Answer:
d.-379 cal/mol
Step-by-step explanation:
ΔG = ΔG⁰ + RT ln K 
for equilibrium ΔG = 0 
ΔG⁰ + RT ln K =0 
ΔG⁰ = - RT ln K 
PG ⇒ PEP 
K = [ PEP ] / [ PG ] 
= .68 / .32 
= 2.125 
ΔG⁰ = - 1.987 x 273 x ln 2.125 
= - 409 Cal / mole 
Option d is the nearest answer .