The Hook: Why You Can’t Unscramble an Egg
Three mysteries of everyday life:
- Drop a glass: It shatters into pieces. Why doesn’t it spontaneously reassemble?
- Melt an ice cube: It becomes water. Why doesn’t water spontaneously freeze at room temperature?
- Scramble an egg: It mixes completely. Why can’t you un-scramble it?
Here’s the puzzle: Some of these processes are exothermic (release heat), some are endothermic (absorb heat). Yet they all happen spontaneously in one direction only!
The deeper question: Why does ice melt at room temperature even though melting is endothermic (ΔH > 0)? Shouldn’t reactions favor releasing heat, not absorbing it?
The answer lies in Entropy - the universe’s tendency toward disorder and randomness.
Movie connection (Oppenheimer): “Now I am become Death, the destroyer of worlds” - just like entropy is the destroyer of order!
The Core Concept
What is Entropy?
Entropy (S) is a measure of disorder, randomness, or number of possible microstates in a system.
$$\text{Entropy} \propto \text{Disorder} \propto \text{Number of arrangements}$$In simple terms: Entropy measures how “spread out” or “mixed up” energy and matter are.
Boltzmann’s Definition (Statistical):
$$S = k_B \ln W$$where:
- $k_B$ = Boltzmann constant ($1.38 \times 10^{-23}$ J/K)
- $W$ = Number of microstates (ways to arrange molecules)
Thermodynamic Definition:
$$\boxed{\Delta S = \frac{q_{rev}}{T}}$$where:
- $q_{rev}$ = Heat exchanged in reversible process
- $T$ = Absolute temperature (Kelvin)
Units: J/K or J/mol·K
Visualizing Entropy: The Card Deck Analogy
Ordered deck (low entropy):
- All cards arranged by suit and rank
- Only 1 specific arrangement = 1 microstate
- Very unlikely to occur by chance
Shuffled deck (high entropy):
- Cards randomly mixed
- Billions of possible arrangements = many microstates
- Very likely to occur
That’s why: Once you shuffle cards, they never spontaneously return to perfect order!
The Second Law of Thermodynamics
Statement of the Second Law
“The total entropy of the universe always increases for a spontaneous process.”
$$\boxed{\Delta S_{universe} = \Delta S_{system} + \Delta S_{surroundings} \geq 0}$$For spontaneous process: $\Delta S_{universe} > 0$
At equilibrium: $\Delta S_{universe} = 0$
For non-spontaneous process: $\Delta S_{universe} < 0$ (will not happen!)
Alternative statements:
- “Heat cannot spontaneously flow from cold to hot body”
- “No process is 100% efficient in converting heat to work”
- “Disorder of the universe always increases”
Why is this the “Arrow of Time”?
The Second Law gives direction to time:
- You can’t unbreak a glass (ΔS_universe would decrease)
- You can’t unmix cream from coffee (ΔS_universe would decrease)
- You age and eventually die (biological entropy increases)
This is why we remember the past but not the future - entropy only increases forward in time!
Entropy of Surroundings
For a process at constant temperature and pressure:
$$\Delta S_{surroundings} = -\frac{\Delta H_{system}}{T}$$Why the negative sign?
- If system releases heat (ΔH < 0, exothermic), surroundings absorb heat (ΔS_surr > 0)
- If system absorbs heat (ΔH > 0, endothermic), surroundings release heat (ΔS_surr < 0)
Total entropy change:
$$\Delta S_{universe} = \Delta S_{system} - \frac{\Delta H_{system}}{T}$$This leads us to Gibbs Energy (next topic)!
Interactive Demo: Visualize Entropy Changes
Watch how disorder increases in reversible and irreversible processes.
Entropy Changes in Different Processes
1. Phase Transitions
$$\Delta S_{fusion} = \frac{\Delta H_{fusion}}{T_m}$$ $$\Delta S_{vaporization} = \frac{\Delta H_{vaporization}}{T_b}$$where $T_m$ = melting point, $T_b$ = boiling point
Example: Water at 0°C:
$$\text{H}_2\text{O}(s) \to \text{H}_2\text{O}(l)$$ $$\Delta S_{fusion} = \frac{6.0 \text{ kJ/mol}}{273 \text{ K}} = \frac{6000}{273} = 22 \text{ J/K·mol}$$Trend: $\Delta S_{vap} > \Delta S_{fusion}$ (much larger disorder increase gas vs liquid)
2. Heating a Substance
At constant pressure:
$$\Delta S = nC_P \ln\frac{T_2}{T_1}$$At constant volume:
$$\Delta S = nC_V \ln\frac{T_2}{T_1}$$Why ln? Because $dS = \frac{dq}{T} = \frac{nC_PdT}{T}$, integrating gives ln.
