Explanation The required reaction is given below. 2 H C l ( g ) + I 2 ( s ) ⇌ 2 H I ( g ) + C l 2 ( g ) , K c = 1.6 × 10 − 34 The expression for equilibrium constant for the above reaction can be written as shown below, K c = [ H I ] 2 [ C l 2 ] [ H C l ] 2 Given that, the 4.00 L o f H C l ( g ) at 1.00 b a r a n d 273 K is transferred to a reaction vessel of volume 12.0 L . The concentration of H C l ( g ) can be calculated as given below. P V = n R T n V = P R T n = P V R T = ( 1.00 b a r ) ( 4.00 L ) ( 8.3145 × 10 − 2 L . b a r . K − 1 . m o l − 1 ) ( 273 K ) n = 0.176 m o l [ H C l ] = M o l e s V o l u m e = 0.176 m o l 12.0 L = 0 . 0 1 4 6 M An equilibrium table can be set up as given below. C o m p o s i t i o n ( M ) H C l H I C l 2 I n i t i a l c o n c e n t r a t i o n 0.0146 0 0 C h a n g e i n c o n c e n t r a t i o n − 2 x + 2 x + x E q u i l i b r i u m c o n c e n t r a t i o n 0.0146 − 2 x 2 x x Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below. K c = ( 2 x ) 2 ( x ) ( 0

Chemical Principles: The Quest for Insight

7th Edition

ISBN: 9781464183959

Author: Peter Atkins, Loretta Jones, Leroy Laverman

Publisher: W. H. Freeman

See similar textbooks

Videos

Question

Chapter 5, Problem 5I.30E

Interpretation Introduction

Interpretation:

The equilibrium concentrations of HCl, HI, and Cl2 has to be calculated. The reaction is given below.

2HCl(g)+I2(s)⇌2HI(g)+Cl2(g), Kc=1.6×10−34

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 5, Problem 5I.29E

Chapter 5, Problem 5I.32E

Students have asked these similar questions

I need the nomenclature of this compound.

I need the nomenclature of this compound

2. Name the following hydrocarbons. (9 marks) a) HHHHHHHH H-C-C- H-O-S b) HCEC-CH3 H H H H H d) c) H C=C- H H H e) CH3 CH3 CH2CH=CH-CH=CHCH3 HHHH H-C-C-C-C-H H HH H f) large CH2CH3 pola H3C section lovels tower, able ocart firs g) Tower H3C-CH2 then in H3C-CH-CH-CH3 enblbano bne noitsidab Copyright © 2008. Durham Continuing Education CH3

Chapter 5 Solutions

Chemical Principles: The Quest for Insight

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.

The equilibrium concentrations of H C l , H I , a n d C l 2 has 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 c = 1.6 × 10 − 34

Solution Summary: The author explains how the equilibrium concentrations of HCl,TextHI, Cl_2 have to be calculated. The reaction is given below.

Videos

Interpretation:

The equilibrium concentrations of HCl, HI, and Cl2 has to be calculated. The reaction is given below.

2HCl(g)+I2(s)⇌2HI(g)+Cl2(g), Kc=1.6×10−34

Want to see the full answer?

Chapter 5 Solutions

The equilibrium concentrations of H C l , H I , a n d C l 2 has 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 c = 1.6 × 10 − 34

Solution Summary: The author explains how the equilibrium concentrations of HCl,TextHI,&#160;Cl_2 have to be calculated. The reaction is given below.

Videos

Interpretation: The equilibrium concentrations of HCl, HI, and Cl2 has to be calculated. The reaction is given below. 2HCl(g)+I2(s)⇌2HI(g)+Cl2(g), Kc=1.6×10−34

Want to see the full answer?

Chapter 5 Solutions

Solution Summary: The author explains how the equilibrium concentrations of HCl,TextHI, Cl_2 have to be calculated. The reaction is given below.

Interpretation:

The equilibrium concentrations of HCl, HI, and Cl2 has to be calculated. The reaction is given below.

2HCl(g)+I2(s)⇌2HI(g)+Cl2(g), Kc=1.6×10−34