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

Short Answer

Expert verified
(a) Rate laws for each elementary reaction: \(rate1 = k_1[\ce{H2O2}][\ce{I-}]\) \(rate2 = k_2[\ce{IO-}][\ce{H2O2}]\) (b) Overall chemical equation: \(\ce{2H2O2} \longrightarrow \ce{2H2O} + \ce{O2}\) (c) Intermediate species identification: The intermediate species is \(\ce{IO-}\). (d) Predicting the rate law for the overall process: \[rate = k_1[\ce{H2O2}][\ce{I-}]\]\[rate = k[\ce{H2O2}][\ce{I-}]\]

Step by step solution

01

(a) Rate laws for each elementary reaction

The rate laws depend on the molecularity of the reaction, so for the slow step, the reaction is bimolecular, as it involves two molecules, \(\ce{H2O2}\) and \(\ce{I-}\), reacting together. For the fast step, it is also bimolecular, as it involves \(\ce{IO-}\) and \(\ce{H2O2}\). For the slow step: \[rate1 = k_1[\ce{H2O2}][\ce{I-}]\] For the fast step: \[rate2 = k_2[\ce{IO-}][\ce{H2O2}]\]
02

(b) Overall chemical equation

Combine both elementary reactions and cancel any common species on both sides: \[\ce{H2O2} + \ce{I-} \longrightarrow \ce{H2O} + \ce{IO-}\] \[\ce{IO-} + \ce{H2O2} \longrightarrow \ce{H2O} + \ce{O2} + \ce{I-}\] Resulting in the overall reaction: \[\ce{2H2O2} \longrightarrow \ce{2H2O} + \ce{O2}\]
03

(c) Intermediate species identification

The intermediate species appear in the mechanism but not in the overall reaction. In this case, the intermediate species is \(\ce{IO-}\).
04

(d) Predicting the rate law for the overall process

Since we assume that the first step is rate-determining, the overall rate depends on the rate of the slow step: \[rate = k_1[\ce{H2O2}][\ce{I-}]\] However, we must express the intermediate, \(\ce{IO-}\), in terms of the reactants. From the fast step: \[rate2 = k_2[\ce{IO-}][\ce{H2O2}]\] Since the fast step is much faster than the slow step, the intermediate concentration is maintained at a steady state, so: \[\frac{d[\ce{IO-}]}{dt} = 0\] Express \(\ce{IO-}\) concentration: \[[\ce{IO-}] = \frac{rate1}{k_2[\ce{H2O2}]}\] Therefore, the overall rate law is: \[rate = k_1[\ce{H2O2}][\ce{I-}]\] For this problem, the rate law remains the same because the assumptions made did not require us to change the rate law. So, the overall rate law for the decomposition of hydrogen peroxide catalyzed by iodide ions is: \[rate = k[\ce{H2O2}][\ce{I-}]\]

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Key Concepts

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

Rate Law
In chemical kinetics, the rate law describes how the rate of a chemical reaction depends on the concentration of its reactants. For the decomposition of hydrogen peroxide in the presence of iodide ions, understanding the rate law helps us predict how changes in concentration affect the reaction speed.

For a bimolecular reaction, where two reactant molecules collide, the rate law is of the form:
  • Rate = k[ Reactant1] [Reactant2]
Here, *k* is the rate constant, a crucial component that influences the reaction rate. Also, the rate of each elementary step in a reaction mechanism depends on the concentration of the reactants involved in that step.

In the given mechanism, the slow step involving hydrogen peroxide and iodide ions dictates the overall reaction rate. Thus, the rate law for the entire process depends on the concentrations of these two species:
  • Rate = k_1[ H鈧侽鈧俔 [I鈦籡
Chemical Equation
A chemical equation provides a concise way to represent a chemical reaction. It shows the reactants transforming into products, often using a balanced equation to reflect the conservation of mass. In this exercise, the elementary steps of the decomposition process split the reaction into more accessible parts.

