Answer :
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:
[tex]\Delta G^o=-RT\times \ln k[/tex]
where,
[tex]\Delta G^o[/tex] = standard Gibbs free energy = +1.67 kJ/mol = +1670 J/mol
R = gas constant = 8.314 J/K.mol
T = temperature = [tex]25.0^oC=273+25.0=298K[/tex]
K = equilibrium constant = ?
Now put all the given values in the above formula, we get:
[tex]+1670J/mol=-(8.314J/K.mol)\times (298K)\times \ln k[/tex]
[tex]k=0.509[/tex]
Now we have to calculate the concentration of fructose-6-phosphate.
The expression of equilibrium constant is:
[tex]K=\frac{[\text{fructose-6-phosphate}]}{[\text{glucose-6-phosphate}]}[/tex]
[tex]0.509=\frac{[\text{fructose-6-phosphate}]}{1.85mM}[/tex]
[tex]\text{fructose-6-phosphate}=0.275mM[/tex]
Therefore, the concentration of fructose-6-phosphate is 0.275 mM