Entropy Calculator

Solve reaction entropy, Gibbs free energy, and isothermal ideal-gas entropy change in either direction.

Supports 13 volume, 12 pressure, 9 energy and 3 temperature units, with the exact SI gas constant.

Updated October 4, 2026

CreatorFrank Zhao

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What this calculator does

Entropy questions arrive in three quite different shapes, and each one has its own working.

You have tabulated entropies for the products and the reactants of a reaction and need the reaction’s entropy change.
You know the enthalpy change and the entropy change of a process and need its Gibbs free energy — or the temperature at which that verdict flips.
An ideal gas expands or compresses at constant temperature, and you need the entropy change that goes with it.

All three are here, and every one of them works in both directions. Leave any one box empty and the calculator solves it, whether that means recovering the reactants, the temperature, the amount of gas, or the final volume. Energy fields accept joules, kilojoules, megajoules, watt hours, kilowatt hours, foot-pounds, kilocalories and electronvolts; volume fields cover thirteen scales from cubic millimetres to UK fluid ounces; pressure fields cover twelve from pascals to gigapascals.

The one thing to know before you start

The Change in entropy box is the same box in both the Gibbs group and the ideal-gas group. It is one quantity, not two, and it carries the units $\mathrm{J\,K^{-1}}$ . The reaction group above it uses a different quantity again — entropy per mole, in $\mathrm{J\,mol^{-1}\,K^{-1}}$ . Mixing those two up is the single most expensive mistake in this topic, so it has a section of its own below.

How to use it, step by step

Start from whatever your problem actually gives you — everything else follows from the boxes you already have.

Pick the group that matches your question

Reaction entropies come from tables and live in the first group. Anything involving free energy, spontaneity or a temperature belongs in the Gibbs group. A volume or pressure ratio for a gas belongs in the ideal-gas group.

For the ideal-gas group, choose the base variable first

The Base variable menu decides whether the entropy change is driven by a volume ratio or a pressure ratio. Expansion and compression are the same physical event described two ways, and the two forms carry opposite signs — so pick the pair you actually measured. The group swaps between Initial volume / Final volume and Initial pressure / Final pressure, and the other pair leaves the calculation.

Enter values in whatever units you have them in

Each field has its own unit menu, and the conversion happens before the formula runs. Set the menu first, then type the number the way you would write it down. Blue text means the calculator produced that value rather than you.

Read the verdict under the Gibbs group

Once the Gibbs group has an answer, a one-line note appears beneath it: the process is spontaneous, at equilibrium, or nonspontaneous. That is the sign of $\Delta G$ , and it is the whole point of computing it.

Worked walkthrough

One mole of ideal gas expands isothermally from 1.00 L to 2.00 L. Leave Base variable on volume, put 1 in Amount of moles, and set both volumes in litres:

Isothermal entropy change from a volume ratio

\Delta S = n R \ln\!\left(\frac{V_2}{V_1}\right)

=

(1)(8.314462618)

\ln\!\left(\frac{2.00}{1.00}\right)

=

5.76315\ \mathrm{J\,K^{-1}}

The gas constant is the exact 2019 SI value, which the NIST Reference on Constants, Units and Uncertainty lists as $8.314\,462\,618\ldots\ \mathrm{J\,mol^{-1}\,K^{-1}}$ . You may see $8.3145$ quoted in older tables; the calculator uses the full value.

What the sign is telling you

Positive, because the gas had more room and therefore more accessible microstates — that is the second law showing up in a single number. The same Change in entropy box now reads 5.76315 J/K in the Gibbs group too. Add an enthalpy change of, say, 0 and a temperature of 298 K there and you get a Gibbs free energy of −1.72 kJ, which is why the expansion runs on its own once the partition is removed.

The three formulas and their reverse directions

Each group contains one relationship, and each relationship can be solved for any of its variables. What differs is how much you have to supply.

Reaction entropy change

\Delta S_{\mathrm{rxn}}

=

\Delta S_{\mathrm{products}}

-

\Delta S_{\mathrm{reactants}}

Give any two of the three and you get the third. A negative result means the products are more ordered than the reactants, which is the usual outcome for reactions that reduce the number of gas molecules.

