Gibbs Free Energy Calculator

What Gibbs free energy measures

Gibbs free energy answers one question about a process: at this temperature, does it want to run forward on its own? The quantity is written G and defined as G = H - T S, where H is enthalpy (the heat side), S is entropy (the disorder side) and T is the absolute temperature. Only changes matter in practice, so every formula here works with delta G, delta H and delta S.

At a fixed temperature and pressure, change in G equals change in H minus T times change in S. The first term pays for the heat the process needs or gives off. The second term is the temperature-weighted entropy the process creates or destroys. Whichever term is bigger sets the sign. A negative delta G means the forward direction is downhill and products are favored, which chemists call spontaneous or exergonic. A positive value means the reverse is favored. Zero means the system sits at equilibrium.

The main formula, and the unit trap inside it

The working equation is delta G = delta H - T delta S. Feed it enthalpy in kJ per mole, entropy in J per mole per kelvin, and temperature in kelvin, and it returns free energy in kJ per mole. The trap is the mismatch: tables list heat changes in kilojoules but entropy in joules, a factor of 1,000 apart. Divide the entropy term by 1,000 before subtracting, or the answer will be nonsense.

Worked example. Limestone breaks down into quicklime and carbon dioxide: CaCO3 gives CaO plus CO2. Standard values are delta H = +178.3 kJ per mole and delta S = +160.5 J per mole per kelvin, which is 0.1605 kJ per mole per kelvin. At 298 K: delta G = 178.3 - 298 x 0.1605 = 178.3 - 47.8 = 130.5 kJ per mole. Positive, so at room temperature the rock holds together and the reaction runs the other way.

The same reaction, hotter. At 1200 K: delta G = 178.3 - 1200 x 0.1605 = 178.3 - 192.6 = -14.3 kJ per mole. The sign has flipped. That is exactly what a lime kiln does when it runs above 900 C, and the calculator returns that same 130.5 at 298 K if you enter 178.3, 160.5 and 298.

The temperature where the sign flips

Set delta G to zero and rearrange: delta H = T delta S, so T = delta H / delta S. That crossover temperature is where a reaction stops being uphill and starts being downhill. It only exists when delta H and delta S have the same sign. If they have opposite signs, the reaction is favored on one side of the thermometer and never changes sides.

Keep the units straight here too: multiply delta H by 1,000 before dividing by delta S in joules per kelvin, or convert delta S to kJ per mole per kelvin. The answer comes out in kelvin, and subtracting 273.15 turns it into degrees C.

The crossover temperatures of four familiar changes, using standard values at 25 C:

The water rows land almost exactly on the real melting and boiling points, a good check on the arithmetic. The small gap comes from treating delta H and delta S as fixed numbers when both drift with temperature. For limestone the calculated 838 C matches what kiln operators see.

Reading the verdict: the four sign cases

Every reaction falls into one of four buckets, and knowing the bucket saves you a calculation. When delta H is negative and delta S is positive, both terms push the same way and the reaction is favored at every temperature.

When delta H is positive and delta S is negative, both terms fight you and the reaction is never favored at any temperature. When the signs are mixed, temperature decides. A negative delta H with a negative delta S, such as water freezing, is favored at low temperature, because the heat term wins when T is small. A positive delta H with a positive delta S, such as ice melting, is favored at high temperature, because the entropy term grows with T until it overtakes the heat term.

This is why an ice cube melts on a kitchen counter without any help: at 22 C the entropy term beats the 6.01 kJ per mole needed to break the lattice. In a freezer at -18 C the heat term wins, so the cube keeps its shape.

From delta G to the equilibrium constant

Standard free energy and the equilibrium constant are two views of the same fact. The bridge is delta G standard = -R T ln K, with R = 8.314462618 J per mole per kelvin (NIST CODATA 2022) and T in kelvin. A negative standard delta G means K is above 1, so products dominate at equilibrium.

At 25 C the whole scale collapses into a handy rule: every 5.7 kJ per mole of standard free energy changes K by a factor of 10. The calculator uses the full formula instead of the shortcut.

ATP hydrolysis is the standard biology example: delta G standard is about -30.5 kJ per mole at 298 K, which puts K near 2.2 x 10^5. Cells spend energy rebuilding ATP so they can spend it again.

One caution: the standard value describes 1 M solutes and 1 bar gases, and a real cell is nowhere near that. The actual value is delta G = delta G standard + R T ln Q, where Q is the current ratio of products to reactants.

Free energies of formation do the bookkeeping

For a reaction you cannot measure directly, tables do the work. The standard free energy of formation, delta Gf, is the free energy change when one mole of a compound forms from its elements in their standard states. Elements in their usual form get a value of zero, which is why oxygen and graphite never appear in the sums.

The rule is the same one used for enthalpy: add the formation values of the products, add the formation values of the reactants, and subtract. Products minus reactants, with each value multiplied by its coefficient.

Worked example. Burn one mole of methane: CH4 + 2 O2 gives CO2 + 2 H2O liquid. Products: -394.4 + 2 x (-237.1) = -868.6 kJ. Reactants: -50.8 + 0 = -50.8 kJ. Then delta G standard = -868.6 - (-50.8) = -817.8 kJ per mole of methane. Methane has a molar mass near 16 g per mole, so that is about 51 kJ per gram of gas, and it explains why a gas furnace heats a house so easily.

Where this shows up outside the classroom

Muscles. ATP hydrolysis releases about 30.5 kJ per mole under standard conditions, and a working muscle burns through kilograms of ATP in a day, rebuilding it continuously. The free energy does not become motion directly; it gets passed through coupled reactions that make the unfavored steps in a muscle fiber possible.

Cement and lime. Limestone decomposition needs 178.3 kJ per mole and only turns favorable near 1,111 K, which is why a cement plant burns fuel at 1,400 C or more. The temperature is not a suggestion; below the crossover the reaction cannot run on its own no matter how long you wait.

Cold packs and warmers. An instant cold pack works because dissolving ammonium nitrate absorbs heat, and the entropy of mixing makes the process favorable even though it gets cold. A hand warmer runs the other way: rusting iron releases heat and the free energy change is negative, so it works without any input.

Batteries. The voltage on a cell is free energy per unit charge, so 1.5 V means 1.5 joules per coulomb pushed through the circuit.

What the calculator leaves out

Every number on this page assumes delta H and delta S stay constant over the temperature range you ask about. They do not, exactly: heat capacities change and phases change. For a rough answer the assumption is fine; for process design, tables of temperature-dependent values replace the single number.

The calculator also reports standard values unless you supply real concentrations. A reaction with a positive standard delta G can still run in a cell if the products are removed fast enough to keep Q small, and the R T ln Q term is what accounts for that.

Finally, a negative delta G does not promise a reaction you can watch. It promises that the products are lower in free energy than the reactants. How fast the system gets there depends on the activation barrier, the temperature, and whether a catalyst or a surface is around.

Values here come from standard thermodynamic tables, the NIST Chemistry WebBook and the NIST CODATA 2022 constants list, which fixes R at 8.314462618 J per mole per kelvin. The OpenStax Chemistry 2e chapter on free energy walks through the same formulas, and the NIST constants database is where the gas constant comes from.