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 12.75QA
Interpretation Introduction

To find:

Grxn0 of given reaction at 1450 0C.

Expert Solution & Answer
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Answer to Problem 12.75QA

Solution:

Grxn0 of given reaction at 1450 0C is -51.5 kJ.

Explanation of Solution

1) Formula:

i. Grxn0=Hrxn0-TSrxn0

ii. Hrxn0=nproducts×Hf,products0-nreactants×Hf,reactants0

iii. Srxn0=nproducts×Sf,products0-nreactants×Sf,reactants0

Here Grxn0 is the standard Gibb’s free energy of reaction, Hrxn0 the standard enthalpy of the reaction, Srxn0 the standard entropy of the reaction, T is the absolute temperature, and n is the number of moles of each reactant or product in the balanced equation for the reaction.

2) Concept:

We can calculate the change in enthalpy that accompanies a chemical reaction under standard conditions, Hrxn0, from the difference between the standard molar enthalpies of n­­reactants, the moles of reactants and the standard molar enthalpies of nproducts, the moles of products. This is expressed mathematically as the formula shown above. Same applies to Srxn0.

The relation between Gibbs free energy (Grxn0), enthalpy (Hrxn0) and entropy (Srxn0) is given by the above formula. We can use this formula to calculate Grxn0 of reaction.

3) Given:

Fe2O3s+3CO(g)2Fe(s)+3CO2g

The reaction is carried out at a temperature T=14500C

Hf0(kJ/mol) Sf0(J/mol.K)
Fe2O3s -824.2 87.4
CO(g) -110.5 197.7
Fe(s) 0.0 27.3
CO2g -393.5 213.8

4) Calculations:

The unit of Sf0 is J/mol.K, therefore we need to convert T from 0C to K.

T(K)=14500C+273.15=1723.15K

Calculation for Hrxn0:

Hrxn0=nproducts×Hf,products0-nreactants×Hf,reactants0

Hrxn0=

2 mol×Hf,Fes0+3 mol×Hf,CO2g0-1 mol×Hf,Fe2O3s0+3 mol×Hf,COg0

=2 mol×0.0kJmol+3 mol×-393.5kJmol-1 mol×-824.2kJmol+3 mol×-110.5kJmol

=[0-1180.5 kJ]-[(-824.2-331.5)]

=(-1180.5 kJ)--1155.7 kJ

=-1180.5+1155.7 kJ

Hrxn0=-24.8 kJ

Calculation for Srxn0:

Srxn0=

2 mol×Sf,Fes0+3 mol×Sf,CO2g0-1 mol×Sf,Fe2O3s0+3 mol×Sf,COg0

=2 mol×27.3Jmol.K+3 mol×213.8Jmol.K-1 mol×87.4Jmol.K+3 mol×197.7Jmol.K

=54.6JK+641.4 JK-87.4JK+593.1 JK

=696JK-680.5JK

Srxn0=15.5 JK

All units of Hrxn0, Srxn0 and Grxn0, must be in same unit, so need to convert Srxn0 from JK to kJK.

Srxn0=15.5 JK×1 kJ1000 J=0.0155kJK

Calculation for Grxn0:

Grxn0=Hrxn0-TSrxn0

Grxn0=-24.8 kJ-1723.15 K×0.0155kJK

Grxn0=-24.8 kJ-26.708825 kJ

Grxn0=-51.5 kJ

After calculating values of Hrxn0 and Srxn0, we plugged these values in Grxn0 formula and calculated it.

Conclusion:

Based on the relation between G, H, and S, we find that Grxn0 for the given reaction at 14500C is-51.5 kJ.

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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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