Explanation The required reaction is given below. H 2 ( g ) + I 2 ( s ) ⇌ 2 H I ( g ) The expression for equilibrium constant for the above reaction can be written as shown below, K c = [ H I ] 2 [ H 2 ] [ I 2 ] Given that, the 0.400 m o l H 2 a n d 1.60 m o l I 2 was introduced into a flask of volume 3.0 L . The concentrations of H 2 ( g ) a n d I 2 ( g ) can be calculated as given below. [ H 2 ] = M o l e s V o l u m e = 0.400 m o l 3.00 L = 0 . 1 3 3 M [ I 2 ] = M o l e s V o l u m e = 1.60 m o l 3.00 L = 0 . 5 3 3 M An equilibrium table can be set up as given below. C o m p o s i t i o n ( M ) H 2 I 2 H I I n i t i a l c o n c e n t r a t i o n 0.133 0.533 0 C h a n g e i n c o n c e n t r a t i o n − x − x + 2 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

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.19E

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Interpretation:

The equilibrium constant for the reaction H2(g)+I2(s)⇌2HI(g) has to be calculated.

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Chapter 5, Problem 5I.18E

Chapter 5, Problem 5I.20E

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Chemical Principles: The Quest for Insight

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The equilibrium constant for the reaction H 2 ( g ) + I 2 ( s ) ⇌ 2 H I ( g ) has to be calculated.

Solution Summary: The author explains how the equilibrium constant for the reaction H_2(g)+I

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Interpretation:

The equilibrium constant for the reaction H2(g)+I2(s)⇌2HI(g) has to be calculated.

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Chapter 5 Solutions

The equilibrium constant for the reaction H 2 ( g ) + I 2 ( s ) ⇌ 2 H I ( g ) has to be calculated.

Solution Summary: The author explains how the equilibrium constant for the reaction H_2(g)+I

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Interpretation: The equilibrium constant for the reaction H2(g)+I2(s)⇌2HI(g) has to be calculated.

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Chapter 5 Solutions

Interpretation:

The equilibrium constant for the reaction H2(g)+I2(s)⇌2HI(g) has to be calculated.