Using values of Δ f H ° and S°, calculate the standard molar free energy of formation, Δ f G °, for each of the following: (a) Ca(OH) 2 (s) (b) Cl(g) (c) Na 2 CO 3 (s) Compare your calculated values of Δ f G° with those listed in Appendix L. Which of these formation reactions are predicted to be product-favored at equilibrium at 25 °C?

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Answer 1

Textbook Question

Chapter 18, Problem 18PS

Using values of Δ_fH° and S°, calculate the standard molar free energy of formation, Δ_fG°, for each of the following:

(a) Ca(OH)₂(s)
(b) Cl(g)
(c) Na₂CO₃(s)

Compare your calculated values of Δ_fG° with those listed in Appendix L. Which of these formation reactions are predicted to be product-favored at equilibrium at 25 °C?

(a)

Expert Solution

Interpretation Introduction

Interpretation:

The the standard molar free energy for formation of $Ca(OH)2(s)$ should be calculated and compared with the values placed in appendix L. It should be identified that whether the reaction is product favored at equilibrium.

Concept introduction:

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

$ΔGo=ΔHo-TΔSo$

The sign of $ΔGo$ should be positive for a product-favored reaction. Thus, spontaneous reactions are referred to those that have negative free energy formation.

Answer to Problem 18PS

The standard molar energy of formation for $Ca(OH)2(s)$ is $−898.486 kJ/mol-rxn$ .

Explanation of Solution

The standard molar energy of formation for $Ca(OH)2(s)$ is calculated below.

Given:

The Appendix L referred for the values of standard entropies and enthalpies.

$Ca(s)+H2(g)+O2(g)→Ca(OH)2(s)ΔfH°(kJ/mol)000-986.09So(J/K×mol)41.59130.7205.0783.39$

The enthalpy change is expressed as,

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)= [(1 mol Ca(OH)2(s)/mol-rxn)ΔfH°[Ca(OH)2(s)]-[(1 mol H2(g)/mol-rxn)ΔfH°[H2(g)]+(1 mol O2(g)/mol-rxn)ΔfH°[O2(g)]+(1 mol Ca(s)/mol-rxn)ΔfH°[Ca(s)]] ]$

Substituting the values,

$ΔrH°= [(1 mol Ca(OH)2(s)/mol-rxn)(-986.09 kJ/mol)[(1 mol H2(g)/mol-rxn)(0 kJ/mol)+(1 mol O2(g)/mol-rxn)(0 kJ/mol)+(1 mol Na(s)/mol-rxn)(0 kJ/mol)] ]=-986.09 kJ/mol-rxn$

The entropy change is expressed as,

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)= [(1 mol Ca(OH)2(s)/mol-rxn)S°[Ca(OH)2(s)]-[(1 mol H2(g)/mol-rxn)S°[H2(g)]+(1 mol O2(g)/mol-rxn)S°[O2(g)]+(1 mol Ca(s)/mol-rxn)S°[Ca(s)]] ]$

Substituting the values,

$ΔrS°= [(1 mol Ca(OH)2(s)/mol-rxn)(83.39 J/K×mol)-[(1 mol H2(g)/mol-rxn)(130.7 J/K×mol)+(1 mol O2(g)/mol-rxn)(205.07 J/K×mol)+(1 mol Ca(s)/mol-rxn)(41.59 J/K×mol)] ]=-293.97 J/K×mol-rxn$

Now,

$ΔGo= ΔHo-TΔSo$

Substitute the value of $ΔrHo$ and $ΔrSo$ .

$ΔGo= - 986.09 kJ/mol-[(298 K)(-293.97 J/K×mol-rxn)](1 kJ1000 J)= - 898.486 kJ/mol-rxn$

The value in Appendix L is $−898.43 kJ/mol-rxn$ .

The value of free energy change is negative. Thus, the reaction is product-favored at equilibrium.

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The the standard molar free energy for formation of $Cl(g)$ should be calculated and compared with the values placed in appendix L. It should be identified that whether the reaction is product favored at equilibrium.

Concept introduction:

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

$ΔGo=ΔHo-TΔSo$

The sign of $ΔGo$ should be positive for a product-favored reaction. Thus, spontaneous reactions are referred to those that have negative free energy formation.

Answer to Problem 18PS

The standard molar energy of formation for $Cl(g)$ is $105.3 kJ/mol-rxn$ .

Explanation of Solution

The standard molar energy of formation for $Cl(g)$ is calculated below.

Given:

The Appendix L referred for the values of standard entropies and enthalpies.

$12Cl2(g)→Cl(g)ΔfH°(kJ/mol)0121.3So(J/K×mol)223.08165.19$

The enthalpy change is expressed as,

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)= [(1 mol Cl(g)/mol-rxn)ΔfH°[Cl(g)]-(0.5 mol Cl2(g)/mol-rxn)ΔfH°[Cl2(g)]]$

Substituting the respective values,

$ΔrH°= [(1 mol Cl(g)/mol-rxn)(121.3 kJ/mol)-(0.5 mol Cl2(g)/mol-rxn)(0 kJ/mol)]= 121.3 kJ/mol-rxn$

The entropy change is expressed as,

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)=[(1 mol Cl(g)/mol-rxn)S°[Cl(g)]-(0.5 mol Cl2(g)/mol-rxn)S°[Cl2(g)]]$

Substituting the respective values,

$ΔrS°= [(1 mol Cl(g)/mol-rxn)(165.19 J/K×mol)-(0.5 mol Cl2(g)/mol-rxn)(223.08 J/K×mol)]= 53.65 J/K×mol-rxn$

Now,

$ΔfGo= ΔfHo- TΔSo$

Substitute the value of $ΔrHo$ and $ΔrSo$ .

$ΔGo= 121.3 kJ/mol-[(298 K)(53.65 J/K×mol-rxn)](1 kJ1000 J)= 105.31 kJ/mol-rxn$

The value in Appendix L is $105.3 kJ/mol-rxn$ .

The value of free energy change is positive. Thus, the reaction is reactant-favored at equilibrium.

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The the standard molar free energy for formation of $Na2CO3(s)$ should be calculated and compared with the values placed in appendix L. It should be identified that whether the reaction is product favored at equilibrium.

Concept introduction:

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

$ΔGo=ΔHo-TΔSo$

The sign of $ΔGo$ should be positive for a product-favored reaction. Thus, spontaneous reactions are referred to those that have negative free energy formation.

Answer to Problem 18PS

The standard molar energy of formation for $Na2CO3(s)$ is $−1047.08 kJ/mol-rxn$ .

Explanation of Solution

The standard molar energy of formation for $Na2CO3(s)$ is calculated below.

Given:

The Appendix L referred for the values of standard entropies and enthalpies.