3. Isothermal Expansion of Ideal Gas
At constant temperature:
$$\Delta S = nR\ln\frac{V_2}{V_1} = -nR\ln\frac{P_2}{P_1}$$Interpretation: Expanding gas → molecules more spread out → higher entropy
Sign check:
- Expansion: $V_2 > V_1$ → $\ln\frac{V_2}{V_1} > 0$ → ΔS > 0 ✓
- Compression: $V_2 < V_1$ → $\ln\frac{V_2}{V_1} < 0$ → ΔS < 0 ✓
4. Mixing of Gases
When two ideal gases mix (both at same T, P):
$$\Delta S_{mixing} = -nR(x_1\ln x_1 + x_2\ln x_2)$$where $x_1, x_2$ are mole fractions
Always positive because mixing increases disorder!
5. Chemical Reactions
$$\boxed{\Delta S°_{rxn} = \sum S°(\text{products}) - \sum S°(\text{reactants})}$$Note: Unlike enthalpy, we use absolute entropies $S°$, not formation entropies.
Why? Third Law of Thermodynamics states: $S = 0$ for perfect crystal at 0 K (absolute zero).
Factors Affecting Entropy
1. State of Matter
$$S_{gas} \gg S_{liquid} > S_{solid}$$Reason: Molecules most free to move in gas, least in solid.
Typical values at 298 K:
- Solids: 50-100 J/mol·K
- Liquids: 100-200 J/mol·K
- Gases: 150-300 J/mol·K
Example:
- $S°[\text{H}_2\text{O}(s)] = 48$ J/mol·K
- $S°[\text{H}_2\text{O}(l)] = 70$ J/mol·K
- $S°[\text{H}_2\text{O}(g)] = 189$ J/mol·K
2. Temperature
Higher temperature → Higher entropy
$$\left(\frac{\partial S}{\partial T}\right)_P = \frac{C_P}{T} > 0$$Why? More thermal energy → molecules move faster → more possible arrangements
3. Volume
Larger volume → Higher entropy
For ideal gas: $S \propto \ln V$
Why? More space → molecules more spread out → more possible positions
4. Number of Moles
More particles → Higher entropy
$S$ is extensive property: $S \propto n$
Example:
$$2\text{H}_2\text{O}(l) \quad \text{has} \quad S = 2 \times 70 = 140 \text{ J/K}$$5. Molecular Complexity
More complex molecules → Higher entropy
Reason: More bonds → more vibrational modes → more ways to store energy
Example (standard entropies at 298 K):
- $S°[\text{CH}_4] = 186$ J/mol·K
- $S°[\text{C}_2\text{H}_6] = 230$ J/mol·K
- $S°[\text{C}_3\text{H}_8] = 270$ J/mol·K
6. Number of Particles in Reaction
$$\Delta S > 0 \quad \text{when} \quad n_{products} > n_{reactants}$$Example:
$$2\text{H}_2\text{O}_2(l) \to 2\text{H}_2\text{O}(l) + \text{O}_2(g)$$- Reactants: 2 liquid molecules
- Products: 2 liquid + 1 gas molecules
- Net: Formation of gas → large increase in entropy
Predicting Sign of ΔS for Reactions
Quick Rules
ΔS > 0 (entropy increases):
- Solid → Liquid → Gas (phase change to higher disorder)
- Few moles → Many moles (especially gases)
- Dissociation reactions (1 molecule → many ions/molecules)
- Dissolving ionic solids (usually, but exceptions exist)
ΔS < 0 (entropy decreases):
- Gas → Liquid → Solid (phase change to lower disorder)
- Many moles → Few moles (especially gases)
- Association reactions (many molecules → 1)
- Condensation, crystallization
ΔS ≈ 0:
- No change in number of gas moles
- No phase change
- Similar molecular complexity
Examples
Large positive ΔS:
$$\text{CaCO}_3(s) \to \text{CaO}(s) + \text{CO}_2(g)$$- Solid produces gas → large entropy increase
- Ordered crystal → dispersed ions in solution
Large negative ΔS:
$$\text{N}_2(g) + 3\text{H}_2(g) \to 2\text{NH}_3(g)$$- 4 gas moles → 2 gas moles (Δn_g = -2)
- 3 moles → 2 moles
Third Law of Thermodynamics
“The entropy of a perfect crystal at absolute zero (0 K) is zero.”
