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Entropy Change Calculator

Chemistry

ΔS_universe = ΔS_system + ΔS_surroundings ≥ 0 for anything that actually happens. Find ΔS° for a reaction from standard entropies, ΔS for heating and phase changes, for ideal-gas processes and for mixing, and see whether the universe's entropy goes up.

Looking for information entropy in bits? Shannon Entropy Calculator →

Quick Examples:

Values from standard tables (298.15 K, 1 bar): NBS tables (Wagman et al., 1982), 298.15 K, 1 bar.

Reaction

N₂(g) + 3 H₂(g) → 2 NH₃(g)

S° in J/(mol·K); ΔfH° in kJ/mol (optional, needed for the surroundings). Picking a species fills both from the table; you can edit them.

Reactants

Products

Feeds the second-law panel only.
Standard reaction entropy, per mole of reaction as written
ΔS° = −198.76 J/(mol·K)
= −0.19876 kJ/(mol·K) = −47.51 cal/(mol·K)
Entropy decreases

ΔS° to 2 decimal places, set by N₂(g) (191.61).

ΔH° = −92.22 kJ/mol (ΔH° to 2 decimal places, set by NH₃(g).)

As expected: the reaction uses up gas (Δn(gas) = −2), and gas carries most of the entropy.

Balanced: every element matches on both sides.

  • Standard states: gases at 1 bar, solutes at 1 mol/kg. Real mixtures differ by −R·ln(activity) per mole.

Where ΔS° comes from

J/(mol·K) · products rise from 0, reactants fall, the last bar is ΔS°

The biggest term is 3 H₂(g) (−392.05 J/(mol·K)). Turning 4 mol of gas into 2 is what makes ΔS° negative.

Second-law ledger: system, surroundings, universe

ΔS_univ > 0: this happens on its own under standard conditions (irreversible).

ΔS, J/(mol·K)

At 298.15 K: ΔS_surr = −ΔH°(T)/T, the surroundings being a large reservoir at the reaction temperature.

Spontaneous below 463.97 K (190.82 °C). The surroundings' gain shrinks as T rises. (rough: assumes ΔH° and ΔS° don't change with temperature)

ΔG° = −T·ΔS_univ = −32.96 kJ/mol at 298.15 K. More free-energy calculations: Gibbs Free Energy Calculator →

Have K at two temperatures instead? Van 't Hoff Equation Calculator →

ΔS_univ against temperature

J/(mol·K)

Temperature (K)

ΔS_sys(T)

ΔS_surr(T) = −ΔH°(T)/T

ΔS_univ(T)

Below 463.97 K the heat released to the surroundings outweighs the system's entropy loss. Above it, it doesn't.

Step-by-Step Solution

ΔS° = Σν·S°(products) − Σν·S°(reactants)

= [2 × 192.45] − [1 × 191.61 + 3 × 130.684]

= 384.900 − 583.662 = −198.762 J/(mol·K) → −198.76 J/(mol·K)

ΔH° = [2 × (−46.11)] − [0 + 0] = −92.22 kJ/mol

ΔS_surr = −ΔH°/T = −(−92 220 J/mol) / 298.15 K = +309.31 J/(mol·K)

ΔS_univ = −198.76 + 309.31 = +110.55 J/(mol·K) > 0

ΔG° = −T·ΔS_univ = −298.15 × 110.545 / 1000 = −32.959 kJ/mol

ΔS_univ = 0 at T = ΔH°/ΔS° = −92 220 / −198.762 = 463.97 K (190.82 °C)

About This Tool

Entropy Change Calculator – ΔS for Reactions, Heating, Gases and Mixing

This entropy change calculator finds ΔS four ways: the standard reaction entropy ΔS° from tabulated S° values, the entropy gained by heating, melting and boiling, ΔS for ideal-gas expansions and compressions, and the entropy of mixing. Every result ends in the same second-law ledger: ΔS_univ = ΔS_sys + ΔS_surr, which is positive for anything that happens on its own.

What entropy measures

Rudolf Clausius named entropy in 1865 as heat transferred reversibly divided by temperature, dS = dq_rev/T. Ludwig Boltzmann later tied it to counting: S = k·ln W, where W is the number of microscopic arrangements consistent with what you see. "Disorder" is a loose metaphor; "how widely energy and matter are spread" is closer. Doubling the volume a gas can roam doubles the positions open to each molecule, which works out to ΔS = R·ln 2 = 5.763 J/(mol·K).

Absolute entropies and the third law

Formation enthalpies are measured relative to the elements, so every element has ΔfH° = 0. Entropy is different: the third law fixes S = 0 for a perfect crystal at 0 K, so S° values are absolute and an element's S° is never zero. O₂(g) carries 205.138 J/(mol·K). The one exception is aqueous ions, which are measured relative to H⁺(aq) = 0; small, highly charged ions such as Ca²⁺ and CO₃²⁻ order the water around them and end up with negative S°.

