Answer:
ΔG° = –7.4 kcal/mole
Step-by-step explanation:
When there are coupled reactions taking place, we can say that their free energy variations are additive. Meaning that if there is a reaction A → B followed by B → C, each reaction will have its own free energy variation, ΔG°₁ and ΔG°₂. As both reactions are secuential, there is a global reaction taking place, which is A → C. This global reaction also has its own free energy variation, ΔG°total, which represents the sum of the individual free energy variations of the coupled reactions, ΔG°₁ + ΔG°₂ = ΔG°total.
For this case we have, in the first reaction of glycolysis, there are two coupled reactions that take place:
(1) glucose + Pi → glucose 6-phosphate + H₂O
(2) ATP + H₂O → ADP + Pi
We know that the total variation of free energy for that reaction is:
ΔG°total = –4.06 kcal/mole
But the individual variations of free energy, ΔG°₁ and ΔG°₂, are unknown, so we can propose the next equation:
ΔG°₁ + ΔG°₂ = –4.06 kcal/mole
Then we have another value given to us, the variation of free energy of the reversed first reaction (1), which would be:
(3) glucose 6-phosphate → glucose + Pi
ΔG°₃ = –3.34 kcal/mole
As this reaction is the reversed reaction of the first one (1), we can assume the next:
ΔG°₁ = (–1) * (ΔG°₃) = (–1) * (–3.34 kcal/mole) = 3.34 kcal/mole
So now that we have the value of ΔG°₁ we can substitute it in the proposed equation to find out the value of ΔG°₂ :
ΔG°₁ + ΔG°₂ = –4.06 kcal/mole
ΔG°₂ = –4.06 kcal/mole – ΔG°₁
ΔG°₂ = –4.06 kcal/mole – 3.34 kcal/mol
ΔG°₂ = –7.4 kcal/mole
So there it is, that is the value of the variation of free energy of the second reaction (2), which is the hydrolisis of ATP.