Explanation The required reaction is given below. 2 H C l ( g ) + I 2 ( s ) ⇌ 2 H I ( g ) + C l 2 ( g ) , K = 3.5 × 10 − 32 The expression for equilibrium constant for the above reaction can be written as shown below, K = ( P H I ) 2 ( P C l 2 ) ( P H C l ) 2 Given that, the initial partial pressure of H C l is 0.012 b a r . An equilibrium table can be set up as given below. C o m p o s i t i o n ( b a r ) H C l H I C l 2 I n i t i a l p r e s s u r e 0.012 0 0 C h a n g e i n p r e s s u r e − 2 x + 2 x + x E q u i l i b r i u m p r e s s u r e 0.012 − 2 x 2 x x Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below. K = ( 2 x ) 2 × ( x ) ( 0.012 − 2 x ) 2 By assuming x ≪ 0.012 , it can be written that 0.012 − x ≈ 0.012 . Now, the above equilibrium expression can be solved for x . K = ( 2 x ) 2 × ( x ) ( 0.012 − 2 x ) 2 ≈ K = ( 2 x ) 2 × ( x ) ( 0

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM

7th Edition

ISBN: 9781319403959

Author: ATKINS

Publisher: MAC HIGHER

See similar textbooks

Videos

Question

Chapter 5, Problem 5I.3BST

Interpretation Introduction

Interpretation:

The equilibrium partial pressures of each gaseous substance in the reaction mixture have to be calculated. The reaction is given below.

2HCl(g)+I2(s)⇌2HI(g)+Cl2(g), K=3.5×10−32

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 5, Problem 5I.3AST

Chapter 5, Problem 5I.4AST

Chapter 5 Solutions

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM

Ch. 5 - Prob. 5A.1AST Ch. 5 - Prob. 5A.1BST Ch. 5 - Prob. 5A.2AST Ch. 5 - Prob. 5A.2BST Ch. 5 - Prob. 5A.3AST Ch. 5 - Prob. 5A.3BST Ch. 5 - Prob. 5A.1E Ch. 5 - Prob. 5A.2E Ch. 5 - Prob. 5A.3E Ch. 5 - Prob. 5A.4E

