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: 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:

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

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}:

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

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!)

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:

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

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