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`.

Answer

`Delta G_(1)^(@) = 15.992 kJ mol^(-1). Delta G_(2)^(@) = 12.312 kJ mol^(-1)`,
`Delta H = O, B gt C lt A; Delta G_(3)^(@) = -3.688 kj mol^(-1)`

Explanation

calculation `Delta G , Delta G^(@) = - 2.303 RT log.(("Product"))/(("Reactant"))`
For the equilibrium reaction `B Harr A`, we get and for the reaction `B hArr C`, we get
`Delta G_(1)^(@) = (-2.3030 xx 8.314 xx 448 log.(1.3)/(95.2))`
`= 15.992 kJ mol^(-1)`
`Delta G_(2)^(@) = (-2.3030 xx 8.314 xx 448 log.(3.5)/(95.2))`
`= 12.312 kJ mol^(-1)`
Again, for the reaction `A hArr C`.
`Delta G_(3)^(@) = (-2.3030 xx 8.314 xx 448 log.(3.5)/(1.3))`
`= -3.688 kJ mol^(-1)`
From the above calcualtions, we have
`B hArr A, Delta G_(1)^(@) = + 15.992 kJ mol^(-1)`
`B hArr C, Delta G_(2)^(@) = + 12.312 kJ mol^(-1)`
`A hArr C, Delta G_(3)^(@) = -3.688 kJ mol^(-1)`
Thus, the correct order of stability `B gt C gt A`