Answer:
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
