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) \[K={{e}^{-\Delta {{G}^{o}}/RT}}\]
Correct Answer: C
Solution :
[c] 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.
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