General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
bartleby

Videos

Question
Book Icon
Chapter 15, Problem 15.67QP

a)

Interpretation Introduction

Interpretation:

An expression for KP in terms of α and P, total pressure for the given reaction has to be derived.

Concept introduction:

Law of mass action: The rate of chemical reaction is directly proportional to the product of concentrations of reactant to products.

aA+bBcC+dDKeq=[C]c[D]d[A]a[B]bforaqueousKeq=(PC)c(PD)d(PA)a(PB)bforgases

Multiple equilibria: If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions.

A+   B    C + D   Kc'C+   D    E + F    Kc''_overallreaction:A+   B   E + FKc_

The equilibrium constant for two separate equilibrium constants are,

Kc'=[C][D][A][B]andKc''=[E][F][C][D]

For overall reaction, the equilibrium constant Kc is,

Kc'Kc''=[C][D][A][B]×[E][F][C][D]=[E][F][A][B]

Therefore, Kc=Kc'×Kc''

Calculating equilibrium concentration:

  • Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration.
  • Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x.
  • Having solved for x, calculate the equilibrium concentrations of all species.

a)

Expert Solution
Check Mark

Explanation of Solution

For the given reaction,

     N2O4 (g)2NO2(g)

Given: Initially 1.0moleofN2O4; at equilibrium α moleofN2O4 has dissociated to form NO2.

Construct ICE table as follows,

N2O4 (g)2NO2(g)Initial:1M0Change:α+2αEquilibrium:1α+2α

The total moles in the system = (molesofN2O4+molesofNO2)= [(1α)+2α]=1+α. If the total pressure in the system is P, then:

PN2O4=1α1+αPandPNO2=2α1+αP

KP=(PNO22)PN2O4=(2α1+α)2P2(1α1+α)P

KP=(PNO22)PN2O4=4α21-α2P

 Hence, expression for KP in terms of α and P is 4α21-α2P

b)

Interpretation Introduction

Interpretation:

The shift in equilibrium due to an increase in P has to be predicted and the prediction has to be agreed with Lee-Chatelier’s principle.

Concept introduction:

Law of mass action: The rate of chemical reaction is directly proportional to the product of concentrations of reactant to products.

aA+bBcC+dDKeq=[C]c[D]d[A]a[B]bforaqueousKeq=(PC)c(PD)d(PA)a(PB)bforgases

Multiple equilibria: If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions.

A+   B    C + D   Kc'C+   D    E + F    Kc''_overallreaction:A+   B   E + FKc_

The equilibrium constant for two separate equilibrium constants are,

Kc'=[C][D][A][B]andKc''=[E][F][C][D]

For overall reaction, the equilibrium constant Kc is,

Kc'Kc''=[C][D][A][B]×[E][F][C][D]=[E][F][A][B]

Therefore, Kc=Kc'×Kc''

Calculating equilibrium concentration:

  • Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration.
  • Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x.
  • Having solved for x, calculate the equilibrium concentrations of all species.

b)

Expert Solution
Check Mark

Explanation of Solution

For the given reaction,

     N2O4 (g)2NO2(g)

Given: Initially 1.0moleofN2O4; at equilibrium α moleofN2O4 has dissociated to form NO2. The obtained expression for KP in terms of α and P is 4α21-α2P

Rearranging KP expression:

KP=21-α2P2P=KP-α2KPα2=KP4P+KPα=KP4P+KP

KP is a constant (at constant temperature). Thus, as P increases α must decreases, indicating that the system shifts to the left. This argument would predict based on Le-Chatelier’s principle.

