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= ?

When 1-pentyne (A) is treated with 4N alcoholic KOH at 175^@C, it is c

1.	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= ?
Answer» Solution :`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