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

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Chapter 5, Problem 5I.30E

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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

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Chapter 5, Problem 5I.29E

Chapter 5, Problem 5I.32E

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Chemical Principles: The Quest for Insight

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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.

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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

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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.

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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

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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