±Chelation: Thermodynamic Aspects Ch elates are polydentate ligands. The formati
ID: 1018199 • Letter: #
Question
±Chelation: Thermodynamic Aspects
Chelates are polydentate ligands. The formation of a complex ion between a metal ion and a chelate is referred to as chelation. In chelation reactions, the ligand wraps itself around the metal ion. (Figure 1) Many important biological reactions involve the formation of complexes involving chelating agents. In general, chelating agents (polydentate ligands) form more stable complexes than monodentate ligands, as evidenced by their very large formation constants, Kf.
Entropic effects
Co(H2O)63+(aq)+EDTA4(aq)CoEDTA(aq)+6H2O(l)
Part A
In which of the following reactions will the entropy change, S, be a positive value?
Check all that apply.
SubmitHintsMy AnswersGive UpReview Part
Correct
The favorable entropy changes of chelation reactions contributes to the high stability of the complex-ion products.
Chelates in biological systems
The porphine molecule(Figure 2) is a tetradentate ligand and is the basis for many porphyrins found in biological systems. Porphyrins are complexes with metal ions and have different substituents attached to the carbons of the porphine ligand. Heme (which contains Fe2+) and chlorophyll (which contains Mg2+) are two common examples.
Part B
M(H2O)mn+(aq)+por(aq)M(por)n+(aq)+mH2O(l)
Express your answer numerically.
1.9•103
SubmitHintsMy AnswersGive UpReview Part
Incorrect; Try Again; 2 attempts remaining
Provide FeedbackContinue
Figure 1Figure 2 of 2
Incorrect. Incorrect; Try Again; 2 attempts remaining. No additional feedback.
Chemistry 106 Summer 2016
Help
Close
Homework Set #7
±Chelation: Thermodynamic Aspects
ResourcesConstantsPeriodic Table
« previous 6 of 25 next »
±Chelation: Thermodynamic Aspects
Chelates are polydentate ligands. The formation of a complex ion between a metal ion and a chelate is referred to as chelation. In chelation reactions, the ligand wraps itself around the metal ion. (Figure 1) Many important biological reactions involve the formation of complexes involving chelating agents. In general, chelating agents (polydentate ligands) form more stable complexes than monodentate ligands, as evidenced by their very large formation constants, Kf.
Entropic effects
The chelate effect can be partially explained in terms of the entropy change involved in chelation reactions. The reaction forming CoEDTA from the hexaaqua complex of Co3+,Co(H2O)63+(aq)+EDTA4(aq)CoEDTA(aq)+6H2O(l)
leads to a net increase in the number of particles in the solution.Part A
In which of the following reactions will the entropy change, S, be a positive value?
Check all that apply.
CaEDTA(aq)+3en(aq)Ca(en)33+(aq)+EDTA4(aq) Cu(H2O)62+(aq)+3NH3(aq)Cu(H2O)3(NH3)32+(aq)+3H2O(l) Co(NH3)4(acac)3+(aq)+2acac(aq)Co(acac)33+(aq)+3NH3(aq) ZnCl2(en)(aq)+4OH(aq)Zn(OH)24(aq)+2Cl(aq)+en(aq) FeCl64(aq)+EDTA4(aq)FeEDTA2(aq)+6Cl(aq)SubmitHintsMy AnswersGive UpReview Part
Correct
The favorable entropy changes of chelation reactions contributes to the high stability of the complex-ion products.
Chelates in biological systems
The porphine molecule(Figure 2) is a tetradentate ligand and is the basis for many porphyrins found in biological systems. Porphyrins are complexes with metal ions and have different substituents attached to the carbons of the porphine ligand. Heme (which contains Fe2+) and chlorophyll (which contains Mg2+) are two common examples.
Part B
Consider the reaction of a generic metal ion (Mn+) with a generic porphine molecule (por):M(H2O)mn+(aq)+por(aq)M(por)n+(aq)+mH2O(l)
Based on the values of standard enthalpy, H, and entropy, S, shown here, H S -27.5 kJ 8.82 J/K what is the value of the formation constant, Kf , for the reaction at 25 C?Express your answer numerically.
Kf =1.9•103
SubmitHintsMy AnswersGive UpReview Part
Incorrect; Try Again; 2 attempts remaining
Provide FeedbackContinue
Figure 1Figure 2 of 2
Explanation / Answer
part A ) answers
Co(NH3)4(acac)3+(aq)+2acac(aq)Co(acac)33+(aq)+3NH3(aq)
FeCl64(aq)+EDTA4(aq)FeEDTA2(aq)+6Cl(aq)
part B)
G = H - T S
= -27.5 - 298 x 8.82 x 10^-3
= -30.13 kJ
G = - R T ln Kf
-30.13 = - 8.314 x 10^-3 x 298 x ln Kf
Kf = 1.91 x 10^5
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.