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 5H.4E
Interpretation Introduction
Interpretation:
The value of
Concept Introduction:
Relations between equilibrium constants:
| Chemical equation | Equilibrium constant |
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Carbon monoxide and water vapor, each at 200. Torr, were introduced into a container of volume 0.250 L. When the mixture reached equilibrium at 700 degrees Celsius, the partial pressure of CO2(g) was 88 Torr. Calculate the value of K for the equilibrium CO (g) + H2O (g)⇋ CO2(g)+H2(g).
Select the reaction for which K, = Kc.
2 Na,0,(s) + 2 CO,(g) =2 Na,CO3(s) + 0,(g)
2 KCIO3 (s) = 2 KCI(s) + 30,(g)
2 H,S(g) + SO,(g)= 3 S(s) + 2 H,O(g)
Br, (g) + Cl,(g) 2 BRCI(g)
Write the equilibrium constant for the reaction CH4(g) + 3 Cl2(g) ⇋ CHCl3(l) + 3 HCl(g), with the gases treated as perfect.
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
Similar questions
- Actually, 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_forwardDescribe 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_forward
- Simple acids such as formic acid, HCOOH, and acetic acid, CH3COOH, are very soluble in water; however, fatty acids such as stearic acid, CH3(CH2)16COOH, and palmitic acid, CH3(CH2)14COOH, are water-insoluble. Based on what you know about the solubility of alcohols, explain the solubility of these organic acids.arrow_forwardDetermine the equilibrium constant for the reaction Sn+Pb2+Sn2++Pbarrow_forwardThe 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) ?arrow_forward
- Select the reaction for which K, = Kc. 2 CO, (g) + 2 CF,(g) =4 COF, (g) NH,NO, (s) = N,O(g) + 2H,O(g) 2 H, S(g) + SO, (g) = 3 S(s) + 2 H,O(g) O 2 Na, O, (s) + 2CO,(g) =2 Na,CO;(s) + 0,(g)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_forwardFor the reaction 2 NH3(g) ⇋ N2(g) + 3H2(g), Kc= 0.395 at 350 degrees Celsius. A sample of NH3 of mass 25.6 g is laced in a reaction vessel of volume 5.00 L and heated to 350 degrees Celsius. What are the equilibrium concentrations of NH3, N2, and H2?arrow_forward
- X(9) + 4Y(9) = 3Z(g) K, = 3.12 x 10 2 bar 2 at a temperature of 287 °C. Calculate the value of Ke.arrow_forwardAt 298 K the reaction N(g) + 3 H(g)<-> 2 NH(g) has K = 4.2 x 108. What is the value of Kc for the reaction 1/3 N(g) + H(g)<-> 2/3 NH(g)?arrow_forwardFor the reaction X(9) + 4Y(g) = 3Z(g) K, = 2.51 × 10-2 bar-2 at a temperature of 361 °C. Calculate the value of Kc. Express your answer numerically.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Physical ChemistryChemistryISBN:9781133958437Author:Ball, David W. (david Warren), BAER, TomasPublisher:Wadsworth Cengage Learning,
Chemistry by OpenStax (2015-05-04)ChemistryISBN:9781938168390Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark BlaserPublisher:OpenStax
Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage Learning
Chemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage Learning
Living By Chemistry: First Edition TextbookChemistryISBN:9781559539418Author:Angelica StacyPublisher:MAC HIGHER
Physical Chemistry
Chemistry
ISBN:9781133958437
Author:Ball, David W. (david Warren), BAER, Tomas
Publisher:Wadsworth 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: 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
Living By Chemistry: First Edition Textbook
Chemistry
ISBN:9781559539418
Author:Angelica Stacy
Publisher:MAC HIGHER