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",overset(KOHalc.)hArr,"Pentyne-2",+,"Pentadiene-1-2"),(,(A),,(B),,(C)),("At eqm.",1.3,,95.2,,3.5):}`
`K_(eq)=([B][C])/([A])=(95.2xx3.5)/(1.3)=256.31`….(`1`)
Now, for `BhArrA`
`K_(1)=([A])/([B])` ….(`2`)
Then from eqs. (`1`) and (`2`),
`K_(1)=([C])/(K_(eq))=(3.5)/(256.31)=0.013`
`:. DeltaG_(1)^(@)= -2.303 RT log_(10)K`
`=-2.303xx8.314xx448 log 0.013=16178 J`
`=16.178 kJ`
Stability order for `A` and `B` is `gt`.
Similarly for `BhArrC`
`K_(2)=([C])/([B])=(K_(eq)[A])/([B]^(2))`
`=(256.31xx1.3)/(95.2xx95.2)=0.037`
`:. DeltaG_(2)^(@)= -2.303RT log_(10) K`
`= -2.303xx8.314xx448 log 0.037`
`=12282 J=12.282 kJ`
Thus, stability order for `B` and `C` is `gt`.
The total stability order is `gt gt `.
`CH_(3)*CH_(2)CH_(2)*C-=CHoverset(KOH alc.)to[CH_(3)CH_(2)CH=C=CH_(2)]to[CH_(3)CH_(2)C-=C-CH_(3)]`