Using values of Δ f H ° and S °, calculate the standard molar free energy of formation, Δ f G °, for each of the following compounds: (a) CS 2 (g) (b) NaOH(s) (c) ICl(g) 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?

Question

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Answer 1

Textbook Question

Chapter 18, Problem 17PS

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

(a) CS₂(g)
(b) NaOH(s)
(c) ICl(g)

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 $CS2(g)$ 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 entropy by the following expression,

$ΔGo = ΔHo - TΔSo$

Here, $ΔHo$ is the change in enthalpy and $ΔSo$ is the change in entropy.

Answer to Problem 17PS

The standard molar free energy of formation for $CS2(g)$ is $+66.6 kJ/mol-rxn$ .

The value of $ΔG∘$ is positive, hence the reaction is reactant-favored at equilibrium.

Explanation of Solution

The standard molar free energy of formation for $CS2(g)$ is calculated below.

Given:

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

$C(s)+2S(s)→CS2(g)ΔfH°(kJ/mol)00+116.7So(J/K×mol)+5.6+32.1+237.8$

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)= [(1 mol CS2(g)/mol-rxn)ΔfH°[CS2(g)]-[(1 mol C(s)/mol-rxn)ΔfH°[C(s)]+(2 mol S(s)/mol-rxn)ΔfH°[S(s)]] ]$

Substituting the respective values,

$ΔrH°= [(1 mol CS2(g)/mol-rxn)(116.7 kJ/mol)-[(1 mol C(s)/mol-rxn)(0 kJ/mol)+(2 mol S(s)/mol-rxn)(0 kJ/mol)] ]= 116.7 kJ/mol-rxn$

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)= [(1 mol CS2(g)/mol-rxn)S°[CS2(g)]-[(1 mol C(s)/mol-rxn)S°[C(s)]+(2 mol S(s)/mol-rxn)S°[S(s)]] ]$

Substituting the respective values,

$ΔrS°= [(1 mol CS2(g)/mol-rxn)(237.8 J/K×mol)-[(1 mol C(s)/mol-rxn)(5.6 J/K×mol)+(2 mol S(s)/mol-rxn)(32.1 J/K×mol)] ]= 168 J/K×mol-rxn$

Now, $ΔGo= ΔHo-TΔSo$

Substituting the value of $ΔHo$ and $ΔSo$ .

$ΔGo= 116.7 kJ/mol-rxn-[(298K)(168 J/K×mol-rxn)](1 kJ1000 J)= 66.6 kJ/mol-rxn$

The value listed in Appendix L is $66.61 kJ/mol-rxn$ . Thus both the calculated value and the value in Appendix L are in agreement with each other.

The value of $ΔG∘$ is positive, hence the reaction is reactant-favored at equilibrium.

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The the standard molar free energy for formation of $NaOH(s)$ 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 entropy by the following expression,

$ΔGo = ΔHo - TΔSo$

Here, $ΔHo$ is the change in enthalpy and $ΔSo$ is the change in entropy.

Answer to Problem 17PS

The standard molar free energy of formation for $NaOH(s)$ is $−379.82 kJ/mol-rxn$ .

The value of $ΔG∘$ is negative, hence the reaction is product-favored at equilibrium.

Explanation of Solution

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

Given:

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

$Na(s)+12H2(g)+12O2(g)→NaOH(s)ΔfH°(kJ/mol)000- 425.93So(J/K×mol)51.21130.7205.0764.46$

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)= [(1 mol NaOH(s)/mol-rxn)ΔfH°[NaOH(s)]-[(0.5 mol H2(g)/mol-rxn)ΔfH°[H2(g)]+(0.5 mol O2(g)/mol-rxn)ΔfH°[O2(g)]+(1 mol Na(s)/mol-rxn)ΔfH°[Na(s)]] ]$

Substituting the values,

$ΔrH°= [(1 mol NaOH(s)/mol-rxn)(-425.93 kJ/mol)[(0.5 mol H2(g)/mol-rxn)(0 kJ/mol)+(0.5 mol O2(g)/mol-rxn)(0 kJ/mol)+(1 mol Na(s)/mol-rxn)(0 kJ/mol)] ]= -425.93 kJ/mol-rxn$

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)= [(1 mol NaOH(s)/mol-rxn)S°[NaOH(s)]-[(0.5 mol H2(g)/mol-rxn)S°[H2(g)]+(0.5 mol O2(g)/mol-rxn)S°[O2(g)]+(1 mol Na(s)/mol-rxn)S°[Na(s)]] ]$

Substituting the values,

$ΔrS°= [(1 mol NaOH(s)/mol-rxn)(64.46 J/K×mol)-[(0.5 mol H2(g)/mol-rxn)(130.7 J/K×mol)+(0.5 mol O2(g)/mol-rxn)(205.07 J/K×mol)+(1 mol Na(s)/mol-rxn)(51.21 J/K×mol)] ]= -154.6 J/K×mol-rxn$

Now, $ΔGo= ΔHo-TΔSo$

Substitute the value of $ΔHo$ and $ΔSo$ .

$ΔGo= - 425.93 kJ/mol-rxn- [(298K)(-154.6 J/K×mol-rxn)](1 kJ1000 J)= - 379.82 kJ/mol-rxn$

The value listed in Appendix L is $−379.75 kJ/mol-rxn$ . Thus both the calculated value and the value in Appendix L are in agreement to each other.

The value of $ΔGo$ is negative, hence the reaction is product-favored at equilibrium.

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The the standard molar free energy for formation of $ICl(g)$ 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 entropy by the following expression,

$ΔGo = ΔHo - TΔSo$

Here, $ΔHo$ is the change in enthalpy and $ΔSo$ is the change in entropy.

Answer to Problem 17PS

The standard molar free energy of formation for $ICl(g)$ is $−5.73 kJ/mol-rxn$ .

The value of $ΔG∘$ is negative, hence the reaction is product-favored at equilibrium.

