Thermodynamics: Entropy and Spontaneity - Detailed Study Notes
The Second Law of Thermodynamics
Focus: The degree of disorder (entropy) of a system.
Entropy: Qualitative Description
Entropy is a measure of how spread out a system’s energy is in space.
For a given system, the greater the volume it occupies, the greater its entropy.
Entropy: Quantitative Description — 1st Approach (Boltzmann)
Boltzmann’s definition: S = k \, \ln W
where S is entropy, k = 1.38 \times 10^{-23}\ \mathrm{J\,K^{-1}} is the Boltzmann constant.
W = number of energetically equivalent ways the molecules in a system can be arranged.
In the slide: W = X^N where
X = number of cells (or microstates per molecule)
N = number of molecules
Concept: More microstates (larger W) -> larger entropy.
Entropy and Microstates (Illustrative Example)
Figure description (conceptual): two-compartment container separated by a barrier.
Before barrier removal: molecules confined to left side; 2 cells, 2 molecules → 4 possible arrangements.
After barrier removal: volume (and number of cells) doubles; new number of possible arrangements is 4^2 = 16, of which eight place molecules on opposite sides (the most probable outcome).
Key takeaway: Increasing volume increases the number of microstates, hence increases entropy.
Entropy: Three Possible States for a Two-Molecule System (Qualitative Count)
Three states discussed:
1) One molecule on each side (8 possible arrangements).
2) Both molecules on the left side (4 possible arrangements).
3) Both molecules on the right side (4 possible arrangements).Principle: The state with the largest number of possible arrangements (microstates) has the greatest entropy.
General rule: Entropy increases with the number of microstates of the system.
Entropy Changes in Reactions: Phase Transitions
Predicting ΔS for reactions:
Melting: S{\text{liquid}} > S{\text{solid}}
Vaporization: S{\text{gas}} > S{\text{liquid}}
Dissolving: S{\text{solution}} > S{\text{solute}} + S_{\text{solvent}}
Sample Exercise 2: Sign of ΔS for Various Processes
a) Decomposition of CaCO₃(s) → CaO(s) + CO₂(g) → ΔS positive.
b) Heating bromine vapor from 450°C to 800°C → ΔS positive.
c) Condensation of water vapor on a cold surface → ΔS negative.
d) NH₂(g) + HCl(g) → NH₄Cl(s) → ΔS negative.
e) Dissolution of sugar in water → ΔS positive.
Answers: +, +, -, -, +.
Sample Exercise 3: Positive ΔS Determination
Which reactions have ΔS > 0?
a) 2 O₃(g) → 3 O₂(g) → likely positive (more gas molecules).
b) 4 Fe(s) + 3 O₂(g) → 2 Fe₂O₃(s) → negative (formation of a solid).
c) 2 H₂O₂(aq) → 2 H₂O(l) + O₂(g) → positive (gas produced, liquids decreased).
d) 2 Li(s) + 2 H₂O(l) → 2 LiOH(aq) + H₂(g) → positive (gas produced + ions in solution).
e) 2 NH₃(g) → N₂(g) + 3 H₂(g) → positive.
Answer: All except b (i.e., a, c, d, e have ΔS > 0).
Entropy: 2nd Approach — Clausius Equation
Rudolph Clausius extended Carnot’s work:
\Delta S = \frac{q_{\text{rev}}}{T}
where q_{\text{rev}} is the heat released or absorbed in a reversible process and T is the temperature (K) at which the heat transfer occurred.
This equation encapsulates the core idea that entropy increases in spontaneous processes (for the universe).
Entropy and the Second Law (Fundamental Statement)
The entropy change of the universe for a reversible process is zero; for an irreversible (spontaneous) process it is positive:
\Delta S{\text{universe}} = \Delta S{\text{sys}} + \Delta S_{\text{surr}}
Reversible: \Delta S_{\text{universe}} = 0
Irreversible: \Delta S_{\text{universe}} > 0
The overarching principle: “The entropy of the universe increases for any spontaneous process.”
