Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
10th Edition
ISBN: 9781337399074
Author: John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher: Cengage Learning
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Textbook Question
Chapter 18, Problem 76GQ

Wet limestone is used to scrub SO2 gas from the exhaust gases of power plants. One possible reaction gives hydrated calcium sulfite:

CaCO3(s) + SO2(g) + ½ H2O() ⇄ CaSO3 · ½ H2O(s) + CO2(g)

Another reaction gives hydrated calcium sulfate:

CaCO3(s) + SO2(g) + ½ H2O() + ½ O2(g) ⇄ CaSO4 · ½H2O(s) + CO2(g)

(a) Which reaction is more product-favored at equilibrium? Use the data in the table below and any other information needed in Appendix L to calculate ΔrG° for each reaction at 25 °C.

Chapter 18, Problem 76GQ, Wet limestone is used to scrub SO2 gas from the exhaust gases of power plants. One possible reaction

(b) Calculate ΔrG° for the reaction

CaSO3 · ½ H2O(s) + ½ O2(g) ⇄ CaSO4 · ½ H2O(s)

Is this reaction product- or reactant-favored at equilibrium?

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Among the given reactions the reactions that is more favoured at product side should be determined.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔrGo. 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 76GQ

The reaction which is more product-favored at equilibrium is reaction 2 because it has greater negative free energy change.

Explanation of Solution

The value of ΔGo for the given reaction is calculated below.

Given:

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

For reaction 1,

CaCO3(s)+SO2(g)+12H2O(l)    CaSO3×(12H2O(s))+CO2(g)ΔfH°(kJ/mol)-1207.6-296.84-285.83-1311.7-393.50So(J/K×mol)91.7248.2169.95121.3213.74

The standard enthalpy change is expressed as,

ΔrH°=fH°(products)fH°(reactants)=[[(1 mol CaSO3×12H2O(s)/mol-rxn)ΔfH°[CaSO3×12H2O(s)]+(1 mol CO2(g)/mol-rxn)ΔfH°[CO2(g)]]-[(1 mol CaCO3(s)/mol-rxn)ΔfH°[CaCO3(s)]+(12mol H2O(l)/mol-rxn)ΔfH°[H2O(l)]+(1 mol SO2(g)/mol-rxn)ΔfH°[SO2(g)]] ] 

Substituting the respective values

ΔrH°=[[(1 mol CaSO3×12H2O(s)/mol-rxn)(-1311.7 kJ/mol)+(1 mol CO2(g)/mol-rxn)(-393.50 kJ/mol)]-[(1 mol CaCO3(s)/mol-rxn)(-1207.6 kJ/mol)+(0.5 mol H2O(l)/mol-rxn)(-285.83 kJ/mol)+(1 mol SO2(g)/mol-rxn)(-296.84 kJ/mol)] ] =-57.845 kJ/mol-rxn 

Also,

ΔrS°nS°(products)-nS°(reactants)=[[(1 mol CaSO3×12H2O(s)/mol-rxn)S°[CaSO3×12H2O(s)]+(1 mol CO2(g)/mol-rxn)S°[CO2(g)]]-[(1 mol CaCO3(s)/mol-rxn)S°[CaCO3(s)]+(12mol H2O(l)/mol-rxn)S°[H2O(l)]+(1 mol SO2(g)/mol-rxn)S°[SO2(g)]] ]   

Substituting the respective values

ΔrS°=[[(1 mol CaSO3×12H2O(s)/mol-rxn)(121.3 J/K×mol)+(1 mol CO2(g)/mol-rxn)(213.74 J/K×mol)]-[(1 mol CaCO3(s)/mol-rxn)(91.7 J/K×mol)+(12mol H2O(l)/mol-rxn)(69.95 J/K×mol)+(1 mol SO2(g)/mol-rxn)(248.21 J/K×mol)] ]   =-39.84 J/K×mol-rxn 

Now, ΔGo= ΔHo- TΔSo

Substituting the value of ΔHo and ΔSo.

ΔGo=-57.845 kJ/mol-rxn-[(298K)(-39.84 J/K×mol- rxn)](1 kJ1000 J)=-45.97 kJ/mol-rxn

For reaction 2,

CaCO3(s)+SO2(g)+12H2O(l)+12O2(g)CaSO4×12H2O(s)+CO2(g)ΔfH°(kJ/mol)-1207.6-296.8-285.80-1574.6-393.5So(J/K×mol)91.7248.269.9205.0134.8213.7

The standard enthalpy change is expressed as,

ΔrH°=fH°(products)fH°(reactants)=[[(1 mol CaSO4×12H2O(s)/mol-rxn)ΔfH°[CaSO4×12H2O(s)]+(1 mol CO2(g)/mol-rxn)ΔfH°[CO2(g)]]-[(1 mol CaCO3(s)/mol-rxn)ΔfH°[CaCO3(s)]+(12mol H2O(l)/mol-rxn)ΔfH°[H2O(l)]+(1 mol SO2(g)/mol-rxn)ΔfH°[SO2(g)]+(0.5 mol O2(g)/mol-rxn)ΔfH°[O2(g)]] ] 

