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Chapter 18, Problem 76GQ
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### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

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### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# 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(ℓ) ⇄ CaSO3 · ½ H2O(s) + CO2(g)Another reaction gives hydrated calcium sulfate:CaCO3(s) + SO2(g) + ½ H2O(ℓ) + 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.(b) Calculate ΔrG° for the reactionCaSO3 · ½ H2O(s) + ½ O2(g) ⇄ CaSO4 · ½ H2O(s)Is this reaction product- or reactant-favored at equilibrium?

(a)

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.

Explanation

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

(b)

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.

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