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

`{:(,"Pentyne"-1,hArr,"Pentyne"-2,+,1","2-"pentadiene"),(,(A),,(B),,(C)),(t_(eq),1.3,,95.2,,3.5):}`
`K_(eq)=([B][C])/([A])=(95.2xx3.5)/(1.3)=256.31`
for `BhArrA`
`K_(1)=([A])/([B])=([C])/(K_(eq))=(3.5)/(256.31)=0.013`
`DeltaG_(1)^(@)=-2.303RT" log "K_(1)`
`=-2.303xx8.314xx448" log "0.013`
`=16178J=16.178kJ`
for `BhArrC`
`K_(2)=([C])/([B])=(K_(eq)[A])/([B]^(2))=(256.31xx1.3)/((95.2)^(2))=0.037`
`DeltaG_(2)^(@)=-2.303RT" log "K_(2)`
`=-2.303xx8.314xx448" log "0.037`
`=12282J=12.282kJ`
Stability will lie in the order
`B gt C gt A`