Standard Molar Entropies

Core Concept

The Third Law of Thermodynamics states that a perfect crystal at absolute zero (0 K) has exactly one microstate, so its entropy is zero. This gives entropy an absolute reference point: unlike enthalpy or Gibbs free energy (which are defined relative to arbitrary formation references), every substance has a measurable, positive standard molar entropy (S°). These tabulated values let us compute the entropy change of any reaction at 25 °C and 1 atm with ΔS° = ΣνS°(products) − ΣνS°(reactants).

Key Ideas

• Absolute zero, one microstate. A perfect crystal at 0 K has W = 1, so S = k ln 1 = 0.
• All S° values are positive. Temperature, phase freedom, and molecular complexity all raise S° above zero.
• No "formation" reference needed. S° is absolute; elements do not have S° = 0 (unlike ΔH°f and ΔG°f).
• Reaction entropy from tables. ΔS°rxn = ΣνS°(products) − ΣνS°(reactants), using stoichiometric coefficients ν.
• Gas moles rule of thumb. More moles of gas on the product side ⇒ ΔS°rxn > 0.

Equations and Variables

ΔS°rxn = ΣνS°(products) − ΣνS°(reactants)

How It Works

1. Look up S° for every reactant and product in a standard entropy table (all positive, in J/(mol·K)).
2. Multiply each S° by its stoichiometric coefficient.
3. Sum the products' contributions and subtract the sum of the reactants' contributions.
4. Interpret the sign: positive means the reaction increases entropy (usually more gas moles or looser phases); negative means it decreases entropy.
5. Keep J (not kJ) so the value plugs directly into ΔG = ΔH − TΔS.

Worked Example

Find ΔS°rxn for N₂(g) + 3 H₂(g) → 2 NH₃(g).

Standard molar entropies: S°(N₂) = 191.6, S°(H₂) = 130.7, S°(NH₃) = 192.8 J/(mol·K).

• Products: 2 × 192.8 = 385.6 J/(mol·K)
• Reactants: 191.6 + 3(130.7) = 191.6 + 392.1 = 583.7 J/(mol·K)
• ΔS°rxn = 385.6 − 583.7 = −198.1 J/(mol·K)

Four moles of gas become two, so the entropy decreases — the reaction "tightens up" the system, as the sign confirms.

How it works

1. Look up S° for every reactant and product in a standard entropy table (all positive, in J/(mol·K)).
2. Multiply each S° by its stoichiometric coefficient.
3. Sum the products' contributions and subtract the sum of the reactants' contributions.
4. Interpret the sign: positive means the reaction increases entropy (usually more gas moles or looser phases); negative means it decreases entropy.
5. Keep J (not kJ) so the value plugs directly into ΔG = ΔH − TΔS.

Common confusions

• "Elements have S° = 0 like they have ΔH°f = 0." — Wrong. ΔH°f and ΔG°f of elements are zero by convention; S° is absolute and positive for every element.
• "S° can be negative." — Wrong. At any T > 0 K, W > 1, so S° > 0.
• "Use kJ/mol for entropy." — Wrong. S° and ΔS° are tabulated in J/(mol·K); only ΔH and ΔG use kJ/mol.
• "Ignore stoichiometric coefficients." — Wrong. Each S° is multiplied by its coefficient in the balanced equation.
• "The Third Law says entropy is zero at 0 °C." — Wrong. It is absolute zero (0 K ≈ −273.15 °C), and only for a perfect crystal.

Quick review

• Third Law: perfect crystal at 0 K has S = 0.
• Standard molar entropies are absolute, positive, and tabulated in J/(mol·K).
• ΔS°rxn = ΣνS°(products) − ΣνS°(reactants).
• Sign of ΔS°rxn tracks changes in gas moles and phase freedom.
• Elements have nonzero S°, unlike their zero ΔH°f and ΔG°f.