Chapter 22, Problem 90SCQ

Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

In this question, we explore the differences between metal coordination by monodentate and bidentate ligands. Formation constants, Kf, for [Ni(NH3)6]2+(aq) and [Ni(en)3]2+(aq) are as follows:Ni2+(aq) + 6 NH3(aq) → [Ni(NH3)6]2+(aq) Kf = 108Ni2+(aq) + 3 en(aq) → [Ni(en)3]2+(aq) Kf = 1018The difference in Kf between these complexes indicates a higher thermodynamic stability for the chelated complex, caused by the chelate effect. Recall that K is related to the standard free energy of the reaction by ΔrG° = −RT ln K and ΔrG° = ΔrH° − TΔrS°. We know from experiment that ΔrH° for the NH3 reaction is −109 kJ/mol-rxn, and ΔrH° for the ethylenediamine reaction is −117 kJ/mol-rxn. Is the difference in ΔrH° sufficient to account for the 1010 difference in Kf? Comment on the role of entropy in the second reaction.

Interpretation Introduction

Interpretation: The role of entropy in the given reaction has to be commented

Concept introduction:

Chelate Effect:  The chelate effect is the enhanced affinity of chelating ligands for a metal ion compared to the affinity of a collection of similar nonchelating (mono dentate) ligands for the same metal.

Entropy: A thermodynamic function that describes the randomness and disorder of molecules based on the number of different arrangements available to them in a given system or reaction.

Gibb’s equation: A thermodynamic equation used for calculating changes in Gibbs energy of a system as a function of temperature.

ΔrGo=ΔrHoTΔrSowhere,ΔrGogibbsfreeenergyTtemperatureΔSEntropychangeΔHEnthalpychange

Explanation

Calculate the Gibbs free energy for the given reaction,

[Ni(NH3)6]2+For,Kf=â€‰108Î”rGo=âˆ’RTlnK=âˆ’â€‰(8.314â€‰J/K.â€‰mol)â€‰â€‰(298â€‰K)(ln108)=â€‰-45.638â€‰kJ/mol[Ni(en)3]2+For,Kf=â€‰1018Î”rGo=âˆ’RTlnK=âˆ’â€‰(8.314â€‰J/K.â€‰mol)â€‰â€‰(298â€‰K)(ln1018)=â€‰-102

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