Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A mechanism for a naturally occurring reaction that destroys ozone is: Step I: \(\mathrm{O}_{3}(\mathrm{~g})+\mathrm{HO}(\mathrm{g}) \rightarrow \mathrm{HO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\) Step II: \(\mathrm{HO}_{2}(\mathrm{~g})+\mathrm{O}(\mathrm{g}) \rightarrow \mathrm{HO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g})\) Which species is a catalyst? a. \(\mathrm{O}\) b. \(\mathrm{O}_{3}\) c. \(\mathrm{HO}_{2}\) d. \(\mathrm{HO}\)

Short Answer

Expert verified
The catalyst is \\(\\mathrm{HO}\\u00a0(g)\\u00a0\).

Step by step solution

01

Understand the Role of a Catalyst

A catalyst is a substance that increases the rate of a reaction without being consumed in the process. It appears in the reactants and is regenerated in the products in a multi-step reaction.
02

Identify the Species' Behavior in Step I

In Step I, \(\mathrm{HO}\u00a0(g)\u00a0\) is a reactant that combines with \(\mathrm{O}_{3}\u00a0(g)\u00a0\) to form \(\mathrm{HO}_{2}\u00a0(g)\u00a0\) and \(\mathrm{O}_{2}\u00a0(g)\u00a0\). Observe that \(\mathrm{HO}\u00a0(g)\u00a0\) is consumed in this step.
03

Observe the Species' Behavior in Step II

In Step II, \(\mathrm{HO}_{2}\u00a0(g)\u00a0\) reacts with \(\mathrm{O}\u00a0(g)\u00a0\) to regenerate \(\mathrm{HO}\u00a0(g)\u00a0\) and form more \(\mathrm{O}_{2}\u00a0(g)\u00a0\). This regeneration indicates that \(\mathrm{HO}\u00a0(g)\u00a0\) is reproduced from the intermediate species.
04

Determine the Catalyst

Since \(\mathrm{HO}\u00a0(g)\u00a0\) is consumed in Step I and regenerated in Step II, it acts as a catalyst by facilitating the reaction without being consumed overall. Therefore, \(\mathrm{HO}\u00a0(g)\u00a0\) is the catalyst.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Catalyst Role
Catalysts play a fascinating role in chemical reactions. They are substances that speed up chemical processes without being consumed in the reaction. This means they participate in one or more steps of a reaction mechanism but emerge unchanged at the end. Because of this unique ability, catalysts are incredibly valuable in various chemical processes.
In the provided exercise, hydroxyl radical (\(\mathrm{HO}\)) acts as a catalyst. In Step I of the mechanism, \(\mathrm{HO}\) reacts with ozone (\(\mathrm{O}_{3}\)) but then is regenerated in Step II when \(\mathrm{HO}_{2}\) reacts with another oxygen atom (\(\mathrm{O}\)).
Catalysts can drastically increase the rate of a reaction, allowing processes to occur that would otherwise take much longer or require much higher temperatures.
  • Catalysts lower the activation energy needed for a reaction
  • They provide an alternative reaction pathway
  • They are not consumed in the chemical reaction
Reaction Mechanisms
Understanding reaction mechanisms is like understanding the steps in a dance. Each step must occur for the overall reaction to take place. A reaction mechanism describes the individual stages that lead to a chemical reaction's conversion from reactants to products.
The mechanism given in the exercise consists of two main steps. In Step I, \(\mathrm{HO}\) reacts with \(\mathrm{O}_{3}\) leading to the formation of \(\mathrm{HO}_{2}\) and \(\mathrm{O}_{2}\). This is an intermediate step, which means \(\mathrm{HO}_{2}\) is not the final product but plays a part in the subsequent step. Step II involves \(\mathrm{HO}_{2}\) and \(\mathrm{O}\) interacting to regenerate \(\mathrm{HO}\) and produce more \(\mathrm{O}_{2}\).
Every chemical reaction has a specific series of steps that describe its mechanism. The complete understanding of these steps helps scientists to devise ways to control or accelerate chemical reactions for industrial purposes.
Chemical Kinetics
Chemical kinetics deals with the rates of chemical reactions and gives insights into how different conditions affect these rates. By studying kinetics, we understand how fast a reaction proceeds and what factors can accelerate or decelerate it.
In the context of the ozone depletion mechanism, chemical kinetics helps explain how quickly the reactions occur. Each step in the reaction mechanism has specific kinetics associated with it.
Some important factors influencing these rates include:
  • Concentration of reactants: Higher concentrations generally increase reaction rates
  • Temperature: Increasing temperature usually speeds up reactions due to higher kinetic energy
  • Presence of a catalyst: Catalysts can dramatically reduce the time needed for reactions to occur by providing an alternative mechanism with a lower activation energy
Understanding these elements helps us predict and, in some cases, control the rate at which ozone depletion occurs, contributing to better environmental protection strategies.
Ozone Chemistry
Ozone chemistry is crucial because ozone plays a significant role in protecting life on Earth by absorbing harmful ultraviolet radiation from the sun. Ozone (\(\mathrm{O}_{3}\)) is continuously being formed and destroyed in the stratosphere through a series of natural reactions.
The reaction mechanism presented involves ozone as one of the reactants. The depletion of ozone occurs through reactions that convert \(\mathrm{O}_{3}\) into molecular oxygen (\(\mathrm{O}_{2}\)). These reactions can be influenced by various factors such as the presence of radicals like \(\mathrm{HO}\) and \(\mathrm{HO}_{2}\).
Important aspects of ozone chemistry include:
  • The balance between ozone creation and destruction affects the ozone layer thickness
  • Chemical reactions involving human-made pollutants can accelerate ozone depletion
  • Understanding these processes helps scientists devise strategies to mitigate ozone layer damage
Through the study of ozone chemistry, we can better understand how to preserve our atmosphere and ensure the ozone layer continues to protect the planet from UV radiation exposure.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The data given below is for the reaction of \(\mathrm{NO}\) and \(\mathrm{Cl}_{2}\) to form \(\mathrm{NOCl}\) at \(295 \mathrm{~K}\) What is the rate law? $$ \begin{array}{lll} \hline\left[\mathrm{Cl}_{2}\right] & {[\mathrm{NO}]} & \begin{array}{l} \text { Initial rate } \\ \left(\mathrm{mol}^{-1} \mathrm{~s}^{-1}\right) \end{array} \\ \hline 0.05 & 0.05 & 1 \times 10^{-3} \\ 0.15 & 0.05 & 3 \times 10^{-3} \\ 0.05 & 0.15 & 9 \times 10^{-3} \\ \hline \end{array} $$ a. \(\mathrm{r}=\mathrm{k}[\mathrm{NO}]\left[\mathrm{Cl}_{2}\right]\) b. \(\mathrm{r}=\mathrm{k}\left[\mathrm{Cl}_{2}\right]^{1}[\mathrm{NO}]^{2}\) c. \(\mathrm{r}=\mathrm{k}\left[\mathrm{Cl}_{2}\right]^{2}[\mathrm{NO}]\) d. \(\mathrm{r}=\mathrm{k}\left[\mathrm{Cl}_{2}\right]^{1}\)

