For a system at equilibriumA. `((dellnK_(p))/(delP))=-(del)/(delP)((DeltaG)/(RT))_(T)`B. `DeltaG=0`C. `G_((P))=G_((R))`D. The free energy of the ststem is minimum

For a system at equilibriumA. `((dellnK_(p))/(delP))=-(del)/(delP)((De

1.	For a system at equilibriumA. `((dellnK_(p))/(delP))=-(del)/(delP)((DeltaG)/(RT))_(T)`B. `DeltaG=0`C. `G_((P))=G_((R))`D. The free energy of the ststem is minimum
Answer» Correct Answer - A::B::C::D `DeltaG=-RTlnK_(p)implies-(DeltaG)/(RT)=lnK_(p)` `implies((dellnK_(p))/(delP))_(T)=-(del)/(delP)((DeltaG)/(RT))_(T)` Hence statement (a) is true At equlibrium `G_(p)=G_(R)` i.e., free energy of products and reactants are equal. `impliesDeltaG=G_(p)-G_(R)=0implies` Statement (b) and(c) are correct. At equlibrium, the free energy of system is minimum. Statement (d) is correct.

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For a system at equilibriumA. `((dellnK_(p))/(delP))=-(del)/(delP)((DeltaG)/(RT))_(T)`B. `DeltaG=0`C. `G_((P))=G_((R))`D. The free energy of the ststem is minimum

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