The entropy of a substance can be tabulated on an absolute scale. The third law of thermodynamics assigns zero entropy to a perfect crystalline substance at absolute zero:

$$S(0,\mathrm K)=0$$

for an ideal perfect crystal. Heating the substance and allowing phase changes adds accessible microscopic states, so ordinary substances at finite temperature have positive absolute entropies.

The standard molar entropy, written $S^\circ$, is the entropy per mole of a substance in its defined standard state at a specified temperature. Typical units are

$$\mathrm{J,mol^{-1},K^{-1}}.$$

Unlike a standard enthalpy of formation, an element in its reference form does not generally have $S^\circ=0$ at ordinary temperature. The zero convention applies to the perfect crystal at $0,\mathrm K$, not to elements in their standard states at $298,\mathrm K$.

For a reaction

$$a\mathrm A+b\mathrm B\rightarrow c\mathrm C+d\mathrm D,$$

the standard reaction entropy is calculated from tabulated molar entropies:

$$\boxed{\Delta S_{\mathrm{rxn}}^\circ =\sum_{\mathrm{products}}\nu_i S_i^\circ -\sum_{\mathrm{reactants}}\nu_i S_i^\circ}.$$

Example

For

$$\mathrm{N_2(g)+3H_2(g)\rightarrow2NH_3(g)},$$

use approximate standard molar entropies at $298,\mathrm K$:

$$S^\circ(\mathrm{N_2})=191.6,$$ $$S^\circ(\mathrm{H_2})=130.7,$$ $$S^\circ(\mathrm{NH_3})=192.8$$

in $\mathrm{J,mol^{-1},K^{-1}}$. Then

$$\begin{aligned} \Delta S_{\mathrm{rxn}}^\circ &=2(192.8)-[191.6+3(130.7)]\ &\approx-198,\mathrm{J,mol^{-1},K^{-1}}. \end{aligned}$$

The negative value is consistent with the gas stoichiometry: four moles of gaseous reactants become two moles of gaseous product, reducing the number of translational arrangements available to the system.

Such qualitative reasoning can help predict a sign, but tabulated $S^\circ$ values provide the quantitative result. Physical state matters: the standard molar entropy of a gas, liquid and solid of the same substance are different.