Calculate Δ G for the reaction H 2 O ( l ) ⇌ H + ( a q ) + OH − ( a q ) at 25°C for the following conditions: (a) [ H + ] = 1.0 × 10 − 7 M , [ OH − ] = 1.0 × 10 − 7 M (b) [ H + ] = 1.0 × 10 − 3 M , [ OH − ] = 1.0 × 10 − 4 M (c) [ H + ] = 1.0 × 10 − 12 M , [ OH − ] = 2.0 × 10 − 8 M (d) [ H + ] = 3.5 M , [ OH − ] = 4.8 × 10 − 4 M

Question

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Answer 1

Textbook Question

Chapter 17, Problem 17.38QP

Calculate ΔG for the reaction

$H 2 O ( l ) ⇌ H + ( a q ) + OH − ( a q )$

at 25°C for the following conditions:

(a) $[ H + ] = 1.0 × 10 − 7 M , [ OH − ] = 1.0 × 10 − 7 M$
(b) $[ H + ] = 1.0 × 10 − 3 M , [ OH − ] = 1.0 × 10 − 4 M$
(c) $[ H + ] = 1.0 × 10 − 12 M , [ OH − ] = 2.0 × 10 − 8 M$
(d) $[ H + ] = 3.5 M , [ OH − ] = 4.8 × 10 − 4 M$

(a)

Expert Solution

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 $(ΔG°rxn)$

$ΔG0 =-RTln KΔG = Free energyΔG0 = Standard−state free energyR = Gas Constant (0.0826 l .atm/K.atm)T = Temprature 273 KK= Equlibrium Constant (KP and KC)$

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.

$ΔG0= -RTlnKwGiven values (R, T) are substituted above equationΔG0= -(8.314 J/K⋅mol) (298 K) ln(1.0 ×10-14) = -2477.572 ×ln(1.0 ×10-14) [Q Here In 1.0 ×10-14=-29.9336]ΔG0= 8.0 ×104 J/mol$

(b)

Expert Solution

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 $(ΔG°rxn)$

$ΔG0 =-RTln KΔG = Free energyΔG0 = Standard−state free energyR = Gas Constant (0.0826 l .atm/K.atm)T = Temprature 273 KK= Equlibrium Constant (KP and KC)$

Explanation of Solution

$Let us consider following reaction (b)b). [H+]= 1.0×10-3M, [OH−]= 1.0×10-4M ΔG0= 8.0 ×104 J/molΔG0= -RTlnQΔG=ΔG0+RTln([H+][OH-])Given values are substituted above equatuionΔG=8.0 ×104 J/mol+(8.314 J/K⋅mol) (298 K) ln(1.0 ×10-3) (1.0 ×10-4)ΔG= 4.0 ×104 J/mol$

(c)

Expert Solution

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 $(ΔG°rxn)$

$ΔG0 =-RTln KΔG = Free energyΔG0 = Standard−state free energyR = Gas Constant (0.0826 l .atm/K.atm)T = Temprature 273 KK= Equlibrium Constant (KP and KC)$

Explanation of Solution

$Let us consider the following reaction (c)H2O(l) ⇌ H+(aq) + OH-(aq)ΔG0= -RTlnQ [H+]= 1.0×10-12M, [OH-]= 2.0×10-8MΔG=ΔG0+RTln([H+][OH-])Given values (R, T) are substituted equationΔG= (8.0×104J/mol)+(8.314 J/K⋅mol) (298 K) ln(1.0×10-12) (2.0×10-8)ΔG= −3.2 ×104 J/mol$

(d)

Expert Solution

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 $(ΔG°rxn)$

$ΔG0 =-RTln KΔG = Free energyΔG0 = Standard−state free energyR = Gas Constant (0.0826 l .atm/K.atm)T = Temprature 273 KK= Equlibrium Constant (KP and KC)$

Explanation of Solution

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

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Answer 2

Textbook Question

Chapter 17, Problem 17.38QP

Calculate ΔG for the reaction

$H 2 O ( l ) ⇌ H + ( a q ) + OH − ( a q )$

at 25°C for the following conditions:

(a) $[ H + ] = 1.0 × 10 − 7 M , [ OH − ] = 1.0 × 10 − 7 M$
(b) $[ H + ] = 1.0 × 10 − 3 M , [ OH − ] = 1.0 × 10 − 4 M$
(c) $[ H + ] = 1.0 × 10 − 12 M , [ OH − ] = 2.0 × 10 − 8 M$
(d) $[ H + ] = 3.5 M , [ OH − ] = 4.8 × 10 − 4 M$

(a)

Expert Solution

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 $(ΔG°rxn)$

$ΔG0 =-RTln KΔG = Free energyΔG0 = Standard−state free energyR = Gas Constant (0.0826 l .atm/K.atm)T = Temprature 273 KK= Equlibrium Constant (KP and KC)$

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.

