Chemical Thermodynamics Question 329
Question: The incorrect expression among the following is
Options:
A) $ \frac{\Delta G_{system}}{\Delta S_{total}}=-T $
B) in isothermal process, $ W_{reversible}=-nRTIn\frac{V_{f}}{V_{i}} $
C) $ In,K=\frac{\Delta H-T\Delta S^{o}}{RT} $
D) $ K={{e}^{-\Delta G^{o}/RT}} $
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Answer:
Correct Answer: C
Solution:
- According to Gibbs-Helmholtz equation, $ \Delta G=\Delta H-T\Delta S $ (1) For a system, total entropy change= $ \Delta S_{total} $
$ \Delta H_{total}=0 $
$ \therefore \Delta G_{system}=-T\Delta S_{total} $
$ \therefore \frac{\Delta G_{system}}{\Delta S_{total}}=-T $ Thus, (1) is correct. (2) For isothermal reversible process, $ \Delta E=0 $ By Frist law of thermodynamics, $ \Delta E=q+W $
$ \therefore W_{reversible}=-q=-\int_{V_{i}}^{V_{f}}{p,dV} $
$ \Rightarrow W_{reversible}=-nRT $ In $ \frac{V_{f}}{V_{i}} $ Thus, (2) is correct, (3) $ \Delta G^{o}=\Delta H^{o}-T\Delta S^{o} $ Also, $ \Delta G^{o}=-RT $ ln K In $ K=\frac{-\Delta G^{o}}{RT} $ ln $ K=\frac{(\Delta H^{o}-T\Delta S^{o})}{RT} $ [from Eq. (1)] Thus, (3) is incorrect. (4) The standard free energy ( $ \Delta G^{o} $ ) is related to equilibrium constant K as $ \Delta G^{o}=-RT $ ln K
$ \therefore $ ln $ K=\frac{\Delta G^{o}}{RT} $
$ \Rightarrow K={{e}^{-\Delta G^{o}/RT}} $ Thus, is also correct.
