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

Keq = 3

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:

[G6P] = 0.8 M

[F8P] = 0.0 M

[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

b is the correct answer

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