Question

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

Answer 1

Answer:

Keq = 19

ΔG° = -7.3kJ/mol

Explanation:

Based on the chemical reaction:

Glucose 1-phosphate ⇄ Glucose 6-phosphate

The equilibrium constant, Keq is defined as:

Keq = [Glucose 6-phosphate] / [Glucose 1-phosphate]

Where [] are equilibrium concentrations of each substance

Replacing:

Keq = [0.19M] / [0.01M]

Keq = 19

Now, standard free energy change, ΔG° is defined as:

ΔG° = -RT ln K

Where R is gas constant 8.314J/molK

T is absolute temperature (25°C + 273.15K = 298.15K)

and K is equilibrium constant = 19

Replacing:

ΔG° = -8.314J/molK*298.15K ln 19

ΔG° = -7299J/mol

ΔG° = -7.3kJ/mol