Combining these steps into a single, overall chemical equation involves adding the sequences together and eliminating intermediates, as they do not appear in the final balanced equation. For the decomposition of hydrogen peroxide catalyzed by iodide ions, the overall balanced equation is:
  • 2 H鈧侽鈧 鈫 2 H鈧侽 + O鈧
This equation tells us that two molecules of hydrogen peroxide decompose to form two water molecules and one molecule of oxygen gas.
Reaction Mechanism
A reaction mechanism is a step-by-step sequence of elementary reactions by which an overall chemical change occurs. It offers insight into the specific molecular processes involved, showing how reactants convert to products.

In the reaction mechanism for hydrogen peroxide decomposition catalyzed by iodide ions, we see two elementary steps: a slow one and a fast one. The slow step is crucial because it acts as the rate-determining step, which means it governs the overall reaction rate. Here, iodide ion starts the reaction by transforming hydrogen peroxide into water and the intermediate hypoiodite ion (IO鈦).

The fast step involves the hypoiodite ion reacting with another hydrogen peroxide molecule to finally form water, oxygen, and regenerated iodide ion. This two-step mechanism gives us a complete view of what occurs on a molecular level during this reaction.
Catalysis
Catalysis is an important aspect of this reaction. A catalyst is a substance that increases the rate of a chemical reaction without being consumed in the process. In the decomposition of hydrogen peroxide, iodide ions act as a catalyst. They enable the reaction to proceed at a faster rate by providing an alternative reaction pathway with a lower activation energy.

During the reaction mechanism, iodide ions facilitate the formation of the intermediate hypoiodite ion, which then quickly transforms hydrogen peroxide into products. Although the catalyst is involved in an intermediate form, it is regenerated by the end of the process.

Therefore, the presence of a catalyst like iodide ions is crucial for enhancing the speed of the reaction, confirming that catalysis plays an essential role in many chemical processes.

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Most popular questions from this chapter

One of the many remarkable enzymes in the human body is carbonic anhydrase, which catalyzes the interconversion of carbonic acid with carbon dioxide and water. If it were not for this enzyme, the body could not rid itself rapidly enough of the \(\mathrm{CO}_{2}\) accumulated by cell metabolism. The enzyme catalyzes the dehydration (release to air) of up to \(10^{7} \mathrm{CO}_{2}\) molecules per second. Which components of this description correspond to the terms enzyme, substrate, and turnover number?

The gas-phase decomposition of \(\mathrm{NO}_{2}, 2 \mathrm{NO}_{2}(g)\) \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)\), is studied at \(383^{\circ} \mathrm{C}\), giving the following data: $$ \begin{array}{ll} \hline \text { Time (s) } & {\left[\mathrm{NO}_{2}\right](M)} \\ \hline 0.0 & 0.100 \\ 5.0 & 0.017 \\ 10.0 & 0.0090 \\ 15.0 & 0.0062 \\ 20.0 & 0.0047 \\ \hline \end{array} $$ (a) Is the reaction first order or second order with respect to the concentration of \(\mathrm{NO}_{2} ?\) (b) What is the value of the rate constant?

(a) Consider the combustion of \(\mathrm{H}_{2}(g): 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)\) \(\longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) .\) If hydrogen is buming at the rate of \(0.85 \mathrm{~mol} / \mathrm{s}\), what is the rate of consumption of oxygen? What is the rate of formation of water vapor? (b) The reaction \(2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NOCl}(g)\) is carried out in a closed vessel. If the partial pressure of \(\mathrm{NO}\) is decreasing at the rate of 23 torr/min, what is the rate of change of the total pressure of the vessel?

Suppose that a certain biologically important reaction is quite slow at physiological temperature \(\left(37^{\circ} \mathrm{C}\right)\) in the absence of a catalyst. Assuming that the collision factor remains the same, by how much must an enzyme lower the activation energy of the reaction in order to achieve a \(1 \times 10^{5}\) -fold increase in the reaction rate?

The reaction \(2 \mathrm{NO}_{2} \longrightarrow 2 \mathrm{NO}+\mathrm{O}_{2}\) has the rate constant \(k=0.63 \mathrm{M}^{-1} \mathrm{~s}^{-1}\). Based on the units for \(k\), is the reaction first or second order in \(\mathrm{NO}_{2} ?\) If the initial concentration of \(\mathrm{NO}_{2}\) is \(0.100 \mathrm{M}\), how would you determine how long it would take for the concentration to decrease to \(0.025 \mathrm{M} ?\)

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