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

`\Delta G^(\circ)=-15902 J/mol`

Step-by-step 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)=([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)=((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(J)/(mol*K)*298K*ln(613.1)`

`\Delta G^(\circ)=-15902 J/mol`