Chapter 15: Entropy and Free Energy

Chapter 15: Entropy and Free Energy

Gibbs Free Energy

• Second Law of Thermodynamics: A process is spontaneous if
  DS{univ} = DS{syst} + DS_{surr} > 0

• To analyze spontaneity based solely on the system:

  • DS{surr} = - \frac{\Delta H{system}}{T}
  • Combine to get:
    DS{universe} = DS{system} + [- \frac{\Delta H_{system}}{T}] > 0
  • Therefore,
    T \cdot DS{universe} = T \cdot DS{system} - \Delta H_{system} > 0
  • Rearranged gives:
    -T \cdot DS{universe} = \Delta H{system} - T \cdot DS_{system} < 0

• Gibbs Free Energy (G) defined as:
  G = \Delta H - T \cdot DS

  • Units of G: kJ mol⁻¹
  • At constant T and P:
  • Process is spontaneous only if \Delta G < 0
    • \Delta G < 0 → spontaneous forwards
    • \Delta G > 0 → NOT spontaneous forwards
    • \Delta G = 0 → equilibrium

Cases for Gibbs Free Energy

1. Case 1:

   • \Delta H < 0 (exothermic) and DS > 0 (increase in entropy)
   • \Delta G = \Delta H - T \cdot DS = (-) - T (+) < 0
   • Always spontaneous (e.g., building collapsing into pile of bricks)

2. Case 2:

   • \Delta H > 0 (endothermic) and DS < 0 (decrease in entropy)
   • \Delta G = \Delta H - T \cdot DS = (+) - T (-) > 0
   • Never spontaneous (e.g., pile of bricks assembling into a building)

3. Case 3:

   • \Delta H > 0 and DS > 0
   • \Delta G = \Delta H - T \cdot DS = (+) - T(+) < 0
   • Spontaneous at high T (e.g., ice melting; water evaporating)

4. Case 4:

   • \Delta H < 0 and DS < 0
   • \Delta G = \Delta H - T \cdot DS = (-) - T (-) < 0
   • Spontaneous at low T (e.g., water freezing; water vapor condensing)

Gibbs Free Energy Table

• Summary of Cases:

Determining Temperature for Spontaneity

• Identify what is meant by "high T" or "low T" :

  • Water freezing:
    • H2O(l) \rightleftharpoons H2O(s) (Spontaneous at low T, below 0 °C)
  • Water evaporating:
    • H2O(l) \rightleftharpoons H2O(g) (Spontaneous at high T, above 100 °C)

• To find the temperature where equilibria occurs:

  • Set \Delta G = 0 (Equilibrium condition):
    0 = \Delta H - T \cdot DS
  • Rearranged gives:
    T = \frac{\Delta H}{DS}
  • Example:
    • For a reaction where
      \Delta H = 125 kJ mol^{-1} and
      DS = 325 J mol^{-1} K^{-1}, find the temperature (in °C) above which the reaction is spontaneous.

Entropy Change for Phase Changes

• For phase changes:

  • \Delta G = \Delta H - T \cdot DS
  • Set equilibrium to zero:

  0 = \Delta H - T \cdot DS

  • Solving gives:
    DS = \frac{\Delta H}{T}

• For example, considering water:

  • H2O(s) \rightleftharpoons H2O(l)
    • where
    • \Delta H = \Delta H_{fus}
    • T = melting point
  • H2O(l) \rightleftharpoons H2O(g)
    • where
    • \Delta H = \Delta H_{vap}
    • T = boiling point