   Chapter 18, Problem 80IL

Chapter
Section
Textbook Problem

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

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

Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started 