Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
8th Edition
ISBN: 9781305079373
Author: William L. Masterton, Cecile N. Hurley
Publisher: Cengage Learning
bartleby

Videos

Textbook Question
Chapter 12, Problem 69QAP

Consider the reaction:

A ( g ) + 2 B ( g ) + C ( s ) 2 D ( g ) At 25°C, only A, B, and C are present. The partial pressures of A, B, and D are given as PA, PB, and PD. Equilibrium is established 18 minutes after the reaction starts. Use terms is less than (LT), is greater than (GT), is equal to (EQ), or insufficient information (X) to answer the following questions.

(a) PD at 5 min PA.

(b) PA at 21 min PA at 27 min.

(c) PB at 7 min PB at 13 min.(d) After 20 min, more B is added. When equilibrium is reestablished, K before the addition K after the addition.

(e) After 20 min, the temperature is increased to 35°C. PA before the temperature increase PA after the temperature increase after equilibrium is reestablished.

Expert Solution
Check Mark
Interpretation Introduction

(a)

Interpretation:

The relation between the partial pressure of D and A at 5 min needs to be determined.

Concept introduction:

The system is said to be in equilibrium if the there is no change in the partial pressure or concentration of reactant and product takes place.

For a general equilibrium reaction as follows:

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

The expression for the equilibrium constant is represented as follows:

K=(PC)(PD)(PA)(PB)

Here, to calculate the equilibrium constant, the values of partial pressure of all the species in reactant and product side are required.

Answer to Problem 69QAP

Partial pressure of D gas at 5 min is greater than (GT) partial pressure of A gas.

Explanation of Solution

The reaction is given as follows:

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

Initially A, B and C are present in the system. If reaction starts at time t, the equilibrium will be established at t+18 min.

The change in partial pressure takes place as follows:

A(g)+2B(g)+C(s)2D(g)PA          PB          -            -PA-P     PB-2P     -            2P

At 5 min, the equilibrium is not established thus, the reaction is moving in the forward direction with same rate.

Since, the reaction is in forward direction, the pressure of product will be more than reactant.

Thus,

Partial pressure of D gas at 5 min is greater than partial pressure of A gas.

Expert Solution
Check Mark
Interpretation Introduction

(b)

Interpretation:

The relation between the partial pressure of A 12 min and at 27 min needs to be determined.

Concept introduction:

The system is said to be in equilibrium if the there is no change in the partial pressure or concentration of reactant and product takes place.

Answer to Problem 69QAP

The partial pressure of A gas at 21 min will be equal to (EQ) the partial pressure of B gas at 27 min.

Explanation of Solution

The reaction is given as follows:

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

Initially A, B and C are present in the system. If reaction starts at time t, the equilibrium will be established at t+18 min.

The change in partial pressure takes place as follows:

A(g)+2B(g)+C(s)2D(g)PA          PB          -            -PA-P     PB-2P     -            2P

Since, equilibrium is established at 18 min, after 18 min the pressure of all the gases become equal.

Thus, the partial pressure of A gas at 21 min will be equal to the partial pressure of B gas at 27 min.

Expert Solution
Check Mark
Interpretation Introduction

(c)

Interpretation:

The relation between the partial pressure of B at 7 min and at 13 min needs to be determined.

Concept introduction:

The system is said to be in equilibrium if the there is no change in the partial pressure or concentration of reactant and product takes place.

For a general equilibrium reaction as follows:

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

The expression for the equilibrium constant is represented as follows:

K=(PC)(PD)(PA)(PB)

Here, to calculate the equilibrium constant, the values of partial pressure of all the species in reactant and product side are required.

Answer to Problem 69QAP

The partial pressure of A gas at 7 min will be greater than (GT) partial pressure of B gas at 13 min.

Explanation of Solution

The reaction is given as follows:

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

Initially A, B and C are present in the system. If reaction starts at time t, the equilibrium will be established at t+18 min.