$2Na(s)+C(s)+32O2(g)→Na2CO3(s)ΔfH°(kJ/mol)000−1130.77S∘(J/K⋅mol)51.215.6205.07134.79$

The enthalpy change is expressed as,

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)=[(1 mol Na2CO3(s)/mol-rxn)ΔfH°[Na2CO3]-[(1 mol C(s)/mol-rxn)ΔfH°[ C(s)]+(1.5 mol O2(g)/mol-rxn)ΔfH°[O2(g)]+(2 mol Na(s)/mol-rxn)ΔfH°[Na(s)]] ]$

Substituting the values,

$ΔrH°=[(1 mol Na2CO3(s)/mol-rxn)(-1130.77 kJ/mol)[(1 mol C(s)/mol-rxn)(0 kJ/mol)+(1.5 mol O2(g)/mol-rxn)(0 kJ/mol)+(2 mol Na(s)/mol-rxn)(0 kJ/mol)] ]=-1130.77 kJ/mol-rxn$

The entropy change is expressed as,

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)=[(1 mol Na2CO3(s)/mol-rxn)S°[Na2CO3]-[(1 mol C(s)/mol-rxn)S°[ C(s)]+(1.5 mol O2(g)/mol-rxn)S°[O2(g)]+(2 mol Na(s)/mol-rxn)S°[Na(s)]] ]$

Substituting the values,

$ΔrS°=[(1 mol Na2CO3(s)/mol-rxn)(134.79 J/K⋅mol)−[(1 mol C(s)/mol-rxn)(5.6 J/K⋅mol)+(1.5 mol O2(g)/mol-rxn)(205.07 J/K⋅mol)+(2 mol Na(s)/mol-rxn)(51.21 J/K⋅mol)] ]=−280.83 J/K⋅mol-rxn$

Now,

$ΔGo=ΔHo-TΔSo$

Substituting the value of $ΔrHo$ and $ΔrSo$ .

$ΔGo= -1130.77 kJ/mol-[(298 K)(-280.83 J/K×mol-rxn)](1 kJ1000 J)= -1047.08 kJ/mol-rxn$

The value in Appendix L is $−1047.08 kJ/mol-rxn$ .

The value of free energy change is negative. Thus, the reaction is product-favored at equilibrium.

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Answer 2

Textbook Question

Chapter 18, Problem 18PS

Using values of Δ_fH° and S°, calculate the standard molar free energy of formation, Δ_fG°, for each of the following:

(a) Ca(OH)₂(s)
(b) Cl(g)
(c) Na₂CO₃(s)

Compare your calculated values of Δ_fG° with those listed in Appendix L. Which of these formation reactions are predicted to be product-favored at equilibrium at 25 °C?

(a)

Expert Solution

Interpretation Introduction

Interpretation:

The the standard molar free energy for formation of $Ca(OH)2(s)$ should be calculated and compared with the values placed in appendix L. It should be identified that whether the reaction is product favored at equilibrium.

Concept introduction:

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

$ΔGo=ΔHo-TΔSo$

The sign of $ΔGo$ should be positive for a product-favored reaction. Thus, spontaneous reactions are referred to those that have negative free energy formation.

Answer to Problem 18PS

The standard molar energy of formation for $Ca(OH)2(s)$ is $−898.486 kJ/mol-rxn$ .

Explanation of Solution

The standard molar energy of formation for $Ca(OH)2(s)$ is calculated below.

Given:

The Appendix L referred for the values of standard entropies and enthalpies.

$Ca(s)+H2(g)+O2(g)→Ca(OH)2(s)ΔfH°(kJ/mol)000-986.09So(J/K×mol)41.59130.7205.0783.39$

The enthalpy change is expressed as,

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)= [(1 mol Ca(OH)2(s)/mol-rxn)ΔfH°[Ca(OH)2(s)]-[(1 mol H2(g)/mol-rxn)ΔfH°[H2(g)]+(1 mol O2(g)/mol-rxn)ΔfH°[O2(g)]+(1 mol Ca(s)/mol-rxn)ΔfH°[Ca(s)]] ]$

Substituting the values,

$ΔrH°= [(1 mol Ca(OH)2(s)/mol-rxn)(-986.09 kJ/mol)[(1 mol H2(g)/mol-rxn)(0 kJ/mol)+(1 mol O2(g)/mol-rxn)(0 kJ/mol)+(1 mol Na(s)/mol-rxn)(0 kJ/mol)] ]=-986.09 kJ/mol-rxn$

The entropy change is expressed as,

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)= [(1 mol Ca(OH)2(s)/mol-rxn)S°[Ca(OH)2(s)]-[(1 mol H2(g)/mol-rxn)S°[H2(g)]+(1 mol O2(g)/mol-rxn)S°[O2(g)]+(1 mol Ca(s)/mol-rxn)S°[Ca(s)]] ]$

Substituting the values,

$ΔrS°= [(1 mol Ca(OH)2(s)/mol-rxn)(83.39 J/K×mol)-[(1 mol H2(g)/mol-rxn)(130.7 J/K×mol)+(1 mol O2(g)/mol-rxn)(205.07 J/K×mol)+(1 mol Ca(s)/mol-rxn)(41.59 J/K×mol)] ]=-293.97 J/K×mol-rxn$

Now,

$ΔGo= ΔHo-TΔSo$

Substitute the value of $ΔrHo$ and $ΔrSo$ .

$ΔGo= - 986.09 kJ/mol-[(298 K)(-293.97 J/K×mol-rxn)](1 kJ1000 J)= - 898.486 kJ/mol-rxn$

The value in Appendix L is $−898.43 kJ/mol-rxn$ .

The value of free energy change is negative. Thus, the reaction is product-favored at equilibrium.

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The the standard molar free energy for formation of $Cl(g)$ should be calculated and compared with the values placed in appendix L. It should be identified that whether the reaction is product favored at equilibrium.

Concept introduction:

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

$ΔGo=ΔHo-TΔSo$

The sign of $ΔGo$ should be positive for a product-favored reaction. Thus, spontaneous reactions are referred to those that have negative free energy formation.

Answer to Problem 18PS

The standard molar energy of formation for $Cl(g)$ is $105.3 kJ/mol-rxn$ .

Explanation of Solution

The standard molar energy of formation for $Cl(g)$ is calculated below.

Given:

The Appendix L referred for the values of standard entropies and enthalpies.

$12Cl2(g)→Cl(g)ΔfH°(kJ/mol)0121.3So(J/K×mol)223.08165.19$

The enthalpy change is expressed as,

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)= [(1 mol Cl(g)/mol-rxn)ΔfH°[Cl(g)]-(0.5 mol Cl2(g)/mol-rxn)ΔfH°[Cl2(g)]]$

Substituting the respective values,

$ΔrH°= [(1 mol Cl(g)/mol-rxn)(121.3 kJ/mol)-(0.5 mol Cl2(g)/mol-rxn)(0 kJ/mol)]= 121.3 kJ/mol-rxn$

The entropy change is expressed as,

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)=[(1 mol Cl(g)/mol-rxn)S°[Cl(g)]-(0.5 mol Cl2(g)/mol-rxn)S°[Cl2(g)]]$

Substituting the respective values,

$ΔrS°= [(1 mol Cl(g)/mol-rxn)(165.19 J/K×mol)-(0.5 mol Cl2(g)/mol-rxn)(223.08 J/K×mol)]= 53.65 J/K×mol-rxn$

Now,

$ΔfGo= ΔfHo- TΔSo$

Substitute the value of $ΔrHo$ and $ΔrSo$ .

$ΔGo= 121.3 kJ/mol-[(298 K)(53.65 J/K×mol-rxn)](1 kJ1000 J)= 105.31 kJ/mol-rxn$

The value in Appendix L is $105.3 kJ/mol-rxn$ .

The value of free energy change is positive. Thus, the reaction is reactant-favored at equilibrium.

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The the standard molar free energy for formation of $Na2CO3(s)$ should be calculated and compared with the values placed in appendix L. It should be identified that whether the reaction is product favored at equilibrium.

