Chemistry: An Atoms First Approach
Chemistry: An Atoms First Approach
2nd Edition
ISBN: 9781305079243
Author: Steven S. Zumdahl, Susan A. Zumdahl
Publisher: Cengage Learning
bartleby

Videos

Textbook Question
Book Icon
Chapter 12, Problem 5ALQ

Consider the reaction A ( g ) + 2 B ( g ) C ( g ) + D ( g ) in a 1.0-L rigid flask. Answer the following questions for each situation (a–d):

i. Estimate a range (as, small as possible) for the requested substance. For example, [A] could be between 95 M and 100 M.

ii. Explain how you decided on the limits for the estimated range.

iii. Indicate what other information would enable you to narrow your estimated range.

iv. Compare the estimated concentrations for a through d. and explain any differences.

a. If at equilibrium [A]= 1 M. and then 1 mole of C is added, estimate the value for [A] once equilibrium is reestablished.

b. If at equilibrium [B] = 1 M, and then 1 mole of C is added, estimate the value for [B] once equilibrium is reestablished.

c. If at equilibrium [C] = 1 M, and then 1 mole of C is added, estimate the value for [C] once equilibrium is reestablished.

d. If at equilibrium [D] = 1 M, and then 1 mole of C is added, estimate the value for [D] once equilibrium is reestablished.

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: Answers for the questions for each of the given situations are to be stated.

Concept introduction: Chemical equilibrium is a state of a system in which the rate of forward and backward reactions is equal. It is affected by various factors such as concentration of reactants or products, temperature and pressure.

To determine: The value of [A] once equilibrium is reestablished if at equilibrium [A]=1 M and then 1 mole of C is added and estimate a range for the requested substance.

Answer to Problem 5ALQ

Answer

The equilibrium concentration of [A]=[A+1C]

Explanation of Solution

Explanation

The given reaction is,

A(g)+2B(g)C(g)+D(g)

The equilibrium constant K for the above reaction is,

K=[C][D][A][B]2

It is given that at equilibrium,   [A]=1 M and 1 mole of C is added. Let the change on concentration is assumed to be x . Therefore, the equilibrium concentrations become,

[A]=1 M+x

[C]=[C+1]

The new equilibrium constant (K') for the reaction is,

K'=[C+1][D][A][B]2

Since the equilibrium is reestablished. Therefore, the new equilibrium constant (K') will be equal to K .

[C][D][A][B]2=[C+1][D][A][B]2=[C+1][D][1+x][B]2C=C+11+xx=1C

Substitute the value of x in the concentration of A . Therefore, the equilibrium concentration of [A]=[A+1C] .

(a)(ii)

To determine: An explanation for deciding the limits for the estimated range.

Answer

The limits were decided for the estimated range.

The equilibrium constant expression is important in deciding the limits for the estimated range. It allows us to predict several important features of the reaction: the tendency of the reaction to occur but it does not tell about the speed of the reaction.

(iii)

To determine: The other information that enable to narrow the estimated range.

Answer

The other information involves temperature and pressure.

The equilibrium constant changes with temperature. If energy is added to a system in equilibrium, the position of equilibrium will be in the direction that consumes energy. Therefore, the equilibrium will shift in reverse direction. When pressure of a system in equilibrium is increased, the equilibrium will shift toward the side having fewer moles of gas.

(iv)

To determine: The comparison of estimated concentrations for a through d and explanation for the differences.

Answer

The comparison of estimated concentrations for a through d is done.

The given reaction is,

A(g)+2B(g)C(g)+D(g)

In each of the given case, one mole of product C is added. Addition of product will shift the equilibrium in reverse direction. According to Le Chatelier principle, when product is added to a system in equilibrium, the equilibrium will shift in reverse direction to balance the effect. Thus concentration of reactants B and A will be higher than products C and D .

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: Answers for the questions for each of the given situations are to be stated.

Concept introduction: Chemical equilibrium is a state of a system in which the rate of forward and backward reactions is equal. It is affected by various factors such as concentration of reactants or products, temperature and pressure.

To determine: The value of [B] once equilibrium is reestablished if at equilibrium [B]=1 M and then 1 mole of C is added and estimate a range for the requested substance.

Answer to Problem 5ALQ

Answer

The equilibrium concentration of [B]=C+11 .

