   Chapter 6, Problem 116AE

Chapter
Section
Textbook Problem

# 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 )

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

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

### 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

## Additional Science Solutions

#### Find more solutions based on key concepts 