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The ozone in the Earth's ozone layer decomposes according to the equation $$ 2 \mathrm{O}_{3}(\mathrm{g}) \rightarrow 3 \mathrm{O}_{2}(\mathrm{g}) $$ The mechanism of the reaction is thought to proceed through an initial fast equilibrium and a slow step: Step 1: Fast, reversible \(\quad \mathrm{O}_{3}(\mathrm{g}) \rightleftharpoons \mathrm{O}_{2}(\mathrm{g})+\mathrm{O}(\mathrm{g})\) Step 2: Slow \(\quad \mathrm{O}_{3}(\mathrm{g})+\mathrm{O}(\mathrm{g}) \rightarrow 2 \mathrm{O}_{2}(\mathrm{g})\) Show that the mechanism agrees with this experimental rate law: $$ \text { Rate }=-(1 / 2) \Delta\left[\mathrm{O}_{3}\right] / \Delta t=k\left[\mathrm{O}_{3}\right]^{2} /\left[\mathrm{O}_{2}\right] $$.

Short Answer

Expert verified
The mechanism supports the rate law when considering equilibrium and slow step.

Step by step solution

01

Understand the rate law to show

We need to demonstrate the given experimental rate law: \( \text{Rate} = -\frac{1}{2} \frac{\Delta[\mathrm{O}_3]}{\Delta t} = k \frac{[\mathrm{O}_3]^2}{[\mathrm{O}_2]} \). This implies finding expressions for individual steps that combine to yield this form.
02

Analyze the Reaction Mechanism

The mechanism includes two steps:1. Step 1: \( \mathrm{O}_3(\mathrm{g}) \rightleftharpoons \mathrm{O}_2(\mathrm{g}) + \mathrm{O}(\mathrm{g}) \) (fast and reversible)2. Step 2: \( \mathrm{O}_3(\mathrm{g}) + \mathrm{O}(\mathrm{g}) \rightarrow 2 \mathrm{O}_2(\mathrm{g}) \) (slow)The slow step is the rate-determining step.
03

Write the Rate Law for the Slow Step

The rate law depends on the rate-determining step:\[ \text{Rate}_{\text{slow}} = k_2 [\mathrm{O}_3][\mathrm{O}] \]
04

Use Equilibrium for the Fast Step

Since Step 1 is in equilibrium, apply the equilibrium constant expression:\[ K_1 = \frac{[\mathrm{O}_2][\mathrm{O}]}{[\mathrm{O}_3]} \]This allows us to express \([\mathrm{O}]\) as:\[ [\mathrm{O}] = \frac{K_1 [\mathrm{O}_3]}{[\mathrm{O}_2]} \]
05

Substitute Equilibrium Expression

Substitute \([\mathrm{O}]\) from Step 4 into the rate law from Step 3:\[ \text{Rate} = k_2 [\mathrm{O}_3] \left( \frac{K_1 [\mathrm{O}_3]}{[\mathrm{O}_2]} \right) = k_2 K_1 \frac{[\mathrm{O}_3]^2}{[\mathrm{O}_2]} \]
06

Match with the Experimental Rate Law

The derived rate law is:\[ \text{Rate} = k_2 K_1 \frac{[\mathrm{O}_3]^2}{[\mathrm{O}_2]} \]This matches the experimental rate law \( k \frac{[\mathrm{O}_3]^2}{[\mathrm{O}_2]} \) when we let \( k = k_2 K_1 \), confirming agreement.

