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

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ISBN: 9781319420994

Author: ATKINS

Publisher: MAC HIGHER

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Chapter 5, Problem 5I.3BST

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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

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Chapter 5, Problem 5I.3AST

Chapter 5, Problem 5I.4AST

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Chapter 5 Solutions

CHEM PRINCIPLES LL W/ACHIEVE ONE-SEM

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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.

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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

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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