Thermodynamics and Gibbs Free Energy Study Notes

Calculation of Standard Entropy Change ($\Delta S^\circ$)

• Entropy Change for Phase Transitions (Condensation):

  • For the phase transition of condensation, the value for $\Delta S^\circ$ is negative. This follows the principles discussed in previous sections regarding how entropy changes during phase changes.
  • Check Your Learning Calculation: The standard entropy change calculated for the specified process in the material is $-120.6\,J\,K^{-1}\,mol^{-1}$.

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

  • Process: The combustion of methanol ($CH_3OH$).
  • Calculation Method: Standard entropy changes are calculated using standard entropies ($S^\circ$) as provided in thermodynamic data tables.
  • Check Your Learning Result: For the reaction provided in this section, the calculated standard entropy change is $24.7\,J/K$.

12.4 Free Energy: Learning Objectives

• By the conclusion of this section, learners should be able to:
  • Define the concept of Gibbs free energy and describe its direct relationship to the spontaneity of a process.
  • Calculate the free energy change ($\Delta G$) for a process by utilizing the free energies of formation ($\Delta G_f^\circ$) for all reactants and products involved.
  • Calculate the free energy change for a process by utilizing the enthalpies of formation and the absolute entropies for the reactants and products.

Introduction to Gibbs Free Energy ($G$)

• Background and Context:
  • The second law of thermodynamics identifies spontaneity by measuring the entropy change of the system ($\Delta S_{sys}$) and the entropy change of the surroundings ($\Delta S_{surr}$).
  • A significant challenge in this approach is the requirement to measure surroundings.
  • Josiah Willard Gibbs: An American mathematician in the late nineteenth century introduced a new approach to identifying spontaneity using only system properties.
  • Definition: This thermodynamic property is named Gibbs free energy ($G$), or simply free energy. It is defined in terms of a system’s enthalpy ($H$), entropy ($S$), and temperature ($T$) through the following equation:     $G = H - TS$

Mathematical Definition and Free Energy Change ($\Delta G$)

• State Function Characteristics:
  • Free energy is a state function, meaning its value is determined only by the state of the system and not the path taken to reach it.
  • At constant temperature ($T$) and pressure ($P$), the free energy change ($\Delta G$) for a process is expressed as:     $\Delta G = \Delta H - T\Delta S$
  • Note: In chemical thermodynamic shorthand, the subscript "sys" is typically omitted from the variables in this equation for simplicity.

Derivation of the Free Energy Spontaneity Indicator

• Connection to the Second Law:
  • The second law identifies spontaneity through the universe’s entropy change:     $\Delta S_{univ} = \Delta S_{sys} + \Delta S_{surr}$
  • From the first law of thermodynamics, for a process at constant pressure: $q_{surr} = -q_{sys}$ and $q_{sys} = \Delta H$.
  • Substituting these into the entropy of surroundings definition ($\Delta S_{surr} = \frac{q_{surr}}{T}$) yields:     $\Delta S_{univ} = \Delta S_{sys} - \frac{\Delta H}{T}$
  • Multiplying both sides of the equation by $-T$ and rearranging the terms result in:     $-T\Delta S_{univ} = \Delta H - T\Delta S_{sys}$
  • Conclusion: Comparing this derived expression to the definition of free energy results in the relationship:     $\Delta G = -T\Delta S_{univ}$
  • Because temperature ($T$) is always positive in the Kelvin scale, the sign of $\Delta G$ is always opposite to the sign of $\Delta S_{univ}$. This makes $\Delta G$ a reliable indicator of spontaneity.

