   # Some metal oxides can be decomposed to the metal and oxygen under reasonable conditions. Is the decomposition of silver(I) oxide product-favored at equilibrium at 25 °C? 2 Ag 2 O(s) → 4 Ag(s) + O 2 (g) If not, can it become so if the temperature is raised? At what temperature does the reaction become product-favored at equilibrium? ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 18, Problem 61IL
Textbook Problem
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## Some metal oxides can be decomposed to the metal and oxygen under reasonable conditions. Is the decomposition of silver(I) oxide product-favored at equilibrium at 25 °C?2 Ag2O(s) → 4 Ag(s) + O2(g)If not, can it become so if the temperature is raised? At what temperature does the reaction become product-favored at equilibrium?

Interpretation Introduction

Interpretation:

It should be identified that the given reaction is product favoured at equilibrium at 25oC. If it does not then raising temperature helps in bringing reaction towards product favoured should be identified.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔrGo. It can be calculated in a similar manner as entropy and enthalpy. The expression for the free energy change is:

ΔrG°fG°(products)fG°(reactants)

It is related to entropy and entropy by the following expression,

ΔrGorHo-TΔrSo

Here, ΔrH° is the change in enthalpy and ΔrS° is the change in entropy.

### Explanation of Solution

The value of ΔrGo for the decompositon of silver oxide to silver and oxygen is calculated

below.

Given:

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

The given reaction is,

2Ag2O(s)4Ag(s)+O2(g)

The ΔrGo for Ag2O(s) is 11.32 kJ/mol.

The ΔrGo for O2(g) is 0 kJ/mol.

The ΔrGo for Ag(s) is 0 kJ/mol.

ΔrG°fG°(products)fG°(reactants)=[[(4 mol Ag(s)/mol-rxn)ΔfG°[Ag(s)]+(1 mol O2(g)/mol-rxn)ΔfG°[O2(g)]]-(2 mol Ag2O(s)/mol-rxn)ΔfG°[Ag2O(s)] ]

Substituting the respective values

ΔrG°=[[(4 mol Ag(s)/mol-rxn)(0 kJ/mol)+(1 mol O2(g)/mol-rxn)(0 kJ/mol)]-(2 mol Ag2O(s)/mol-rxn)(-11.32 kJ/mol) ]=+22.64 kJ/mol-rxn

The sign of free energy change is positive, thus the reaction is reactant-favored at equilibrium. Therefore, the reaction is not product-favoured at 25 °C.

The value of ΔrH° and ΔrS° is calculated to predict the temperature at which the reaction becomes product-favored.

2Ag2O(s)4Ag(s)+O2(g)ΔfH°(kJ/mol)-31.100So(J/K×mol)121.342.55205

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