The reaction shown below is involved in the refining of iron. (The table that follows provides all of the
(a) Find
(b)
(c) Calculate
(d) At what temperatures would this reaction be spontaneous?
Compound |
|
S° (kJ mol-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 |
Interpretation: ΔSo for the given reaction is 557.98 J/K. Find ΔSo for Fe2O3(s).
Concept Introduction:
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
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
Interpretation: ΔGo for the given reaction at the standard temperature of 298K
Concept Introduction:
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
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.
Answer to Problem 10.91PAE
Solution: The temperature at which this reaction will be spontaneous is 5353.5K
Explanation of Solution
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Chapter 10 Solutions
Chemistry for Engineering Students
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