Chemistry: The Science in Context (Fifth Edition)
Chemistry: The Science in Context (Fifth Edition)
5th Edition
ISBN: 9780393614046
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster, Stacey Lowery Bretz, Geoffrey Davies
Publisher: W. W. Norton & Company
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Chapter 17, Problem 17.47QP
Interpretation Introduction

Interpretation: The value of ΔGο for the reduction of iron ore for given thermodynamic values at 25°C is to be calculated.

Concept introduction: The free energy change (ΔGο) is calculated by the formula,

ΔGο=ΔHrxnοTΔSrxnο

To determine: The value of ΔGο for the reduction of iron ore for given thermodynamic values at 25°C .

Expert Solution & Answer
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Answer to Problem 17.47QP

Solution

The value of ΔGο for the reduction of iron ore is 56.54kJ_ .

Explanation of Solution

Explanation

Given

The enthalpy change for formation (ΔHfο) of Fe2O3(s) is 824.2kJ/mol .

The enthalpy change for formation (ΔHfο) of Fe(s) is 0kJ/mol .

The enthalpy change for formation (ΔHfο) of H2(g) is 0kJ/mol .

The enthalpy change for formation (ΔHfο) of H2O(g) is 241.8kJ/mol .

The standard molar entropy (Sο) of Fe2O3(s) is 87.4Jmol1K1 .

The standard molar entropy (Sο) of Fe(s) is 27.3Jmol1K1 .

The standard molar entropy (Sο) of H2(g) is 130.6Jmol1K1 .

The standard molar entropy (Sο) of H2O(g) is 188.8Jmol1K1 .

The reaction at 25°C temperature (T) is,

Fe2O3(s)+3H2(g)2Fe(s)+3H2O(g) (1)

The free energy change (ΔGο) is calculated by the formula,

ΔGο=ΔHrxnοTΔSrxnο (2)

The equation (1) is balanced reaction.

The enthalpy change for a reaction (ΔHrxnο) is calculated by the formula,

ΔHrxnο=n×ΔHfο(Products)m×ΔHfο(Reactants) (3)

Where,

  • ΔHfο(Products) is standard enthalpy of formation of the products.
  • ΔHfο(Reactants) is of standard enthalpy of formation of the reactants.
  • n is number of moles of products.
  • m is number of moles of reactants.

In the balanced chemical equation (1) the,

  • Number of moles of product Fe(s) is 2 .
  • Number of moles of product H2O(g) is 3 .
  • Number of moles of reactant Fe2O3(s) is 1 .
  • Number of moles of reactant H2(g) is 3 .

The n×ΔHfο(Products) for the chemical reaction (1) is expressed as,

n×ΔHfο(Products)=2mol×ΔHf,Fe(s)ο+3mol×ΔHf,H2O(g)ο (4)

Where,

  • ΔHf,Fe(s)ο is the enthalpy change for formation (ΔHfο) of Fe(s)
  • ΔHf,H2O(g)ο is the enthalpy change for formation (ΔHfο) of H2O(g) .

Substitute the values of ΔHf,Fe(s)ο and ΔHf,H2O(g)ο in equation (4).

n×ΔHfο(Products)=2mol×0kJ/mol+3mol×241.8kJ/mol=725.4kJ

The m×ΔHfο(Reactants) for the balanced chemical reaction (1) is expressed as,

m×ΔHfο(Reactants)=1mol×ΔHf,Fe2O3(s)ο+3mol×ΔHf,H2(g)ο (5)

Where,

  • ΔHf,Fe2O3(s)ο is the enthalpy change for formation (ΔHfο) of Fe2O3(s) .
  • ΔHf,H2(g)ο is the enthalpy change for formation (ΔHfο) of H2(g) .

Substitute the values of ΔHf,Fe2O3(s)ο and ΔHf,H2(g)ο in equation (5).

m×ΔHfο(Reactants)=1mol×824.2kJ/mol+3mol×0kJ/mol=824.2kJ

Substitute the values of n×ΔHfο(Products) and m×ΔHfο(Reactants) in equation (3).

ΔHrxnο=725.4kJ(824.2kJ)=725.4kJ+824.2kJ=98.8kJ

The standard entropy change (ΔSο) is calculated by the formula,

ΔSrxnο=nproducts×Sο(products)mreactants×Sο(reactants) (6)

Where,

  • nproducts is the number of moles of products.
  • mreactants is the number of moles of of reactants.
  • Sο(products) is the standard molar entropy of products.
  • Sο(reactants) is the standard molar entropy of reactants.

In the balanced chemical equation (1) the,

  • Number of moles of product Fe(s) is 2 .
  • Number of moles of product H2O(g) is 3 .
  • Number of moles of reactant Fe2O3(s) is 1 .
  • Number of moles of reactant H2(g) is 3 .

The nproducts×Sο(products) is expressed as,

nproducts×Sο(products)=2mol×SFe(s)ο+3mol×SH2O(g)ο (7)

Where,

  • SFe(s)ο is the standard molar entropy of Fe(s) .
  • SH2O(g)ο is the standard molar entropy of H2O(g) .

Substitute the values of SFe(s)ο and SH2O(g)ο in equation (7).

nproducts×Sο(products)=2mol×27.3Jmol1K1+3mol×188.8Jmol1K1=54.6JK1+566.4JK1=621JK1

The mreactants×Sο(reactants) is expressed as,

mreactants×Sο(reactants)=(1mol×SFe2O3(s)ο)+(3mol×SH2(g)ο) (8)

Where,

  • SFe2O3(s)ο is the standard molar entropy of Fe2O3(s) .
  • SH2(g)ο is the standard molar entropy of H2(g) .

