/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 91 For the reaction:$$3 \mathrm{O}_... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 91

For the reaction:$$3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{O}_{3}(g)$$. \(K=1.8 \times 10^{-7}\) at a certain temperature. If at equilibrium \(\left[\mathrm{O}_{2}\right]=0.062 \mathrm{M},\) calculate the equilibrium \(\mathrm{O}_{3}\) concentration.

Short Answer

Expert verified
The equilibrium concentration of \(O_3\) is approximately \(3.26\times10^{-4}\mathrm{M}\).

Step by step solution

01

Write the expression for the equilibrium constant (K)

The expression for K can be obtained from the balanced equation. For a general equilibrium reaction, it can be written as:$$K=\frac{([C]^{c}[D]^{d})}{([A]^{a}[B]^{b})}$$In our case, the reaction is: $$3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{O}_{3}(g)$$. We can write the expression for K as:$$K=\frac{[\mathrm{O}_{3}]^{2}}{[\mathrm{O}_{2}]^{3}}$$
02

Substitute the given values

We are given that $$ K=1.8 \times 10^{-7}$$ and the equilibrium concentration of $$O_2$$ is $$0.062 \mathrm{M}$$. We can substitute these values into the expression for K we found in Step 1.$$1.8\times10^{-7}=\frac{[\mathrm{O}_{3}]^{2}}{(0.062)^{3}}$$
03

Solve for the concentration of $$O_3$$

Now, we need to solve for the equilibrium concentration of $$O_3$$. We will first simplify the given equation and then find the value for $$[\mathrm{O}_{3}]$$. $$[\mathrm{O}_{3}]^{2}=(1.8\times10^{-7})(0.062^{3})$$Now, take the square root of both sides to solve for $$[\mathrm{O}_{3}]$$.$$[\mathrm{O}_{3}]=\sqrt{(1.8\times10^{-7})(0.062^{3})}$$
04

Calculate the equilibrium concentration of $$O_3$$

Now, use a calculator to find the equilibrium concentration of $$[\mathrm{O}_{3}]$$.$$[\mathrm{O}_{3}]=\sqrt{(1.8\times10^{-7})(0.062^{3})} \approx 3.26\times10^{-4} \mathrm{M}$$ So, the equilibrium concentration of $$O_3$$ is $$3.26\times10^{-4}\mathrm{M}$$.

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 91影视!

Key Concepts

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

Ozone Concentration
In chemical reactions involving gases, it is essential to monitor the concentration of the species involved. For this exercise, we focus on ozone, i.e., \( \mathrm{O}_3 \), which is formed from dioxygen, \( \mathrm{O}_2 \), through a reversible reaction:\[3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{O}_{3}(g)\]The concentration of ozone at equilibrium helps us understand the extent to which this reaction moves forward or backward under a given set of conditions. The equilibrium concentration tells us about the dynamic balance between the formation and decomposition of ozone under specific conditions, such as pressure and temperature.
Chemical Equilibrium
Chemical equilibrium is a state in a reversible chemical reaction where the rate of the forward reaction equals the rate of the backward reaction. At this point, the concentrations of all reactants and products remain constant over time.
  • In our example, the ozone equation reaches equilibrium when the amount of \( \mathrm{O}_2 \) converting to \( \mathrm{O}_3 \) is equal to the amount of \( \mathrm{O}_3 \) decomposing back into \( \mathrm{O}_2 \).
  • This does not mean the amounts of reactants and products are equal, only that their rates of change are balanced.
It is essential to recognize that reaching equilibrium does not stop the reaction but keeps it in a stable dynamic state where concentrations do not fluctuate anymore.
Reaction Stoichiometry
Reaction stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. Understanding this concept helps us set up our calculations correctly when working with equilibrium problems.For the reaction: \[3 \mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{O}_3(g)\]
  • The coefficients in the balanced equation (3 for \( \mathrm{O}_2 \) and 2 for \( \mathrm{O}_3 \)) are crucial. They indicate the ratio in which reactants combine and products are formed.
  • These coefficients also play a significant role while writing the equilibrium expression and eventually solving for unknown concentrations.
Using stoichiometry, we identify that for every 3 moles of \( \mathrm{O}_2 \) consumed, 2 moles of \( \mathrm{O}_3 \) are produced, which is essential in deriving consistent equilibrium calculations.
Equilibrium Expression
An equilibrium expression shows the relationship between the concentrations of reactants and products at equilibrium. It utilizes the coefficients from the balanced equation, as powers, in a specific mathematical formula.For our reaction:\[3 \mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{O}_3(g)\]The equilibrium constant \( K \) is expressed as:\[K = \frac{[\mathrm{O}_3]^2}{[\mathrm{O}_2]^3}\]
  • This formula helps calculate the concentration of any species in the reaction if the \( K \) value and other species' concentrations are known.
  • The exponents (2 for \( \mathrm{O}_3 \) and 3 for \( \mathrm{O}_2 \)) come from the stoichiometric coefficients of the balanced equation.
By substituting known values into this expression, we can solve for unknown concentrations, like that of \( \mathrm{O}_3 \), and understand the dynamics of the equilibrium state.

