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

The correct answer is option D.

Step-by-step 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)* 298K* \ln (\frac{\text{[fructose 6-phosphate]}}{\text{[glucose 6-phosphate]}})`

`\frac{\text{[fructose 6-phosphate]}}{\text{[glucose 6-phosphate]}}=0.499\approx = 0.5=(1)/(2)`

`\frac{\text{[fructose 6-phosphate]}}{\text{[glucose 6-phosphate]}}=(1)/(2)`

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

`\frac{\text{[glucose 6-phosphate]}}{\text{[fructose 6-phosphate]}}=(2)/(1)`