Answer:
The concentration of fructose-6-phosphate is 1.50x10⁻³M
Explanation:
Gibbs Free energy, ΔG°, is defined as:
ΔG° = -RTln K
Where ΔG° = 1670J/mol
R is gas constant = 8.314J/molK
T is absolute temperature (25°C + 273.15 = 298.15K)
Thus, equilibrium constant, k, is:
1670J/mol = -8.314J/molK* 298.15K ln K
-0.6737 = ln K
0.5098 = K
K of the reaction:
G6P ⇄ F6P
is:
K = 0.5098 = [F6P] / [G6P]
As [G6P] is 2.95x10⁻³M:
0.5098 = [F6P] / [2.95x10⁻³M]
[F6P] = 1.50x10⁻³M
