This very low concentration of the desired product would be unfavorable for glycolysis. In fact the reaction is coupled to ATP hydrolysis to give the overall reaction: ATP + glucose → glucose-6-phosphate + ADP + H+ What is the ΔGo' for the coupled reaction?

ΔG'° for the coupled reaction = -16.7 kJ/mol

Note: The question is missing some parts. The complete question is as follows:

The first reaction in glycolysis is the phosphorylation of glucose:

Pi+glucose⟶glucose−6−phosphate+H2O

This is a thermodynamically unfavorable process, with ΔG∘′= +13.8kJ/mol. In a liver cell at 37 ∘C the concentrations of both phosphate and glucose are normally maintained at about 5 mM each.

This very low concentration of the desired product would be unfavorable for glycolysis. In fact the reaction is coupled to ATP hydrolysis to give the overall reaction: ATP + glucose → glucose-6-phosphate + ADP + H+

What is the ΔG'° for the coupled reaction?

Explanation:

The coupling of ATP hydrolysis which is a thermodynamically favourable reaction to the phosphorylation of glucose makes it favourable. Since the two reactions constitute a sequential reaction, their standard free energy changes are additive.

For ATP hydrolysis: ATP + H₂O ---> ADP + Pi ; ΔG'°  = -30.5KJ/mol

For phosphorylation of glucose:  Pi + glucose⟶glucose−6−phosphate + H2O  ; ΔG'° = +13.8 kJ/Mol

For the overall reaction: ATP + glucose → glucose-6-phosphate + ADP + H⁺ ; ΔG'° = 13.8 + (-30.5) kJ/mol = -16.7 kJ/mol

Therefore, ΔG'° for the coupled reaction = -16.7 kJ/mol

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