Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 107

Consider the equilibrium system \(3 \mathrm{~A} \rightleftharpoons \mathrm{B}\). Sketch the changes in the concentrations of \(\mathrm{A}\) and \(\mathrm{B}\) over time for the following situations: (a) Initially only A is present. (b) Initially only B is present. (c) Initially both A and B are present (with A in higher concentration). In each case, assume that the concentration of \(\mathrm{B}\) is higher than that of \(\mathrm{A}\) at equilibrium.

Short Answer

Expert verified
Under different initial conditions, the changes in concentrations of A and B differ as the system reaches equilibrium. When only A is initially present, A gets consumed forming B until equilibrium. If only B is present initially, B gets converted back into A till equilibrium. In the case where both are present but A is in higher concentration, A gets consumed forming more of B until equilibrium is met. All scenarios eventually reach equilibrium where the concentrations of A and B remain constant over time.

Step by step solution

01

Understand the Equilibrium System

The given system is a reversible reaction where 3 molecules of substance A produce a molecule of substance B (3A ↔ B). At equilibrium, the rates of the forward and reverse reactions are equal, meaning the concentrations of A and B are not changing anymore.
02

Sketch for initial condition: Only A is present

Initially, if only A is present, the reaction will proceed in the forward direction, converting A into B. This means the concentration of A will gradually decrease while the concentration of B will increase until equilibrium is reached. On a graph, plot time on the x-axis, concentration on the y-axis. Draw a declining curve for A and an inclining curve for B, where they reach constant values (equilibrium) at the same time.
03

Sketch for initial condition: Only B is present

If initially only B is present, the reaction will have to proceed in the reverse direction to achieve equilibrium. This means the concentration of B will decrease and the concentration of A will increase. Draw a declining curve for B and an inclining curve for A, again keeping in mind they should reach constant values (equilibrium) at the same time.
04

Sketch for initial condition: Both A and B are present

In this case, depending on the initial concentrations of A and B, the reaction may proceed in either direction until equilibrium is reached. As it's stated that A is initially in higher concentration, the concentration of A will decrease and the concentration of B will increase until they reach equilibrium. Plot on a graph the corresponding changes.

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

When heated at high temperatures, iodine vapor dissociates as follows: $$\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)$$ In one experiment, a chemist finds that when 0.054 mole of \(\mathrm{I}_{2}\) was placed in a flask of volume \(0.48 \mathrm{~L}\) at \(587 \mathrm{~K},\) the degree of dissociation (that is, the fraction of \(\mathrm{I}_{2}\) dissociated) was \(0.0252 .\) Calculate \(K_{\mathrm{c}}\) and \(K_{P}\) for the reaction at this temperature.

Consider the following equilibrium system involving \(\mathrm{SO}_{2}, \mathrm{Cl}_{2},\) and \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) (sulfuryl dichloride): $$\mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2} \mathrm{Cl}_{2}(g)$$ Predict how the equilibrium position would change if (a) \(\mathrm{Cl}_{2}\) gas were added to the system; (b) \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) were removed from the system; (c) \(\mathrm{SO}_{2}\) were removed from the system. The temperature remains constant.

Consider the following equilibrium systems: (a) \(A \Longrightarrow 2 B\) \(\Delta H^{\circ}=20.0 \mathrm{~kJ} / \mathrm{mol}\) (b) \(\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}\) \(\Delta H^{\circ}=-5.4 \mathrm{~kJ} / \mathrm{mol}\) (c) \(A \Longrightarrow B\) \(\Delta H^{\circ}=0.0 \mathrm{~kJ} / \mathrm{mol}\) Predict the change in the equilibrium constant \(K_{\mathrm{c}}\) that would occur in each case if the temperature of the reacting system were raised.

Consider the following reaction: $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)$$ If the equilibrium partial pressures of \(\mathrm{N}_{2}, \mathrm{O}_{2},\) and NO are 0.15 atm, 0.33 atm, and 0.050 atm, respectively, at \(2200^{\circ} \mathrm{C},\) what is \(K_{P} ?\)

Iodine is sparingly soluble in water but much more so in carbon tetrachloride \(\left(\mathrm{CCl}_{4}\right) .\) The equilibrium constant, also called the partition coefficient, for the distribution of \(\mathrm{I}_{2}\) between these two phases $$\mathrm{I}_{2}(a q) \rightleftharpoons \mathrm{I}_{2}\left(\mathrm{CCl}_{4}\right)$$ is 83 at \(20^{\circ} \mathrm{C}\). (a) A student adds \(0.030 \mathrm{~L}\) of \(\mathrm{CCl}_{4}\) to \(0.200 \mathrm{~L}\) of an aqueous solution containing \(0.032 \mathrm{~g}\) \(\mathrm{I}_{2} .\) The mixture is shaken and the two phases are then allowed to separate. Calculate the fraction of \(\mathrm{I}_{2}\) remaining in the aqueous phase. (b) The student now repeats the extraction of \(\mathrm{I}_{2}\) with another \(0.030 \mathrm{~L}\) of \(\mathrm{CCl}_{4} .\) Calculate the fraction of the \(\mathrm{I}_{2}\) from the original solution that remains in the aqueous phase. (c) Compare the result in (b) with a single extraction using \(0.060 \mathrm{~L}\) of \(\mathrm{CCl}_{4}\). Comment on the difference.
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Physical Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation

Nuclear 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