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 .

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

Explanation:

Glucose-6-phosphate ⇄ Fructose-6-phosphate

The equilibrium constant will be given by =

Relation between standard Gibbs free energy and equilibrium constant follows:

where,

= standard Gibbs free energy = 1.67 kJ/mol = 1670 J/mol  (Conversion factor: 1kJ = 1000J)

R = Gas constant =

T = temperature =

On solving above equation :

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

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i would say b

explanation:

but i think i would need more information than that