Explanation of Solution

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

Given:

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

$12I2(s)+12Cl2(g)→ICl(g)ΔfH°(kJ/mol)00+17.51So(J/K×mol)+116.135+223.08+247.56$

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)= [(1 mol ICl(g)/mol-rxn)ΔfH°[ICl(g)]-[(0.5 mol I2(s)/mol-rxn)ΔfH°[I2(s)]+(0.5 mol Cl2(g)/mol-rxn)ΔfH°[Cl2(g)]] ]$

Substituting the respective values,

$ΔrH°= [(1 mol ICl(g)/mol-rxn)(17.51 kJ/mol)-[(0.5 mol I2(s)/mol-rxn)(0 kJ/mol)+(0.5 mol Cl2(g)/mol-rxn)(0 kJ/mol)] ]= 17.51 kJ/mol-rxn$

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)= [(1 mol ICl(g)/mol-rxn)S°[ICl(g)]-[(0.5 mol I2(s)/mol-rxn)S°[I2(s)]+(0.5 mol Cl2(g)/mol-rxn)S°[Cl2(g)]] ]$

Substituting the respective values,

$ΔrS°= [(1 mol ICl(g)/mol-rxn)(247.56 J/K×mol)-[(0.5 mol I2(s)/mol-rxn)(116.135 J/K×mol)+(0.5 mol Cl2(g)/mol-rxn)(223.08 J/K×mol)] ] = 77.95 J/K×mol-rxn$

Now, $ΔGo= ΔHo-TΔSo$

Substituting the value of $ΔHo$ and $ΔSo$ .

$ΔGo= 17.51 kJ/mol-rxn-[(298K)(77.95 J/K×mol- rxn)](1 kJ1000 J)= -5.73 kJ/mol-rxn$

The value listed in Appendix L is $−5.73 kJ/mol-rxn$ . Thus both the calculated value and the value in Appendix L are in agreement to each other.

The value of $ΔG∘$ is negative, hence the reaction is product-favored at equilibrium.

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Answer 2

Textbook Question

Chapter 18, Problem 17PS

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

(a) CS₂(g)
(b) NaOH(s)
(c) ICl(g)

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 $CS2(g)$ 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 entropy by the following expression,

$ΔGo = ΔHo - TΔSo$

Here, $ΔHo$ is the change in enthalpy and $ΔSo$ is the change in entropy.

Answer to Problem 17PS

The standard molar free energy of formation for $CS2(g)$ is $+66.6 kJ/mol-rxn$ .

The value of $ΔG∘$ is positive, hence the reaction is reactant-favored at equilibrium.

Explanation of Solution

The standard molar free energy of formation for $CS2(g)$ is calculated below.

Given:

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

$C(s)+2S(s)→CS2(g)ΔfH°(kJ/mol)00+116.7So(J/K×mol)+5.6+32.1+237.8$

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)= [(1 mol CS2(g)/mol-rxn)ΔfH°[CS2(g)]-[(1 mol C(s)/mol-rxn)ΔfH°[C(s)]+(2 mol S(s)/mol-rxn)ΔfH°[S(s)]] ]$

Substituting the respective values,

$ΔrH°= [(1 mol CS2(g)/mol-rxn)(116.7 kJ/mol)-[(1 mol C(s)/mol-rxn)(0 kJ/mol)+(2 mol S(s)/mol-rxn)(0 kJ/mol)] ]= 116.7 kJ/mol-rxn$

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)= [(1 mol CS2(g)/mol-rxn)S°[CS2(g)]-[(1 mol C(s)/mol-rxn)S°[C(s)]+(2 mol S(s)/mol-rxn)S°[S(s)]] ]$

Substituting the respective values,

$ΔrS°= [(1 mol CS2(g)/mol-rxn)(237.8 J/K×mol)-[(1 mol C(s)/mol-rxn)(5.6 J/K×mol)+(2 mol S(s)/mol-rxn)(32.1 J/K×mol)] ]= 168 J/K×mol-rxn$

Now, $ΔGo= ΔHo-TΔSo$

Substituting the value of $ΔHo$ and $ΔSo$ .

$ΔGo= 116.7 kJ/mol-rxn-[(298K)(168 J/K×mol-rxn)](1 kJ1000 J)= 66.6 kJ/mol-rxn$

The value listed in Appendix L is $66.61 kJ/mol-rxn$ . Thus both the calculated value and the value in Appendix L are in agreement with each other.

The value of $ΔG∘$ is positive, hence the reaction is reactant-favored at equilibrium.

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The the standard molar free energy for formation of $NaOH(s)$ 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 entropy by the following expression,

$ΔGo = ΔHo - TΔSo$

Here, $ΔHo$ is the change in enthalpy and $ΔSo$ is the change in entropy.

Answer to Problem 17PS

The standard molar free energy of formation for $NaOH(s)$ is $−379.82 kJ/mol-rxn$ .

The value of $ΔG∘$ is negative, hence the reaction is product-favored at equilibrium.

Explanation of Solution

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

Given:

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

$Na(s)+12H2(g)+12O2(g)→NaOH(s)ΔfH°(kJ/mol)000- 425.93So(J/K×mol)51.21130.7205.0764.46$

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)= [(1 mol NaOH(s)/mol-rxn)ΔfH°[NaOH(s)]-[(0.5 mol H2(g)/mol-rxn)ΔfH°[H2(g)]+(0.5 mol O2(g)/mol-rxn)ΔfH°[O2(g)]+(1 mol Na(s)/mol-rxn)ΔfH°[Na(s)]] ]$

Substituting the values,

$ΔrH°= [(1 mol NaOH(s)/mol-rxn)(-425.93 kJ/mol)[(0.5 mol H2(g)/mol-rxn)(0 kJ/mol)+(0.5 mol O2(g)/mol-rxn)(0 kJ/mol)+(1 mol Na(s)/mol-rxn)(0 kJ/mol)] ]= -425.93 kJ/mol-rxn$

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)= [(1 mol NaOH(s)/mol-rxn)S°[NaOH(s)]-[(0.5 mol H2(g)/mol-rxn)S°[H2(g)]+(0.5 mol O2(g)/mol-rxn)S°[O2(g)]+(1 mol Na(s)/mol-rxn)S°[Na(s)]] ]$

Substituting the values,

$ΔrS°= [(1 mol NaOH(s)/mol-rxn)(64.46 J/K×mol)-[(0.5 mol H2(g)/mol-rxn)(130.7 J/K×mol)+(0.5 mol O2(g)/mol-rxn)(205.07 J/K×mol)+(1 mol Na(s)/mol-rxn)(51.21 J/K×mol)] ]= -154.6 J/K×mol-rxn$

Now, $ΔGo= ΔHo-TΔSo$

Substitute the value of $ΔHo$ and $ΔSo$ .

$ΔGo= - 425.93 kJ/mol-rxn- [(298K)(-154.6 J/K×mol-rxn)](1 kJ1000 J)= - 379.82 kJ/mol-rxn$

The value listed in Appendix L is $−379.75 kJ/mol-rxn$ . Thus both the calculated value and the value in Appendix L are in agreement to each other.

The value of $ΔGo$ is negative, hence the reaction is product-favored at equilibrium.

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The the standard molar free energy for formation of $ICl(g)$ 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 entropy by the following expression,

$ΔGo = ΔHo - TΔSo$

Here, $ΔHo$ is the change in enthalpy and $ΔSo$ is the change in entropy.

