Entropy and the Second Law

Understand entropy as disorder, predict entropy changes, and learn the second and third laws of thermodynamics.

Entropy and the Second Law

What is Entropy?

Entropy (S): Measure of disorder or randomness in a system

  • Units: J/(mol·K)
  • Symbol: S (absolute), ΔS (change)

Higher entropy = more disorder:

  • More ways to arrange particles
  • More dispersed energy
  • More random motion

Nature favors increasing entropy

Second Law of Thermodynamics

For universe:

ΔSuniverse=ΔSsystem+ΔSsurroundings>0\Delta S_{\text{universe}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} > 0

Spontaneous processes: ΔS_{universe} > 0

  • Entropy of universe always increases
  • Can't reverse without increasing entropy elsewhere

Predicting Entropy Changes

ΔS > 0 (entropy increases):

  1. Phase changes: solid → liquid → gas
  2. Dissolving: Solid/liquid dissolves in solvent
  3. More gas molecules: Δn_{gas} > 0
  4. Temperature increase: Same substance, higher T
  5. Larger molecules: More complex structure

ΔS < 0 (entropy decreases):

  • Opposite of above
  • Gas → liquid → solid
  • Precipitation
  • Fewer gas molecules

Comparing Entropy

States of matter (same substance):

Sgas>Sliquid>SsolidS_{\text{gas}} > S_{\text{liquid}} > S_{\text{solid}}

Example: S(H₂O, g) > S(H₂O, l) > S(H₂O, s)

Temperature:

  • Higher T → higher S
  • More thermal motion

Molecular complexity:

  • More atoms → more S
  • Example: S(C₃H₈) > S(CH₄)

Number of particles:

  • More moles → more S
  • Example: 2 mol > 1 mol

Calculating ΔS°_{rxn}

Standard entropy change:

ΔS°rxn=∑nS°(products)−∑nS°(reactants)\Delta S°_{\text{rxn}} = \sum nS°(\text{products}) - \sum nS°(\text{reactants})

Similar to ΔH°, but:

  • S° values never zero (even for elements)
  • Third Law: S = 0 at 0 K for perfect crystal

Example values:

  • H₂O(l): S° = 69.9 J/(mol·K)
  • CO₂(g): S° = 213.7 J/(mol·K)
  • O₂(g): S° = 205.0 J/(mol·K)

Rules for Predicting Sign of ΔS

Check for gas molecules:

If Δn_{gas} > 0: ΔS° > 0 (likely) If Δn_{gas} < 0: ΔS° < 0 (likely) If Δn_{gas} = 0: Check phase changes, complexity

Examples:

  1. CaCO₃(s) → CaO(s) + CO₂(g)

    • 0 gas → 1 gas
    • Δn_{gas} = +1
    • ΔS° > 0 ✓

  2. 2H₂(g) + O₂(g) → 2H₂O(l)

    • 3 gas → 0 gas
    • Δn_{gas} = -3
    • ΔS° < 0 ✓

  3. N₂(g) + 3H₂(g) → 2NH₃(g)

    • 4 gas → 2 gas
    • Δn_{gas} = -2
    • ΔS° < 0 ✓

Third Law of Thermodynamics

Third Law: Perfect crystal at 0 K has S = 0

Consequences:

  • Absolute entropies can be measured
  • S° always ≥ 0
  • Can't reach absolute zero (Nernst)

Entropy and Temperature

Heating increases entropy:

ΔS=qrevT\Delta S = \frac{q_{\text{rev}}}{T}

At constant pressure:

ΔS=nCpΔTT\Delta S = \frac{nC_p\Delta T}{T}

Phase changes:

ΔSphase=ΔHphaseT\Delta S_{\text{phase}} = \frac{\Delta H_{\text{phase}}}{T}

Example: Melting ice at 0°C (273 K)

ΔSfus=ΔHfusT=6.01 kJ/mol273 K=22.0 J/(mol\cdotpK)\Delta S_{\text{fus}} = \frac{\Delta H_{\text{fus}}}{T} = \frac{6.01 \text{ kJ/mol}}{273 \text{ K}} = 22.0 \text{ J/(mol·K)}

📚 Practice Problems

1Problem 1easy

❓ Question:

Predict the sign of ΔS° for each: (a) 2SO₂(g) + O₂(g) → 2SO₃(g), (b) NH₄NO₃(s) → NH₄⁺(aq) + NO₃⁻(aq), (c) CO₂(s) → CO₂(g)

💡 Show Solution

Strategy: Look for phase changes and gas molecule changes

(a) 2SO₂(g) + O₂(g) → 2SO₃(g)

Count gas molecules:

  • Reactants: 2 + 1 = 3 moles gas
  • Products: 2 moles gas
  • Δn_{gas} = 2 - 3 = -1

Fewer gas molecules → less disorder

ΔS° < 0 (negative)

