The entropy change calculator above does the two jobs that the word entropy usually means in a chemistry course. It computes the standard entropy change of a reaction, ΔS°, by subtracting the summed standard molar entropies of the reactants from those of the products, each weighted by its stoichiometric coefficient. It also computes the entropy change of a physical process from the heat transferred reversibly at a constant temperature, which is the q/T definition that everything else is built on.
Arb Digital publishes free calculators that give one specific answer with no sign-up and no limits. This one exists because the arithmetic is short but the sign convention is where marks and lab reports go wrong. Entropy values are tabulated in joules per mole per kelvin while enthalpies are tabulated in kilojoules per mole, and mixing the two by a factor of a thousand is the single most common error in the whole of introductory thermodynamics.
What This Entropy Change Calculator Does
You type the standard molar entropy of each product and each reactant, with its coefficient from the balanced equation. The tool multiplies each entropy by its coefficient, sums both sides, and subtracts. The headline number is ΔS° in joules per mole per kelvin for the reaction as written. Underneath, a bar shows how each side of the equation contributes, so you can see immediately whether the reaction is creating disorder or destroying it.
The temperature box drives a second output: the −TΔS° term, converted to kilojoules per mole. That is the piece of the Gibbs energy expression that entropy contributes, and printing it in kilojoules is deliberate, because that is the unit it has to be in before you can add it to an enthalpy. Our Gibbs free energy calculator consumes a ΔS value you already have; this page is the one that produces it, which is the boundary between the two tools.
The lower pair of boxes handle the other definition. For a reversible change at constant temperature — melting, boiling, an isothermal expansion — the entropy change is simply the heat divided by the absolute temperature. The default figures melt one mole of ice: 6,010 joules absorbed at 273.15 K gives about 22.0 J/K, which is the textbook entropy of fusion of water.
Nothing here is an equilibrium calculation. If you want to know which way a gas-phase reaction sits, the equilibrium constant calculator is the tool for that. This page stops at the entropy term.
How to Use It
- Balance the equation first. The coefficients you type must be the ones from a balanced equation. If you are not sure, the chemical equation balancer will produce them.
- Look up S° for every species in the right physical state. Liquid water and gaseous water have very different standard entropies — 70.0 against 188.8 J/(mol·K) — and using the wrong one changes the sign of many answers.
- Enter each species as coefficient*value. Three moles of a substance with S° of 205.2 is written 3*205.2. A bare number is treated as a coefficient of one.
- Set the temperature if you want the −TΔS° term at something other than 298.15 K.
- Use the q/T boxes independently for a phase change or any other isothermal transfer. They do not interact with the reaction calculation.
The Formula and How It Is Calculated
For a reaction, the standard entropy change is a products-minus-reactants sum: ΔS° = Σ n S°(products) − Σ n S°(reactants), where n is the stoichiometric coefficient. Standard molar entropies are absolute rather than relative, which is what makes this different from the enthalpy version. The third law of thermodynamics fixes the entropy of a perfect crystal at absolute zero as zero, so every substance has a measurable absolute entropy above that point, and the tables list positive numbers for everything. Enthalpies of formation, by contrast, are defined relative to the elements and can be negative or zero. This is set out clearly in the LibreTexts treatment of entropy changes in chemical reactions.
Worked through with the default values, methane combustion gives products of 213.8 + 2 × 70.0 = 353.8 and reactants of 186.3 + 2 × 205.2 = 596.7, so ΔS° = −242.9 J/(mol·K). Three moles of gas become one mole of gas plus two moles of liquid, and the entropy falls sharply as a result. The −TΔS° term at 298.15 K is +72.4 kJ/mol, which opposes the reaction — and yet methane burns readily, because the enthalpy release of roughly −890 kJ/mol overwhelms it.
For a reversible isothermal process the definition is ΔS = qrev / T. The subscript matters. Only the heat exchanged along a reversible path gives the entropy change; an irreversible path between the same two states transfers less heat but produces the same entropy change, because entropy is a state function. Experimental standard molar entropies come from integrating measured heat capacity divided by temperature from near absolute zero upwards, and thermochemical data of that kind is collected in the NIST Chemistry WebBook.
Why Counting Moles of Gas Predicts the Sign
Before you look up a single number, you can usually predict the sign of ΔS° by counting gas molecules on each side. Gases have standard entropies an order of magnitude above solids, so the gas count dominates the sum. If the number of moles of gas increases, ΔS° is almost certainly positive; if it falls, negative; if it is unchanged, the answer is small and could go either way.
The Haber process, N₂ + 3H₂ → 2NH₂, takes four moles of gas to two, and its ΔS° is about −198 J/(mol·K). Decomposing calcium carbonate to calcium oxide and carbon dioxide creates a mole of gas from a solid, and its ΔS° is about +160 J/(mol·K). Neither number needs a table to sign correctly.
The prediction fails in exactly the cases worth remembering. When the gas count is equal on both sides, you are down to the difference between similar molecules and the sign genuinely depends on molecular complexity: more atoms and more available vibrational modes mean higher entropy per mole. Dissolving a salt is the other trap. It looks like an increase in disorder because an ordered lattice breaks apart, but small, highly charged ions organise a shell of water molecules around themselves so tightly that the net entropy change can be negative. Aluminium and magnesium salts do this routinely.
The Second Law Is About the Universe, Not Your Beaker
A reaction with a negative ΔS° happens all the time. Water freezes. Ammonia forms. Every crystallisation in a lab is a system losing entropy. The second law does not forbid any of this, because it constrains the total: ΔSuniverse = ΔSsystem + ΔSsurroundings must be greater than or equal to zero.
