When `1`pentyne `(A)` is treated with `4N` alcoholic `KOH` at `175^(@)C`, it is slowly converted into an equilibrium mixture of `1.3%` of `1`pentyne `(A), 95.2% 2`-pentyne `(B)` and `3.5%` of `1,2`-pentandiene `(C )`. The equilibrium was maintained at `175^(@)C`. calculate `DeltaG^(Theta)` for the following equilibria:
`B hArr A, DeltaG^(Theta)underset(1) = ?`
`B hArr C, DeltaG^(Theta)underset(2) =?`
From the calculated value of `DeltaG^(Theta)underset(1)`and `DeltaG^(Theta)underset(2)`, indicate the order of stability of `A,B` and `C`.

Explanation

`T=175^(@)C=175+273=448K`
Concentration of 1 - pentype [A] = 1.3%
Concentration of 2 - pentype =95.2%
Concentration of 1,2 pentadiene [C]=3.5%
At equilibrium
`BhArrA`
`95.2%1.3%rArr`
`K_(1)=(1.3)/(95.2)=0.0136`
`BhArrC`
`95.2%3.5%rArr`
`K_(2)=(3.5)/(95.2)=0.0367`
`impliesDeltaG_(1)^(@)=-2.303RTlogK_(1)`
`DeltaG_(1)^(@)=-2.303xx8.314xx448xxlog0.0136`
`DeltaG_(1)^(@)=+16010J =+16KJ`
`impliesDeltaG_(2)^(@)=-2.303RTlogK_(2)`
`DeltaG_(2)^(@)=-2.303xx8.314xx448xxlog0.0367`
`DeltaG_(2)^(@)=+12312J`
`DeltaG_(2)^(@)=+12.312kJ.`