The change in partial pressure takes place as follows:

A(g)+2B(g)+C(s)2D(g)PA          PB          -            -PA-P     PB-2P     -            2P

The reaction is moving in forward direction, the partial pressure of B decreases with time till the system reaches equilibrium. Thus, partial pressure of A gas at 7 min will be greater than partial pressure of B gas at 13 min.

Expert Solution
Check Mark
Interpretation Introduction

(d)

Interpretation:

If more B is added in the system after 20 min, the change in K needs to be determined.

Concept introduction:

According to Le Chatelier’s principle, if at equilibrium, any change in temperature, concentration or pressure is applied to a system, the shift in equilibrium takes place to counteract the change.

Also, on compressing a system, the total pressure of the system increases thus, reaction shifts to decrease the total pressure and number of moles of gaseous species.

Opposite to this on expansion, the total pressure of the system decreases thus, reaction shifts to increase the total pressure and number of moles of gaseous species.

Only change in temperature can change the value of K, in other cases the value of K remains the same.

There are 3 conditions that can change the equilibrium direction in a system:

  1. Addition and removal of gaseous species.
  2. Expansion and compression of the system.
  3. Change in temperature of the system.

Answer to Problem 69QAP

The value of K before the addition of B will be less than (LT) K after the addition.

Explanation of Solution

The reaction is given as follows:

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

Initially A, B and C are present in the system. If reaction starts at time t, the equilibrium will be established at t+18 min.

The change in partial pressure takes place as follows:

A(g)+2B(g)+C(s)2D(g)PA          PB          -            -PA-P     PB-2P     -            2P

After 20 min, the system is in equilibrium and if more B is added the equilibrium will shift in the direction to decrease the partial pressure of B and the reaction moves in forward direction.

Thus, after the addition of B, K increases and the value of K before the addition of B will be less than K after the addition.

Expert Solution
Check Mark
Interpretation Introduction

(e)

Interpretation:

If temperature of the system is increased after 20 min, the change in partial pressure of A needs to be determined.

Concept introduction:

According to Le Chatelier’s principle, if at equilibrium, any change in temperature, concentration or pressure is applied to a system, the shift in equilibrium takes place to counteract the change.

Also, on compressing a system, the total pressure of the system increases thus, reaction shifts to decrease the total pressure and number of moles of gaseous species.

Opposite to this on expansion, the total pressure of the system decreases thus, reaction shifts to increase the total pressure and number of moles of gaseous species.

Only change in temperature can change the value of K, in other cases the value of K remains the same.

There are 3 conditions that can change the equilibrium direction in a system:

  1. Addition and removal of gaseous species.
  2. Expansion and compression of the system.
  3. Change in temperature of the system.

Answer to Problem 69QAP

The change in value of K or partial pressure of A cannot be determined due to increase in temperature. (X)

Explanation of Solution

The reaction is given as follows:

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

Initially A, B and C are present in the system. If reaction starts at time t, the equilibrium will be established at t+18 min.

The change in partial pressure takes place as follows:

A(g)+2B(g)+C(s)2D(g)PA          PB          -            -PA-P     PB-2P     -            2P

After 20 min, if temperature is increased to 35 C, the change in pressure can be determined due to insufficient information.

The value of K depends on the temperature but it also depends on the sign of the change in enthalpy of the reaction which depends on the type of reaction as if it is endothermic or exothermic reaction.

For exothermic reaction, the value of change in enthalpy is negative and for such reaction, the value of K decreases with increase in temperature.

And, for endothermic reaction, the value of change in enthalpy is positive and for such reaction, the value of K increases with increase in temperature.

Since, any information related to type of reaction or change in enthalpy, the change in value of K or partial pressure of A cannot be determined due to increase in temperature.