Sample Exercise 4: Condensation of Ethanol at its Boiling Point
Given: Normal boiling point of ethanol, C2H5OH, is 78.30^{\circ}C; enthalpy of vaporization \Delta H_{vap} = 38.56\ \text{kJ/mol}; 68.3 g of ethanol at 1 atm condenses to liquid at the normal boiling point.
Steps (as shown):
Moles of ethanol: n = \frac{68.3\ \text{g}}{46.07\ \text{g/mol}} \approx 1.4835\ \text{mol}
Heat released reversibly (condensation): q{rev} = -n \Delta H{vap} = -1.4835 \times 38.56\ \text{kJ} \approx -57.2\ \text{kJ}
Temperature at condensation: T = 351.3\ \text{K}
Entropy change of system: \Delta S{sys} = \frac{q{rev}}{T} \approx \frac{-57.2\ \text{kJ}}{351.3\ \text{K}} \approx -0.163\ \text{kJ K}^{-1}
Therefore: \Delta S_{system} \approx -0.163\ \text{kJ K}^{-1}
Note: Some slides show units as kJ/mol, but standard practice for entropy is J/(mol·K).
Entropy Changes in the Universe: Surroundings and System
Definitions:
The system is the portion of the universe being studied; the surroundings are everything else.
The universe is the sum: \Delta S{\text{universe}} = \Delta S{\text{system}} + \Delta S_{\text{surroundings}}
For a reversible process: \Delta S_{\text{universe}} = 0
For an irreversible process: \Delta S_{\text{universe}} > 0
Insight: These relationships underpin the second law: the total entropy of the universe increases for spontaneous processes.
Sample Exercise 5: Entropy Change and Surroundings
Reaction: Na(s) + Cl₂(g) → NaCl(s). Spontaneous with a decrease in the entropy of the system.
Conclusion: Since the reaction is spontaneous but reduces system entropy, the entropy change of the surroundings must increase: \Delta S_{\text{surroundings}} > 0.
Answer: a. ΔSsurroundings increases.
Entropy Changes in Everyday Processes (Universe Perspective)
Ice melts at room temperature (endothermic):
Heat flows from surroundings to system.
Surroundings temperature decreases; \Delta S{sys} > 0; \Delta S{surroundings} < 0.
Hot cocoa cools spontaneously to room temperature (exothermic):
Heat flows from system to surroundings.
Surroundings temperature increases; \Delta S{sys} < 0; \Delta S{surroundings} > 0.
ΔS°rxn (Standard Entropy of Reaction)
Definition: \Delta S^{\circ}{\mathrm{rxn}} = \sum\limits{\text{products}} \nui S^{\circ}{i,\text{products}} - \sum\limits{\text{reactants}} \nuj S^{\circ}_{j,\text{reactants}}
Note: S^{\circ} are standard molar entropies (units: J mol⁻¹ K⁻¹).
Hess’s Law can be applied to standard entropies of reaction.
Sample Exercise 6: Be(OH)₂(s) → BeO(s) + H₂O(g)
Given: S^{\circ} Be(OH)₂(s) = 50.21 J mol⁻¹ K⁻¹, BeO(s) = 13.77 J mol⁻¹ K⁻¹, H₂O(g) = 188.83 J mol⁻¹ K⁻¹.
Calculation:
\Delta S^{\circ}_{\mathrm{rxn}} = [1\cdot S^{\circ}(\text{BeO}) + 1\cdot S^{\circ}(\text{H₂O(g)})] - [1\cdot S^{\circ}(\text{Be(OH)₂(s)})]
= (13.77 + 188.83) - 50.21 = 152.39\ \text{J K}^{-1}\text{ mol}^{-1}
Result: \Delta S^{\circ}_{\mathrm{rxn}} \approx +152.4\ \text{J K}^{-1}\text{ mol}^{-1}
Sample Exercise 7: CO₂(g) + H₂O(g) → CH₃OH(g) + O₂(g)
Given: S° values (units shown as kJ/mol in slides, but treated as J mol⁻¹ K⁻¹):
CO₂(g) = 213.6; H₂O(g) = 188.83; CH₃OH(g) = 237.6; O₂(g) = 205.0.
