Chemistry: An Atoms-Focused Approach (Second Edition)
Chemistry: An Atoms-Focused Approach (Second Edition)
2nd Edition
ISBN: 9780393614053
Author: Thomas R. Gilbert, Rein V. Kirss, Stacey Lowery Bretz, Natalie Foster
Publisher: W. W. Norton & Company
Question
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Chapter 12, Problem 12.68QA
Interpretation Introduction

To find:

a. Go for the given reaction at 298 K.

b. If the reaction is spontaneous at 298 K

c. G for the given reaction at 1500 K

d. If the reaction is spontaneous at 1500 K

e. If the reverse reaction is spontaneous at the temperature of the kiln (i.e. 1500 K)

Expert Solution & Answer
Check Mark

Answer to Problem 12.68QA

Solution:

a. At 298 K, the value of Go= -131.1 kJ/mol.

b. Yes, the reaction is spontaneous at 298 K.

c. At 1500 K, the value of G=60 kJ/mol.

d. No, the reaction is not spontaneous at 1500 K.

e. Yes, the reverse reaction is spontaneous at the temperature of the kiln (i.e. 1500 K).

Explanation of Solution

1) Concept:

 We have been asked to calculate the Go  at 298 K by using appendix 4. We can calculate Go from the Hrxno  and So values. We can determine the spontaneity of the reaction from the sign of the Go values.

2) Given information:

Substance CaOs CO2g CaCO3s
HfokJ/mol -634.9 -393.5 -1206.9
SoJ/mol.K 38.1 213.8 92.9

3) Formulae:

i. Hrxno= nproduct  Hf,  producto- nreactant  Hf,reactanto

ii. Srxno= nproduct  Sproducto- nreactant  Sreactanto

iii. Grxno=Hrxno-TSrxno

4) Calculations:

We will calculate the Go at 298 K from given data.

a. Go for the given reaction at 298 K:

CaOs+CO2(g)CaCO3(s)

Hrxno(Enthalpy change for the reaction) can be calculated from the difference between the standard molar enthalpies of nreactant moles of reactants and the standard molar enthalpies of nproduct moles of products,

Therefore,

Hrxno= nproduct  Hf,producto- nreactant  Hf,reactanto

Hrxno= nCaCO3s H CaCO3so-n CaOs  H CaOso+n CO2g  H CO2go

Hrxno=1 mol ×-1206.9 kJ/mol-1 mol ×-634.9 kJ/mol+1 mol ×-393.5 kJ/ mol

Hrxno=-1206.9 kJ+1028.4 kJ

Hrxno=-178.5 kJ

Change in entropy of reaction i.e. Srxno:

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 of nproduct moles of products,

Therefore,

Srxno= nproduct  Sproducto- nreactant  Sreactanto

Srxno= nCaCO3s S CaCO3so-n CaOs  S CaOso+n CO2g  S CO2go

Srxno=1 mol ×92.9J/mol.K-1 mol ×38.1 J/mol.K+1 mol ×213.8 J/mol.K

Srxno=92.9 J/K-251.9J/K

Srxno=-159.0 J/K

The Gibb’s free energy change for reaction, Grxno at 298 K:

Hrxno=-178.5 kJ

Srxno=-159.0 J/K= -0.1590 kJ/K

T = 298 K

Therefore, Grxno=Hrxno-TSrxno

Grxno=(-178.5 kJ- 298 K×-0.1590 kJ/K

Grxno=(-178.5 kJ+47.382 kJ)

Grxno=-131.1 kJ

b. Grxno=-131.1 kJ at 298K, as the value of Grxno   is negative. Thus, the reaction is spontaneous.

c. G for the given reaction at 1500 K:

Grxn=Hrxn-TS rxn

Grxn=(-178.5 kJ- 1500 K×-0.1590 kJ/K

Grxn=(-178.5 kJ+238.5 kJ)

Grxn=60.0 kJ

d. Grxn=60.0 kJ at 1500 K, as the value of Grxn is positive. Thus, the reaction is nonspontaneous.

e. Grxn for reverse reaction at the temperature of kiln (i.e. 1500 K):

For the given reaction,

CaOs+CO2(g)CaCO3(s); Grxn=60.0 kJ

Therefore, for reverse reaction,

CaCO3sCaOs+CO2g; Grxn=-60.0 kJ

Since the value of Grxn is negative for reverse reaction, the reaction is spontaneous at 1500 K.

Conclusion:

We can calculate the values of Hrxn0 and Srxn0 using the standard values from the appendix and  Grxno can be calculate using this equation,  Grxn=Hrxn-TS rxn. With that, we can conclude the following:

a. At 298 K, the value of Go= -131.1 kJ/mol.

b. Yes, the reaction is spontaneous at 298 K.

c. At 1500 K, the value of G=60 kJ/mol.

d. No, the reaction is not spontaneous at 1500 K.

e. Yes, the reverse reaction is spontaneous at the temperature of the kiln (i.e. 1500 K).

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Chapter 12 Solutions

Chemistry: An Atoms-Focused Approach (Second Edition)

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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