CHEMISTRY >CUSTOM<
CHEMISTRY >CUSTOM<
14th Edition
ISBN: 9781259137815
Author: Julia Burdge
Publisher: McGraw-Hill Education
bartleby

Videos

Question
Book Icon
Chapter 22, Problem 76AP
Interpretation Introduction

Interpretation:

The standard Gibbs free energy and the equilibrium constant for the given reaction are to be calculated with the formation constant for the reaction.

Concept introduction:

The change in free energy is called Gibb’s free energy and is denoted as ΔGo.

If ΔGo is negative, it indicates that the electrical work has been done by the system on the surroundings and if it is positive then it means the electrical work has been done on the system by the surroundings.

The value of K1 for the reverse of reaction is given as:

K1=1K, K is the formation constant for the reaction.

The expression of equilibrium constant (Kc) is given as:

Kc=K1× K2

The relation between free energy change and standard free energy change is as:

ΔG=ΔGo+RTlnQ

Here, ΔG is free energy change, ΔGo is standard free energy change, Q is the reaction quotient, R is the gas constant, and T is the temperature.

At equilibrium the above equation is reduced to the expression:  ΔGo=TlnK 

Expert Solution & Answer
Check Mark

Answer to Problem 76AP

Solution: Kc = 6.7×1013 and ΔG°= 7.89×104J/mol  

Explanation of Solution

The overall reaction is obtained by adding the two half reactions as follows:    anode(oxidation):    Ag(NH3)2+(aq)  Ag+(aq) + 2NH3(aq)                    cathode(oxidation):Ag+(aq) +2CN  Ag(CN)2(aq)]                                             _____________________________________________                    overall: Ag(NH3)2+(aq) +2CN(aq) Ag(CN)2 + 2NH3(aq) 

The value of K1 for the reverse of reaction is given as follows:

K1=11.5×107=6.7×108

The expression of equilibrium constant (Kc) is given as follows:

Kc=K1× K2

Substitute 6.7×108 for K1 and 1.0×1021 for K2 in the above equation as follows:

Kc=(6.7 × 108)(1.0 × 1021) = 6.7×1013

The change in free energy or Gibbs free energy is calculated as follows:

ΔGo= RT ln Kc

Now, substitute  6.7×1013 for Kc, 8.314J/Kmol for R, and 298 K for T in the above equation as follows:

ΔGo=(8.314 J/mol. K)(298 K)ln(6.7 × 1013) = 7.89×104J/mol

Conclusion

The standard Gibbs free energy and equilibrium constant for the given reaction are 7.89×104J/mol and  6.7×1013.

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
You are trying to decide if there is a single reagent you can add that will make the following synthesis possible without any other major side products: xi 1. ☑ 2. H₂O хе i Draw the missing reagent X you think will make this synthesis work in the drawing area below. If there is no reagent that will make your desired product in good yield or without complications, just check the box under the drawing area and leave it blank. Click and drag to start drawing a structure. There is no reagent that will make this synthesis work without complications. : ☐ S ☐
Predict the major products of this organic reaction: H OH 1. LiAlH4 2. H₂O ? Note: be sure you use dash and wedge bonds when necessary, for example to distinguish between major products with different stereochemistry. Click and drag to start drawing a structure. G C टे
For each reaction below, decide if the first stable organic product that forms in solution will create a new C-C bond, and check the appropriate box. Next, for each reaction to which you answered "Yes" to in the table, draw this product in the drawing area below. Note for advanced students: for this problem, don't worry if you think this product will continue to react under the current conditions - just focus on the first stable product you expect to form in solution. NH2 CI MgCl ? Will the first product that forms in this reaction create a new CC bond? Yes No MgBr ? Will the first product that forms in this reaction create a new CC bond? Yes No G टे