Concept introduction:

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

$ΔGo=ΔHo-TΔSo$

The sign of $ΔGo$ should be positive for a product-favored reaction. Thus, spontaneous reactions are referred to those that have negative free energy formation.

Answer to Problem 18PS

The standard molar energy of formation for $Na2CO3(s)$ is $−1047.08 kJ/mol-rxn$ .

Explanation of Solution

The standard molar energy of formation for $Na2CO3(s)$ is calculated below.

Given:

The Appendix L referred for the values of standard entropies and enthalpies.

$2Na(s)+C(s)+32O2(g)→Na2CO3(s)ΔfH°(kJ/mol)000−1130.77S∘(J/K⋅mol)51.215.6205.07134.79$

The enthalpy change is expressed as,

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)=[(1 mol Na2CO3(s)/mol-rxn)ΔfH°[Na2CO3]-[(1 mol C(s)/mol-rxn)ΔfH°[ C(s)]+(1.5 mol O2(g)/mol-rxn)ΔfH°[O2(g)]+(2 mol Na(s)/mol-rxn)ΔfH°[Na(s)]] ]$

Substituting the values,

$ΔrH°=[(1 mol Na2CO3(s)/mol-rxn)(-1130.77 kJ/mol)[(1 mol C(s)/mol-rxn)(0 kJ/mol)+(1.5 mol O2(g)/mol-rxn)(0 kJ/mol)+(2 mol Na(s)/mol-rxn)(0 kJ/mol)] ]=-1130.77 kJ/mol-rxn$

The entropy change is expressed as,

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)=[(1 mol Na2CO3(s)/mol-rxn)S°[Na2CO3]-[(1 mol C(s)/mol-rxn)S°[ C(s)]+(1.5 mol O2(g)/mol-rxn)S°[O2(g)]+(2 mol Na(s)/mol-rxn)S°[Na(s)]] ]$

Substituting the values,

$ΔrS°=[(1 mol Na2CO3(s)/mol-rxn)(134.79 J/K⋅mol)−[(1 mol C(s)/mol-rxn)(5.6 J/K⋅mol)+(1.5 mol O2(g)/mol-rxn)(205.07 J/K⋅mol)+(2 mol Na(s)/mol-rxn)(51.21 J/K⋅mol)] ]=−280.83 J/K⋅mol-rxn$

Now,

$ΔGo=ΔHo-TΔSo$

Substituting the value of $ΔrHo$ and $ΔrSo$ .

$ΔGo= -1130.77 kJ/mol-[(298 K)(-280.83 J/K×mol-rxn)](1 kJ1000 J)= -1047.08 kJ/mol-rxn$

The value in Appendix L is $−1047.08 kJ/mol-rxn$ .

The value of free energy change is negative. Thus, the reaction is product-favored at equilibrium.

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Answer 3

Textbook Question

Chapter 18, Problem 18PS

Using values of Δ_fH° and S°, calculate the standard molar free energy of formation, Δ_fG°, for each of the following:

(a) Ca(OH)₂(s)
(b) Cl(g)
(c) Na₂CO₃(s)

Compare your calculated values of Δ_fG° with those listed in Appendix L. Which of these formation reactions are predicted to be product-favored at equilibrium at 25 °C?

(a)

Expert Solution

Interpretation Introduction

Interpretation:

The the standard molar free energy for formation of $Ca(OH)2(s)$ should be calculated and compared with the values placed in appendix L. It should be identified that whether the reaction is product favored at equilibrium.

Concept introduction:

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

$ΔGo=ΔHo-TΔSo$

The sign of $ΔGo$ should be positive for a product-favored reaction. Thus, spontaneous reactions are referred to those that have negative free energy formation.

Answer to Problem 18PS

The standard molar energy of formation for $Ca(OH)2(s)$ is $−898.486 kJ/mol-rxn$ .

Explanation of Solution

The standard molar energy of formation for $Ca(OH)2(s)$ is calculated below.

Given:

The Appendix L referred for the values of standard entropies and enthalpies.

$Ca(s)+H2(g)+O2(g)→Ca(OH)2(s)ΔfH°(kJ/mol)000-986.09So(J/K×mol)41.59130.7205.0783.39$

The enthalpy change is expressed as,

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)= [(1 mol Ca(OH)2(s)/mol-rxn)ΔfH°[Ca(OH)2(s)]-[(1 mol H2(g)/mol-rxn)ΔfH°[H2(g)]+(1 mol O2(g)/mol-rxn)ΔfH°[O2(g)]+(1 mol Ca(s)/mol-rxn)ΔfH°[Ca(s)]] ]$

Substituting the values,

$ΔrH°= [(1 mol Ca(OH)2(s)/mol-rxn)(-986.09 kJ/mol)[(1 mol H2(g)/mol-rxn)(0 kJ/mol)+(1 mol O2(g)/mol-rxn)(0 kJ/mol)+(1 mol Na(s)/mol-rxn)(0 kJ/mol)] ]=-986.09 kJ/mol-rxn$

The entropy change is expressed as,

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)= [(1 mol Ca(OH)2(s)/mol-rxn)S°[Ca(OH)2(s)]-[(1 mol H2(g)/mol-rxn)S°[H2(g)]+(1 mol O2(g)/mol-rxn)S°[O2(g)]+(1 mol Ca(s)/mol-rxn)S°[Ca(s)]] ]$

Substituting the values,

$ΔrS°= [(1 mol Ca(OH)2(s)/mol-rxn)(83.39 J/K×mol)-[(1 mol H2(g)/mol-rxn)(130.7 J/K×mol)+(1 mol O2(g)/mol-rxn)(205.07 J/K×mol)+(1 mol Ca(s)/mol-rxn)(41.59 J/K×mol)] ]=-293.97 J/K×mol-rxn$

Now,

$ΔGo= ΔHo-TΔSo$

Substitute the value of $ΔrHo$ and $ΔrSo$ .

$ΔGo= - 986.09 kJ/mol-[(298 K)(-293.97 J/K×mol-rxn)](1 kJ1000 J)= - 898.486 kJ/mol-rxn$

The value in Appendix L is $−898.43 kJ/mol-rxn$ .

The value of free energy change is negative. Thus, the reaction is product-favored at equilibrium.

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The the standard molar free energy for formation of $Cl(g)$ should be calculated and compared with the values placed in appendix L. It should be identified that whether the reaction is product favored at equilibrium.

Concept introduction:

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

$ΔGo=ΔHo-TΔSo$

The sign of $ΔGo$ should be positive for a product-favored reaction. Thus, spontaneous reactions are referred to those that have negative free energy formation.

Answer to Problem 18PS

The standard molar energy of formation for $Cl(g)$ is $105.3 kJ/mol-rxn$ .

Explanation of Solution

The standard molar energy of formation for $Cl(g)$ is calculated below.

Given:

The Appendix L referred for the values of standard entropies and enthalpies.

