# Write an equation for the reaction of Fe 2 O 3 (s) and C(s) to give Fe(s) and CO(g) (one of the reactions occurring in a blast furnace). How does Δ r G ° vary with temperature? Is there a temperature at which the reaction is product-favored at equilibrium?

### Chemistry & Chemical Reactivity

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

Chapter
Section

### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 18, Problem 82SCQ
Textbook Problem
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## Write an equation for the reaction of Fe2O3(s) and C(s) to give Fe(s) and CO(g) (one of the reactions occurring in a blast furnace). How does ΔrG° vary with temperature? Is there a temperature at which the reaction is product-favored at equilibrium?

Interpretation Introduction

Interpretation:

The reaction between Fe2O3(s) and C(s) should be written. The response of ΔrG° with respect to temperature and the temperature at which reaction will be product favoured at equilibrium should be identified.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔGo. It is related to entropy and entropy by the following expression,

ΔGo= ΔHo-TΔSo

The sign of ΔrGo should be negative for a product-favored reaction. Thus, spontaneous reactions are referred to those that have negative free energy formation.

### Explanation of Solution

The temperature at which reaction is product favored at equilibrium is calculated below.

Given: The reaction of Fe2O3(s) and C(s) to give Fe(s) and CO(g)

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

Fe2O3(s)+3C(s)2Fe(s)+3CO(g)ΔfH°(kJ/mol)-825.500-110.525So(J/K×mol)87.405.627.78197.674

The ΔrH° is calculated as,

ΔrH°=fH°(products)fH°(reactants)=[[(2 mol Fe(s)/mol-rxn)ΔfH°[Fe(s)]+(3 mol CO(g)/mol-rxn)ΔfH°[CO(g)]]-[(1 mol Fe2O3(s)/mol-rxn)ΔfH°[Fe2O3(s)]+(3 mol C(s)/mol-rxn)ΔfH°[C(s)]] ]

Substituting the values,

ΔrH°[[(2 mol Fe(s)/mol-rxn)(0 kJ/mol)+(3 mol CO(g)/mol-rxn)(-110.525 kJ/mol)]-[(1 mol Fe2O3(s)/mol-rxn)(-825.5 kJ/mol)+(3 mol C(s)/mol-rxn)(0 kJ/mol)] ] =  493

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