Answer to Problem 17PS

The standard molar free energy of formation for $ICl(g)$ is $−5.73 kJ/mol-rxn$ .

The value of $ΔG∘$ is negative, hence the reaction is product-favored at equilibrium.

Explanation of Solution

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

Given:

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

$12I2(s)+12Cl2(g)→ICl(g)ΔfH°(kJ/mol)00+17.51So(J/K×mol)+116.135+223.08+247.56$

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)= [(1 mol ICl(g)/mol-rxn)ΔfH°[ICl(g)]-[(0.5 mol I2(s)/mol-rxn)ΔfH°[I2(s)]+(0.5 mol Cl2(g)/mol-rxn)ΔfH°[Cl2(g)]] ]$

Substituting the respective values,

$ΔrH°= [(1 mol ICl(g)/mol-rxn)(17.51 kJ/mol)-[(0.5 mol I2(s)/mol-rxn)(0 kJ/mol)+(0.5 mol Cl2(g)/mol-rxn)(0 kJ/mol)] ]= 17.51 kJ/mol-rxn$

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)= [(1 mol ICl(g)/mol-rxn)S°[ICl(g)]-[(0.5 mol I2(s)/mol-rxn)S°[I2(s)]+(0.5 mol Cl2(g)/mol-rxn)S°[Cl2(g)]] ]$

Substituting the respective values,

$ΔrS°= [(1 mol ICl(g)/mol-rxn)(247.56 J/K×mol)-[(0.5 mol I2(s)/mol-rxn)(116.135 J/K×mol)+(0.5 mol Cl2(g)/mol-rxn)(223.08 J/K×mol)] ] = 77.95 J/K×mol-rxn$

Now, $ΔGo= ΔHo-TΔSo$

Substituting the value of $ΔHo$ and $ΔSo$ .

$ΔGo= 17.51 kJ/mol-rxn-[(298K)(77.95 J/K×mol- rxn)](1 kJ1000 J)= -5.73 kJ/mol-rxn$

The value listed in Appendix L is $−5.73 kJ/mol-rxn$ . Thus both the calculated value and the value in Appendix L are in agreement to each other.

The value of $ΔG∘$ is negative, hence the reaction is product-favored at equilibrium.

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Answer 3

Textbook Question

Chapter 18, Problem 17PS

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

(a) CS₂(g)
(b) NaOH(s)
(c) ICl(g)

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 $CS2(g)$ 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 entropy by the following expression,

$ΔGo = ΔHo - TΔSo$

Here, $ΔHo$ is the change in enthalpy and $ΔSo$ is the change in entropy.

Answer to Problem 17PS

The standard molar free energy of formation for $CS2(g)$ is $+66.6 kJ/mol-rxn$ .

The value of $ΔG∘$ is positive, hence the reaction is reactant-favored at equilibrium.

Explanation of Solution

The standard molar free energy of formation for $CS2(g)$ is calculated below.

Given:

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

$C(s)+2S(s)→CS2(g)ΔfH°(kJ/mol)00+116.7So(J/K×mol)+5.6+32.1+237.8$

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)= [(1 mol CS2(g)/mol-rxn)ΔfH°[CS2(g)]-[(1 mol C(s)/mol-rxn)ΔfH°[C(s)]+(2 mol S(s)/mol-rxn)ΔfH°[S(s)]] ]$

Substituting the respective values,

$ΔrH°= [(1 mol CS2(g)/mol-rxn)(116.7 kJ/mol)-[(1 mol C(s)/mol-rxn)(0 kJ/mol)+(2 mol S(s)/mol-rxn)(0 kJ/mol)] ]= 116.7 kJ/mol-rxn$

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)= [(1 mol CS2(g)/mol-rxn)S°[CS2(g)]-[(1 mol C(s)/mol-rxn)S°[C(s)]+(2 mol S(s)/mol-rxn)S°[S(s)]] ]$

Substituting the respective values,

$ΔrS°= [(1 mol CS2(g)/mol-rxn)(237.8 J/K×mol)-[(1 mol C(s)/mol-rxn)(5.6 J/K×mol)+(2 mol S(s)/mol-rxn)(32.1 J/K×mol)] ]= 168 J/K×mol-rxn$

Now, $ΔGo= ΔHo-TΔSo$

Substituting the value of $ΔHo$ and $ΔSo$ .

$ΔGo= 116.7 kJ/mol-rxn-[(298K)(168 J/K×mol-rxn)](1 kJ1000 J)= 66.6 kJ/mol-rxn$

The value listed in Appendix L is $66.61 kJ/mol-rxn$ . Thus both the calculated value and the value in Appendix L are in agreement with each other.

The value of $ΔG∘$ is positive, hence the reaction is reactant-favored at equilibrium.

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The the standard molar free energy for formation of $NaOH(s)$ 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 entropy by the following expression,

$ΔGo = ΔHo - TΔSo$

Here, $ΔHo$ is the change in enthalpy and $ΔSo$ is the change in entropy.

Answer to Problem 17PS

The standard molar free energy of formation for $NaOH(s)$ is $−379.82 kJ/mol-rxn$ .

The value of $ΔG∘$ is negative, hence the reaction is product-favored at equilibrium.

Explanation of Solution

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

Given:

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

$Na(s)+12H2(g)+12O2(g)→NaOH(s)ΔfH°(kJ/mol)000- 425.93So(J/K×mol)51.21130.7205.0764.46$

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)= [(1 mol NaOH(s)/mol-rxn)ΔfH°[NaOH(s)]-[(0.5 mol H2(g)/mol-rxn)ΔfH°[H2(g)]+(0.5 mol O2(g)/mol-rxn)ΔfH°[O2(g)]+(1 mol Na(s)/mol-rxn)ΔfH°[Na(s)]] ]$

Substituting the values,

$ΔrH°= [(1 mol NaOH(s)/mol-rxn)(-425.93 kJ/mol)[(0.5 mol H2(g)/mol-rxn)(0 kJ/mol)+(0.5 mol O2(g)/mol-rxn)(0 kJ/mol)+(1 mol Na(s)/mol-rxn)(0 kJ/mol)] ]= -425.93 kJ/mol-rxn$

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)= [(1 mol NaOH(s)/mol-rxn)S°[NaOH(s)]-[(0.5 mol H2(g)/mol-rxn)S°[H2(g)]+(0.5 mol O2(g)/mol-rxn)S°[O2(g)]+(1 mol Na(s)/mol-rxn)S°[Na(s)]] ]$

Substituting the values,

$ΔrS°= [(1 mol NaOH(s)/mol-rxn)(64.46 J/K×mol)-[(0.5 mol H2(g)/mol-rxn)(130.7 J/K×mol)+(0.5 mol O2(g)/mol-rxn)(205.07 J/K×mol)+(1 mol Na(s)/mol-rxn)(51.21 J/K×mol)] ]= -154.6 J/K×mol-rxn$

Now, $ΔGo= ΔHo-TΔSo$

Substitute the value of $ΔHo$ and $ΔSo$ .