$12Cl2(g)→Cl(g)ΔfH°(kJ/mol)0121.3So(J/K×mol)223.08165.19$

The enthalpy change is expressed as,

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)= [(1 mol Cl(g)/mol-rxn)ΔfH°[Cl(g)]-(0.5 mol Cl2(g)/mol-rxn)ΔfH°[Cl2(g)]]$

Substituting the respective values,

$ΔrH°= [(1 mol Cl(g)/mol-rxn)(121.3 kJ/mol)-(0.5 mol Cl2(g)/mol-rxn)(0 kJ/mol)]= 121.3 kJ/mol-rxn$

The entropy change is expressed as,

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)=[(1 mol Cl(g)/mol-rxn)S°[Cl(g)]-(0.5 mol Cl2(g)/mol-rxn)S°[Cl2(g)]]$

Substituting the respective values,

$ΔrS°= [(1 mol Cl(g)/mol-rxn)(165.19 J/K×mol)-(0.5 mol Cl2(g)/mol-rxn)(223.08 J/K×mol)]= 53.65 J/K×mol-rxn$

Now,

$ΔfGo= ΔfHo- TΔSo$

Substitute the value of $ΔrHo$ and $ΔrSo$ .

$ΔGo= 121.3 kJ/mol-[(298 K)(53.65 J/K×mol-rxn)](1 kJ1000 J)= 105.31 kJ/mol-rxn$

The value in Appendix L is $105.3 kJ/mol-rxn$ .

The value of free energy change is positive. Thus, the reaction is reactant-favored at equilibrium.

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The the standard molar free energy for formation of $Na2CO3(s)$ should be calculated and compared with the values placed in appendix L. It should be identified that whether the reaction is product favored at equilibrium.

Concept introduction:

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

$ΔGo=ΔHo-TΔSo$

The sign of $ΔGo$ should be positive for a product-favored reaction. Thus, spontaneous reactions are referred to those that have negative free energy formation.

Answer to Problem 18PS

The standard molar energy of formation for $Na2CO3(s)$ is $−1047.08 kJ/mol-rxn$ .

Explanation of Solution

The standard molar energy of formation for $Na2CO3(s)$ is calculated below.

Given:

The Appendix L referred for the values of standard entropies and enthalpies.

$2Na(s)+C(s)+32O2(g)→Na2CO3(s)ΔfH°(kJ/mol)000−1130.77S∘(J/K⋅mol)51.215.6205.07134.79$

The enthalpy change is expressed as,

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)=[(1 mol Na2CO3(s)/mol-rxn)ΔfH°[Na2CO3]-[(1 mol C(s)/mol-rxn)ΔfH°[ C(s)]+(1.5 mol O2(g)/mol-rxn)ΔfH°[O2(g)]+(2 mol Na(s)/mol-rxn)ΔfH°[Na(s)]] ]$

Substituting the values,

$ΔrH°=[(1 mol Na2CO3(s)/mol-rxn)(-1130.77 kJ/mol)[(1 mol C(s)/mol-rxn)(0 kJ/mol)+(1.5 mol O2(g)/mol-rxn)(0 kJ/mol)+(2 mol Na(s)/mol-rxn)(0 kJ/mol)] ]=-1130.77 kJ/mol-rxn$

The entropy change is expressed as,

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)=[(1 mol Na2CO3(s)/mol-rxn)S°[Na2CO3]-[(1 mol C(s)/mol-rxn)S°[ C(s)]+(1.5 mol O2(g)/mol-rxn)S°[O2(g)]+(2 mol Na(s)/mol-rxn)S°[Na(s)]] ]$

Substituting the values,

$ΔrS°=[(1 mol Na2CO3(s)/mol-rxn)(134.79 J/K⋅mol)−[(1 mol C(s)/mol-rxn)(5.6 J/K⋅mol)+(1.5 mol O2(g)/mol-rxn)(205.07 J/K⋅mol)+(2 mol Na(s)/mol-rxn)(51.21 J/K⋅mol)] ]=−280.83 J/K⋅mol-rxn$

Now,

$ΔGo=ΔHo-TΔSo$

Substituting the value of $ΔrHo$ and $ΔrSo$ .

$ΔGo= -1130.77 kJ/mol-[(298 K)(-280.83 J/K×mol-rxn)](1 kJ1000 J)= -1047.08 kJ/mol-rxn$

The value in Appendix L is $−1047.08 kJ/mol-rxn$ .

The value of free energy change is negative. Thus, the reaction is product-favored at equilibrium.

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Answer 4

Textbook Question

Chapter 18, Problem 18PS

Using values of Δ_fH° and S°, calculate the standard molar free energy of formation, Δ_fG°, for each of the following:

(a) Ca(OH)₂(s)
(b) Cl(g)
(c) Na₂CO₃(s)

Compare your calculated values of Δ_fG° with those listed in Appendix L. Which of these formation reactions are predicted to be product-favored at equilibrium at 25 °C?

(a)

Expert Solution

Interpretation Introduction

Interpretation:

The the standard molar free energy for formation of $Ca(OH)2(s)$ should be calculated and compared with the values placed in appendix L. It should be identified that whether the reaction is product favored at equilibrium.

Concept introduction:

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

$ΔGo=ΔHo-TΔSo$

The sign of $ΔGo$ should be positive for a product-favored reaction. Thus, spontaneous reactions are referred to those that have negative free energy formation.

Answer to Problem 18PS

The standard molar energy of formation for $Ca(OH)2(s)$ is $−898.486 kJ/mol-rxn$ .

Explanation of Solution

The standard molar energy of formation for $Ca(OH)2(s)$ is calculated below.

Given:

The Appendix L referred for the values of standard entropies and enthalpies.

$Ca(s)+H2(g)+O2(g)→Ca(OH)2(s)ΔfH°(kJ/mol)000-986.09So(J/K×mol)41.59130.7205.0783.39$

The enthalpy change is expressed as,

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)= [(1 mol Ca(OH)2(s)/mol-rxn)ΔfH°[Ca(OH)2(s)]-[(1 mol H2(g)/mol-rxn)ΔfH°[H2(g)]+(1 mol O2(g)/mol-rxn)ΔfH°[O2(g)]+(1 mol Ca(s)/mol-rxn)ΔfH°[Ca(s)]] ]$

Substituting the values,

$ΔrH°= [(1 mol Ca(OH)2(s)/mol-rxn)(-986.09 kJ/mol)[(1 mol H2(g)/mol-rxn)(0 kJ/mol)+(1 mol O2(g)/mol-rxn)(0 kJ/mol)+(1 mol Na(s)/mol-rxn)(0 kJ/mol)] ]=-986.09 kJ/mol-rxn$

The entropy change is expressed as,

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)= [(1 mol Ca(OH)2(s)/mol-rxn)S°[Ca(OH)2(s)]-[(1 mol H2(g)/mol-rxn)S°[H2(g)]+(1 mol O2(g)/mol-rxn)S°[O2(g)]+(1 mol Ca(s)/mol-rxn)S°[Ca(s)]] ]$

Substituting the values,

$ΔrS°= [(1 mol Ca(OH)2(s)/mol-rxn)(83.39 J/K×mol)-[(1 mol H2(g)/mol-rxn)(130.7 J/K×mol)+(1 mol O2(g)/mol-rxn)(205.07 J/K×mol)+(1 mol Ca(s)/mol-rxn)(41.59 J/K×mol)] ]=-293.97 J/K×mol-rxn$

Now,

$ΔGo= ΔHo-TΔSo$

Substitute the value of $ΔrHo$ and $ΔrSo$ .

$ΔGo= - 986.09 kJ/mol-[(298 K)(-293.97 J/K×mol-rxn)](1 kJ1000 J)= - 898.486 kJ/mol-rxn$

The value in Appendix L is $−898.43 kJ/mol-rxn$ .

