   # Copper(II) oxide, CuO, can be reduced to copper metal with hydrogen at higher temperatures. CuO(s) + H 2 (g) → Cu(s) + H 2 O(g) Is this reaction product- or reactant-favored at equilibrium at 298 K? ### 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 62IL
Textbook Problem
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## Copper(II) oxide, CuO, can be reduced to copper metal with hydrogen at higher temperatures.CuO(s) + H2(g) → Cu(s) + H2O(g)Is this reaction product- or reactant-favored at equilibrium at 298 K?

Interpretation Introduction

Interpretation:

It should be predicted that the given reduction of copper oxide is product-favored or reactant-favored at equilibrium at 298 K.

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)

### Explanation of Solution

The ΔrG° for the given reaction is calculated below.

Given: CuO(s) + H2(g)Cu(s)+H2O(g)

The Appendix L referred for the values of standard free energy values.

The ΔrG° for CuO(s) is 128.3 kJ/mol.

The ΔrG° for H2(g) is 0 kJ/mol.

The ΔrG° for Cu(s) is 0 kJ/mol.

The ΔrG° for H2O(g) is 228.59 kJ/mol.

ΔrG°fG°(products)-fG°(reactants)=[[(1 mol Cu(s)/mol-rxn)ΔfG°[Cu(s)]+(1 mol H2O(g)/mol-rxn

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