   Chapter 18.7, Problem 1.1ACP

Chapter
Section
Textbook Problem

Consider the hydrolysis reactions of creatine phosphate and adenosine-5′-monophosphate.Creatine Phosphate + H2O → Creatine + HPiΔrG°’ = −43.3 KJ/mol-rxnAdenosine-5′-Honophosphate + H2O → Adenosine + HPiΔrG°’ = −9.2 KJ/mol-rxnWhich of the following combinations produces a reaction that is product-favored at equilibrium: for creatine phosphate to transfer phosphate to adenosine or for adenosine-5′-monophosphate to transfer phosphate to creatine?

Interpretation Introduction

Interpretation:

Of the given combinations the one that produces product favoured reaction at equilibrium should be identified.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔrGo. It can be calculated in a similar manner as entropy and enthalpy.  The expression for the free energy change is:

ΔrG°fG°(products)fG°(reactants)

The reaction is product favored at equilibrium if the calculated ΔrGo value is negative.

The reaction is reactant favored at equilibrium if the calculated ΔrGo value is positive.

Explanation

The ΔrGo for the net reaction is calculated below,

Given:

The given reactions are:

Creatine Phosphate + H2OCreatine + HPi

The ΔrGo for the reaction is 43.3 kJ/mol-rxn.

The ΔrG°' for the reaction is 9.2 kJ/mol-rxn.

The sign of ΔrG°' is reversed if the equation is reversed

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