Entropy Calculator
Choose a calculation mode and get the entropy change in joules per kelvin instantly. The heat-transfer mode uses ΔS = Q/T for any reversible process or phase change. The reaction mode subtracts the sum of standard molar entropies of reactants from products. The ideal-gas mode applies ΔS = nR·ln(V2/V1) for an isothermal expansion or compression. The Gibbs mode rearranges ΔG = ΔH - TΔS to find the entropy change from enthalpy and free energy data. Every mode shows a full worked solution.
Formula
Worked example
Mode 1 (heat/temp): 2,000 J at 300 K gives ΔS = 2000/300 = 6.667 J/K. Mode 2 (reaction): H2(g) combustion, ΣS° products = 188.8, ΣS° reactants = 130.6, ΔS = 58.2 J/(K·mol). Mode 3 (ideal gas): 1 mol expanding from 1 L to 2 L: ΔS = 1 × 8.314 × ln(2) = 5.76 J/K. Mode 4 (Gibbs): water formation at 298.15 K, ΔH = -286,000 J, ΔG = -237,100 J: ΔS = (-286,000 - (-237,100)) / 298.15 = -163.4 J/K.
Four ways to calculate entropy change
Entropy (S) quantifies how many microscopic arrangements are compatible with a macroscopic state. The more arrangements, the higher the entropy and the less "useful" the system energy. Depending on what data you have, four formulas are available. The heat-transfer formula ΔS = Q/T applies to any reversible process where Q joules of heat cross a boundary at temperature T kelvin. The reaction formula ΔS = ΣS° products - ΣS° reactants uses tabulated standard molar entropies and works directly from the chemical equation. The ideal-gas formula ΔS = nR·ln(V2/V1) applies to an isothermal (constant-temperature) expansion or compression. The Gibbs rearrangement ΔS = (ΔH - ΔG)/T back-calculates the entropy contribution once you know both the enthalpy and free energy of a reaction.
Heat transfer mode: ΔS = Q/T and phase changes
Adding heat to a system at absolute temperature T increases its entropy by ΔS = Q/T. The kelvin denominator means the same heat causes a larger entropy rise at low temperatures than at high ones. Sign follows the heat: positive Q (absorbed) raises entropy, negative Q (released) lowers it. Phase changes are a special case of this formula. Melting ice at 273.15 K absorbs roughly 6,010 J per mole of latent heat, giving ΔS = 6,010 / 273.15 = 22.0 J/(K·mol). Vaporizing liquid water at 373.15 K absorbs about 40,700 J per mole, giving ΔS = 40,700 / 373.15 = 109.1 J/(K·mol). Enter the latent heat in the Q field and the transition temperature in T to handle any phase change.
Reaction mode: standard molar entropies
Every substance has a measurable absolute entropy at standard conditions (298.15 K, 1 bar). Unlike enthalpies of formation, these values are not zero for elements in their standard states; all pure substances have positive S° values because they have more than one accessible microstate at any temperature above absolute zero. The entropy change of a reaction is simply ΣS° (products, weighted by stoichiometry) minus ΣS° (reactants, weighted by stoichiometry). For the combustion of hydrogen, H2(g) + 1/2 O2(g) -> H2O(g), the products contribute 188.8 J/(K·mol) and the reactants 130.7 + 1/2 × 205.2 = 233.3 J/(K·mol), giving ΔS = -44.5 J/(K·mol), a small entropy decrease because one mole of gaseous product replaces 1.5 moles of gaseous reactants.
Isothermal ideal-gas mode: expansion and compression
When an ideal gas expands at constant temperature, its molecules explore more volume and the number of accessible microstates grows. The entropy change is ΔS = nR·ln(V2/V1), where n is moles, R = 8.314 J/(mol·K) is the universal gas constant, and the ratio V2/V1 captures how much larger the final volume is. Because PV = nRT and T is fixed, you can substitute pressure: V2/V1 = P1/P2, so ΔS = -nR·ln(P2/P1). Expansion (V2 > V1 or P2 < P1) gives a positive entropy change; compression gives a negative one. The temperature does not appear explicitly because the isothermal constraint fixes it.
