
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