$ΔGo= - 425.93 kJ/mol-rxn- [(298K)(-154.6 J/K×mol-rxn)](1 kJ1000 J)= - 379.82 kJ/mol-rxn$

The value listed in Appendix L is $−379.75 kJ/mol-rxn$ . Thus both the calculated value and the value in Appendix L are in agreement to each other.

The value of $ΔGo$ is negative, hence the reaction is product-favored at equilibrium.

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The the standard molar free energy for formation of $ICl(g)$ 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 entropy by the following expression,

$ΔGo = ΔHo - TΔSo$

Here, $ΔHo$ is the change in enthalpy and $ΔSo$ is the change in entropy.

Answer to Problem 17PS

The standard molar free energy of formation for $ICl(g)$ is $−5.73 kJ/mol-rxn$ .

The value of $ΔG∘$ is negative, hence the reaction is product-favored at equilibrium.

Explanation of Solution

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

Given:

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

$12I2(s)+12Cl2(g)→ICl(g)ΔfH°(kJ/mol)00+17.51So(J/K×mol)+116.135+223.08+247.56$

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)= [(1 mol ICl(g)/mol-rxn)ΔfH°[ICl(g)]-[(0.5 mol I2(s)/mol-rxn)ΔfH°[I2(s)]+(0.5 mol Cl2(g)/mol-rxn)ΔfH°[Cl2(g)]] ]$

Substituting the respective values,

$ΔrH°= [(1 mol ICl(g)/mol-rxn)(17.51 kJ/mol)-[(0.5 mol I2(s)/mol-rxn)(0 kJ/mol)+(0.5 mol Cl2(g)/mol-rxn)(0 kJ/mol)] ]= 17.51 kJ/mol-rxn$

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)= [(1 mol ICl(g)/mol-rxn)S°[ICl(g)]-[(0.5 mol I2(s)/mol-rxn)S°[I2(s)]+(0.5 mol Cl2(g)/mol-rxn)S°[Cl2(g)]] ]$

Substituting the respective values,

$ΔrS°= [(1 mol ICl(g)/mol-rxn)(247.56 J/K×mol)-[(0.5 mol I2(s)/mol-rxn)(116.135 J/K×mol)+(0.5 mol Cl2(g)/mol-rxn)(223.08 J/K×mol)] ] = 77.95 J/K×mol-rxn$

Now, $ΔGo= ΔHo-TΔSo$

Substituting the value of $ΔHo$ and $ΔSo$ .

$ΔGo= 17.51 kJ/mol-rxn-[(298K)(77.95 J/K×mol- rxn)](1 kJ1000 J)= -5.73 kJ/mol-rxn$

The value listed in Appendix L is $−5.73 kJ/mol-rxn$ . Thus both the calculated value and the value in Appendix L are in agreement to each other.

The value of $ΔG∘$ is negative, hence the reaction is product-favored at equilibrium.

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Answer 4

Textbook Question

Chapter 18, Problem 17PS

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

(a) CS₂(g)
(b) NaOH(s)
(c) ICl(g)

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 $CS2(g)$ 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 entropy by the following expression,

$ΔGo = ΔHo - TΔSo$

Here, $ΔHo$ is the change in enthalpy and $ΔSo$ is the change in entropy.

Answer to Problem 17PS

The standard molar free energy of formation for $CS2(g)$ is $+66.6 kJ/mol-rxn$ .

The value of $ΔG∘$ is positive, hence the reaction is reactant-favored at equilibrium.

Explanation of Solution

The standard molar free energy of formation for $CS2(g)$ is calculated below.

Given:

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

$C(s)+2S(s)→CS2(g)ΔfH°(kJ/mol)00+116.7So(J/K×mol)+5.6+32.1+237.8$

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)= [(1 mol CS2(g)/mol-rxn)ΔfH°[CS2(g)]-[(1 mol C(s)/mol-rxn)ΔfH°[C(s)]+(2 mol S(s)/mol-rxn)ΔfH°[S(s)]] ]$

Substituting the respective values,

$ΔrH°= [(1 mol CS2(g)/mol-rxn)(116.7 kJ/mol)-[(1 mol C(s)/mol-rxn)(0 kJ/mol)+(2 mol S(s)/mol-rxn)(0 kJ/mol)] ]= 116.7 kJ/mol-rxn$

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)= [(1 mol CS2(g)/mol-rxn)S°[CS2(g)]-[(1 mol C(s)/mol-rxn)S°[C(s)]+(2 mol S(s)/mol-rxn)S°[S(s)]] ]$

Substituting the respective values,

$ΔrS°= [(1 mol CS2(g)/mol-rxn)(237.8 J/K×mol)-[(1 mol C(s)/mol-rxn)(5.6 J/K×mol)+(2 mol S(s)/mol-rxn)(32.1 J/K×mol)] ]= 168 J/K×mol-rxn$

Now, $ΔGo= ΔHo-TΔSo$

Substituting the value of $ΔHo$ and $ΔSo$ .

$ΔGo= 116.7 kJ/mol-rxn-[(298K)(168 J/K×mol-rxn)](1 kJ1000 J)= 66.6 kJ/mol-rxn$

The value listed in Appendix L is $66.61 kJ/mol-rxn$ . Thus both the calculated value and the value in Appendix L are in agreement with each other.

The value of $ΔG∘$ is positive, hence the reaction is reactant-favored at equilibrium.

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The the standard molar free energy for formation of $NaOH(s)$ 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 entropy by the following expression,

$ΔGo = ΔHo - TΔSo$

Here, $ΔHo$ is the change in enthalpy and $ΔSo$ is the change in entropy.

Answer to Problem 17PS

The standard molar free energy of formation for $NaOH(s)$ is $−379.82 kJ/mol-rxn$ .

The value of $ΔG∘$ is negative, hence the reaction is product-favored at equilibrium.

