When 1-pentyne (A) is treated with 4N alcoholic KOH at `175^@C`, it is converted slowly 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 was maintained at `175^@C`, calculate `DeltaG^0` for the following equilibria.
`B hArr A " " DeltaG_1^0`=?
`B hArr C " " DeltaG_2^0`= ?

Explanation

`T=175^@C`=175+273 =448 K
Concentration of 1 pentyne [A] =1.3%
Concentration of 2-pentyne [B] =95.2%
Concentration of 1,2 - pentadiene [C] =3.5%
At equilibrium
`B hArr A`
`95.2 % " " 1.3% rArr`
`K_1=3.5/95.2`=0.0136
`B hArrC `
`95.2% " " 3.5% rArr`
`K_2=1.3/95.2`=0.0367
`rArr DeltaG_1^0=-2.303 RT log K_1`
`DeltaG_1^0`=-2.303 x 8.314 x 448 x log 0.0136
`DeltaG_1^0` =+16010 J
`DeltaG_1^0`=+16 kJ
`rArr DeltaG_2^0=-2.303 RT log K_2`
`DeltaG_2^0`=-2.303 x 8.314 x 448 x log 0.0367
`DeltaG_2^0`=+12312 J
`DeltaG_2^0` =+12.312 kJ