Explanation of Solution

Explanation

The given reaction is,

A(g)+2B(g)C(g)+D(g)

The equilibrium constant K for the above reaction is,

K=[C][D][A][B]2

It is given that at equilibrium, [B]=1 M and 1 mole of C is added. Let the change on concentration is assumed to be x . Therefore, the equilibrium concentrations become,

[B]=(1 M+x)2

[C]=[C+1]

The new equilibrium constant (K') for the reaction is,

K'=[C+1][D][A][B]2

Since the equilibrium is reestablished. Therefore, the new equilibrium constant (K') will be equal to K .

[C+1][D][A][1+x]2=[C][D][A][B]2C+1(1+x)2=[C]B=C+11

Therefore, the equilibrium concentration of [B]=C+11 .

(ii)

To determine: An explanation for deciding the limits for the estimated range.

Solution

The limits were decided for the estimated range.

The equilibrium constant expression is important in deciding the limits for the estimated range. It allows us to predict several important features of the reaction: the tendency of the reaction to occur but it does not tell about the speed of the reaction.

(iii)

To determine: The other information that enable to narrow the estimated range.

Answer

The other information involves temperature and pressure.

The equilibrium constant changes with temperature. If energy is added to a system in equilibrium, the position of equilibrium will be in the direction that consumes energy. Therefore, the equilibrium will shift in reverse direction. When pressure of a system in equilibrium is increased, the equilibrium will shift toward the side having fewer moles of gas.

(iv)

To determine: The comparison of estimated concentrations for a through d and explanation for the differences.

Answer

The comparison of estimated concentrations for a through d is done.

The given reaction is,

A(g)+2B(g)C(g)+D(g)

In each of the given case, one mole of product C is added. Addition of product will shift the equilibrium in reverse direction. According to Le Chatelier principle, when product is added to a system in equilibrium, the equilibrium will shift in reverse direction to balance the effect. Thus concentration of reactants B and A will be higher than products C and D .

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: Answers for the questions for each of the given situations are to be stated.

Concept introduction: Chemical equilibrium is a state of a system in which the rate of forward and backward reactions is equal. It is affected by various factors such as concentration of reactants or products, temperature and pressure.

To determine: The value of [C] once equilibrium is reestablished if at equilibrium [C]=1 M and then 1 mole of C is added and estimate a range for the requested substance.

Answer to Problem 5ALQ

Answer

There will be a change in concentration of C .

Explanation of Solution

Explanation

The given reaction is,

A(g)+2B(g)C(g)+D(g)

The equilibrium constant K for the above reaction is,

K=[C][D][A][B]2

It is given that at equilibrium, [C]=1 M and 1 mole of C is added. Let the change on concentration is assumed to be x . Therefore, the equilibrium concentrations become,

[C]=[C+1]

Therefore, the equilibrium concentration of [C]=[C+1] .

(ii)

To determine: An explanation for deciding the limits for the estimated range.

Answer

The limits were decided for the estimated range.

The equilibrium constant expression is important in deciding the limits for the estimated range. It allows us to predict several important features of the reaction: the tendency of the reaction to occur but it does not tell about the speed of the reaction.

(iii)

To determine: The other information that enable to narrow the estimated range.

Answer

The other information involves temperature and pressure.

The equilibrium constant changes with temperature. If energy is added to a system in equilibrium, the position of equilibrium will be in the direction that consumes energy. Therefore, the equilibrium will shift in reverse direction. When pressure of a system in equilibrium is increased, the equilibrium will shift toward the side having fewer moles of gas.

(iv)

To determine: The comparison of estimated concentrations for a through d and explanation for the differences.

Answer

The comparison of estimated concentrations for a through d is done.

The given reaction is,

A(g)+2B(g)C(g)+D(g)

In each of the given case, one mole of product C is added. Addition of product will shift the equilibrium in reverse direction. According to Le Chatelier principle, when product is added to a system in equilibrium, the equilibrium will shift in reverse direction to balance the effect. Thus concentration of reactants B and A will be higher than products C and D .

(d)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: Answers for the questions for each of the given situations are to be stated.

Concept introduction: Chemical equilibrium is a state of a system in which the rate of forward and backward reactions is equal. It is affected by various factors such as concentration of reactants or products, temperature and pressure.

(i)

To determine: The value of [D] once equilibrium is reestablished if at equilibrium [D]=1 M and then 1 mole of C is added and estimate a range for the requested substance.

Answer to Problem 5ALQ

Answer

There will be a change in concentration of D .

