Introduction to Chemistry
4th Edition
ISBN: 9780073523002
Author: Rich Bauer, James Birk Professor Dr., Pamela S. Marks
Publisher: McGraw-Hill Education
expand_more
expand_more
format_list_bulleted
Question
Chapter 12, Problem 103QP
Interpretation Introduction
Interpretation:
The effect of temperature change on the value of the equilibrium constant when the temperature change causes the reaction to shift to the right is to be determined.
Concept Introduction:
According to Le Chatelier`s principle, the equilibrium can shift, if a system at equilibrium is disturbed.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionChapter 12 Solutions
Introduction to Chemistry
Ch. 12 - Prob. 1QCCh. 12 - Prob. 2QCCh. 12 - Prob. 3QCCh. 12 - Prob. 4QCCh. 12 - Prob. 5QCCh. 12 - Prob. 6QCCh. 12 - Prob. 1PPCh. 12 - Prob. 2PPCh. 12 - Prob. 3PPCh. 12 - Prob. 4PP
Ch. 12 - Prob. 5PPCh. 12 - Prob. 6PPCh. 12 - Prob. 7PPCh. 12 - Prob. 8PPCh. 12 - Prob. 9PPCh. 12 - Prob. 10PPCh. 12 - Consider the following equilibrium:...Ch. 12 - Prob. 12PPCh. 12 - Prob. 1QPCh. 12 - Match the key terms with the descriptions...Ch. 12 - Prob. 3QPCh. 12 - Prob. 4QPCh. 12 - Prob. 5QPCh. 12 - Prob. 6QPCh. 12 - Prob. 7QPCh. 12 - Prob. 8QPCh. 12 - Prob. 9QPCh. 12 - Prob. 10QPCh. 12 - Prob. 11QPCh. 12 - Prob. 12QPCh. 12 - Prob. 13QPCh. 12 - Prob. 14QPCh. 12 - Prob. 15QPCh. 12 - Prob. 16QPCh. 12 - Prob. 17QPCh. 12 - Prob. 18QPCh. 12 - Prob. 19QPCh. 12 - Prob. 20QPCh. 12 - Prob. 21QPCh. 12 - Prob. 22QPCh. 12 - Prob. 23QPCh. 12 - Prob. 24QPCh. 12 - Prob. 25QPCh. 12 - Prob. 26QPCh. 12 - Prob. 27QPCh. 12 - Prob. 28QPCh. 12 - Prob. 29QPCh. 12 - Prob. 30QPCh. 12 - Prob. 31QPCh. 12 - Prob. 32QPCh. 12 - Prob. 33QPCh. 12 - Prob. 34QPCh. 12 - Prob. 35QPCh. 12 - Prob. 36QPCh. 12 - Prob. 37QPCh. 12 - Prob. 38QPCh. 12 - Prob. 39QPCh. 12 - Prob. 40QPCh. 12 - Prob. 41QPCh. 12 - Prob. 42QPCh. 12 - Prob. 43QPCh. 12 - Prob. 44QPCh. 12 - Prob. 45QPCh. 12 - Prob. 46QPCh. 12 - Prob. 47QPCh. 12 - Prob. 48QPCh. 12 - Prob. 49QPCh. 12 - Prob. 50QPCh. 12 - Prob. 51QPCh. 12 - Prob. 52QPCh. 12 - Prob. 53QPCh. 12 - Prob. 54QPCh. 12 - Prob. 55QPCh. 12 - Prob. 56QPCh. 12 - Prob. 57QPCh. 12 - Prob. 58QPCh. 12 - Prob. 59QPCh. 12 - Prob. 60QPCh. 12 - Prob. 61QPCh. 12 - Prob. 62QPCh. 12 - Prob. 63QPCh. 12 - Prob. 64QPCh. 12 - Prob. 65QPCh. 12 - Prob. 66QPCh. 12 - Prob. 67QPCh. 12 - Prob. 68QPCh. 12 - Prob. 69QPCh. 12 - Prob. 70QPCh. 12 - Prob. 71QPCh. 12 - Prob. 72QPCh. 12 - Prob. 73QPCh. 12 - Prob. 74QPCh. 12 - Prob. 75QPCh. 12 - Prob. 76QPCh. 12 - Prob. 77QPCh. 12 - Prob. 78QPCh. 12 - Prob. 79QPCh. 12 - Prob. 80QPCh. 12 - Prob. 81QPCh. 12 - Prob. 82QPCh. 12 - Prob. 83QPCh. 12 - Prob. 84QPCh. 12 - Prob. 85QPCh. 12 - Prob. 86QPCh. 12 - Prob. 87QPCh. 12 - Prob. 88QPCh. 12 - Prob. 89QPCh. 12 - Prob. 90QPCh. 12 - Prob. 91QPCh. 12 - Prob. 92QPCh. 12 - Prob. 93QPCh. 12 - Prob. 94QPCh. 12 - Prob. 95QPCh. 12 - Prob. 96QPCh. 12 - Prob. 97QPCh. 12 - Prob. 98QPCh. 12 - Prob. 99QPCh. 12 - Prob. 100QPCh. 12 - Prob. 101QPCh. 12 - Prob. 102QPCh. 12 - Prob. 103QPCh. 12 - Prob. 104QPCh. 12 - Prob. 105QPCh. 12 - Prob. 106QPCh. 12 - Prob. 107QPCh. 12 - Prob. 108QPCh. 12 - Prob. 109QPCh. 12 - Prob. 110QPCh. 12 - Prob. 111QPCh. 12 - Prob. 112QPCh. 