Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 15: Problem 14

The equilibrium constant for the dissociation of molecular iodine, \(\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g),\) at 800 \(\mathrm{K}\) is \(K_{c}=3.1 \times 10^{-5} .\) (a) Which species predominates at equilibrium \(\mathrm{I}_{2}\) or \(\mathrm{I}\) ? (b) Assuming both forward and reverse reactions are elementary processes, which reaction has the larger rate constant, the forward or the reverse reaction?

Short Answer

Expert verified
(a) I2 predominates at equilibrium. (b) The reverse reaction has the larger rate constant.

Step by step solution

01

(a) Determine predominant species

Let's first write the balanced equilibrium equation: \[I_2(g) \rightleftharpoons 2 I(g)\] Let's denote the equilibrium concentrations as: - [I2] = equilibrium concentration of I2 - [I] = equilibrium concentration of I The equilibrium constant Kc is given by: \[K_c = \frac{[\mathrm{I}]^2}{[\mathrm{I}_2]}\] We are given Kc = 3.1 x 10^(-5). Now we need to determine the predominant species (I2 or I) at equilibrium. If [I2] > [I], then Kc < 1; If [I2] < [I], then Kc > 1. In our case, Kc < 1, since Kc = 3.1 x 10^(-5). Thus, [I2] > [I], and I2 is the predominant species at equilibrium. Answer to (a): I2 predominates at equilibrium.
02

(b) Compare rate constants of forward and reverse reactions

Since both forward and reverse reactions are elementary processes, we can use the expressions for the rate constants kf (forward reaction) and kr (reverse reaction). The relationship between the equilibrium constant Kc, the rate constants kf and kr is given by: \[K_c = \frac{k_f}{k_r}\] We need to compare kf and kr to determine which reaction has the larger rate constant. If Kc < 1, then kf < kr; If Kc > 1, then kf > kr. As Kc = 3.1 x 10^(-5) < 1, the forward reaction has a smaller rate constant compared to the reverse reaction. Answer to (b): The reverse reaction has the larger rate constant.

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

(a) If \(Q_{c}>K_{c}\) how must the reaction proceed to reach equilibrium? (b) At the start of a certain reaction, only reactants are present; no products have been formed. What is the value of \(Q_{c}\) at this point in the reaction?

For the reaction \(\mathrm{I}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \operatorname{IBr}(g), K_{c}=280\) at \(150^{\circ} \mathrm{C} .\) Suppose that 0.500 \(\mathrm{mol} \mathrm{IBr}\) in a 2.00 -L flask is allowed to reach equilibrium at \(150^{\circ} \mathrm{C}\) . What are the equilibrium concentrations of IBr, I \(_{2},\) and \(\mathrm{Br}_{2} ?\)

The equilibrium constant for the reaction $$2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$$ is \(K_{c}=1.3 \times 10^{-2}\) at 1000 \(\mathrm{K}\) . (a) At this temperature does the equilibrium favor \(\mathrm{NO}\) and \(\mathrm{Br}_{2},\) or does it favor NOBr? (b) Calculate \(K_{c}\) for \(2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)\) (c) Calculate \(K_{c}\) for \(\operatorname{NOBr}(g) \rightleftharpoons \mathrm{NO}(g)+\frac{1}{2} \mathrm{Br}_{2}(g)\)

A mixture of 1.374 \(\mathrm{g}\) of \(\mathrm{H}_{2}\) and 70.31 \(\mathrm{g}\) of \(\mathrm{Br}_{2}\) is heated in a 2.00 -L vessel at 700 \(\mathrm{K}\) . These substances react according to $$\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g)$$ At equilibrium, the vessel is found to contain 0.566 \(\mathrm{g}\) of \(\mathrm{H}_{2}\) (a) Calculate the equilibrium concentrations of \(\mathrm{H}_{2}, \mathrm{Br}_{2},\) and \(\mathrm{HBr}\) . (b ) Calculate \(K_{c} .\)

The reaction \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g)\) has \(K_{p}=\) 0.0870 at \(300^{\circ} \mathrm{C} .\) A flask is charged with 0.50 atm \(\mathrm{PCl}_{3}, 0.50 \mathrm{atm} \mathrm{Cl}_{2},\) and 0.20 atm \(\mathrm{PCl}_{5}\) at this temperature. (a) Use the reaction quotient to determine the direction the reaction must proceed to reach equilibrium. (b) Calculate the equilibrium partial pressures of the gases. (c) What effect will increasing the volume of the system have on the mole fraction of \(\mathrm{Cl}_{2}\) in the equilibrium mixture? (d) The reaction is exothermic. What effect will increasing the temperature of the system have on the mole fraction of \(\mathrm{Cl}_{2}\) in the equilibrium mixture?
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Chemical Reactions

Read Explanation

Physical Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Kinetics

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