In spite of the high equilibrium constant (K = 1025) for the reaction givenhere, the reaction proceeds very slowly, reaching equilibrium only afterseveral days at room temperature.[Co(NH3)6]3+ + 6 H3O* = [Co(H20)6]3+ + 6 NH,*The observation is consistent with:the activation energy for the forward reaction being low.[Co(NH3)6]3+ being unstable in strong acid.[Co(NH3)6]3+ being more labile than [Co(H2O)6]3+.the classification of [Co(NH3)6]3+ as an inert complex.

Ammonia is a neutral ligand. The oxidation state of cobalt in the complex [Co(NH3)6]3+ is +3. The…

In spite of the high equilibrium constant (K = 1025) for the reaction given here, the reaction proceeds very slowly, reaching equilibrium only after several days at room temperature. [Co(NH3)6]3+ + 6 H3O* = [Co(H20)6]3+ + 6 NH4* The observation is consistent with: the activation energy for the forward reaction being low. [Co(NH3)6]3+ being unstable in strong acid. [Co(NH3)6]3+ being more labile than [Co(H2O)6]3+. the classification of [Co(NH3)6]3+ as an inert complex.

In spite of the high equilibrium constant (K = 1025) for the reaction given here, the reaction proceeds very slowly, reaching equilibrium only after several days at room temperature. [Co(NH3)6]3+ + 6 H3O* = [Co(H20)6]3+ + 6 NH4* The observation is consistent with: the activation energy for the forward reaction being low. [Co(NH3)6]3+ being unstable in strong acid. [Co(NH3)6]3+ being more labile than [Co(H2O)6]3+. the classification of [Co(NH3)6]3+ as an inert complex.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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