$ΔG0= -RTlnKwGiven values (R, T) are substituted above equationΔG0= -(8.314 J/K⋅mol) (298 K) ln(1.0 ×10-14) = -2477.572 ×ln(1.0 ×10-14) [Q Here In 1.0 ×10-14=-29.9336]ΔG0= 8.0 ×104 J/mol$

(b)

Expert Solution

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 $(ΔG°rxn)$

$ΔG0 =-RTln KΔG = Free energyΔG0 = Standard−state free energyR = Gas Constant (0.0826 l .atm/K.atm)T = Temprature 273 KK= Equlibrium Constant (KP and KC)$

Explanation of Solution

$Let us consider following reaction (b)b). [H+]= 1.0×10-3M, [OH−]= 1.0×10-4M ΔG0= 8.0 ×104 J/molΔG0= -RTlnQΔG=ΔG0+RTln([H+][OH-])Given values are substituted above equatuionΔG=8.0 ×104 J/mol+(8.314 J/K⋅mol) (298 K) ln(1.0 ×10-3) (1.0 ×10-4)ΔG= 4.0 ×104 J/mol$

(c)

Expert Solution

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 $(ΔG°rxn)$

$ΔG0 =-RTln KΔG = Free energyΔG0 = Standard−state free energyR = Gas Constant (0.0826 l .atm/K.atm)T = Temprature 273 KK= Equlibrium Constant (KP and KC)$

Explanation of Solution

$Let us consider the following reaction (c)H2O(l) ⇌ H+(aq) + OH-(aq)ΔG0= -RTlnQ [H+]= 1.0×10-12M, [OH-]= 2.0×10-8MΔG=ΔG0+RTln([H+][OH-])Given values (R, T) are substituted equationΔG= (8.0×104J/mol)+(8.314 J/K⋅mol) (298 K) ln(1.0×10-12) (2.0×10-8)ΔG= −3.2 ×104 J/mol$

(d)

Expert Solution

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 $(ΔG°rxn)$

$ΔG0 =-RTln KΔG = Free energyΔG0 = Standard−state free energyR = Gas Constant (0.0826 l .atm/K.atm)T = Temprature 273 KK= Equlibrium Constant (KP and KC)$

Explanation of Solution

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

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Answer 3

Textbook Question

Chapter 17, Problem 17.38QP

Calculate ΔG for the reaction

$H 2 O ( l ) ⇌ H + ( a q ) + OH − ( a q )$

at 25°C for the following conditions:

(a) $[ H + ] = 1.0 × 10 − 7 M , [ OH − ] = 1.0 × 10 − 7 M$
(b) $[ H + ] = 1.0 × 10 − 3 M , [ OH − ] = 1.0 × 10 − 4 M$
(c) $[ H + ] = 1.0 × 10 − 12 M , [ OH − ] = 2.0 × 10 − 8 M$
(d) $[ H + ] = 3.5 M , [ OH − ] = 4.8 × 10 − 4 M$

(a)

Expert Solution

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 $(ΔG°rxn)$

$ΔG0 =-RTln KΔG = Free energyΔG0 = Standard−state free energyR = Gas Constant (0.0826 l .atm/K.atm)T = Temprature 273 KK= Equlibrium Constant (KP and KC)$

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.

$ΔG0= -RTlnKwGiven values (R, T) are substituted above equationΔG0= -(8.314 J/K⋅mol) (298 K) ln(1.0 ×10-14) = -2477.572 ×ln(1.0 ×10-14) [Q Here In 1.0 ×10-14=-29.9336]ΔG0= 8.0 ×104 J/mol$

(b)

Expert Solution

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 $(ΔG°rxn)$

$ΔG0 =-RTln KΔG = Free energyΔG0 = Standard−state free energyR = Gas Constant (0.0826 l .atm/K.atm)T = Temprature 273 KK= Equlibrium Constant (KP and KC)$

