   Chapter 17, Problem 81AE

Chapter
Section
Textbook Problem

# Using Appendix 4 and the following data, determine S° for Fe(CO)5(g). Fe ( s ) + 5 CO ( g )   → Fe ( CO ) 5 ( g )           Δ S ∘ = ? Fe ( CO ) 5 ( l )   → Fe ( CO ) 5 ( g )           Δ S ∘ = 107   J/K Fe ( s ) + 5 CO ( g )   → Fe ( CO ) 5 ( l )           Δ S ∘ = − 677   J/K

Interpretation Introduction

Interpretation: The value of ΔS° is given for the formation reactions of Fe(CO)5(g) and Fe(CO)5(l) . The value of S° for Fe(CO)5(g) by using Appendix 4 , is to be calculated.

Concept introduction: The expression for standard entropy change, ΔS° , is,

ΔS°=npΔS°(product)nfΔS°(reactant)

Explanation

Explanation

The reaction for which ΔS° is to be calculated is,

Fe(s)+5CO(g)Fe(CO)5(g)

For the reaction,

Fe(CO)5(l)Fe(CO)5(g) (1)

The value of ΔS° is 107J/K .

For the reaction,

Fe(s)+5CO(g)Fe(CO)5(l) (2)

The value of ΔS° is 677J/K .

Add equation (1) and (2) and cancel the like terms on both sides to get the required reaction.

Fe(s)+5CO(g)Fe(CO)5(g)

The resultant value of ΔS° for this reaction is calculated by the formula,

ΔS°=(107677)J/K=570J/K_

The value of ΔS° for the reaction Fe(s)+5CO(g)Fe(CO)5(g) is 570J/K .

Refer Appendix 4

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