When a mixture of glucose 6-phosphate and fructose 6-phosphate is incubated with the enzyme phosphohexose isomerase (which catalyzes the interconversion of these two compounds) until equilibrium is reached, the final mixture contains twice as much glucose 6-phosphate as fructose 6-phosphate. Which one of the following statements is best applied to this reaction outlined below? (R = 8.315 J/mol · K; T = 298 K) glucose 6-phosphate $\rightarrow$fructose 6-phosphate
A) $\Delta$G'° is incalculably large and negative.
B) $\Delta$G'° is -1.72 kJ/mol.
C) $\Delta$G'° is zero.
D) $\Delta$G'° is +1.72 kJ/mol.
E) $\Delta$G'° is incalculably large and positive.

Question

Quizplus · Accepted Answer

Since the equilibrium mixture contains twice as much glucose 6-phosphate as fructose 6-phosphate, we can set up the equilibrium expression as follows:
K_eq = [glucose 6-phosphate] / [fructose 6-phosphate] = 2
From the equation given, we can see that one mole of glucose 6-phosphate is converted into one mole of fructose 6-phosphate during the reaction. Therefore, there is no net change in the number of moles of gas, and we can assume that Δn = 0. Using the equation ΔG° = -RTlnK_eq and substituting in the given values, we get:
ΔG° = -8.315 J/mol · K * 298 K * ln(2) / 1000 J/kJ = -1.72 kJ/mol
Therefore, the answer is (D), and the value of ΔG° is -1.72 kJ/mol.

When a Mixture of Glucose 6-Phosphate and Fructose 6-Phosphate Is →\rightarrow→

When a Mixture of Glucose 6-Phosphate and Fructose 6-Phosphate Is $\rightarrow$