Question

The standard Gibbs free energy change, \(\Delta G^\circ\) is related to equilibrium constant, \(K_p\), as

• A. \(K_p = -RT\ln \Delta G^\circ\)
• B. \(K_p = \dfrac{e}{RT}\Delta G^\circ\)
• C. \(K_p = -\dfrac{\Delta G}{RT}\)
• D. \(K_p = e^{-\Delta G^\circ/RT}\)

Hint:

Always remember: \(\Delta G^\circ = -RT\ln K\). If \(\Delta G^\circ\) is negative, \(K\) becomes greater than 1 (reaction spontaneous).

Answer 1

The Correct Option is D

Step 1: Fundamental relation.
Thermodynamics gives:
\[ \Delta G^\circ = -RT\ln K_p \] Step 2: Rearranging for \(K_p\).
\[ \ln K_p = -\frac{\Delta G^\circ}{RT} \] Taking exponential:
\[ K_p = e^{-\Delta G^\circ/RT} \] Final Answer: \[ \boxed{K_p = e^{-\Delta G^\circ/RT}} \]