Chapter 18, Problem 42PS

### Chemistry & Chemical Reactivity

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

Chapter
Section

### Chemistry & Chemical Reactivity

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

# Calculate ΔrH° and ΔrS° for the reaction of tin(IV) oxide with carbon.SnO2(s) + C(s) → Sn(s) + CO2(g) (a) Is the reaction product-favored at equilibrium at 298 K? (b) Is the reaction predicted to be product-favored at equilibrium at higher temperatures?

(a)

Interpretation Introduction

Interpretation:

The values of ΔHo and ΔSo for the given reaction should be calculated.

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 ΔHo, ΔSo and ΔrGo is calculated below.

Given:

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

SnO2(s)+C(s)Sn(s)+CO2(g)ΔfH°(kJ/mol)577.6300393.509S( J/Kmol)49.0469.955.6213.74

ΔrH°=fH°(products)fH°(reactants)=[[(1 mol Sn(s)/mol-rxn)ΔfH°[Sn(s)]+(1 mol CO2(g)/mol-rxn)ΔfH°[CO2(g)] ][(1 mol SnO2(s)/mol-rxn)ΔfH°[SnO2(s)]+(1 mol C(s)/mol-rxn)ΔfH°[C(s)]] ]

Substituting the respective values

ΔrH°=[[(1 mol Sn(s)/mol-rxn)(0 kJ/mol)+(1 mol CO2(g)/mol-rxn)(-393.509 kJ/mol) ]-[(1 mol SnO2(s)/mol-rxn)(-577.63 kJ/mol)+(1 mol C(s)/mol-rxn)(0 kJ/mol)] ]=184

(b)

Interpretation Introduction

Interpretation:

It should be predicted that whether the given reaction is product-favoured at equilibrium at higher temperatures.

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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