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.3AST
Interpretation Introduction
Interpretation:
The response of compression on equilibrium composition for given equation
Expert Solution & Answer
![Check Mark](/static/check-mark.png)
Want to see the full answer?
Check out a sample textbook solution![Blurred answer](/static/blurred-answer.jpg)
Students have asked these similar questions
The equilibrium constant for the reaction
NO(g) + 1/2 O2(g) NO2(g)
has a value of Kc = 1.23 at a certain temperature. What is the value of Kc for the reaction
2 NO2(g)2 NO(g) + O2(g) ?
Write the equilibrium expression for the reaction
CO (g) + 3 H2 (g) CH4 (g) + H2O (g).
› Calculate the value of the equilibrium constant, Kỵ, for the reaction
Q(g) + X(g) — 2M(g) + N(g)
given that
Kc
=
M(g) — Z(g)
6R(g)
2 N(g) + 4 Z(g)
3 X(g) + 3 Q(g) = 9R(g)
Kc1 = 3.48
Kc2
Kc3
= 0.587
=
= 13.6
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_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_forwardWrite an equation for an equilibrium system that would lead to the following expressions (ac) for K. (a) K=(Pco)2 (PH2)5(PC2H6)(PH2O)2 (b) K=(PNH3)4 (PO2)5(PNO)4 (PH2O)6 (c) K=[ ClO3 ]2 [ Mn2+ ]2(Pcl2)[ MNO4 ]2 [ H+ ]4 ; liquid water is a productarrow_forward
- Nitrogen and oxygen react at high temperatures.(a) Write the expression for the equilibrium constant (Kc) for the reversible reactionN2(g) + O2(g) ⇌ 2NO(g) ΔH = 181 kJarrow_forwardCalculate the value of the equilibrium constant, Kc, for the reaction Q(g) +X(g) 2M(g) + N(g) given that Kc = M(g)Z(g) Kcl = 3.14 6R(g) 2N(g) + 4Z(g) Kc2 = 0.561 3X(g) +3Q(g) 9R(g) Kc3 = 11.9arrow_forwardCalculate the value of the equilibrium constant (Keg) for the following reaction: 2 CO (g) + 6 H2 (g) → 2 CH4 (g) + 2 H2O (g) The following two equilibrium reactions should be helpful in determining this value: CO (g) + 2 H2S (g) <→ CS2 (g) + H2O (g) + H2 (g) ½ CH4 (g) + H2S (g) <→ ½ CS2 (g) + 2 H2 (g) Reaction 1: K1 = 1.3 x 105 Reaction 2: K2 = 180 (A) 2.3 x 107 (B) 1.9 x108 (C) (D) 16.1 722 (E) 1.5 x 10-8arrow_forward
- What is the symbolic expression for the equilibrium constant for the reaction: 2 C2H6(g) + 7 O2(g) ⇌ 4 CO2(g) + 6 H2O(g)?arrow_forward3) Based on the information below, determine the value of the unknown equilibrium constant, Kc. 4 CO2 (g) + 2 H2O (g) « 4 O2 (g) + 2 CH2CO (g)Kc = 3.7 x 1017 CO2 (g) + 2 H2O (g) « CH4 (g) + 2 O2 (g)Kc = 8.3 x 10-15 CH4 (g) + CO2 (g) « CH2CO (g) + H2O (g)Kc = ?arrow_forwardCalculate the equilibrium constant of the reaction CO(g) + 2 H2S(g) CS2(g) + H2O(g) + H2(g) from the following information: CO(g) + 3 H2(g) CH4(g) + H2O(g) K = 9.17 x 10-2 CH4(g) + 2 H2S(g) CS2(g) + H2(g) K = 3.3 x 104 Answer: [3.03 x 103 ]arrow_forward
- Write the equilibrium constant expression (Kc) for the reaction below 2P2O5(g) + 10Cl2(g) ⇌ 4PCl5(aq) + 5O2(g)arrow_forwardWhat is the expression for the equilibrium constant for the reaction: 2 C2H6(g) + 7 O2(g) ⇌ 4 CO2(g) + 6 H2O(g)?arrow_forwardSuppose the reaction system CH4 (9) + 202 (g) = CO2(g) + 2H20(1) has already reached equilibrium. The position of the equilibrium was shifted by performing the reaction in a metal cylinder fitted with a piston, which is compressed to decrease the total volume of the system. Choose the correct effect on equilibrium (it will shift to the right, will shift to the left, or it will not be affected). It will shift to the right. O It will shift to the left. O It will not be affected.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Chemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage LearningChemistry by OpenStax (2015-05-04)ChemistryISBN:9781938168390Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark BlaserPublisher:OpenStaxChemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage Learning
- Principles of Modern ChemistryChemistryISBN:9781305079113Author:David W. Oxtoby, H. Pat Gillis, Laurie J. ButlerPublisher:Cengage LearningPhysical ChemistryChemistryISBN:9781133958437Author:Ball, David W. (david Warren), BAER, TomasPublisher:Wadsworth Cengage Learning,Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage Learning
![Text book image](https://www.bartleby.com/isbn_cover_images/9780534420123/9780534420123_smallCoverImage.gif)
Chemistry: Principles and Practice
Chemistry
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Cengage Learning
![Text book image](https://www.bartleby.com/isbn_cover_images/9781938168390/9781938168390_smallCoverImage.gif)
Chemistry by OpenStax (2015-05-04)
Chemistry
ISBN:9781938168390
Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
Publisher:OpenStax
![Text book image](https://www.bartleby.com/isbn_cover_images/9781305079373/9781305079373_smallCoverImage.gif)
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
![Text book image](https://www.bartleby.com/isbn_cover_images/9781305079113/9781305079113_smallCoverImage.gif)
Principles of Modern Chemistry
Chemistry
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:Cengage Learning
![Text book image](https://www.bartleby.com/isbn_cover_images/9781133958437/9781133958437_smallCoverImage.gif)
Physical Chemistry
Chemistry
ISBN:9781133958437
Author:Ball, David W. (david Warren), BAER, Tomas
Publisher:Wadsworth Cengage Learning,
![Text book image](https://www.bartleby.com/isbn_cover_images/9781285199047/9781285199047_smallCoverImage.gif)
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
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