Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Chapter 17, Problem 96AE

Consider the following reaction:

H 2 O ( g ) + CI 2 O ( g ) 2 HOCI ( g ) K 298 = 0.090

For Cl2O(g),

Δ G f o = 97.9 KJ/mol Δ H f o = 80.3 KJ/mol S = 266.1 J / K mol

a. Calculate ∆G° for the reaction using the equation ∆G°=−RT In(K).

b. Use bond energy values (Table 3-3) to estimate ∆H° for the reaction.

c. Use the results from parts a and b to estimate ∆S° for the reaction.

d. Estimate Δ H f o and S° for HOCI(g).

e. Estimate the value of K at 500. K.

f. Calculate ∆G at 25°C when P H 2 O = 18 torr , P CI 2 O = 2.0 torr , a n d P HOCI = 0.10 torr .

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The reaction corresponding to the formation of HOCl(g) , its equilibrium constant at 298K and values of ΔGf°,ΔHf°,S° are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, K , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

ΔG=0Q=K

The expression for free energy change is,

ΔG°=RTln(K)

The formula of ΔS° is,

ΔS°=npΔS°(product)nfΔS°(reactant)

To determine: The value of ΔG° for the given reaction, using the equation ΔG°=RTln(K) .

Explanation of Solution

Explanation

Given

Temperature is 298K .

The value of equilibrium constant is 0.090 .

The stated reaction is,

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

Formula

The expression for free energy change is,

ΔG°=RTln(K)

Where,

  • ΔG° is the standard Gibbs free energy change.
  • R is the gas law constant (8.3145J/Kmol) .
  • T is the absolute temperature.
  • K is the equilibrium constant.

Substitute the values of R,T and K in the above expression.

ΔG°=RTln(K)=(8.3145J/Kmol)(298K)ln(0.090)=6.0×103J/mol

The conversion of joule per mole (J/mol) into kilo-joule per mole (kJ/mol) is done as,

1J/mol=103kJ/mol

Hence,

The conversion of 6.0×103J/mol into kilo-joule per mole is,

6.0×103J/mol=(6.0×103×103)kJ/mol=6.0kJ/mol_

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The reaction corresponding to the formation of HOCl(g) , its equilibrium constant at 298K and values of ΔGf°,ΔHf°,S° are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, K , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

ΔG=0Q=K

The expression for free energy change is,

ΔG°=RTln(K)

The formula of ΔS° is,

ΔS°=npΔS°(product)nfΔS°(reactant)

To determine: The value of ΔG° for the given reaction, using the equation ΔG°=RTln(K) .

Explanation of Solution

Explanation

The reaction that takes place is,

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

Refer to Table 3-3 .

The value of bond energy, D°(kJ/mol) , for the given reactant and product is,

Bonds D°(kJ/mol)
OH 467
ClO 203
OCl 203

The formula of ΔH° is,

ΔH°=npD°(bondsbroken)nfD°(bondsformed)

Where,

  • np is the number of moles of each product.
  • nr is the number of moles each reactant.
  • D°(bonds broken) is the bond energy of broken bonds.
  • D°(bondsformed) is the bond energy of bonds formed.

Substitute all values from the table in the above equation.

ΔH°=npD°(bondsbroken)nfD°(bondsformed)=(2×DOH+2×DClO)(2×DOH+2×DOCl)=[(2×467)+(2×203){(2×467)+(2×203)}]kJ/mol

Simplify the above equation.

ΔH°=[(2×467)+(2×203){(2×467)+(2×203)}]kJ/mol=0kJ/mol_

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The reaction corresponding to the formation of HOCl(g) , its equilibrium constant at 298K and values of ΔGf°,ΔHf°,S° are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, K , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

ΔG=0Q=K

The expression for free energy change is,

ΔG°=RTln(K)

The formula of ΔS° is,

ΔS°=npΔS°(product)nfΔS°(reactant)

To determine: The value of ΔG° for the given reaction, using the equation ΔG°=RTln(K) .

Explanation of Solution

Explanation

The value of ΔG° is 6.0×103J/mol .

The value of ΔH° is 0J/mol .

Temperature is 298K .

The formula of ΔG° is,

ΔG°=ΔH°TΔS°

Where,

  • ΔH° is the standard enthalpy of reaction.
  • ΔG° is the free energy change.
  • T is the given temperature.
  • ΔS° is the standard entropy of the reaction.

Substitute the values of ΔG°,ΔH° and T in equation (1).

ΔG°=ΔH°TΔS°6.0×103J/mol=0298(ΔS°)ΔS°=20.13J/Kmol_

(d)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The reaction corresponding to the formation of HOCl(g) , its equilibrium constant at 298K and values of ΔGf°,ΔHf°,S° are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, K , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

ΔG=0Q=K

The expression for free energy change is,

ΔG°=RTln(K)

The formula of ΔS° is,

ΔS°=npΔS°(product)nfΔS°(reactant)

To determine: The value of ΔG° for the given reaction, using the equation ΔG°=RTln(K) .

Explanation of Solution

Explanation

Refer Appendix 4 .

The value of standard enthalpy of H2O is 241.82kJ/mol .

The value of standard enthalpy of Cl2O is 80.3kJ/mol .

