Chemistry: Atoms First
Chemistry: Atoms First
3rd Edition
ISBN: 9781259923142
Author: Burdge
Publisher: MCG
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 25, Problem 25.63QP
Interpretation Introduction

Interpretation:

For the formation of NO2 , the values of ΔGo , Kp and Kc have to be calculated.

Expert Solution & Answer
Check Mark

Answer to Problem 25.63QP

  • Gibb’s free energy change for the given reaction

ΔGo=-198.3kJmol

  • For the given reaction the value of Kp

Kp=6×1034

  • For the given reaction the value of Kc

KC=6×1034

Explanation of Solution

To calculate: Gibb’s free energy change for the formation of NO2

Formation reaction of NO2

NO(g)+O3(g)NO2(g)+O2(g)

General equation for calculation of Gibb’s free energy change of the given reaction

ΔGo=ΔGfo(products)-ΔGfo(reactants)ΔGo-freeenergychangeofthereactionΔGfo-freeenergychangeofformation

Apply the formula for the given reaction

Substances in elemental form (here oxygen) have Gibb’s free energy of formation value is zero.

ΔGo=ΔGfo(NO2)+ΔGfo(O2)-[ΔGfo(NO)+ΔGfo(O3)]ΔGo=((1)(51.8)+(0)-[(1)(86.7)+(1)(163.4)])kJmol=-198.3kJmol

Gibb’s free energy change for the formation of NO2 is calculated as -198.3kJmol

Gibb’s free energy change for the formation of NO2 molecule is calculated by the application of general equation of free energy change.

To calculate: The Kp value for the given reaction

The relationship between ΔGo and Kp is given by

ΔGo=-RTlnKpR-gas constant; T - temperature; Kp - equilibrium constant at partial pressure

Modify the above equation

lnKp=-ΔGoRT

Substitute the values of ΔGo , T and R

lnKp=-198.3×103Jmol(8.314Jmol)(298K)Kp=6×1034

By the use of relationship between ΔGo and Kp .  The value of Kp is calculated as 6×1034

To calculate: The Kc value for the given reaction

The relationship between Kc and Kp is given by

Kp=Kc(0.08206T)ΔnΔn-changeinnumberofmoles

Here no change in number of moles, therefore, Δn=0

Kp=Kc=6×1034

By the use of relationship between Kc and Kp .  The value of Kc is calculated as 6×1034

Conclusion

For the formation of NO2 , the values of ΔGo , Kp and Kc were calculated.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
8. Draw all the resonance forms for each of the fling molecules or ions, and indicate the major contributor in each case, or if they are equivalent (45) (2) -PH2 سمة مد
A J то گای ه +0 Also calculate the amount of starting materials chlorobenzaldehyde and p-chloroacetophenone required to prepare 400 mg of the given chalcone product 1, 3-bis(4-chlorophenyl)prop-2-en-1-one molar mass ok 1,3-bis(4-Chlorophenyl) prop-2-en-1-one = 277.1591m01 number of moles= 0.400/277.15 = 0.00144 moles 2 x 0.00 144=0.00288 moves arams of acetophenone = 0.00144 X 120.16 = 0.1739 0.1739x2=0.3469 grams of benzaldehyde = 0.00144X106.12=0.1539 0.1539x2 = 0.3069 Starting materials: 0.3469 Ox acetophenone, 0.3069 of benzaldehyde 3
1. Answer the questions about the following reaction: (a) Draw in the arrows that can be used make this reaction occur and draw in the product of substitution in this reaction. Be sure to include any relevant stereochemistry in the product structure. + SK F Br + (b) In which solvent would this reaction proceed the fastest (Circle one) Methanol Acetone (c) Imagine that you are working for a chemical company and it was your job to perform a similar reaction to the one above, with the exception of the S atom in this reaction being replaced by an O atom. During the reaction, you observe the formation of three separate molecules instead of the single molecule obtained above. What is the likeliest other products that are formed? Draw them in the box provided.

