Thermodynamics: The Second and Third Laws and Free Energy

Predicting Signs of Entropy Change

• General Rules for Entropy ($\Delta S$) Prediction:
  • Negative Entropy Change ($\Delta S < 0$):
    • Reduction in the number of ions or particles in a solution.
    • Decreased dispersal of matter.
    • Net decrease in the amount of gaseous species.
    • Phase transition from liquid to solid (freezing), as the liquid becomes a more ordered solid.
  • Positive Entropy Change ($\Delta S > 0$):
    • Phase transition from solid to liquid (melting), resulting in a net increase in the dispersal of matter.
    • A solid dissolving to produce an increase of mobile ions in a solution.
    • A relatively ordered solid transitioning into a gas (sublimation).
    • A net increase in the amount of gaseous species during a chemical reaction.

The Second Law of Thermodynamics

• Core Concept: The second law of thermodynamics identifies entropy ($S$) as the property used to reliably predict the spontaneity of a process.
• Entropy of the Universe: To determine spontaneity, entropy changes must include both the system and the surroundings. In thermodynamic models, the universe encompasses everything (system + surroundings):
  • $\\Delta S_{univ} = \\Delta S_{sys} + \\Delta S_{surr}$
• Spontaneity Criteria:
  • Spontaneous Process: The total entropy of the universe increases ($\\Delta S_{univ} > 0$).
  • Nonspontaneous Process: The total entropy of the universe decreases ($\\Delta S_{univ} < 0$). Such a process is spontaneous in the opposite direction.
  • Equilibrium: The total entropy change of the universe is zero ($\\Delta S_{univ} = 0$).

Heat Flow and Entropy Scenarios

• Scenario 1: Spontaneous Heat Flow (Hot to Cold):
  • Heat flows from a hotter object ($T_{sys}$) to a cooler object ($T_{surr}$).
  • The loss of heat by the system ($-q_{rev}$) equals the gain of heat by the surroundings ($q_{rev}$).
  • Since $T_{sys} > T_{surr}$, the magnitude of entropy decrease in the system is less than the entropy increase in the surroundings.
  • Result: $\\Delta S_{univ} > 0$.
• Scenario 2: Nonspontaneous Heat Flow (Cold to Hot):
  • Heat flows from a cooler object to a hotter object.
  • This is never observed to occur spontaneously.
  • The magnitude of entropy change for the surroundings is greater than that of the system, but the direction of heat flow yields a negative value for the universe.
  • Result: $\\Delta S_{univ} < 0$.
• Scenario 3: Equilibrium ($T_{sys} \\approx T_{surr}$):
  • The magnitudes of entropy changes for the system and surroundings are essentially the same.
  • Result: $\\Delta S_{univ} = 0$.

Thermodynamic Approximations for Surroundings

• In many applications, the surroundings are vastly larger than the system (e.g., fuel combustion in the Earth's atmosphere).
• The heat gained or lost by the surroundings ($q_{surr}$) is a tiny fraction of its total thermal energy.
• Approximately, the heat of the surroundings is the negative of the heat of the system: $q_{surr} = -q_{rev}$.
• The entropy change of the surroundings is calculated as:
  • $\\Delta S_{surr} = \\frac{q_{surr}}{T}$

Example 12.4: Spontaneity of Ice Melting

• Process: $H_2O(s) \\rightarrow H_2O(l)$
• System Data: $\\Delta S_{sys} = 22.1\\,J/K$. The process requires surroundings to transfer $6.00\\,kJ$ ($6000\\,J$) of heat to the system ($q_{surr} = -6.00\\,kJ$).
• At $-10.00\\,^{\\circ}C$ ($263.15\\,K$):
  • $\\Delta S_{univ} = \\Delta S_{sys} + \\frac{q_{surr}}{T}$
  • $\\Delta S_{univ} = 22.1\\,J/K + \\frac{-6000\\,J}{263.15\\,K} = 22.1\\,J/K - 22.8\\,J/K = -0.7\\,J/K$
  • Conclusion: Nonspontaneous ($\\Delta S_{univ} < 0$).
• At $+10.00\\,^{\\circ}C$ ($283.15\\,K$):
  • $\\Delta S_{univ} = 22.1\\,J/K + \\frac{-6000\\,J}{283.15\\,K} = 22.1\\,J/K - 21.2\\,J/K = +0.9\\,J/K$
  • Conclusion: Spontaneous ($\\Delta S_{univ} > 0$).
• Opposite Process (Freezing):
  • Entropy is a state function: $\\Delta S_{freezing} = -\\Delta S_{melting} = -22.1\\,J/K$.
  • Heat transfer: $q_{surr} = +6.00\\,kJ$.
  • Conclusion at $-10.00\\,^{\\circ}C$: Spontaneous ($+0.7\\,J/K$).
  • Conclusion at $+10.00\\,^{\\circ}C$: Nonspontaneous ($-0.9\\,J/K$).