Substitute the values of SFe2O3(s)ο and SH2(g)ο in equation (8).

mreactants×Sο(reactants)=(1mol×87.4Jmol1K1)+(3mol×130.6Jmol1K1)=87.4JK1+391.8JK1=479.2JK1

Substitute the values of mreactants×Sο(reactants) and nproducts×Sο(products) in equation (6).

ΔSrxnο=621JK1479.2JK1=141.8JK1

To change the value of ΔSrxnο in kJK1 , use conversion factor.

1J=103kJ141.8JK1=141.8×103kJK1

To change the value of temperature (T) in Kelvin (K) , use conversion factor.

0°C=(0+273)K25°C=(25+273)K=298K

Substitute the values of ΔSrxnο (kJK1) , ΔHrxnο and T (K) in equation (2).

ΔGο=ΔHrxnοTΔSrxnο=98.8kJ(298K×141.8×103kJK1)=98.8kJ42.25kJ=56.54kJ

Thus, the value of ΔGο for the reduction of iron ore is 56.54kJ_ .

Conclusion

The value of ΔGο for the reduction of iron ore is 56.54kJ_

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Chapter 17 Solutions

Chemistry: The Science in Context (Fifth Edition)

Ch. 17 - Prob. 17.3VPCh. 17 - Prob. 17.4VPCh. 17 - Prob. 17.5VPCh. 17 - Prob. 17.6VPCh. 17 - Prob. 17.7VPCh. 17 - Prob. 17.8VPCh. 17 - Prob. 17.9QPCh. 17 - Prob. 17.10QPCh. 17 - Prob. 17.11QPCh. 17 - Prob. 17.12QPCh. 17 - Prob. 17.13QPCh. 17 - Prob. 17.14QPCh. 17 - Prob. 17.15QPCh. 17 - Prob. 17.16QPCh. 17 - Prob. 17.17QPCh. 17 - Prob. 17.18QPCh. 17 - Prob. 17.19QPCh. 17 - Prob. 17.20QPCh. 17 - Prob. 17.21QPCh. 17 - Prob. 17.22QPCh. 17 - Prob. 17.23QPCh. 17 - Prob. 17.24QPCh. 17 - Prob. 17.25QPCh. 17 - Prob. 17.26QPCh. 17 - Prob. 17.27QPCh. 17 - Prob. 17.28QPCh. 17 - Prob. 17.29QPCh. 17 - Prob. 17.30QPCh. 17 - Prob. 17.31QPCh. 17 - Prob. 17.32QPCh. 17 - Prob. 17.33QPCh. 17 - Prob. 17.34QPCh. 17 - Prob. 17.35QPCh. 17 - Prob. 17.36QPCh. 17 - Prob. 17.37QPCh. 17 - Prob. 17.38QPCh. 17 - Prob. 17.39QPCh. 17 - Prob. 17.40QPCh. 17 - Prob. 17.41QPCh. 17 - Prob. 17.42QPCh. 17 - Prob. 17.43QPCh. 17 - Prob. 17.44QPCh. 17 - Prob. 17.45QPCh. 17 - Prob. 17.46QPCh. 17 - Prob. 17.47QPCh. 17 - Prob. 17.48QPCh. 17 - Prob. 17.49QPCh. 17 - Prob. 17.50QPCh. 17 - Prob. 17.51QPCh. 17 - Prob. 17.52QPCh. 17 - Prob. 17.53QPCh. 17 - Prob. 17.54QPCh. 17 - Prob. 17.55QPCh. 17 - Prob. 17.56QPCh. 17 - Prob. 17.57QPCh. 17 - Prob. 17.58QPCh. 17 - Prob. 17.59QPCh. 17 - Prob. 17.60QPCh. 17 - Prob. 17.61QPCh. 17 - Prob. 17.62QPCh. 17 - Prob. 17.63QPCh. 17 - Prob. 17.64QPCh. 17 - Prob. 17.65QPCh. 17 - Prob. 17.66QPCh. 17 - Prob. 17.67QPCh. 17 - Prob. 17.68QPCh. 17 - Prob. 17.69QPCh. 17 - Prob. 17.70QPCh. 17 - Prob. 17.71QPCh. 17 - Prob. 17.72QPCh. 17 - Prob. 17.73QPCh. 17 - Prob. 17.74QPCh. 17 - Prob. 17.75QPCh. 17 - Prob. 17.76QPCh. 17 - Prob. 17.77QPCh. 17 - Prob. 17.78QPCh. 17 - Prob. 17.79QPCh. 17 - Prob. 17.80QPCh. 17 - Prob. 17.81QPCh. 17 - Prob. 17.82QPCh. 17 - Prob. 17.83QPCh. 17 - Prob. 17.84QPCh. 17 - Prob. 17.85QPCh. 17 - Prob. 17.86QPCh. 17 - Prob. 17.87QPCh. 17 - Prob. 17.88APCh. 17 - Prob. 17.89APCh. 17 - Prob. 17.90APCh. 17 - Prob. 17.91APCh. 17 - Prob. 17.92APCh. 17 - Prob. 17.93APCh. 17 - Prob. 17.94APCh. 17 - Prob. 17.95APCh. 17 - Prob. 17.96APCh. 17 - Prob. 17.97APCh. 17 - Prob. 17.98APCh. 17 - Prob. 17.99APCh. 17 - Prob. 17.100APCh. 17 - Prob. 17.101APCh. 17 - Prob. 17.102APCh. 17 - Prob. 17.103APCh. 17 - Prob. 17.104APCh. 17 - Prob. 17.105AP
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