In the following redox reaction with an equilibrium constant K=2.0 × 10⁸ [Ru (NH₃)₆]²⁺+ [Fe (H₂ O₆)]³⁺ → [Ru… · ABC26IN1093

Subject: Inorganic Chemistry · Chapter: Coordination Chemistry · Topic: Reaction Mechanisms Substitution · Exam: CSIR NET DEC 2016 · Marks: 2 · Difficulty: Medium

In the following redox reaction with an equilibrium constant $\mathrm{K}=2.0 \times 10^{8}$
\[ \left[\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}+\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}_{6}\right)\right]^{3+} \rightarrow\left[\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}+\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+} \]
the self exchange rates for oxidant and reductant are $5.0\ \mathrm{M}^{-1}\mathrm{s}^{-1}$ and $4.0 \times 10^{3}\ \mathrm{M}^{-1}\mathrm{s}^{-1}$ respectively. The approximate rate constant $\left(\mathrm{M}^{-1} \mathrm{s}^{-1}\right)$ for the reaction is
(a)$3.16 \times 10^{6}$
(b)$2.0 \times 10^{6}$
(c)$6.32 \times 10^{6}$
(d)$3.16 \times 10^{6}$
Answer
Answer: B ✓ checked by 4AB · confidence high
Explanation
Marcus cross relation: k₁₂ ≈ √(k₁₁k₂₂K₁₂) = √(5 × 4×10³ × 2×10⁸) = √(4×10¹²) = 2×10⁶ M⁻¹ s⁻¹.

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