Question

The value of ΔG°′ 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 2.05 mM, what is the concentration of fructose-6-phosphate? Assume a temperature of 25.0°C.

Answer 1

1.04 mM

Step-by-step explanation:

The conversion reaction given is reversible, and for reversible reactions, the free-energy can be calculated by:

ΔG = -RTlnK

Where R is the constant of the gases(8.3145 J/mol.K), T is the temperature( 25°C + 273 = 298 K), and K is the equilibrium constant.

K = [F6P]/[glucose-6-phosphate]

Because T = 25ºC, ΔG = ΔG°' = 1670 J/mol

1670 = -8.3145*298*ln[F6P]/2.05

-2477.721*ln[F6P]/2.05 = 1670

ln[F6P]/2.05 = -0.6740

[F6P]/2.05 = 0.50966

[F6P] = 1.04 mM