Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 15: Problem 37

Equilibrium is established in a 2.50 L flask at \(250^{\circ} \mathrm{C}\) for the reaction $$\mathrm{PCl}_{5}(\mathrm{g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \quad K_{\mathrm{c}}=3.8 \times 10^{-2}$$ How many moles of \(\mathrm{PCl}_{5}, \mathrm{PCl}_{3},\) and \(\mathrm{Cl}_{2}\) are present at equilibrium, if (a) 0.550 mol each of \(\mathrm{PCl}_{5}\) and \(\mathrm{PCl}_{3}\) are initially introduced into the flask? (b) \(0.610 \mathrm{mol} \mathrm{PCl}_{5}\) alone is introduced into the flask?

Short Answer

Expert verified
The solutions to equation at Step 3 and Step 6 will give the equilibrium mole values in Cases A and B, respectively. You need to solve this equation to get the numerical answer.

Step by step solution

01

Analyzing Scenario (A)

Case (a) begins with 0.550 mol each of \( \mathrm{PCl}_{5} \) and \( \mathrm{PCl}_{3} \). At equilibrium, \( \mathrm{PCl}_{5} \) decreases by \( x \), \( \mathrm{PCl}_{3} \) increases by \( x \), and \( \mathrm{Cl}_{2} \) also increases by \( x \). The equilibrium concentration of each compound can be calculated by dividing these moles by the volume of the flask (2.50 L).
02

Setting Up Equilibrium Expression for Scenario (A)

Set up an equation for \( K_c \) using the concentrations from Step 1. \(\ K_c = [\mathrm{PCl}_{3}][\mathrm{Cl}_{2}]/[\mathrm{PCl}_{5}] = (0.550+x)(0+x)/(0.550-x)\). This simplifies to \( K_c = \frac{x^2}{0.550-x}\) as \( 0 + x \) just becomes \( x \).
03

Solving for 'x' in Scenario (A)

Now solve this equation for \( x \). Substitute known \( K_c = 3.8 \times 10^{-2} \) into the equation obtained from Step 2 and solve for \( x \). Therefore, \( 3.8 \times 10^{-2} = \frac{x^2}{0.550-x}\).
04

Analyzing Scenario (B)

In case (b), only \(0.610 \, \mathrm{mol} \, \mathrm{PCl}_{5} \) is initially introduced. By a similar reasoning as in Scenario A, at equilibrium, \( \mathrm{PCl}_{5} \) decreases by \( x \), \( \mathrm{PCl}_{3} \) and \( \mathrm{Cl}_{2} \) both increase by \( x \).
05

Setting Up Equilibrium Expression for Scenario (B)

Similarly to step 2, write the equilibrium equation. This time, \(K_c = [\mathrm{PCl}_{3}][\mathrm{Cl}_{2}]/[\mathrm{PCl}_{5}] = (0 +x)(0+x)/(0.610 -x)\). This simplifies to \( K_c = \frac{x^2}{0.610-x} \).
06

Solving for 'x' in Scenario (B)

Substitute the known \( K_c = 3.8 \times 10^{-2} \) into the equation and solve for \( x \) to find the equilibrium number of moles. \( 3.8 \times 10^{-2} = \frac{x^2}{0.610-x} \).

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

\(1.00 \times 10^{-3} \mathrm{mol} \mathrm{PCl}_{5}\) is introduced into a \(250.0 \mathrm{mL}\) flask, and equilibrium is established at \(284^{\circ} \mathrm{C}\) : \(\mathrm{PCl}_{5}(\mathrm{g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) .\) The quantity of \(\mathrm{Cl}_{2}(\mathrm{g})\) present at equilibrium is found to be \(9.65 \times 10^{-4} \mathrm{mol}\) What is the value of \(K_{c}\) for the dissociation reaction at \(284^{\circ} \mathrm{C} ?\)

One important reaction in the citric acid cycle is citrate(aq) \(\rightleftharpoons\) aconitate \((\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad K=0.031\) Write the equilibrium constant expression for the above reaction. Given that the concentrations of \([\text { citrate }(\mathrm{aq})]=0.00128 \mathrm{M},[\text { aconitate }(\mathrm{aq})]=4.0 \times\) \(10^{-5} \mathrm{M},\) and \(\left[\mathrm{H}_{2} \mathrm{O}\right]=55.5 \mathrm{M},\) calculate the reaction quotient. Is this reaction at equilibrium? If not, in which direction will it proceed?

Is a mixture of \(0.0205 \mathrm{mol} \mathrm{NO}_{2}(\mathrm{g})\) and \(0.750 \mathrm{mol}\) \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g})\) in a \(5.25 \mathrm{L}\) flask at \(25^{\circ} \mathrm{C},\) at equilibrium? If not, in which direction will the reaction proceed toward products or reactants? $$\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g}) \quad K_{\mathrm{c}}=4.61 \times 10^{-3} \mathrm{at} 25^{\circ} \mathrm{C}$$

The volume of the reaction vessel containing an equilibrium mixture in the reaction \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons\) \(\mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})\) is increased. When equilibrium is re-established, (a) the amount of \(\mathrm{Cl}_{2}\) will have increased; (b) the amount of \(\mathrm{SO}_{2}\) will have decreased; (c) the amounts of \(\mathrm{SO}_{2}\) and \(\mathrm{Cl}_{2}\) will have remained the same; (d) the amount of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) will have increased.

A mixture consisting of \(0.150 \mathrm{mol} \mathrm{H}_{2}\) and \(0.150 \mathrm{mol} \mathrm{I}_{2}\) is brought to equilibrium at \(445^{\circ} \mathrm{C},\) in a 3.25 L flask. What are the equilibrium amounts of \(\mathrm{H}_{2}, \mathrm{I}_{2},\) and HI? $$\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) \quad K_{\mathrm{c}}=50.2\ \mathrm\ {at}\ 445^{\circ} \mathrm{C}$$
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Making Measurements

Read Explanation

Kinetics

Read Explanation

Inorganic 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.

Sign-up for free