<u>Answer:</u> The equilibrium constant of the reaction is 0.0526 and standard Gibbs free energy of the reaction is -7.296 kJ/mol
<u>Explanation:</u>
For the given chemical equation:

The expression of
for above equation follows:
![K_{eq}=\frac{[\text{Glucose-1-phosphate}]}{[\text{Glucose-6-phosphate}]}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5B%5Ctext%7BGlucose-1-phosphate%7D%5D%7D%7B%5B%5Ctext%7BGlucose-6-phosphate%7D%5D%7D)
We are given:
![[\text{Glucose-1-phosphate}]=0.01M](https://tex.z-dn.net/?f=%5B%5Ctext%7BGlucose-1-phosphate%7D%5D%3D0.01M)
![[\text{Glucose-6-phosphate}]=0.19M](https://tex.z-dn.net/?f=%5B%5Ctext%7BGlucose-6-phosphate%7D%5D%3D0.19M)
Putting values in above expression, we get:

The equation used to calculate standard Gibbs free energy of the reaction follows:

where,
= Standard Gibbs free energy
R = Gas constant = 8.314 J/K mol
T = Temperature = ![25^oC=[25+273]K=298K](https://tex.z-dn.net/?f=25%5EoC%3D%5B25%2B273%5DK%3D298K)
= Equilibrium constant of the reaction = 0.0526
Putting values in above equation, we get:

Hence, the equilibrium constant of the reaction is 0.0526 and standard Gibbs free energy of the reaction is -7.296 kJ/mol