$$S = 0 \quad \text{at} \quad T = 0 \text{ K (perfect crystal)}$$Consequence: We can calculate absolute entropies $S°$ (not just ΔS).
This is different from enthalpy! We can only calculate ΔH, not absolute H values.
Why this works:
- At 0 K: No molecular motion, only one possible arrangement (perfect order)
- $W = 1$ (one microstate)
- $S = k_B \ln(1) = 0$
Practical use: Standard entropy tables give $S°_{298}$ values for all substances.
Spontaneity and Entropy
Why Do Endothermic Reactions Sometimes Happen?
The ice melting paradox:
$$\text{H}_2\text{O}(s) \to \text{H}_2\text{O}(l) \quad \text{at 25°C}$$- $\Delta H = +6.0$ kJ/mol (endothermic - unfavorable!)
- But it happens spontaneously!
Resolution: Look at total entropy change:
$$\Delta S_{system} = \frac{6000}{298} = +20.1 \text{ J/K·mol} \quad \text{(favorable!)}$$ $$\Delta S_{surroundings} = -\frac{\Delta H}{T} = -\frac{6000}{298} = -20.1 \text{ J/K·mol}$$But wait, at 25°C:
$$\Delta S_{system} > |\Delta S_{surroundings}|$$Actually, let’s recalculate properly:
At 0°C (273 K), equilibrium:
$$\Delta S_{universe} = \Delta S_{system} + \Delta S_{surroundings} = +22 - 22 = 0$$At 25°C (298 K), spontaneous melting:
$$\Delta S_{system} = +22 \text{ J/K·mol (approximately)}$$ $$\Delta S_{surroundings} = -\frac{6000}{298} = -20.1 \text{ J/K·mol}$$ $$\Delta S_{universe} = 22 - 20.1 = +1.9 \text{ J/K·mol} > 0 \quad \checkmark$$Conclusion: Even though melting absorbs heat (unfavorable enthalpy), the large entropy increase dominates, making the process spontaneous!
Entropy vs Enthalpy Competition
| ΔH | ΔS | Result | Example |
|---|---|---|---|
| - (exo) | + (disorder increases) | Always spontaneous | Combustion |
| + (endo) | - (disorder decreases) | Never spontaneous | Freezing at high T |
| - (exo) | - (disorder decreases) | Spontaneous at low T | Freezing water |
| + (endo) | + (disorder increases) | Spontaneous at high T | Melting ice |
This leads directly to Gibbs Energy!
Memory Tricks & Patterns
Mnemonic for Entropy
“Entropy = Disorder”
“MESS” = More Entropy, Spread Out, Scattered
“Gases are Gossips” (spread information everywhere) → High entropy
“Solids are Secretive” (keep things organized) → Low entropy
Pattern Recognition
Entropy increases when:
- S → L → G (phase change upward)
- Few → Many (particles increase)
- Organized → Messy (disorder increases)
- Temperature increases (more molecular motion)
- Volume increases (more space to spread out)
Mnemonic for phase entropy:
$$S_{ice} < S_{water} < S_{steam}$$“Ice Skaters Slide” → I < W < S (alphabetically!)
Quick Formula Summary
| Process | Formula | Quick Check |
|---|---|---|
| Phase change | $\Delta S = \frac{\Delta H}{T}$ | Always use equilibrium T |
| Heating | $\Delta S = nC_P\ln\frac{T_2}{T_1}$ | T₂ > T₁ → ΔS > 0 |
| Expansion | $\Delta S = nR\ln\frac{V_2}{V_1}$ | V₂ > V₁ → ΔS > 0 |
| Reaction | $\Delta S° = \sum S°_{\text{prod}} - \sum S°_{\text{react}}$ | Use absolute S° values |
Common Mistakes to Avoid
Wrong: “ΔS > 0, so reaction is spontaneous”
Right: “ΔS_universe > 0, so reaction is spontaneous”
A reaction can have ΔS_system < 0 but still be spontaneous if ΔS_surroundings is sufficiently positive!