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!
Students have asked these similar questions
Carbon tetrachloride, an important industrial solvent, is prepared by the chlorination of methane at 850 K.CH4(g) + 4Cl2(g) ⟶ CCl4(g) + 4HCl(g)What is the equilibrium constant for the reaction at 850 K? Would the reaction vessel need to be heated or cooled to keep the temperature of the reaction constant?
Calculate the percentage change in the equilibrium constant K of the reaction     CH3OH(g) + NOCl(g) ↔ HCl(g) + CH3NO2(g)    When the total pressure p is increased from 1 bar to 2 bar at constant temperature.
The chemical equation for the water-gas shift reaction is CO (g) + H2O (g) ⇌ CO2 (g) + H2 (g). What is the value of molal equilibrium constant Kp at 700 K if the partial pressures in an equilibrium mixture at 700 K are 1.31 atm of CO, 10.0 atm of water, 6.12 atm of carbon dioxide, and 20.3 atm of hydrogen gas?

Chapter 12 Solutions

Chemistry: Principles and Reactions

Ch. 12 - Write an equation for an equilibrium system that...Ch. 12 - Write a chemical equation for an equilibrium...Ch. 12 - Consider the following reaction at 250C:...Ch. 12 - Consider the following reaction at 1000 C:...Ch. 12 - At 627C, K=0.76 for the reaction...Ch. 12 - At 800C, K=2.2104 for the following reaction...Ch. 12 - Prob. 17QAPCh. 12 - Given the following data at 25C...Ch. 12 - Given the following data at a certain temperature,...Ch. 12 - Consider the following hypothetical reactions and...Ch. 12 - When one mole of carbon disulfide gas reacts with...Ch. 12 - Calculate K for the formation of methyl alcohol at...Ch. 12 - Ammonium carbamate solid (NH4CO2NH2) decomposes at...Ch. 12 - Consider the decomposition at 25C of one mole of...Ch. 12 - Consider the decomposition of ammonium hydrogen...Ch. 12 - A sealed flask has 0.541 atm of SO3 at 1000 K. The...Ch. 12 - A gaseous reaction mixture contains 0.30 atm SO2,...Ch. 12 - For the system PCl5(g)PCl3(g)+Cl2(g)K is 26 at...Ch. 12 - The reversible reaction between hydrogen chloride...Ch. 12 - The reversible reaction between hydrogen chloride...Ch. 12 - A compound, X, decomposes at 131C according to the...Ch. 12 - Consider the following reaction at 75C:...Ch. 12 - Consider the reaction between nitrogen and steam:...Ch. 12 - At 500C, k for the for the formation of ammonia...Ch. 12 - At a certain temperature, K is 4.9 for the...Ch. 12 - At a certain temperature, K=0.29 for the...Ch. 12 - For the reaction N2(g)+2H2O(g)2NO(g)+2H2(g) K is...Ch. 12 - Nitrogen dioxide can decompose to nitrogen oxide...Ch. 12 - Consider the following reaction:...Ch. 12 - Consider the hypothetical reaction at 325C...Ch. 12 - At a certain temperature, the equilibrium constant...Ch. 12 - At 460C, the reaction SO2(g)+NO2(g)NO(g)+SO3(g)...Ch. 12 - Solid ammonium iodide decomposes to ammonia and...Ch. 12 - Consider the following decomposition at 80C....Ch. 12 - Hydrogen cyanide, a highly toxic gas, can...Ch. 12 - At 800 K, hydrogen iodide can decompose into...Ch. 12 - For the following reactions, predict whether the...Ch. 12 - Follow the directions of Question 47 for the...Ch. 12 - Consider the system SO3(g)SO2(g)+12 O2(g)H=98.9kJ...Ch. 12 - Consider the system...Ch. 12 - Predict the direction in which each of the...Ch. 12 - Predict the direction in which each of the...Ch. 12 - At a certain temperature, nitrogen and oxygen...Ch. 12 - Consider the following hypothetical reaction:...Ch. 12 - Iodine chloride decomposes at high temperatures to...Ch. 12 - Sulfur oxychloride, SO2Cl2, decomposes to sulfur...Ch. 12 - For the following reaction C(s)+2H2(g)CH4(g)...Ch. 12 - For the system 2SO3(g)2SO2(g)+O2(g) K=1.32 at 627....Ch. 12 - For a certain reaction, H is +33 kJ. What is the...Ch. 12 - Prob. 60QAPCh. 