Answer:
The correct answer is option D.
Explanation:
The chemical equation for the conversion follows:
[tex]\text{ glucose 6-phosphate}\rightleftharpoons \text{fructose 6-phosphate}[/tex]
The expression for [tex]K_{eq}[/tex] of above equation is:
[tex]K_{eq}=\frac{\text{[fructose 6-phosphate]}}{\text{[glucose 6-phosphate]}}[/tex]
[tex]\Delta G^o=-RT\ln K_[eq}[/tex]
where,
[tex]\Delta G^o[/tex] = standard Gibbs free energy = 1.72 kJ/mol = 1720 J/mol (Conversion factor: 1kJ = 1000J)
R = Gas constant = [tex]8.315J/K mol[/tex] (given)
T = temperature =298K
Putting values in above equation, we get:
[tex]1720 J/mol=-(8.315J/Kmol)\times 298K\times \ln (\frac{\text{[fructose 6-phosphate]}}{\text{[glucose 6-phosphate]}})[/tex]
[tex]\frac{\text{[fructose 6-phosphate]}}{\text{[glucose 6-phosphate]}}=0.499\approx = 0.5=\frac{1}{2}[/tex]
[tex]\frac{\text{[fructose 6-phosphate]}}{\text{[glucose 6-phosphate]}}=\frac{1}{2}[/tex]
The ratio of glucose 6-phosphate to fructose 6-phosphate at equilibrium :
[tex]\frac{\text{[glucose 6-phosphate]}}{\text{[fructose 6-phosphate]}}=\frac{2}{1}[/tex]
