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

In the reaction \(2 \mathrm{O}_{3}(\mathrm{g}) \rightarrow 3 \mathrm{O}_{2}(\mathrm{g}),\) the rate of formation of \(\mathrm{O}_{2}\) is \(1.5 \times 10^{-3} \mathrm{mol} / \mathrm{L} \cdot\) s. What is the rate of decomposition of \(\mathrm{O}_{3} ?\)

Short Answer

Expert verified
The rate of decomposition of \(\mathrm{O}_3\) is \(1.0 \times 10^{-3} \, \mathrm{mol/L} \cdot \mathrm{s}\).

Step by step solution

01

Understand the Reaction

We are given a chemical reaction \(2 \mathrm{O}_3(\mathrm{g}) \rightarrow 3 \mathrm{O}_2(\mathrm{g})\). This reaction tells us that 2 moles of ozone (\(\mathrm{O}_3\)) decompose to form 3 moles of oxygen gas (\(\mathrm{O}_2\)).
02

Identify Given Rate of Formation

The rate of formation of \(\mathrm{O}_2\) is provided as \(1.5 \times 10^{-3} \, \mathrm{mol/L} \cdot \mathrm{s}\). This means that \(\mathrm{O}_2\) is being produced at this rate in moles per liter per second.
03

Relate Rates with Stoichiometry

The stoichiometry of the reaction tells us that 3 moles of \(\mathrm{O}_2\) are produced for every 2 moles of \(\mathrm{O}_3\) that decompose. Therefore, the rate of decomposition of \(\mathrm{O}_3\) is related to the rate of formation of \(\mathrm{O}_2\) by the ratio of their coefficients.
04

Calculate the Rate of Decomposition

Using stoichiometry, the rate of decomposition of \(\mathrm{O}_3\) is given by \(\frac{2}{3}\) times the rate of formation of \(\mathrm{O}_2\). Thus, the rate of decomposition of \(\mathrm{O}_3\) is \(\frac{2}{3} \times 1.5 \times 10^{-3} = 1.0 \times 10^{-3} \, \mathrm{mol/L} \cdot \mathrm{s}\).

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.

Stoichiometry
Stoichiometry is a fundamental concept in chemistry that helps us understand the proportions in which substances react with each other. It acts like a recipe, telling us how much of each ingredient we need to make a product. In our reaction, where 2 moles of ozone (\(\mathrm{O}_3\)) decompose to form 3 moles of oxygen gas (\(\mathrm{O}_2\)), stoichiometry helps us determine the relationship between the reactants and products.

By examining the balanced chemical equation, we can see this ratio and use it to relate the amounts and rates at which reactants and products are consumed and produced. This is crucial for calculating rates of reactions, as it tells us how changes in the amount of one substance will affect others involved in the reaction.
Rate of Formation
The rate of formation is a measure of how quickly a product is formed in a chemical reaction. In the given problem, the rate of formation of oxygen (\(\mathrm{O}_2\)) is specified as \(1.5 \times 10^{-3} \text{ mol/L} \cdot \text{s}\).

This tells us that oxygen is being produced at a rate of \(1.5 \times 10^{-3}\) moles per liter every second. Understanding the rate of formation is important for practical applications, as it can influence how a reaction is controlled or used in industrial processes.
  • The units "mol/L \cdot s" help us understand the concentration change over time.
  • Knowing the rate can help predict how much product can be formed in a given time.
Rate of Decomposition
The rate of decomposition refers to how quickly a reactant breaks down in a chemical reaction. In our exercise, we want to find out the rate at which ozone (\(\mathrm{O}_3\)) decomposes. Since we know the rate of formation of \(\mathrm{O}_2\), we can use stoichiometry to relate these rates.

Given that \(3\) moles of \(\mathrm{O}_2\) are produced for every \(2\) moles of \(\mathrm{O}_3\) that decompose, the rate of decomposition of \(\mathrm{O}_3\) can be calculated using the ratio of these coefficients. The calculation involves multiplying the rate of formation of \(\mathrm{O}_2\) by \(\frac{2}{3}\).

In our specific case, this leads to a rate of decomposition of \(1.0 \times 10^{-3}\) mol/L \cdot s, which shows us how quickly \(\mathrm{O}_3\) is breaking down in this reaction.
Ozone Decomposition
Ozone decomposition is an important chemical reaction with significant environmental implications. Ozone (\(\mathrm{O}_3\)) decomposition refers to the process where ozone molecules break down into oxygen gas (\(\mathrm{O}_2\)).

This reaction is critical in maintaining atmospheric balance and protecting life on Earth. Ozone exists in the Earth's atmosphere and acts as a shield that absorbs and scatters ultraviolet solar radiation. However, it is also a reactive molecule that can readily decompose, contributing to complex interactions within the atmosphere.
  • Decomposition of ozone affects the concentration of ozone and oxygen in the atmosphere.
  • Various factors, including sunlight and pollutants, can influence the rate of ozone decomposition.
Understanding the dynamics of ozone decomposition helps atmospheric scientists and environmentalists monitor air quality and assess environmental policies.

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

Give the relative rates of disappearance of reactants and formation of products for each of the following reactions. (a) \(2 \mathrm{NO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NOBr}(\mathrm{g})\) (b) \(\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{g})\)

Calculate the activation energy, \(E_{\alpha}\), for the reaction $$2 \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \rightarrow 4 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})$$ from the observed rate constants: \(k\) at \(25^{\circ} \mathrm{C}=\) \(3.46 \times 10^{-5} s^{-1}\) and \(k\) at \(55^{\circ} \mathrm{C}=1.5 \times 10^{-3} \mathrm{s}^{-1}.\)

A reaction has the following experimental rate equation: Rate \(=k[\mathrm{A}]^{2}[\mathrm{B}] .\) If the concentration of \(\mathrm{A}\) is doubled and the concentration of \(\mathrm{B}\) is halved, what happens to the reaction rate?

The mechanism for the reaction of \(\mathrm{CH}_{3} \mathrm{OH}\) and \(\mathrm{HBr}\) is believed to involve two steps. The overall reaction is exothermic. Step 1: Fast, endothermic $$\mathrm{CH}_{3} \mathrm{OH}+\mathrm{H}^{+} \rightleftarrows \mathrm{CH}_{3} \mathrm{OH}_{2}^{+}$$ Step 2: Slow $$\mathrm{CH}_{3} \mathrm{OH}_{2}^{+}+\mathrm{Br}^{-} \rightarrow \mathrm{CH}_{3} \mathrm{Br}+\mathrm{H}_{2} \mathrm{O}$$ (a) Write an equation for the overall reaction. (b) Draw a reaction coordinate diagram for this reaction. (c) Show that the rate law for this reaction is Rate \(=k\left[\mathrm{CH}_{3} \mathrm{OH}\right]\left[\mathrm{H}^{+}\right]\left[\mathrm{Br}^{-}\right]\)

The compound \(\mathrm{Xe}\left(\mathrm{CF}_{3}\right)_{2}\) decomposes in a first-order reaction to elemental Xe with a half-life of \(30 .\) minutes. If you place \(7.50 \mathrm{mg}\) of \(\mathrm{Xe}\left(\mathrm{CF}_{3}\right)_{2}\) in a flask, how long must you wait until only 0.25 mg of \(\mathrm{Xe}\left(\mathrm{CF}_{3}\right)_{2}\) remains?

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