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A reactant Aⁿ+ is oxidised to A^(n+4)+ in solution. To follow the rate, a fixed volume of the solution is titrated with a reducing reagent that reacts with both Aⁿ+ (reducing it to A^(n-2)+) and A^(n+4)+ (reducing it to A^(n-1)+). Volume of reagent consumed: 30 mL at t = 0 and 45 mL at t = 10 min. Assuming first order kinetics for Aⁿ+ -> A^(n+4)+, find the rate constant (min⁻¹). The n-factor of the reagent is the same toward both species. Given ln(3/2) = 0.4.
- 0.040 min⁻¹
- 0.020 min⁻¹
- 0.004 min⁻¹
- 0.400 min⁻¹
Correct answer: 0.040 min⁻¹
Solution
Equivalents change from 30 to 45 as Aⁿ+ (n-factor 2 toward reagent) converts to A^(n+4)+ (n-factor 5); solving gives [Aⁿ+]t/[Aⁿ+]0 = 2/3, so k = (1/t) ln(3/2) = 0.04 min⁻¹.
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