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 15 Solutions

General Chemistry

Ch. 15.4 - Practice Exercise At 430°C. the equilibrium...Ch. 15.4 - Prob. 2PECh. 15.4 - Prob. 1RCCh. 15.4 - Prob. 2RCCh. 15.4 - Prob. 3PECh. 15 - Prob. 15.1QPCh. 15 - 15.2 Explain the difference between physical...Ch. 15 - 15.3 Briefly describe the importance of...Ch. 15 - 15.4 Consider the equilibrium system 3A ⇌ B....Ch. 15 - 15.5 Define homogeneous equilibrium and...Ch. 15 - Prob. 15.6QPCh. 15 - 15.7 Write equilibrium constant expressions for...Ch. 15 - 15.8 Write the expressions for the equilibrium...Ch. 15 - 15.9 Write the equilibrium constant expressions...Ch. 15 - 15.10 Write the equation relating Kc and KP and...Ch. 15 - Prob. 15.11QPCh. 15 - Prob. 15.12QPCh. 15 - 15.13 The equilibrium constant (Kc) for the...Ch. 15 - Prob. 15.14QPCh. 15 - 15.15 What is the KP at 1273°C for the...Ch. 15 - 15.16 The equilibrium constant KP for the...Ch. 15 - Prob. 15.17QPCh. 15 - Prob. 15.18QPCh. 15 - Prob. 15.19QPCh. 15 - Prob. 15.20QPCh. 15 - Prob. 15.21QPCh. 15 - 15.22 Ammonium carbamate, NH4CO2NH2, decomposes...Ch. 15 - Prob. 15.23QPCh. 15 - 15.24 Pure phosgene gas (COCl2), 3.00 × 10−2...Ch. 15 - Prob. 15.25QPCh. 15 - 15.26 A 2.50-mol quantity of NOCl was initially...Ch. 15 - 15.27 Define reaction quotient. How does it...Ch. 15 - Prob. 15.28QPCh. 15 - Prob. 15.29QPCh. 15 - Prob. 15.30QPCh. 15 - Prob. 15.31QPCh. 15 - Prob. 15.32QPCh. 15 - Prob. 15.33QPCh. 15 - Prob. 15.34QPCh. 15 - Prob. 15.35QPCh. 15 - Prob. 15.36QPCh. 15 - Prob. 15.37QPCh. 15 - Prob. 15.38QPCh. 15 - Prob. 15.39QPCh. 15 - Prob. 15.40QPCh. 15 - Prob. 15.41QPCh. 15 - Prob. 15.42QPCh. 15 - Prob. 15.43QPCh. 15 - Prob. 15.44QPCh. 15 - Prob. 15.45QPCh. 15 - 15.46 What effect does an increase in pressure...Ch. 15 - Prob. 15.47QPCh. 15 - Prob. 15.48QPCh. 15 - 15.49 Consider the reaction Comment on the...Ch. 15 - Prob. 15.50QPCh. 15 - Prob. 15.51QPCh. 15 - Prob. 15.53QPCh. 15 - Prob. 15.54QPCh. 15 - Prob. 15.55QPCh. 15 - Prob. 15.56QPCh. 15 - Prob. 15.57QPCh. 15 - 15.58 Baking soda (sodium bicarbonate) undergoes...Ch. 15 - 15.59 Consider the following reaction at...Ch. 15 - Prob. 15.60QPCh. 15 - Prob. 15.61QPCh. 15 - Prob. 15.62QPCh. 15 - Prob. 15.64QPCh. 15 - Prob. 15.65QPCh. 15 - Prob. 15.66QPCh. 15 - Prob. 15.67QPCh. 15 - Prob. 15.68QPCh. 15 - Prob. 15.69QPCh. 15 - Prob. 15.70QPCh. 15 - Prob. 15.71QPCh. 15 - Prob. 15.72QPCh. 15 - Prob. 15.73QPCh. 15 - Prob. 15.74QPCh. 15 - Prob. 15.75QPCh. 15 - Prob. 15.76QPCh. 15 - Prob. 15.78QPCh. 15 - Prob. 15.79QPCh. 15 - Prob. 15.81QPCh. 15 - Prob. 15.82QPCh. 15 - Prob. 15.83QPCh. 15 - Prob. 15.84QPCh. 15 - Prob. 15.85QPCh. 15 - Prob. 15.86QPCh. 15 - Prob. 15.89QPCh. 15 - Prob. 15.90QPCh. 15 - Prob. 15.91QPCh. 15 - Prob. 15.92QPCh. 15 - Prob. 15.93QPCh. 15 - 15.94 Consider the decomposition of ammonium...Ch. 15 - Prob. 15.95QPCh. 15 - 15.96 In 1899 the German chemist Ludwig Mond...Ch. 15 - Prob. 15.98QPCh. 15 - Prob. 15.99QPCh. 15 - Prob. 15.100QPCh. 15 - Prob. 15.101QPCh. 15 - Prob. 15.102QPCh. 15 - Prob. 15.103SPCh. 15 - Prob. 15.104SPCh. 15 - Prob. 15.105SPCh. 15 - Prob. 15.106SPCh. 15 - Prob. 15.107SPCh. 15 - Prob. 15.110SPCh. 15 - Prob. 15.111SPCh. 15 - Prob. 15.112SP
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.
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
Chemical Equilibria and Reaction Quotients; Author: Professor Dave Explains;https://www.youtube.com/watch?v=1GiZzCzmO5Q;License: Standard YouTube License, CC-BY