General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
bartleby

Videos

Question
Book Icon
Chapter 18, Problem 18.74QP

(a)

Interpretation Introduction

Interpretation:

The equilibrium constant Kp value should be derived given the spontaneous reaction at 25°C.

Concept Introduction:

Spontaneous reaction: This reaction is said to be spontaneous it occurs without being drive by some of outside force. The two driving forces for all chemical reactions, first one is enthalpy and second one is entropy, spontaneous reactions occurs without outside intervention.

Equilibrium constant: The concentration at equilibrium always combine in the manner below where the products are in the numerator and the reactant are in the denominator to produce the (K) value regardless the initial concentration of species.

Keq=[C]c[D]d[A]a[B]b

Free energy: is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter G.  All spontaneous process is associated with the decrease of free energy in the system.  The free energy change (ΔGrxn) is the difference in free energy of the reactants and products in their standard state.

ΔG0=-RTlnKΔG=Free energyΔG0=Standard-state free energyR=GasConstant(8.314J/K×mol/K.mol)T=Temprature273KK=EqulibriumConstant(KPandKC)ln=(ve(log)State Function)

(a)

Expert Solution
Check Mark

Explanation of Solution

First calculate the free energy (ΔG°) value for given reaction

Fe(s)+2H+(aq)Fe2+(aq)+H2(g)ΔGrxn°=nΔGf°(Products)-mΔGf°(Reactants)ΔG°=ΔGfo(H2)+ΔGfo(Fe2+)-[ΔGfo(Fe)-2ΔGfo(H+)](ΔGfoValues referredfromAppendix-2)ΔG°=(1)(0)+(1)(-84.9kJ/mol)-[(1)(0)-(2)(0)]ΔG0=-84.9kJ/mol

Next calculate the equilibrium constant (Kp) value of given reaction (a),

Thefreeenergyequationis,ΔG0=-RTlnK-84.9×103J/mol=-(8.314J/molK)(298K)InKRewrite the above freeenergy ΔG°equation InKp=ΔG0-RT=-84.9×103J/mol-(8.314J/mol)(298K)InKp=34.152Kp=e34.152Kp=6.79×1014

(b)

Interpretation Introduction

Interpretation:

The equilibrium constant Kp value should be derived given the non-spontaneous reaction at 25°C.

Concept Introduction:

Equilibrium constant: The concentration at equilibrium always combine in the manner below where the products are in the numerator and the reactant are in the denominator to produce the (K) value regardless the initial concentration of species.

Keq=[C]c[D]d[A]a[B]b

Free energy: is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter G.  All spontaneous process is associated with the decrease of free energy in the system.  The free energy change (ΔGrxn) is the difference in free energy of the reactants and products in their standard state.

ΔG0=-RTlnKΔG=Free energyΔG0=Standard-state free energyR=GasConstant(8.314J/K×mol/K.mol)T=Temprature273KK=EqulibriumConstant(KPandKC)ln=(ve(log)State Function)

(b)

Expert Solution
Check Mark

Explanation of Solution

First calculate the free energy (ΔG°) value for given reaction

Cu(s)+2H+(aq)Cu2+(aq)+H2(g)ΔGrxn°=nΔGf°(Products)-mΔGf°(Reactants)ΔG°=ΔGfo(H2)+ΔGfo(Cu2+)-[ΔGfo(Cu)-2ΔGfo(H+)]ΔG°=(1)(0)+(1)(64.98kJ/mol)(1)(0)(2)(0)ΔG°=64.98kJ/mol

Next we calculate the equilibrium constant (Kp) value of given reaction (b),

ΔG°=-RTlnKThe respactive values are substituted for above equlibrium reaction (2)64.98×103J/mol=-(8.314J/Kmol)(298K)InKRewrite the above equation InKp=ΔG0-RT=64.98×103J/mol(8.314J/Kmol)(298K)InKp=26.231K=e26.231Kp=4.05×1012

In conclusion the activity series is correct, the large value of K for reaction (a) it is indicate that products are highly favoured, whereas the small value of K for reaction (b) indicates that reactants are highly favoured.

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!