Explanation of Solution

$Let us consider following reaction (b)b). [H+]= 1.0×10-3M, [OH−]= 1.0×10-4M ΔG0= 8.0 ×104 J/molΔG0= -RTlnQΔG=ΔG0+RTln([H+][OH-])Given values are substituted above equatuionΔG=8.0 ×104 J/mol+(8.314 J/K⋅mol) (298 K) ln(1.0 ×10-3) (1.0 ×10-4)ΔG= 4.0 ×104 J/mol$

(c)

Expert Solution

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 $(ΔG°rxn)$

$ΔG0 =-RTln KΔG = Free energyΔG0 = Standard−state free energyR = Gas Constant (0.0826 l .atm/K.atm)T = Temprature 273 KK= Equlibrium Constant (KP and KC)$

Explanation of Solution

$Let us consider the following reaction (c)H2O(l) ⇌ H+(aq) + OH-(aq)ΔG0= -RTlnQ [H+]= 1.0×10-12M, [OH-]= 2.0×10-8MΔG=ΔG0+RTln([H+][OH-])Given values (R, T) are substituted equationΔG= (8.0×104J/mol)+(8.314 J/K⋅mol) (298 K) ln(1.0×10-12) (2.0×10-8)ΔG= −3.2 ×104 J/mol$

(d)

Expert Solution

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 $(ΔG°rxn)$

$ΔG0 =-RTln KΔG = Free energyΔG0 = Standard−state free energyR = Gas Constant (0.0826 l .atm/K.atm)T = Temprature 273 KK= Equlibrium Constant (KP and KC)$

Explanation of Solution

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

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Textbook Question

Chapter 17, Problem 17.38QP

Calculate ΔG for the reaction

$H 2 O ( l ) ⇌ H + ( a q ) + OH − ( a q )$

at 25°C for the following conditions:

(a) $[ H + ] = 1.0 × 10 − 7 M , [ OH − ] = 1.0 × 10 − 7 M$
(b) $[ H + ] = 1.0 × 10 − 3 M , [ OH − ] = 1.0 × 10 − 4 M$
(c) $[ H + ] = 1.0 × 10 − 12 M , [ OH − ] = 2.0 × 10 − 8 M$
(d) $[ H + ] = 3.5 M , [ OH − ] = 4.8 × 10 − 4 M$

(a)

Expert Solution

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 $(ΔG°rxn)$

$ΔG0 =-RTln KΔG = Free energyΔG0 = Standard−state free energyR = Gas Constant (0.0826 l .atm/K.atm)T = Temprature 273 KK= Equlibrium Constant (KP and KC)$

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.

$ΔG0= -RTlnKwGiven values (R, T) are substituted above equationΔG0= -(8.314 J/K⋅mol) (298 K) ln(1.0 ×10-14) = -2477.572 ×ln(1.0 ×10-14) [Q Here In 1.0 ×10-14=-29.9336]ΔG0= 8.0 ×104 J/mol$

(b)

Expert Solution

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 $(ΔG°rxn)$

$ΔG0 =-RTln KΔG = Free energyΔG0 = Standard−state free energyR = Gas Constant (0.0826 l .atm/K.atm)T = Temprature 273 KK= Equlibrium Constant (KP and KC)$

Explanation of Solution

$Let us consider following reaction (b)b). [H+]= 1.0×10-3M, [OH−]= 1.0×10-4M ΔG0= 8.0 ×104 J/molΔG0= -RTlnQΔG=ΔG0+RTln([H+][OH-])Given values are substituted above equatuionΔG=8.0 ×104 J/mol+(8.314 J/K⋅mol) (298 K) ln(1.0 ×10-3) (1.0 ×10-4)ΔG= 4.0 ×104 J/mol$

(c)

Expert Solution

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 $(ΔG°rxn)$

$ΔG0 =-RTln KΔG = Free energyΔG0 = Standard−state free energyR = Gas Constant (0.0826 l .atm/K.atm)T = Temprature 273 KK= Equlibrium Constant (KP and KC)$

Explanation of Solution

$Let us consider the following reaction (c)H2O(l) ⇌ H+(aq) + OH-(aq)ΔG0= -RTlnQ [H+]= 1.0×10-12M, [OH-]= 2.0×10-8MΔG=ΔG0+RTln([H+][OH-])Given values (R, T) are substituted equationΔG= (8.0×104J/mol)+(8.314 J/K⋅mol) (298 K) ln(1.0×10-12) (2.0×10-8)ΔG= −3.2 ×104 J/mol$

(d)

Expert Solution

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 $(ΔG°rxn)$

$ΔG0 =-RTln KΔG = Free energyΔG0 = Standard−state free energyR = Gas Constant (0.0826 l .atm/K.atm)T = Temprature 273 KK= Equlibrium Constant (KP and KC)$

Explanation of Solution

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

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

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 $(ΔG°rxn)$

$ΔG0 =-RTln KΔG = Free energyΔG0 = Standard−state free energyR = Gas Constant (0.0826 l .atm/K.atm)T = Temprature 273 KK= Equlibrium Constant (KP and KC)$

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.