Substituting the respective values

ΔrH°=[[(1 mol CaSO3×12H2O(s)/mol-rxn)(-1574.6 kJ/mol)+(1 mol CO2(g)/mol-rxn)(-393.50 kJ/mol)]-[(1 mol CaCO3(s)/mol-rxn)(-1207.6 kJ/mol)+(0.5 mol H2O(l)/mol-rxn)(-285.83 kJ/mol)+(1 mol SO2(g)/mol-rxn)(-296.84 kJ/mol)+(0.5 mol O2(g)/mol-rxn)(0 kJ/mol)] ] =-320.75 kJ/mol-rxn 

Also,

ΔrS°nS°(products)-nS°(reactants)=[[(1 mol CaSO4×12H2O(s)/mol-rxn)S°[CaSO4×12H2O(s)]+(1 mol CO2(g)/mol-rxn)S°[CO2(g)]]-[(1 mol CaCO3(s)/mol-rxn)S°[CaCO3(s)]+(12mol H2O(l)/mol-rxn)S°[H2O(l)]+(1 mol SO2(g)/mol-rxn)S°[SO2(g)]+(0.5 mol O2(g)/mol-rxn)S°[O2(g)]] ]    

Substituting the respective values

ΔrS°=[[(1 mol CaSO3×12H2O(s)/mol-rxn)(134.8 J/K×mol)+(1 mol CO2(g)/mol-rxn)(213.7 J/K×mol)]-[(1 mol CaCO3(s)/mol-rxn)(91.7 J/K×mol)+(0.5 mol H2O(l)/mol-rxn)(69.9 J/K×mol)+(1 mol SO2(g)/mol-rxn)( 248.2 J/K×mol)+(0.5 mol O2(g)/mol-rxn)(205.08 J/K×mol)] ] =-219.61 kJ/mol-rxn 

Now, ΔGo= ΔHo- TΔSo

Substituting the value of ΔHo and ΔSo.

ΔGo= -320.75 kJ/mol-rxn-[(298K)(-219.61 J/K×mol- rxn)](1 kJ1000 J)= -255.30 kJ/mol-rxn

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The ΔrGo value for the given reaction should be calculated and identified that the reaction is product or reactant favoured.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔrGo. 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 76GQ

The ΔrG° for the given reaction is 236.41 kJ/mol-rxn.

Thus, the reaction is product-favored at equilibrium.

Explanation of Solution

The value of ΔGo is calculated below.

Given:

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

CaSO3×12H2O(s)+12O2(g)CaSO4×12H2O(s)ΔfH°(kJ/mol)-1311.70-1574.65So(J/K×mol)121.3205.07134.8

The standard enthalpy change is expressed as,

ΔrH°=fH°(products)fH°(reactants)=[(1 mol CaSO4×12H2O(s)/mol-rxn)ΔfH°[CaSO4×12H2O(s)]-[(1 mol CaSO3×12H2O(s)/mol-rxn)ΔfH°[CaSO3×12H2O(s)]+(0.5 mol O2(g)/mol-rxn)ΔfH°[O2(g)]] ] 

Substituting the respective values

ΔrH°=[(1 mol CaSO4×12H2O(s)/mol-rxn)(-1574.65 kJ/mol)-[(1 mol CaSO3×12H2O(s)/mol-rxn)(-1311.7 kJ/mol)+(0.5 mol O2(g)/mol-rxn)(0 kJ/mol)] ] =-262.95 kJ/mol-rxn

Also,

ΔrS°=nS°(products)nS°(reactants)=[(1 mol CaSO4×12H2O(s)/mol-rxn)S°[CaSO4×12H2O(s)]-[(1 mol CaSO3×12H2O(s)/mol-rxn)S°[CaSO3×12H2O(s)]+(0.5 mol O2(g)/mol-rxn)S°[O2(g)]] ] 

Substituting the values,

ΔrS°=[(1 mol CaSO4×12H2O(s)/mol-rxn)(134.8 J/K×mol)-[(1 mol CaSO3×12H2O(s)/mol-rxn)(121.3 J/K×mol)+(0.5 mol O2(g)/mol-rxn)(205.07 J/K×mol)] ] =-89.03 J/K×mol-rxn

Now, ΔGo= ΔHo- TΔSo

Substitute the value of ΔHo and ΔSo.

ΔGo= -262.95 kJ/mol-rxn-[(298K)(-89.03 J/K×mol- rxn)](1 kJ1000 J)= -236.41 kJ/mol-rxn

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Chapter 18 Solutions

Chemistry & Chemical Reactivity

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