Lecture Notes: Free Energy, Reaction Quotient, and Equilibrium

Key Concepts

• Delta-G (ΔG): Measure of free energy change in a reaction. Its sign indicates if a reaction is spontaneous.
• Reaction Quotient (Q): Determines the direction of the reaction shift.
• Equilibrium Constant (K): At equilibrium, Q = K. ΔG = 0, indicating no net change in free energy.
• Standard Free Energy Change (ΔG⁰): Calculated under standard conditions, related to K.

Important Equations

1. ΔG = ΔG⁰ + RT ln(Q):
   • At equilibrium, ΔG = 0, Q = K, so:
   • ΔG⁰ = -RT ln(K)
   • Variables:
     • R: Gas constant
     • T: Temperature in Kelvin
     • K: Equilibrium constant

Understanding ΔG and K

• ΔG = 0: Reaction is at equilibrium.
• Negative ΔG: Reaction is spontaneous, K > 1, products favored.
• Positive ΔG: Reaction is non-spontaneous, K < 1, reactants favored.

Application Examples

Example 1: Ammonia Synthesis at 298 K

• Conditions:

  • ΔG⁰ = -33.0 kJ
  • T = 298 K
  • Equation: ΔG⁰ = -RT ln(K)

• Calculation:

  • Convert ΔG⁰ to joules: -33.0 kJ = -33,000 J
  • Substitute:
    • ΔG⁰ = -8.314 J/mol·K * 298 K * ln(K)
  • Solve for K:
    • ln(K) = 13.32
    • K ≈ 6.1 x 10⁵
  • Conclusion:
    • K > 1, products are favored.

Example 2: Ammonia at 464 K

• Conditions:

  • ΔG⁰ = 0
  • T = 464 K

• Calculation:

  • ΔG⁰ = -RT ln(K)
  • ln(K) = 0
  • K = 1
  • Conclusion:
    • Products and reactants equally favored.

Example 3: Ammonia at 1000 K

• Conditions:

  • ΔG⁰ = +106.5 kJ
  • T = 1000 K

• Calculation:

  • Convert ΔG⁰ to joules: 106.5 kJ = 106,500 J
  • Substitute:
    • ΔG⁰ = -8.314 J/mol·K * 1000 K * ln(K)
  • Solve for K:
    • ln(K) = -12.81
    • K ≈ 2.7 x 10⁻⁶
  • Conclusion:
    • K < 1, reactants are favored.

Summary

• ΔG and K Relationship: Helps in understanding the favorability of products vs. reactants in a reaction.
• Using ΔG⁰ and K: Allows prediction of reaction direction and equilibrium composition at given temperatures.