CHEM:ATOM FOC 2E CL (TEXT)
CHEM:ATOM FOC 2E CL (TEXT)
2nd Edition
ISBN: 9780393284218
Author: Stacey Lowery Bretz, Natalie Foster, Thomas R. Gilbert, Rein V. Kirss
Publisher: WW Norton & Co
Question
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Chapter 12, Problem 12.39QA
Interpretation Introduction

To calculate:

The entropy change for the given reactions using the standard molar entropies.

Expert Solution & Answer
Check Mark

Answer to Problem 12.39QA

Solution:

a) N2g+O2g     2 NOgSrxno=24.9 J/K

b) 2 NOg+O2g    2 NO2g; Srxno=-146.4 J/K

c) NOg+12O2g    NO2gSrxno=- 73.2 J/K

d) 2 NO2g N2O4(g ; Srxno=-175.8 J/K

Explanation of Solution

1) Concept:

Srxno(Entropy change for the reaction) can be calculated from the difference between the standard molar entropies of nreactant moles of reactants and the standard molar entropies nproduct moles of products.

2) Formula:

Srxno= nproduct  Sproducto- nreactant  Sreactanto

3) Given:

Substance S0 [J/(mol.K)]
N2g 191.5
O2g 205.0
NOg 210.7
NO2g 240.0
N2O4(g) 304.2

4) Calculations:

a) N2g+O2g     2 NOg 

The values of standard molar entropies are,

So for N2g=191.5 J/mol.K

So for O2g=205.0 J/mol.K

So for NOg=210.7 J/mol.K

Therefore,

Srxno=(n NOg  S NOgo)-[nN2g  SN2go+(n O2g  S O2go)]

Srxno=2 mol×210.7 J/mol.K-1 mol×191.5 J/mol.K+1 mol×205.0 J/mol.K

Srxno=421.4 J/K  -396.5 J/K

Srxno=24.9 J/K

The value of Srxno is 24.9 J/K.

b) 2 NOg+O2g    2 NO2g 

The values of standard molar entropies are,

So for O2g=205.0 J/mol.K

So for NOg=210.7 J/mol.K

So for NO2g=240.0 J/mol.K

Srxno= nNO2g  SNO2go-n O2g  S O2go+n NOg  S NOgo

Srxno=2 mol×240.0 J/mol.K-2 mol×210.7 J/mol.K+1 mol×205.0 J/mol.K

Srxno=480.0 J/K -626.4 J/K

Srxno=-146.4 J/K

The value of Srxno is -146.4 J/K.

c)  NOg+12O2g    NO2g 

The values of standard molar entropies are,

So for O2g=205.0 J/mol.K

So for NOg=210.7 J/mol.K

So for NO2g=240.0 J/mol.K

Srxno= nNO2g  SNO2go-n O2g  S O2go+n NOg  S NOgo

Srxno=1 mol×240.0 J/mol.K-1/2 mol×205.0 J/mol.K+ 1 mol×210.7 J/mol.K

Srxno=240.0 J/K -313.2 J/K

Srxno=- 73.2 J/K

The value of Srxno is -73.2 J/K.

d) 2 NO2g N2O4(g) 

The values of standard molar entropies are,

So for NO2g=240.0 J/mol.K

So for N2O4g= 304.2 J/mol.K

Srxno= nN2O4(g)    SN2O4(g)  o-n NO2g  S NO2go

Srxno=1 mol×304.2 J/mol.K-2 mol×240.0 J/mol.K

Srxno=304.2 J/K -480.0 J/K

Srxno=-175.8 J/K

The value of Srxno is -175.8 J/K.

Conclusion:

Srxno(entropy change for the reaction) can be calculated from the difference between the standard molar entropies of nreactant moles of reactants and the standard molar entropies nproduct moles of products,

Srxno= nproduct  Sproducto- nreactant  Sreactanto

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Chapter 12 Solutions

CHEM:ATOM FOC 2E CL (TEXT)

Ch. 12 - Prob. 12.11QACh. 12 - Prob. 12.12QACh. 12 - Prob. 12.13QACh. 12 - Prob. 12.14QACh. 12 - Prob. 12.15QACh. 12 - Prob. 12.16QACh. 12 - Prob. 12.17QACh. 12 - Prob. 12.18QACh. 12 - Prob. 12.19QACh. 12 - Prob. 12.20QACh. 12 - Prob. 12.21QACh. 12 - Prob. 12.22QACh. 12 - Prob. 12.23QACh. 12 - Prob. 12.24QACh. 12 - Prob. 12.25QACh. 12 - Prob. 12.26QACh. 12 - Prob. 12.27QACh. 12 - Prob. 12.28QACh. 12 - Prob. 12.29QACh. 12 - Prob. 12.30QACh. 12 - Prob. 12.31QACh. 12 - Prob. 12.32QACh. 12 - Prob. 12.33QACh. 12 - Prob. 12.34QACh. 12 - Prob. 12.35QACh. 12 - Prob. 12.36QACh. 12 - Prob. 12.37QACh. 12 - Prob. 12.38QACh. 12 - Prob. 12.39QACh. 12 - Prob. 12.40QACh. 12 - Prob. 12.41QACh. 12 - Prob. 12.42QACh. 12 - Prob. 12.43QACh. 12 - Prob. 12.44QACh. 12 - Prob. 12.45QACh. 12 - Prob. 12.46QACh. 12 - Prob. 12.47QACh. 12 - Prob. 12.48QACh. 12 - Prob. 12.49QACh. 12 - Prob. 12.50QACh. 12 - Prob. 12.51QACh. 12 - Prob. 12.52QACh. 12 - Prob. 12.53QACh. 12 - Prob. 12.54QACh. 12 - Prob. 12.55QACh. 12 - Prob. 12.56QACh. 12 - Prob. 12.57QACh. 12 - Prob. 12.58QACh. 12 - Prob. 12.59QACh. 12 - Prob. 12.60QACh. 12 - Prob. 12.61QACh. 12 - Prob. 12.62QACh. 12 - Prob. 12.63QACh. 12 - Prob. 12.64QACh. 12 - Prob. 12.65QACh. 12 - Prob. 12.66QACh. 12 - Prob. 12.67QACh. 12 - Prob. 12.68QACh. 12 - Prob. 12.69QACh. 12 - Prob. 12.70QACh. 12 - Prob. 12.71QACh. 12 - Prob. 12.72QACh. 12 - Prob. 12.73QACh. 12 - Prob. 12.74QACh. 12 - Prob. 12.75QACh. 12 - Prob. 12.76QACh. 12 - Prob. 12.77QACh. 12 - Prob. 12.78QACh. 12 - Prob. 12.79QACh. 12 - Prob. 12.80QACh. 12 - Prob. 12.81QACh. 12 - Prob. 12.82QACh. 12 - Prob. 12.83QACh. 12 - Prob. 12.84QACh. 12 - Prob. 12.85QACh. 12 - Prob. 12.86QACh. 12 - Prob. 12.87QACh. 12 - Prob. 12.88QACh. 12 - Prob. 12.89QACh. 12 - Prob. 12.90QACh. 12 - Prob. 12.91QACh. 12 - Prob. 12.92QA
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