Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 60

Lexan is a plastic used to make compact discs, eyeglass lenses, and bullet- proof glass. One of the compounds used to make Lexan is phosgene \(\left(\mathrm{COCl}_{2}\right)\), an extremely poisonous gas. Phosgene decomposes by the reaction $$ \mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) $$ for which \(K_{\mathrm{p}}=6.8 \times 10^{-9}\) at \(100^{\circ} \mathrm{C}\). If pure phosgene at an initial pressure of \(1.0\) atm decomposes, calculate the equilibrium pressures of all species.

Short Answer

Expert verified
At equilibrium, the pressures for all species involved in the decomposition of phosgene \(\left(\mathrm{COCl}_{2}\right)\) are: \(P_{COCl_2} \approx 1.00\) atm, \(P_{CO} \approx 2.61 \times 10^{-5}\) atm, and \(P_{Cl_2} \approx 2.61 \times 10^{-5}\) atm.

Step by step solution

01

Write the balanced chemical equation and the expression for Kp

\[ \mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g) + \mathrm{Cl}_{2}(g) \] The equilibrium constant expression is: \[ K_p = \frac{P_{CO} P_{Cl_2}}{P_{COCl_2}} \]
02

Set up the ICE table

Construct an ICE table using the initial pressure of phosgene (1.0 atm) and assume some amount of phosgene dissociates 'x'. Then, determine the change and equilibrium pressures for each species. | | COCl2 | CO | Cl2 | |-------|-------|----|-----| | I | 1.0 | 0 | 0 | | C | -x | +x | +x | | E | 1-x | x | x |
03

Substitute the equilibrium pressures into the Kp expression

Substitute the equilibrium pressures from the ICE table into the Kp expression: \[ K_p = \frac{x^2}{1-x} \]
04

Solve for x

Now, substitute the given value of Kp in the equation and solve for x: \[ 6.8 \times 10^{-9} = \frac{x^2}{1-x} \] To simplify the calculations, we can make the assumption that x in the denominator is negligible and simplify the equation as follows: \[ 6.8 \times 10^{-9} \approx \frac{x^2}{1} \] Now, solve for x: \[ x = \sqrt{6.8 \times 10^{-9}} \approx 2.61 \times 10^{-5} \]
05

Calculate the equilibrium pressures for each species

Using the value of x found in Step 4, determine the equilibrium pressures for each species: \[ P_{COCl_2} = 1-x = 1 - 2.61 \times 10^{-5} \approx 1.00 \text{ atm} \] \[ P_{CO} = P_{Cl_2} = x = 2.61 \times 10^{-5} \text{ atm} \] At equilibrium, the pressures for all species are: \(P_{COCl_2} \approx 1.00\) atm, \(P_{CO} \approx 2.61 \times 10^{-5}\) atm, and \(P_{Cl_2} \approx 2.61 \times 10^{-5}\) atm.

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

At a particular temperature, \(K_{\mathrm{p}}=0.25\) for the reaction $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ a. A flask containing only \(\mathrm{N}_{2} \mathrm{O}_{4}\) at an initial pressure of \(4.5\) atm is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases. b. A flask containing only \(\mathrm{NO}_{2}\) at an initial pressure of \(9.0\) atm is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases. c. From your answers to parts a and \(\mathrm{b}\), does it matter from which direction an equilibrium position is reached?

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 \(1.0 \mathrm{~atm}\) (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 \(=1.00 \mathrm{~atm}\) )?

Consider the following reactions. \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \longrightarrow 2 \mathrm{HI}(g)\) and \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(s) \longrightarrow 2 \mathrm{HI}(g)\) List two property differences between these two reactions that relate to equilibrium.

Suppose \(K=4.5 \times 10^{-3}\) at a certain temperature for the reaction $$ \mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) $$ If it is found that the concentration of \(\mathrm{PCl}_{5}\) is twice the concentration of \(\mathrm{PCl}_{3}\), what must be the concentration of \(\mathrm{Cl}_{2}\) under these conditions?

Consider the reaction $$ \mathrm{P}_{4}(g) \longrightarrow 2 \mathrm{P}_{2}(g) $$ 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 \(1.00 \mathrm{~atm} .\) Calculate the equilibrium pressures of \(\mathrm{P}_{4}(g)\) and \(\mathrm{P}_{2}(g) .\) Calculate the fraction (by moles) of \(\mathrm{P}_{4}(g)\) that has dissociated to reach equilibrium.
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Organic 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