If a solution of glucose 1-phosphate is incubated with an enzyme, the glucose 1-

Question

If a solution of glucose 1-phosphate is incubated with an enzyme, the glucose 1-phosphate is transformed to glucose 6-phosphate until equilibrium is reached. At equilibrium, the concentration of glucose 1-phosphate is 4.5 times 10^-3 M and that of glucose 6-phosphate is 8.6 times 10^-2 M Set up the expressions for the calculation of K_eg and Delta G degree for this reaction (in the direction of glucose 6-phosphate formation) then compute the numerical values. You do not need to explain your steps. (R = 8.315 J/mol.K; T = 298 K)

Explanation / Answer

Given reaction is

glucose-1-phosphate <-----------------> glucose-6-phosphate

Keq:

Keq =   Concentration of products at equilibrium/ Concentration of reactants at equilibrium

Given that at equilibrium

[glucose-1- phosphate] = 4.5 x 10-3 M and [glucose-6- phosphate] = 8.6 x 10-2 M

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

   = 8.6 x 10-2 M / 4.5 x 10-3 M

      = 19.11

Keq = 19.11

Go

Given that

Temperature T = 298 K

R = universal gas constant = 8.315 J/K/mol

Then,

Go = - RT In Keq

       = - (8.315 J/K/mol) (298 K) In (19.11)

       = -7310 J/mol

  Go = -7310 J/mol

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