The value of free energy change is negative. Thus, the reaction is product-favored at equilibrium.

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The the standard molar free energy for formation of $Cl(g)$ should be calculated and compared with the values placed in appendix L. It should be identified that whether the reaction is product favored at equilibrium.

Concept introduction:

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

$ΔGo=ΔHo-TΔSo$

The sign of $ΔGo$ should be positive for a product-favored reaction. Thus, spontaneous reactions are referred to those that have negative free energy formation.

Answer to Problem 18PS

The standard molar energy of formation for $Cl(g)$ is $105.3 kJ/mol-rxn$ .

Explanation of Solution

The standard molar energy of formation for $Cl(g)$ is calculated below.

Given:

The Appendix L referred for the values of standard entropies and enthalpies.

$12Cl2(g)→Cl(g)ΔfH°(kJ/mol)0121.3So(J/K×mol)223.08165.19$

The enthalpy change is expressed as,

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)= [(1 mol Cl(g)/mol-rxn)ΔfH°[Cl(g)]-(0.5 mol Cl2(g)/mol-rxn)ΔfH°[Cl2(g)]]$

Substituting the respective values,

$ΔrH°= [(1 mol Cl(g)/mol-rxn)(121.3 kJ/mol)-(0.5 mol Cl2(g)/mol-rxn)(0 kJ/mol)]= 121.3 kJ/mol-rxn$

The entropy change is expressed as,

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)=[(1 mol Cl(g)/mol-rxn)S°[Cl(g)]-(0.5 mol Cl2(g)/mol-rxn)S°[Cl2(g)]]$

Substituting the respective values,

$ΔrS°= [(1 mol Cl(g)/mol-rxn)(165.19 J/K×mol)-(0.5 mol Cl2(g)/mol-rxn)(223.08 J/K×mol)]= 53.65 J/K×mol-rxn$

Now,

$ΔfGo= ΔfHo- TΔSo$

Substitute the value of $ΔrHo$ and $ΔrSo$ .

$ΔGo= 121.3 kJ/mol-[(298 K)(53.65 J/K×mol-rxn)](1 kJ1000 J)= 105.31 kJ/mol-rxn$

The value in Appendix L is $105.3 kJ/mol-rxn$ .

The value of free energy change is positive. Thus, the reaction is reactant-favored at equilibrium.

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The the standard molar free energy for formation of $Na2CO3(s)$ should be calculated and compared with the values placed in appendix L. It should be identified that whether the reaction is product favored at equilibrium.

Concept introduction:

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

$ΔGo=ΔHo-TΔSo$

The sign of $ΔGo$ should be positive for a product-favored reaction. Thus, spontaneous reactions are referred to those that have negative free energy formation.

Answer to Problem 18PS

The standard molar energy of formation for $Na2CO3(s)$ is $−1047.08 kJ/mol-rxn$ .

Explanation of Solution

The standard molar energy of formation for $Na2CO3(s)$ is calculated below.

Given:

The Appendix L referred for the values of standard entropies and enthalpies.

$2Na(s)+C(s)+32O2(g)→Na2CO3(s)ΔfH°(kJ/mol)000−1130.77S∘(J/K⋅mol)51.215.6205.07134.79$

The enthalpy change is expressed as,

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)=[(1 mol Na2CO3(s)/mol-rxn)ΔfH°[Na2CO3]-[(1 mol C(s)/mol-rxn)ΔfH°[ C(s)]+(1.5 mol O2(g)/mol-rxn)ΔfH°[O2(g)]+(2 mol Na(s)/mol-rxn)ΔfH°[Na(s)]] ]$

Substituting the values,

$ΔrH°=[(1 mol Na2CO3(s)/mol-rxn)(-1130.77 kJ/mol)[(1 mol C(s)/mol-rxn)(0 kJ/mol)+(1.5 mol O2(g)/mol-rxn)(0 kJ/mol)+(2 mol Na(s)/mol-rxn)(0 kJ/mol)] ]=-1130.77 kJ/mol-rxn$

The entropy change is expressed as,

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)=[(1 mol Na2CO3(s)/mol-rxn)S°[Na2CO3]-[(1 mol C(s)/mol-rxn)S°[ C(s)]+(1.5 mol O2(g)/mol-rxn)S°[O2(g)]+(2 mol Na(s)/mol-rxn)S°[Na(s)]] ]$

Substituting the values,

$ΔrS°=[(1 mol Na2CO3(s)/mol-rxn)(134.79 J/K⋅mol)−[(1 mol C(s)/mol-rxn)(5.6 J/K⋅mol)+(1.5 mol O2(g)/mol-rxn)(205.07 J/K⋅mol)+(2 mol Na(s)/mol-rxn)(51.21 J/K⋅mol)] ]=−280.83 J/K⋅mol-rxn$

Now,

$ΔGo=ΔHo-TΔSo$

Substituting the value of $ΔrHo$ and $ΔrSo$ .

$ΔGo= -1130.77 kJ/mol-[(298 K)(-280.83 J/K×mol-rxn)](1 kJ1000 J)= -1047.08 kJ/mol-rxn$

The value in Appendix L is $−1047.08 kJ/mol-rxn$ .

The value of free energy change is negative. Thus, the reaction is product-favored at equilibrium.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

Interpretation Introduction

Interpretation:

The the standard molar free energy for formation of $Ca(OH)2(s)$ should be calculated and compared with the values placed in appendix L. It should be identified that whether the reaction is product favored at equilibrium.

Concept introduction:

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

$ΔGo=ΔHo-TΔSo$

The sign of $ΔGo$ should be positive for a product-favored reaction. Thus, spontaneous reactions are referred to those that have negative free energy formation.

Answer to Problem 18PS

The standard molar energy of formation for $Ca(OH)2(s)$ is $−898.486 kJ/mol-rxn$ .

Explanation of Solution

The standard molar energy of formation for $Ca(OH)2(s)$ is calculated below.

Given:

The Appendix L referred for the values of standard entropies and enthalpies.

$Ca(s)+H2(g)+O2(g)→Ca(OH)2(s)ΔfH°(kJ/mol)000-986.09So(J/K×mol)41.59130.7205.0783.39$

The enthalpy change is expressed as,

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)= [(1 mol Ca(OH)2(s)/mol-rxn)ΔfH°[Ca(OH)2(s)]-[(1 mol H2(g)/mol-rxn)ΔfH°[H2(g)]+(1 mol O2(g)/mol-rxn)ΔfH°[O2(g)]+(1 mol Ca(s)/mol-rxn)ΔfH°[Ca(s)]] ]$

Substituting the values,

$ΔrH°= [(1 mol Ca(OH)2(s)/mol-rxn)(-986.09 kJ/mol)[(1 mol H2(g)/mol-rxn)(0 kJ/mol)+(1 mol O2(g)/mol-rxn)(0 kJ/mol)+(1 mol Na(s)/mol-rxn)(0 kJ/mol)] ]=-986.09 kJ/mol-rxn$

The entropy change is expressed as,

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)= [(1 mol Ca(OH)2(s)/mol-rxn)S°[Ca(OH)2(s)]-[(1 mol H2(g)/mol-rxn)S°[H2(g)]+(1 mol O2(g)/mol-rxn)S°[O2(g)]+(1 mol Ca(s)/mol-rxn)S°[Ca(s)]] ]$

Substituting the values,

$ΔrS°= [(1 mol Ca(OH)2(s)/mol-rxn)(83.39 J/K×mol)-[(1 mol H2(g)/mol-rxn)(130.7 J/K×mol)+(1 mol O2(g)/mol-rxn)(205.07 J/K×mol)+(1 mol Ca(s)/mol-rxn)(41.59 J/K×mol)] ]=-293.97 J/K×mol-rxn$

Now,

$ΔGo= ΔHo-TΔSo$

Substitute the value of $ΔrHo$ and $ΔrSo$ .

