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Entropy Calculator

Chemistry

Calculate standard entropy change ΔS° from enthalpy change ΔH° and Gibbs free energy ΔG° at a given temperature using the thermodynamic relationship ΔS = (ΔH − ΔG) / T.

-286 kJ/mol
kJ/mol
-237 kJ/mol
kJ/mol
298 K
K

Entropy Change ΔS° (J/mol·K)

-164.43
ΔS° (kJ/mol·K)
-0.164
Entropy Contribution
Entropy-opposed (ΔS < 0, order increases)

This calculator computes your Entropy Change ΔS° (J/mol·K), ΔS° (kJ/mol·K), Entropy Contribution from the values you enter.

Inputs
Enthalpy Change (ΔH°)Gibbs Free Energy Change (ΔG°)Temperature
Outputs
Entropy Change ΔS° (J/mol·K)ΔS° (kJ/mol·K)Entropy Contribution

What is a Entropy?

The Entropy Calculator determines the standard entropy change (ΔS°) for a chemical reaction by rearranging the Gibbs–Helmholtz equation: ΔG° = ΔH° − TΔS°, solving for ΔS° = (ΔH° − ΔG°) / T. Given the standard enthalpy change, standard Gibbs free energy change, and temperature, it returns ΔS° in J/mol·K — the primary thermodynamic quantity measuring the change in disorder between reactants and products.

Entropy (S) is one of the three fundamental thermodynamic quantities governing chemical spontaneity. While enthalpy change (ΔH°) captures energy release or absorption, entropy change (ΔS°) captures whether the reaction moves toward more disorder or more order. The Gibbs free energy change (ΔG°), computed by the Gibbs Free Energy Calculator, combines both into the single spontaneity criterion. The Entropy Calculator closes the thermodynamic triangle: given any two of ΔG°, ΔH°, and ΔS°, you can find the third.

For phase transitions at the equilibrium temperature (melting, boiling, sublimation), ΔG = 0 by definition, so this formula reduces to ΔS = ΔH/T. The entropy of vaporisation of water at 100°C (373 K) is ΔS = 40,700 J/mol / 373 K = 109 J/mol·K — a result known as Trouton's rule predicts should be about 85–90 J/mol·K for most non-polar liquids, with water being higher due to hydrogen bonding structure in liquid water.

How to use this Entropy calculator

  1. Find the standard enthalpy change ΔH° for your reaction from a thermochemical table, NIST database, or calorimetric measurement. Enter it in kJ/mol in the Enthalpy Change (ΔH°) field — use negative values for exothermic reactions.
  2. Find the standard Gibbs free energy change ΔG° from the same source. Enter it in kJ/mol in the Gibbs Free Energy Change (ΔG°) field. For phase transition calculations at the equilibrium temperature, enter 0 for ΔG°.
  3. Enter the temperature in Kelvin in the Temperature field. Standard conditions use 298.15 K (25°C); for phase transitions, use the equilibrium temperature.
  4. Read ΔS° (J/mol·K) — check the sign and order of magnitude against expectations for your reaction type.
  5. Use the ΔS° result as an input to the Gibbs Free Energy Calculator to predict ΔG at any temperature, showing whether spontaneity changes with temperature.

Formula & Methodology

Gibbs–Helmholtz equation (rearranged for ΔS):

ΔG° = ΔH° − TΔS° ΔS° = (ΔH° − ΔG°) / T

Special case — phase transition at equilibrium temperature (ΔG = 0):

ΔS_transition = ΔH_transition / T_eq

Worked example — standard entropy change for water formation:

Reaction: H₂(g) + ½ O₂(g) → H₂O(l) at 25°C (T = 298.15 K)

Known values: ΔH° = −285.8 kJ/mol, ΔG° = −237.1 kJ/mol

ΔS° = (ΔH° − ΔG°) / T      = (−285,800 J/mol − (−237,100 J/mol)) / 298.15 K      = (−285,800 + 237,100) / 298.15      = −48,700 / 298.15      = −163.3 J/mol·K

The negative ΔS° confirms that forming liquid water from gases reduces disorder — two gas-phase molecules combine into one liquid-phase product, a large decrease in accessible microstates. Despite this entropy penalty, the large negative ΔH° (−285.8 kJ/mol) drives the reaction strongly spontaneous (ΔG° = −237.1 kJ/mol) at 25°C.

