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The mechanism for catalytic destruction of ozone by chlorine radicals is: (1) \(2\left[\mathrm{Cl}+\mathrm{O}_{3} \longrightarrow \mathrm{O}_{2}+\mathrm{ClO}\right]\) Faster (2) \(2 \mathrm{ClO} \longrightarrow \mathrm{Cl}_{2} \mathrm{O}_{2} \quad\) Slower, Rate-determining step (3) \(\mathrm{Cl}_{2} \mathrm{O}_{2}+h v \longrightarrow 2 \mathrm{Cl}+\mathrm{O}_{2}\) Faster \(2 \mathrm{O}_{3}+h v \longrightarrow 3 \mathrm{O}_{2}\) Overall (a) Write the rate law for the rate-determining step. (b) The rate constant for this reaction at \(190 \mathrm{~K}\) is \(7.2 \times 10^{-13} \mathrm{~cm}^{3} /\) molecule \(\cdot \mathrm{s} .\) Calculate the rate of reaction for Step 2 when the concentration of \(\mathrm{ClO}\) is \(2.4 \times 10^{9}\) molecules \(/ \mathrm{cm}^{3}\). (c) Calculate the rate of ozone loss, which is determined by the rate- determining step.

Short Answer

Expert verified
(a) Rate law: \(\text{Rate} = k[\text{ClO}]^2\). (b) Reaction rate: \(4.1472 \times 10^6\) molecules/cm³/s. (c) Ozone loss rate: \(4.1472 \times 10^6\) molecules/cm³/s.

Step by step solution

01

Identify the Rate-Determining Step

The rate-determining step, also known as the slowest step in a reaction mechanism, is the second step in the given mechanism: \(2 \text{ClO} \rightarrow \text{Cl}_2\text{O}_2\). This step is slower compared to others and dictates the overall reaction rate.
02

Write the Rate Law for the Rate-Determining Step

The rate law can be expressed based on the rate-determining step. Since step 2 is \(2 \text{ClO} \rightarrow \text{Cl}_2\text{O}_2\), the rate law is given by \(\text{Rate} = k[\text{ClO}]^2\), where \(k\) is the rate constant for this step and \([\text{ClO}]\) is the concentration of chlorine monoxide.
03

Substitute Values to Calculate the Rate of Reaction

Using the rate constant \(k = 7.2 \times 10^{-13} \text{ cm}^3/\text{molecule} \cdot \text{s}\) and the concentration \([\text{ClO}] = 2.4 \times 10^{9}\) molecules/cm³, substitute into the rate law: \[\text{Rate} = (7.2 \times 10^{-13}) (2.4 \times 10^{9})^2\]. Calculate to find the specific rate of reaction.
04

Calculate the Rate Expression

Continuing from the substitution:\[\text{Rate} = (7.2 \times 10^{-13}) \times (5.76 \times 10^{18})\]. Perform the multiplication to get \(\text{Rate} = 4.1472 \times 10^{6}\) molecules/cm³/s.
05

Relate Rate of Reaction to Ozone Loss

For the loss of ozone, recognize that each occurrence of the rate-determining step involves the depletion of 2 moles of ozone molecules, as reflected in the stoichiometry of the steps. Therefore, the rate of ozone loss equals the rate of the rate-determining step: \(\text{Rate of ozone loss} = 4.1472 \times 10^{6}\) \text{ molecules/cm}^3/\text{s}.

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Key Concepts

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

Catalytic Destruction
Catalytic destruction of ozone involves reactions where radicals, like chlorine, act as catalysts to remove ozone molecules. A catalyst speeds up a chemical reaction without undergoing permanent change. In the atmosphere, chlorine radicals come from compounds like chlorofluorocarbons (CFCs). These CFCs release chlorine when they are broken down by ultraviolet light. Once free, chlorine radicals can hurry up the transformation of ozone into oxygen molecules, causing depletion. This process is concerning because the ozone layer protects life on Earth by blocking dangerous UV radiation.
Reaction Mechanism
A reaction mechanism in chemistry describes the step-by-step process by which reactants transform into products. It includes all intermediate steps and species formed along the way. For catalytic destruction of ozone by chlorine radicals, the mechanism has several steps:
  • First, chlorine reacts with ozone to form chlorine monoxide and oxygen.
  • Next, chlorine monoxide combines to produce dimer (Clâ‚‚Oâ‚‚).
  • Then, the dimer breaks down under UV light back to chlorine radicals and oxygen.
The overall reaction must align with observed data and follows the conservation of mass and atoms throughout.
Rate Law
The rate law is an equation that links the reaction rate with the concentration of reactants. It expresses how fast a reaction proceeds. For the rate-determining step in ozone catalytic destruction, which is the slowest step, the rate law is given by the equation:\[\text{Rate} = k[\text{ClO}]^2\]Here, \(k\) is the rate constant for that step, and \([\text{ClO}]\) is the concentration of chlorine monoxide. The rate law helps chemists understand how changes in reactant concentrations affect the overall speed of a reaction. You need to know the rate constant and the concentrations of reactants to calculate exact rates.
Rate-Determining Step
The rate-determining step in a reaction mechanism is the slowest step that controls the overall reaction speed. Think of it as the bottleneck of a series of reactions. In the catalytic destruction of ozone, the second step, where two chlorine monoxide molecules combine, is the rate-determining step. This step is slower than the others, and its rate directly affects the overall depletion of ozone. Understanding which step is the rate-determining step allows scientists to focus on it to change or improve reaction conditions for desired outcomes. Since this step controls the pace, scientists may target it to slow down unwanted reactions, like ozone depletion, by examining and adjusting the factors that make this step slow.