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

At a given temperature, \(K=1.3 \times 10^{-2}\) for the reaction \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\).Calculate values of \(K\) for the following reactions at this temperature. a. \(\frac{1}{2} \mathrm{N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{NH}_{3}(g)\). b. \(2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)\) c. \(\mathrm{NH}_{3}(g) \rightleftharpoons \frac{1}{2} \mathrm{N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g)\). d. \(2 \mathrm{N}_{2}(g)+6 \mathrm{H}_{2}(g) \rightleftharpoons 4 \mathrm{NH}_{3}(g)\).

Consider the reaction $$\mathbf{P}_{4}(g) \longrightarrow 2 \mathbf{P}_{2}(g)$$, where \(K_{\mathrm{p}}=1.00 \times 10^{-1}\) at \(1325 \mathrm{K}\). In an experiment where \(\mathrm{P}_{4}(g)\) is placed into a container at \(1325 \mathrm{K},\) the equilibrium mixture of \(\mathrm{P}_{4}(g)\) and \(\mathrm{P}_{2}(g)\) has a total pressure of 1.00 atm. Calculate the equilibrium pressures of \(\mathrm{P}_{4}(g)\) and \(\mathrm{P}_{2}(g) .\) Calculate the fraction (by moles) of \(P_{4}(g)\) that has dissociated to reach equilibrium.

The following equilibrium pressures were observed at a certain temperature for the reaction $$\begin{array}{c}\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) \\\P_{\mathrm{NH}}=3.1 \times 10^{-2} \mathrm{atm} \\\P_{\mathrm{N}_{2}}=8.5 \times 10^{-1} \mathrm{atm} \\\P_{\mathrm{H}_{2}}=3.1 \times 10^{-3} \mathrm{atm}\end{array}$$.Calculate the value for the equilibrium constant \(K_{\mathrm{p}}\) at this temperature. If \(P_{\mathrm{N}_{2}}=0.525\) atm, \(P_{\mathrm{NH},}=0.0167\) atm, and \(P_{\mathrm{H}_{2}}=0.00761\) atm, does this represent a system at equilibrium?

Consider an equilibrium mixture of four chemicals \((\mathrm{A}, \mathrm{B}, \mathrm{C}\) and \(\mathrm{D},\) all gases) reacting in a closed flask according to the equation:$$\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)+\mathrm{D}(g)$$. a. You add more \(A\) to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer. b. You have the original setup at equilibrium, and you add more \(\mathrm{D}\) to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer.

A sample of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) is placed in an empty cylinder at \(25^{\circ} \mathrm{C}\). After equilibrium is reached the total pressure is 1.5 atm and \(16 \%\) (by moles) of the original \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) has dissociated to \(\mathrm{NO}_{2}(g)\) a. Calculate the value of \(K_{\mathrm{p}}\) for this dissociation reaction at \(25^{\circ} \mathrm{C}\) b. If the volume of the cylinder is increased until the total pressure is 1.0 atm (the temperature of the system remains constant), calculate the equilibrium pressure of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) and \(\mathrm{NO}_{2}(g)\) c. What percentage (by moles) of the original \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) is dissociated at the new equilibrium position (total pressure \(=\) \(1.00 \mathrm{atm}) ?\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Reactions

Read Explanation

Organic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Chemical Analysis

Read Explanation
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.