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a
∆Gsystem∆Gtotal=-T
b
in isothermal process, Wreversible=-nRTln VfVi
c
ln K = ∆H-T∆S0RT
d
K=e-∆G0/RT
answer is C.
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Detailed Solution
According to Gibbs Helmholtz equation,∆G=∆H-T∆S(a) For a system, total entropy change = ∆Stotal∆Stotal=0∴ ∆Gsystem=-T∆Stotal ∴ ∆Gsystem∆Stotal=-TThus, (a) is correct.(b) For isothermal reversible process, ∆E=0By first law of thermodynamics, ∆E=q+W∴ Wreversible =-q=-∫ViVfpdV ⇒ Wreversible =-nRT ln VfViThus, (b) is correct(c) ∆G0=∆H0-T∆S0 ....(1) Also, ∆G0= -RT ln K ln K = ∆H0-T∆S0RT ln K = ∆G0RT [from Eq. (i)] Thus, (c) is incorrect(d) The standard free energy ∆G0 is related to equilibrium constant K as∆G0=-RT ln K ∴ ln K = ∆G0RT ⇒ K = -e-∆G0 Thus, (d) is also correct.