40. Match standard free energy of the reaction with the corresponding equilibrium constant.

A. G>0

1. K>1

B. G<0

2. K=1

C. G=0

3. K=0


4. K<1

A.(4) B.(1) C.(2)
As we know that, G=-RT ln K
A. If G>0, i.e., G° is positive, then ln K is negative i.e., K<1
B. If G<0, i.e., G° is negative then ln K is positive i.e., K>1
C. If G=0, ln K=0, i.e., K=1.