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Question #44571

Difficulty: 8/10 | 🟢 Free

The Question

A metalloenzyme features a rigid tetrahedral binding site that acts as a perfect 'lock and key' for d10d^{10} metal ions. The standard reduction potential for the aqueous couple \ce{Cu^{2+}{(aq)} / Cu^+{(aq)}} is +0.15 V+0.15 \text{ V} at 300 K300 \text{ K}.

The binding of \ce{Cu^+_{(aq)}} to the apoenzyme (E) has ΔH∘=−40 kJ mol−1\Delta H^\circ = -40 \text{ kJ mol}^{-1} and ΔS∘=−50 J K−1 mol−1\Delta S^\circ = -50 \text{ J K}^{-1} \text{ mol}^{-1}.

The binding of \ce{Cu^{2+}_{(aq)}} to the apoenzyme has ΔH∘=−70 kJ mol−1\Delta H^\circ = -70 \text{ kJ mol}^{-1}. However, because \ce{Cu^{2+}} is a d9d^9 ion, it undergoes Jahn-Teller distortion, forcing a conformational change in the rigid enzyme. This induced fit results in an additional conformational entropy penalty, making its total binding entropy ΔS∘\Delta S^\circ more negative by 36 J K−1 mol−136 \text{ J K}^{-1} \text{ mol}^{-1} compared to the ideal d10d^{10} binding.

Calculate the standard reduction potential (E∘E^\circ) of the enzyme-bound copper couple (\ce{E-Cu^{2+} / E-Cu^+}) at 300 K300 \text{ K}. (Assume Faraday's constant F=96000 C mol−1F = 96000 \text{ C mol}^{-1})

Options

  • A.

    -0.050 V

  • B.

    -0.163 V

  • C.

    +0.350 V

  • D.

    -0.200 V

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Numerical_Precision

Improve calculation stamina and avoid rounding errors in high-stakes competitive examinations.

Concepts Behind This Question

E° (M²⁺/M) Anomalies|Reaction Progress Coordinate|Enzyme Specificity 'Lock and Key'

Why This Question Is Tricky

Common Mistakes

This problem tests the 'Entatic State' principle in bioinorganic chemistry via a thermodynamic cycle.

Students often calculate ΔG\Delta G using standard enthalpy and entropy values, failing to account for the hidden variable: the conformational entropy loss (ΔSJT\Delta S_{\text{JT}}) induced by the Jahn-Teller distortion of the d9d^9 \ce{Cu^{2+}} ion in a rigid site designed for d10d^{10} ions.

By forcing a distorted geometry, the enzyme entropically destabilizes the oxidized \ce{Cu^{2+}} state.

Ignoring this entropy penalty (Trap B) leads to an overly stabilized \ce{Cu^{2+}} state and an incorrectly low reduction potential (-0.163 V).

The problem rigorously bridges coordination chemistry anomalies with thermodynamic reaction coordinates.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterD and F Block Elements - Class 12
ConceptE° (M²⁺/M) Anomalies
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree