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Entropy Change Calculator

ΔS°rxn from absolute entropies — where elements are not zero.

Work out Entropy Change. ΔS°rxn from absolute entropies — where elements are not zero. Names the sign error before you make it.

Written and maintained by Mohit PatelLast checked August 4, 2026How we build these

Absolute entropies in J/(mol·K), written as value:coefficient.

Elements are NOT zero here — every substance has a positive S° at 298 K.

Standard entropy of reaction

-242.9 J/(mol·K)

Entropy decreases — unfavourable on its own

ΔS°rxn-242.9 J/(mol·K)
Σ products353.8 J/(mol·K)
Σ reactants596.7 J/(mol·K)
In kJ, for use with ΔG-0.2429 kJ/(mol·K)

The trap here is assuming elements have S° = 0, by analogy with enthalpies of formation. They do not. The third law puts zero entropy at a perfect crystal at absolute zero, so every substance at 298 K has a positive absolute entropy — oxygen gas is 205.2 J/(mol·K), not nothing. The sign is usually predictable before you calculate: more moles of gas on the right means positive ΔS, fewer means negative. A reaction consuming gas to make a liquid or solid almost always loses entropy. Note the units. Entropy is tabulated in J/(mol·K) while enthalpy is in kJ/mol, and mixing them in ΔG = ΔH − TΔS gives an answer out by a factor of a thousand.

How the Entropy Change Calculator works

Enter each product and reactant as its absolute entropy and coefficient. The trap here is assuming elements have S° = 0 by analogy with enthalpies of formation — they do not, because the third law puts zero at a perfect crystal at absolute zero.

Also known as: delta s calculator · standard entropy of reaction calculator · entropy calculator for chemistry · how to calculate entropy change

Frequently asked questions

How do I calculate entropy change?

ΔS°rxn = Σ n·S°(products) − Σ n·S°(reactants), using absolute entropies in J/(mol·K). Same structure as the enthalpy calculation, different table.

Why is the entropy of an element not zero?

Because S° is an absolute entropy, measured from the third-law zero of a perfect crystal at 0 K. Every substance at 298 K has a positive value — oxygen gas is 205.2 J/(mol·K).

How can I predict the sign of ΔS?

Count moles of gas. More gas on the right means entropy increases; fewer means it falls. A reaction consuming gas to make a liquid or solid almost always loses entropy.

Why do the units matter here?

Because entropy is tabulated in J/(mol·K) while enthalpy is in kJ/mol. Combining them in ΔG = ΔH − TΔS without converting gives an answer out by a factor of a thousand.

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