Chemistry: An Atoms First Approach
Chemistry: An Atoms First Approach
2nd Edition
ISBN: 9781305079243
Author: Steven S. Zumdahl, Susan A. Zumdahl
Publisher: Cengage Learning
Question
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Chapter 20, Problem 66E

(a)

Interpretation Introduction

Interpretation: The reaction between Fe2O3 and CO is given. The value of ΔH,ΔS and ΔG are to be calculated for the given options.

Concept introduction: The expression for ΔH is,

ΔH=npΔH(product)nfΔH(reactant)

The expression for ΔS is,

ΔS=npΔS(product)nfΔS(reactant)

The expression for standard Gibbs free energy, ΔG, is,

ΔG=ΔHTΔS

To determine: The value of ΔH and ΔS for the given reaction.

(a)

Expert Solution
Check Mark

Answer to Problem 66E

Solution

The value of ΔS for the given reaction is 38J/K_. The value of ΔH for the given reaction is 39kJ_.

Explanation of Solution

Explanation

The reaction that takes place is,

3Fe2O3(s)+CO(g)2Fe3O4(s)+CO2(g)

Refer to Appendix 4.

The value of ΔH(kJ/mol) for the given reactant and product is,

Molecules ΔH0(kJ/mol)
Fe3O4(s) 1117
CO2(g) 393.5
Fe2O3(s) 826
CO(g) 110.5

The formula of ΔH is,

ΔH=npΔH(product)nfΔH(reactant)

Where,

  • ΔH is the standard enthalpy of reaction.
  • np is the number of moles of each product.
  • nr is the number of moles each reactant.
  • ΔH(product) is the standard enthalpy of product at a pressure of 1atm.
  • ΔH(reactant) is the standard enthalpy of reactant at a pressure of 1atm.

Substitute all values from the table in the above equation.

ΔH=npΔH(product)nfΔH(reactant)=[2(1117)+2(393.5){3(826)+(110.5)}]kJ=39kJ_

The value of ΔS(J/Kmol) for the given reactant and product is,

Molecules ΔS0(kJ/mol)
Fe3O4(s) 146
CO2(g) 214
Fe2O3(s) 90
CO(g) 198

The formula of ΔS is,

ΔS=npΔS(product)nfΔS(reactant)

Where,

  • ΔS standard enthalpy of reaction.
  • np is number of moles of each product.
  • nr is number of moles each reactant.
  • ΔS(product) is standard enthalpy of product at a pressure of 1atm.
  • ΔS(reactant) is standard enthalpy of reactant at a pressure of 1atm.

Substitute all values from the table in the above equation.

ΔS=npΔS(product)nfΔS(reactant)=[2(146)+1(214){3(90)+(198)}]J/K=38J/K_

(b)

Interpretation Introduction

Interpretation: The reaction between Fe2O3 and CO is given. The value of ΔH,ΔS and ΔG are to be calculated for the given options.

Concept introduction: The expression for ΔH is,

ΔH=npΔH(product)nfΔH(reactant)

The expression for ΔS is,

ΔS=npΔS(product)nfΔS(reactant)

The expression for standard Gibbs free energy, ΔG, is,

ΔG=ΔHTΔS

To determine: The value of ΔG at 800°C for the given reaction.

(b)

Expert Solution
Check Mark

Answer to Problem 66E

Answer

The value of ΔG for the given reaction is 80kJ_.

Explanation of Solution

Explanation

Given

Temperature is 800°C.

The value of ΔH is 39kJ.

The value of ΔS is 38J/K.

The conversion of degree Celsius (°C) into Kelvin (K) is done as,

T(K)=T(°C)+273

Hence,

The conversion of 800°C into Kelvin is,

T(K)=T(°C)+273T(K)=(800+273)K=1073K

The conversion of joule (J) into kilo-joule (kJ) is done as,

1J=103kJ

Hence,

The conversion of 38J/K into kilo-joule is,

38J=(38×103)kJ=38×103kJ

Formula

The expression for standard Gibbs free energy, ΔG, is,

ΔG=ΔHTΔS

Where,

  • ΔH is the standard enthalpy of reaction.
  • ΔG is the free energy change.
  • T is the given temperature.
  • ΔS is the entropy of reaction.

Substitute the values of ΔH and ΔS in the above equation.

ΔG=ΔHTΔS=39kJ1073K(38×103kJ/K)=80kJ_

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Chapter 20 Solutions

Chemistry: An Atoms First Approach

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