$ΔGo= - 986.09 kJ/mol-[(298 K)(-293.97 J/K×mol-rxn)](1 kJ1000 J)= - 898.486 kJ/mol-rxn$

The value in Appendix L is $−898.43 kJ/mol-rxn$ .

The value of free energy change is negative. Thus, the reaction is product-favored at equilibrium.

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The the standard molar free energy for formation of $Cl(g)$ should be calculated and compared with the values placed in appendix L. It should be identified that whether the reaction is product favored at equilibrium.

Concept introduction:

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

$ΔGo=ΔHo-TΔSo$

The sign of $ΔGo$ should be positive for a product-favored reaction. Thus, spontaneous reactions are referred to those that have negative free energy formation.

Answer to Problem 18PS

The standard molar energy of formation for $Cl(g)$ is $105.3 kJ/mol-rxn$ .

Explanation of Solution

The standard molar energy of formation for $Cl(g)$ is calculated below.

Given:

The Appendix L referred for the values of standard entropies and enthalpies.

$12Cl2(g)→Cl(g)ΔfH°(kJ/mol)0121.3So(J/K×mol)223.08165.19$

The enthalpy change is expressed as,

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)= [(1 mol Cl(g)/mol-rxn)ΔfH°[Cl(g)]-(0.5 mol Cl2(g)/mol-rxn)ΔfH°[Cl2(g)]]$

Substituting the respective values,

$ΔrH°= [(1 mol Cl(g)/mol-rxn)(121.3 kJ/mol)-(0.5 mol Cl2(g)/mol-rxn)(0 kJ/mol)]= 121.3 kJ/mol-rxn$

The entropy change is expressed as,

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)=[(1 mol Cl(g)/mol-rxn)S°[Cl(g)]-(0.5 mol Cl2(g)/mol-rxn)S°[Cl2(g)]]$

Substituting the respective values,

$ΔrS°= [(1 mol Cl(g)/mol-rxn)(165.19 J/K×mol)-(0.5 mol Cl2(g)/mol-rxn)(223.08 J/K×mol)]= 53.65 J/K×mol-rxn$

Now,

$ΔfGo= ΔfHo- TΔSo$

Substitute the value of $ΔrHo$ and $ΔrSo$ .

$ΔGo= 121.3 kJ/mol-[(298 K)(53.65 J/K×mol-rxn)](1 kJ1000 J)= 105.31 kJ/mol-rxn$

The value in Appendix L is $105.3 kJ/mol-rxn$ .

The value of free energy change is positive. Thus, the reaction is reactant-favored at equilibrium.

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The the standard molar free energy for formation of $Na2CO3(s)$ should be calculated and compared with the values placed in appendix L. It should be identified that whether the reaction is product favored at equilibrium.

Concept introduction:

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

$ΔGo=ΔHo-TΔSo$

The sign of $ΔGo$ should be positive for a product-favored reaction. Thus, spontaneous reactions are referred to those that have negative free energy formation.

Answer to Problem 18PS

The standard molar energy of formation for $Na2CO3(s)$ is $−1047.08 kJ/mol-rxn$ .

Explanation of Solution

The standard molar energy of formation for $Na2CO3(s)$ is calculated below.

Given:

The Appendix L referred for the values of standard entropies and enthalpies.

$2Na(s)+C(s)+32O2(g)→Na2CO3(s)ΔfH°(kJ/mol)000−1130.77S∘(J/K⋅mol)51.215.6205.07134.79$

The enthalpy change is expressed as,

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)=[(1 mol Na2CO3(s)/mol-rxn)ΔfH°[Na2CO3]-[(1 mol C(s)/mol-rxn)ΔfH°[ C(s)]+(1.5 mol O2(g)/mol-rxn)ΔfH°[O2(g)]+(2 mol Na(s)/mol-rxn)ΔfH°[Na(s)]] ]$

Substituting the values,

$ΔrH°=[(1 mol Na2CO3(s)/mol-rxn)(-1130.77 kJ/mol)[(1 mol C(s)/mol-rxn)(0 kJ/mol)+(1.5 mol O2(g)/mol-rxn)(0 kJ/mol)+(2 mol Na(s)/mol-rxn)(0 kJ/mol)] ]=-1130.77 kJ/mol-rxn$

The entropy change is expressed as,

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)=[(1 mol Na2CO3(s)/mol-rxn)S°[Na2CO3]-[(1 mol C(s)/mol-rxn)S°[ C(s)]+(1.5 mol O2(g)/mol-rxn)S°[O2(g)]+(2 mol Na(s)/mol-rxn)S°[Na(s)]] ]$

Substituting the values,

$ΔrS°=[(1 mol Na2CO3(s)/mol-rxn)(134.79 J/K⋅mol)−[(1 mol C(s)/mol-rxn)(5.6 J/K⋅mol)+(1.5 mol O2(g)/mol-rxn)(205.07 J/K⋅mol)+(2 mol Na(s)/mol-rxn)(51.21 J/K⋅mol)] ]=−280.83 J/K⋅mol-rxn$

Now,

$ΔGo=ΔHo-TΔSo$

Substituting the value of $ΔrHo$ and $ΔrSo$ .

$ΔGo= -1130.77 kJ/mol-[(298 K)(-280.83 J/K×mol-rxn)](1 kJ1000 J)= -1047.08 kJ/mol-rxn$

The value in Appendix L is $−1047.08 kJ/mol-rxn$ .

The value of free energy change is negative. Thus, the reaction is product-favored at equilibrium.

Answer 8

(a)

Expert Solution

Interpretation Introduction

Interpretation:

The the standard molar free energy for formation of $Ca(OH)2(s)$ should be calculated and compared with the values placed in appendix L. It should be identified that whether the reaction is product favored at equilibrium.

Concept introduction:

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

$ΔGo=ΔHo-TΔSo$

The sign of $ΔGo$ should be positive for a product-favored reaction. Thus, spontaneous reactions are referred to those that have negative free energy formation.

Answer to Problem 18PS

The standard molar energy of formation for $Ca(OH)2(s)$ is $−898.486 kJ/mol-rxn$ .

Explanation of Solution

The standard molar energy of formation for $Ca(OH)2(s)$ is calculated below.

Given:

The Appendix L referred for the values of standard entropies and enthalpies.