Relation between Process Spontaneity and Signs of Thermodynamic Properties

• The following table summarizes how the signs of $\Delta S_{univ}$ and $\Delta G$ define the nature of a chemical or physical process:

Physical Interpretation: The Concept of "Free" in Gibbs Free Energy

• Useful Work ($w$):
  • The free energy change indicates the amount of "useful work" that can be extracted from a spontaneous process.
  • Consider a spontaneous, exothermic process that leads to a decrease in entropy. In the equation $\Delta G = \Delta H - T\Delta S$:
    • $\Delta H$ represents the total energy produced by the process.
    • $T\Delta S$ represents the energy lost to the surroundings (energy that must be paid as an "entropy tax").
    • $\Delta G$ represents the remaining energy available, or "free," to do work.
• Maximum Expansion Work:
  • If a process occurs under conditions of thermodynamic reversibility, the work extracted would be at its maximum theoretical limit:     $\Delta G = w_{max}$
  • In this context, $w_{max}$ refers to all types of work excluding expansion (pressure-volume) work.
• Practical Limitations:
  • Thermodynamically reversible conditions are not realistic in practice.
  • Technologies and mechanical systems (e.g., batteries) are never 100% efficient; therefore, actual work done is always less than the theoretical maximum ($\Delta G$).
• Nonspontaneous Processes:
  • For processes that are nonspontaneous ($\Delta G > 0$), the value of $\Delta G$ represents the absolute minimum amount of work that must be performed on the system by an external source to force the process to occur.

Calculation of Standard Free Energy Change ($\Delta G^\circ$)

• Methodology:
  • Since free energy is a state function, calculating the change depends only on the initial and final states.
  • Commonly, calculations utilize compiled standard state thermodynamic data (often found in references like "Appendix G").
  • The calculation method uses standard enthalpy and entropy data as follows:     $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$

Example 12.7: Calculating $\Delta G^\circ$ for the Vaporization of Water at Room Temperature

• Problem Statement:
  • Determine the standard free energy change ($\Delta G^\circ$) for the vaporization of water at room temperature ($298\,K$) and conclude if the process is spontaneous.
• Process Equation:$H_2O(l) \rightarrow H_2O(g)$
• Thermodynamic Data (from Appendix G):
  • For $H_2O(l)$:
    • Standard Molar Enthalpy of Formation ($\Delta H_f^\circ$): $-285.83\,kJ/mol$
    • Standard Molar Entropy ($S^\circ$): $70.0\,J/K \cdot mol$
  • For $H_2O(g)$:
    • Standard Molar Enthalpy of Formation ($\Delta H_f^\circ$): $-241.82\,kJ/mol$
    • Standard Molar Entropy ($S^\circ$): $188.8\,J/K \cdot mol$
• Step 1: Calculate Standard Enthalpy Change ($\Delta H^\circ$):$\Delta H^\circ = \Delta H_f^\circ(H_2O(g)) - \Delta H_f^\circ(H_2O(l))$$\Delta H^\circ = (-241.82\,kJ) - (-285.83\,kJ) = 44.01\,kJ$
• Step 2: Calculate Standard Entropy Change ($\Delta S^\circ$):$\Delta S^\circ = S^\circ(H_2O(g)) - S^\circ(H_2O(l))$$\Delta S^\circ = 188.8\,J/K - 70.0\,J/K = 118.8\,J/K$
• Step 3: Calculate $\Delta G^\circ$ at $298\,K$:$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$\Delta G^\circ = 44.01\,kJ - (298\,K \times 118.8\,J/K \times \frac{1\,kJ}{1000\,J})$$\Delta G^\circ = 44.01\,kJ - 35.40\,kJ = 8.61\,kJ$
• Conclusion:
  • Because $\Delta G^\circ$ is positive ($8.61\,kJ$) at $298\,K$ ($25\,^{\circ}C$), the boiling of water (vaporization) is nonspontaneous at room temperature.

Questions & Discussion

• Check Your Learning (Vaporization): Use the data in Appendix G to calculate $\Delta G^\circ$ for a specific reaction at $298\,K$ and interpret the spontaneity based on the sign of the result. (Calculation parameters are noted to be identical in methodology to Example 12.7).