Absolute Entropy and Entropy Change绝对熵与熵变

Big Idea · Energy (ENE)AP Science Practice 5 · Mathematical RoutinesAP link · Unit 9~40 min

Learning goals

  • Explain what standard molar entropy S° means, and compare S° values for different substances (phase, molecular size, molar mass, structure of solids)
  • Calculate the standard entropy change of a reaction with ΔS° = Σn·S°(products) − Σn·S°(reactants), and compare it with a qualitative prediction
  • Scale an entropy change to the actual amount reacting, and convert between J/K and kJ/K

Before you start

  • Enthalpy of formation (U6.8)
  • Introduction to entropy (U9.1)

Absolute entropy and standard molar entropy

Only changes in enthalpy can be measured, which is why the enthalpy of formation of an element is defined as zero (U6.8). Entropy is different: a perfect crystal at absolute zero has zero entropy (the third law of thermodynamics热力学第三定律), and starting from there, the absolute entropy of any substance at a given temperature can be measured.

The entropy of 1 mol of a substance in its standard state (25 °C, 1 bar) is its standard molar entropy标准摩尔熵 S°, in J/(mol·K). Note that the S° of an element is not zero; for O₂(g), for example, S° = 205.2 J/(mol·K).

S° values follow these patterns:

  • Phase: for the same substance, S°(g) ≫ S°(l) > S°(s).
  • Molecular size: among gases, the more atoms and the more complex the molecule, the larger the S°.
  • Molar mass: for substances with similar structures, the larger the molar mass, the larger the S°.
  • Structure of solids: the more tightly bonded and rigid the structure, the smaller the S°.

Calculating the standard entropy change of a reaction

Just as enthalpies of formation give the enthalpy change of a reaction, S° values give its standard entropy change:

ΔSrxn∘=∑n S∘(products)−∑n S∘(reactants)\begin{aligned} \Delta S^\circ_\text{rxn} = &\sum n\,S^\circ(\text{products}) \\ &- \sum n\,S^\circ(\text{reactants}) \end{aligned}

Multiply each S° by the substance’s coefficient n, and include the elements too. ΔS° is in J/K (for the amounts in the equation).

Compare the S° values of several groups, then choose a reaction to calculate ΔS° and compare it with the prediction from counting gas molecules (data: OpenStax Appendix G):

Try it: standard molar entropy and entropy change

Example 1: ΔS° for the combustion of ethanol

Find ΔS° for C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l). S° in J/(mol·K): C₂H₅OH(l) 160.7, O₂(g) 205.2, CO₂(g) 213.8, H₂O(l) 70.0.

Step-by-step solution

  1. Predict the sign first. The gas coefficients go from 3 to 2, and liquid water forms, so ΔS° should be negative.

  2. Write the formula. Sum of the products’ S° minus the reactants’, each times its coefficient (in J/K):

    ΔS∘=∑nS∘(prod)−∑nS∘(react)\begin{aligned} \Delta S^\circ = {} &\textstyle\sum nS^\circ(\text{prod}) \\ &- \textstyle\sum nS^\circ(\text{react}) \end{aligned}

  3. Products.

    2×213.8+3×70.0=637.6\begin{aligned} &2 \times 213.8 + 3 \times 70.0 \\ &= 637.6 \end{aligned}

  4. Reactants. Note that S° of O₂ is not 0 (it’s ΔHf° of an element that is 0):

    160.7+3×205.2=776.3\begin{aligned} &160.7 + 3 \times 205.2 \\ &= 776.3 \end{aligned}

  5. Subtract.

    ΔS∘=637.6−776.3=−138.7 J/K\begin{aligned} \Delta S^\circ &= 637.6 - 776.3 \\ &= \boxed{-138.7\ \text{J/K}} \end{aligned}

  6. Check. Negative, as predicted in step 1.

Example 2: scaling to the actual amount

For I₂(s) → I₂(g), ΔS° = +144.6 J/K (per mole of I₂). What is the entropy change when 10.0 g of iodine sublimes? M(I₂) = 253.8 g/mol.

Step-by-step solution

  1. What ΔS° means. +144.6 J/K is the entropy change per mole of I₂ subliming; scale it to the actual amount.

  2. Moles of iodine.

    n=10.0253.8=0.0394 moln = \frac{10.0}{253.8} = 0.0394\ \text{mol}

  3. Scale to the actual entropy change. In J/K:

    ΔS=0.0394×144.6=5.70 J/K\begin{aligned} \Delta S &= 0.0394 \times 144.6 \\ &= \boxed{5.70\ \text{J/K}} \end{aligned}

  4. Check the sign and size. A solid becomes a gas, so ΔS > 0, which makes sense; 10.0 g is less than 0.04 mol, so the entropy change is only about 4% of the per-mole value, which also makes sense.