Explanation of Solution

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

Given:

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

$Na(s)+12H2(g)+12O2(g)→NaOH(s)ΔfH°(kJ/mol)000- 425.93So(J/K×mol)51.21130.7205.0764.46$

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)= [(1 mol NaOH(s)/mol-rxn)ΔfH°[NaOH(s)]-[(0.5 mol H2(g)/mol-rxn)ΔfH°[H2(g)]+(0.5 mol O2(g)/mol-rxn)ΔfH°[O2(g)]+(1 mol Na(s)/mol-rxn)ΔfH°[Na(s)]] ]$

Substituting the values,

$ΔrH°= [(1 mol NaOH(s)/mol-rxn)(-425.93 kJ/mol)[(0.5 mol H2(g)/mol-rxn)(0 kJ/mol)+(0.5 mol O2(g)/mol-rxn)(0 kJ/mol)+(1 mol Na(s)/mol-rxn)(0 kJ/mol)] ]= -425.93 kJ/mol-rxn$

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)= [(1 mol NaOH(s)/mol-rxn)S°[NaOH(s)]-[(0.5 mol H2(g)/mol-rxn)S°[H2(g)]+(0.5 mol O2(g)/mol-rxn)S°[O2(g)]+(1 mol Na(s)/mol-rxn)S°[Na(s)]] ]$

Substituting the values,

$ΔrS°= [(1 mol NaOH(s)/mol-rxn)(64.46 J/K×mol)-[(0.5 mol H2(g)/mol-rxn)(130.7 J/K×mol)+(0.5 mol O2(g)/mol-rxn)(205.07 J/K×mol)+(1 mol Na(s)/mol-rxn)(51.21 J/K×mol)] ]= -154.6 J/K×mol-rxn$

Now, $ΔGo= ΔHo-TΔSo$

Substitute the value of $ΔHo$ and $ΔSo$ .

$ΔGo= - 425.93 kJ/mol-rxn- [(298K)(-154.6 J/K×mol-rxn)](1 kJ1000 J)= - 379.82 kJ/mol-rxn$

The value listed in Appendix L is $−379.75 kJ/mol-rxn$ . Thus both the calculated value and the value in Appendix L are in agreement to each other.

The value of $ΔGo$ is negative, hence the reaction is product-favored at equilibrium.

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The the standard molar free energy for formation of $ICl(g)$ 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 entropy by the following expression,

$ΔGo = ΔHo - TΔSo$

Here, $ΔHo$ is the change in enthalpy and $ΔSo$ is the change in entropy.

Answer to Problem 17PS

The standard molar free energy of formation for $ICl(g)$ is $−5.73 kJ/mol-rxn$ .

The value of $ΔG∘$ is negative, hence the reaction is product-favored at equilibrium.

Explanation of Solution

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

Given:

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

$12I2(s)+12Cl2(g)→ICl(g)ΔfH°(kJ/mol)00+17.51So(J/K×mol)+116.135+223.08+247.56$

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)= [(1 mol ICl(g)/mol-rxn)ΔfH°[ICl(g)]-[(0.5 mol I2(s)/mol-rxn)ΔfH°[I2(s)]+(0.5 mol Cl2(g)/mol-rxn)ΔfH°[Cl2(g)]] ]$

Substituting the respective values,

$ΔrH°= [(1 mol ICl(g)/mol-rxn)(17.51 kJ/mol)-[(0.5 mol I2(s)/mol-rxn)(0 kJ/mol)+(0.5 mol Cl2(g)/mol-rxn)(0 kJ/mol)] ]= 17.51 kJ/mol-rxn$

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)= [(1 mol ICl(g)/mol-rxn)S°[ICl(g)]-[(0.5 mol I2(s)/mol-rxn)S°[I2(s)]+(0.5 mol Cl2(g)/mol-rxn)S°[Cl2(g)]] ]$

Substituting the respective values,

$ΔrS°= [(1 mol ICl(g)/mol-rxn)(247.56 J/K×mol)-[(0.5 mol I2(s)/mol-rxn)(116.135 J/K×mol)+(0.5 mol Cl2(g)/mol-rxn)(223.08 J/K×mol)] ] = 77.95 J/K×mol-rxn$

Now, $ΔGo= ΔHo-TΔSo$

Substituting the value of $ΔHo$ and $ΔSo$ .

$ΔGo= 17.51 kJ/mol-rxn-[(298K)(77.95 J/K×mol- rxn)](1 kJ1000 J)= -5.73 kJ/mol-rxn$

The value listed in Appendix L is $−5.73 kJ/mol-rxn$ . Thus both the calculated value and the value in Appendix L are in agreement to each other.

The value of $ΔG∘$ is negative, hence 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 $CS2(g)$ 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 entropy by the following expression,

$ΔGo = ΔHo - TΔSo$

Here, $ΔHo$ is the change in enthalpy and $ΔSo$ is the change in entropy.

Answer to Problem 17PS

The standard molar free energy of formation for $CS2(g)$ is $+66.6 kJ/mol-rxn$ .

The value of $ΔG∘$ is positive, hence the reaction is reactant-favored at equilibrium.

Explanation of Solution

The standard molar free energy of formation for $CS2(g)$ is calculated below.

Given:

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

$C(s)+2S(s)→CS2(g)ΔfH°(kJ/mol)00+116.7So(J/K×mol)+5.6+32.1+237.8$

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)= [(1 mol CS2(g)/mol-rxn)ΔfH°[CS2(g)]-[(1 mol C(s)/mol-rxn)ΔfH°[C(s)]+(2 mol S(s)/mol-rxn)ΔfH°[S(s)]] ]$

Substituting the respective values,

$ΔrH°= [(1 mol CS2(g)/mol-rxn)(116.7 kJ/mol)-[(1 mol C(s)/mol-rxn)(0 kJ/mol)+(2 mol S(s)/mol-rxn)(0 kJ/mol)] ]= 116.7 kJ/mol-rxn$

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)= [(1 mol CS2(g)/mol-rxn)S°[CS2(g)]-[(1 mol C(s)/mol-rxn)S°[C(s)]+(2 mol S(s)/mol-rxn)S°[S(s)]] ]$

Substituting the respective values,

$ΔrS°= [(1 mol CS2(g)/mol-rxn)(237.8 J/K×mol)-[(1 mol C(s)/mol-rxn)(5.6 J/K×mol)+(2 mol S(s)/mol-rxn)(32.1 J/K×mol)] ]= 168 J/K×mol-rxn$

Now, $ΔGo= ΔHo-TΔSo$

Substituting the value of $ΔHo$ and $ΔSo$ .

$ΔGo= 116.7 kJ/mol-rxn-[(298K)(168 J/K×mol-rxn)](1 kJ1000 J)= 66.6 kJ/mol-rxn$

The value listed in Appendix L is $66.61 kJ/mol-rxn$ . Thus both the calculated value and the value in Appendix L are in agreement with each other.

The value of $ΔG∘$ is positive, hence the reaction is reactant-favored at equilibrium.

