EBK GENERAL CHEMISTRY: THE ESSENTIAL CO
EBK GENERAL CHEMISTRY: THE ESSENTIAL CO
7th Edition
ISBN: 8220106637203
Author: Chang
Publisher: YUZU
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Chapter 18, Problem 18.38QP

(a)

Interpretation Introduction

Interpretation:

To calculate the free energy (ΔG) values for given aqueous phase equilibrium reaction at 25°C.

Concept Information:

Thermodynamics is the branch of science that relates heat and energy in a system.  The laws of thermodynamics explain the fundamental quantities such as temperature, energy and randomness in a system.  Entropy is the measure of randomness in a system.  For a spontaneous process there is always a positive change in entropy. Free energy (Gibbs 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 equation given below helps us to calculate the change in free energy in a system.

ΔG = ΔΗ- TΔS

Where,

  ΔG  is the change in free energy of the system

  ΔΗ is the change in enthalpy of the system

  T is the absolute value of the temperature

  ΔS is the change in entropy in the system

Free energy (Gibbs 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 standard free energy change (ΔG°rxn)

    ΔG0=-RTlnKΔG=Free energyΔG0=Standardstate free energyR=GasConstant(0.0826l.atm/K.atm)T=Temprature273KK=EqulibriumConstant(KPandKC)

(a)

Expert Solution
Check Mark

Explanation of Solution

In each part of this problem we can use the following equation to calculate ΔG

ΔG=ΔG0+RTlnQΔG=ΔG0+RTln([H+][OH-])

Given the concentrations are equilibrium concentrations at 250C, Since the reaction is at equilibrium ΔG=0. This is advantage, because it allows us to calculate free energy ΔG°, the equilibrium Q=K, so we can write fallowing equation.  

ΔG0=-RTlnKwGivenvalues(R,T)aresubstitutedaboveequationΔG0=-(8.314J/Kmol)(298K)ln(1.0×10-14)=-2477.572×ln(1.0×10-14)[QHereIn1.0×10-14=-29.9336]ΔG0=8.0×104J/mol

(b)

Interpretation Introduction

Interpretation:

To calculate the free energy (ΔG) values for given aqueous phase equilibrium reaction at 25°C.

Concept Information:

Free energy (Gibbs 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 standard free energy change (ΔG°rxn) is the difference in free energy of the reactants and products in their standard state.

ΔG°rxn=nΔGf°(Products)-nΔGf°(Reactants)

Thermodynamics is the branch of science that relates heat and energy in a system.  The four laws of thermodynamics explain the fundamental quantities such as temperature, energy and randomness in a system.  Entropy is the measure of randomness in a system.  For a spontaneous process there is always a positive change in entropy. Free energy (Gibbs 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 equation given below helps us to calculate the change in free energy in a system.

ΔG = ΔΗ- TΔS

Where,

  ΔG  is the change in free energy of the system

  ΔΗ is the change in enthalpy of the system

  T is the absolute value of the temperature

  ΔS is the change in entropy in the system

Free energy (Gibbs 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 standard free energy change (ΔG°rxn)

      ΔG0=-RTlnKΔG=Free energyΔG0=Standardstate free energyR=GasConstant(0.0826l.atm/K.atm)T=Temprature273KK=EqulibriumConstant(KPandKC)

(b)

Expert Solution
Check Mark

Explanation of Solution

Letus consider following reaction (b)b).[H+]=1.0×10-3M,[OH]=1.0×10-4MΔG0=8.0×104J/molΔG0=-RTlnQΔG=ΔG0+RTln([H+][OH-])Given values are substituted aboveequatuionΔG=8.0×104J/mol+(8.314J/Kmol)(298K)ln(1.0×10-3)(1.0×10-4)ΔG=4.0×104J/mol 

(c)

Interpretation Introduction

Interpretation:

To calculate the free energy (ΔG) values for given aqueous phase equilibrium reaction at 25°C

Concept Information:

Free energy (Gibbs 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 standard free energy change (ΔG°rxn)

    ΔG0=-RTlnKΔG=Free energyΔG0=Standardstate free energyR=GasConstant(0.0826l.atm/K.atm)T=Temprature273KK=EqulibriumConstant(KPandKC)

(c)

Expert Solution
Check Mark

Explanation of Solution

Let us consider the following reaction (c)H2O(l)H+(aq)+OH-(aq)ΔG0=-RTlnQ[H+]=1.0×10-12M,[OH-]=2.0×10-8MΔG=ΔG0+RTln([H+][OH-])Givenvalues(R,T)aresubstitutedequationΔG=(8.0×104J/mol)+(8.314J/Kmol)(298K)ln(1.0×10-12)(2.0×10-8)ΔG=3.2×104J/mol

(d)

Interpretation Introduction

Interpretation:

To calculate the free energy (ΔG) values for given aqueous phase equilibrium reaction at 25°C

Concept Information:

Free energy (Gibbs 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 standard free energy change (ΔG°rxn)

      ΔG0=-RTlnKΔG=Free energyΔG0=Standardstate free energyR=GasConstant(0.0826l.atm/K.atm)T=Temprature273KK=EqulibriumConstant(KPandKC)

(d)

Expert Solution
Check Mark

Explanation of Solution

Let us consider the following reaction (d)[H+]=3.5M,[OH]=4.8×10-4MΔG0=-RTlnQΔG=ΔG0+RTln([H+][OH-])Given values are substituted equatuionΔG=8.0×104J/mol+(8.314J/Kmol)(298K)ln(3.5)(4.8×10-4)ΔG=6.4×104J/mol

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Chapter 18 Solutions

EBK GENERAL CHEMISTRY: THE ESSENTIAL CO

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...
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