Relationship Between Equilibrium Constant (K), Reaction Quotient (Q), and Gibbs Energy (G)

1. What Gibbs Energy Tells Us 🔥

  • If ΔG is negative → Reaction is spontaneous! It moves forward. ⏩
  • If ΔG is positive → Reaction is not spontaneous. It prefers the reverse direction. ⏪
  • If ΔG = 0 → Reaction is at equilibrium. No net change! ⚖️

2. The Magic Equation ✨

Gibbs energy (ΔG) connects to the reaction quotient (Q) with:

\( \Delta G = \Delta G^\circ + RT \ln Q \)

Where:
– \( \Delta G^\circ \) = standard Gibbs energy
– R = gas constant (8.314 J/mol·K)
– T = temperature in Kelvin
– Q = reaction quotient (like K, but for any point in the reaction)

3. At Equilibrium… ⚖️

When ΔG = 0 and Q = K (equilibrium constant), the equation becomes:

\( \Delta G^\circ = -RT \ln K \)

Rearrange to find K:

\( K = e^{-\Delta G^\circ / RT} \)

4. What K Tells Us (via ΔG°) 🔍

  • If \( \Delta G^\circ < 0 \) → K > 1
    Reaction is spontaneous and favors products! 🎉
  • If \( \Delta G^\circ > 0 \) → K < 1
    Reaction is NOT spontaneous and favors reactants. 🛑

5. Practice Problems 🧠

Problem 1:

Phosphorylation of glucose has \( \Delta G^\circ = 13.8 \) kJ/mol. Find Kc at 298 K.

Solution:
\( \Delta G^\circ = 13.8 \times 10^3 \) J/mol
Using \( \Delta G^\circ = -RT \ln K_c \):
\( \ln K_c = \frac{-\Delta G^\circ}{RT} = \frac{-13.8 \times 10^3}{(8.314 \times 298)} = -5.569 \)
\( K_c = e^{-5.569} = 3.81 \times 10^{-3} \)

Problem 2:

Sucrose hydrolysis has Kc = 2 × 1013 at 300 K. Find ΔG°.

Solution:
Using \( \Delta G^\circ = -RT \ln K_c \):
\( \Delta G^\circ = -8.314 \times 300 \times \ln(2 \times 10^{13}) \)
\( \Delta G^\circ = -7.64 \times 10^4 \) J/mol

Important for NEET! 🚀

  1. ΔG° = -RT ln K is GOLD! Memorize it. 🥇
  2. K > 1 means ΔG° < 0 (spontaneous forward reaction).
  3. K < 1 means ΔG° > 0 (non-spontaneous forward reaction).
  4. Units matter! Convert kJ ↔ J in ΔG° calculations.
  5. At equilibrium: ΔG = 0 and Q = K.