Frequently Asked Questions

Entropy (S) is a thermodynamic state function that measures the degree of disorder or randomness in a system. A high-entropy system has many accessible microstates — its atoms and molecules can be arranged in many equivalent ways. A low-entropy system is highly ordered with few accessible arrangements. In chemical thermodynamics, it is the change in entropy (ΔS) that determines whether a process increases or decreases disorder, and together with enthalpy change (ΔH), it governs whether a reaction is spontaneous at a given temperature.
The standard entropy change for a chemical reaction is ΔS° = (ΔH° − ΔG°) / T, derived by rearranging the Gibbs equation ΔG° = ΔH° − TΔS°. This gives ΔS° in J/mol·K when ΔH° and ΔG° are in J/mol and T is in Kelvin. For a physical process like melting or vaporisation at the equilibrium temperature, ΔS = ΔH/T, since ΔG = 0 at equilibrium.
ΔS° is the standard entropy change — the entropy change when reactants in their standard states (1 bar pressure, specified temperature, usually 298.15 K) convert to products in their standard states. ΔS is the entropy change under actual reaction conditions, which includes a concentration-dependent term: ΔS = ΔS° − R ln Q. For calculations at standard conditions, ΔS° is used directly. The Gibbs equation relating ΔG°, ΔH°, and ΔS° holds only for standard-state quantities.
A positive ΔS means entropy increases — the products are more disordered than the reactants. This is thermodynamically favourable (it contributes to making the reaction spontaneous). Positive ΔS is typically associated with reactions that produce more moles of gas than they consume, dissolving solids into solution, or heating a substance. The −TΔS term in ΔG = ΔH − TΔS becomes negative when ΔS is positive, lowering ΔG and favouring spontaneity.
A negative ΔS means entropy decreases — the products are more ordered than the reactants. This thermodynamically opposes spontaneity (the −TΔS term adds to ΔG, making it less negative or more positive). Reactions that form precipitates from solution, reduce gas-phase moles, or form highly ordered products (e.g., polymerisation, crystallisation) typically have negative ΔS. A reaction with negative ΔH and negative ΔS may still be spontaneous at low temperatures where the |ΔH| term dominates.
For a reaction with positive ΔH (endothermic) and positive ΔS, the reaction is non-spontaneous at low temperature (ΔG = ΔH − TΔS > 0 when T is small) but becomes spontaneous above a crossover temperature T_cross = ΔH/ΔS. For a reaction with negative ΔH (exothermic) and negative ΔS, the reaction is spontaneous at low temperature but becomes non-spontaneous above T_cross. The [Gibbs Free Energy Calculator](/gibbs-free-energy-calculator/) computes ΔG at any temperature to show this crossover.
Enter the standard enthalpy change ΔH° in kJ/mol, the standard Gibbs free energy change ΔG° in kJ/mol, and the temperature T in Kelvin. The calculator rearranges ΔG° = ΔH° − TΔS° to give ΔS° = (ΔH° − ΔG°) / T, returning the result in J/mol·K and kJ/mol·K along with an entropy favourability interpretation.
Standard entropy changes span a wide range: gas-producing reactions (e.g., decomposition of CaCO₃ to CaO + CO₂) have ΔS° ≈ +160 J/mol·K. Reactions that consume gas (e.g., synthesis of ammonia N₂ + 3H₂ → 2NH₃) have ΔS° ≈ −198 J/mol·K. Aqueous-phase reactions typically show smaller changes (±50 to ±100 J/mol·K). Phase transitions at the equilibrium temperature (e.g., ice melting, ΔHfus = 6.01 kJ/mol at 273 K) give ΔS = 6010/273 = 22.0 J/mol·K.
Entropy contributes to the equilibrium constant through the Gibbs equation: ΔG° = −RT ln(Kc). Since ΔG° = ΔH° − TΔS°, we get ln(Kc) = −ΔH°/RT + ΔS°/R. A large positive ΔS° (increasing disorder) shifts Kc to larger values, favouring products. A large negative ΔS° reduces Kc. The entropy and enthalpy contributions are temperature-dependent in different ways: the ΔH°/RT term diminishes at high temperature, making the ΔS° term increasingly dominant.
Yes — entropy considerations govern several major Indian industries. In the Haber process for ammonia (critical for Indian fertiliser production at IFFCO, NFL), the strongly negative ΔS° (−198 J/mol·K) means equilibrium shifts backward at high temperature, explaining why industrial operation at 400–500°C requires high pressure to maintain acceptable conversion. In pharmaceutical manufacturing, crystallisation processes (negative ΔS) must be optimised for temperature to maximise yield of ordered crystalline product.
The third law states that the entropy of a perfect crystalline substance at absolute zero (0 K) is exactly zero — there is only one possible arrangement (perfect order) at 0 K. This provides an absolute reference point, unlike energy where only differences (ΔH, ΔG) are meaningful. Standard molar entropies S° are tabulated as absolute values (J/mol·K) by integrating heat capacity data from 0 K to 298 K. These tabulated values allow ΔS° to be calculated as ΔS° = ΣS°(products) − ΣS°(reactants).