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The the standard molar free energy for formation of $NaOH(s)$ 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 entropy by the following expression,

$ΔGo = ΔHo - TΔSo$

Here, $ΔHo$ is the change in enthalpy and $ΔSo$ is the change in entropy.

Answer to Problem 17PS

The standard molar free energy of formation for $NaOH(s)$ is $−379.82 kJ/mol-rxn$ .

The value of $ΔG∘$ is negative, hence the reaction is product-favored at equilibrium.

Explanation of Solution

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

Given:

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

$Na(s)+12H2(g)+12O2(g)→NaOH(s)ΔfH°(kJ/mol)000- 425.93So(J/K×mol)51.21130.7205.0764.46$

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)= [(1 mol NaOH(s)/mol-rxn)ΔfH°[NaOH(s)]-[(0.5 mol H2(g)/mol-rxn)ΔfH°[H2(g)]+(0.5 mol O2(g)/mol-rxn)ΔfH°[O2(g)]+(1 mol Na(s)/mol-rxn)ΔfH°[Na(s)]] ]$

Substituting the values,

$ΔrH°= [(1 mol NaOH(s)/mol-rxn)(-425.93 kJ/mol)[(0.5 mol H2(g)/mol-rxn)(0 kJ/mol)+(0.5 mol O2(g)/mol-rxn)(0 kJ/mol)+(1 mol Na(s)/mol-rxn)(0 kJ/mol)] ]= -425.93 kJ/mol-rxn$

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)= [(1 mol NaOH(s)/mol-rxn)S°[NaOH(s)]-[(0.5 mol H2(g)/mol-rxn)S°[H2(g)]+(0.5 mol O2(g)/mol-rxn)S°[O2(g)]+(1 mol Na(s)/mol-rxn)S°[Na(s)]] ]$

Substituting the values,

$ΔrS°= [(1 mol NaOH(s)/mol-rxn)(64.46 J/K×mol)-[(0.5 mol H2(g)/mol-rxn)(130.7 J/K×mol)+(0.5 mol O2(g)/mol-rxn)(205.07 J/K×mol)+(1 mol Na(s)/mol-rxn)(51.21 J/K×mol)] ]= -154.6 J/K×mol-rxn$

Now, $ΔGo= ΔHo-TΔSo$

Substitute the value of $ΔHo$ and $ΔSo$ .

$ΔGo= - 425.93 kJ/mol-rxn- [(298K)(-154.6 J/K×mol-rxn)](1 kJ1000 J)= - 379.82 kJ/mol-rxn$

The value listed in Appendix L is $−379.75 kJ/mol-rxn$ . Thus both the calculated value and the value in Appendix L are in agreement to each other.

The value of $ΔGo$ is negative, hence the reaction is product-favored at equilibrium.

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The the standard molar free energy for formation of $ICl(g)$ 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 entropy by the following expression,

$ΔGo = ΔHo - TΔSo$

Here, $ΔHo$ is the change in enthalpy and $ΔSo$ is the change in entropy.

Answer to Problem 17PS

The standard molar free energy of formation for $ICl(g)$ is $−5.73 kJ/mol-rxn$ .

The value of $ΔG∘$ is negative, hence the reaction is product-favored at equilibrium.

Explanation of Solution

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

Given:

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

$12I2(s)+12Cl2(g)→ICl(g)ΔfH°(kJ/mol)00+17.51So(J/K×mol)+116.135+223.08+247.56$

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)= [(1 mol ICl(g)/mol-rxn)ΔfH°[ICl(g)]-[(0.5 mol I2(s)/mol-rxn)ΔfH°[I2(s)]+(0.5 mol Cl2(g)/mol-rxn)ΔfH°[Cl2(g)]] ]$

Substituting the respective values,

$ΔrH°= [(1 mol ICl(g)/mol-rxn)(17.51 kJ/mol)-[(0.5 mol I2(s)/mol-rxn)(0 kJ/mol)+(0.5 mol Cl2(g)/mol-rxn)(0 kJ/mol)] ]= 17.51 kJ/mol-rxn$

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)= [(1 mol ICl(g)/mol-rxn)S°[ICl(g)]-[(0.5 mol I2(s)/mol-rxn)S°[I2(s)]+(0.5 mol Cl2(g)/mol-rxn)S°[Cl2(g)]] ]$

Substituting the respective values,

$ΔrS°= [(1 mol ICl(g)/mol-rxn)(247.56 J/K×mol)-[(0.5 mol I2(s)/mol-rxn)(116.135 J/K×mol)+(0.5 mol Cl2(g)/mol-rxn)(223.08 J/K×mol)] ] = 77.95 J/K×mol-rxn$

Now, $ΔGo= ΔHo-TΔSo$

Substituting the value of $ΔHo$ and $ΔSo$ .

$ΔGo= 17.51 kJ/mol-rxn-[(298K)(77.95 J/K×mol- rxn)](1 kJ1000 J)= -5.73 kJ/mol-rxn$

The value listed in Appendix L is $−5.73 kJ/mol-rxn$ . Thus both the calculated value and the value in Appendix L are in agreement to each other.

The value of $ΔG∘$ is negative, hence the reaction is product-favored at equilibrium.

Answer 8

(a)

Expert Solution

Interpretation Introduction

Interpretation:

The the standard molar free energy for formation of $CS2(g)$ 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 entropy by the following expression,

$ΔGo = ΔHo - TΔSo$

Here, $ΔHo$ is the change in enthalpy and $ΔSo$ is the change in entropy.

Answer to Problem 17PS

The standard molar free energy of formation for $CS2(g)$ is $+66.6 kJ/mol-rxn$ .

The value of $ΔG∘$ is positive, hence the reaction is reactant-favored at equilibrium.

Explanation of Solution

The standard molar free energy of formation for $CS2(g)$ is calculated below.

Given:

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

$C(s)+2S(s)→CS2(g)ΔfH°(kJ/mol)00+116.7So(J/K×mol)+5.6+32.1+237.8$

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)= [(1 mol CS2(g)/mol-rxn)ΔfH°[CS2(g)]-[(1 mol C(s)/mol-rxn)ΔfH°[C(s)]+(2 mol S(s)/mol-rxn)ΔfH°[S(s)]] ]$

Substituting the respective values,

$ΔrH°= [(1 mol CS2(g)/mol-rxn)(116.7 kJ/mol)-[(1 mol C(s)/mol-rxn)(0 kJ/mol)+(2 mol S(s)/mol-rxn)(0 kJ/mol)] ]= 116.7 kJ/mol-rxn$

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)= [(1 mol CS2(g)/mol-rxn)S°[CS2(g)]-[(1 mol C(s)/mol-rxn)S°[C(s)]+(2 mol S(s)/mol-rxn)S°[S(s)]] ]$

Substituting the respective values,

$ΔrS°= [(1 mol CS2(g)/mol-rxn)(237.8 J/K×mol)-[(1 mol C(s)/mol-rxn)(5.6 J/K×mol)+(2 mol S(s)/mol-rxn)(32.1 J/K×mol)] ]= 168 J/K×mol-rxn$

Now, $ΔGo= ΔHo-TΔSo$

Substituting the value of $ΔHo$ and $ΔSo$ .

