An enzymatic hydrolysis of fructose-1-P Fructose-1-Pag)H2O) » Fructose(a) Pi (dg

Question

An enzymatic hydrolysis of fructose-1-P Fructose-1-Pag)H2O) » Fructose(a) Pi (dg) was allowed to proceed to equilibrium at 298 K. The original concentration of fructose-1-P was 0.2 M, but when the system had reached equilibrium the concentration of fructose-1-P was only 6.52x105 M. Calculate the equilibrium constant and G°. R = 8.314 . Hint: Do not use water in the equilibrium expression. mol K tea B. Keg= 1.63x10-3, 2G" = 15.9 kJ/mol C. Kea = 3066; AG=-19.9 kJ/mol D. Keg = 3.26, AG° = 19.9 kJ/mol

Explanation / Answer

For fructose-1-P + H2O D-fructose + Pi

The equilibrium constant, Keq, is given by

Keq= [fructose]eq [Pi]eq / [fructose-1-P]eq

At 25°C or 298K (273 + 25), = [fructose-1-P]eq = 6.52 x 10-5M.

Initially= [fructose-1-P] = 0.2 M.

The amount of the fructose produced is given by:= 0.2 M - 6.52 x 10-5M.

And, since an equal amount of [Pi] is produced, Keq may be written as follows:

Keq=(0.2M-6.52×10-5)(0.2M-6.52×10-5) / 6.52×10-5

Keq= 613 M

²G°= -RTlnKeq

²G°= -(8.314 J/mol K) × 298 K × ln613

²G°= -15.9 kJ/mol

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