Chapter 22 Solutions

CHEMISTRY >CUSTOM<

Ch. 22.3 - Prob. 1PPACh. 22.3 - Prob. 1PPBCh. 22.3 - Prob. 1PPCCh. 22.3 - Prob. 1CPCh. 22.3 - Prob. 2CPCh. 22.4 - Prob. 1PPACh. 22.4 - Practice Problem BUILD Transition metal ions can...Ch. 22.4 - Prob. 1PPCCh. 22 - Prob. 1QPCh. 22 - Prob. 2QPCh. 22 - 22.3 Explain why atomic radii decrease very...Ch. 22 - Prob. 4QPCh. 22 - Write the electron configurations of the following...Ch. 22 - Prob. 6QPCh. 22 - Prob. 7QPCh. 22 - Prob. 8QPCh. 22 - Describe the interaction between a donor atom and...Ch. 22 - Prob. 10QPCh. 22 - Complete the following statements for the complex...Ch. 22 - Prob. 12QPCh. 22 - Prob. 13QPCh. 22 - Prob. 14QPCh. 22 - Prob. 15QPCh. 22 - Prob. 16QPCh. 22 - Prob. 17QPCh. 22 - Prob. 18QPCh. 22 - Prob. 19QPCh. 22 - Prob. 20QPCh. 22 - Prob. 21QPCh. 22 - The complex ion [Ni ( CN ) 2 Br 2 ] 2- has a...Ch. 22 - Prob. 23QPCh. 22 - Prob. 24QPCh. 22 - Prob. 25QPCh. 22 - Prob. 26QPCh. 22 - Prob. 27QPCh. 22 - Prob. 28QPCh. 22 - Prob. 29QPCh. 22 - Prob. 30QPCh. 22 - Prob. 31QPCh. 22 - Prob. 32QPCh. 22 - Prob. 33QPCh. 22 - Prob. 34QPCh. 22 - Prob. 35QPCh. 22 - Prob. 36QPCh. 22 - Prob. 37QPCh. 22 - Prob. 38QPCh. 22 - Prob. 39QPCh. 22 - Prob. 40QPCh. 22 - Prob. 41QPCh. 22 - A concentrated aqueous copper(II) chloride...Ch. 22 - Prob. 43QPCh. 22 - Prob. 44QPCh. 22 - Prob. 45APCh. 22 - Prob. 46APCh. 22 - Prob. 47APCh. 22 - Prob. 48APCh. 22 - Prob. 49APCh. 22 - Draw qualitative diagrams for the crystal held...Ch. 22 - Prob. 51APCh. 22 - Prob. 52APCh. 22 - Prob. 53APCh. 22 - Prob. 54APCh. 22 - Prob. 55APCh. 22 - Prob. 56APCh. 22 - Prob. 57APCh. 22 - Prob. 58APCh. 22 - 22.59 The -porphyrin complex is more stable than...Ch. 22 - Prob. 60APCh. 22 - Prob. 61APCh. 22 - Prob. 62APCh. 22 - Prob. 63APCh. 22 - Prob. 64APCh. 22 - Prob. 65APCh. 22 - Prob. 66APCh. 22 - Prob. 67APCh. 22 - Prob. 68APCh. 22 - Locate the transition metal atoms and ions in the...Ch. 22 - Prob. 70APCh. 22 - Prob. 71APCh. 22 - Copper is known to exist in the +3 oxidation...Ch. 22 - 22.73 Chemical analysis shows that hemoglobin...Ch. 22 - Prob. 74APCh. 22 - Prob. 75APCh. 22 - Prob. 76APCh. 22 - 22.77 Commercial silver-plating operations...Ch. 22 - Prob. 78APCh. 22 - 22.79 (a) The free Cu(I) ion is unstable in...Ch. 22 - Prob. 1SEPPCh. 22 - Prob. 2SEPPCh. 22 - Prob. 3SEPPCh. 22 - Prob. 4SEPP
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
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
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: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
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
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:Cengage Learning
Lanthanoids and its Position in Periodic Table - D and F Block Elements - Chemistry Class 12; Author: Ekeeda;https://www.youtube.com/watch?v=ZM04kRxm6tY;License: Standard YouTube License, CC-BY