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

For the equilibrium `B Leftrightarrow A`,
`K=(1.3)/(95.2)=0.0136`
`triangleG_(1)^(0)=-2.303 RT log K`
`=-2.303 xx 8.314 xx 448 xx log .0.0136`
=+16010 J
=+16.01kJ.
For the equilibrium `B Leftrightarrow C`,
`K=(3.5)/(95.2) =0.0367`
`triangleG_(2)^(0)=-2.303 xx 8.314 xx 448 xx log 0.0367`
`=+12312J`
=+12.312kJ
Further negative value of `triangleG^(0)` favours the forward reaction
for `B Leftrightarrow A, triangleG^(0)=+16.01 kJ` stability order is `B gt A`
And for, `B Leftrightarrow C, triangleG^(0)=+12.312kJ, ` stability order is `B gt C`
To determine the order of stability between A and C,
`A Leftrightarrow B, triangleG^(0)=-16.01kJ`
`B Leftrightarrow C, triangleG^(0)=+12.312kJ`
Add : `A Leftrightarrow C, triangleG^(0)=-3.6998`kJ stability order is `C gt A`
Thus, the decreasing order of stability is `B gt C gt A`.