   Chapter 13, Problem 80AE

Chapter
Section
Textbook Problem

# For the following reaction at a certain temperature H 2 ( g ) + F 2 ( g ) ⇌ 2 HF ( g ) it is found that the equilibrium concentrations in a 5.00-L rigid container are [H2 = 0.0500 M, [F2] = 0.0100 M, and [HF] = 0.400 M. If 0.200 mole of F2 is added to this equilibrium mixture, calculate the concentrations of all gases once equilibrium is reestablished.

Interpretation Introduction

Interpretation: The equilibrium concentration of each species, volume of container and the reaction for the production of HF(g) is given. The concentration of all species of this reaction is to be calculated when given moles of Fe2 is added to this equilibrium mixture.

Concept introduction: The state when the reactants involved in a chemical reaction and the products formed in the reaction exist in concentrations having no further tendency to change is known as an equilibrium state of the reaction.

For a reaction,

jA+kBlC+mD

The equilibrium constant is given by,

K=[C]l[D]m[A]j[B]k

To determine: The equilibrium concentration of [H2] , [F2] and [HF] when 0.200 moles of Fe2 is added to the given equilibrium reaction.

The equilibrium constant of the given reaction is 320_ .

Explanation

Explanation

Given

The given concentration of each species is,

[H2]=0.0500M[F2]=0.0100M[HF]=0.400M

The given reaction is,

H2(g)+F2(g)2HF(g)

Formula

The equilibrium constant for the given reaction is,

K=[HF]2[H2][F2]

Where,

• K is the equilibrium constant.

Substitute the values of concentration of each species in the above expression.

K=[HF]2[H2][F2]=[0.400]2[0.0500][0.0100]=320_

The equilibrium concentration of [H2] is 0.025M_ .

Given

The volume of container is 5.00L .

As 0.200mole are added to the reaction mixture in container therefore, the concentration of [F2] increased by,

[F2]=0.200mole5L=0.04M

Now, the total concentration of F2 in the container is,

[F2]=(0.010+0.04)M=0.50M

It is assumed that the concentration of reactants changed by x .

Make the ICE table for the given reaction.

H2(g)F2(g)2HF(g)Initial(M):0.050.050.40Change(M):-x-x2xEquilibrium(M):0

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