What is the relationship between the standard free-energy change,...

Question

What is the relationship between the standard free-energy change, ∆G°, for a reaction and the equilibrium constant, K? What is the sign of ∆G° when: (a) K > 1? (b) K = 1? (c) K < 1?

Accepted Answer

Understand the relationship between the standard free-energy change ($ \Delta G^\circ $) and the equilibrium constant ($ K $) using the equation: $ \Delta G^\circ = -RT \ln K $, where $ R $ is the universal gas constant and $ T $ is the temperature in Kelvin.. Analyze the equation $ \Delta G^\circ = -RT \ln K $ to determine the sign of $ \Delta G^\circ $ based on the value of $ K $.. For case (a), when $ K > 1 $, $ \ln K $ is positive, making $ \Delta G^\circ $ negative, indicating that the reaction is spontaneous under standard conditions.. For case (b), when $ K = 1 $, $ \ln K $ is zero, resulting in $ \Delta G^\circ = 0 $, meaning the system is at equilibrium under standard conditions.. For case (c), when $ K < 1 $, $ \ln K $ is negative, making $ \Delta G^\circ $ positive, indicating that the reaction is non-spontaneous under standard conditions.