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Key Concepts

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

Reaction mechanism
A reaction mechanism is a detailed step-by-step description of how a chemical reaction occurs. It provides insight into the individual steps, known as elementary reactions, that make up the overall process. In the context of ozone decomposition, the reaction mechanism involves both fast and slow steps. The first step is fast and involves the reversible breakdown of ozone (\( \text{O}_3 \)) into oxygen molecule (\( \text{O}_2 \)) and oxygen atom (\( \text{O} \)). The second step is slow and combines ozone with the oxygen atom to produce two oxygen molecules. Understanding the mechanism helps in explaining the experimental rate law observed for ozone decomposition. Mechanisms allow chemists to interpret how intermediates form and react, providing a complete picture of the chemical process occurring at a molecular level.
Studying mechanisms is essential for predicting reaction behavior and for designing methods to control reaction rates, which is useful in both industrial applications and environmental settings.
Rate law
The rate law defines how the rate of a chemical reaction depends on the concentration of its reactants. For a reaction, the rate law is usually expressed in terms of the rate constant (\( k \)) and the concentrations of reactants raised to some power. In the case of ozone decomposition, the experimentally determined rate law is given by: \[ \text{Rate} = k \frac{[\mathrm{O}_3]^2}{[\mathrm{O}_2]} \] This shows that the reaction rate is directly proportional to the square of the ozone concentration and inversely proportional to the concentration of oxygen molecules.
The rate law not only informs us of how the concentration of reactants influences the speed of the reaction, but it also suggests the presence of certain reaction steps or intermediates, as seen in the mechanism analysis. Understanding the rate law is crucial for predicting the speed of the reaction under various conditions and aids in kinetic studies of complex reactions.
Ozone decomposition
Ozone decomposition is an important environmental process that involves the breakdown of ozone \( \mathrm{O}_3 \) into oxygen molecules \( \mathrm{O}_2 \). This reaction is fundamental in balancing the amount of ozone in the Earth's atmosphere, serving as a natural equilibrium between ozone formation and decomposition. The decomposition process, as investigated in the exercise, occurs via a series of steps involving fast equilibrium and a slow reaction step.
  • Step 1: Ozone first rapidly decomposes into an oxygen molecule and a free oxygen atom.
  • Step 2: The free oxygen atom then reacts slowly with another ozone molecule to form two oxygen molecules.
Understanding the details of ozone decomposition is critical for atmospheric chemistry. This knowledge helps scientists to predict changes in ozone levels, which impact both environmental conditions and human health. Monitoring and understanding these reactions help in addressing and mitigating the effects of ozone layer depletion.
Rate-determining step
The rate-determining step is the slowest step in a reaction mechanism and thus dictates the overall rate of the reaction. In ozone decomposition, the second step, where \( \mathrm{O}_3 \) reacts with \( \mathrm{O} \) to form \( 2 \mathrm{O}_2 \), is the rate-determining step. This step effectively controls how fast the entire reaction proceeds.
The slowest step is crucial because the overall rate law of the reaction is generally determined by the rate law of the rate-determining step. In kinetic studies, identifying this step helps chemists to understand the bottlenecks in the process and to modify conditions or catalysts to improve reaction rates if necessary.
Smart understanding of the rate-determining step allows for the optimization of industrial processes and environmental reactions. By knowing which step limits the reaction pace, solutions can be designed to enhance efficiency and performance.

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Most popular questions from this chapter

Gaseous azomethane, \(\mathrm{CH}_{3} \mathrm{N}=\mathrm{NCH}_{3,}\) decomposes in a first-order reaction when heated: $$ \mathrm{CH}_{3} \mathrm{N}=\mathrm{NCH}_{3}(\mathrm{g}) \rightarrow \mathrm{N}_{2}(\mathrm{g})+\mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{g}) $$ The rate constant for this reaction at \(600 \mathrm{K}\) is \(0.0216 \mathrm{min}^{-1} .\) If the initial quantity of azomethane in the flask is \(2.00 \mathrm{g}\), how much remains after 0.0500 hour? What mass of \(\mathrm{N}_{2}\) is formed in this time?

Hydrogen iodide decomposes when heated, forming \(\mathrm{H}_{2}(\mathrm{g})\) and \(\mathrm{I}_{2}(\mathrm{g}) .\) The rate law for this reaction is \(-\Delta|\mathrm{HI}| / \Delta t=k|\mathrm{HI}|^{2} \cdot \mathrm{At} 443^{\circ} \mathrm{C}\) \(k=30 . \mathrm{L} / \mathrm{mol} \cdot\) min. If the initial HI(g) concentration is \(1.5 \times 10^{-2} \mathrm{mol} / \mathrm{L},\) what concentration of \(\mathrm{HI}(\mathrm{g})\) will remain after \(10 .\) minutes?

The reaction cyclopropane \(\rightarrow\) propene occurs on a platinum metal surface at \(200^{\circ} \mathrm{C}\). (The platinum is a catalyst.) The reaction is first-order in cyclopropane. Indicate how the following quantities change (increase, decrease, or no change) as this reaction progresses, assuming constant temperature. (a) [cyclopropane] (b) Ipropene] (c) [catalyst] (d) the rate constant, \(k\) (e) the order of the reaction (f) the half-life of cyclopropane

The decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) to \(\mathrm{SO}_{2}\) and \(\mathrm{Cl}_{2}\) is first-order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) $$ \begin{array}{c} \mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \\ \text { Rate }=k\left[\mathrm{SO}_{2} \mathrm{Cl}_{2}\right] \text { where } k=0.17 / \mathrm{hr} \end{array} $$ (a) What is the rate of decomposition when \(\left[\mathrm{SO}_{2} \mathrm{Cl}_{2}\right]=0.010 \mathrm{M} ?\) (b) What is the half-life of the reaction? (c) If the initial pressure of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) in a flask is \(0.050 \mathrm{atm},\) what is the pressure of all gases (i.e., the total pressure) in the flask after the reaction has proceeded for one half-life?

The conversion of cyclopropane to propene (Example \(14.5)\) occurs with a first-order rate constant of \(2.42 \times 10^{-2} \mathrm{h}^{-1} .\) How long will it take for the concentration of cyclopropane to decrease from an initial concentration of \(0.080 \mathrm{mol} / \mathrm{L}\) to \(0.020 \mathrm{mol} / \mathrm{L} ?\)

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