The enzyme, phosphoglucomutase, catalyzes the interconversion
of glucose 1-phosphate and glucose 6-phosphate,

Glucose 1-phosphate image from custom entry tool Glucose 6-phosphate

If at chemical equilibrium at 25° C, 95% glucose 6-phospate is
present, calculate:

a. K’eq and ΔG°’ of the reaction forming glucose 6-phosphate;
b.ΔG’ of the reaction under cellular conditions where the glucose 1-phosphate concentration is 1.090 x 10-2 M and the concentration of glucose 6-phosphate is 1.395 x 10-4 M.

= 19

ΔG° of the reaction forming glucose 6-phosphate =  -7295.06 J

ΔG° of the reaction  under cellular conditions = 10817.46 J

Explanation:

Glucose 1-phosphate     ⇄     Glucose 6-phosphate

Given that: at equilibrium, 95% glucose 6-phospate is  present, that implies that we 5% for glucose 1-phosphate

So, the equilibrium constant can be calculated as:

= 19

The formula for calculating ΔG° is shown below as:

ΔG° = - RTinK

ΔG° = - (8.314 Jmol⁻¹ k⁻¹ × 298 k ×  1n(19))

ΔG° = 7295.05957 J

ΔG°≅ - 7295.06 J

b)

Given that; the concentration  for  glucose 1-phosphate = 1.090 x 10⁻² M

the concentration of glucose 6-phosphate is 1.395 x 10⁻⁴ M

Equilibrium constant   can be calculated as:

0.01279816514  M

0.0127 M

ΔG° = - RTinK

ΔG° = -(8.314*298*In(0.0127)

ΔG° = 10817.45913 J

ΔG° = 10817.46 J

answer: c. the experiment shows that the red substance experienced a chemical change.

explanation: chemical change is the change in which chemical composition of the substance changes. physical change is the change in which there is no change in chemical composition.

chemical changes are accompanied by following changes:

a) precipitate formation

b) energy is released or absorbed

c) formation of bubbles

d) change in color

as it is given that the red color has disappeared and there is production of gas, it means the red substance has undergone a chemical change.

e

i think it is true since you can test the answer and can use observations to complete it.