Question

The enzyme phosphohexose isomerase catalyzes the interconversion glucose 6-phosphate and fructose 6-phosphate. Given that the ΔG'° for the reaction below is +1.72 kJ/mol, what is ratio of glucose 6-phosphate to fructose 6-phosphate at equilibrium? (R = 8.315 J/mol·K; T = 298 K) Glucose 6-phosphate → fructose 6-phosphate A) 1:1. B) 1:2 C) 1:3 D) 2:1 E) 3:1.

Answer 1

Answer:

The correct answer is option D.

Explanation:

The chemical equation for the conversion follows:

$\text{ glucose 6-phosphate}\rightleftharpoons \text{fructose 6-phosphate}$

The expression for $K_{eq}$ of above equation is:

$K_{eq}=\frac{\text{[fructose 6-phosphate]}}{\text{[glucose 6-phosphate]}}$

$\Delta G^o=-RT\ln K_[eq}$

where,

$\Delta G^o$ = standard Gibbs free energy = 1.72 kJ/mol = 1720 J/mol  (Conversion factor: 1kJ = 1000J)

R = Gas constant = $8.315J/K mol$ (given)

T = temperature =298K

Putting values in above equation, we get:

$1720 J/mol=-(8.315J/Kmol)\times 298K\times \ln (\frac{\text{[fructose 6-phosphate]}}{\text{[glucose 6-phosphate]}})$

$\frac{\text{[fructose 6-phosphate]}}{\text{[glucose 6-phosphate]}}=0.499\approx = 0.5=\frac{1}{2}$

$\frac{\text{[fructose 6-phosphate]}}{\text{[glucose 6-phosphate]}}=\frac{1}{2}$

The ratio of glucose 6-phosphate to fructose 6-phosphate at equilibrium :

$\frac{\text{[glucose 6-phosphate]}}{\text{[fructose 6-phosphate]}}=\frac{2}{1}$