ACHIEVE/CHEMICAL PRINCIPLES ACCESS 1TERM
7th Edition
ISBN: 9781319399849
Author: ATKINS
Publisher: MAC HIGHER
expand_more
expand_more
format_list_bulleted
Question
Chapter 5, Problem 5J.4BST
Interpretation Introduction
Interpretation:
The response of equilibrium composition on lowering the temperature has to be identified.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionStudents have asked these similar questions
Consider the following equilibrium:
2 NO2(g) + Cl2(g) = 2 NO2CI(g)
1.67 bar of gaseous NO2 and 1.78 bar of gaseous C12 are added to a sealed
container at a fixed temperature. After the reaction has reached dynamic equilibrium
at this temperature, it was determined that the partial pressure of gaseous NO2CI is
1.06 bar. What is the equilibrium constant for the above reaction at this
temperature?
Nitrogen and hydrogen react to form ammonia, like this:
N2(g)+3H2(g)→ 2NH3(g)
Suppose a mixture of N2, H2 and NH3 has come to equilibrium in a closed reaction vessel. Predict what change, if any, the perturbations in the table below will cause in the composition of the mixture in the vessel. Also decide whether the equilibrium shifts to the right or left.
Consider the equilibrium system described by the chemical reaction below, which has a value of Kc equal to 4.10 × 10⁻⁴ at a high temperature. If 0.20 mol N₂ and 0.15 mol O₂ react in a 1.0 L vessel, what will the equilibrium concentration of O₂ be?
N₂(g) + O₂(g) ⇌ 2 NO(g)
Set up the expression for Qc and then evaluate it to determine the direction of the reaction. Do not combine or simplify terms.
Qc= (_______) / (_______) = (______)
Based on the given values and your value for Qc, set up ICE table in order to determine the unknown.
N₂(g) + O₂(g) ⇌ 2 NO(g)
Initial (M)
Change (M)
Equilibrium (M)
Based on your ICE table, set up the expression for Kc in order to determine the unknown. Do not combine or simplify terms.
Kc = (________) / (_______)= 4.10 × 10⁻⁴
Based on your ICE table and expression for Kc, solve for the…
Chapter 5 Solutions
ACHIEVE/CHEMICAL PRINCIPLES ACCESS 1TERM
Ch. 5 - Prob. 5A.1ASTCh. 5 - Prob. 5A.1BSTCh. 5 - Prob. 5A.2ASTCh. 5 - Prob. 5A.2BSTCh. 5 - Prob. 5A.3ASTCh. 5 - Prob. 5A.3BSTCh. 5 - Prob. 5A.1ECh. 5 - Prob. 5A.2ECh. 5 - Prob. 5A.3ECh. 5 - Prob. 5A.4E
Ch. 5 - Prob. 5A.5ECh. 5 - Prob. 5A.6ECh. 5 - Prob. 5A.7ECh. 5 - Prob. 5A.8ECh. 5 - Prob. 5A.11ECh. 5 - Prob. 5B.1ASTCh. 5 - Prob. 5B.1BSTCh. 5 - Prob. 5B.2ASTCh. 5 - Prob. 5B.2BSTCh. 5 - Prob. 5B.3ASTCh. 5 - Prob. 5B.3BSTCh. 5 - Prob. 5B.1ECh. 5 - Prob. 5B.2ECh. 5 - Prob. 5B.3ECh. 5 - Prob. 5B.5ECh. 5 - Prob. 5B.7ECh. 5 - Prob. 5C.1ASTCh. 5 - Prob. 5C.1BSTCh. 5 - Prob. 5C.2ASTCh. 5 - Prob. 5C.2BSTCh. 5 - Prob. 5C.3ASTCh. 5 - Prob. 5C.3BSTCh. 5 - Prob. 5C.1ECh. 5 - Prob. 5C.3ECh. 5 - Prob. 5C.4ECh. 5 - Prob. 5C.5ECh. 5 - Prob. 5C.6ECh. 5 - Prob. 5C.7ECh. 5 - Prob. 5C.8ECh. 5 - Prob. 5C.9ECh. 5 - Prob. 5C.10ECh. 5 - Prob. 5C.11ECh. 5 - Prob. 5C.12ECh. 5 - Prob. 5C.15ECh. 5 - Prob. 5C.16ECh. 5 - Prob. 5D.1ASTCh. 