Example: Freezing water at -10°C
- ΔS_system < 0 (disorder decreases)
- ΔS_surroundings > 0 (heat released to surroundings)
- ΔS_universe > 0 (spontaneous!)
Wrong: $\Delta S°_{rxn} = \sum \Delta_f S°(\text{products}) - \sum \Delta_f S°(\text{reactants})$
Right: $\Delta S°_{rxn} = \sum S°(\text{products}) - \sum S°(\text{reactants})$
Unlike enthalpy, we use absolute entropies $S°$, not “formation entropies”!
Why? Third Law gives us absolute zero reference point for entropy.
Wrong: Using °C in $\Delta S = \frac{q}{T}$
Right: Always use Kelvin (K) for temperature in entropy calculations!
$$\Delta S = \frac{6000 \text{ J}}{25°\text{C}} \quad \text{← WRONG!}$$ $$\Delta S = \frac{6000 \text{ J}}{298 \text{ K}} \quad \text{← Correct!}$$Common assumption: Dissolving always increases entropy (solid → dispersed ions)
Reality: Sometimes ΔS < 0 for dissolving!
Example: $\text{LiCl}(s) \to \text{Li}^+(aq) + \text{Cl}^-(aq)$ has ΔS < 0
Why? Small Li⁺ ion heavily hydrates water molecules, creating more order!
Rule: Entropy of dissolution depends on ion size and hydration effects.
Wrong: Using 298 K for all phase change entropy calculations
Right: Use the actual transition temperature (melting point or boiling point)
$$\Delta S_{fusion} = \frac{\Delta H_{fusion}}{T_m} \quad \text{(use melting point!)}$$Example: For water melting:
- Use T = 273 K (0°C), not 298 K
- $\Delta S = \frac{6000}{273} = 22$ J/K·mol
Practice Problems
Level 1: Foundation (NCERT)
Question: Calculate the entropy of fusion of ice given:
- $\Delta H_{fusion} = 6.0$ kJ/mol
- Melting point = 273 K
Solution:
At melting point (equilibrium):
$$\Delta S_{fusion} = \frac{\Delta H_{fusion}}{T_m}$$ $$\Delta S_{fusion} = \frac{6000 \text{ J/mol}}{273 \text{ K}} = 22.0 \text{ J/K·mol}$$Answer: 22.0 J/K·mol
Interpretation: Positive ΔS confirms disorder increases when solid → liquid.
Question: Predict the sign of ΔS for: (a) $\text{CaCO}_3(s) \to \text{CaO}(s) + \text{CO}_2(g)$ (b) $\text{N}_2(g) + 3\text{H}_2(g) \to 2\text{NH}_3(g)$ (c) $\text{H}_2\text{O}(l) \to \text{H}_2\text{O}(g)$
Solution:
(a) $\text{CaCO}_3(s) \to \text{CaO}(s) + \text{CO}_2(g)$
- Solid produces gas
- 1 mole → 2 moles (including 1 gas)
- ΔS > 0 (large increase)
(b) $\text{N}_2(g) + 3\text{H}_2(g) \to 2\text{NH}_3(g)$
- 4 gas moles → 2 gas moles
- Fewer particles
- ΔS < 0 (decrease)
(c) $\text{H}_2\text{O}(l) \to \text{H}_2\text{O}(g)$
- Liquid → gas
- Large increase in disorder
- ΔS > 0 (very large increase)
Answer: (a) positive, (b) negative, (c) positive (large)
Question: One mole of ideal gas expands isothermally from 10 L to 20 L at 300 K. Calculate ΔS.
