Question

The value of Δ G ° ' for the conversion of glucose-6-phosphate to fructose-6-phosphate (F6P) is + 1.67 kJ/mol . If the concentration of glucose-6-phosphate at equilibrium is 2.95 mM , what is the concentration of fructose-6-phosphate? Assume a temperature of 25.0 ° C .

Answer 1

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 = $K_c$

$K_c=\frac{[\text{Fructose-6-phosphate}]}{[\text{Glucose-6-phosphate}]}$

$K_c=\frac{[\text{Fructose-6-phosphate}]}{2.95 mM}$

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K_c$

where,

$\Delta G^o$ = standard Gibbs free energy = 1.67 kJ/mol = 1670 J/mol  (Conversion factor: 1kJ = 1000J)

R = Gas constant = $8.314 J/K mol$

T = temperature = $25.0^oC=[273+25.0]K=298.0 K$

$1670 J/mol=-8.314 J/mol K\times 298.0 K\times \ln \frac{[\text{Fructose-6-phosphate}]}{2.95 mM}$

On solving above equation :

$[\text{Fructose-6-phosphate}]=1.503 mM$

1.503 mM is the concentration of fructose-6-phosphate.