InterviewSolution
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InterviewSolution
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The incorrect expression among the following isA. `(DeltaG_("system"))/(DeltaG_("total"))=-T`B. in isothermal process, `W_("reversible")=-nRT"in"(V_(f))/(V_(i))`C. In `K=(DeltaH-TDeltaS^(@))/(RT)`D. `K=e^(-DeltaG^(@)//RT)` |
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Answer» Correct Answer - C According to Gibbs Helmholtz equation, `DeltaG=DeltaH-TDeltaS` (a) For a system, total entropy change `=DeltaS_("total")` `Delta_("total")=0` `:. DeltaG_("system")=-TDeltaS_("total")` `:. (DeltaG_("system"))/(DeltaS_("total"))=-T` Thus, `(a)` is correct. (b) For isothermal reversible process, `DeltaE=0` By first law of thermodynamics, `DeltaE=q+W` `:. W_("reversible")=-q=-int_(v_(i))^(v_(f))pdV` `implies W_("reversible")=-RT` In `(V_(f))/(V_(i))` Thus, `(b)` is correct (c) `DeltaG^(@)+DeltaH^(@)-TDeltaS^(@)` Also, `DeltaG^(@)=-RT` In `K` In `K=((DeltaH^(@)-TDeltaS^(@)))/(RT)` In `K=(DeltaG^(@))/(RT)` [from Eq. (i)] (d) The standard free energy `(DeltaG^(@))` is related to equilibrium constant `K` as `DeltaG^(@)=-RT` In `K` `:.` In `K=-(DeltaG^(@))/(RT)` `k=-e^(-DeltaG^(@)//Rt)` Thus, `(d)` also correct. |
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