12 - Hemoglobin (Hb) binds to both oxygen and carbon...Ch. 12 - Mustard gas, used in chemical warfare in World War...Ch. 12 - Prob. 63QAPCh. 12 - For the decomposition of CaCO3 at 900C, K=1.04....Ch. 12 - Isopropyl alcohol is the main ingredient in...Ch. 12 - Consider the equilibrium H2(g)+S(s)H2S(g)When this...Ch. 12 - Prob. 67QAPCh. 12 - The following data apply to the unbalanced...Ch. 12 - Consider the reaction: A(g)+2B(g)+C(s)2D(g)At 25C,...Ch. 12 - For the reaction C(s)+CO2(g)2CO(g) K=168 at 1273...Ch. 12 - Consider the system A(g)+2B(g)+C(g)2D(g)at 25C. At...Ch. 12 - The graph below is similar to that of Figure 12.2....Ch. 12 - Prob. 73QAPCh. 12 - The figures below represent the following reaction...Ch. 12 - Prob. 75QAPCh. 12 - Prob. 76QAPCh. 12 - Consider the following reaction at a certain...Ch. 12 - Prob. 78QAPCh. 12 - Ammonia can decompose into its constituent...Ch. 12 - Hydrogen iodide gas decomposes to hydrogen gas and...Ch. 12 - For the system SO3(g)SO2(g)+12 O2(g)at 1000 K,...Ch. 12 - A student studies the equilibrium I2(g)2I(g)at a...Ch. 12 - At a certain temperature, the reaction...Ch. 12 - Benzaldehyde, a flavoring agent, is obtained by...Ch. 12 - Prob. 85QAPCh. 12 - Prob. 86QAP
Knowledge Booster
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
    Recommended textbooks for you
  • Introductory Chemistry: A Foundation
    Chemistry
    ISBN:9781285199030
    Author:Steven S. Zumdahl, Donald J. DeCoste
    Publisher:Cengage Learning
    Chemistry: The Molecular Science
    Chemistry
    ISBN:9781285199047
    Author:John W. Moore, Conrad L. Stanitski
    Publisher:Cengage Learning
    Chemistry: Principles and Reactions
    Chemistry
    ISBN:9781305079373
    Author:William L. Masterton, Cecile N. Hurley
    Publisher:Cengage Learning
  • Chemistry for Engineering Students
    Chemistry
    ISBN:9781337398909
    Author:Lawrence S. Brown, Tom Holme
    Publisher:Cengage Learning
    Chemistry by OpenStax (2015-05-04)
    Chemistry
    ISBN:9781938168390
    Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
    Publisher:OpenStax
    Chemistry: Principles and Practice
    Chemistry
    ISBN:9780534420123
    Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
    Publisher:Cengage Learning
  • Introductory Chemistry: A Foundation
    Chemistry
    ISBN:9781285199030
    Author:Steven S. Zumdahl, Donald J. DeCoste
    Publisher:Cengage Learning
    Chemistry: The Molecular Science
    Chemistry
    ISBN:9781285199047
    Author:John W. Moore, Conrad L. Stanitski
    Publisher:Cengage Learning
    Chemistry: Principles and Reactions
    Chemistry
    ISBN:9781305079373
    Author:William L. Masterton, Cecile N. Hurley
    Publisher:Cengage Learning
    Chemistry for Engineering Students
    Chemistry
    ISBN:9781337398909
    Author:Lawrence S. Brown, Tom Holme
    Publisher:Cengage Learning
    Chemistry by OpenStax (2015-05-04)
    Chemistry
    ISBN:9781938168390
    Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
    Publisher:OpenStax
    Chemistry: Principles and Practice
    Chemistry
    ISBN:9780534420123
    Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
    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