/*! 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 65 Lexan is a plastic used to make ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 65

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
The equilibrium pressures for COCl\(_2\), CO, and Cl\(_2\) in the decomposition of phosgene are approximately \(0.999974 \, \mathrm{atm}\), \(2.61 \times 10^{-5} \, \mathrm{atm}\), and \(2.61 \times 10^{-5} \, \mathrm{atm}\), respectively.

Step by step solution

01

Define the change in pressure

Let the change in pressure during the reaction be represented by \(x\). This means that the phosgene will decrease by \(x\) atm, and both CO and Cl\(_2\) will increase by \(x\) atm. Then, the equilibrium pressures of COCl\(_2\), CO, and Cl\(_2\) will be \((1.0-x)\) atm, \(x\) atm, and \(x\) atm respectively.
02

Write the expression for \(K_{p}\) and solve for \(x\)

Now we can write the expression for \(K_{p}\): \[K_{p} = \frac{[\mathrm{CO}][\mathrm{Cl}_{2}]}{[\mathrm{COCl}_2]}\] Since we are working with pressures, the brackets represent the partial pressures of the species: \[K_{p} = \frac{(x)(x)}{(1 - x)}\] Given the value of \(K_{p} = 6.8 \times 10^{-9}\) at \(100^{\circ} \mathrm{C}\), we can now substitute: \[\begin{equation} 6.8 \times 10^{-9} = \frac{x^{2}}{(1 - x)} \end{equation}\] This is a quadratic equation in \(x\). We can solve it by rearranging and simplifying: \[\begin{equation} x^2 + (6.8 \times 10^{-9})(x) - 6.8 \times 10^{-9} = 0 \end{equation}\] Now we can solve for \(x\) using the Quadratic Formula: \[\begin{equation} x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \end{equation}\] where \(a = 1\), \(b = 6.8 \times 10^{-9}\), and \(c = -6.8 \times 10^{-9}\). Plugging these values, we only consider the positive root since the pressure cannot be negative: \[x \approx 2.61 \times 10^{-5}\]
03

Determine the equilibrium pressures

Using the value of \(x\), we can now determine the equilibrium pressures for each species in the reaction: \[\mathrm{COCl}_{2}: \: 1.0 - 2.61 \times 10^{-5} \approx 0.999974 \, \mathrm{atm}\] \[\mathrm{CO}: \: 2.61 \times 10^{-5} \, \mathrm{atm}\] \[\mathrm{Cl}_{2}: \: 2.61 \times 10^{-5} \, \mathrm{atm}\] Hence, the equilibrium pressures for COCl\(_2\), CO, and Cl\(_2\) are \(0.999974 \, \mathrm{atm}\), \(2.61 \times 10^{-5} \, \mathrm{atm}\), and \(2.61 \times 10^{-5} \, \mathrm{atm}\), respectively.

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.

Chemical Equilibrium
Chemical equilibrium refers to a state where the concentrations or pressures of the reactants and products remain constant over time. This occurs when the forward and reverse reactions balance each other out. In our example, phosgene decomposes into CO and Cl\(_2\). Initially, only phosgene is present at 1.0 atm, and as it decomposes, both CO and Cl\(_2\) start forming until equilibrium is reached.

Key points about chemical equilibrium include:
  • It is dynamic, meaning reactions continue to occur at the molecular level.
  • No net change in pressures or concentrations is observed at equilibrium.
  • The equilibrium constant, like our given \(K_{p} = 6.8 \times 10^{-9}\), quantifies the ratio of product pressures to reactant pressures at equilibrium.
Understanding chemical equilibrium helps predict how reactions behave under different conditions and is crucial for controlling industrial processes, ensuring safety, and optimizing product yields.
Quadratic Equation
The quadratic equation is a vital tool in solving equilibrium problems, especially when dealing with unknown pressures or concentrations. In the decomposition of phosgene, we derived a quadratic equation to find the equilibrium pressures. The initial setup was based on the expression for \(K_{p}\):

\[6.8 \times 10^{-9} = \frac{x^{2}}{(1 - x)}\]

This equation represents a typical quadratic form \(ax^2 + bx + c = 0\). Using the solutions of quadratic equations, particularly the quadratic formula:

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

We solved for \(x\) considering our specific values of \(a\), \(b\), and \(c\), discarding the negative root since pressure cannot be negative. Solving quadratic equations in this way helps us determine how the system behaves and reach the correct equilibrium pressures. Thus, mastering this math skill is essential when dealing with equilibrium scenarios.
Le Chatelier's Principle
Le Chatelier's Principle is a principle used to predict the effect of a change in conditions on chemical equilibria. It states that if a dynamic equilibrium is disturbed by changing the conditions, the system will respond to counteract the change and re-establish equilibrium.