12 - Prob. 113QPCh. 12 - Prob. 114QPCh. 12 - Prob. 115QPCh. 12 - Prob. 116QPCh. 12 - Prob. 117QPCh. 12 - Prob. 118QPCh. 12 - Prob. 119QPCh. 12 - Prob. 120QPCh. 12 - Prob. 121QPCh. 12 - Prob. 122QPCh. 12 - Prob. 123QPCh. 12 - Prob. 124QPCh. 12 - Prob. 125QPCh. 12 - Prob. 126QPCh. 12 - Prob. 127QPCh. 12 - Prob. 128QPCh. 12 - Prob. 129QPCh. 12 - Prob. 130QPCh. 12 - Prob. 131QPCh. 12 - Prob. 132QPCh. 12 - Prob. 133QPCh. 12 - Prob. 134QPCh. 12 - Prob. 135QPCh. 12 - Prob. 136QPCh. 12 - Prob. 137QPCh. 12 - Prob. 138QPCh. 12 - Prob. 139QPCh. 12 - Prob. 140QPCh. 12 - Prob. 141QPCh. 12 - Prob. 142QPCh. 12 - Prob. 143QP
Knowledge Booster
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
- Suppose a reaction has the equilibrium constant K = 1.3 108. What does the magnitude of this constant tell you about the relative concentrations of products and reactants that will be present once equilibrium is reached? Is this reaction likely to be a good source of the products?arrow_forwardWrite a chemical equation for an equilibrium system that would lead to the following expressions (ad) for K. (a) K=(PH2S)2 (PO2)3(PSO2)2 (PH2O)2 (b) K=(PF2)1/2 (PI2)1/2PIF (c) K=[ Cl ]2(Pcl2)[ Br ]2 (d) K=(PNO)2 (PH2O)4 [ Cu2+ ]3[ NO3 ]2 [ H+ ]8arrow_forwardMethanol, a common laboratory solvent, poses a threat of blindness or death if consumed in sufficient amounts. Once in the body, the substance is oxidized to produce formaldehyde (embalming fluid) and eventually formic acid. Both of these substances are also toxic in varying levels. The equilibrium between methanol and formaldehyde can be described as follows: CH3OH(aq)H2CO(aq)+H2(aq) Assuming the value of K for this reaction is 3.7 1010, what are the equilibrium concentrations of each species if you start with a 1.24 M solution of methanol? What will happen to the concentration of methanol as the formaldehyde is further converted to formic acid?arrow_forward
- 12.103 Methanol, CH3OH, can be produced by the reaction of CO with H2, with the liberation of heat. All species in the reaction are gaseous. What effect will each of the following have on the equilibrium concentration of CO? (a) Pressure is increased, (b) volume of the reaction container is decreased, (c) heat is added, (d) the concentration of CO is increased, (e) some methanol is removed from the container, and (f) H2 is added.arrow_forwardBased on the diagrams, chemical reaction, and reaction conditions depicted in Problem 9-81, for which of the diagrams is the numerical value of the equilibrium constant the smallest?arrow_forwardWrite a balanced chemical equation for a totally gaseous equilibrium system that would lead to the following equilibrium constant expression. Keq=[N2]2[H2O]6[NH3]4[O2]3arrow_forward