Gibbs free energy

\Delta G

=

\Delta H

-

T \cdot \Delta S

Here you need three of the four boxes. Two are not enough, because one equation cannot pin down two unknowns — leave two empty and nothing is solved. With three supplied, the fourth follows:

Reverse directions

\Delta H = \Delta G + T \cdot \Delta S

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

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

The temperature version is the interesting one

$T = (\Delta H - \Delta G)/\Delta S$ answers a practical question: at what temperature does this process stop being spontaneous? For a reaction where both $\Delta H$ and $\Delta S$ are negative, the quotient is a real temperature above absolute zero, and above it the reaction runs on its own. If the calculator hands back a negative temperature, the inputs describe a combination that cannot happen — and it will say so rather than quietly returning a number.

Isothermal ideal-gas entropy change

\Delta S

=

n \, R \, \ln

\left(\frac{V_2}{V_1}\right)

\Delta S

=

-n \, R \, \ln

\left(\frac{P_2}{P_1}\right)

The two forms describe the same process, which is why the signs differ: pressure and volume move in opposite directions. Compression raises pressure, lowers volume, and gives a negative entropy change in both formulations.

Again, supply three of the four and the fourth is solved — the amount of gas, the final volume, the initial volume, or the final pressure.

A useful sanity check

Real entropy-of-vaporization values sit in the tens of $\mathrm{J\,mol^{-1}\,K^{-1}}$ . The NIST Chemistry WebBook gives methane an entropy of vaporization of 85.58 J/(mol·K) at 99.54 K. If your own answer comes back in the thousands, a unit is wrong — almost always calories where joules were expected.

Which entropy goes in which box

Two of the three groups carry a box labelled Change in entropy, and they are not the same quantity. This is the part worth two minutes.

Molar entropy versus total entropy

Reaction group

Entropy per mole. Tabulated values are written per mole of substance, so the units end in $\mathrm{J\,mol^{-1}\,K^{-1}}$

This group also offers calories in its unit menu, because molar entropies are still tabulated that way in a lot of older reference data.

Gibbs and ideal-gas groups

Entropy of the actual sample, in $\mathrm{J\,K^{-1}}$

This group omits calories, since that scale is far too small for whole-sample energies.

Copying a molar entropy straight from the reaction group into the Gibbs group understates the entropy term by however many moles are actually present — a factor of 1000 for a single mole. Nothing in the calculator can catch that for you, because the two boxes differ by a factor the calculator has no way of knowing. Check the unit suffix next to each box before you accept a result.

One entropy, two groups

The Gibbs group and the ideal-gas group share a single Change in entropy box. It appears in both places, and the two always show the same number. That is deliberate: the isothermal expansion entropy and the $\Delta S$ in $\Delta G = \Delta H - T\Delta S$ are the same physical quantity for the same process.

Fill in moles and the volume or pressure pair and the entropy is solved for you. Because the same value sits in the Gibbs group, adding an enthalpy change and a temperature there completes $\Delta G$ with no copying.
Type into Change in entropy yourself and the ideal-gas group rotates: it solves for the amount of gas, or for whichever volume or pressure you left blank.
If you go back and change an input the ideal-gas group depends on, a value you typed into Change in entropy turns blue and is replaced by the solved one. Blue is the signal that the calculator is now driving that box rather than you.

Worked examples

Reaction entropy from tabulated values

Suppose your table gives total product entropies of 353.6 J/(mol·K) and total reactant entropies of 596.7 J/(mol·K). Enter both and the reaction entropy follows:

\Delta S_{\mathrm{rxn}}

=

353.6

-

596.7

=

-243.1

Negative, so the products are more ordered than the reactants. For anything that reduces the number of gas molecules — two moles of gas becoming one, say — that negative sign is the expected result rather than a warning sign.

