Question

Identify from following, the relation between standard free energy change in a reaction and equilibrium constant \(K_c\).

• A. \(△G^{\circ} = RTlnK_c\)
• B. \(-△G^{\circ} = \frac{RT}{lnK_c}\)
• C. \(△G^{\circ} = \frac{lnK_c}{RT}\)
• D. \(-△G^{\circ} = RTlnK_c\)

Hint:

Use the standard relation delta G = minus RT ln K.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
At equilibrium the reaction quotient equals the equilibrium constant. Under standard conditions the free energy change is tied to \(K_c\) by a logarithmic relation.

Step 2: Key Formula:
From \(\Delta G = \Delta G^{\circ} + RT\ln Q\) and \(\Delta G = 0\), \(Q = K_c\) at equilibrium:
\[ 0 = \Delta G^{\circ} + RT\ln K_c \]
\[ -\Delta G^{\circ} = RT\ln K_c \]

Step 3: Interpret:
If \(K_c > 1\), then \(\ln K_c > 0\) and \(\Delta G^{\circ}\) is negative, so the reaction favours products. This matches common sense.

Step 4: Check the Other Options:
(A) lacks the minus sign, which would give the wrong direction. (B) and (C) put \(\ln K_c\) or \(RT\) in the denominator, but the relation is a product and not a ratio. So (D) is correct.

Final Answer:
The relation is \(-\Delta G^{\circ} = RT\ln K_c\), option (D). \[ \boxed{\text{(D) } -\Delta G^{\circ} = RT\ln K_c} \]