Solution:
For isothermal expansion:
$$\Delta S = nR\ln\frac{V_2}{V_1}$$ $$\Delta S = 1 \times 8.314 \times \ln\frac{20}{10}$$ $$\Delta S = 8.314 \times \ln 2$$ $$\Delta S = 8.314 \times 0.693 = 5.76 \text{ J/K}$$Answer: +5.76 J/K
Check: Expansion → ΔS > 0 ✓
Level 2: JEE Main
Question: Calculate $\Delta S°$ for the reaction:
$$\text{CH}_4(g) + 2\text{O}_2(g) \to \text{CO}_2(g) + 2\text{H}_2\text{O}(l)$$Given standard entropies:
- $S°[\text{CH}_4(g)] = 186$ J/mol·K
- $S°[\text{O}_2(g)] = 205$ J/mol·K
- $S°[\text{CO}_2(g)] = 214$ J/mol·K
- $S°[\text{H}_2\text{O}(l)] = 70$ J/mol·K
Solution:
$$\Delta S° = \sum S°(\text{products}) - \sum S°(\text{reactants})$$Products:
$$S°_{\text{products}} = S°[\text{CO}_2] + 2 \times S°[\text{H}_2\text{O}]$$ $$= 214 + 2(70) = 354 \text{ J/K}$$Reactants:
$$S°_{\text{reactants}} = S°[\text{CH}_4] + 2 \times S°[\text{O}_2]$$ $$= 186 + 2(205) = 596 \text{ J/K}$$ $$\Delta S° = 354 - 596 = -242 \text{ J/K}$$Answer: -242 J/K
Interpretation:
- Negative because 3 gas moles → 1 gas mole + 2 liquid
- Significant decrease in disorder
- Yet combustion is spontaneous! Why? Large negative ΔH dominates (see Gibbs Energy)
Question: For the melting of ice at 10°C, calculate ΔS_universe and determine if spontaneous.
Given:
- $\Delta H_{fusion} = 6.0$ kJ/mol
- $T = 283$ K (10°C)
Solution:
System (ice → water):
$$\Delta S_{system} = \frac{\Delta H_{fusion}}{T_{melting}} = \frac{6000}{273} = 22.0 \text{ J/K·mol}$$Note: We use melting point 273 K for the phase transition entropy.
Surroundings: At 283 K (actual temperature):
$$\Delta S_{surroundings} = -\frac{\Delta H_{system}}{T_{actual}} = -\frac{6000}{283} = -21.2 \text{ J/K·mol}$$Universe:
$$\Delta S_{universe} = 22.0 + (-21.2) = +0.8 \text{ J/K·mol}$$Answer: ΔS_universe = +0.8 J/K·mol > 0, therefore spontaneous ✓
Insight: At 10°C (above melting point), entropy gain of system > entropy loss of surroundings, so ice melts spontaneously despite being endothermic!
Question: Calculate entropy change when 2 moles of water are heated from 300 K to 350 K at constant pressure. ($C_P = 75.3$ J/mol·K)
Solution:
$$\Delta S = nC_P\ln\frac{T_2}{T_1}$$ $$\Delta S = 2 \times 75.3 \times \ln\frac{350}{300}$$ $$\Delta S = 150.6 \times \ln(1.167)$$ $$\Delta S = 150.6 \times 0.154$$ $$\Delta S = 23.2 \text{ J/K}$$Answer: +23.2 J/K
Check: Heating → temperature increases → entropy increases → ΔS > 0 ✓
Level 3: JEE Advanced
Question: Calculate the total entropy change when 1 mole of ice at -10°C is converted to steam at 110°C.
Given:
- $C_P[\text{ice}] = 37.7$ J/mol·K
- $C_P[\text{water}] = 75.3$ J/mol·K
- $C_P[\text{steam}] = 33.6$ J/mol·K
- $\Delta H_{fusion} = 6.0$ kJ/mol at 273 K
- $\Delta H_{vaporization} = 40.7$ kJ/mol at 373 K
Solution:
Step 1: Heat ice from -10°C (263 K) to 0°C (273 K)
$$\Delta S_1 = nC_P\ln\frac{T_2}{T_1} = 1 \times 37.7 \times \ln\frac{273}{263}$$ $$= 37.7 \times \ln(1.038) = 37.7 \times 0.0373 = 1.41 \text{ J/K}$$Step 2: Melt ice at 0°C (273 K)
$$\Delta S_2 = \frac{\Delta H_{fusion}}{T_m} = \frac{6000}{273} = 22.0 \text{ J/K}$$Step 3: Heat water from 0°C (273 K) to 100°C (373 K)
$$\Delta S_3 = 1 \times 75.3 \times \ln\frac{373}{273}$$ $$= 75.3 \times \ln(1.366) = 75.3 \times 0.312 = 23.5 \text{ J/K}$$Step 4: Vaporize water at 100°C (373 K)
$$\Delta S_4 = \frac{\Delta H_{vap}}{T_b} = \frac{40700}{373} = 109.1 \text{ J/K}$$Step 5: Heat steam from 100°C (373 K) to 110°C (383 K)
$$\Delta S_5 = 1 \times 33.6 \times \ln\frac{383}{373}$$ $$= 33.6 \times \ln(1.027) = 33.6 \times 0.0266 = 0.89 \text{ J/K}$$Total entropy change:
$$\Delta S_{total} = 1.41 + 22.0 + 23.5 + 109.1 + 0.89 = 156.9 \text{ J/K}$$Answer: +156.9 J/K
Insights:
- Vaporization contributes most (109.1 J/K) - huge disorder increase!
