Question

Phosphoglucomutase catalyzes the reaction of glucose 6-phosphate (G6P) to fructose 6-phosphate (F8P). You are starting the reaction in a test tube with the 0.8M substrate (G6P), and you let the reaction reach equilibrium. The product (F6P) concentration at equilibrium is 0.6 M. There are no intermediates in this reaction and no products at the beginning. The Keq for this reaction is_____

Answer 1

Keq = 3

Step-by-step explanation:

The reaction can be expressed as:

G6P → F8P

Keq = [F8P] / [G6P]

Because there are no products at the beginning, and there are no intermediates in this reaction, we can express the concentrations of the substances as followed:

• At the beginning of the reaction:

[G6P] = 0.8 M

[F8P] = 0.0 M

• At equilibrium, some G6P has turned into F8P, so its concentration can be expressed as a substraction of its original value:

[G6P] = 0.8 M - x

[F8P] = x = 0.6 M

We can rewrite [G6P]:

[G6P] = 0.8 M - 0.6 M = 0.2 M

Now that we know the values of [G6P] and [F8P] we can calculate Keq:

Keq = [F8P] / [G6P] = 0.6 / 0.2 = 3