Entropy and the Second Law

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

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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:

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

Spontaneous processes: ΔS_{universe} > 0

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

Predicting Entropy Changes

ΔS > 0 (entropy increases):

Phase changes: solid → liquid → gas
Dissolving: Solid/liquid dissolves in solvent
More gas molecules: Δn_{gas} > 0
Temperature increase: Same substance, higher T
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:

CaCO₃(s) → CaO(s) + CO₂(g)
- 0 gas → 1 gas
- Δn_{gas} = +1
- ΔS° > 0 ✓
2H₂(g) + O₂(g) → 2H₂O(l)
- 3 gas → 0 gas
- Δn_{gas} = -3
- ΔS° < 0 ✓
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)

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

📚 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:

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

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!)

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:

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

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

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Gibbs Free Energy and Spontaneity

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