Chemistry: An Atoms First Approach
Chemistry: An Atoms First Approach
2nd Edition
ISBN: 9781305079243
Author: Steven S. Zumdahl, Susan A. Zumdahl
Publisher: Cengage Learning
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Chapter 12, Problem 54E

At 25°c, K = 0.090 for the reaction

H 2 O ( g ) + Cl 2 O ( g ) 2 HOCI ( g )

Calculate the concentrations of all species at equilibrium for each of the following cases.

a. 1.0 g H2O and 2.0 g Cl2O are mixed in a 1.0-L flask.

b. 1.0 mole of pure HOCI is placed in a 2.0-L flask.

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The value of the equilibrium constant for the reaction between H2O and Cl2O and the initial mass of the species involved is given. The equilibrium concentrations of the three gases involved in the system are to be calculated.

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. When the equilibrium constant is expressed in terms of concentration, it is represented as K .

To determine: The equilibrium concentrations of the three gases involved in the system

Answer to Problem 54E

The equilibrium concentrations of the three gases are calculated as follows.

Explanation of Solution

Given

The stated reaction is,

H2O(g)+Cl2O(g)2HOCl(g)

The initial mass of H2O(g) is 1.0g .

The initial mass of Cl2O(g) is 2.0g .

The equilibrium constant (K) value is 0.090 .

The volume of the container is 1.00L .

The molar mass of H2O=2H+O=2(1)+16=18g/mol

The molar mass of Cl2O=2Cl+O=2(35.5)+16=87g/mol

Formula

The number of moles of a substance is calculated by the formula,

Numberofmoles=GivenmassMolarmass

Substitute the values of the given mass and molar mass of H2O and Cl2O in the above expression.

For H2O ,

NumberofmolesofH2O=1.0g18.0g/mol=0.056mol

For Cl2O ,

NumberofmolesofCl2O=2.0g87.0g/mol=0.023mol

The concentration of a reactant is calculated by the formula,

Concentration=MolesVolume(L)

The initial concentration of H2O is calculated by the formula,

ConcentrationofH2O=MolesofH2OVolumeoftheflask(L)

Substitute the given values of the number of moles of H2O and the volume of the flask in the above expression.

ConcentrationofH2O=0.056moles1.00L=0.056M

The initial concentration of Cl2O is calculated by the formula,

ConcentrationofCl2O=MolesofCl2OVolumeoftheflask(L)

Substitute the given values of the number of moles of Cl2O and the volume of the flask in the above expression.

ConcentrationofCl2O=0.023moles1.00L=0.023M

To determine: The equilibrium concentrations of the three gases involved in the system.

Explanation:

The concentration of H2O and Cl2O consumed is assumed to be x .

The equilibrium concentrations are represented as,

H2O(g)+Cl2O(g)2HOCl(g)Initialconcentration0.0560.0230Change-x-x+2xEquilibriumconcentration0.056-x0.023-x2x

At equilibrium, the equilibrium ratio is expressed by the formula,

K=ConcentrationofproductsConcentrationofreactants

Where,

  • K is the equilibrium constant.

The equilibrium ratio for the given reaction is,

K=[HOCl]2[H2O][Cl2O] (1)

According to the formulated ICE table,

The equilibrium concentration of H2O(g) is (0.056x)M .

The equilibrium initial concentration of Cl2O(g) is (0.023x)M .

The equilibrium initial concentration of HOCl(g) is (2x)M .

Substitute these values in equation (1).

K=[HOCl]2[H2O][Cl2O]K=[2x]2[0.056x][0.023x]

The given value of K is 0.090 .

Substitute the value of K in the above expression.

0.090=[2x]2[0.056x][0.023x]

Simplify the above expression.

0.090=[2x]2[0.056x][0.023x](0.09x20.00711x+0.00011592)=4x2(3.91x20.007x+0.00011592)=0

Comparing this with the general quadratic equation, ax2+bx+c=0 .

In the obtained equation,

  • a is 3.91 .
  • b is 0.007 .
  • c is 0.00011592 .

