What relationship connects the standard equilibrium constant K to standard Gibbs free energy change Delta G^0 at temperature T?
Answer and explanation
Correct answer: D. Delta G^0 = -R T ln(K)
- Step 1: From chemical thermodynamics, Delta G = Delta G^0 + R T ln(Q). At equilibrium, Delta G = 0 and Q = K. Step 2. Setting Delta G = 0 gives 0 = Delta G^0 + R T ln(K) => Delta G^0 = -R T ln(K). *Answer: Delta G^0 = -R T ln(K). Why not the others:* When K > 1, ln(K) > 0, making Delta G^0 negative (thermodynamically favorable under standard conditions).
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