The surroundings term is the one people skip. When a reaction releases heat, that heat flows into the surroundings and raises their entropy by roughly −ΔH/T. An exothermic reaction with a modest entropy penalty in the system therefore still increases the entropy of the universe. Combine the two terms and multiply through by −T and you get exactly the Gibbs free energy criterion, ΔG = ΔH − TΔS. Gibbs energy is not a separate law; it is the second law rewritten so that you only have to think about the system.
That is why the −TΔS° box on this page prints kilojoules. It is the quantity you add to ΔH, and its size relative to the enthalpy is what decides whether temperature can flip a reaction from unfavourable to favourable.
When the Standard Entropy Change Stops Being Enough
Standard values assume every species is in its standard state at one bar and, by convention, at 298.15 K. Three things break that assumption in real work.
First, temperature. Standard entropies themselves change with temperature, following the integral of heat capacity over temperature. Over a modest range the change in ΔS° for a reaction is often small enough to ignore, because the product and reactant heat capacities partially cancel, and that approximation is what lets a first-year course use 298 K values at 500 K. Over hundreds of kelvin, or across a phase change of any participant, it fails badly.
Second, concentration and pressure. Standard entropy assumes pure substances or one-molar solutions. A gas at ten bar has a lower molar entropy than the same gas at one bar, by R ln(10) or about 19 J/(mol·K). That correction is small compared to a typical reaction ΔS° but it is not negligible near equilibrium.
Third, the entropy of mixing. Combining two ideal gases or forming an ideal solution increases entropy even when nothing reacts, and this term is why dissolution is often spontaneous with a near-zero enthalpy change. Our molarity calculator and solution dilution calculator handle the concentration bookkeeping that sits underneath these corrections.
Entropy, Disorder, and a Better Mental Picture
Describing entropy as disorder gets students to the right answer often enough to survive an exam and then quietly misleads them. A tidy room and a messy room have essentially the same entropy; the analogy has no thermodynamic content. The statistical definition is more useful: entropy counts the number of microscopic arrangements consistent with what you can measure about the system.
Under that picture, a gas has high entropy because its molecules have an enormous number of available positions and momenta. A large molecule has higher entropy than a small one at the same temperature because it has more vibrational modes to distribute energy across. Heating anything raises its entropy because more energy levels become accessible. Every one of those statements is about counting accessible states, and every one of them predicts the tabulated numbers correctly, which the disorder analogy does not.
Reading the Result Without Overstating It
Tabulated standard molar entropies are usually quoted to one decimal place, which supports about four significant figures. The subtraction at the heart of this calculation destroys precision, because two large numbers of similar size are being differenced. If your product and reactant sums are both near 600 and differ by 3, you have perhaps one meaningful figure left in the answer even though the inputs looked precise. Our significant figures calculator will keep a reported value honest, and the percent error calculator is useful when comparing a calculated ΔS to a measured one.
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Browse All Free Tools Suggest a ToolCommon Mistakes to Avoid
- Mixing joules and kilojoules — entropies are tabulated in J/(mol·K) and enthalpies in kJ/mol. Divide ΔS by 1,000 before adding it to a ΔH.
- Ignoring the physical state — H₂O(l) is 70.0 and H₂O(g) is 188.8 J/(mol·K). Picking the wrong one can reverse the sign of the whole reaction.
- Forgetting the coefficients — a species with a coefficient of three contributes three times its standard entropy, not one.
- Using Celsius in q/T — the temperature in an entropy expression is always absolute. At 0 °C you must divide by 273.15, not by zero.
- Assuming a negative ΔS° forbids the reaction — the second law applies to the universe. Exothermic reactions raise the entropy of the surroundings and routinely proceed with a negative system entropy change.
Related Free Tools From Arb Digital
Feed the result straight into the Gibbs free energy calculator to decide spontaneity, or work out how fast the reaction goes with the activation energy calculator. The specific heat calculator gives you the q that the second half of this page needs, and the temperature converter handles Celsius to kelvin. For the composition side of a problem, use the molar mass calculator and the moles to grams calculator. Everything else is listed on the free online tools hub.
Frequently Asked Questions
Standard entropy change is the sum of the standard molar entropies of the products minus the sum for the reactants, each multiplied by its stoichiometric coefficient. The result is in joules per mole per kelvin for the reaction exactly as it is balanced.
The third law of thermodynamics sets the entropy of a perfect crystal at absolute zero to zero, so every substance above that temperature has a positive absolute entropy. This is unlike enthalpy of formation, which is measured relative to the elements and can be negative.
It means the system has fewer accessible arrangements after the change than before, usually because moles of gas were consumed or a liquid or solid was formed. It does not prevent the reaction, because the surroundings can gain more entropy than the system loses.
For a reversible change at constant temperature, divide the heat transferred in joules by the absolute temperature in kelvin. Melting one mole of ice absorbs about 6,010 joules at 273.15 K, giving an entropy of fusion near 22 joules per kelvin per mole.
Usually yes, by counting moles of gas. Gases carry far more entropy per mole than liquids or solids, so if the gas count rises the entropy change is positive, and if it falls the change is negative. The shortcut fails when the gas count is unchanged.
Yes. Absolute entropies rise with temperature following the integral of heat capacity divided by temperature. The reaction entropy change is less sensitive because product and reactant heat capacities partly cancel, which is why 298 K values are often reused over moderate ranges.
Entropy is defined as heat divided by absolute temperature, so its unit is energy per unit temperature. The per-mole part simply scales the value to one mole of substance, which is how thermodynamic tables report it.
Not really. Entropy counts the number of microscopic arrangements consistent with the measured state of a system. The disorder analogy gets simple cases right but fails for dissolution of small ions, where broken lattice order is offset by tightly organised water molecules.
This calculator is provided for education and general reference. It describes how an entropy change is computed and is not laboratory or safety guidance; follow the procedures and risk assessments issued by your own institution.
