Consider the reaction: Glucose 6 phosphate <==> Fructose 6-phosphate Q 1: If the

ID: 887145 • Letter: C

Question

Consider the reaction:

Glucose 6 phosphate <==> Fructose 6-phosphate

Q 1: If the Keq for this reaction is 0.5 what does it imply about it?

A) Under standard conditions, the formation of fructose 6-phosphate is favored

B) Under standard conditions, the formation of glucose 6-phosphate is favored

C) Under standard conditions equal amounts of both gluose 6-phosphate and fructose 6-phosphate will be made

Q2:

What will the standard free energy change be for this reactions? Assume temp = 25C (show work)

Q 1 ) Answer : B) Under standard conditions, the formation of glucose 6-phosphate is favored

explanation : Keq < 1 --------------> reaction favoured to backward direction

Keq > 1 --------------> reaction favoured to forward direction

Keq = 1 -------------------> at equilibrium

here Keq = 0.5 means reaction favoured to backward direction

Q2)

R = 8.314 x 10^-3 kJ / mol -K

T = 25 + 273 = 298 K

delta Go = - RT ln Keq

= - 2.303 R T log Keq

= - 2.303 x 8.314 x 10^-3 x 298 x log (0.5)

= 1.72 kJ /mol

delta Go = 1.72 kJ / mol