$Ca(s)+H2(g)+O2(g)→Ca(OH)2(s)ΔfH°(kJ/mol)000-986.09So(J/K×mol)41.59130.7205.0783.39$

The enthalpy change is expressed as,

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)= [(1 mol Ca(OH)2(s)/mol-rxn)ΔfH°[Ca(OH)2(s)]-[(1 mol H2(g)/mol-rxn)ΔfH°[H2(g)]+(1 mol O2(g)/mol-rxn)ΔfH°[O2(g)]+(1 mol Ca(s)/mol-rxn)ΔfH°[Ca(s)]] ]$

Substituting the values,

$ΔrH°= [(1 mol Ca(OH)2(s)/mol-rxn)(-986.09 kJ/mol)[(1 mol H2(g)/mol-rxn)(0 kJ/mol)+(1 mol O2(g)/mol-rxn)(0 kJ/mol)+(1 mol Na(s)/mol-rxn)(0 kJ/mol)] ]=-986.09 kJ/mol-rxn$

The entropy change is expressed as,

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)= [(1 mol Ca(OH)2(s)/mol-rxn)S°[Ca(OH)2(s)]-[(1 mol H2(g)/mol-rxn)S°[H2(g)]+(1 mol O2(g)/mol-rxn)S°[O2(g)]+(1 mol Ca(s)/mol-rxn)S°[Ca(s)]] ]$

Substituting the values,

$ΔrS°= [(1 mol Ca(OH)2(s)/mol-rxn)(83.39 J/K×mol)-[(1 mol H2(g)/mol-rxn)(130.7 J/K×mol)+(1 mol O2(g)/mol-rxn)(205.07 J/K×mol)+(1 mol Ca(s)/mol-rxn)(41.59 J/K×mol)] ]=-293.97 J/K×mol-rxn$

Now,

$ΔGo= ΔHo-TΔSo$

Substitute the value of $ΔrHo$ and $ΔrSo$ .

$ΔGo= - 986.09 kJ/mol-[(298 K)(-293.97 J/K×mol-rxn)](1 kJ1000 J)= - 898.486 kJ/mol-rxn$

The value in Appendix L is $−898.43 kJ/mol-rxn$ .

The value of free energy change is negative. Thus, the reaction is product-favored at equilibrium.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Interpretation Introduction

Interpretation:

The the standard molar free energy for formation of $Ca(OH)2(s)$ should be calculated and compared with the values placed in appendix L. It should be identified that whether the reaction is product favored at equilibrium.

Concept introduction:

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

$ΔGo=ΔHo-TΔSo$

The sign of $ΔGo$ should be positive for a product-favored reaction. Thus, spontaneous reactions are referred to those that have negative free energy formation.

Answer 13

Answer to Problem 18PS

The standard molar energy of formation for $Ca(OH)2(s)$ is $−898.486 kJ/mol-rxn$ .

Answer 14

Explanation of Solution

The standard molar energy of formation for $Ca(OH)2(s)$ is calculated below.

Given:

The Appendix L referred for the values of standard entropies and enthalpies.

$Ca(s)+H2(g)+O2(g)→Ca(OH)2(s)ΔfH°(kJ/mol)000-986.09So(J/K×mol)41.59130.7205.0783.39$

The enthalpy change is expressed as,

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)= [(1 mol Ca(OH)2(s)/mol-rxn)ΔfH°[Ca(OH)2(s)]-[(1 mol H2(g)/mol-rxn)ΔfH°[H2(g)]+(1 mol O2(g)/mol-rxn)ΔfH°[O2(g)]+(1 mol Ca(s)/mol-rxn)ΔfH°[Ca(s)]] ]$

Substituting the values,

$ΔrH°= [(1 mol Ca(OH)2(s)/mol-rxn)(-986.09 kJ/mol)[(1 mol H2(g)/mol-rxn)(0 kJ/mol)+(1 mol O2(g)/mol-rxn)(0 kJ/mol)+(1 mol Na(s)/mol-rxn)(0 kJ/mol)] ]=-986.09 kJ/mol-rxn$

The entropy change is expressed as,

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)= [(1 mol Ca(OH)2(s)/mol-rxn)S°[Ca(OH)2(s)]-[(1 mol H2(g)/mol-rxn)S°[H2(g)]+(1 mol O2(g)/mol-rxn)S°[O2(g)]+(1 mol Ca(s)/mol-rxn)S°[Ca(s)]] ]$

Substituting the values,

$ΔrS°= [(1 mol Ca(OH)2(s)/mol-rxn)(83.39 J/K×mol)-[(1 mol H2(g)/mol-rxn)(130.7 J/K×mol)+(1 mol O2(g)/mol-rxn)(205.07 J/K×mol)+(1 mol Ca(s)/mol-rxn)(41.59 J/K×mol)] ]=-293.97 J/K×mol-rxn$

Now,

$ΔGo= ΔHo-TΔSo$

Substitute the value of $ΔrHo$ and $ΔrSo$ .

$ΔGo= - 986.09 kJ/mol-[(298 K)(-293.97 J/K×mol-rxn)](1 kJ1000 J)= - 898.486 kJ/mol-rxn$

The value in Appendix L is $−898.43 kJ/mol-rxn$ .

The value of free energy change is negative. Thus, the reaction is product-favored at equilibrium.

Answer 15

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The the standard molar free energy for formation of $Cl(g)$ should be calculated and compared with the values placed in appendix L. It should be identified that whether the reaction is product favored at equilibrium.

Concept introduction:

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

$ΔGo=ΔHo-TΔSo$

The sign of $ΔGo$ should be positive for a product-favored reaction. Thus, spontaneous reactions are referred to those that have negative free energy formation.

Answer to Problem 18PS

The standard molar energy of formation for $Cl(g)$ is $105.3 kJ/mol-rxn$ .

Explanation of Solution

The standard molar energy of formation for $Cl(g)$ is calculated below.

Given:

The Appendix L referred for the values of standard entropies and enthalpies.

$12Cl2(g)→Cl(g)ΔfH°(kJ/mol)0121.3So(J/K×mol)223.08165.19$

The enthalpy change is expressed as,

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)= [(1 mol Cl(g)/mol-rxn)ΔfH°[Cl(g)]-(0.5 mol Cl2(g)/mol-rxn)ΔfH°[Cl2(g)]]$

Substituting the respective values,

$ΔrH°= [(1 mol Cl(g)/mol-rxn)(121.3 kJ/mol)-(0.5 mol Cl2(g)/mol-rxn)(0 kJ/mol)]= 121.3 kJ/mol-rxn$

The entropy change is expressed as,

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)=[(1 mol Cl(g)/mol-rxn)S°[Cl(g)]-(0.5 mol Cl2(g)/mol-rxn)S°[Cl2(g)]]$

Substituting the respective values,

$ΔrS°= [(1 mol Cl(g)/mol-rxn)(165.19 J/K×mol)-(0.5 mol Cl2(g)/mol-rxn)(223.08 J/K×mol)]= 53.65 J/K×mol-rxn$

Now,

$ΔfGo= ΔfHo- TΔSo$

Substitute the value of $ΔrHo$ and $ΔrSo$ .

$ΔGo= 121.3 kJ/mol-[(298 K)(53.65 J/K×mol-rxn)](1 kJ1000 J)= 105.31 kJ/mol-rxn$

The value in Appendix L is $105.3 kJ/mol-rxn$ .

The value of free energy change is positive. Thus, the reaction is reactant-favored at equilibrium.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Interpretation Introduction

Interpretation:

The the standard molar free energy for formation of $Cl(g)$ should be calculated and compared with the values placed in appendix L. It should be identified that whether the reaction is product favored at equilibrium.

Concept introduction:

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

$ΔGo=ΔHo-TΔSo$

The sign of $ΔGo$ should be positive for a product-favored reaction. Thus, spontaneous reactions are referred to those that have negative free energy formation.

