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Chapter 14: Problem 72

The decomposition of hydrogen peroxide is catalyzed by iodide ion. The catalyzed reaction is thought to proceed by a two-step mechanism: $$ \begin{array}{c}{\mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{IO}^{-}(a q) \text { (slow) }} \\ {\mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(\mathrm{g})+\mathrm{I}^{-}(a q) \text { (fast) }}\end{array} $$ \(\begin{array}{l}{\text { (a) Write the chemical equation for the overall process. }} \\ {\text { (b) Identify the intermediate, if any, in the mechanism. }} \\ {\text { (c) Assuming that the first step of the mechanism is rate }} \\ {\text { determining, predict the rate law for the overall process. }}\end{array}\)

Short Answer

Expert verified
The overall chemical equation for the decomposition of hydrogen peroxide catalyzed by the iodide ion is \(\ce{2H2O2(aq) -> 2H2O(l) + O2(g)}\). The intermediate in this reaction mechanism is the iodide oxide ion (\(\ce{IO^{-}}\)). Considering the first step as rate-determining, the rate law for the overall process is \(Rate = k \times [\ce{H2O2}] \times [\ce{I^{-}]\).

Step by step solution

01

Writing the overall chemical equation

To find the overall chemical equation for the decomposition of hydrogen peroxide, we need to add both steps given in the problem statement: Step 1: \(\ce{H2O2(aq)} + \ce{I^{-}(aq)} -> \ce{H2O(l)} + \ce{IO^{-}(aq)}\) (slow) Step 2: \(\ce{IO^{-}(aq)} + \ce{H2O2(aq)} -> \ce{H2O(l)} + \ce{O2(g)} + \ce{I^{-}(aq)}\) (fast) Now, we can add these two equations to get the overall chemical equation: \[\ce{2H2O2(aq) -> 2H2O(l) + O2(g)}\]
02

Identifying the intermediate

The intermediate in a reaction mechanism is a species that is produced in one step but consumed in a subsequent step. In this case, the iodide oxide ion (\(\ce{IO^{-}}\)) is produced in Step 1 as a product and then consumed in Step 2 as a reactant. Therefore, \(\ce{IO^{-}}\) is the intermediate in this mechanism.
03

Predicting the rate law for the overall process

If the first step is rate-determining, the rate law for the overall process will depend on the rate law of the first step. The rate law for a reaction can be written as: \[Rate = k \times [A]^m \times [B]^n\] In this case, the first step involves the reaction between \(\ce{H2O2}\) and \(\ce{I^{-}}\). We can then write the rate law for this step: \[Rate = k \times [\ce{H2O2}]^m \times [\ce{I^{-}]^n\] Since the first step is bimolecular (two reacting particles), we have \(m = 1\) and \(n = 1\). Therefore, the rate law for the overall process becomes: \[Rate = k \times [\ce{H2O2}] \times [\ce{I^{-}]\]

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Key Concepts

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

Reaction Mechanisms
Understanding chemical kinetics involves studying how reactions occur and the series of steps that constitute the transformation from reactants to products, known as the reaction mechanism. This mechanism details the sequence of elementary steps, which can be simple individual reactions or complex multiple processes, leading to the overall chemical reaction. For instance, the decomposition of hydrogen peroxide catalyzed by iodide ion is thought to follow a two-step mechanism.

In the first step, hydrogen peroxide reacts with an iodide ion to produce water and iodine oxide ion, while the second step sees the iodine oxide ion reacting with more hydrogen peroxide to yield water, oxygen, and regenerate the iodide ion. By understanding the individual steps, we can better predict the kinetics of the reaction, including the rate at which it will proceed under various conditions.
Rate-Determining Step
The rate-determining step (RDS) is the slowest step in a reaction mechanism and significantly affects the speed of the overall reaction. This concept is critical in chemical kinetics as it helps to identify which step controls the reaction rate. Think of it like a bottleneck in a busy highway - no matter how fast the other areas flow, the overall speed is limited by the slowest point.

In our example of hydrogen peroxide decomposition, the first step is slower than the second and is, therefore, rate-determining. This means the reaction's speed is mainly influenced by how quickly the iodide ion and hydrogen peroxide react to form iodine oxide ion and water. By recognizing the RDS, chemists can formulate a rate law - a mathematical expression relating the reaction rate to the concentration of the reactants involved in the RDS. If the RDS involves one molecule of hydrogen peroxide and one iodide ion, the rate law can be expressed as \( Rate = k \times [H_2O_2] \times [I^{-}] \), where \( k \) is the rate constant and the reactant concentrations are raised to the power of their stoichiometric coefficients in the RDS.
Reaction Intermediates
Reaction intermediates are species that appear in the course of a reaction mechanism but are neither reactants nor final products. They are essential puzzle pieces in the transformation process, created in one elementary step and consumed in another. These intermediates can often provide insight into the mechanism by which the overall reaction occurs.

