Question

For the reaction: Fructose 6-phosphate to glucose 6-phosphate, delta G is -0.4 kcla/mol. Starting with a 0.50 M solution of fructose 6-phosphate, what is the equilibrium concentration of both fructose 6- phosphate and glucose 6-phosphate?

Answer 1

Calculate K_c for the reaction using the relation,

G^o = -RTlnK_c

Given G^o = 0.4kcal/ mol.........substituting the value ofG^o , R = 1.987 x 10^-3 kcal /mole K , T =298 K , the value of K_c can be calculated as,

-0.4 = - (1.987 x 10^-3  x 298 ln K_c )

hence K_c  = anti ln of  [ 0.4 / (1.987 x 10^-3 x 298 )

................= 1.9651

Now , K_c   = [ glucose 6-phosphate ] / [ Fructose 6-phosphate ]

Let the concentration of glucose 6-phosphate at equilibrium be = x moles / L

then concentratiion of Fructose 6- phosphate = ( 0.5 - x ) moles / L

K_c = [ Fructose 6-phosphate ] / [ Glucose 6-phosphate ]

substituting the values we can calculate x (ie. equilibrium concentration of Glucose 6-phosphate ), there from the concentration of Fructose 6- phosphate at equilibrium can also be calculated. Thus,

1.9651 = x / (0.5 - x )

....x = 0.3314

.....or, = 3.31 x 10^-1 moles/ L

This represents equilibrium concentration of Glucose 6- phosphate = 3.31 x 10^-1 moles / L

.......................& equilibrium concentration of Fructose 6-phosphate = ( 0.5 - 0.3314 )

......................................................................................................... = 0.1686 moles / L

..................... ...............................................................................or, = 1.67 x 10^-1 moles / L