Elements Of Physical Chemistry
Elements Of Physical Chemistry
7th Edition
ISBN: 9780198796701
Author: ATKINS, P. W. (peter William), De Paula, Julio
Publisher: Oxford University Press
Question
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Chapter 3, Problem 3.12P

(a)

Interpretation Introduction

Interpretation:

The change in standard reaction entropy of the reaction given has to be calculated.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given reaction: 2H2(g)+O2(g)2H2O(g)

Given: Cp,mlinear=72R;Cp,mnon-linear=4R

At temperature 283K, the reaction becomes 2H2(g)+O2(g)ΔS283o2H2O(g)

At temperature 273K, the reaction becomes 2H2(g)+O2(g)ΔS273o2H2O(g)

Entropy calculation for each reactant and products in both reactions are as follows,

ΔSH2=nCp,mlinearln273283=2(72R)ln273283ΔSO2=nCp,mlinearln273283=(72R)ln273283ΔSH2O=nCp,mnon-linearln273283=2(4R)ln273283

Change in entropy obtained as follows,

ΔS283o-ΔS273o=ΔSH2+ΔSO2+ΔSH2O(or)ΔrCpln283273=ΔSH2+ΔSO2+ΔSH2O

According to Kirchhoff’s law,

ΔrCp=ΔSH2+ΔSO2+ΔSH2O=2(4R)-2(72R)-(72R)=-2.5R=-20.785JK-1mol-1

Change in entropy becomes,

ΔS283o-ΔS273o=-20.785ln283273=-0.75JK-1mol-1

Therefore, the change in standard reaction entropy of the reaction obtained is -0.75JK-1mol-1

(b)

Interpretation Introduction

Interpretation:

The change in standard reaction entropy of the reaction given has to be calculated.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given reaction: CH4(g)+2O2(g)CO2(g)+2H2O(g)

Given: Cp,mlinear=72R;Cp,mnon-linear=4R

At temperature 283K, the reaction becomes CH4(g)+2O2(g)ΔS283oCO2(g)+2H2O(g)

At temperature 273K, the reaction becomes CH4(g)+2O2(g)ΔS273oCO2(g)+2H2O(g)

Entropy calculation for each reactant and products in both reactions are as follows,

ΔSCH4=nCp,mnon-linearln273283=1(4R)ln273283ΔSO2=nCp,mlinearln273283=2(72R)ln273283ΔSCO2=nCp,mlinearln273283=(72R)ln273283ΔSH2O=nCp,mnon-linearln273283=2(4R)ln273283

Change in entropy obtained as follows,

ΔS283o-ΔS273o=ΔSCH4+ΔSO2+ΔSCO2+ΔSH2O(or)ΔrCpln283273=ΔSCH4+ΔSO2+ΔSCO2+ΔSH2O

According to Kirchhoff’s law,

ΔrCp=ΔSCH4+ΔSO2+ΔSCO2+ΔSH2O=1(4R)-2(72R)-(72R)+2(4R)=(12R)-3(72R)=R[12-212]=1.5R=12.47JK-1mol-1

Change in entropy becomes,

ΔS283o-ΔS273o=12.47JK-1mol-1ln283273=0.4486 JK-1mol-1

Therefore, the change in standard reaction entropy of the reaction obtained is 0.4486 JK-1mol-1

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