   Chapter 15, Problem 43GQ

Chapter
Section
Textbook Problem

Phosphorus pentachloride decomposes at elevated temperatures.PCl5(g) ⇄ PCl3(g) + Cl2(g)An equilibrium mixture at some temperature consists of 3.120 g of PCl5, 3.845 g of PCl3, and 1.787 g of Cl2 in a 10.0-L flask. If you add 1.418 g of Cl2, how will the equilibrium be affected? What will the concentrations of PCl5, PCl3, and Cl2 be when equilibrium is reestablished?

Interpretation Introduction

Interpretation:

The effect on the equilibrium 1.418g of Cl2 is added in the reaction of heating of barium carbonate is to be given. The equilibrium concentration of PCl5,PCl3,andCl2 after the addition of 1.418g of Cl2 has to be given.

Concept Introduction:

Equilibrium constant in terms of concentration[Kc]: Equilibrium constant can be expressed in terms of concentration.

aA(g)+bB(g)cC(g)+dD(g)Kc=KCc×KDdKAa×KBb

In order to calculate number of moles of a substance:

No.ofmoles=massmolarmass

Explanation

Given:

PCl5(g)PCl3(g)+Cl2(g)

The equilibrium mixture consist of

From the given data we can find out the number of moles of each species.

No.ofmoles=massmolarmass

No.ofmolesofPCl5=3.120g208.24g/mol=0.015molNo.ofmolesofPCl3=3.845g137.33g/mol=0.028molNo.ofmolesofCl2=1.787g70.91g/mol=0.025mol

Using this equilibrium constant can be calculated.

aA(g)+bB(g)cC(g)+dD(g)K=KCc×KDdKAa×KBb

For the reaction PCl5(g)PCl3(g)+Cl2(g),

K=[PCl3][Cl2][PCl5]=0.028×0.0250.015=0.047

Given that 1.418g of Cl2 is added to the mixture.

According to Le Chatelier’s principle, if the concentration of product increases then the equilibrium will shift towards reactant side or left side in order to cancel the effect.

Therefore,

When 1.418g of Cl2 is added to the mixture the equilibrium will shift towards reactant side

Number of moles of chlorine can be calculated,

No.ofmolesofCl2=1.418g70.91g/mol=0

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