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

Explanation

`{:("","Pentyne",overset(KOH)hArr,2-"Pentyne"+1","2,-"Pentadiene"),("",(A),"",(B),(C )),("At eqm"% ,1.3,"",95.5,3.5):}`
`K_c=([B][C])/([A])=(95.2xx3.5)/(1.3)=256.31`
For eqm. `B hArr A`
`K_1=([A])/([B])`
From Eqs. (i) and (ii) ,`K_1=([C])/(K_c)`
`=(3.5)/(256.31)=0.013`
`DeltaG^2=-2.303 RTlog_(10)K`
`=-2.303 xx 8.314xx448 log _(10)0.013`
=161787
=16.178 kJ
Stability order for A and B is `B gt A`
Similarly `B hArr C`
`K_2=([C])/([B])`
`=(K_cxx[A])/([B]^2)`
`=(256.31xx1.3)/(95.2xx95.2)`
`therefore DeltaG_2^@=-2.303 RT log _(10)k`
`=-2.303 xx 8.314 xx448 log _(10) 0.037`
=12282 J
=12.282 kJ
Thus stability order for B and C is `B gt C`
Total ordeer of stability is `B gt C gt A`