5 - Prob. 5D.1BSTCh. 5 - Prob. 5D.1ECh. 5 - Prob. 5D.2ECh. 5 - Prob. 5D.3ECh. 5 - Prob. 5D.4ECh. 5 - Prob. 5D.5ECh. 5 - Prob. 5D.6ECh. 5 - Prob. 5D.7ECh. 5 - Prob. 5D.8ECh. 5 - Prob. 5D.9ECh. 5 - Prob. 5D.10ECh. 5 - Prob. 5D.11ECh. 5 - Prob. 5D.12ECh. 5 - Prob. 5D.13ECh. 5 - Prob. 5D.14ECh. 5 - Prob. 5D.15ECh. 5 - Prob. 5D.16ECh. 5 - Prob. 5D.18ECh. 5 - Prob. 5D.19ECh. 5 - Prob. 5D.20ECh. 5 - Prob. 5E.1ASTCh. 5 - Prob. 5E.1BSTCh. 5 - Prob. 5E.2ASTCh. 5 - Prob. 5E.2BSTCh. 5 - Prob. 5E.1ECh. 5 - Prob. 5E.2ECh. 5 - Prob. 5E.11ECh. 5 - Prob. 5E.12ECh. 5 - Prob. 5F.1ASTCh. 5 - Prob. 5F.1BSTCh. 5 - Prob. 5F.2ASTCh. 5 - Prob. 5F.2BSTCh. 5 - Prob. 5F.3ASTCh. 5 - Prob. 5F.3BSTCh. 5 - Prob. 5F.4ASTCh. 5 - Prob. 5F.4BSTCh. 5 - Prob. 5F.5ASTCh. 5 - Prob. 5F.5BSTCh. 5 - Prob. 5F.1ECh. 5 - Prob. 5F.2ECh. 5 - Prob. 5F.3ECh. 5 - Prob. 5F.5ECh. 5 - Prob. 5F.7ECh. 5 - Prob. 5F.9ECh. 5 - Prob. 5F.10ECh. 5 - Prob. 5F.11ECh. 5 - Prob. 5F.12ECh. 5 - Prob. 5F.13ECh. 5 - Prob. 5F.14ECh. 5 - Prob. 5F.15ECh. 5 - Prob. 5F.16ECh. 5 - Prob. 5G.1ASTCh. 5 - Prob. 5G.1BSTCh. 5 - Prob. 5G.2ASTCh. 5 - Prob. 5G.2BSTCh. 5 - Prob. 5G.3ASTCh. 5 - Prob. 5G.3BSTCh. 5 - Prob. 5G.4ASTCh. 5 - Prob. 5G.4BSTCh. 5 - Prob. 5G.5ASTCh. 5 - Prob. 5G.5BSTCh. 5 - Prob. 5G.1ECh. 5 - Prob. 5G.2ECh. 5 - Prob. 5G.3ECh. 5 - Prob. 5G.4ECh. 5 - Prob. 5G.7ECh. 5 - Prob. 5G.8ECh. 5 - Prob. 5G.9ECh. 5 - Prob. 5G.11ECh. 5 - Prob. 5G.12ECh. 5 - Prob. 5G.13ECh. 5 - Prob. 5G.14ECh. 5 - Prob. 5G.15ECh. 5 - Prob. 5G.16ECh. 5 - Prob. 5G.17ECh. 5 - Prob. 5G.19ECh. 5 - Prob. 5G.20ECh. 5 - Prob. 5G.21ECh. 5 - Prob. 5G.22ECh. 5 - Prob. 5H.1ASTCh. 5 - Prob. 5H.1BSTCh. 5 - Prob. 5H.2ASTCh. 5 - Prob. 5H.2BSTCh. 5 - Prob. 5H.1ECh. 5 - Prob. 5H.2ECh. 5 - Prob. 5H.3ECh. 5 - Prob. 5H.4ECh. 5 - Prob. 5H.5ECh. 5 - Prob. 5H.6ECh. 5 - Prob. 5I.1ASTCh. 5 - Prob. 5I.1BSTCh. 5 - Prob. 5I.2ASTCh. 5 - Prob. 5I.2BSTCh. 5 - Prob. 5I.3ASTCh. 5 - Prob. 5I.3BSTCh. 5 - Prob. 5I.4ASTCh. 5 - Prob. 5I.4BSTCh. 5 - Prob. 5I.1ECh. 5 - Prob. 5I.2ECh. 5 - Prob. 5I.3ECh. 5 - Prob. 5I.4ECh. 5 - Prob. 5I.5ECh. 5 - Prob. 5I.6ECh. 5 - Prob. 5I.7ECh. 5 - Prob. 5I.9ECh. 5 - Prob. 5I.10ECh. 5 - Prob. 5I.11ECh. 5 - Prob. 5I.12ECh. 5 - Prob. 5I.13ECh. 5 - Prob. 5I.14ECh. 5 - Prob. 5I.15ECh. 5 - Prob. 5I.16ECh. 5 - Prob. 5I.17ECh. 5 - Prob. 5I.18ECh. 5 - Prob. 5I.19ECh. 5 - Prob. 5I.20ECh. 5 - Prob. 5I.21ECh. 5 - Prob. 5I.22ECh. 5 - Prob. 5I.23ECh. 5 - Prob. 5I.24ECh. 5 - Prob. 5I.25ECh. 5 - Prob. 5I.26ECh. 5 - Prob. 5I.27ECh. 5 - Prob. 5I.28ECh. 5 - Prob. 5I.29ECh. 5 - Prob. 5I.30ECh. 5 - Prob. 5I.32ECh. 5 - Prob. 5I.33ECh. 5 - Prob. 5I.34ECh. 5 - Prob. 5I.35ECh. 5 - Prob. 5I.36ECh. 5 - Prob. 5J.1ASTCh. 5 - Prob. 5J.1BSTCh. 5 - Prob. 5J.3ASTCh. 5 - Prob. 5J.3BSTCh. 5 - Prob. 5J.4ASTCh. 5 - Prob. 5J.4BSTCh. 5 - Prob. 5J.5ASTCh. 5 - Prob. 5J.5BSTCh. 