Phosphoglucoisomerase interconverts glucose 6‑phosphate to, and from, glucose 1‑phosphate.

phosphate phosphoglucoisomerase

glucose 6 - ⇌ glucose 1 - phosphate After mixing equal amounts of the two molecules, the solution achieved equilibrium at 25 ∘ C. The concentration at equilibrium of glucose 1‑phosphate is 0.01 M, and the concentration at equilibrium of glucose 6‑phosphate is 0.19 M. Calculate the equilibrium constant, K eq, and the standard free energy change, Δ G ∘, of the reaction mixture.

Question

Kolby York

Chemistry

13 October, 07:29

Phosphoglucoisomerase interconverts glucose 6‑phosphate to, and from, glucose 1‑phosphate.

phosphate phosphoglucoisomerase

glucose 6 - ⇌ glucose 1 - phosphate After mixing equal amounts of the two molecules, the solution achieved equilibrium at 25 ∘ C. The concentration at equilibrium of glucose 1‑phosphate is 0.01 M, and the concentration at equilibrium of glucose 6‑phosphate is 0.19 M. Calculate the equilibrium constant, K eq, and the standard free energy change, Δ G ∘, of the reaction mixture.

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Answers (1)

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Gianna Johnson · Answer 1

Keq = 0.053

7.3 kJ/mol

Explanation:

Let's consider the following isomerization reaction.

glucose 6‑phosphate ⇄ glucose 1 - phosphate

The concentrations at equilibrium are:

[G6P] = 0.19 M

[G1P] = 0.01 M

The concentration equilibrium constant (Keq) is:

Keq = [G1P] / [G6P]

Keq = 0.01 / 0.19

Keq = 0.053

We can find the standard free energy change, ΔG°, of the reaction mixture using the following expression.

ΔG° = - R * T * lnKeq

ΔG° = - 8.314 J/mol. K * 298 K * ln0.053

ΔG° = 7.3 * 10³ J/mol = 7.3 kJ/mol