- Fusion contributes less (22.0 J/K)
- Heating contributions are smaller
Question: A Carnot engine operates between 500 K (hot reservoir) and 300 K (cold reservoir).
(a) Calculate maximum efficiency (b) If engine absorbs 1000 J from hot reservoir, calculate entropy change of universe
Solution:
(a) Maximum efficiency (Carnot efficiency):
$$\eta_{max} = 1 - \frac{T_C}{T_H} = 1 - \frac{300}{500} = 1 - 0.6 = 0.4 = 40\%$$(b) Entropy changes:
Heat absorbed from hot reservoir: $q_H = 1000$ J
Work done: $w = \eta \times q_H = 0.4 \times 1000 = 400$ J
Heat rejected to cold reservoir: $q_C = q_H - w = 1000 - 400 = 600$ J
Hot reservoir:
$$\Delta S_H = -\frac{q_H}{T_H} = -\frac{1000}{500} = -2.0 \text{ J/K}$$Cold reservoir:
$$\Delta S_C = +\frac{q_C}{T_C} = +\frac{600}{300} = +2.0 \text{ J/K}$$Universe (for reversible Carnot engine):
$$\Delta S_{universe} = \Delta S_H + \Delta S_C = -2.0 + 2.0 = 0$$Answer:
- (a) Maximum efficiency = 40%
- (b) ΔS_universe = 0 (reversible process!)
Key insight: For reversible Carnot cycle, ΔS_universe = 0 (equilibrium condition). Real engines are irreversible and have ΔS_universe > 0, meaning lower efficiency.
Question: One mole of ideal gas at 300 K expands freely (into vacuum) from 10 L to 20 L.
(a) Calculate ΔS_system (b) Calculate ΔS_surroundings (c) Show this is irreversible
Solution:
(a) System entropy change:
Even though process is irreversible, entropy is a state function:
$$\Delta S_{system} = nR\ln\frac{V_2}{V_1} = 1 \times 8.314 \times \ln 2$$ $$= 8.314 \times 0.693 = 5.76 \text{ J/K}$$(b) Surroundings:
Free expansion: $q = 0$ (no heat exchange)
Therefore:
$$\Delta S_{surroundings} = 0$$(c) Universe:
$$\Delta S_{universe} = \Delta S_{system} + \Delta S_{surroundings}$$ $$= 5.76 + 0 = 5.76 \text{ J/K} > 0$$Answer:
- (a) ΔS_system = +5.76 J/K
- (b) ΔS_surroundings = 0
- (c) ΔS_universe = +5.76 J/K > 0 → Irreversible!
Deep insight:
- Free expansion is spontaneous (ΔS_universe > 0)
- But does no work (w = 0) despite volume change
- This is irreversible - you can’t spontaneously recompress the gas
- Maximum work (reversible) would be: $w_{max} = -nRT\ln 2 = -1729$ J
- Free expansion wastes this work potential - this is why entropy increases!
Comparison:
| Process | w (J) | q (J) | ΔS_sys (J/K) | ΔS_surr (J/K) | ΔS_univ (J/K) |
|---|---|---|---|---|---|
| Free expansion | 0 | 0 | +5.76 | 0 | +5.76 |
| Reversible isothermal | -1729 | +1729 | +5.76 | -5.76 | 0 |
The reversible process extracts maximum work and has ΔS_univ = 0 (equilibrium). The free expansion is wasteful and has ΔS_univ > 0 (irreversible).
