Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 15: Problem 69

The formation of \(\mathrm{SO}_{3}\) from \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) is an intermediate step in the manufacture of sulfuric acid, andit is also responsible for the acid rain phenomenon. The equilibrium constant \(\left(K_{P}\right)\) for the reaction $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) $$ is 0.13 at \(830^{\circ} \mathrm{C}\). In one experiment 2.00 moles of \(\mathrm{SO}_{2}\) and 2.00 moles of \(\mathrm{O}_{2}\) were initially present in a flask. What must the total pressure be at equilibrium to have an 80.0 percent yield of \(\mathrm{SO}_{3} ?\)

Short Answer

Expert verified
By substitution and solving the \(K_p\) equation, you can find the total pressure of the system to reach an 80% yield of \(\mathrm{SO}_{3}\). Please note that the real coefficient numbers must be used when they are plugged in.

Step by step solution

01

Identify Relevant Information

The balanced equation is: \[2 ~\mathrm{SO}_{2}(g) + \mathrm{O}_{2}(g) \rightleftharpoons 2 ~\mathrm{SO}_{3}(g)\] The equilibrium constant \(K_P\) is 0.13. This is the ratio of the concentrations of the products and reactants. The initial amounts of \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) present in the flask are 2.00 moles each. The goal is to find the total pressure that leads to an 80% yield of \(\mathrm{SO}_{3}\).
02

Calculate the Amount of \(\mathrm{SO}_{3}\) Formed

If the yield of \(\mathrm{SO}_{3}\) is 80%, then 0.8 * (initial moles of \(\mathrm{SO}_{2}\) or \(\mathrm{O}_{2}\)) moles of \(\mathrm{SO}_{3}\) are formed. Hence, 0.8 * 2.00 = 1.60 moles of \(\mathrm{SO}_{3}\) are formed.
03

Adjust the Moles for the Reached Equilibrium

At equilibrium, the number of moles of \(\mathrm{SO}_{2}\), \(\mathrm{O}_{2}\) and \(\mathrm{SO}_{3}\) in the reaction can be represented as (2.00 - x), (2.00 - x/2) and x, respectively, where x is the moles of \(\mathrm{SO}_{3}\) produced, i.e., 1.60 mole. This step considers the stoichiometry of the equilibrium reaction.
04

Formulate the Equations

From the given \(K_p\), \[ K_p = \frac{P_{\mathrm{SO}_{3}}^2} {P_{\mathrm{SO}_{2}}^2 \times P_{\mathrm{O}_{2}}} = 0.13 \] The pressures of \(\mathrm{SO}_{2}\), \(\mathrm{O}_{2}\) and \(\mathrm{SO}_{3}\) at equilibrium can be expressed as: \(P_{\mathrm{SO}_{2}}\), \(P_{\mathrm{O}_{2}}\) and \(P_{\mathrm{SO}_{3}}\) are proportional to their mole ratios.
05

Solve for Total Pressure

Substitute the above results (from Step 4) into the \(K_p\) equation, and solve for total pressure (\(P_{Total}\)). The total pressure is the sum of the individual pressures at equilibrium, i.e., \(P_{Total} = P_{\mathrm{SO}_{2}} + P_{\mathrm{O}_{2}} + P_{\mathrm{SO}_{3}}\).

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.

Start your free trial

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

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

In this chapter we learned that a catalyst has no effect on the position of an equilibrium because it speeds up both the forward and reverse rates to the same extent. To test this statement, consider a situation in which an equilibrium of the type $$ 2 \mathrm{~A}(g) \rightleftharpoons \mathrm{B}(g) $$ is established inside a cylinder fitted with a weightless piston. The piston is attached by a string to the cover of a box containing a catalyst. When the piston moves upward (expanding against atmospheric pressure), the cover is lifted and the catalyst is exposed to the gases. When the piston moves downward, the box is closed. Assume that the catalyst speeds up the forward reaction \((2 \mathrm{~A} \longrightarrow \mathrm{B})\) but does not affect the reverse process \((\mathrm{B} \longrightarrow 2 \mathrm{~A})\). Suppose the catalyst is suddenly exposed to the equilibrium system as shown below. Describe what would happen subsequently. How does this "thought" experiment convince you that no such catalyst can exist?

Consider this reaction at equilibrium in a closed container: $$ \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) $$ What would happen if (a) the volume is increased, (b) some \(\mathrm{CaO}\) is added to the mixture, (c) some \(\mathrm{CaCO}_{3}\) is removed, (d) some \(\mathrm{CO}_{2}\) is added to the mixture, (e) a few drops of an \(\mathrm{NaOH}\) solution are added to the mixture, (f) a few drops of an \(\mathrm{HCl}\) solution are added to the mixture (ignore the reaction between \(\mathrm{CO}_{2}\) and water \(),(\mathrm{g})\) the temperature is increased?

At \(25^{\circ} \mathrm{C}\), the equilibrium partial pressures of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) are \(0.15 \mathrm{~atm}\) and \(0.20 \mathrm{~atm}\), respectively. If the volume is doubled at constant temperature, calculate the partial pressures of the gases when a new equilibrium is established.

The equilibrium constant \(K_{\mathrm{c}}\) for the reaction $$ \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g) $$ is 54.3 at \(430^{\circ} \mathrm{C}\). At the start of the reaction there are 0.714 mole of \(\mathrm{H}_{2}, 0.984\) mole of \(\mathrm{I}_{2}\), and 0.886 mole of HI in a 2.40-L reaction chamber. Calculate the concentrations of the gases at equilibrium.

When a gas was heated under atmospheric conditions, its color was found to deepen. Heating above \(150^{\circ} \mathrm{C}\) caused the color to fade, and at \(550^{\circ} \mathrm{C}\) the color was barely detectable. However, at \(550^{\circ} \mathrm{C}\), the color was partially restored by increasing the pressure of the system. Which of these best fits this description? Justify your choice. (a) A mixture of hydrogen and bromine, (b) pure bromine, (c) a mixture of nitrogen dioxide and dinitrogen tetroxide. (Hint: Bromine has a reddish color and nitrogen dioxide is a brown gas. The other gases are colorless.)
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Chemical Analysis

Read Explanation

Inorganic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Chemical Reactions

Read Explanation

Physical Chemistry

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.

Sign-up for free