$ΔGo= 116.7 kJ/mol-rxn-[(298K)(168 J/K×mol-rxn)](1 kJ1000 J)= 66.6 kJ/mol-rxn$

The value listed in Appendix L is $66.61 kJ/mol-rxn$ . Thus both the calculated value and the value in Appendix L are in agreement with each other.

The value of $ΔG∘$ is positive, hence the reaction is reactant-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 $CS2(g)$ 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 entropy by the following expression,

$ΔGo = ΔHo - TΔSo$

Here, $ΔHo$ is the change in enthalpy and $ΔSo$ is the change in entropy.

Answer 13

Answer to Problem 17PS

The standard molar free energy of formation for $CS2(g)$ is $+66.6 kJ/mol-rxn$ .

The value of $ΔG∘$ is positive, hence the reaction is reactant-favored at equilibrium.

Answer 14

Explanation of Solution

The standard molar free energy of formation for $CS2(g)$ is calculated below.

Given:

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

$C(s)+2S(s)→CS2(g)ΔfH°(kJ/mol)00+116.7So(J/K×mol)+5.6+32.1+237.8$

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)= [(1 mol CS2(g)/mol-rxn)ΔfH°[CS2(g)]-[(1 mol C(s)/mol-rxn)ΔfH°[C(s)]+(2 mol S(s)/mol-rxn)ΔfH°[S(s)]] ]$

Substituting the respective values,

$ΔrH°= [(1 mol CS2(g)/mol-rxn)(116.7 kJ/mol)-[(1 mol C(s)/mol-rxn)(0 kJ/mol)+(2 mol S(s)/mol-rxn)(0 kJ/mol)] ]= 116.7 kJ/mol-rxn$

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)= [(1 mol CS2(g)/mol-rxn)S°[CS2(g)]-[(1 mol C(s)/mol-rxn)S°[C(s)]+(2 mol S(s)/mol-rxn)S°[S(s)]] ]$

Substituting the respective values,

$ΔrS°= [(1 mol CS2(g)/mol-rxn)(237.8 J/K×mol)-[(1 mol C(s)/mol-rxn)(5.6 J/K×mol)+(2 mol S(s)/mol-rxn)(32.1 J/K×mol)] ]= 168 J/K×mol-rxn$

Now, $ΔGo= ΔHo-TΔSo$

Substituting the value of $ΔHo$ and $ΔSo$ .

$ΔGo= 116.7 kJ/mol-rxn-[(298K)(168 J/K×mol-rxn)](1 kJ1000 J)= 66.6 kJ/mol-rxn$

The value listed in Appendix L is $66.61 kJ/mol-rxn$ . Thus both the calculated value and the value in Appendix L are in agreement with each other.

The value of $ΔG∘$ is positive, hence the reaction is reactant-favored at equilibrium.

Answer 15

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The the standard molar free energy for formation of $NaOH(s)$ 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 entropy by the following expression,

$ΔGo = ΔHo - TΔSo$

Here, $ΔHo$ is the change in enthalpy and $ΔSo$ is the change in entropy.

Answer to Problem 17PS

The standard molar free energy of formation for $NaOH(s)$ is $−379.82 kJ/mol-rxn$ .

The value of $ΔG∘$ is negative, hence the reaction is product-favored at equilibrium.

Explanation of Solution

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

Given:

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

$Na(s)+12H2(g)+12O2(g)→NaOH(s)ΔfH°(kJ/mol)000- 425.93So(J/K×mol)51.21130.7205.0764.46$

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)= [(1 mol NaOH(s)/mol-rxn)ΔfH°[NaOH(s)]-[(0.5 mol H2(g)/mol-rxn)ΔfH°[H2(g)]+(0.5 mol O2(g)/mol-rxn)ΔfH°[O2(g)]+(1 mol Na(s)/mol-rxn)ΔfH°[Na(s)]] ]$

Substituting the values,

$ΔrH°= [(1 mol NaOH(s)/mol-rxn)(-425.93 kJ/mol)[(0.5 mol H2(g)/mol-rxn)(0 kJ/mol)+(0.5 mol O2(g)/mol-rxn)(0 kJ/mol)+(1 mol Na(s)/mol-rxn)(0 kJ/mol)] ]= -425.93 kJ/mol-rxn$

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)= [(1 mol NaOH(s)/mol-rxn)S°[NaOH(s)]-[(0.5 mol H2(g)/mol-rxn)S°[H2(g)]+(0.5 mol O2(g)/mol-rxn)S°[O2(g)]+(1 mol Na(s)/mol-rxn)S°[Na(s)]] ]$

Substituting the values,

$ΔrS°= [(1 mol NaOH(s)/mol-rxn)(64.46 J/K×mol)-[(0.5 mol H2(g)/mol-rxn)(130.7 J/K×mol)+(0.5 mol O2(g)/mol-rxn)(205.07 J/K×mol)+(1 mol Na(s)/mol-rxn)(51.21 J/K×mol)] ]= -154.6 J/K×mol-rxn$

Now, $ΔGo= ΔHo-TΔSo$

Substitute the value of $ΔHo$ and $ΔSo$ .

$ΔGo= - 425.93 kJ/mol-rxn- [(298K)(-154.6 J/K×mol-rxn)](1 kJ1000 J)= - 379.82 kJ/mol-rxn$

The value listed in Appendix L is $−379.75 kJ/mol-rxn$ . Thus both the calculated value and the value in Appendix L are in agreement to each other.

The value of $ΔGo$ is negative, hence the reaction is product-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 $NaOH(s)$ 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 entropy by the following expression,

$ΔGo = ΔHo - TΔSo$

Here, $ΔHo$ is the change in enthalpy and $ΔSo$ is the change in entropy.

Answer 20

Answer to Problem 17PS

The standard molar free energy of formation for $NaOH(s)$ is $−379.82 kJ/mol-rxn$ .

The value of $ΔG∘$ is negative, hence the reaction is product-favored at equilibrium.

