9.2 Absolute Entropy

Absolute Entropy & ΔS°

Pick a reaction and watch coefficient-scaled S° bars stack on a shared J/mol·K axis. Signed subtraction gives ΔS°rxn, with phase-colored bars hinting at signs.

ΔS° = Σn·S°S° Table10 ReactionsPhase Trends
Topic 9.2

Absolute Entropy and Entropy Change

Calculate the standard entropy change for a chemical or physical process based on the absolute entropies (standard molar entropies) of the species involved.

The entropy change for a process can be calculated from the absolute entropies of the species involved before and after the process occurs. The equation is on the AP sheet:

ΔS°reaction = Σ S°(products) − Σ S°(reactants)

Same "products minus reactants" pattern as ΔH°f (6.8) and ΔG°f (9.3), each weighted by the stoichiometric coefficient.

One critical difference from enthalpy: entropies are absolute, not relative to a reference. A perfect crystal at 0 K has S = 0, and every substance above that has a positive absolute entropy. So unlike ΔH°f, S° for an element in its standard state is not zero — S° for O₂(g) is 205.0 J·mol⁻¹·K⁻¹, and it must be included in your sum.

Units. Entropies are quoted in J·mol⁻¹·K⁻¹, while enthalpies are in kJ/mol. That factor of 1000 is the single most common arithmetic error in Unit 9, because ΔG° = ΔH° − TΔS° requires them in the same units. Convert before subtracting.

Patterns in tabulated S° values that let you sanity-check an answer:

  • Gases ≫ liquids > solids for comparable substances.
  • Larger, more complex molecules have higher S° — more ways to store energy in vibrations and rotations.
  • Heavier atoms have higher S° than lighter ones in the same group.
  • Hard, highly ordered solids like diamond have very low S° (2.4 J·mol⁻¹·K⁻¹).

Key points

  • ΔS° = Σ S°(products) − Σ S°(reactants), weighted by coefficients.
  • S° for an element in its standard state is NOT zero — unlike ΔH°_f.
  • Entropies are in J·mol⁻¹·K⁻¹; convert to kJ before combining with ΔH°.
  • Gases have far larger S° than condensed phases.

Equations

  • on the exam sheetPrinted on the AP equation sheet.
    • standard molar (absolute) entropy, J mol⁻¹ K⁻¹

Common mistakes

  • Setting S° of an element to zero. Only ΔH°_f and ΔG°_f are zero for elements.
  • Forgetting the unit mismatch when computing ΔG° = ΔH° − TΔS°.
  • Omitting coefficients.
  • Expecting a negative absolute entropy. S° is always positive for any substance above 0 K.

Worked example

Calculate ΔS° for 2 NO(g) + O₂(g) → 2 NO₂(g). S° values (J·mol⁻¹·K⁻¹): NO(g) = 210.8, O₂(g) = 205.0, NO₂(g) = 240.1. Does the sign agree with a qualitative prediction?

Qualitative prediction first: gas moles go from 2 + 1 = 3 to 2. Fewer gas particles means less dispersal, so ΔS° should be negative.

Products: 2 × 240.1 = 480.2 J·mol⁻¹·K⁻¹
Reactants: 2 × 210.8 + 1 × 205.0 = 421.6 + 205.0 = 626.6 J·mol⁻¹·K⁻¹

ΔS° = 480.2 − 626.6 = −146.4 J·mol⁻¹·K⁻¹

The sign matches the prediction ✓. Note that O₂ contributed 205.0 rather than zero — a reminder that absolute entropies of elements are not zero.

To use this in ΔG° = ΔH° − TΔS°, first convert: ΔS° = −0.1464 kJ·mol⁻¹·K⁻¹.

Full notes for topic 9.2 →