Question

Consider the following isomerization reactions of some simple sugars and values for their standard Gibbs free energy ΔG∘:

reaction A:glucose-1-phosphate⟶ glucose-6-phosphate, ΔG∘=−7.28 kJ/mol

Reaction B: fructose-6-phosphate⟶⟶glucose-6-phosphate,ΔG∘=−1.67 kJ/mol

Calculate the equilibrium constant K for the isomerization of glucose-1-phosphate to fructose-6-phosphate at 298 K. Express your answer numerically using two significant figures.

Answer 1

Reuqired reaction   glucose-1-phosphate ------------> fructose-6-phosphate

This can be achieved by

Eq (1) + Reverse of Eq (2)

Eq (1) ------- glucose-1-phosphate⟶ glucose-6-phosphate ΔG∘ = −7.28 kJ/mol

Reverse of Eq (2) -- glucose-6-phosphate ----> fructose-6-phosphate   ΔG∘= +1.67 kJ/mol

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add above two equations and cancel similar terms on bothe sides

The resultant equation is

glucose-1-phosphate ------------> fructose-6-phosphate

ΔG∘ = −7.28 kJ/mol + 1.67 kJ/mol

= -5.61 kJ/mol

= -5610 J/mol

ΔG∘ = -5610 J/mol

We know that

  ΔG∘ = -RT InK

K = e ^-ΔG∘/RT

= e ^{-[-5610 J/mol/ 8.314 J/K/mol x 298 K ]}

   = 9.6

K = 9.6

Therefore,

Equilibrium constant K = 9.6