Answer 21

Explanation of Solution

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

Given:

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

$Na(s)+12H2(g)+12O2(g)→NaOH(s)ΔfH°(kJ/mol)000- 425.93So(J/K×mol)51.21130.7205.0764.46$

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)= [(1 mol NaOH(s)/mol-rxn)ΔfH°[NaOH(s)]-[(0.5 mol H2(g)/mol-rxn)ΔfH°[H2(g)]+(0.5 mol O2(g)/mol-rxn)ΔfH°[O2(g)]+(1 mol Na(s)/mol-rxn)ΔfH°[Na(s)]] ]$

Substituting the values,

$ΔrH°= [(1 mol NaOH(s)/mol-rxn)(-425.93 kJ/mol)[(0.5 mol H2(g)/mol-rxn)(0 kJ/mol)+(0.5 mol O2(g)/mol-rxn)(0 kJ/mol)+(1 mol Na(s)/mol-rxn)(0 kJ/mol)] ]= -425.93 kJ/mol-rxn$

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)= [(1 mol NaOH(s)/mol-rxn)S°[NaOH(s)]-[(0.5 mol H2(g)/mol-rxn)S°[H2(g)]+(0.5 mol O2(g)/mol-rxn)S°[O2(g)]+(1 mol Na(s)/mol-rxn)S°[Na(s)]] ]$

Substituting the values,

$ΔrS°= [(1 mol NaOH(s)/mol-rxn)(64.46 J/K×mol)-[(0.5 mol H2(g)/mol-rxn)(130.7 J/K×mol)+(0.5 mol O2(g)/mol-rxn)(205.07 J/K×mol)+(1 mol Na(s)/mol-rxn)(51.21 J/K×mol)] ]= -154.6 J/K×mol-rxn$

Now, $ΔGo= ΔHo-TΔSo$

Substitute the value of $ΔHo$ and $ΔSo$ .

$ΔGo= - 425.93 kJ/mol-rxn- [(298K)(-154.6 J/K×mol-rxn)](1 kJ1000 J)= - 379.82 kJ/mol-rxn$

The value listed in Appendix L is $−379.75 kJ/mol-rxn$ . Thus both the calculated value and the value in Appendix L are in agreement to each other.

The value of $ΔGo$ is negative, hence the reaction is product-favored at equilibrium.

Answer 22

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The the standard molar free energy for formation of $ICl(g)$ 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 entropy by the following expression,

$ΔGo = ΔHo - TΔSo$

Here, $ΔHo$ is the change in enthalpy and $ΔSo$ is the change in entropy.

Answer to Problem 17PS

The standard molar free energy of formation for $ICl(g)$ is $−5.73 kJ/mol-rxn$ .

The value of $ΔG∘$ is negative, hence the reaction is product-favored at equilibrium.

Explanation of Solution

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

Given:

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

$12I2(s)+12Cl2(g)→ICl(g)ΔfH°(kJ/mol)00+17.51So(J/K×mol)+116.135+223.08+247.56$

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)= [(1 mol ICl(g)/mol-rxn)ΔfH°[ICl(g)]-[(0.5 mol I2(s)/mol-rxn)ΔfH°[I2(s)]+(0.5 mol Cl2(g)/mol-rxn)ΔfH°[Cl2(g)]] ]$

Substituting the respective values,

$ΔrH°= [(1 mol ICl(g)/mol-rxn)(17.51 kJ/mol)-[(0.5 mol I2(s)/mol-rxn)(0 kJ/mol)+(0.5 mol Cl2(g)/mol-rxn)(0 kJ/mol)] ]= 17.51 kJ/mol-rxn$

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)= [(1 mol ICl(g)/mol-rxn)S°[ICl(g)]-[(0.5 mol I2(s)/mol-rxn)S°[I2(s)]+(0.5 mol Cl2(g)/mol-rxn)S°[Cl2(g)]] ]$

Substituting the respective values,

$ΔrS°= [(1 mol ICl(g)/mol-rxn)(247.56 J/K×mol)-[(0.5 mol I2(s)/mol-rxn)(116.135 J/K×mol)+(0.5 mol Cl2(g)/mol-rxn)(223.08 J/K×mol)] ] = 77.95 J/K×mol-rxn$

Now, $ΔGo= ΔHo-TΔSo$

Substituting the value of $ΔHo$ and $ΔSo$ .

$ΔGo= 17.51 kJ/mol-rxn-[(298K)(77.95 J/K×mol- rxn)](1 kJ1000 J)= -5.73 kJ/mol-rxn$

The value listed in Appendix L is $−5.73 kJ/mol-rxn$ . Thus both the calculated value and the value in Appendix L are in agreement to each other.

The value of $ΔG∘$ is negative, hence 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 $ICl(g)$ 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 entropy by the following expression,

$ΔGo = ΔHo - TΔSo$

Here, $ΔHo$ is the change in enthalpy and $ΔSo$ is the change in entropy.

Answer 27

Answer to Problem 17PS

The standard molar free energy of formation for $ICl(g)$ is $−5.73 kJ/mol-rxn$ .

The value of $ΔG∘$ is negative, hence the reaction is product-favored at equilibrium.

Answer 28

Explanation of Solution

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

Given:

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

$12I2(s)+12Cl2(g)→ICl(g)ΔfH°(kJ/mol)00+17.51So(J/K×mol)+116.135+223.08+247.56$

$ΔrH°= ∑nΔfH°(products)- ∑nΔfH°(reactants)= [(1 mol ICl(g)/mol-rxn)ΔfH°[ICl(g)]-[(0.5 mol I2(s)/mol-rxn)ΔfH°[I2(s)]+(0.5 mol Cl2(g)/mol-rxn)ΔfH°[Cl2(g)]] ]$

Substituting the respective values,

$ΔrH°= [(1 mol ICl(g)/mol-rxn)(17.51 kJ/mol)-[(0.5 mol I2(s)/mol-rxn)(0 kJ/mol)+(0.5 mol Cl2(g)/mol-rxn)(0 kJ/mol)] ]= 17.51 kJ/mol-rxn$

$ΔrS°= ∑nS°(products)- ∑nS°(reactants)= [(1 mol ICl(g)/mol-rxn)S°[ICl(g)]-[(0.5 mol I2(s)/mol-rxn)S°[I2(s)]+(0.5 mol Cl2(g)/mol-rxn)S°[Cl2(g)]] ]$

Substituting the respective values,

$ΔrS°= [(1 mol ICl(g)/mol-rxn)(247.56 J/K×mol)-[(0.5 mol I2(s)/mol-rxn)(116.135 J/K×mol)+(0.5 mol Cl2(g)/mol-rxn)(223.08 J/K×mol)] ] = 77.95 J/K×mol-rxn$