📘 IB Understanding
The equilibrium constant K and the standard Gibbs energy change ΔG° can both be used to measure the position of an equilibrium reaction. At equilibrium, ΔG = 0, giving the relationship ΔG° = −RT ln K.
The Key Equations
At any point during a reaction (not necessarily at equilibrium):
At equilibrium, ΔG = 0 and Q = K, so:
What Does the Sign of ΔG° Tell Us?
ΔG°, K, and Spontaneity
Worked Example: Calculating K from ΔG°
Problem: At 25 °C, ΔG° = −4.38 kJ mol⁻¹ for a reaction. Calculate K.
Step 1: Convert T to Kelvin: T = 25 + 273.15 = 298.15 K
Step 2: Convert ΔG° to J: ΔG° = −4380 J mol⁻¹
Step 3: Rearrange: \(\ln K = \frac{-\Delta G^\circ}{RT}\)
Step 4: Substitute: \(\ln K = \frac{-(-4380)}{8.314 \times 298.15} = \frac{4380}{2478.8} = 1.767\)
Step 5: Solve: \(K = e^{1.767} = \textbf{5.85}\)
Conclusion: K > 1 and ΔG° is negative, so the equilibrium lies to the right (products favoured). ✅
📋 Exam Tips
- Unit mismatch trap: ΔG° is usually in kJ but R = 8.314 J K⁻¹ mol⁻¹. Always convert to the same unit (×1000 or ÷1000)
- Remember: ln is the natural log (base e), not log₁₀
- A large negative ΔG° means a very large K (reaction essentially goes to completion)
Think About It
If ΔG° is positive, does it mean the reaction can never happen?
No! A positive ΔG° means the reaction is non-spontaneous under standard conditions. The actual ΔG depends on the concentrations (via Q). Even with a positive ΔG°, if Q is very small (low product concentrations), ΔG can still be negative and the forward reaction will proceed — it just won't go all the way to completion.