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):

$S_{\text{gas}} > S_{\text{liquid}} > S_{\text{solid}}$

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:

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

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:

$\Delta S = \frac{q_{\text{rev}}}{T}$

At constant pressure:

$\Delta S = \frac{nC_p\Delta T}{T}$

Phase changes:

$\Delta S_{\text{phase}} = \frac{\Delta H_{\text{phase}}}{T}$

Example: Melting ice at 0°C (273 K)

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

(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: