Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
9th Edition
ISBN: 9781133949640
Author: John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher: Cengage Learning
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Question
Chapter 18, Problem 84SCQ

(a)

Interpretation Introduction

Interpretation:

The  ΔSo(system) for reaction between NO(g) and Cl2(g) should be determined.

Concept introduction:

The universe consists of two parts, systems and surroundings. The entropy change for the universe is the sum of entropy change for the system and for surroundings.

  ΔSo(universe)= ΔSo(system)+ΔSo(surroundings)

The ΔSo(universe) should be greater than zero for a spontaneous process.

The  ΔSo(system) can be calculated by the following expression,

  ΔSo(system)=ΔrS°nS°(products)-nS°(reactants)

The ΔSo(surroundings) can be calculated by the following expression,

  ΔSo(surroundings)=rHoT

Here, ΔrH° is the enthalpy change for the reaction.

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔGo. It is related to entropy and entropy by the following expression,

  ΔGo= ΔHo- TΔSo

ΔGo is also related to the equilibrium constant K by the equation,

  ΔrGo= -RTlnKp

The rearranged expression is,

  Kp= eΔrGoRT

(a)

Expert Solution
Check Mark

Answer to Problem 84SCQ

The  ΔSo(system) for the reaction is 60.5 J/Kmol-rxn.

Explanation of Solution

The  ΔSo(system) for the reaction is calculated below.

Given:

Refer to Appendix L for the values of standard entropies.

The standard entropy of NOCl(g) is 261.8 J/Kmol.

The standard entropy of NO(g) is 210.76 J/Kmol.

The standard entropy of Cl2(g) is 223.08 J/Kmol.

The balanced chemical equation is:

  NO(g)+12Cl2(g)NOCl(g)

The  ΔSo(system) can be calculated by the following expression,

  ΔSo(system)=ΔrS°nS°(products)-nS°(reactants)=[(1 mol NOCl(g)/mol-rxn)S°[NOCl(g)]-[(1 mol NO(g)/mol-rxn)S°[NO(g)]+(0.5 mol Cl2(g)/mol-rxn)S°[Cl2(g)]]]

Substitute the values,

  ΔSo(system)[(1 mol NOCl(g)/mol-rxn)(261.8 J/K×mol)-[(1 mol NO(g)/mol-rxn)(210.76 J/K×mol)+(0.5 mol Cl2(g)/mol-rxn)(223.08 J/K×mol)]]ΔSo= -60.5 J/K×mol-rxn

(b)

Interpretation Introduction

Interpretation:

It should be identified that  ΔSo(system) changes with temperature or not.

Concept introduction:

The universe consists of two parts, systems and surroundings. The entropy change for the universe is the sum of entropy change for the system and for surroundings.

  ΔSo(universe)= ΔSo(system)+ΔSo(surroundings)

The ΔSo(universe) should be greater than zero for a spontaneous process.

The  ΔSo(system) can be calculated by the following expression,

  ΔSo(system)=ΔrS°nS°(products)-nS°(reactants)

The ΔSo(surroundings) can be calculated by the following expression,

  ΔSo(surroundings)=rHoT

Here, ΔrH° is the enthalpy change for the reaction.

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔGo. It is related to entropy and entropy by the following expression,

  ΔGo= ΔHo- TΔSo

ΔGo is also related to the equilibrium constant K by the equation,

  ΔrGo= -RTlnKp

The rearranged expression is,

  Kp= eΔrGoRT

(b)

Expert Solution
Check Mark

Answer to Problem 84SCQ

The value of ΔSo(system) changes with temperature.

Explanation of Solution

The entropy of the system is dependent upon temperature. Entropy of any system increases with increase in the temperature due to the heat which is added to the system at higher temperatures.

(c)

Interpretation Introduction

Interpretation:

It should be identified that ΔSo(surroundings) changes with temperature.

Concept introduction:

The universe consists of two parts, systems and surroundings. The entropy change for the universe is the sum of entropy change for the system and for surroundings.

  ΔSo(universe)= ΔSo(system)+ΔSo(surroundings)

The ΔSo(universe) should be greater than zero for a spontaneous process.

The  ΔSo(system) can be calculated by the following expression,

  ΔSo(system)=ΔrS°nS°(products)-nS°(reactants)

The ΔSo(surroundings) can be calculated by the following expression,

  ΔSo(surroundings)=rHoT

Here, ΔrH° is the enthalpy change for the reaction.

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔGo. It is related to entropy and entropy by the following expression,

  ΔGo= ΔHo- TΔSo

ΔGo is also related to the equilibrium constant K by the equation,

  ΔrGo= -RTlnKp

The rearranged expression is,

  Kp= eΔrGoRT

(c)

Expert Solution
Check Mark

Answer to Problem 84SCQ

The ΔSo(surroundings) for any reaction changes with temperature.

Explanation of Solution

The ΔSo(surroundings) can be calculated by the following expression,

  ΔSo(surroundings)rHoT

Here, ΔrHo is the enthalpy change for the reaction.

Thus, ΔSo(surroundings) is inversely related to temperature.

(d)

Interpretation Introduction

Interpretation:

It should be identified that ΔSo(surroundings) changes with increase in temperature or not.

Concept introduction:

The universe consists of two parts, systems and surroundings. The entropy change for the universe is the sum of entropy change for the system and for surroundings.

  ΔSo(universe)= ΔSo(system)+ΔSo(surroundings)

The ΔSo(universe) should be greater than zero for a spontaneous process.

The  ΔSo(system) can be calculated by the following expression,

  ΔSo(system)=ΔrS°nS°(products)-nS°(reactants)

The ΔSo(surroundings) can be calculated by the following expression,

  ΔSo(surroundings)=rHoT

Here, ΔrH° is the enthalpy change for the reaction.

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔGo. It is related to entropy and entropy by the following expression,

  ΔGo= ΔHo- TΔSo

ΔGo is also related to the equilibrium constant K by the equation,

  ΔrGo= -RTlnKp

The rearranged expression is,

  Kp= eΔrGoRT

(d)

Expert Solution
Check Mark

Answer to Problem 84SCQ

The value of ΔSo(universe) changes with an increase in temperature.

Explanation of Solution

The entropy change for the universe is the sum of entropy change for the system and for surroundings.

  ΔSo(universe)= ΔSo(system)+ΔSo(surroundings)

Both ΔSo(system) and ΔSo(surroundings) are dependent upon temperatures. Thus, the value of ΔSo(universe) changes with an increase in temperature.

(e)

Interpretation Introduction

Interpretation:

It should be identified that does exothermic reaction will always results in positive ΔSo(universe) value.

Concept introduction:

The universe consists of two parts, systems and surroundings. The entropy change for the universe is the sum of entropy change for the system and for surroundings.

  ΔSo(universe)= ΔSo(system)+ΔSo(surroundings)

The ΔSo(universe) should be greater than zero for a spontaneous process.

The  ΔSo(system) can be calculated by the following expression,

  ΔSo(system)=ΔrS°nS°(products)-nS°(reactants)

The ΔSo(surroundings) can be calculated by the following expression,

  ΔSo(surroundings)=rHoT

Here, ΔrH° is the enthalpy change for the reaction.

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔGo. It is related to entropy and entropy by the following expression,

  ΔGo= ΔHo- TΔSo

ΔGo is also related to the equilibrium constant K by the equation,

  ΔrGo= -RTlnKp

The rearranged expression is,

  Kp= eΔrGoRT

(e)

Expert Solution
Check Mark

Answer to Problem 84SCQ

The exothermic reaction does not necessarily leads to a positive value of ΔSo(universe).

Explanation of Solution

The exothermic reaction have negative value of free energy change which means that the ΔSo(surroundings) will be positive. However, if the value of ΔSo(system) is negative and has greater magnitude than the ΔSo(surroundings) than the sign of ΔSo(universe) is negative. The reaction in such a case is non-spontaneous.

(f)

Interpretation Introduction

Interpretation:

It should be identified that reaction of NO(g) and Cl2(g) is spontaneous or not.

Concept introduction:

The universe consists of two parts, systems and surroundings. The entropy change for the universe is the sum of entropy change for the system and for surroundings.

  ΔSo(universe)= ΔSo(system)+ΔSo(surroundings)

The ΔSo(universe) should be greater than zero for a spontaneous process.

The  ΔSo(system) can be calculated by the following expression,

  ΔSo(system)=ΔrS°nS°(products)-nS°(reactants)

The ΔSo(surroundings) can be calculated by the following expression,

  ΔSo(surroundings)=rHoT

Here, ΔrH° is the enthalpy change for the reaction.

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔGo. It is related to entropy and entropy by the following expression,

  ΔGo= ΔHo- TΔSo

ΔGo is also related to the equilibrium constant K by the equation,

  ΔrGo= -RTlnKp

The rearranged expression is,

  Kp= eΔrGoRT

(f)

Expert Solution
Check Mark

Answer to Problem 84SCQ

The reaction is spontaneous at 298 K.

The reaction is not spontaneous at 700 K.

Explanation of Solution

The free energy change for the given reaction is calculated below.

Given:

Refer to Appendix L for the values of standard entropies.

The standard enthalpy of NOCl(g) is 51.71 kJ/mol.

The standard enthalpy of NO(g) is 90.29 kJ/mol.

The standard enthalpy of Cl2(g) is 0 kJ/mol.

The balanced chemical equation is:

  NO(g)+12Cl2(g)NOCl(g)

The ΔrHo can be calculated by the following expression,

  ΔrH°=fH°(products)fH°(reactants)=[(1 mol NOCl(g)/mol-rxn)ΔfH°[NOCl(g)]-[(1 mol NO(g)/mol-rxn)ΔfH°[NO(g)]+(0.5 mol Cl2(g)/mol-rxn)ΔfH°[Cl2(g)]]]

Substitute the values,

  ΔrH°=[(1 mol NOCl(g)/mol-rxn)(51.71 kJ/mol)-[(1 mol NO(g)/mol-rxn)(90.29 kJ/mol)+(0.5 mol Cl2(g)/mol-rxn)(0 kJ/mol)]]ΔrH°= -38.58 J/K×mol-rxn

Now,

  ΔGo=ΔHo-TΔSo

Substitute the values at temperature 298 K,

  ΔGo=-38.58 kJ/mol-rxn-[(298K)(-60.5 J/K×mol-rxn)](1 kJ1000 J)=-20.551 kJ/mol-rxn

Thus, the reaction is spontaneous at 298 K.

Substitute the values at temperature 700 K,

ΔGo= -38.58 kJ/mol-rxn-[(700K)(-60.5 J/K×mol-rxn)](1 kJ1000 J)= +3.77 kJ/mol-rxn

Thus, the reaction is not spontaneous at 700 K.

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Chapter 18 Solutions

Chemistry & Chemical Reactivity

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