Explanation of Solution

Explanation

The given reaction is,

A(g)+2B(g)C(g)+D(g)

It is given that at equilibrium, [D]=1 M and 1 mole of C is added. Let the change on concentration is assumed to be x . Therefore, the equilibrium concentrations become,

[D]=1 M+x

[C]=[C+1]

The new equilibrium constant (K') for the reaction is,

K'=[C+1][1+x][A][B]2

Since the equilibrium is reestablished. Therefore, the new equilibrium constant (K') will be equal to K .

[C+1][1+x][A][B]2=[C][D][A][B]2(C+1)(1+x)=C×[D][D]=C(C+1)(1+x)

Therefore, the equilibrium concentration of [D]=C(C+1)(1+x) .

(ii)

To determine: An explanation for deciding the limits for the estimated range.

Answer

The limits were decided for the estimated range.

The equilibrium constant expression is important in deciding the limits for the estimated range. It allows us to predict several important features of the reaction: the tendency of the reaction to occur but it does not tell about the speed of the reaction.

(iii)

To determine: The other information that enable to narrow the estimated range.

Answer

The other information involves temperature and pressure.

The equilibrium constant changes with temperature. If energy is added to a system in equilibrium, the position of equilibrium will be in the direction that consumes energy. Therefore, the equilibrium will shift in reverse direction. When pressure of a system in equilibrium is increased, the equilibrium will shift toward the side having fewer moles of gas.

(iv)

To determine: The comparison of estimated concentrations for a through d and explanation for the differences.

Answer

The comparison of estimated concentrations for a through d is done.

The given reaction is,

A(g)+2B(g)C(g)+D(g)

In each of the given case, one mole of product C is added. Addition of product will shift the equilibrium in reverse direction. According to Le Chatelier principle, when product is added to a system in equilibrium, the equilibrium will shift in reverse direction to balance the effect. Thus concentration of reactants B and A will be higher than products C and D .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 12 Solutions

