4AB › Chemistry PYQ › CSIR-NET › December 2016 › Q52Q52 · CSIR-NET Chemistry, December 2016 Paper: CSIR-NET December 2016 · Subject: Inorganic Chemistry · Chapter: Coordination Chemistry · Topic: Reaction Mechanisms Substitution · 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.
← Q51 · All questions in this paper · Coordination Chemistry chapter list · Q53 → Study loop for Coordination Chemistry
Open in whiteboard · Browse this chapter in the app