Chapter 18 Solutions

General Chemistry

Ch. 18.6 - Practice Exercise Calculate the equilibrium...Ch. 18.6 - Prob. 2PECh. 18.6 - Prob. 3PECh. 18.6 - Prob. 1RCCh. 18 - Prob. 18.1QPCh. 18 - Prob. 18.2QPCh. 18 - Prob. 18.3QPCh. 18 - Prob. 18.4QPCh. 18 - Prob. 18.5QPCh. 18 - Prob. 18.7QPCh. 18 - Prob. 18.8QPCh. 18 - Prob. 18.9QPCh. 18 - 18.10 Arrange the following substances (1 mole...Ch. 18 - Prob. 18.11QPCh. 18 - Prob. 18.12QPCh. 18 - Prob. 18.13QPCh. 18 - 18.14 State whether the sign of the entropy...Ch. 18 - 18.15 Define free energy. What are its units? Ch. 18 - 18.16 Why is it more convenient to predict the...Ch. 18 - 18.17 Calculate ΔG° for the following reactions at...Ch. 18 - 18.18 Calculate ΔG° for the following reactions at...Ch. 18 - Prob. 18.19QPCh. 18 - Prob. 18.20QPCh. 18 - Prob. 18.21QPCh. 18 - Prob. 18.22QPCh. 18 - Prob. 18.23QPCh. 18 - 18.24 For the autoionization of water at...Ch. 18 - Prob. 18.25QPCh. 18 - Prob. 18.26QPCh. 18 - Prob. 18.27QPCh. 18 - Prob. 18.28QPCh. 18 - Prob. 18.29QPCh. 18 - Prob. 18.30QPCh. 18 - Prob. 18.31QPCh. 18 - Prob. 18.32QPCh. 18 - Prob. 18.33QPCh. 18 - Prob. 18.34QPCh. 18 - Prob. 18.35QPCh. 18 - Prob. 18.36QPCh. 18 - Prob. 18.37QPCh. 18 - Prob. 18.38QPCh. 18 - Prob. 18.39QPCh. 18 - Prob. 18.40QPCh. 18 - Prob. 18.41QPCh. 18 - Prob. 18.42QPCh. 18 - Prob. 18.43QPCh. 18 - Prob. 18.44QPCh. 18 - Prob. 18.45QPCh. 18 - Prob. 18.46QPCh. 18 - 18.47 Calculate the equilibrium pressure of CO2...Ch. 18 - Prob. 18.48QPCh. 18 - 18.49 Referring to Problem 18.48, explain why the...Ch. 18 - Prob. 18.50QPCh. 18 - Prob. 18.51QPCh. 18 - Prob. 18.52QPCh. 18 - Prob. 18.53QPCh. 18 - Prob. 18.54QPCh. 18 - Prob. 18.55QPCh. 18 - 18.56 Crystallization of sodium acetate from a...Ch. 18 - Prob. 18.57QPCh. 18 - Prob. 18.58QPCh. 18 - Prob. 18.59QPCh. 18 - Prob. 18.60QPCh. 18 - Prob. 18.61QPCh. 18 - Prob. 18.62QPCh. 18 - Prob. 18.63QPCh. 18 - Prob. 18.64QPCh. 18 - Prob. 18.65QPCh. 18 - Prob. 18.66QPCh. 18 - Prob. 18.67QPCh. 18 - Prob. 18.68QPCh. 18 - Prob. 18.69QPCh. 18 - Prob. 18.70QPCh. 18 - Prob. 18.71QPCh. 18 - Prob. 18.72QPCh. 18 - 18.73 (a) Over the years there have been numerous...Ch. 18 - Prob. 18.74QPCh. 18 - 18.75 Shown here are the thermodynamic data for...Ch. 18 - Prob. 18.76QPCh. 18 - Prob. 18.77QPCh. 18 - Prob. 18.78QPCh. 18 - Prob. 18.79QPCh. 18 - Prob. 18.80QPCh. 18 - Prob. 18.81QPCh. 18 - Prob. 18.82QPCh. 18 - Prob. 18.83QPCh. 18 - 18.84 Large quantities of hydrogen are needed for...Ch. 18 - Prob. 18.85QPCh. 18 - Prob. 18.86QPCh. 18 - Prob. 18.87QPCh. 18 - Prob. 18.88QPCh. 18 - Prob. 18.89QPCh. 18 - Prob. 18.90QPCh. 18 - Prob. 18.91QPCh. 18 - Prob. 18.92QPCh. 18 - Prob. 18.93QPCh. 18 - Prob. 18.94QPCh. 18 - Prob. 18.95QPCh. 18 - Prob. 18.96QPCh. 18 - Prob. 18.98QPCh. 18 - Prob. 18.100SPCh. 18 - Prob. 18.101SPCh. 18 - Prob. 18.102SPCh. 18 - Prob. 18.103SPCh. 18 - Prob. 18.104SPCh. 18 - Prob. 18.105SPCh. 18 - Prob. 18.106SPCh. 18 - Prob. 18.107SPCh. 18 - Prob. 18.108SPCh. 18 - 18.109 The boiling point of diethyl ether is...
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.
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
The Laws of Thermodynamics, Entropy, and Gibbs Free Energy; Author: Professor Dave Explains;https://www.youtube.com/watch?v=8N1BxHgsoOw;License: Standard YouTube License, CC-BY