The standard free energy change G is related to K (equilibrium constant) as — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

# Relationship Between ΔG° and K **Step 1: Recall the fundamental relationship** At equilibrium, the standard free energy change is related to the equilibrium constant through: $$\Delta G° = -RT \ln K$$ where: - $R$ = gas constant (8.314 J/mol·K) - $T$ = absolute temperature (K) - $K$ = equilibrium constant - $\ln$ = natural logarithm **Step 2: Understanding the sign relationship** - When $K > 1$: $\ln K > 0$, so $\Delta G° < 0$ (spontaneous reaction) - When $K < 1$: $\ln K < 0$, so $\Delta G° > 0$ (non-spontaneous reaction) - When $K = 1$: $\ln K = 0$, so $\Delta G° = 0$ (equilibrium) **Step 3: Why this is correct** This equation directly connects thermodynamic spontaneity ($\Delta G°$) to the position of equilibrium ($K$). A larger equilibrium constant means the reaction proceeds further toward products, corresponding to a more negative free energy change. **Why other options fail:** - Linear relationships like $\Delta G° = -K$ or $\Delta G° = RT·K$ don't account for the exponential nature of equilibrium - Logarithmic relationships with positive signs contradict the spontaneity principle **Answer: A is** $$\boxed{\Delta G° = -RT \ln K}$$