Gibbs free energy mode: back-calculating ΔS
The Gibbs free energy equation ΔG = ΔH - TΔS links three key thermodynamic quantities. Rearranging gives ΔS = (ΔH - ΔG) / T, which lets you extract the entropy contribution when ΔH and ΔG are known from calorimetry or electrochemistry. For the formation of liquid water at 298.15 K, ΔH = -286,000 J/mol and ΔG = -237,100 J/mol, so ΔS = (-286,000 - (-237,100)) / 298.15 = -163.4 J/(K·mol). A negative value here reflects that a liquid is more ordered than a mixture of two gases. If ΔG is negative the process is spontaneous; if positive it requires energy input.
Standard molar entropies at 298.15 K and 1 bar
| Substance | Phase | S° (J/K·mol) |
|---|---|---|
| H2 | gas | 130.7 |
| O2 | gas | 205.2 |
| N2 | gas | 191.6 |
| H2O | gas | 188.8 |
| H2O | liquid | 70.0 |
| CO2 | gas | 213.8 |
| CO | gas | 197.7 |
| CH4 | gas | 186.3 |
| C (graphite) | solid | 5.7 |
| C (diamond) | solid | 2.4 |
| Fe | solid | 27.3 |
| NaCl | solid | 72.1 |
Reference values for common substances. Multiply by stoichiometric coefficients and subtract reactants from products to get the reaction entropy.
Frequently asked questions
What is the formula for entropy change?
There are several depending on the scenario. For a reversible heat transfer, ΔS = Q/T (heat in joules divided by absolute temperature in kelvin). For a chemical reaction using tabulated data, ΔS = ΣS° products - ΣS° reactants. For an isothermal ideal-gas process, ΔS = nR·ln(V2/V1) or equivalently -nR·ln(P2/P1). When ΔH and ΔG are known, rearranging the Gibbs equation gives ΔS = (ΔH - ΔG) / T.
Why must temperature be in kelvin?
Entropy is defined on the absolute thermodynamic scale, where zero corresponds to a single perfectly ordered microstate (the third law). Celsius and Fahrenheit place zero at arbitrary reference points, so dividing heat by a Celsius temperature produces a result with no physical meaning. This calculator accepts Celsius and Fahrenheit and converts them to kelvin automatically before computing.
Can entropy change be negative?
Yes. When heat leaves a system (negative Q), when a reaction produces fewer, more ordered molecules, when a gas is compressed, or when a substance freezes, the entropy decreases and ΔS is negative. The second law of thermodynamics only requires that the total entropy of the system plus surroundings never decreases; the system alone can lose entropy as long as the surroundings gain at least as much.
How do I calculate the entropy change of a chemical reaction?
Look up the standard molar entropy S° of each reactant and product (at 298.15 K and 1 bar). Multiply each value by its stoichiometric coefficient, sum the products and sum the reactants, then subtract: ΔS = ΣS° products - ΣS° reactants. The reference table on this page lists S° values for common substances. For reactions at temperatures other than 298.15 K, corrections require heat capacity data.
What is the entropy change for an ideal gas expansion?
For an isothermal (constant-temperature) expansion, ΔS = nR·ln(V2/V1), where n is moles, R = 8.314 J/(mol·K), and V2/V1 is the ratio of final to initial volume. Because PV = constant at fixed temperature, you can also write ΔS = -nR·ln(P2/P1). Expansion increases entropy (positive ΔS); compression decreases it.
How does the Gibbs free energy relate to entropy?
The Gibbs equation ΔG = ΔH - TΔS ties together enthalpy, entropy and free energy at constant temperature and pressure. A reaction is spontaneous when ΔG is negative. Rearranging to ΔS = (ΔH - ΔG) / T lets you back-calculate the entropy change from experimental ΔG and ΔH values, which is often more accurate than summing tabulated S° data.