Explanation The required reaction is given below. P C l 5 ( g ) ⇌ P C l 3 ( g ) + C l 2 ( g ) , K c = 1.80 The expression for equilibrium constant for the above reaction can be written as shown below, K c = [ P C l 3 ] [ C l 2 ] [ P C l 5 ] Given that, the 0.724 m o l P C l 5 is placed in a reaction vessel of volume 0.500 L . The concentration P C l 5 can be calculated as given below. [ P C l 5 ] = m o l e s v o l u m e = 0.724 m o l 0.500 L = 1 . 4 5 M An equilibrium table can be set up as given below. C o m p o s i t i o n ( M ) P C l 5 P C l 3 C l 2 I n i t i a l c o n c e n t r a t i o n 1.45 0 0 C h a n g e i n c o n c e n t r a t i o n − x + 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 1.45 − x x x Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below. K c = ( x ) ( x ) ( 1

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ISBN: 9781319420994

Author: ATKINS

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Chapter 5, Problem 5I.28E

Interpretation Introduction

Interpretation:

The equilibrium concentrations of PCl5, PCl3, and Cl2 in a reaction vessel of volume 0.500 L that initially held 0.724 mol PCl5 have to be calculated. The reaction is given below.

PCl5(g)⇌PCl3(g)+Cl2(g), Kc=1.80

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Chapter 5, Problem 5I.27E

Chapter 5, Problem 5I.29E

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Chapter 5 Solutions

CHEM PRINCIPLES LL W/ACHIEVE ONE-SEM

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The equilibrium concentrations of P C l 5 , P C l 3 , a n d C l 2 in a reaction vessel of volume 0.500 L that initially held 0.724 m o l P C l 5 have to be calculated. The reaction is given below. P C l 5 ( g ) ⇌ P C l 3 ( g ) + C l 2 ( g ) , K c = 1.80

Solution Summary: The author explains how the equilibrium concentrations of PCl_5,molesvolume=

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Interpretation:

The equilibrium concentrations of PCl5, PCl3, and Cl2 in a reaction vessel of volume 0.500 L that initially held 0.724 mol PCl5 have to be calculated. The reaction is given below.

PCl5(g)⇌PCl3(g)+Cl2(g), Kc=1.80

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Chapter 5 Solutions

The equilibrium concentrations of P C l 5 , P C l 3 , a n d C l 2 in a reaction vessel of volume 0.500 L that initially held 0.724 m o l P C l 5 have to be calculated. The reaction is given below. P C l 5 ( g ) ⇌ P C l 3 ( g ) + C l 2 ( g ) , K c = 1.80

Solution Summary: The author explains how the equilibrium concentrations of PCl_5,molesvolume=

Videos

Interpretation: The equilibrium concentrations of PCl5, PCl3, and Cl2 in a reaction vessel of volume 0.500 L that initially held 0.724 mol PCl5 have to be calculated. The reaction is given below. PCl5(g)⇌PCl3(g)+Cl2(g), Kc=1.80

Want to see the full answer?

Chapter 5 Solutions

Interpretation:

The equilibrium concentrations of PCl5, PCl3, and Cl2 in a reaction vessel of volume 0.500 L that initially held 0.724 mol PCl5 have to be calculated. The reaction is given below.

PCl5(g)⇌PCl3(g)+Cl2(g), Kc=1.80