Question

Fructose-1-P is hydrolyzed according to: Fructose-1-P + H2O → Fructose + Pi If a 0.2 M aqueous solution of Froctose-1-P is allowed to reach equilibrium, its final concentration is 6.52 × 10-5 M.
What is the standard free energy of Froctose-1-P hydrolysis?

Answer 1

Answer:

$\Delta G^{\circ}=-15902 J/mol$

Explanation:

In this problem we only have information of the equilibrium, so we need to find a expression of the free energy in function of the constant of equilireium (Keq):

$\Delta G^{\circ}=-R*T*ln(K_{eq})$

Being Keq:

$K_{eq}=\frac{[fructose][Pi]}{[Fructose-1-P]}$

Initial conditions:

$[Fructose-1-P]=0.2M$

$[Fructose]=0M$

$[Pi]=0M$

Equilibrium conditions:

$[Fructose-1-P]=6.52*10^{-5}M$

$[Fructose]=0.2M-6.52*10^{-5}M$

$[Pi]=0.2M-6.52*10^{-5}M$

$K_{eq}=\frac{(0.2M-6.52*10^{-5}M)*(0.2M-6.52*10^{-5}M)}{6.52*10^{-5}M}$

$K_{eq}=613.1$

Free-energy for T=298K (standard):

$\Delta G^{\circ}=-8.314\frac{J}{mol*K}*298K*ln(613.1)$

$\Delta G^{\circ}=-15902 J/mol$