Chemistry: An Atoms-Focused Approach
Chemistry: An Atoms-Focused Approach
14th Edition
ISBN: 9780393912340
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster
Publisher: W. W. Norton & Company
Question
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Chapter 12, Problem 12.72QA
Interpretation Introduction

To find:

The overall ΔG0rxn value of the iron reduction reaction at 14500C.

Expert Solution & Answer
Check Mark

Answer to Problem 12.72QA

Solution:

The overall Grxn0 value of the iron reduction reaction at 14500C  is -1690 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 the absolute temperature, and n the moles of each specie.

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 moles of reactants and the standard molar enthalpies of nproducts moles of products. Therefore, we can get the above formula.  Same is applied for 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.

Formation of carbon monoxide from coal as follow:

2C(s)+O2g2CO(g)(1)

3) Given information:

Fe2O3s+3CO(g)2Fe(s)+3CO2g  reaction equation of problem 12.752

T=1450 

Hf0(kJ/mol) Sf0(J/mol.K)
Fe2O3s -824.2 87.4
C(s) 0.0 5.7
O2g 0.0 205.0
Fe(s) 0.0 27.3
CO2g -393.5 213.8

4) Calculations:

Reduction of iron from coal is the overall reaction of the two reactions above.

2C(s)+O2g2CO(g)(1)

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

To get the same number of moles of  CO(g), we need to multiply reaction (1) by 3 and reaction (2) by 2.

6Cs+3O2g6COg3

2Fe2O3s+6CO(g)4Fe(s)+6CO2g (4)

Adding reactions (3) and (4), we are getting overall reaction:

2Fe2O3s+6Cs+3O2g4Fe(s)+6CO2g

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

T=1450 +273.15=1723.15 K

Calculation for Hrxn0:

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

Hrxn0=

4 mol×Hf,Fes0+6 mol×Hf,CO2g0-2 mol×Hf,Fe2O3s0+6 mol×Hf,Cs0+3 mol×Hf,O2g0

=4 mol×0.0kJmol+6 mol×-393.5kJmol-2 mol×-824.2kJmol+6 mol×0.0kJmol+3 mol×0.0kJmol

=[0-2361 kJ]-[(-1648-0)]

=(-2361 kJ)--1648 kJ

=-2361 +1648 kJ

Hrxn0=-712.6 kJ

Calculation for Srxn0:

Srxn0=

4 mol×Sf,Fes0+6 mol×Sf,CO2g0-2 mol×Sf,Fe2O3s0+6 mol×Sf,Cs0+3 mol×Sf,O2g0

=4 mol×27.3Jmol.K+6 mol×213.8Jmol.K-2 mol×87.4Jmol.K+6 mol×5.7kJmol+3 mol×205.0kJmol

=109.2JK+1282.8 JK-174.8JK+34.2 JK+615 JK

=1392JK-824JK

Srxn0=568 JK

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

Srxn0=568 JK×1 kJ1000 J=0.568kJK

Calculation for Grxn0:

Grxn0=Hrxn0-TSrxn0

Grxn0=-712.6 kJ-1723.15 K×0.568kJK

Grxn0=-712.6 kJ-978.7492 kJ

Grxn0=-1690 kJ

Conclusion:

Based on the values of Hf0 and Sf0, the Grxn0  for the reduction of iron at 1450  with coke is  -1690 kJ.

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Chapter 12 Solutions

Chemistry: An Atoms-Focused Approach

Ch. 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. 12.86QA
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