Aqueous thiosulfate ion,

CHEMISTRY · Multi-part · AP · Chemistry · Kinetics · Kinetics

Problem

Aqueous thiosulfate ion, S2O32−S_2O_3^{2-}, is known to reduce the ferricyanide ion, Fe(CN)63−Fe(CN)_6^{3-}, to ferrocyanide, Fe(CN)64−Fe(CN)_6^{4-}, according to the balanced equation 2 Fe(CN)63−(aq)+S2O32−(aq)→2 Fe(CN)64−(aq)+S4O62−(aq)2\,Fe(CN)_6^{3-}(aq) + S_2O_3^{2-}(aq) \to 2\,Fe(CN)_6^{4-}(aq) + S_4O_6^{2-}(aq) Kinetic studies at 25 ∘C25\,^{\circ}\text{C} yield the experimental rate law rate=kexp[Fe(CN)63−]2[S2O32−]\text{rate} = k_{\text{exp}}[Fe(CN)_6^{3-}]^2[S_2O_3^{2-}] A chemist proposes the two-step mechanism below: Step 1: Fe(CN)63−+S2O32−⇌[Fe(CN)63−⋅S2O32−]−(fast equilibrium)Fe(CN)_6^{3-} + S_2O_3^{2-} \rightleftharpoons [Fe(CN)_6^{3-}\cdot S_2O_3^{2-}]^{-} \quad\text{(fast equilibrium)} Step 2: [Fe(CN)63−⋅S2O32−]−+Fe(CN)63−→2 Fe(CN)64−+S2O3∙−(slow)[Fe(CN)_6^{3-}\cdot S_2O_3^{2-}]^{-} + Fe(CN)_6^{3-} \to 2\,Fe(CN)_6^{4-} + S_2O_3^{\bullet-} \quad\text{(slow)} (The radical product S2O3∙−S_2O_3^{\bullet-} quickly dimerizes in a very fast step that does not affect the rate law.) (a) Identify the intermediate in the proposed mechanism. (b) Write the rate law for the slow elementary step using the concentration of the intermediate. (c) Using the equilibrium in Step 1, express the concentration of the intermediate in terms of [Fe(CN)63−][Fe(CN)_6^{3-}], [S2O32−][S_2O_3^{2-}], and an equilibrium constant KcK_c. (d) Derive the overall rate law predicted by the mechanism and simplify it as much as possible. (e) Compare the derived rate law with the experimental rate law given. Discuss whether the proposed mechanism is consistent with the experimental observations, giving specific reasons for agreement or disagreement.

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