Referring back to our hydrogen peroxide decomposition, the iodine oxide ion (\( IO^{-} \)) after being formed in the first step - the rate-determining step - is consumed in the subsequent fast step. It's crucial to identify intermediates like \( IO^{-} \) as they can sometimes be the target of strategies to enhance or inhibit a reaction. Understanding these intermediates allows chemists to influence a reaction's pathway, speed, and selectivity by adding substances that stabilize or destabilize them, or by changing the reaction conditions.

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Most popular questions from this chapter

The addition of NO accelerates the decomposition of \(\mathrm{N}_{2} \mathrm{O},\) possibly by the following mechanism: $$\begin{array}{c}{\mathrm{NO}(g)+\mathrm{N}_{2} \mathrm{O}(g) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{NO}_{2}(g)} \\ {2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)}\end{array}$$ (a) What is the chemical equation for the overall reaction? Show how the two steps can be added to give the overall equation. (b) Is NO serving as a catalyst or an intermediate in this reaction? (c) If experiments show that during the decomposition of \(N_{2} \mathrm{O}, \mathrm{NO}_{2}\) does not accumulate in measurable quantities, does this rule out the proposed mechanism?

The reaction between ethyl bromide \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}\right)\) and hydroxide ion in ethyl alcohol at 330 \(\mathrm{K}\) , \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}(a l c)+\mathrm{OH}^{-}(a l c) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{Br}^{-}(a l c)\) is first order each in ethyl bromide and hydroxide ion. When \(\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}\right]\) is 0.0477 \(\mathrm{M}\) and \(\left[\mathrm{OH}^{-}\right]\) is \(0.100 \mathrm{M},\) the rate of disappearance of ethyl bromide is \(1.7 \times 10^{-7} \mathrm{M} / \mathrm{s}\) (a) What is the value of the rate constant? (b) What are the units of the rate constant? (c) How would the rate of disappearance of ethyl bromide change if the solution were diluted by adding an equal volume of pure ethyl alcohol to the solution?

Indicate whether each statement is true or false. \(\begin{array}{l}{\text { (a) If you measure the rate constant for a reaction at different}} \\ {\text { temperatures, you can calculate the overall }} \\ {\text { enthalpy change for the reaction. }} \\ {\text { (b) Exothermic reactions are faster than endothermic }} \\ {\text { reactions. }} \\ {\text { (c) If you double the temperature for a reaction, you cut }} \\ {\text { the activation energy in half. }}\end{array}\)

The enzyme urease catalyzes the reaction of urea, \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right),\) with water to produce carbon dioxide and ammonia. In water, without the enzyme, the reaction proceeds with a first-order rate constant of \(4.15 \times 10^{-5} \mathrm{s}^{-1}\) at \(100^{\circ} \mathrm{C} .\) In the presence of the enzyme in water, the reaction proceeds with a rate constant of \(3.4 \times 10^{4} \mathrm{s}^{-1}\) at \(21^{\circ} \mathrm{C}\) . (a) Write out the balanced equation for the reaction catalyzed by urease. (b) If the rate of the catalyzed reaction were the same at \(100^{\circ} \mathrm{C}\) as it is at \(21^{\circ} \mathrm{C},\) what would be the difference in the activation energy between the catalyzed and uncatalyzed reactions? (c) In actuality, what would you expect for the rate of the catalyzed reaction at \(100^{\circ} \mathrm{Cas} \mathrm{com}-\) pared to that at \(21^{\circ} \mathrm{C} ?(\mathbf{d})\) On the basis of parts \((\mathrm{c})\) and \((\mathrm{d}),\) what can you conclude about the difference in activation energies for the catalyzed and uncatalyzed reactions?

(a) In which of the following reactions would you expect the orientation factor to be least important in leading to reaction: \(\mathrm{NO}+\mathrm{O} \longrightarrow \mathrm{NO}_{2}\) or \(\mathrm{H}+\mathrm{Cl} \longrightarrow \mathrm{HCl}\) ? (b) Does the orientation factor depend on temperature?
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