Chemistry for Engineering Students
Chemistry for Engineering Students
4th Edition
ISBN: 9781337398909
Author: Lawrence S. Brown, Tom Holme
Publisher: Cengage Learning
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Chapter 10, Problem 10.91PAE

The reaction shown below is involved in the refining of iron. (The table that follows provides all of the thermodynamic data you should need for this problem.)

   2Fe 2 O 3 ( s ) + 3C ( s , graphite ) 4Fe ( s ) + 3CO 2 ( g )

(a) Find Δ H ° for the reaction.

(b) Δ S ° for the reaction above is 557.98 J/K. Find S° for Fe2O3(s).

(c) Calculate Δ G ° for the reaction at the standard temperature of 298 K. (There are two ways that you could do this.)

(d) At what temperatures would this reaction be spontaneous?

    Compound    Δ H f ° (kJ mol-1)
    (kJ mol-1)
       Δ G f ° (J mol-1 K-1)
    Fe2O3(s) -824.2 ? -742.2
    C(s, graphite) 0 5.740 0
    Fe(s) 0 27.3 0
    CO2(g) -393.5 213.6 -394.4.83

Expert Solution
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Interpretation Introduction

Interpretation: ΔSo for the given reaction is 557.98 J/K. Find ΔSo for Fe2O3(s).

Concept Introduction: ΔSreaction= mSproducts nSreactants

where m and n stands for the moles of products and reactants respectively

Answer to Problem 10.91PAE

Solution: ΔSo for Fe2O3(s) = 87.4 JK-1

Explanation of Solution

2 Fe2O3(s) + 3C(s)4Fe(s) + 3CO2(g)

ΔSo for reaction =557.98 JK-1, ΔSo for C(s) = 5.740 JK-1, ΔSo for Fe(s) = 27.3 JK-1, ΔSo for CO2(g) = 213.6 JK-1

ΔSreaction= mSproducts nSreactants

ΔSreaction = ( 4 x ΔS for Fe( s )  + 3 x ΔS for CO 2 ( g ))

 – ( 2 x ΔS for Fe 2 O 3 ( s )  + 3 x ΔS for C( s ))

557 .98 JK -1( 4 x 27 .3 JK -1 +3 x 213 .6 JK -1 )

 – ( 2 x ΔS for Fe 2 O 3 ( s ) + 3 x 5 .740 JK -1 )

557 .98 JK -1 = 750 JK -1( 2 x ΔS for Fe 2 O 3 ( s ) + 17 .22 JK -1 )

2 x ΔS for Fe2O3(s)  = 750 JK -1-557 .98 JK -1-17 .22 JK -1 2 x ΔS for Fe2O3(s) = 174 .8 JK -1 ΔS for Fe2O3(s) = 87 .4JK -1

Expert Solution
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Interpretation Introduction

Interpretation: ΔGo for the given reaction at the standard temperature of 298K

Concept Introduction: ΔGfreaction= mGfproducts nGfreactants

where m and n stands for the moles of products and reactants respectively

Answer to Problem 10.91PAE

Solution: [G] for the given reaction is 301.2 kJ mol-1

Explanation of Solution

ΔGfo for Fe2O3 = -742.2 kJ mol-1, ΔGf° for C(s)= 0, ΔGf° for Fe(s) = 0, ΔGf° for CO2(g) = -394.4 kJ mol-1

ΔGfreaction= mGfproducts nGfreactants

ΔGreaction = ( 4 x ΔG for Fe( s )  + 3 x ΔG for CO 2 ( g ))

 – ( 2 x ΔG for Fe 2 O 3 ( s )  + 3 x ΔG for C( s )) ΔGreaction

( 4 x 0 kJ mol -1 +3 x 394.4  kJ mol -1 )

 – ( 2 x 742.2  kJ mol -1 + 3 x 0 kJ mol -1 ) =

1183.2 kJ mol -1 +1484.4 kJ mol -1 = 301 .2 kJ mol -1           

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: the temperature at which this reaction be spontaneous

Concept Introduction: For the reaction to be spontaneous ΔG must be negative. Using the Gibbs free energy equation, ΔG = ΔH − TΔS, we can find the temperature at which this reaction be spontaneous.

ΔG =ΔH  TΔS 0 =ΔH  TΔST = ΔH ΔS 

Answer to Problem 10.91PAE

Solution: The temperature at which this reaction will be spontaneous is 5353.5K

Explanation of Solution

T =ΔHΔS=467.9kJmol187.4JK1=467.9x1000Jmol187.4JK1=5353.5K

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Chapter 10 Solutions

Chemistry for Engineering Students

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