The Third Law of Thermodynamics

• Definition: The entropy of a pure, perfectly crystalline substance at absolute zero ($0\\,K$) is zero.
• Theoretical Basis:
  • A perfectly crystalline solid at $0\\,K$ possesses no kinetic energy.
  • The system is described by a single microstate ($W = 1$) because there is only one possible location for each identical atom or molecule.
  • According to Boltzmann’s equation ($S = k\\ln(W)$), if $W=1$, then $S = k\\ln(1) = 0$.
• Standard Entropies ($S^{\\circ}$):
  • Measured for $1\\,mole$ of substance under standard conditions ($1\\,bar$ pressure, $298.15\\,K$).
  • Standard Entropies allow for the calculation of the absolute entropy change of chemical reactions.

Calculating Standard Entropy Changes ($\\Delta S^{\\circ}$)

• Mathematical Formula:
  • $\\Delta S^{\\circ} = \\sum \\nu S^{\\circ}(\\text{products}) - \\sum \\nu S^{\\circ}(\\text{reactants})$
  • $\\nu$ represents the stoichiometric coefficients from the balanced chemical equation.
• Standard Entropy Values (Table 12.2 at $298.15\\,K$):
  • $C(s, graphite) = 5.740\\,J\\,mol^{-1}K^{-1}$
  • $C(s, diamond) = 2.38\\,J\\,mol^{-1}K^{-1}$
  • $CO(g) = 197.7\\,J\\,mol^{-1}K^{-1}$
  • $CO_2(g) = 213.8\\,J\\,mol^{-1}K^{-1}$
  • $CH_4(g) = 186.3\\,J\\,mol^{-1}K^{-1}$
  • $C_2H_4(g) = 219.3\\,J\\,mol^{-1}K^{-1}$
  • $C_2H_6(g) = 229.2\\,J\\,mol^{-1}K^{-1}$
  • $CH_3OH(l) = 126.8\\,J\\,mol^{-1}K^{-1}$
  • $C_2H_5OH(l) = 160.7\\,J\\,mol^{-1}K^{-1}$
  • $H_2(g) = 130.7\\,J\\,mol^{-1}K^{-1}$
  • $H(g) = 114.7\\,J\\,mol^{-1}K^{-1}$
  • $H_2O(g) = 188.8\\,J\\,mol^{-1}K^{-1}$
  • $H_2O(l) = 70.0\\,J\\,mol^{-1}K^{-1}$
  • $HCl(g) = 186.8\\,J\\,mol^{-1}K^{-1}$
  • $H_2S(g) = 205.7\\,J\\,mol^{-1}K^{-1}$
  • $O_2(g) = 205.2\\,J\\,mol^{-1}K^{-1}$

Example 12.5: Determination of $\\Delta S^{\\circ}$ for Phase Transition

• Process: $H_2O(g) \\rightarrow H_2O(l)$
• Calculation:
  • $\\Delta S^{\\circ} = S^{\\circ}(H_2O(l)) - S^{\\circ}(H_2O(g))$
  • $\\Delta S^{\\circ} = 70.0\\,J\\,mol^{-1}K^{-1} - 188.8\\,J\\,mol^{-1}K^{-1} = -118.8\\,J\\,mol^{-1}K^{-1}$
• Observation: The negative sign is expected for condensation (gas to liquid transition).

Example 12.6: Determination of $\\Delta S^{\\circ}$ for Combustion

• Reaction: $2CH_3OH(l) + 3O_2(g) \\rightarrow 2CO_2(g) + 4H_2O(l)$
• Calculation:
  • $\\Delta S^{\\circ} = [2 \\times S^{\\circ}(CO_2(g)) + 4 \\times S^{\\circ}(H_2O(l))] - [2 \\times S^{\\circ}(CH_3OH(l)) + 3 \\times S^{\\circ}(O_2(g))]$
  • $\\Delta S^{\\circ} = [2 \\times 213.8 + 4 \\times 70.0] - [2 \\times 126.8 + 3 \\times 205.2]$
  • $\\Delta S^{\\circ} = 707.6 - 869.2 = -161.6\\,J/K$

Introduction to Gibbs Free Energy

• Origin: Developed by American mathematician Josiah Willard Gibbs in the late 19th century.
• Purpose: To determine spontaneity using system properties ONLY, avoiding the need for surroundings measurements required by the second law.
• Definition: Gibbs free energy ($G$) is defined in terms of system enthalpy ($H$), temperature ($T$), and entropy ($S$):
  • $G = H - TS$