Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 15: Problem 24

For the dissociation of \(\mathrm{I}_{2}(\mathrm{g})\) at about \(1200^{\circ} \mathrm{C}\) \(\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g}), K_{\mathrm{c}}=1.1 \times 10^{-2} .\) What volume flask should we use if we want 0.37 mol I to be present for every \(1.00 \mathrm{mol} \mathrm{I}_{2}\) at equilibrium?

Short Answer

Expert verified
A volume of 80.4 Liters is needed to reach equilibrium under the given conditions.

Step by step solution

01

Understanding the Chemical Equilibrium

Write the balanced chemical equation. \[\mathrm{I}_2(g) \rightleftharpoons 2\mathrm{I}(g)\]This represents the equilibrium of the dissociation of Iodine gas.
02

Equilibrium Expression and Substituting given Values

For this reaction, \(Kc\) is given by \[Kc = {[I]^2 \over [I_2]} \] Substitute the given values into the equilibrium expression, we have: \[1.1 * 10^{-2} = {([0.37]^2) \over [I_2]} \] Solve for \([I_2]\), we find \([I_2] = 12.43 M \]
03

Recipe Interpretation

The given recipe states that at equilibrium, for every 1.00 mol \(\mathrm{I}_2\), there are 0.37 mol I, therefore the concentration of \(\mathrm{I}_2\) is \([I_2] = \frac{1.00 mol}{Volume}\). Therefore we can set equal the two equations for \([I_2]\) and solve for the Volume : \[Volume = \frac{1.00 mol}{12.43 M} = 0.0804 m^3 = 80.4 L \]

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

The following reaction represents the binding of oxygen by the protein hemoglobin (Hb): $$\mathrm{Hb}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{aq}) \rightleftharpoons \mathrm{Hb}: \mathrm{O}_{2}(\mathrm{aq}) \quad \Delta H<0$$ Explain how each of the following affects the amount of \(\mathrm{Hb}: \mathrm{O}_{2}:\) (a) increasing the temperature; (b) decreasing the pressure of \(\mathrm{O}_{2} ;\) (c) increasing the amount of 6 hemoglobin.

Would you expect that the amount of \(\mathrm{N}_{2}\) to increase, decrease, or remain the same in a scuba diver's body as he or she descends below the water surface?

The following is an approach to establishing a relationship between the equilibrium constant and rate constants mentioned in the section on page 660 \(\bullet\)Work with the detailed mechanism for the reaction. \(\bullet\) Use the principle of microscopic reversibility, the idea that every step in a reaction mechanism is reversible. (In the presentation of elementary reactions in Chapter \(14,\) we treated some reaction steps as reversible and others as going to completion. However, as noted in Table \(15.3,\) every reaction has an equilibrium constant even though a reaction is generally considered to go to completion if its equilibrium constant is very large.) \(\bullet\) Use the idea that when equilibrium is attained in an overall reaction, it is also attained in each step of its mechanism. Moreover, we can write an equilibrium constant expression for each step in the mechanism, similar to what we did with the steady-state assumption in describing reaction mechanisms. \(\bullet\)Combine the \(K_{\mathrm{c}}\) expressions for the elementary steps into a \(K_{\mathrm{c}}\) expression for the overall reaction. The numerical value of the overall \(K_{c}\) can thereby be expressed as a ratio of rate constants, \(k\) Use this approach to establish the equilibrium constant expression for the overall reaction, $$ \mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) $$ The mechanism of the reaction appears to be the following: Fast: \(\quad \mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g})\) Slow: \(\quad 2 \mathrm{I}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\)

Briefly describe each of the following ideas or phenomena: (a) dynamic equilibrium; (b) direction of a net chemical change; (c) Le Châtelier's principle; (d) effect of a catalyst on equilibrium.

In the reaction \(2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g}), 0.455\) \(\mathrm{mol} \mathrm{SO}_{2}, 0.183 \mathrm{mol} \mathrm{O}_{2},\) and \(0.568 \mathrm{mol} \mathrm{SO}_{3}\) are introduced simultaneously into a 1.90 L vessel at \(1000 \mathrm{K}\). (a) If \(K_{c}=2.8 \times 10^{2},\) is this mixture at equilibrium? (b) If not, in which direction will a net change occur?
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Making Measurements

Read Explanation

Inorganic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Chemistry Branches

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