Question

The incorrect expression among the following is :

• A. \( \frac{\Delta G_{\text{System}}}{\Delta S_{\text{Total}}} = -T \) (at constant P)
• B. For isothermal process \( w_{\text{reversible}} = -nRT \ln \frac{V_f}{V_i} \)
• C. \( \ln K = \frac{\Delta H^\circ - T\Delta S^\circ}{RT} \)
• D. \( K = e^{-\Delta G^\circ/RT} \)

Hint:

Always remember: \( \Delta G^\circ = -RT \ln K \). A common trap in exams is flipping the signs or the position of terms in the resulting \( \ln K \) expression.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This question tests fundamental thermodynamic relationships involving Gibbs free energy (\( \Delta G \)), entropy (\( S \)), work (\( w \)), and the equilibrium constant (\( K \)).
Step 2: Key Formula or Approach:
1. \( \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \)
2. \( \Delta G^\circ = -RT \ln K \)
Step 3: Detailed Explanation:
(A) We know \( \Delta S_{\text{Total}} = \Delta S_{\text{System}} + \Delta S_{\text{Surroundings}} \).
At constant P, \( \Delta S_{\text{Surroundings}} = \frac{-\Delta H_{\text{System}}}{T} \).
So, \( \Delta S_{\text{Total}} = \Delta S_{\text{System}} - \frac{\Delta H_{\text{System}}}{T} = \frac{T\Delta S_{\text{System}} - \Delta H_{\text{System}}}{T} \).
Since \( \Delta G_{\text{System}} = \Delta H_{\text{System}} - T\Delta S_{\text{System}} \), then \( \Delta S_{\text{Total}} = \frac{-\Delta G_{\text{System}}}{T} \).
Rearranging, \( \frac{\Delta G_{\text{System}}}{\Delta S_{\text{Total}}} = -T \). This is correct.
(B) For a reversible isothermal expansion of an ideal gas, work is given by \( w = - \int P dV = -nRT \ln \frac{V_f}{V_i} \). This is correct.
(C) From \( \Delta G^\circ = -RT \ln K \), we have \( \ln K = \frac{-\Delta G^\circ}{RT} \).
Substituting \( \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \):
\( \ln K = \frac{-(\Delta H^\circ - T\Delta S^\circ)}{RT} = \frac{T\Delta S^\circ - \Delta H^\circ}{RT} \).
The expression in option (C) is missing the negative sign. Thus, it is incorrect.
(D) Rearranging \( \Delta G^\circ = -RT \ln K \):
\( \ln K = \frac{-\Delta G^\circ}{RT} \implies K = e^{-\Delta G^\circ/RT} \). This is correct.
Step 4: Final Answer:
Expression (C) is incorrect.