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Most popular questions from this chapter

The light-stimulated conversion of 11 -cis-retinal to 11 -transretinal is central to the vision process in humans. This reaction also occurs (more slowly) in the absence of light. At \(80.0^{\circ} \mathrm{C}\) in heptane solution, the reaction is first order with a rate constant of \(1.02 \times 10^{-5} / \mathrm{s}\) (a) What is the molarity of 11 -cis-retinal after \(6.00 \mathrm{~h}\) if its initial concentration is \(3.50 \times 10^{-3} \mathrm{M} ?\) (b) How many minutes does it take for \(25 \%\) of the 11 -cis-retinal to react? (c) How many hours does it take for the concentration of 11 -trans-retinal to reach \(3.15 \times 10^{-3} \mathrm{M} ?\)

The half-life for the first-order decomposition of \(\mathrm{N}_{2} \mathrm{O}_{4}\) is \(1.3 \times 10^{-5} \mathrm{~s}\) $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) $$ If \(\mathrm{N}_{2} \mathrm{O}_{4}\) is introduced into an evacuated flask at a pressure of \(17.0 \mathrm{~mm} \mathrm{Hg}\), how many seconds are required for the pressure of \(\mathrm{NO}_{2}\) to reach \(1.3 \mathrm{~mm} \mathrm{Hg} ?\)

A \(1.50 \mathrm{~L}\) sample of gaseous HI having a density of \(0.0101 \mathrm{~g} / \mathrm{cm}^{3}\) is heated at \(410^{\circ} \mathrm{C}\). As time passes, the HI decomposes to gaseous \(\mathrm{H}_{2}\) and \(\mathrm{I}_{2}\). The rate law is \(-\Delta[\mathrm{HI}] / \Delta t=k[\mathrm{HI}]^{2}\), where \(k=0.031 /(\mathrm{M} \cdot \mathrm{min})\) at \(410^{\circ} \mathrm{C}\) (a) What is the initial rate of production of \(\mathrm{I}_{2}\) in molecules \(/ \mathrm{min}\) ? (b) What is the partial pressure of \(\mathrm{H}_{2}\) after a reaction time of \(8.00 \mathrm{~h}\) ?

What fraction of the molecules in a gas at \(300 \mathrm{~K}\) collide with an energy equal to or greater than \(E_{a}\) when \(E_{a}\) equals \(50 \mathrm{~kJ} / \mathrm{mol} ?\) What is the value of this fraction when \(E_{a}\) is \(100 \mathrm{~kJ} / \mathrm{mol} ?\)

Initial rate data at \(25^{\circ} \mathrm{C}\) are listed in the table for the reaction \(\mathrm{NH}_{4}{ }^{+}(a q)+\mathrm{NO}_{2}^{-}(a q) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) \begin{tabular}{cccc} \hline & Initial & Initial & Initial Reaction \\ Experiment & {\(\left[\mathrm{NH}_{4}{ }^{+}\right]\)} & {\(\left[\mathrm{NO}_{2}^{-}\right]\)} & Rate \((\mathrm{M} / \mathrm{s})\) \\ \hline 1 & \(0.24\) & \(0.10\) & \(7.2 \times 10^{-6}\) \\ 2 & \(0.12\) & \(0.10\) & \(3.6 \times 10^{-6}\) \\ 3 & \(0.12\) & \(0.15\) & \(5.4 \times 10^{-6}\) \\ \hline \end{tabular} (a) What is the rate law? (b) What is the value of the rate constant? (c) What is the initial rate when the initial concentrations are \(\left[\mathrm{NH}_{4}{ }^{+}\right]=0.39 \mathrm{M}\) and \(\left[\mathrm{NO}_{2}^{-}\right]=0.052 \mathrm{M} ?\)

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