The formula of ΔH° is,

ΔH°=npΔH°(product)nfΔH°(reactant)

Where,

  • ΔH° is the standard enthalpy of reaction.
  • np is the number of moles of each product.
  • nr is the number of moles each reactant.
  • ΔH°(product) is the standard enthalpy of product at a pressure of 1atm .
  • ΔH°(reactant) is the standard enthalpy of reactant at a pressure of 1atm .

Substitute all values from the table in the above equation.

ΔH°=npΔH°(product)nfΔH°(reactant)=[2(ΔH°HOCl)(241.82)(80.3)]kJ/mol=80.76kJ/mol_

Refer Appendix 4 .

The value of standard entropy of H2O is 188.83J/Kmol .

The value of standard entropy of Cl2O is 266.1J/Kmol .

The formula of ΔS° is,

ΔS°=npΔS°(product)nfΔS°(reactant)

Where,

  • ΔS° is the standard enthalpy of reaction.
  • np is the number of moles of each product.
  • nr is the number of moles each reactant.
  • ΔS°(product) is the standard entropy of the product at a pressure of 1atm .
  • ΔS°(reactant) is the standard entropy of the reactant at a pressure of 1atm .

Substitute all values from the table in the above equation.

ΔS°=npΔS°(product)nfΔS°(reactant)20.13J/Kmol=[2(ΔS°HOCl)(188.83)(266.1)]J/KmolΔS°HOCl=217.4J/Kmol_

(e)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The reaction corresponding to the formation of HOCl(g) , its equilibrium constant at 298K and values of ΔGf°,ΔHf°,S° are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, K , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

ΔG=0Q=K

The expression for free energy change is,

ΔG°=RTln(K)

The formula of ΔS° is,

ΔS°=npΔS°(product)nfΔS°(reactant)

To determine: The value of ΔG° for the given reaction, using the equation ΔG°=RTln(K) .

Explanation of Solution

Explanation

The value of ΔS° is 20.3J/Kmol .

The value of ΔH° is 0J/mol .

Temperature is 500K .

It is assumed that ΔH° and ΔS° are independent of T .

The formula of ΔG° is,

ΔG°=ΔH°TΔS°

Where,

  • ΔH° is the standard enthalpy of reaction.
  • ΔG° is the free energy change.
  • T is the given temperature.
  • ΔS° is the standard entropy of the reaction.

Substitute the values of ΔG°,ΔH° and T in equation (1).

ΔG°=ΔH°TΔS°=0500(20.13J/Kmol)=1.01×104J/mol_

Formula

The expression for free energy change is,

ΔG°=RTln(K)ln(K)=ΔG°RT

Where,

  • ΔG is the free energy change for a reaction at specified pressure.
  • R is the gas law constant (8.3145J/Kmol) .
  • K is the equilibrium constant.
  •   T is the absolute temperature.

Substitute the values of R,ΔG° and T in the above expression.

ln(K)=ΔG°RT=1.01×104J/mol(8.3145J/Kmol)(500K)K=0.088_

(f)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The reaction corresponding to the formation of HOCl(g) , its equilibrium constant at 298K and values of ΔGf°,ΔHf°,S° are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, K , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

ΔG=0Q=K

The expression for free energy change is,

ΔG°=RTln(K)

The formula of ΔS° is,

ΔS°=npΔS°(product)nfΔS°(reactant)

To determine: The value of ΔG° for the given reaction, using the equation ΔG°=RTln(K) .

Explanation of Solution

Explanation

Given

Partial pressure of H2O , PH2O , is 18torr .

Partial pressure of Cl2O , PCl2O , is 2.0torr .

Partial pressure of HOCl , PHOCl , is 0.10torr .

The stated reaction is,

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

If the partial pressure of reactant and product is given, then equilibrium pressure is expressed as Kp and its expression is written as,

Kp=(Partialpressure of product)a(Partialpressure of reactant)b

Where,

  • a is the number of moles of product.
  • b is the number of moles of reactant.

The equilibrium constant expression for the given reaction is,

Kp=(PHOCl)2(PH2O)(PCl2O)

Substitute the values of PCl2O,PH2O and PHOCl in the above expression.

Kp=(PHOCl)2(PH2O)(PCl2O)=(0.10torr)2(18torr)(2.0torr)=2.7×104_

The value of Kp is 2.7×104 .

Temperature is 298K .

The expression for free energy change is,

ΔG=ΔG°+RTln(Kp)

Where,

  • ΔG is the free energy change for a reaction at specified pressure.
  • R is the gas law constant (8.3145J/Kmol) .
  • T is the absolute temperature.
  • Kp is the equilibrium constant.

Substitute the values of R,Kp and T in the equation.

ΔG=ΔG°+RTln(Kp)=[6.0×103+(8.3145)(298)ln(2.7×104)]J/mol=14.3×103J/mol

The conversion of joule-per mole (J/mol) into kilo-joule per mole (kJ/mol) is done as,

1J/mol=103kJ/mol

Hence,

The conversion of 14.3×103J/mol into kilo- joule per mole is,

14.3×103J/mol=(14.3×103×103)kJ/mol=14.3kJ/mol_

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