Stoichiometry shown in slides (note: actual balancing in slides appears inconsistent; use the exact given calculation for consistency):
Products: 2 CH₃OH and 3 O₂ → 2\cdot 237.6 + 3\cdot 205.0 = 1090.2\,\text{J mol}^{-1}\text{K}^{-1}
Reactants: 2 CO₂ and 4 H₂O → 2\cdot 213.6 + 4\cdot 188.83 = 1182.52\,\text{J mol}^{-1}\text{K}^{-1}
\Delta S^{\circ}_{\mathrm{rxn}} = 1090.2 - 1182.52 = -92.3\ \text{J K}^{-1}\text{ mol}^{-1}
Note: The exact stoichiometric balancing for the listed transformation may differ; the slide’s calculation yields \Delta S^{\circ}_{\mathrm{rxn}} \approx -92.3\ \text{J K}^{-1}\text{ mol}^{-1} for the given coefficients.
Unfolded vs Folded; Active vs Inactive (Biological Context)
Slides show a schematic of enzyme catalysis: unfolded (active catalysis) → folded state (active catalysis) → chain of enzyme conformations; highlights that entropy changes in immediate surroundings for biological processes can be difficult to determine.
Key takeaway: In biology, local entropy changes can be hard to measure; overall universe considerations still apply.
Gibbs Free Energy and Spontaneity
Relationships that connect enthalpy, entropy, and spontaneity at constant temperature and pressure:
Entropy of the universe: \Delta S{\mathrm{univ}} = \Delta S{\mathrm{sys}} + \Delta S_{\mathrm{surr}}
Surroundings at constant pressure: \Delta S{\mathrm{surr}} = \frac{q{\mathrm{sys}}}{T} = \frac{-\Delta H_{\mathrm{sys}}}{T}
Therefore: \Delta S{\mathrm{univ}} = \Delta S{\mathrm{sys}} - \frac{\Delta H_{\mathrm{sys}}}{T}
Rearrange to connect with Gibbs free energy:
-T\,\Delta S{\mathrm{univ}} = \Delta H{\mathrm{sys}} - T\,\Delta S{\mathrm{sys}} = \Delta G{\mathrm{sys}}
Thus, \Delta G = \Delta H - T\Delta S
Condition for spontaneity at constant temperature and pressure:
If \Delta G < 0, the process is spontaneous in the forward direction.
If \Delta G > 0, the process is nonspontaneous in the forward direction (the reverse is spontaneous).
If \Delta G = 0, the system is at equilibrium.
Sign Table: How AH and AS Affect Spontaneity
A compact summary (signs of \Delta H and \Delta S determine spontaneity at different temperatures):
If \Delta H > 0 and \Delta S > 0: spontaneous at high T, nonspontaneous at low T.
If \Delta H < 0 and \Delta S < 0: spontaneous at low T, nonspontaneous at high T.
If one sign is positive and the other negative, spontaneity depends on the magnitude of T.
The formula to use is: \Delta G = \Delta H - T\Delta S.
Sample Exercise 1: Melting Ice at Room Temperature
Given: \Delta H{sys} = 53.6\ \text{kJ}, \Delta S{sys} = 132\ \text{J K}^{-1}\text{mol}^{-1}, room temperature T = 25^{\circ}C = 298\ \text{K}.
Compute \Delta G = \Delta H - T\Delta S:
Convert entropy to kJ: \Delta S = 0.132\ \text{kJ K}^{-1}\text{mol}^{-1}.
\Delta G = 53.6 - (298\ \text{K})(0.132) \approx 53.6 - 39.3 \approx 14.3\ \text{kJ}
Result: Nonspontaneous at room temperature (positive \Delta G).
