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Given the following data Fe 2 O 3 ( s ) + 3 CO ( g ) → 2 Fe ( s ) + 3 CO 2 ( g ) Δ H ° = − 23 KJ 3 Fe 2 O 3 ( s ) + CO ( g ) → 2 Fe 3 O 4 ( s ) + CO 2 ( g ) Δ H ° = − 39 KJ Fe 3 O 4 ( s ) + CO ( g ) → 3 FeO ( s ) + CO 2 ( g ) Δ H ° = 18 KJ Calculate ∆ H° for this reaction FeO ( s ) + CO ( g ) → Fe ( s ) + CO 2 ( g )

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Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

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BuyFindarrow_forward

Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 7, Problem 106AE
Textbook Problem
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Given the following data

Fe 2 O 3 ( s ) + 3 CO ( g ) 2 Fe ( s ) + 3 CO 2 ( g ) Δ H ° = 23 KJ 3 Fe 2 O 3 ( s ) + CO ( g ) 2 Fe 3 O 4 ( s ) + CO 2 ( g ) Δ H ° = 39 KJ Fe 3 O 4 ( s ) + CO ( g ) 3 FeO ( s ) + CO 2 ( g ) Δ H ° = 18 KJ

Calculate ∆ for this reaction

FeO ( s ) + CO ( g ) Fe ( s ) + CO 2 ( g )

Interpretation Introduction

Interpretation:  Standard enthalpy change to be calculate by using given data.

Concept introduction

Standard Enthalpy change ( ΔH0 ): The heat change when molar quantities of reactants as specified by chemical equation to form a product at standard conditions.

Standard condition: 250C and 1 atmosphere  pressure.

Explanation of Solution

Explanation

The standard enthalpy changes for given reactions are

Fe2O3(s)+3CO(g)2Fe(s)+3CO2(g)  ΔH0= -23kJ

3Fe2O3(s)+CO(g)2Fe3O4(s)+CO2(g)         ΔH0= -39kJ

Fe3O4(s)+CO(g)3FeO(s)+CO2(g)  ΔH0= 18kJ

6FeO+2CO22FeO3+42CO                ΔH0= -2(18kJ)

2Fe3O4+CO22Fe2O3+CO  ΔH0= -(-39kJ)

3Fe2O3+9CO6Fe+9CO2(g)  ΔH0= 3(-23kJ)

6FeO(s)+6CO(g)6Fe(s)+6CO2(g)  ΔH0= -66kJ

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Chemistry: An Atoms First Approach
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