$ΔG0= -RTlnKwGiven values (R, T) are substituted above equationΔG0= -(8.314 J/K⋅mol) (298 K) ln(1.0 ×10-14) = -2477.572 ×ln(1.0 ×10-14) [Q Here In 1.0 ×10-14=-29.9336]ΔG0= 8.0 ×104 J/mol$

(b)

Expert Solution

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 $(ΔG°rxn)$

$ΔG0 =-RTln KΔG = Free energyΔG0 = Standard−state free energyR = Gas Constant (0.0826 l .atm/K.atm)T = Temprature 273 KK= Equlibrium Constant (KP and KC)$

Explanation of Solution

$Let us consider following reaction (b)b). [H+]= 1.0×10-3M, [OH−]= 1.0×10-4M ΔG0= 8.0 ×104 J/molΔG0= -RTlnQΔG=ΔG0+RTln([H+][OH-])Given values are substituted above equatuionΔG=8.0 ×104 J/mol+(8.314 J/K⋅mol) (298 K) ln(1.0 ×10-3) (1.0 ×10-4)ΔG= 4.0 ×104 J/mol$

(c)

Expert Solution

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 $(ΔG°rxn)$

$ΔG0 =-RTln KΔG = Free energyΔG0 = Standard−state free energyR = Gas Constant (0.0826 l .atm/K.atm)T = Temprature 273 KK= Equlibrium Constant (KP and KC)$

Explanation of Solution

$Let us consider the following reaction (c)H2O(l) ⇌ H+(aq) + OH-(aq)ΔG0= -RTlnQ [H+]= 1.0×10-12M, [OH-]= 2.0×10-8MΔG=ΔG0+RTln([H+][OH-])Given values (R, T) are substituted equationΔG= (8.0×104J/mol)+(8.314 J/K⋅mol) (298 K) ln(1.0×10-12) (2.0×10-8)ΔG= −3.2 ×104 J/mol$

(d)

Expert Solution

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 $(ΔG°rxn)$

$ΔG0 =-RTln KΔG = Free energyΔG0 = Standard−state free energyR = Gas Constant (0.0826 l .atm/K.atm)T = Temprature 273 KK= Equlibrium Constant (KP and KC)$

Explanation of Solution

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

Answer 8

(a)

Expert Solution

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 $(ΔG°rxn)$

$ΔG0 =-RTln KΔG = Free energyΔG0 = Standard−state free energyR = Gas Constant (0.0826 l .atm/K.atm)T = Temprature 273 KK= Equlibrium Constant (KP and KC)$

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.

$ΔG0= -RTlnKwGiven values (R, T) are substituted above equationΔG0= -(8.314 J/K⋅mol) (298 K) ln(1.0 ×10-14) = -2477.572 ×ln(1.0 ×10-14) [Q Here In 1.0 ×10-14=-29.9336]ΔG0= 8.0 ×104 J/mol$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

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 $(ΔG°rxn)$

$ΔG0 =-RTln KΔG = Free energyΔG0 = Standard−state free energyR = Gas Constant (0.0826 l .atm/K.atm)T = Temprature 273 KK= Equlibrium Constant (KP and KC)$

Answer 13

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.