Quick Revision Box
Essential Formulas
| Concept | Formula | Condition |
|---|---|---|
| Entropy definition | $\Delta S = \frac{q_{rev}}{T}$ | Reversible process |
| Second Law | $\Delta S_{universe} \geq 0$ | Always |
| Surroundings | $\Delta S_{surr} = -\frac{\Delta H_{sys}}{T}$ | Constant T, P |
| Phase transition | $\Delta S = \frac{\Delta H}{T_{transition}}$ | At equilibrium T |
| Heating | $\Delta S = nC_P\ln\frac{T_2}{T_1}$ | Constant P |
| Expansion (ideal gas) | $\Delta S = nR\ln\frac{V_2}{V_1}$ | Isothermal |
| Reaction | $\Delta S° = \sum S°_{prod} - \sum S°_{react}$ | Use absolute S° |
Entropy Trends
| Factor | Effect on S |
|---|---|
| Solid → Liquid → Gas | Increases (S_g » S_l > S_s) |
| Temperature increases | Increases |
| Volume increases | Increases |
| Particles increase | Increases |
| Molecular complexity increases | Increases |
| Dissolution (usually) | Increases |
Spontaneity Quick Guide
| Condition | Spontaneity |
|---|---|
| ΔS_universe > 0 | Spontaneous |
| ΔS_universe = 0 | Equilibrium |
| ΔS_universe < 0 | Non-spontaneous |
Standard Entropy Values (298 K)
| Substance | S° (J/mol·K) |
|---|---|
| H₂(g) | 131 |
| O₂(g) | 205 |
| N₂(g) | 192 |
| H₂O(l) | 70 |
| H₂O(g) | 189 |
| CO₂(g) | 214 |
| CH₄(g) | 186 |
Teacher’s Summary
1. Entropy Measures Disorder
$$S = k_B \ln W, \quad \Delta S = \frac{q_{rev}}{T}$$Higher entropy = more disorder, more randomness, more possible microstates.
2. Second Law: The Arrow of Time
$$\Delta S_{universe} = \Delta S_{system} + \Delta S_{surroundings} \geq 0$$For spontaneous processes, total entropy of universe always increases. This gives direction to time.
3. Entropy Depends on State
- Gases » Liquids > Solids
- Higher T → Higher S
- Larger V → Higher S
- More complex molecules → Higher S
4. Predicting ΔS
- Phase change to higher state → ΔS > 0
- More gas moles in products → ΔS > 0
- Dissociation/dissolving → Usually ΔS > 0
5. Spontaneity Requires ΔS_universe > 0
Not just ΔS_system! Must consider surroundings too:
$$\Delta S_{surr} = -\frac{\Delta H_{sys}}{T}$$This competition leads to Gibbs Energy (next topic).
6. Entropy is a State Function
Even for irreversible processes, ΔS depends only on initial and final states, not path.
“The universe tends toward maximum disorder - entropy is destiny.”
JEE Weightage: 2-3 questions per paper. Often combined with Gibbs Energy for spontaneity predictions.
Time-saving tip: For MCQs, quickly check if gas moles increase or decrease to predict sign of ΔS without calculation!
Cross-Links to Related Concepts
Prerequisites:
Builds on:
- States of Matter - Phase transitions
- Atomic Structure - Microstates, energy levels
Leads to:
- Gibbs Energy - Combining ΔH and ΔS for spontaneity
- Equilibrium - Entropy at equilibrium
Applications:
- Chemical Kinetics - Activation entropy
- Electrochemistry - Entropy change in cells
Related in Physics:
- Statistical Mechanics - Microscopic basis of entropy
- Heat Engines - Carnot cycle, efficiency
What’s Next?
Gibbs Free Energy - The ultimate spontaneity predictor! Learn how to combine enthalpy and entropy into one powerful criterion, master the equation ΔG = ΔH - TΔS, and predict equilibrium constants without doing experiments!
Teaser:
- We know WHAT energy changes (ΔH)
- We know WHY things happen (ΔS)
- Now we’ll learn WHEN exactly they happen (ΔG)!
The magic equation that combines everything:
$$\Delta G = \Delta H - T\Delta S$$When ΔG < 0, reaction is spontaneous! When ΔG = 0, system is at equilibrium! When ΔG > 0, reaction is non-spontaneous!