Predicting the sign of ΔS°

Gases carry most of the entropy, so the sign of ΔS° usually follows Δn(gas): ammonia synthesis turns 4 mol of gas into 2 and has ΔS° = −198.76 J/(mol·K). When no gas changes, smaller effects decide: solid → liquid → solution, more atoms per molecule, and the water that ions order or release. That is why calcite precipitating from Ca²⁺(aq) and CO₃²⁻(aq) raises the entropy by 202.9 J/(mol·K).

The second law in practice

A reaction that releases heat raises the surroundings' entropy by ΔS_surr = −ΔH/T. Adding the two gives ΔS_univ = −ΔG/T, so the entropy verdict and the Gibbs verdict always agree. Ammonia synthesis is favoured below about 464 K because its heat outweighs the lost gas; limestone only decomposes above about 1110 K, where the gained CO₂ outweighs the heat it costs. To work in free energies directly, use the Gibbs Free Energy Calculator.

ΔS° is per mole of reaction as written
Doubling every coefficient doubles ΔS°. Enter an amount of reaction ξ to get the entropy change in J/K for a real batch.

Heating, melting and boiling

Warming a substance adds m·c·ln(T₂/T₁); each phase change adds ΔH_trans/T_trans, always with temperatures in kelvin. On a T–S diagram the area under a reversible path equals the heat, which makes the large plateaus easy to see: boiling supplies 68 % of the entropy gained between −20 °C ice and 120 °C steam. Most liquids gain about 88 J/(mol·K) on boiling (Trouton's rule); water gains 109 J/(mol·K) because hydrogen bonds order the liquid. Vapor pressures and ΔH_vap fits belong in the Clausius-Clapeyron Calculator.

Irreversible changes

The system's ΔS depends only on its end states, so the tool finds it along a reversible route even when the real process is irreversible: supercooled water freezing at −10 °C, a gas rushing into a vacuum, or hot copper dropped into cold water. What depends on the path is the entropy generated, the part that lands in ΔS_univ. A free expansion and a reversible isothermal expansion give the gas the same +5.763 J/K; only the second leaves the universe unchanged.

Mixing and the Gibbs paradox

When ideal gases mix, each one expands into the total volume, giving ΔS_mix = −R·Σ nᵢ·ln xᵢ, which peaks at R ln 2 for a 50:50 binary mixture. Mixing a gas with more of the same gas creates nothing, which is the resolution of Gibbs's paradox: identical molecules have no "which one is where" to count. The same −Σ p·ln p form measures information, as the Shannon Entropy Calculator shows in bits.

Model limits
Standard values are for 298.15 K and 1 bar, heat capacities are constant, gases are ideal and mixtures are ideal. The calculator flags inputs where those assumptions start to bite.

Frequently Asked Questions

Is the Entropy Change Calculator free?

Yes, Entropy Change Calculator is totally free :)

Can I use the Entropy Change Calculator offline?

Yes, you can install the webapp as PWA.

Is it safe to use Entropy Change Calculator?

Yes, any data related to Entropy Change Calculator only stored in your browser (if storage required). You can simply clear browser cache to clear all the stored data. We do not store any data on server.

How do I calculate the entropy change of a reaction?

Use ΔS° = Σν·S°(products) − Σν·S°(reactants), multiplying each standard molar entropy by its coefficient. For N₂ + 3H₂ → 2NH₃ that is 2 × 192.45 − (191.61 + 3 × 130.684) = −198.76 J/(mol·K). The calculator fills S° from a built-in standard table and rounds by the addition rule.

Why isn't the standard entropy of an element zero?

S° is an absolute value measured up from 0 K (the third law), not a value relative to the elements like ΔfH°. Only a perfect crystal at 0 K has zero entropy, so O₂(g) has S° = 205.138 J/(mol·K). Aqueous ions can be negative because they are measured relative to H⁺(aq) = 0.

How do I calculate the entropy change when heating or melting a substance?

Split the path into steps. Each temperature change adds m·c·ln(T₂/T₁) and each phase change adds ΔH_trans/T_trans, with temperatures in kelvin. Taking 1 kg of ice at −20 °C to steam at 120 °C gives +8.849 kJ/K, and boiling alone is 68 % of it.

What is the entropy change for an ideal gas expansion?

ΔS = n·C_v,m·ln(T₂/T₁) + n·R·ln(V₂/V₁) for any path, because entropy is a state function. Doubling the volume of 1 mol at constant temperature gives R·ln 2 = +5.763 J/K whether the expansion is reversible, against a constant pressure or into a vacuum; only the surroundings' share changes.

How can a reaction with a negative ΔS be spontaneous?

The second law applies to the universe, not the system alone. An exothermic reaction heats its surroundings, raising their entropy by −ΔH/T. Burning methane lowers the system's entropy by 243 J/(mol·K) but raises the surroundings' by about 2986, so ΔS_univ is positive, which is the same verdict as ΔG < 0.

How accurate are the results?

The arithmetic is exact; the limits are the model. S° and ΔfH° are tabulated at 298.15 K, so far from that temperature add ΔCp° or treat the result as an estimate. Heat capacities are taken as constant, gases as ideal and mixtures as ideal, and the tool warns when an input pushes past those assumptions.