Isothermal compression, done through pressure

Two moles of gas are compressed isothermally from 2 bar to 8 bar. Set Base variable to pressure, enter 2 for the amount of moles, and 2 and 8 for the pressures:

\Delta S = -nR\ln\!\left(\frac{P_2}{P_1}\right)

=

-(2)(8.314462618)

\ln\!\left(\frac{8}{2}\right)

=

-23.0526\ \mathrm{J\,K^{-1}}

Had you run the same compression as a volume ratio — one fifth of the starting volume — you would get exactly the same negative answer from the first form. If you had entered it the other way round and got a positive number, the base variable is on the wrong setting.

An ideal gas at constant temperature has no enthalpy change

For an ideal gas, enthalpy depends only on temperature. An isothermal process therefore has $\Delta H = 0$ , and the Gibbs equation collapses to $\Delta G = -T\Delta S$ . Enter 0 for the enthalpy change, 298 K for the temperature, and let the 1.00 L to 2.00 L expansion from the earlier walkthrough supply the entropy:

\Delta G

=

0 - (298)(5.76315)

=

-1717.42\ \mathrm{J}

Negative, so the expansion is spontaneous. This is the cleanest demonstration that the two groups describe one process: the same entropy number now produces a Gibbs free energy with no extra input beyond the temperature.

Working backwards to find a temperature

Phase changes give a useful check. A process sits at equilibrium when $\Delta G = 0$ , so its equilibrium temperature is just $T = \Delta H / \Delta S$ . Take an enthalpy change of 8.517 kJ and an entropy change of 85.58 J/K, set the temperature unit to kelvin, and leave the temperature and Gibbs free energy boxes to be solved:

T = \frac{8517\ \mathrm{J}}{85.58\ \mathrm{J\,K^{-1}}}

=

99.5209\ \mathrm{K}

A little below the freezing point of water, and a sensible temperature at which liquid and vapour methane are in equilibrium. Going the other way is just as easy: fix the temperature at that value and the calculator returns an enthalpy change of 8.51863 kJ for the same entropy. If you want the full boiling-point calculation rather than the entropy part, the Boiling Point Calculator works from tabulated heat and pressure data instead.

Limitations and disclaimers

This is a working tool for textbook and coursework thermodynamics, and for quick sanity checks on real numbers. It evaluates relationships that hold cleanly; it does not model the systems behind them.

What this model leaves out

Ideal gases only. The ideal-gas group uses $PV = nRT$ . Near the critical point or at high pressure, real gases deviate enough that the isothermal entropy change is not this simple logarithm. Check a compressibility factor before trusting a result at extreme pressure.
Isothermal means isothermal. The formula has no temperature term because it assumes the temperature does not change. Adding heat to the gas invalidates it; use the enthalpy change instead.
One temperature at a time. $\Delta H$ and $\Delta S$ are treated as constants evaluated at the temperature you supply. Heat capacities are not built in, so you cannot ask how $\Delta G$ varies between two temperatures without supplying values for both.
You supply the tabulated values. The calculator cannot look up standard molar entropies or enthalpies of formation. It also draws no distinction between a standard-state quantity and one measured at other conditions — if your problem needs standard-state corrections, apply them before entering anything.
Unchanged state gives zero, but an unchanging ratio can make a reverse solve impossible. Equal initial and final volumes correctly give an entropy change of zero. The opposite case bites: if you type an entropy change while the two states are identical, the calculator has to divide by $\ln(1) = 0$ and will report the result as undefined rather than invent a number.
Very large values are flagged, not hidden. Results far outside any real range stay on screen with a warning, so you can see the number you got and what to check — usually a unit or a misplaced decimal point.

Before you rely on the result

For process design, safety assessment or any decision with money or equipment behind it, check your inputs against current experimental data and your own plant or laboratory documentation. Mixing, activity coefficients, non-ideal behaviour and heat-transfer effects are all outside what this tool covers.

If you have a mass of gas and need the amount in moles before you can use the ideal-gas group, the Grams to Moles Calculator covers 25 common compounds, and Molar Mass of Gas Calculator gets you there from pressure, volume, temperature and mass instead. To turn measured equilibrium compositions into a constant of equilibrium, the Equilibrium Constant Calculator is the next step.

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