The formula to calculate the value of x is,

x=b±b24ac2a

Substitute the values of a,b and c in the above expression.

x=(0.007)±(0.007)24(3.91)(0.00011592)2(3.91)x=0.0046M

The equilibrium concentration of H2O(g) is calculated by the formula,

EquilibriumconcentrationofH2O(g)=(0.056x)

Substitute the calculated value of x in the above expression.

EquilibriumconcentrationofH2O(g)=(0.056x)=(0.05600.0046)M=0.0514M_

The equilibrium concentration of Cl2O(g) is calculated by the formula,

EquilibriumconcentrationofCl2O(g)=(0.023x)

Substitute the calculated value of x in the above expression.

EquilibriumconcentrationofCl2O(g)=(0.023x)=(0.0230.0046)M=0.0184M_

The equilibrium concentration of HOCl(g) is calculated by the formula,

EquilibriumconcentrationofHOCl(g)=(2x)

Substitute the calculated value of x in the above expression.

EquilibriumconcentrationofHOCl(g)=(2x)=(2(0.0046))M=0.0092M_

Conclusion

The equilibrium concentration of H2O(g) 0.0514M_ and Cl2O(g) is 0.0184M_ and of HOCl(g) is 0.0092M_ .

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The value of the equilibrium constant for the reaction between H2O and Cl2O and the initial mass of the species involved is given. The equilibrium concentrations of the three gases involved in the system are to be calculated.

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. When the equilibrium constant is expressed in terms of concentration, it is represented as K .

To determine: The equilibrium concentrations of the three gases involved in the system.

Answer to Problem 54E

The equilibrium concentrations of the three gases are calculated as follows.

Explanation of Solution

Given

The stated reaction is,

2HOCl(g)H2O(g)+Cl2O(g)

The initial moles of HOCl(g) are 1.0mole .

The value of K for this reaction will be 10.090 .

The volume of the container is 2.0L .

The concentration of a reactant is calculated by the formula,

Concentration=MolesVolume(L)

The initial concentration of HOCl is calculated by the formula,

ConcentrationofHOCl=MolesofHOClVolumeoftheflask(L)

Substitute the given values of the number of moles of HOCl and the volume of the flask in the above expression.

ConcentrationofHOCl=1.0mole2.00L=0.5M

The concentration of HOCl consumed is assumed to be 2x .

The equilibrium concentrations are represented as,

2HOCl(g)H2O(g)+Cl2O(g)Initialconcentration0.500Change-2x+x+xEquilibriumconcentration0.5-2xxx

At equilibrium, the equilibrium ratio is expressed by the formula,

K=ConcentrationofproductsConcentrationofreactants

Where,

  • K is the equilibrium constant.

The equilibrium ratio for the given reaction is,

K=[H2O][Cl2O][HOCl]2 (1)

According to the formulated ICE table,

The equilibrium concentration of H2O(g) is (x)M .

The equilibrium initial concentration of Cl2O(g) is (x)M .

The equilibrium initial concentration of HOCl(g) is (0.52x)M .

Substitute these values in equation (1).

K=[H2O][Cl2O][HOCl]2K=[x][x][0.52x]2

The value of K for this reaction will be 10.090 .

Substitute the value of K in the above expression.

10.090=[x]2[0.52x]2

Taking square root on both sides,

10.3=[x][0.52x]

Simplify the above expression.

0.3x=0.52xx=0.217

The equilibrium concentration of H2O(g) is equal to the equilibrium concentration of Cl2O(g) . These are equal to x .

The equilibrium concentration of H2O(g) is equal to the equilibrium concentration of Cl2O(g) , that is, 0.217M_ .

The equilibrium concentration of HOCl(g) is calculated by the formula,

EquilibriumconcentrationofHOCl(g)=(0.52x)

Substitute the calculated value of x in the above expression.

EquilibriumconcentrationofHOCl(g)=(0.52x)=(0.52(0.217))M=0.066M_

Conclusion

The equilibrium concentration of H2O(g) and Cl2O(g) is 0.217M_ and of HOCl(g) is 0.066M_

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Chapter 12 Solutions

Chemistry: An Atoms First Approach

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