In the phosgene decomposition example, if we were to increase the pressure or add more CO or Cl\(_2\), the equilibrium would shift to counteract this change. For instance, adding more Cl\(_2\) would drive the reaction towards forming more phosgene, reducing the disturbance.

Key applications of Le Chatelier's Principle include:
  • Predicting how changes in pressure, temperature, or concentration will affect the equilibrium position.
  • Understanding the reversibility of reactions and how to manipulate product yields.
  • It is especially helpful in designing chemical processes and optimizing industrial reactions.
Le Chatelier's Principle is an indispensable concept for chemists and engineers, allowing them to control and optimize various chemical systems.

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 a study of the reaction $$3 \mathrm{Fe}(s)+4 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+4 \mathrm{H}_{2}(g)$$ at 1200 \(\mathrm{K}\) it was observed that when the equilibrium partial pressure of water vapor is 15.0 torr, the total pressure at equilibrium is 36.3 torr. Calculate the value of \(K_{\mathrm{p}}\) for this reaction at 1200 \(\mathrm{K}\) . Hint: Apply Dalton's law of partial pressures.)

Methanol, a common laboratory solvent, poses a threat of blindness or death if consumed in sufficient amounts. Once in the body, the substance is oxidized to produce formaldehyde (embalming fluid) and eventually formic acid. Both of these substances are also toxic in varying levels. The equilibrium between methanol and formaldehyde can be described as follows: $$\mathrm{CH}_{3} \mathrm{OH}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}(a q)+\mathrm{H}_{2}(a q)$$ Assuming the value of \(K\) for this reaction is \(3.7 \times 10^{-10},\) what are the equilibrium concentrations of each species if you start with a 1.24\(M\) solution of methanol? What will happen to the concentration of methanol as the formaldehyde is further converted to formic acid?

The creation of shells by mollusk species is a fascinating process. By utilizing the \(\mathrm{Ca}^{2+}\) in their food and aqueous environment, as well as some complex equilibrium processes, a hard calcium carbonate shell can be produced. One important equilibrium reaction in this complex process is $$\mathrm{HCO}_{3}^{-}(a q) \leftrightharpoons \mathrm{H}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q) K=5.6 \times 10^{-11}$$ If 0.16 mole of \(\mathrm{HCO}_{3}^{-}\) is placed into 1.00 \(\mathrm{L}\) of solution, what will be the equilibrium concentration of \(\mathrm{CO}_{3}^{2-2}\) ?

For the reaction \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g),\) consider two possibilities: (a) you mix 0.5 mole of each reactant, allow the system to come to equilibrium, and then add another mole of \(\mathrm{H}_{2}\) and allow the system to reach equilibrium again, or \((b)\) you \(\operatorname{mix} 1.5\) moles of \(\mathrm{H}_{2}\) and 0.5 mole of \(\mathrm{I}_{2}\) and allow the system to reach equilibrium. Will the final equilibrium mixture be different for the two procedures? Explain.

A sample of gaseous nitrosyl bromide (NOBr) was placed in a container fitted with a frictionless, massless piston, where it decomposed at \(25^{\circ} \mathrm{C}\) according to the following equation: $$2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)$$ The initial density of the system was recorded as 4.495 \(\mathrm{g} / \mathrm{L}\) . After equilibrium was reached, the density was noted to be 4.086 \(\mathrm{g} / \mathrm{L}\) . a. Determine the value of the equilibrium constant \(K\) for the reaction. b. If \(\operatorname{Ar}(g)\) is added to the system at equilibrium at constant temperature, what will happen to the equilibrium position? What happens to the value of \(K ?\) Explain each answer.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Chemical Reactions

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Kinetics

Read Explanation

The Earths Atmosphere

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.