- At room temperature, the equilibrium constant Kc for the reaction 2 NO(g) ⇌ N2(g) + O2(g) is 1.4 × 1030. Is this reaction product-favored or reactant-favored? Explain your answer. In the atmosphere at room temperature the concentration of N2 is 0.33 mol/L, and the concentration of O2 is about 25% of that value. Calculate the equilibrium concentration of NO in the atmosphere produced by the reaction of N2 and O2. How does this affect your answer to Question 11?arrow_forwardAn equilibrium involving the carbonate and bicarbonate ions exists in natural waters: HCO5_(aq) «=* H+(aq) + COf-(aq) Assuming that the reactions in both directions are elementary' processes: Write rate expressions for the forward and reverse reactions. Write an expression for the equilibrium constant based on the rates of the forward and reverse reactions.arrow_forwardWrite equilibrium constant expressions for the following generalized reactions. a. 2X(g)+3Y(g)2Z(g) b. 2X(g)+3Y(s)2Z(g) c. 2X(s)+3Y(s)2Z(g) d. 2X(g)+3Y(g)2Z(s)arrow_forward
- . For the reaction 3O2(g)2O3(g)The equilibrium constant, K, has the value 1.121054at a particular temperature. a. What does the very small equilibrium constant indicate about the extent to which oxygen gas, O2(g), is converted to ozone gas, O3(g), at this temperature? b. If the equilibrium mixture is analyzed and [O2(g)]is found to be 3.04102M, what is the concentration of O3(g) in the mixture’?arrow_forwardUse Le Chteliers principle to predict the direction of the equilibrium shift in the following equilibria when the indicated stress is applied: a. Ag+(aq)+Cl(aq)AgCl(s); some Ag+ is removed. b. 2HI(g)+heatH2(g)+I2(g); the system is heated. c. 6Cu(s)+N2(g)+heat2Cu3N(s); the system is cooled and some N2 is removed.arrow_forwardThe value of the equilibrium constant, K, is dependent on which of the following? (There may be more than one answer.) a. the initial concentrations of the reactants b. the initial concentrations of the products c. the temperature of the system d. the nature of the reactants and products Explain.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage LearningChemistry for Engineering StudentsChemistryISBN:9781337398909Author:Lawrence S. Brown, Tom HolmePublisher:Cengage LearningChemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage Learning
- World of Chemistry, 3rd editionChemistryISBN:9781133109655Author:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCostePublisher:Brooks / Cole / Cengage LearningChemistry: Matter and ChangeChemistryISBN:9780078746376Author:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl WistromPublisher:Glencoe/McGraw-Hill School Pub CoChemistry for Today: General, Organic, and Bioche...ChemistryISBN:9781305960060Author:Spencer L. Seager, Michael R. Slabaugh, Maren S. HansenPublisher:Cengage Learning
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:Cengage Learning
Chemistry for Engineering Students
Chemistry
ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
Publisher:Cengage Learning
Chemistry: Principles and Practice
Chemistry
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Cengage Learning
World of Chemistry, 3rd edition
Chemistry
ISBN:9781133109655
Author:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCoste
Publisher:Brooks / Cole / Cengage Learning
Chemistry: Matter and Change
Chemistry
ISBN:9780078746376
Author:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Publisher:Glencoe/McGraw-Hill School Pub Co
Chemistry for Today: General, Organic, and Bioche...
Chemistry
ISBN:9781305960060
Author:Spencer L. Seager, Michael R. Slabaugh, Maren S. Hansen
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