• 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:

$\Delta S_{\text{universe}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} > 0$

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):

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

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)

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

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}:

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

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!)

At equilibrium (melting point): process is reversible

Calculate ΔS_{fus}:

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

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

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

For 1.00 mol:

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

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:

$\Delta S_{\text{sub}} > \Delta S_{\text{vap}} > \Delta S_{\text{fus}} > 0$

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