Ch. 5 - Prob. 5A.5E Ch. 5 - Prob. 5A.6E Ch. 5 - Prob. 5A.7E Ch. 5 - Prob. 5A.8E Ch. 5 - Prob. 5A.11E Ch. 5 - Prob. 5B.1AST Ch. 5 - Prob. 5B.1BST Ch. 5 - Prob. 5B.2AST Ch. 5 - Prob. 5B.2BST Ch. 5 - Prob. 5B.3AST Ch. 5 - Prob. 5B.3BST Ch. 5 - Prob. 5B.1E Ch. 5 - Prob. 5B.2E Ch. 5 - Prob. 5B.3E Ch. 5 - Prob. 5B.5E Ch. 5 - Prob. 5B.7E Ch. 5 - Prob. 5C.1AST Ch. 5 - Prob. 5C.1BST Ch. 5 - Prob. 5C.2AST Ch. 5 - Prob. 5C.2BST Ch. 5 - Prob. 5C.3AST Ch. 5 - Prob. 5C.3BST Ch. 5 - Prob. 5C.1E Ch. 5 - Prob. 5C.3E Ch. 5 - Prob. 5C.4E Ch. 5 - Prob. 5C.5E Ch. 5 - Prob. 5C.6E Ch. 5 - Prob. 5C.7E Ch. 5 - Prob. 5C.8E Ch. 5 - Prob. 5C.9E Ch. 5 - Prob. 5C.10E Ch. 5 - Prob. 5C.11E Ch. 5 - Prob. 5C.12E Ch. 5 - Prob. 5C.15E Ch. 5 - Prob. 5C.16E Ch. 5 - Prob. 5D.1AST Ch. 5 - Prob. 5D.1BST Ch. 5 - Prob. 5D.1E Ch. 5 - Prob. 5D.2E Ch. 5 - Prob. 5D.3E Ch. 5 - Prob. 5D.4E Ch. 5 - Prob. 5D.5E Ch. 5 - Prob. 5D.6E Ch. 5 - Prob. 5D.7E Ch. 5 - Prob. 5D.8E Ch. 5 - Prob. 5D.9E Ch. 5 - Prob. 5D.10E Ch. 5 - Prob. 5D.11E Ch. 5 - Prob. 5D.12E Ch. 5 - Prob. 5D.13E Ch. 5 - Prob. 5D.14E Ch. 5 - Prob. 5D.15E Ch. 5 - Prob. 5D.16E Ch. 5 - Prob. 5D.18E Ch. 5 - Prob. 5D.19E Ch. 5 - Prob. 5D.20E Ch. 5 - Prob. 5E.1AST Ch. 5 - Prob. 5E.1BST Ch. 5 - Prob. 5E.2AST Ch. 5 - Prob. 5E.2BST Ch. 5 - Prob. 5E.1E Ch. 5 - Prob. 5E.2E Ch. 5 - Prob. 5E.11E Ch. 5 - Prob. 5E.12E Ch. 5 - Prob. 5F.1AST Ch. 5 - Prob. 5F.1BST Ch. 5 - Prob. 5F.2AST Ch. 5 - Prob. 5F.2BST Ch. 5 - Prob. 5F.3AST Ch. 5 - Prob. 5F.3BST Ch. 5 - Prob. 5F.4AST Ch. 5 - Prob. 5F.4BST Ch. 5 - Prob. 5F.5AST Ch. 5 - Prob. 5F.5BST Ch. 5 - Prob. 5F.1E Ch. 5 - Prob. 5F.2E Ch. 5 - Prob. 5F.3E Ch. 5 - Prob. 5F.5E Ch. 5 - Prob. 5F.7E Ch. 5 - Prob. 5F.9E Ch. 5 - Prob. 5F.10E Ch. 5 - Prob. 5F.11E Ch. 5 - Prob. 5F.12E Ch. 5 - Prob. 5F.13E Ch. 5 - Prob. 5F.14E Ch. 5 - Prob. 5F.15E Ch. 5 - Prob. 5F.16E Ch. 5 - Prob. 5G.1AST Ch. 5 - Prob. 5G.1BST Ch. 5 - Prob. 5G.2AST Ch. 5 - Prob. 5G.2BST Ch. 5 - Prob. 5G.3AST Ch. 5 - Prob. 5G.3BST Ch. 5 - Prob. 5G.4AST Ch. 5 - Prob. 5G.4BST Ch. 5 - Prob. 5G.5AST Ch. 5 - Prob. 5G.5BST Ch. 5 - Prob. 5G.1E Ch. 5 - Prob. 5G.2E Ch. 5 - Prob. 5G.3E Ch. 5 - Prob. 5G.4E Ch. 5 - Prob. 5G.7E Ch. 5 - Prob. 5G.8E Ch. 5 - Prob. 5G.9E Ch. 5 - Prob. 5G.11E Ch. 5 - Prob. 5G.12E Ch. 5 - Prob. 5G.13E Ch. 5 - Prob. 5G.14E Ch. 5 - Prob. 5G.15E Ch. 5 - Prob. 5G.16E Ch. 5 - Prob. 5G.17E Ch. 5 - Prob. 5G.19E Ch. 5 - Prob. 5G.20E Ch. 5 - Prob. 5G.21E Ch. 5 - Prob. 5G.22E Ch. 5 - Prob. 5H.1AST Ch. 5 - Prob. 5H.1BST Ch. 5 - Prob. 5H.2AST Ch. 5 - Prob. 5H.2BST Ch. 5 - Prob. 5H.1E Ch. 5 - Prob. 5H.2E Ch. 5 - Prob. 5H.3E Ch. 5 - Prob. 5H.4E Ch. 5 - Prob. 5H.5E Ch. 5 - Prob. 5H.6E Ch. 5 - Prob. 5I.1AST Ch. 5 - Prob. 5I.1BST Ch. 5 - Prob. 5I.2AST Ch. 5 - Prob. 5I.2BST Ch. 5 - Prob. 5I.3AST Ch. 5 - Prob. 5I.3BST Ch. 5 - Prob. 5I.4AST Ch. 5 - Prob. 5I.4BST Ch. 5 - Prob. 5I.1E Ch. 5 - Prob. 5I.2E Ch. 5 - Prob. 5I.3E Ch. 5 - Prob. 5I.4E Ch. 5 - Prob. 5I.5E Ch. 5 - Prob. 5I.6E Ch. 5 - Prob. 5I.7E Ch. 5 - Prob. 5I.9E Ch. 5 - Prob. 5I.10E Ch. 5 - Prob. 5I.11E Ch. 5 - Prob. 5I.12E Ch. 5 - Prob. 5I.13E Ch. 5 - Prob. 5I.14E Ch. 5 - Prob. 5I.15E Ch. 5 - Prob. 5I.16E Ch. 5 - Prob. 5I.17E Ch. 5 - Prob. 5I.18E Ch. 5 - Prob. 5I.19E Ch. 5 - Prob. 5I.20E Ch. 5 - Prob. 5I.21E Ch. 5 - Prob. 5I.22E Ch. 5 - Prob. 5I.23E Ch. 5 - Prob. 5I.24E Ch. 5 - Prob. 5I.25E Ch. 5 - Prob. 5I.26E Ch. 5 - Prob. 5I.27E Ch. 5 - Prob. 5I.28E Ch. 5 - Prob. 5I.29E Ch. 5 - Prob. 5I.30E Ch. 5 - Prob. 5I.32E Ch. 5 - Prob. 5I.33E Ch. 5 - Prob. 5I.34E Ch. 5 - Prob. 5I.35E Ch. 5 - Prob. 5I.36E Ch. 5 - Prob. 5J.1AST Ch. 5 - Prob. 5J.1BST Ch. 5 - Prob. 5J.3AST Ch. 5 - Prob. 5J.3BST Ch. 5 - Prob. 5J.4AST Ch. 5 - Prob. 5J.4BST Ch. 5 - Prob. 5J.5AST Ch. 5 - Prob. 5J.5BST Ch. 5 - Prob. 5J.1E Ch. 5 - Prob. 5J.2E Ch. 5 - Prob. 5J.3E Ch. 5 - Prob. 5J.4E Ch. 5 - Prob. 5J.5E Ch. 5 - Prob. 5J.6E Ch. 5 - Prob. 5J.9E Ch. 5 - Prob. 5J.10E Ch. 5 - Prob. 5J.11E Ch. 5 - Prob. 5J.12E Ch. 5 - Prob. 5J.13E Ch. 5 - Prob. 5J.17E Ch. 5 - Prob. 5.1E Ch. 5 - Prob. 5.2E Ch. 5 - Prob. 5.3E Ch. 5 - Prob. 5.4E Ch. 5 - Prob. 5.5E Ch. 5 - Prob. 5.6E Ch. 5 - Prob. 5.7E Ch. 5 - Prob. 5.8E Ch. 5 - Prob. 5.9E Ch. 5 - Prob. 5.10E Ch. 5 - Prob. 5.11E Ch. 5 - Prob. 5.12E Ch. 5 - Prob. 5.13E Ch. 5 - Prob. 5.14E Ch. 5 - Prob. 5.15E Ch. 5 - Prob. 5.16E Ch. 5 - Prob. 5.17E Ch. 5 - Prob. 5.19E Ch. 5 - Prob. 5.23E Ch. 5 - Prob. 5.24E Ch. 5 - Prob. 5.25E Ch. 5 - Prob. 5.26E Ch. 5 - Prob. 5.27E Ch. 5 - Prob. 5.28E Ch. 5 - Prob. 5.29E Ch. 5 - Prob. 5.30E Ch. 5 - Prob. 5.31E Ch. 5 - Prob. 5.32E Ch. 5 - Prob. 5.33E Ch. 5 - Prob. 5.35E Ch. 5 - Prob. 5.37E Ch. 5 - Prob. 5.38E Ch. 5 - Prob. 5.41E Ch. 5 - Prob. 5.43E Ch. 5 - Prob. 5.44E Ch. 5 - Prob. 5.45E Ch. 5 - Prob. 5.46E Ch. 5 - Prob. 5.47E Ch. 5 - Prob. 5.49E Ch. 5 - Prob. 5.51E Ch. 5 - Prob. 5.53E Ch. 5 - Prob. 5.55E Ch. 5 - Prob. 5.57E Ch. 5 - Prob. 5.58E Ch. 5 - Prob. 5.61E Ch. 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

