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Absolute entropy and entropy change

Unlike enthalpy, entropy has a true zero — a perfect crystal at absolute zero. That means every substance carries a real, positive entropy you can look up, and reaction entropy changes are just products minus reactants.

§1

Absolute entropy and ΔS°.

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Every substance has an absolute (standard molar) entropy S°, a positive value in J·mol⁻¹·K⁻¹. Because entropy has a true zero (a perfect crystal at 0 K), even elements have nonzero S° — unlike the standard enthalpy of formation, which is zero for elements.

A reaction's standard entropy change is the familiar products-minus-reactants sum: ΔS° = Σ S°(products) − Σ S°(reactants), with each S° weighted by its coefficient.

Because gases carry much larger S° than liquids or solids, the sign of ΔS° usually tracks the change in moles of gas: more gas on the product side means a positive ΔS°.

UNIT 9 TOPIC 9.2 • ABSOLUTE ENTROPY AND ENTROPY CHANGE ENTROPY CHANGE Standard entropy change (ΔS°rxn) comes from standard molar entropies (S°) of products and reactants. CALCULATING STANDARD ENTROPY CHANGE ΔS°rxn = Σ n S°products − Σ n S°reactants • n = stoichiometric coefficient from the balanced equation • ΣnS° = sum of (coefficient × S°) for every species units: J·mol⁻¹·K⁻¹ WORKED EXAMPLE N₂O₄(g) → 2 NO₂(g) GIVEN S° VALUES (298 K) S°(N₂O₄) = 304.3 J·mol⁻¹·K⁻¹ reactant S°(NO₂) = 240.1 J·mol⁻¹·K⁻¹ product S° is ALWAYS positive (Third Law: S = 0 at 0 K for a perfect crystal). CALCULATION Σ S°prod = 2 × S°(NO₂) = 2(240.1) = 480.2 Σ S°react = 1 × S°(N₂O₄) = 304.3 ΔS°rxn = 2(240.1) − 304.3 = +175.9 J·mol⁻¹·K⁻¹ More gas particles form (2 mol vs 1 mol) → ΔS°rxn is positive WHAT INCREASES ENTROPY ΔS° increases (S° gets larger) when… PHASE CHANGE · solid → liquid → gas solid liquid gas MORE MOLES OF GAS fewer more HIGHER TEMPERATURE lower T higher T also: mixing & dissolving raise S° KEY IDEAS & COMMON CORRECTIONS • Standard molar entropies S° are always POSITIVE — larger for more accessible microstates. • Favorability is set by ΔG, not ΔS alone: ΔG° = ΔH° − TΔS° — a reaction can be nonspontaneous even when ΔS° > 0 if ΔH° is large & positive. AP Chemistry · Unit 9 · Applications of Thermodynamics
Fig. 9.2.1 Every substance, elements included, has a positive absolute entropy S°. The reaction entropy change is ΔS° = Σ S°(products) − Σ S°(reactants), each term weighted by its coefficient.
§2

Computing ΔS°.

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Sum products minus reactants, with coefficients.

  1. Look up each S°. Every substance, including elements, has a positive standard molar entropy.
  2. Weight by coefficients. Multiply each S° by the stoichiometric coefficient in the balanced equation.
  3. Subtract in order. ΔS° = Σ S°(products) − Σ S°(reactants), products first.
  4. Sanity-check the sign. More moles of gas on the right usually means a positive ΔS°.
§3

The pieces you'll meet.

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Absolute entropies and the products-minus-reactants rule.

S°
Standard molar entropy
Positive; in J·mol⁻¹·K⁻¹, elements included.
ΔS°
ΔS°
Σ S°(products) − Σ S°(reactants).
coeff
Coefficients
Each S° is weighted by its coefficient.
gas
Moles of gas
Usually drives the sign of ΔS°.
order
Order
Products minus reactants, not the reverse.
elements
Elements
Have nonzero S° (unlike ΔH°f).
§4

Worked example: ΔS° from tabulated S°.

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Reaction. N₂(g) + 3 H₂(g) → 2 NH₃(g). Use S° values (J·mol⁻¹·K⁻¹): N₂ 192, H₂ 131, NH₃ 193.

Products. 2 × 193 = 386.

Reactants. 192 + 3 × 131 = 192 + 393 = 585.

ΔS°. 386 − 585 = −199 J·mol⁻¹·K⁻¹. Negative, consistent with 4 moles of gas collapsing to 2 — fewer gas moles, less entropy.

§5

Mistakes that cost real points.

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Pitfall · 01

"Elements have zero standard entropy, like their formation enthalpy."

Standard molar entropy is not zero for elements — only the standard enthalpy of formation is. Every substance, elements included, has a real positive S° that must be included in the sum. Skipping element entropies gives the wrong ΔS°.

Fix. Include every substance's S° (elements too); only ΔH°f, not S°, is zero for elements.

Pitfall · 02

"You can add the entropies without the coefficients."

Each S° must be multiplied by its stoichiometric coefficient before summing. Forgetting the coefficients (for example, using S° of H₂ once instead of three times) gives a wrong ΔS°.

Fix. Weight every S° by its coefficient, then take products minus reactants.

Pitfall · 03

"A gas has about the same S° as a solid, so moles of gas do not matter."

Gases have much larger absolute entropies than solids or liquids, so a change in the moles of gas dominates ΔS°. Ignoring the gas count leads you to mispredict both the size and the sign of ΔS°.

Fix. Track the change in moles of gas; it usually sets the sign of ΔS°.

§6

Skill Check.

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