Which of the following is incorrect about order of reaction? a. it is calculated experimentally b. it is sum of powers of concentration in rate law expression c. the order of reaction cannot be fractional d. there is not necessarily a connection between order and stoichiometry of a reaction.

Match the following: (Here \(\mathrm{a}=\) Initial concentration of the reactant, \(\mathrm{p}=\) Initial pressure of the reactant) List I List II A. \(t \frac{1}{2}=\) constant (p) Zero order B. \(\mathrm{t} \frac{1}{2} \alpha \mathrm{a}\) (q) First order C. \(\mathrm{t} 1 / 2 \alpha \mathrm{l} / \mathrm{a}\) (r) Second order D. \(t^{1 / 2} \alpha p^{-1}\) (s) Pseudo first order

The equation of tris(1,10-phenanthroline) iron(II) in acid solution takes place according to the equation: \(\mathrm{Fe}(\text { phen })_{3}^{2+}+3 \mathrm{H}_{3} \mathrm{O}^{+}+3 \mathrm{H}_{2} \mathrm{O} \rightarrow\) \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}+3\) (phen) \(\mathrm{H}^{+}\) If the activation energy (Ea) is \(126 \mathrm{~kJ} / \mathrm{mol}\) and the rate constant at \(30^{\circ} \mathrm{C}\) is \(9.8 \times 10^{-3} \mathrm{~min}^{-1}\), what is the rate constant at \(50^{\circ} \mathrm{C}\) ? a. \(2.2 \times 10^{-1} \mathrm{~min}^{-1}\) b. \(3.4 \times 10^{-2} \mathrm{~min}^{-1}\) c. \(0.23 \times 10^{-1} \mathrm{~min}^{-1}\) d. \(1.2 \times 10^{-1} \min ^{-1}\)

In Arrhenius equation, \(\mathrm{k}=\mathrm{A} \exp (-\mathrm{Ea} / \mathrm{RT})\). A may be regarded as the rate constant at a. Very high temperature b. Very low temperature c. High activation energy d. Zero activation energy

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free