Question

The value of ΔG°′ΔG°′ for the conversion of glucose-6-phosphate to fructose-6-phosphate (F6P) is +1.67 kJ/mol+1.67 kJ/mol . If the concentration of glucose-6-phosphate at equilibrium is 2.65 mM2.65 mM , what is the concentration of fructose-6-phosphate? Assume a temperature of 25.0°C25.0°C .

Answer 1

The concentration of fructose-6-phosphate F6P ≅ 1.35 mM

Step-by-step explanation:

Given that:

ΔG°′ is the conversion of glucose-6-phosphate to fructose-6-phosphate (F6P) = +1.67 kJ/mol = 1670 J/mol

concentration of glucose-6-phosphate at equilibrium = 2.65 mM

Assuming temperature = 25.0°C

=( 25 + 273)K

= 298 K

We are to find the concentration of fructose-6-phosphate

Using the relation;

ΔG' = -RT In K_c

where;

R = 8.314 J/K/mol

1670 = - (8.314 × 298 ) In K_c

1670 = -2477.572 × In K_c

1670/ 2477.572 = In K_c

0.67 = In K_c

`K_c = e^(-0.67)`

`K_c =` 0.511

Now using the equilibrium constant
`K_c`

`K_c = ([F6P])/([G6P])`

`0.511 = ([F6P])/([2.65])`

F6P = 0.511 × 2.65

F6P = 1.35415

F6P ≅ 1.35 mM