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The reaction \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}+2 \mathrm{I}^{-} \longrightarrow 2 \mathrm{SO}_{4}^{2-}+\mathrm{I}_{2}\) proceeds slowly in aqueous solution, but it can be catalyzed by the \(\mathrm{Fe}^{3+}\) ion. Given that \(\mathrm{Fe}^{3+}\) can oxidize \(\mathrm{I}^{-}\) and \(\mathrm{Fe}^{2+}\) can reduce \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-},\) write a plausible two-step mechanism for this reaction. Explain why the uncatalyzed reaction is slow.

Short Answer

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The uncatalyzed reaction is slow due to high activation energy. Catalysis by \( \mathrm{Fe}^{3+} \) occurs via two steps: \( \mathrm{Fe}^{3+} \) oxidizes \( \mathrm{I}^{-} \) to \( \mathrm{I}_{2} \), and then \( \mathrm{Fe}^{2+} \) reduces \( \mathrm{S}_{2} \mathrm{O}_{8}^{2-} \) back to \( \mathrm{Fe}^{3+} \), thus accelerating the reaction.

Step by step solution

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01

Understanding the Uncatalyzed Reaction

The reaction without a catalyst involves shaking hands between reactants that have similar charges or are strongly attracted to each other, which doesn't usually happen fast in solutions. This mutual repulsion or sluggish reactivity makes the reaction slow. The persulfate ion \( \mathrm{S}_{2} \mathrm{O}_{8}^{2-} \) has to oxidize multiple iodide ions \( \mathrm{I}^{-} \), each requiring a collision with the right orientation and enough energy to break bonds and form \( \mathrm{I}_{2} \). These conditions in a solution without any help (like a catalyst) make the whole process slow due to the high activation energy needed.
02

Introducing the Catalyst Effect

The role of a catalyst here is like a matchmaker that helps these species react by creating intermediates that react more easily. \( \mathrm{Fe}^{3+} \) takes up electrons to become \( \mathrm{Fe}^{2+} \), making iodine \( \mathrm{I}_{2} \). Meanwhile, \( \mathrm{Fe}^{2+} \) can give up electrons to the persulfate ion, reverting back to \( \mathrm{Fe}^{3+} \) and sulfates \( \mathrm{SO}_{4}^{2-} \). This cycle makes the reaction pathway smoother and faster by lowering the energy barriers.
03

Writing the Catalyzed Reaction Mechanism

Let's outline the steps using \( \mathrm{Fe}^{3+} \) as a catalyst: 1. \( \mathrm{Fe}^{3+} + \mathrm{I}^{-} \rightarrow \mathrm{Fe}^{2+} + \mathrm{I}_{2} \) - Here, \( \mathrm{Fe}^{3+} \) oxidizes \( \mathrm{I}^{-} \) to \( \mathrm{I}_{2} \), which is a fast step due to the catalytic effect.2. \( \mathrm{Fe}^{2+} + \mathrm{S}_{2} \mathrm{O}_{8}^{2-} \rightarrow \mathrm{Fe}^{3+} + 2\mathrm{SO}_{4}^{2-} \) - Next, \( \mathrm{Fe}^{2+} \) reduces \( \mathrm{S}_{2} \mathrm{O}_{8}^{2-} \) back to \( \mathrm{Fe}^{3+} \), completing the catalytic cycle.
04

Explanation of the Catalysis

The slow uncatalyzed reaction becomes faster due to catalysis because each step in the catalyzed series has a lower activation energy compared to the direct reaction. The catalyst \( \mathrm{Fe}^{3+} \) continuously changes from its oxidized to reduced state and back again, facilitating the transformation of reactants to products without being consumed itself, thus speeding up the overall process.

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Catalysis
Catalysis is a fascinating phenomenon where a substance, called a catalyst, speeds up a reaction without undergoing any permanent chemical change itself. In the reaction provided, the presence of the \(\text{Fe}^{3+}\) ion serves this purpose effectively.