Recommended textbooks for you

Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Chemistry: An Atoms First Approach
Chemistry
ISBN:9781305079243
Author:Steven S. Zumdahl, Susan A. Zumdahl
Publisher:Cengage Learning
Chemistry
Chemistry
ISBN:9781133611097
Author:Steven S. Zumdahl
Publisher:Cengage Learning
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
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

Chemistry: Principles and Reactions

Chemistry

ISBN:9781305079373

Author:William L. Masterton, Cecile N. Hurley

Publisher:Cengage Learning

Chemistry: An Atoms First Approach

Chemistry

ISBN:9781305079243

Author:Steven S. Zumdahl, Susan A. Zumdahl

Publisher:Cengage Learning

Chemistry

ISBN:9781133611097

Author:Steven S. Zumdahl

Publisher:Cengage Learning

Chemistry

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Cengage Learning

Chemistry: The Molecular Science

Chemistry

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

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

SEE MORE TEXTBOOKS

Chemical Equilibria and Reaction Quotients; Author: Professor Dave Explains;https://www.youtube.com/watch?v=1GiZzCzmO5Q;License: Standard YouTube License, CC-BY

The equilibrium partial pressures of each gaseous substance in the reaction mixture have to be calculated. The reaction is given below. 2 H C l ( g ) + I 2 ( s ) ⇌ 2 H I ( g ) + C l 2 ( g ) , K = 3.5 × 10 − 32

Solution Summary: The author explains how the equilibrium partial pressures of each gaseous substance in the reaction mixture have to be calculated.

Videos

Interpretation:

The equilibrium partial pressures of each gaseous substance in the reaction mixture have to be calculated. The reaction is given below.

2HCl(g)+I2(s)⇌2HI(g)+Cl2(g), K=3.5×10−32

Want to see the full answer?

Chapter 5 Solutions

The equilibrium partial pressures of each gaseous substance in the reaction mixture have to be calculated. The reaction is given below. 2 H C l ( g ) + I 2 ( s ) ⇌ 2 H I ( g ) + C l 2 ( g ) , K = 3.5 × 10 − 32

Solution Summary: The author explains how the equilibrium partial pressures of each gaseous substance in the reaction mixture have to be calculated.

Videos

Interpretation: The equilibrium partial pressures of each gaseous substance in the reaction mixture have to be calculated. The reaction is given below. 2HCl(g)+I2(s)⇌2HI(g)+Cl2(g), K=3.5×10−32

Want to see the full answer?

Chapter 5 Solutions

Interpretation:

The equilibrium partial pressures of each gaseous substance in the reaction mixture have to be calculated. The reaction is given below.

2HCl(g)+I2(s)⇌2HI(g)+Cl2(g), K=3.5×10−32