Practice: which has the larger S°? (1 mol each, 25 °C)
  1. H₂O(g) or H₂O(l)
  2. C₂H₆(g) or CH₄(g)
  3. Cl₂(g) or I₂(g)
  4. Si(s) or Pb(s)
  5. O₃(g) or O₂(g)
  6. CO₂(g) at 300 K or CO₂(g) at 600 K
  7. NaCl(s) or dissolved Na⁺(aq) + Cl⁻(aq)
  8. Br₂(l) or Br₂(g)

Common mistakes

Mistake 1: “Elements have S° = 0.” ΔfH° and ΔfG° of elements are defined as zero, but S° is not. When calculating ΔS°, include the S° of elements such as O₂ and H₂.

Mistake 2: “Just subtract the S° values without the coefficients.” Multiply each S° by the substance’s coefficient in the equation.

Mistake 3: “ΔS° has the same units as ΔH°, kJ.” ΔS° is usually given in J/K. When using it with ΔH° (kJ), divide by 1000 to convert it to kJ/K first.

Quick check

For O₂(g), S° = 205.2 J/(mol·K). Which statement is correct?

S° in J/(mol·K): H₂(g) 130.7, I₂(s) 116.14, HI(g) 206.59. What is ΔS° for H₂(g) + I₂(s) → 2HI(g)?

What is the order of S° for CH₄(g), C₂H₆(g) and C₃H₈(g), from smallest to largest?

For H₂(g) + Cl₂(g) → 2HCl(g), ΔS° is only +20.0 J/K, while for NH₄Cl(s) → NH₃(g) + HCl(g) it is +285.1 J/K. The main reason for the difference is:

A reaction has ΔS° = −188 J/K. In kJ/K, this is:

Most solids have ΔS > 0 when they dissolve, but some ions (such as Mg²⁺ and Al³⁺) have negative S° in water. The most reasonable explanation is:

AP-style free response

AP-style free response

TiCl₄ is a colorless liquid that gives off dense white smoke in moist air by the reaction TiCl₄(l) + 2H₂O(l) → TiO₂(s) + 4HCl(g). S° in J/(mol·K): TiCl₄(l) 252.4, H₂O(l) 70.0, TiO₂(s) 50.6, HCl(g) 186.9 (OpenStax Appendix G).

(a) Without calculating, predict the sign of ΔS° for this reaction, and justify your prediction.

(b) Calculate ΔS°.

(c) Calculate the entropy change, in kJ/K, when 1.00 kg of TiCl₄ (M = 189.7 g/mol) reacts completely.

(d) TiCl₄(g) has S° = 353.2 J/(mol·K), larger than TiCl₄(l). Explain at the particle level.

(e) For this reaction ΔH° = +62.6 kJ, so it is endothermic. Using your result from (b), explain why it is still spontaneous at room temperature.

Model answer and scoring

(a) ΔS° > 0. The reactants are liquids, and the products include 4 mol of gas.

(b) Products: 50.6 + 4 × 186.9 = 798.2 J/K; reactants: 252.4 + 2 × 70.0 = 392.4 J/K; ΔS° = 798.2 − 392.4 = +405.8 J/K.

(c) n = 1000 ÷ 189.7 = 5.27 mol, so ΔS = 5.27 × 405.8 = 2.14 × 10³ J/K = 2.14 kJ/K.

(d) Gas molecules can move and rotate freely through a much larger space, so they have far more possible arrangements and a larger S°.

(e) ΔS° is large and positive. At room temperature, TΔS° (298 × 0.4058 ≈ 121 kJ) is larger than ΔH° (62.6 kJ), so the effect of the entropy increase outweighs the unfavorable heat absorption and the reaction is spontaneous (U9.3 uses ΔG° to judge this quantitatively).

Scoring: (a) reasoning from the gas formed; (b) calculation; (c) conversion of amount and units; (d) particle explanation; (e) comparison of TΔS° with ΔH°.

Summary

  • A perfect crystal at 0 K has zero entropy, so S° is an absolute value; elements also have S° > 0.
  • S°: gas ≫ liquid > solid; the more complex the molecule and the larger its molar mass, the larger the S°; the more rigid the structure, the smaller the S°.
  • ΔS° = Σn·S°(products) − Σn·S°(reactants), in J/K; convert to kJ/K when using it with ΔH°.

AP tip: When calculating ΔS°, write the coefficients into the setup and list the elements too; afterward, check that the sign makes sense from the change in gas molecules.

Practice this chapter: Unit 9 · Thermodynamics and Electrochemistry · 12 questionsPractice this topic: U9.2 · Absolute Entropy and Entropy Change · 6 questions

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