5 - Prob. 5J.1ECh. 5 - Prob. 5J.2ECh. 5 - Prob. 5J.3ECh. 5 - Prob. 5J.4ECh. 5 - Prob. 5J.5ECh. 5 - Prob. 5J.6ECh. 5 - Prob. 5J.9ECh. 5 - Prob. 5J.10ECh. 5 - Prob. 5J.11ECh. 5 - Prob. 5J.12ECh. 5 - Prob. 5J.13ECh. 5 - Prob. 5J.17ECh. 5 - Prob. 5.1ECh. 5 - Prob. 5.2ECh. 5 - Prob. 5.3ECh. 5 - Prob. 5.4ECh. 5 - Prob. 5.5ECh. 5 - Prob. 5.6ECh. 5 - Prob. 5.7ECh. 5 - Prob. 5.8ECh. 5 - Prob. 5.9ECh. 5 - Prob. 5.10ECh. 5 - Prob. 5.11ECh. 5 - Prob. 5.12ECh. 5 - Prob. 5.13ECh. 5 - Prob. 5.14ECh. 5 - Prob. 5.15ECh. 5 - Prob. 5.16ECh. 5 - Prob. 5.17ECh. 5 - Prob. 5.19ECh. 5 - Prob. 5.23ECh. 5 - Prob. 5.24ECh. 5 - Prob. 5.25ECh. 5 - Prob. 5.26ECh. 5 - Prob. 5.27ECh. 5 - Prob. 5.28ECh. 5 - Prob. 5.29ECh. 5 - Prob. 5.30ECh. 5 - Prob. 5.31ECh. 5 - Prob. 5.32ECh. 5 - Prob. 5.33ECh. 5 - Prob. 5.35ECh. 5 - Prob. 5.37ECh. 5 - Prob. 5.38ECh. 5 - Prob. 5.41ECh. 5 - Prob. 5.43ECh. 5 - Prob. 5.44ECh. 5 - Prob. 5.45ECh. 5 - Prob. 5.46ECh. 5 - Prob. 5.47ECh. 5 - Prob. 5.49ECh. 5 - Prob. 5.51ECh. 5 - Prob. 5.53ECh. 5 - Prob. 5.55ECh. 5 - Prob. 5.57ECh. 5 - Prob. 5.58ECh. 5 - Prob. 5.61ECh. 5 - Prob. 5.62E
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
- Describe a nonchemical system that is not in equilibrium, and explain why equilibrium has not been achieved.arrow_forwardShow that the complete chemical equation, the total ionic equation, and the net ionic equation for the reaction represented by the equation KI(aq)+I2(aq)KI3(aq) give the same expression for the reaction quotient. KI3 is composed of the ions K+ and I3-.arrow_forwardActually, the carbon in CO2(g) is thermodynamically unstable with respect to the carbon in calcium carbonate(limestone). Verify this by determining the standardGibbs free energy change for the reaction of lime,CaO(s), with CO2(g) to make CaCO3(s).arrow_forward
- Hydrogen gas and iodine gas react to form hydrogen iodide. If 0.500 mol H2 and 1.00 mol I2 are placed in a closed 10.0-L vessel, what is the mole fraction of HI in the mixture when equilibrium is reached at 205C? Use data from Appendix C and any reasonable approximations to obtain K.arrow_forwardConsider a metal ion A2+ and its nitrate salt, In an experiment, 35.00 mL of a 0.217 M solution of A(NO3)2 is made to react with 25.00 mL of 0.195 M NaOH. A precipitate, A(OH)2, forms. Along with the precipitation, the temperature increases from 24.8C to 28.2C. What is H for the precipitation of A(OH)2? The following assumptions can be made. • The density of the solution is 1.00 g/mL. • Volumes are additive. • The specific heat of the solution is 4.18 J/g C.arrow_forwardAt a certain temperature, K=0.29 for the decomposition of two moles of iodine trichloride, ICl3(s), to chlorine and iodine gases. The partial pressure of chlorine gas at equilibrium is three times that of iodine gas. What are the partial pressures of iodine and chlorine at equilibrium?arrow_forward