Catalysts offer an alternate pathway for the reaction with lower activation energy. This means that even though the individual particles need less energy to react, they combine faster.
  • The catalyst facilitates a series of intermediate steps.
  • These steps are easier than the direct pathway of the uncatalyzed reaction.
In this exercise, \(\text{Fe}^{3+}\) is helping to change the usual reaction dynamics. It assists in forming and breaking bonds more efficiently by creating significant intermediates.

This effectively lowers energy barriers between reactants and products, giving a boost to reaction speed.
Activation Energy
Activation energy is the minimum amount of energy required to start a chemical reaction. It's like the initial push needed to start rolling a boulder up a hill.

In our given reaction, the uncatalyzed path involves large initial energy, making it slow. This is because reactants such as \(\text{S}_2\text{O}_8^{2-}\) and \(\text{I}^{-}\) have to collide with the correct orientation and energy to initiate the reaction.

Without a catalyst, each molecule needs to overcome a significant barrier before reacting effectively.
  • High activation energy means fewer molecules can collide with enough force to react.
  • This results in slower reaction rates.
Adding \(\text{Fe}^{3+}\) acts as a bridge, lowering the activation energy. So more molecules can react at normal energy levels they already have, significantly enhancing the speed of reaction.
Oxidation-Reduction Reactions
Oxidation-reduction reactions, or redox reactions, are processes where electrons are transferred between substances. In the context of this reaction, we have both oxidation and reduction happening in tandem.

To understand the dynamics:
  • Oxidation is the loss of electrons.
  • Reduction is the gain of electrons.
In the given reaction, \(\text{Fe}^{3+}\) plays a pivotal role by acting as an electron shuttle:
- It oxidizes iodide \(\text{I}^{-}\) to iodine \(\text{I}_2\), capturing electrons and becoming \(\text{Fe}^{2+}\).
- Once reduced to \(\text{Fe}^{2+}\), it transfers electrons to \(\text{S}_2\text{O}_8^{2-}\) catalyzing its breakdown to \(\text{SO}_4^{2-}\), and regenerating \(\text{Fe}^{3+}\).

Each cycle of the catalyst involves redox reactions at its core, making the process of transforming reactants to products highly efficient without the catalyst being consumed.

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Most popular questions from this chapter

The rate constants of some reactions double with every \(10^{\circ}\) rise in temperature. Assume that a reaction takes place at \(295 \mathrm{~K}\) and \(305 \mathrm{~K}\). What must the activation energy be for the rate constant to double as described?

Write the equation relating the half-life of a secondorder reaction to the rate constant. How does it differ from the equation for a first-order reaction?

Consider the reaction: $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) $$ Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of \(0.082 \mathrm{M} / \mathrm{s}\). (a) At what rate is ammonia being formed? (b) At what rate is molecular nitrogen reacting?

The reaction \(2 \mathrm{~A}+3 \mathrm{~B} \longrightarrow \mathrm{C}\) is first order with respect to \(\mathrm{A}\) and \(\mathrm{B}\). When the initial concentrations are \([\mathrm{A}]=1.6 \times 10^{-2} M\) and \([\mathrm{B}]=2.4 \times 10^{-3} M,\) the rate is \(4.1 \times 10^{-4} \mathrm{M} / \mathrm{s} .\) Calculate the rate constant of the reaction.

The rate law for the decomposition of ozone to molecular oxygen: $$ 2 \mathrm{O}_{3}(g) \longrightarrow 3 \mathrm{O}_{2}(g) $$ is $$ \text { rate }=k \frac{\left[\mathrm{O}_{3}\right]^{2}}{\left[\mathrm{O}_{2}\right]} $$ The mechanism proposed for this process is: $$ \begin{array}{l} \mathrm{O}_{3} \stackrel{k_{1}}{\rightleftarrows} \mathrm{O}+\mathrm{O}_{2} \\\ \mathrm{O}+\mathrm{O}_{3} \stackrel{k_{2}}{\longrightarrow} 2 \mathrm{O}_{2} \end{array} $$ Derive the rate law from these elementary steps. Clearly state the assumptions you use in the derivation. Explain why the rate decreases with increasing \(\mathrm{O}_{2}\) concentration.

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