Chemistry: An Atoms First Approach

Ch. 12 - Consider an equilibrium mixture of four chemicals...Ch. 12 - The boxes shown below represent a set of initial...Ch. 12 - For the reactionH2(g)+I2(g)2HI(g), consider two...Ch. 12 - Prob. 4ALQCh. 12 - Consider the reaction A(g)+2B(g)C(g)+D(g) in a...Ch. 12 - Consider the reactionA(g)+B(g)C(g)+D(g). A friend...Ch. 12 - Prob. 7ALQCh. 12 - Prob. 8ALQCh. 12 - Prob. 9ALQCh. 12 - Prob. 10QCh. 12 - Consider the following reaction:...Ch. 12 - Prob. 12QCh. 12 - Suppose a reaction has the equilibrium constant K...Ch. 12 - Prob. 14QCh. 12 - Consider the following reaction at some...Ch. 12 - Prob. 16QCh. 12 - Prob. 17QCh. 12 - Prob. 18QCh. 12 - For a typical equilibrium problem, the value of K...Ch. 12 - Prob. 20QCh. 12 - Write the equilibrium expression (K) for each of...Ch. 12 - Write the equilibrium expression (Kp) for each...Ch. 12 - Prob. 23ECh. 12 - For the reaction H2(g)+Br2(g)2HBr(g) Kp = 3.5 104...Ch. 12 - Prob. 25ECh. 12 - At high temperatures, elemental nitrogen and...Ch. 12 - At a particular temperature, a 3.0-L flask...Ch. 12 - At a particular temperature a 2.00-L flask at...Ch. 12 - Prob. 29ECh. 12 - Prob. 30ECh. 12 - Prob. 31ECh. 12 - Prob. 32ECh. 12 - Prob. 33ECh. 12 - Write expressions for Kp for the following...Ch. 12 - Prob. 35ECh. 12 - Prob. 36ECh. 12 - Prob. 37ECh. 12 - In a study of the reaction...Ch. 12 - The equilibrium constant is 0.0900 at 25C for the...Ch. 12 - The equilibrium constant is 0.0900 at 25C for the...Ch. 12 - At 900c, Kp = 1.04 for the reaction...Ch. 12 - Ethyl acetate is synthesized in a nonreacting...Ch. 12 - For the reaction 2H2O(g)2H2(g)+O2(g) K = 2.4 103...Ch. 12 - The reaction 2NO(g)+Br2(g)2NOBr(g) has Kp = 109 at...Ch. 12 - A 1.00-L flask was filled with 2.00 moles of...Ch. 12 - Prob. 46ECh. 12 - Prob. 47ECh. 12 - Prob. 48ECh. 12 - Prob. 49ECh. 12 - Nitrogen gas (N2) reacts with hydrogen gas (H2) to...Ch. 12 - Prob. 51ECh. 12 - Prob. 52ECh. 12 - Prob. 53ECh. 12 - At 25c, K = 0.090 for the reaction...Ch. 12 - Prob. 55ECh. 12 - Prob. 56ECh. 12 - Prob. 57ECh. 12 - At o particular temperature, K = 4 .0 107 for the...Ch. 12 - Prob. 59ECh. 12 - Lexan is a plastic used to make compact discs,...Ch. 12 - At 25C, Kp. = 2.9 103 for the reaction...Ch. 12 - A sample of solid ammonium chloride was placed in...Ch. 12 - Prob. 63ECh. 12 - Predict the shift in the equilibrium position that...Ch. 12 - An important reaction in the commercial production...Ch. 12 - What will happen to the number of moles of SO3 in...Ch. 12 - Prob. 67ECh. 12 - Hydrogen for use in ammonia production is produced...Ch. 12 - Old-fashioned smelling salts consist of ammonium...Ch. 12 - Ammonia is produced by the Haber process, in which...Ch. 12 - Prob. 71AECh. 12 - Given the following equilibrium constants at...Ch. 12 - Consider the decomposition of the compound C5H6O3...Ch. 12 - Prob. 74AECh. 12 - The gas arsine, AsH3, decomposes as follows:...Ch. 12 - At a certain temperature, K = 9.1 10-4 for the...Ch. 12 - At a certain temperature, K = 1.1 l03 for the...Ch. 12 - Prob. 78AECh. 12 - At 25C, gaseous SO2Cl2 decomposes to SO2(g) and...Ch. 12 - For the following reaction at a certain...Ch. 12 - Prob. 81AECh. 12 - Consider the reaction Fe3+(aq)+SCN(aq)FeSCN2+(aq)...Ch. 12 - Chromium(VI) forms two different oxyanions, the...Ch. 12 - Prob. 84AECh. 12 - Prob. 85AECh. 12 - For the reaction below, Kp = 1.16 at 800C....Ch. 12 - Many sugars undergo a process called mutarotation,...Ch. 12 - Peptide decomposition is one of the key processes...Ch. 12 - The creation of shells by mollusk species is a...Ch. 12 - Methanol, a common laboratory solvent, poses a...Ch. 12 - Prob. 91CWPCh. 12 - Prob. 92CWPCh. 12 - Prob. 93CWPCh. 12 - Prob. 94CWPCh. 12 - Prob. 95CWPCh. 12 - Prob. 96CWPCh. 12 - Consider the following exothermic reaction at...Ch. 12 - For the following endothermic reaction at...Ch. 12 - Prob. 99CPCh. 12 - A 4.72-g sample of methanol (CH3OH) was placed in...Ch. 12 - At 35C, K = 1.6 105 for the reaction...Ch. 12 - Nitric oxide and bromine at initial partial...Ch. 12 - At 25C. Kp = 5.3 105 for the reaction...Ch. 12 - Prob. 104CPCh. 12 - The partial pressures of an equilibrium mixture of...Ch. 12 - At 125C, KP = 0.25 for the reaction...Ch. 12 - A mixture of N2, H2, and NH3 is at equilibrium...Ch. 12 - Prob. 108CPCh. 12 - Prob. 109CPCh. 12 - Prob. 110CPCh. 12 - Prob. 111CPCh. 12 - A sample of N2O4(g) is placed in an empty cylinder...Ch. 12 - A sample of gaseous nitrosyl bromide (NOBr) was...Ch. 12 - Prob. 114CPCh. 12 - For the reaction NH3(g)+H2S(g)NH4HS(s) K = 400. at...Ch. 12 - Prob. 116IPCh. 12 - In a solution with carbon tetrachloride as the...Ch. 12 - Prob. 118IPCh. 12 - A gaseous material XY(g) dissociates to some...
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry: An Atoms First Approach
Chemistry
ISBN:9781305079243
Author:Steven S. Zumdahl, Susan A. Zumdahl
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781133611097
Author:Steven S. Zumdahl
Publisher:Cengage Learning
Text book image
Chemistry for Engineering Students
Chemistry
ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
Publisher:Cengage Learning
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Chemical Equilibria and Reaction Quotients; Author: Professor Dave Explains;https://www.youtube.com/watch?v=1GiZzCzmO5Q;License: Standard YouTube License, CC-BY