- 5.12. True or false: If all the partial pressures of reactants and products drop by half, the value of Q drops by half. Give an example of a chemical reaction to support your answer.arrow_forward5.49. Consider the following equilibrium: What is the effect on the equilibrium of each of the following changes? (You may need to calculate some standard enthalpy or Gibbs energy changes to answer these.) (a) The pressure is increased by decreasing the volume. (b) The temperature is decreased. (c) The pressure is increased by the addition of nitrogen gas, .arrow_forwardDistinguish between the terms equilibrium constant and reaction quotient. When Q = K, what does this say about a reaction? When Q K, what does this say about a reaction? When Q K. what does this say about a reaction?arrow_forward
- Predict whether each of the following processes results in an increase in entropy in the system. (Define reactants and products as the system.) (a) Water vapor condenses to liquid water at 90 C and 1 atm pressure. (b) The exothermic reaction of Na(s) and Cl2(g) forms NaCl(s). (c) The endothermic reaction of H2 and I2 produces an equilibrium mixture of H2(g), I2(g), and HI(g). (d) Solid NaCl dissolves in water forming a saturated solution.arrow_forwardCalculate H when a 38-g sample of glucose, C6H12O6(s), burns in excess O2(g) to form CO2(g) and H2O() in a reaction at constant pressure and 298.15 K.arrow_forwardAt 1 atm and 25 C, NO2 with an initial concentration of 1.00 M is 3.3103 decomposed into NO and O2. Calculate the value of the equilibrium constant for the reaction. 2NO2(g)2NO(g)+O2(g)arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Principles of Modern ChemistryChemistryISBN:9781305079113Author:David W. Oxtoby, H. Pat Gillis, Laurie J. ButlerPublisher:Cengage LearningChemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage LearningChemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage Learning
- Chemistry & Chemical ReactivityChemistryISBN:9781133949640Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage LearningGeneral Chemistry - Standalone book (MindTap Cour...ChemistryISBN:9781305580343Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; DarrellPublisher:Cengage LearningChemistry by OpenStax (2015-05-04)ChemistryISBN:9781938168390Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark BlaserPublisher:OpenStax
Principles of Modern Chemistry
Chemistry
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:Cengage Learning
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
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
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781133949640
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
General Chemistry - Standalone book (MindTap Cour...
Chemistry
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
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
Chemical Equilibria and Reaction Quotients; Author: Professor Dave Explains;https://www.youtube.com/watch?v=1GiZzCzmO5Q;License: Standard YouTube License, CC-BY