Answer : The concentration of fructose-6-phosphate is 0.275 mM
Explanation :
First we have to calculate the value of equilibrium constant.
The relation between the equilibrium constant and standard Gibbs free energy is:

where,
= standard Gibbs free energy = +1.67 kJ/mol = +1670 J/mol
R = gas constant = 8.314 J/K.mol
T = temperature =

K = equilibrium constant = ?
Now put all the given values in the above formula, we get:


Now we have to calculate the concentration of fructose-6-phosphate.
The expression of equilibrium constant is:
![K=\frac{[\text{fructose-6-phosphate}]}{[\text{glucose-6-phosphate}]}](https://img.qammunity.org/2020/formulas/chemistry/college/6tx2jfu7eflk0klojkgjtyezerh0xtmc8h.png)
![0.509=\frac{[\text{fructose-6-phosphate}]}{1.85mM}](https://img.qammunity.org/2020/formulas/chemistry/college/t45w9krcl4l7wtkmjvs626c33vd6xxw1xs.png)

Therefore, the concentration of fructose-6-phosphate is 0.275 mM