In the first reaction of glycolysis (the pathway that begins the oxidative breakdown of sugars), the enzyme hexokinase uses atp to catalyze the phosphorylation of glucose to glucose 6-phosphate and adp. the δg° of this reaction is a favorable –4.06 kcal/mole. another sometimes active enzyme, called glucose 6-phosphatase, effectively "reverses" this reaction, hydrolyzing glucose 6-phosphate back to glucose and releasing a phosphate. the δg° of this reaction is –3.34 kcal/mole. based on these values, what is the δg° for the hydrolysis of atp: atp + h2o → adp + pi?

ΔG° = –7.4 kcal/mole

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.

o conteúdo de calor de um sistema é igual à entalpia apenas para um sistema que é constante