(b) NH₄NO₃(s) → NH₄⁺(aq) + NO₃⁻(aq)

Phase change: solid → aqueous ions

Considerations:

  • Solid is ordered crystal
  • Aqueous ions are mobile, dispersed
  • 1 particle → 2 particles (more disorder)

Dissolving increases disorder

ΔS° > 0 (positive)

(c) CO₂(s) → CO₂(g)

Phase change: solid → gas (sublimation)

Solid: Ordered, fixed positions Gas: Random motion, highly dispersed

Major increase in disorder

ΔS° > 0 (positive, large)

Summary:

| Reaction | ΔS° Sign | Reason | |----------|----------|--------| | (a) 2SO₂ + O₂ → 2SO₃ | Negative | 3 gas → 2 gas (Δn = -1) | | (b) NH₄NO₃(s) → ions(aq) | Positive | Solid → dissolved ions | | (c) CO₂(s) → CO₂(g) | Positive | Solid → gas (large increase) |

2Problem 2medium

❓ Question:

Calculate ΔS°_{rxn} for: 2H₂(g) + O₂(g) → 2H₂O(l). Given S° values (J/mol·K): H₂(g) = 130.6, O₂(g) = 205.0, H₂O(l) = 69.9

💡 Show Solution

Reaction: 2H₂(g) + O₂(g) → 2H₂O(l)

Given S° values (J/mol·K):

  • H₂(g): 130.6
  • O₂(g): 205.0
  • H₂O(l): 69.9

Formula:

ΔS°rxn=∑nS°(products)−∑nS°(reactants)\Delta S°_{\text{rxn}} = \sum nS°(\text{products}) - \sum nS°(\text{reactants})

Calculate products:

2 mol H₂O(l): 2(69.9) = 139.8 J/K

Sum products: 139.8 J/K

Calculate reactants:

2 mol H₂(g): 2(130.6) = 261.2 J/K 1 mol O₂(g): 1(205.0) = 205.0 J/K

Sum reactants: 261.2 + 205.0 = 466.2 J/K

Calculate ΔS°_{rxn}:

ΔS°rxn=139.8−466.2\Delta S°_{\text{rxn}} = 139.8 - 466.2 ΔS°rxn=−326.4 J/K\Delta S°_{\text{rxn}} = -326.4 \text{ J/K}

Answer: ΔS°_{rxn} = -326 J/K

Interpretation:

Negative ΔS°:

  • Makes sense! 3 gas molecules → 2 liquid molecules
  • Δn_{gas} = 0 - 3 = -3
  • Large decrease in disorder
  • Gases (high entropy) → liquid (lower entropy)

Yet reaction is spontaneous:

  • Very exothermic (ΔH° = -572 kJ)
  • Entropy decrease offset by large energy release
  • ΔG° determines spontaneity (next topic!)

3Problem 3hard

❓ Question:

Calculate the entropy change when 1.00 mol of ice melts at 0°C. ΔH_{fus} = 6.01 kJ/mol.

💡 Show Solution

Given:

  • n = 1.00 mol ice
  • T = 0°C = 273 K (constant during phase change)
  • ΔH_{fus} = 6.01 kJ/mol = 6010 J/mol

Process: H₂O(s) → H₂O(l) at 273 K

For phase change at constant T:

ΔS=qrevT=ΔHphaseT\Delta S = \frac{q_{\text{rev}}}{T} = \frac{\Delta H_{\text{phase}}}{T}

At equilibrium (melting point): process is reversible

Calculate ΔS_{fus}:

ΔSfus=ΔHfusT\Delta S_{\text{fus}} = \frac{\Delta H_{\text{fus}}}{T}

ΔSfus=6010 J/mol273 K\Delta S_{\text{fus}} = \frac{6010 \text{ J/mol}}{273 \text{ K}}

ΔSfus=22.0 J/(mol\cdotpK)\Delta S_{\text{fus}} = 22.0 \text{ J/(mol·K)}

For 1.00 mol:

ΔS=1.00 mol×22.0 J/(mol\cdotpK)=22.0 J/K\Delta S = 1.00 \text{ mol} \times 22.0 \text{ J/(mol·K)} = 22.0 \text{ J/K}

Answer: ΔS = 22.0 J/K

Interpretation:

Positive ΔS:

  • Expected! Solid → liquid
  • Ice: ordered crystal lattice
  • Water: molecules can move more freely
  • Disorder increases

Magnitude:

  • ΔS_{fus} always positive
  • ΔS_{vap} much larger (liquid → gas)
  • ΔS_{sub} = ΔS_{fus} + ΔS_{vap}

General pattern for H₂O:

ΔSsub>ΔSvap>ΔSfus>0\Delta S_{\text{sub}} > \Delta S_{\text{vap}} > \Delta S_{\text{fus}} > 0

All phase changes from more ordered → less ordered have ΔS > 0

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