The standard free-energy changes for the reactions below are given. phosphocreatine → creatine + pi ∆ g'° = –43.0 kj/molatp → adp + pi ∆ g'° = –30.5 kj/mol
what is the overall ∆ g'° for the following reaction?
phosphocreatine + adp → creatine + atp

Gibbs free-energy of the reaction = (–12.5 kJ/mol)

Explanation:

The Gibbs free-energy of a reaction predicts the spontaneity or feasibility of a given chemical reaction.

Given the standard Gibbs free energy changes:

Phosphocreatine → creatine + Pi,  ∆G° = –43.0 kJ/mol     ...(1)

ATP → ADP + Pi , ∆G° = –30.5 kJ/mol      (2)

Now to calculate the Gibbs free-energy of the given chemical reaction: Phosphocreatine + ADP → creatine + ATP; the equation (2) is reversed to give:

ADP + Pi  → ATP, ∆G° = + 30.5 kJ/mol      ...(3)

Now the equation (3) and (1) are added, to give:

Phosphocreatine + ADP + Pi→ creatine + ATP + Pi

⇒ Phosphocreatine + ADP → creatine + ATP  

Therefore, to calculate the Gibbs free-energy of the reaction, the standard Gibbs free energy changes of the equations (1) and (3) are added similarly:

Gibbs free-energy of the reaction: ∆G° = (–43.0 kJ/mol) + ( + 30.5 kJ/mol) = (–12.5 kJ/mol)

Therefore, the Gibbs free-energy of the reaction = (–12.5 kJ/mol)

43.452

step-by-step explanation:

284 times .153 = 43.452