For the given set of reactions in problem 20.52 the standard free energy Δ G o value have to be calculated using the Δ H f o a n d S o values. Concept introduction: Enthalpy is the amount energy absorbed or released in a process. The enthalpy change in a system (Δ Η sys ) can be calculated by the following equation. ΔH rxn = ∑ ΔH ° produdcts - ∑ ΔH ° reactants Where, ΔH r o ( reactants ) is the standard enthalpy of the reactants ΔH r o ( produdcts ) is the standard enthalpy of the products Entropy is the measure of randomness in the system. Standard entropy change in a reaction is the difference in entropy of the products and reactants. (ΔS ° rxn ) can be calculated by the following equation. ΔS ° rxn = ∑ m S ° Products - ∑ n S ° reactants Where, S ° reactants is the standard entropy of the reactants S ° Products is the standard entropy of the products Standard entropy change in a reaction and entropy change in the system are same. 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 = Δ Η - T Δ S Where, ΔG is the change in free energy of the system Δ Η is the change in enthalpy of the system T is the absolute value of the temperature Δ S 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 ) is the difference in free energy of the reactants and products in their standard state. ΔG ° rxn = ∑ mΔG f ° (Products)- ∑ nΔG f ° (Reactants) Where, nΔG f ° ( Reactants ) is the standard entropy of the reactants mΔG f ° ( products ) is the standard free energy of the products

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Answer 1

Question

Chapter 20, Problem 20.54P

Interpretation Introduction

Interpretation:

For the given set of reactions in problem 20.52 the standard free energy $ΔGo$ value have to be calculated using the $ΔHfo and So$ values.

Concept introduction:

Enthalpy is the amount energy absorbed or released in a process.

The enthalpy change in a system $(ΔΗsys)$ can be calculated by the following equation.

$ΔHrxn = ∑ΔH°produdcts-∑ΔH°reactants$

Where,

$ΔHro(reactants)$ is the standard enthalpy of the reactants

$ΔHro(produdcts)$ is the standard enthalpy of the products

Entropy is the measure of randomness in the system. Standard entropy change in a reaction is the difference in entropy of the products and reactants. $(ΔS°rxn)$ can be calculated by the following equation.

$ΔS°rxn = ∑m S°Products- ∑n S°reactants$

Where,

$S°reactants$ is the standard entropy of the reactants

$S°Products$ is the standard entropy of the products

Standard entropy change in a reaction and entropy change in the system are same.

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

$ΔG°rxn=∑mΔGf°(Products)-∑nΔGf°(Reactants)$

Where,

$nΔGf°(Reactants)$ is the standard entropy of the reactants

$mΔGf°(products)$ is the standard free energy of the products

Expert Solution & Answer

Answer to Problem 20.54P

Reaction-A the standard free energy value is $ΔGrxno= 2.4 kJ_$

Reaction-B the standard free energy value is $ΔGrxno= -48.4 kJ_$

Reaction-C the standard free energy value is $ΔGrxno= 91.2 kJ_$

Explanation of Solution

Reaction-A

Given chemical reaction is,

$H2 (g)+ I2(g) → 2HI(g)$

The number of particle also decreases, indicating a decrease in entropy.

So, given reaction $ΔGfo$ value is zero.

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

$ΔH°rxn = ∑m ΔH°f(Products)- ∑n ΔH°f(reactants)$

$ΔH°rxn = [(2 mol HI) ( ΔH°fof HI)] −[(1 mol H2)(ΔH°f of H2) + (1 mol I2)(ΔH°f of I2)] ΔH°rxn = [(2 mol HI)(25.9 kJ/mol )]− [(1 mol H2)(0 kJ/mol ) +(1 mol I2)(0 kJ/mol )] ΔH°rxn = -51.8 kJ$

The enthalpy change is negative.

Hence, the enthalpy $(ΔH°rxn)$ value is $-51.8 kJ_$

Entropy change $ΔS°system$

Standard entropy change equation is,

$ΔS°rxn = ∑m S°Products- ∑n S°reactants$

Where, $(m) and (n)$ are the stoichiometric co-efficient.

