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31 May, 09:53

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

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  1. 31 May, 11:29
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    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)
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