Sample Exercise 2: Free Energy of a Reaction
Given: \Delta H{sys} = -35.4\ \text{kJ}, \Delta S{sys} = -85.5\ \text{J K}^{-1}\text{mol}^{-1}, T = 25^{\circ}C = 298\ \text{K}.
Compute \Delta G = \Delta H - T\Delta S:
Convert entropy to kJ: \Delta S = -0.0855\ \text{kJ K}^{-1}\text{mol}^{-1}.
\Delta G = -35.4 - (298)(-0.0855) \approx -35.4 + 25.5 \approx -9.92\ \text{kJ}
Result: Spontaneous reaction.
ΔG°rxn and Formation Energies
Standard Gibbs energy change of reaction:
\Delta G^{\circ}{\mathrm{rxn}} = \sum{\text{products}} \nui \Delta G^{\circ}{f,i} - \sum{\text{reactants}} \nuj \Delta G^{\circ}_{f,j}
where \Delta G^{\circ}_{f} is the standard free energy of formation (in J mol⁻¹).
Important note: For pure solids, pure liquids, and elements in their standard states, \Delta G^{\circ}_{f} = 0 by convention (Hess’s Law applied to standard free energies).
Sample Exercise 3: Standard Gibbs Energy Changes
Reaction: \mathrm{CH}4(g) + 2\ \mathrm{O}2(g) \rightarrow \mathrm{CO}2(g) + 2\ \mathrm{H}2\mathrm{O}(l)
Given: \Delta G^{\circ}{f}(\mathrm{CH4}) = -50.4\ \text{kJ/mol}, \Delta G^{\circ}{f}(\mathrm{CO2}) = -394.4\ \text{kJ/mol}, \Delta G^{\circ}{f}(\mathrm{H2O(l)}) = -237.1\ \text{kJ/mol}
Compute: \Delta G^{\circ}_{\mathrm{rxn}} = [(-394.4) + 2(-237.1)] - [(-50.4) + 2(0)]
Result (given): -818.2\ \text{kJ/mol}; spontaneous.
Sample Exercise 4: MgO Decomposition
Reaction: \mathrm{MgO(s)} \rightarrow \mathrm{Mg(s)} + \mathrm{O_2(g)}
Given: \Delta G^{\circ}_{f}(\mathrm{MgO(s)}) = -569.3\ \text{kJ/mol}; Mg(s) and O₂(g) have 0 formation energy in their standard states.
Result (given): \Delta G^{\circ}_{\mathrm{rxn}} = +1139\ \text{kJ/mol}; nonspontaneous.
Pop Quiz: Haber Process (N₂ + 3 H₂ ⇌ 2 NH₃)
Given: \Delta H^{\circ} = -92.38\ \text{kJ}, \Delta S^{\circ} = -198.3\ \text{J K}^{-1}.
Assume these don’t change with temperature.
Tasks:
a) Predict how \Delta G^{\circ} changes with increasing temperature.
b) Compute \Delta G^{\circ} at 25°C (298 K).
c) Compute \Delta G^{\circ} at 5000°C (5273 K).
Answers (from slides):
a) \Delta G^{\circ} becomes more positive or less negative with increasing temperature (i.e., less favorable at higher T).
b) \Delta G^{\circ} \approx -33.3\ \text{kJ} (spontaneous at 25°C).
c) \Delta G^{\circ} \approx 61\ \text{kJ} (nonspontaneous at 5000°C).
End of Module 1: Thermochemistry
This module covers the Second Law, entropy, and Gibbs free energy with qualitative and quantitative frameworks, key equations, and multiple worked examples to illustrate spontaneity and the role of temperature.
Real-world relevance includes phase transitions, dissolution, biological entropy questions, and energetic considerations in chemical processes.
Notes on notation and units
In several worked examples, standard entropy values were shown with units of kJ mol⁻¹ K⁻¹ in slides, but the conventional unit for standard molar entropies is J mol⁻¹ K⁻¹.
Always verify unit consistency when performing computations: convert to a consistent unit before calculating \Delta S or $$\Delta G).