![Chemistry for Engineering Students](https://www.bartleby.com/isbn_cover_images/9781285199023/9781285199023_largeCoverImage.gif)
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 |
![Check Mark](/static/check-mark.png)
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
![Check Mark](/static/check-mark.png)
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
![Check Mark](/static/check-mark.png)
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
Want to see more full solutions like this?
Chapter 10 Solutions
Chemistry for Engineering Students
- Nonearrow_forward3. Consider the compounds below and determine if they are aromatic, antiaromatic, or non-aromatic. In case of aromatic or anti-aromatic, please indicate number of I electrons in the respective systems. (Hint: 1. Not all lone pair electrons were explicitly drawn and you should be able to tell that the bonding electrons and lone pair electrons should reside in which hybridized atomic orbital 2. You should consider ring strain- flexibility and steric repulsion that facilitates adoption of aromaticity or avoidance of anti- aromaticity) H H N N: NH2 N Aromaticity (Circle) Aromatic Aromatic Aromatic Aromatic Aromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic nonaromatic nonaromatic nonaromatic nonaromatic nonaromatic aromatic TT electrons Me H Me Aromaticity (Circle) Aromatic Aromatic Aromatic Aromatic Aromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic nonaromatic nonaromatic nonaromatic nonaromatic nonaromatic aromatic πT electrons H HH…arrow_forwardA chemistry graduate student is studying the rate of this reaction: 2 HI (g) →H2(g) +12(g) She fills a reaction vessel with HI and measures its concentration as the reaction proceeds: time (minutes) [IH] 0 0.800M 1.0 0.301 M 2.0 0.185 M 3.0 0.134M 4.0 0.105 M Use this data to answer the following questions. Write the rate law for this reaction. rate = 0 Calculate the value of the rate constant k. k = Round your answer to 2 significant digits. Also be sure your answer has the correct unit symbol.arrow_forward
- 1. For the four structures provided, Please answer the following questions in the table below. a. Please draw π molecular orbital diagram (use the polygon-and-circle method if appropriate) and fill electrons in each molecular orbital b. Please indicate the number of π electrons c. Please indicate if each molecule provided is anti-aromatic, aromatic, or non- aromatic TT MO diagram Number of π e- Aromaticity Evaluation (X choose one) Non-aromatic Aromatic Anti-aromatic || ||| + IVarrow_forward1.3 grams of pottasium iodide is placed in 100 mL of o.11 mol/L lead nitrate solution. At room temperature, lead iodide has a Ksp of 4.4x10^-9. How many moles of precipitate will form?arrow_forwardQ3: Circle the molecules that are optically active: ДДДДarrow_forward
- General Chemistry - Standalone book (MindTap Cour...ChemistryISBN:9781305580343Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; DarrellPublisher:Cengage LearningChemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage LearningChemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage Learning
- Chemistry for Engineering StudentsChemistryISBN:9781337398909Author:Lawrence S. Brown, Tom HolmePublisher:Cengage LearningChemistry for Today: General, Organic, and Bioche...ChemistryISBN:9781305960060Author:Spencer L. Seager, Michael R. Slabaugh, Maren S. HansenPublisher:Cengage Learning
![Text book image](https://www.bartleby.com/isbn_cover_images/9781305580343/9781305580343_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9780534420123/9780534420123_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781285199047/9781285199047_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781337398909/9781337398909_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781305960060/9781305960060_smallCoverImage.gif)