Question

Assume that you have a solution of 0.1 M glucose 6-phosphate. To this solution, you add the enzyme Phosphoglucomutase, which catalyzes the following reaction:

The ΔG0’ for the reaction is +1.8 kcal mol-1. (a) Does the reaction proceed as written? If so, what are the final concentrations of glucose 6-phosphate and glucose 1-phosphate?

Answer 1

(a)   

∆G is expressed as change in gibbs free energy of the system that occurs during a reaction which is expressed as

∆G = ∆H - T∆S

and when this reaction occur under standard conditions the change in gibbs free energy is termed as Standard-state free energy of reaction (∆G⁰)

∆G⁰ =  ∆H^{0 -} T∆S⁰

In the given reaction, the change is gibbs free energy is expressed as

ΔG⁰ = 1.8 kcal / mol

Glucose-6-P <-----> Glucose-1-P (ΔG⁰ = +1.8 kcal/mol)

Here ΔG⁰ has positive value of 1.8 kcal/mol which signifies that reaction is nonspontaneous

The given reaction requires the input of energy to proceed the reaction in the normal forward direction.

b)

ΔG = ΔG⁰ + RT ln Q

At equilibrium the change in the product to reactant will fall under standard condition therefore ΔG = 0 and Q will be expressed as Keq

So the equation ΔG= ΔG⁰ + RT ln Q will be expressed as

0 = ΔG⁰+ RT ln Keq

ΔG⁰ = - RT ln Keq

1.8 kcal/mol (convert it to joule/ mol by multiplying it with 4184, because 1 calorie is equivalent to 4184 joule; 7531.2 J/mol), R is 8.314 J/mol K and T is 298 K so

In Keq= -(7531.2 J/mol)/(8.314j/mol K x 298K)

In Keq= -3.039

2.303 x log (Keq) = -3.039

Log (Keq) = -(3.039/2.303)

Keq= 0.0479

Q= =Keq

Glucose 6 phosphate is 0.1 M there fore

Glucose 1 phospahte is=Keq x [glucose 6 phosphate]

                                     =0.0479 x0.1

                                     =4.79 X 10 ^-3