Question

Please explain in detail "Biochemistry"

Given: [R= 8.315 J/mol . K] Consider the reaction at standard temperature: glucose-1-P -> glucose-6-P This reaction has a K' eq = 21

Answer the following questions:

a. Is the concentration of glucose-6-P equal, higher or lower than that of glucose-1-P at equilibrium? Justify your answer using the expression for K' eq.

b. What is ∆G'° for the reaction? Write down the equation used in the calculations and the units in your answer.

c. If the concentrations of glucose-6-P and glucose-1-P are 100 M and 1 M, respectively, what is ∆G at 25 oC? Under this condition is the reaction endergonic or exergonic

Answer 1

a) Glucose - 1-P <-----> Glucose - 6 - P

K'eq = [ Glucose - 6 - P ]/[ Glucose - 1 - P ] =21

Where,

[ Glucose -6 - P ] = Concentration of Glucose - 6 - P

[ Glucose - 1 - P ] = Concentration of Glucose - 1 - P

From this expression , we know that [ Glucose - 6 - P ] should be greater than [ Glucose - 1 - P ] , so that we can get K'eq greater than 1. Therefore, at equillibrium

[ Glucose - 6 - P ] > [ Glucose - 1 - P ]

b) The relation between standard free energy change and equillibrium constant is

   ∆G°' = - RTlnK'eq

where,

   R = gas constant , 8.314J/K mol

T = Kelvin tempearature, 298.15K (standard condition)

   Therefore,

   ∆G°' = - (8.314 (J/K mol ) × 298.15K × ln21)

   = - ( 8.314 ( J /K mol) × 298.15K × 2.303 log21)

   = -7548J/mol

= - 7.54kJ/mol

c) ∆G = - RTlnK'eq + RTlnQ

   Q = [ G6P ]/ [ G1P ]

   = 100M/1M

   = 100

Therefore,

∆G = -7.54kJ/mol+(8.314(J/K mol)×298.15K×2.303× log100

   = - 7.54kJ/mol + 11.42kJ/mol

   = 3.88kJ/mol

∆G is positive , so the reaction is entergonic