$ΔG0= -RTlnKwGiven values (R, T) are substituted above equationΔG0= -(8.314 J/K⋅mol) (298 K) ln(1.0 ×10-14) = -2477.572 ×ln(1.0 ×10-14) [Q Here In 1.0 ×10-14=-29.9336]ΔG0= 8.0 ×104 J/mol$

Answer 14

(b)

Expert Solution

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 $(ΔG°rxn)$

$ΔG0 =-RTln KΔG = Free energyΔG0 = Standard−state free energyR = Gas Constant (0.0826 l .atm/K.atm)T = Temprature 273 KK= Equlibrium Constant (KP and KC)$

Explanation of Solution

$Let us consider following reaction (b)b). [H+]= 1.0×10-3M, [OH−]= 1.0×10-4M ΔG0= 8.0 ×104 J/molΔG0= -RTlnQΔG=ΔG0+RTln([H+][OH-])Given values are substituted above equatuionΔG=8.0 ×104 J/mol+(8.314 J/K⋅mol) (298 K) ln(1.0 ×10-3) (1.0 ×10-4)ΔG= 4.0 ×104 J/mol$

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

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 $(ΔG°rxn)$

$ΔG0 =-RTln KΔG = Free energyΔG0 = Standard−state free energyR = Gas Constant (0.0826 l .atm/K.atm)T = Temprature 273 KK= Equlibrium Constant (KP and KC)$

Answer 19

Explanation of Solution

$Let us consider following reaction (b)b). [H+]= 1.0×10-3M, [OH−]= 1.0×10-4M ΔG0= 8.0 ×104 J/molΔG0= -RTlnQΔG=ΔG0+RTln([H+][OH-])Given values are substituted above equatuionΔG=8.0 ×104 J/mol+(8.314 J/K⋅mol) (298 K) ln(1.0 ×10-3) (1.0 ×10-4)ΔG= 4.0 ×104 J/mol$

Answer 20

(c)

Expert Solution

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 $(ΔG°rxn)$

$ΔG0 =-RTln KΔG = Free energyΔG0 = Standard−state free energyR = Gas Constant (0.0826 l .atm/K.atm)T = Temprature 273 KK= Equlibrium Constant (KP and KC)$

Explanation of Solution

$Let us consider the following reaction (c)H2O(l) ⇌ H+(aq) + OH-(aq)ΔG0= -RTlnQ [H+]= 1.0×10-12M, [OH-]= 2.0×10-8MΔG=ΔG0+RTln([H+][OH-])Given values (R, T) are substituted equationΔG= (8.0×104J/mol)+(8.314 J/K⋅mol) (298 K) ln(1.0×10-12) (2.0×10-8)ΔG= −3.2 ×104 J/mol$

Answer 21

(c)

Expert Solution

Answer 22

(c)

Expert Solution

Answer 23

Expert Solution

Answer 24

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 $(ΔG°rxn)$

$ΔG0 =-RTln KΔG = Free energyΔG0 = Standard−state free energyR = Gas Constant (0.0826 l .atm/K.atm)T = Temprature 273 KK= Equlibrium Constant (KP and KC)$

Answer 25

Explanation of Solution

$Let us consider the following reaction (c)H2O(l) ⇌ H+(aq) + OH-(aq)ΔG0= -RTlnQ [H+]= 1.0×10-12M, [OH-]= 2.0×10-8MΔG=ΔG0+RTln([H+][OH-])Given values (R, T) are substituted equationΔG= (8.0×104J/mol)+(8.314 J/K⋅mol) (298 K) ln(1.0×10-12) (2.0×10-8)ΔG= −3.2 ×104 J/mol$

Answer 26

(d)

Expert Solution

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 $(ΔG°rxn)$

$ΔG0 =-RTln KΔG = Free energyΔG0 = Standard−state free energyR = Gas Constant (0.0826 l .atm/K.atm)T = Temprature 273 KK= Equlibrium Constant (KP and KC)$

Explanation of Solution

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

Answer 27

(d)

Expert Solution

Answer 28

(d)

Expert Solution

Answer 29

Expert Solution

Answer 30

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 $(ΔG°rxn)$

$ΔG0 =-RTln KΔG = Free energyΔG0 = Standard−state free energyR = Gas Constant (0.0826 l .atm/K.atm)T = Temprature 273 KK= Equlibrium Constant (KP and KC)$

Answer 31

Explanation of Solution

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

Answer 32

Ch. 17.3 - How does the entropy of a system change for each...Ch. 17.3 - For which of the following physical changes is S...Ch. 17.3 - Determine if the sign for S is positive or...Ch. 17.4 - Prob. 2PE Ch. 17.4 - Prob. 3PE Ch. 17.4 - Calculate the standard entropy change for the...Ch. 17.4 - Consider the gas-phase reaction of A2 (blue) and...Ch. 17.5 - Prob. 4PE Ch. 17.5 - The molar heats of fusion and vaporization of...Ch. 17.5 - Under what circumstances will an endothermic...

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