Chapter 25 Solutions

Chemistry: Atoms First

Ch. 25 - Prob. 25.11QPCh. 25 - Prob. 25.12QPCh. 25 - Elements 17 and 20 form compounds with hydrogen....Ch. 25 - Prob. 25.14QPCh. 25 - Prob. 25.15QPCh. 25 - Prob. 25.16QPCh. 25 - Prob. 25.17QPCh. 25 - Prob. 25.18QPCh. 25 - Prob. 25.19QPCh. 25 - Prob. 25.20QPCh. 25 - Prob. 25.21QPCh. 25 - Prob. 25.22QPCh. 25 - Prob. 25.23QPCh. 25 - Prob. 25.24QPCh. 25 - Prob. 25.25QPCh. 25 - Prob. 25.26QPCh. 25 - Prob. 25.27QPCh. 25 - Prob. 25.28QPCh. 25 - Prob. 25.29QPCh. 25 - Prob. 25.30QPCh. 25 - Prob. 25.31QPCh. 25 - Prob. 25.32QPCh. 25 - Prob. 25.33QPCh. 25 - Prob. 25.34QPCh. 25 - Prob. 25.35QPCh. 25 - Prob. 25.36QPCh. 25 - Prob. 25.37QPCh. 25 - Prob. 25.38QPCh. 25 - Prob. 25.39QPCh. 25 - Prob. 25.40QPCh. 25 - Prob. 25.41QPCh. 25 - At 620 K, the vapor density of ammonium chloride...Ch. 25 - Prob. 25.43QPCh. 25 - Prob. 25.44QPCh. 25 - Prob. 25.45QPCh. 25 - Prob. 25.46QPCh. 25 - Prob. 25.47QPCh. 25 - Prob. 25.48QPCh. 25 - Prob. 25.49QPCh. 25 - Prob. 25.50QPCh. 25 - Prob. 25.51QPCh. 25 - Prob. 25.52QPCh. 25 - Prob. 25.53QPCh. 25 - Prob. 25.54QPCh. 25 - Prob. 25.55QPCh. 25 - Prob. 25.56QPCh. 25 - Prob. 25.57QPCh. 25 - Prob. 25.58QPCh. 25 - Prob. 25.59QPCh. 25 - Prob. 25.60QPCh. 25 - Prob. 25.61QPCh. 25 - Prob. 25.62QPCh. 25 - Prob. 25.63QPCh. 25 - Prob. 25.64QPCh. 25 - Prob. 25.65QPCh. 25 - Prob. 25.66QPCh. 25 - Prob. 25.67QPCh. 25 - Prob. 25.68QPCh. 25 - Prob. 25.69QPCh. 25 - Prob. 25.70QPCh. 25 - Prob. 25.71QPCh. 25 - Prob. 25.72QPCh. 25 - What are the oxidation numbers of O and F in HFO?Ch. 25 - Prob. 25.74QPCh. 25 - Prob. 25.75QPCh. 25 - Prob. 25.76QPCh. 25 - Prob. 25.77QPCh. 25 - Prob. 25.78QPCh. 25 - Prob. 25.79QPCh. 25 - Prob. 25.80QPCh. 25 - Prob. 25.81QPCh. 25 - Prob. 25.82QPCh. 25 - Prob. 25.83QPCh. 25 - Prob. 25.84QPCh. 25 - Iodine pentoxide (I2O5) is sometimes used to...Ch. 25 - Prob. 25.86QPCh. 25 - Prob. 25.87QPCh. 25 - Prob. 25.88QPCh. 25 - Prob. 25.89QPCh. 25 - Prob. 25.90QPCh. 25 - Prob. 25.91QPCh. 25 - Prob. 25.92QPCh. 25 - Prob. 25.93QPCh. 25 - Prob. 25.94QPCh. 25 - Prob. 25.95QPCh. 25 - Prob. 25.96QPCh. 25 - Prob. 25.97QPCh. 25 - Prob. 25.98QPCh. 25 - Prob. 25.99QPCh. 25 - Prob. 25.100QPCh. 25 - Prob. 25.101QPCh. 25 - Prob. 25.102QPCh. 25 - Prob. 25.103QPCh. 25 - Prob. 25.104QP
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:Cengage Learning
Text book image
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
Text book image
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781133949640
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
Text book image
Principles of Modern Chemistry
Chemistry
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:Cengage Learning
Text book image
Chemistry: Principles and Practice
Chemistry
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Cengage Learning
Text book image
General Chemistry - Standalone book (MindTap Cour...
Chemistry
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Cengage Learning
Electrolysis; Author: Tyler DeWitt;https://www.youtube.com/watch?v=dRtSjJCKkIo;License: Standard YouTube License, CC-BY