Respuesta :
The ΔG° at 298 K is 1.665 kJ/mol or 1665.45 J/mol in key step in the metabolism of glucose for energy is the isomerization of glucose-6-phosphate (G6P) to fructose6-phosphate (F6P): G6P ⇄ F6P; K = 0.510 at 298 K.
Calculation ,
Formula used for for energy is the isomerization of glucose-6-phosphate (G6P) to fructose6-phosphate (F6P):
: ΔG° = - RT㏑K ( i )
Given equilibrium constant ( K ) = 0.510
Temperature in Kelvin ( T ) = 298 K
Universal gas constant ( R ) = 8.3 J/K.mol
Putting the value of temperature ( T ) , equilibrium constant ( K ) and R in equation ( i ) we get ,
ΔG° = - RT㏑K
ΔG° = - 8.3 J/ K.mol × 298 K ×㏑0.510 = - 8.3 J/mol × 298 K × ( - 0.6 )
ΔG° = 1665.45 J/mol = 1.665 kJ/mol
To learn about glucose-6-phosphate here
https://brainly.com/question/16411744
#SPJ4
