Answer:
1.503 mM is the concentration of fructose-6-phosphate.
Explanation:
Glucose-6-phosphate ⇄ Fructose-6-phosphate
The equilibrium constant will be given by = [tex]K_c[/tex]
[tex]K_c=\frac{[\text{Fructose-6-phosphate}]}{[\text{Glucose-6-phosphate}]}[/tex]
[tex]K_c=\frac{[\text{Fructose-6-phosphate}]}{2.95 mM}[/tex]
Relation between standard Gibbs free energy and equilibrium constant follows:
[tex]\Delta G^o=-RT\ln K_c[/tex]
where,
[tex]\Delta G^o[/tex] = standard Gibbs free energy = 1.67 kJ/mol = 1670 J/mol (Conversion factor: 1kJ = 1000J)
R = Gas constant = [tex]8.314 J/K mol[/tex]
T = temperature = [tex]25.0^oC=[273+25.0]K=298.0 K[/tex]
[tex]1670 J/mol=-8.314 J/mol K\times 298.0 K\times \ln \frac{[\text{Fructose-6-phosphate}]}{2.95 mM}[/tex]
On solving above equation :
[tex][\text{Fructose-6-phosphate}]=1.503 mM[/tex]
1.503 mM is the concentration of fructose-6-phosphate.