Answer 20

Answer to Problem 18PS

The standard molar energy of formation for $Cl(g)$ is $105.3 kJ/mol-rxn$ .

Answer 21

Explanation of Solution

The standard molar energy of formation for $Cl(g)$ is calculated below.

Given:

The Appendix L referred for the values of standard entropies and enthalpies.

$12Cl2(g)→Cl(g)ΔfH°(kJ/mol)0121.3So(J/K×mol)223.08165.19$

The enthalpy change is expressed as,

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)= [(1 mol Cl(g)/mol-rxn)ΔfH°[Cl(g)]-(0.5 mol Cl2(g)/mol-rxn)ΔfH°[Cl2(g)]]$

Substituting the respective values,

$ΔrH°= [(1 mol Cl(g)/mol-rxn)(121.3 kJ/mol)-(0.5 mol Cl2(g)/mol-rxn)(0 kJ/mol)]= 121.3 kJ/mol-rxn$

The entropy change is expressed as,

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)=[(1 mol Cl(g)/mol-rxn)S°[Cl(g)]-(0.5 mol Cl2(g)/mol-rxn)S°[Cl2(g)]]$

Substituting the respective values,

$ΔrS°= [(1 mol Cl(g)/mol-rxn)(165.19 J/K×mol)-(0.5 mol Cl2(g)/mol-rxn)(223.08 J/K×mol)]= 53.65 J/K×mol-rxn$

Now,

$ΔfGo= ΔfHo- TΔSo$

Substitute the value of $ΔrHo$ and $ΔrSo$ .

$ΔGo= 121.3 kJ/mol-[(298 K)(53.65 J/K×mol-rxn)](1 kJ1000 J)= 105.31 kJ/mol-rxn$

The value in Appendix L is $105.3 kJ/mol-rxn$ .

The value of free energy change is positive. Thus, the reaction is reactant-favored at equilibrium.

Answer 22

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The the standard molar free energy for formation of $Na2CO3(s)$ should be calculated and compared with the values placed in appendix L. It should be identified that whether the reaction is product favored at equilibrium.

Concept introduction:

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

$ΔGo=ΔHo-TΔSo$

The sign of $ΔGo$ should be positive for a product-favored reaction. Thus, spontaneous reactions are referred to those that have negative free energy formation.

Answer to Problem 18PS

The standard molar energy of formation for $Na2CO3(s)$ is $−1047.08 kJ/mol-rxn$ .

Explanation of Solution

The standard molar energy of formation for $Na2CO3(s)$ is calculated below.

Given:

The Appendix L referred for the values of standard entropies and enthalpies.

$2Na(s)+C(s)+32O2(g)→Na2CO3(s)ΔfH°(kJ/mol)000−1130.77S∘(J/K⋅mol)51.215.6205.07134.79$

The enthalpy change is expressed as,

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)=[(1 mol Na2CO3(s)/mol-rxn)ΔfH°[Na2CO3]-[(1 mol C(s)/mol-rxn)ΔfH°[ C(s)]+(1.5 mol O2(g)/mol-rxn)ΔfH°[O2(g)]+(2 mol Na(s)/mol-rxn)ΔfH°[Na(s)]] ]$

Substituting the values,

$ΔrH°=[(1 mol Na2CO3(s)/mol-rxn)(-1130.77 kJ/mol)[(1 mol C(s)/mol-rxn)(0 kJ/mol)+(1.5 mol O2(g)/mol-rxn)(0 kJ/mol)+(2 mol Na(s)/mol-rxn)(0 kJ/mol)] ]=-1130.77 kJ/mol-rxn$

The entropy change is expressed as,

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)=[(1 mol Na2CO3(s)/mol-rxn)S°[Na2CO3]-[(1 mol C(s)/mol-rxn)S°[ C(s)]+(1.5 mol O2(g)/mol-rxn)S°[O2(g)]+(2 mol Na(s)/mol-rxn)S°[Na(s)]] ]$

Substituting the values,

$ΔrS°=[(1 mol Na2CO3(s)/mol-rxn)(134.79 J/K⋅mol)−[(1 mol C(s)/mol-rxn)(5.6 J/K⋅mol)+(1.5 mol O2(g)/mol-rxn)(205.07 J/K⋅mol)+(2 mol Na(s)/mol-rxn)(51.21 J/K⋅mol)] ]=−280.83 J/K⋅mol-rxn$

Now,

$ΔGo=ΔHo-TΔSo$

Substituting the value of $ΔrHo$ and $ΔrSo$ .

$ΔGo= -1130.77 kJ/mol-[(298 K)(-280.83 J/K×mol-rxn)](1 kJ1000 J)= -1047.08 kJ/mol-rxn$

The value in Appendix L is $−1047.08 kJ/mol-rxn$ .

The value of free energy change is negative. Thus, the reaction is product-favored at equilibrium.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Interpretation Introduction

Interpretation:

The the standard molar free energy for formation of $Na2CO3(s)$ should be calculated and compared with the values placed in appendix L. It should be identified that whether the reaction is product favored at equilibrium.

Concept introduction:

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

$ΔGo=ΔHo-TΔSo$

The sign of $ΔGo$ should be positive for a product-favored reaction. Thus, spontaneous reactions are referred to those that have negative free energy formation.

Answer 27

Answer to Problem 18PS

The standard molar energy of formation for $Na2CO3(s)$ is $−1047.08 kJ/mol-rxn$ .

Answer 28

Explanation of Solution

The standard molar energy of formation for $Na2CO3(s)$ is calculated below.

Given:

The Appendix L referred for the values of standard entropies and enthalpies.

$2Na(s)+C(s)+32O2(g)→Na2CO3(s)ΔfH°(kJ/mol)000−1130.77S∘(J/K⋅mol)51.215.6205.07134.79$

The enthalpy change is expressed as,

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)=[(1 mol Na2CO3(s)/mol-rxn)ΔfH°[Na2CO3]-[(1 mol C(s)/mol-rxn)ΔfH°[ C(s)]+(1.5 mol O2(g)/mol-rxn)ΔfH°[O2(g)]+(2 mol Na(s)/mol-rxn)ΔfH°[Na(s)]] ]$

Substituting the values,

$ΔrH°=[(1 mol Na2CO3(s)/mol-rxn)(-1130.77 kJ/mol)[(1 mol C(s)/mol-rxn)(0 kJ/mol)+(1.5 mol O2(g)/mol-rxn)(0 kJ/mol)+(2 mol Na(s)/mol-rxn)(0 kJ/mol)] ]=-1130.77 kJ/mol-rxn$

The entropy change is expressed as,

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)=[(1 mol Na2CO3(s)/mol-rxn)S°[Na2CO3]-[(1 mol C(s)/mol-rxn)S°[ C(s)]+(1.5 mol O2(g)/mol-rxn)S°[O2(g)]+(2 mol Na(s)/mol-rxn)S°[Na(s)]] ]$

Substituting the values,

$ΔrS°=[(1 mol Na2CO3(s)/mol-rxn)(134.79 J/K⋅mol)−[(1 mol C(s)/mol-rxn)(5.6 J/K⋅mol)+(1.5 mol O2(g)/mol-rxn)(205.07 J/K⋅mol)+(2 mol Na(s)/mol-rxn)(51.21 J/K⋅mol)] ]=−280.83 J/K⋅mol-rxn$