1.

For a system at equilibrium

Answer»

`((dellnK_(p))/(delP))=-(del)/(delP)((DeltaG)/(RT))_(T)`
`DeltaG=0`
`G_((P))=G_((R))`
The free energy of the ststem is minimum

Solution :`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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