$ΔS°rxn = [(2 mol HI)(206.33 J/mol×K) ]− [(2 mol H2)(130.6 J/mol×K) + (1 mol I2)(116.14 J/mol×K) ] ΔS°rxn =165.92 J/K$

Therefore, the $(ΔS°rxn)$ of the reaction is $165.92 J/K_$

Next calculate the Free enrgy change $ΔGrxno$

Standared Free energy change equation iss,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Free energy change $ΔGfo$

Calcualted enthalpy and entropy values are

$ΔΗrxno=-51.8 kJ$

$ΔSrxno=165.92 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= 51.8 kJ - [(298 K)(165.92 J/K)(1 kJ/103J)]ΔGrxno= 2.3558 kJ$

Therefore, the standard free energy value is $ΔGrxno= 2.4 kJ_$

Reaction-B

Given reaction is,

$MnO2 (s)+ 2CO(g) → Mn(s) + 4CO2(g)$

The number of particle also increases, indicating a increase in entropy.

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

$ΔH°rxn = ∑m ΔH°f(Products)- ∑n ΔH°f(reactants)$

$ΔΗrxno = [(1 mol Mn)(0 kJ/mol ) + (2 mol CO2)(−393.5 kJ/mol )]− [(1 mol MnO2)(−520.9 kJ/mol ) +(2 mol CO)(−110.5 kJ/mol )] ΔΗrxno = -45.1 kJ$

The enthalpy change is negative.

Hence, the enthalpy $(ΔH°rxn)$ value is $-45.1 kJ_$

Entropy change $ΔS°system$

Standard entropy change equation is,

$ΔS°rxn = ∑m S°Products- ∑n S°reactants$

Where, $(m) and (n)$ are the stoichiometric co-efficient.

$ΔSrxno = [(1 mol Mn)(31.8 J/mol⋅K ) + (2 mol CO2)(213.7 J/mol⋅K ) ]− [(1 mol MnO2)(53.1 J/mol⋅K ) +(2 mol CO)(197.5 J/mol⋅K )] ΔSrxno = 11.1 J/K$

Therefore, the $(ΔS°rxn)$ of the reaction is $-91.28 J/K_$

Next calculate the Free energy change $ΔGrxno$

Standared Free energy change equation iss,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Free enrgy change $ΔGfo$

Calcualted enthalpy and entropy values are

$ΔΗrxno=-45.1 kJ$

$ΔSrxno=-91.28 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= −45.1 kJ - [(298 K)(−11.1 J/K)(1 kJ/103J)]ΔGrxno= −48.4078 kJ$

Therefore, the standard free energy value is $ΔGrxno= -48.4 kJ_$

Reaction-C

Given reaction is,

$NH4Cl (s) → NH3(g) + HCl(g)$

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

$ΔH°rxn = ∑m ΔH°f(Products)- ∑n ΔH°f(reactants)$

$ΔΗrxno = [(1 mol NH3)(−45.9 kJ/mol ) + (1 mol HCl)(−92.3 kJ/mol ) ]− [(1 mol NH4Cl)(−314.4 kJ/mol ) ] ΔΗrxno = 176.2 kJ$

The enthalpy change is positive.

Hence, the enthalpy $(ΔH°rxn)$ value is $176.2 kJ_$

Entropy change $ΔS°system$

Standard entropy change equation is,

$ΔS°rxn = ∑m S°Products- ∑n S°reactants$

Where, $(m) and (n)$ are the stoichiometric co-efficient.

$ΔSrxno = [(1 mol NH3 )(193 J/mol⋅K ) + (1 mol HCl)(186.79 J/mol⋅K )] − [(1 mol NH4Cl )(94.6 J/mol⋅K ) ] ΔSrxno = 285.19 J/K$

Therefore, the $(ΔS°rxn)$ of the reaction is $285.19 J/K_$

Finally calculate the Free enrgy change $ΔGrxno$

Standared Free energy change equation iss,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Free energy change $ΔGfo$

Calcualted enthalpy and entropy values are

$ΔΗrxno=176.2 kJ$

$ΔSrxno=285.19 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= 176.2 kJ - [(298 K)(285.19 J/K)(1 kJ/103J)]ΔGrxno= 91.213 kJ$

Therefore, the standard free energy value is $ΔGrxno= 91.2 kJ_$

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Answer 2

Question

Chapter 20, Problem 20.54P

Interpretation Introduction

Interpretation:

For the given set of reactions in problem 20.52 the standard free energy $ΔGo$ value have to be calculated using the $ΔHfo and So$ values.

Concept introduction:

Enthalpy is the amount energy absorbed or released in a process.

The enthalpy change in a system $(ΔΗsys)$ can be calculated by the following equation.

$ΔHrxn = ∑ΔH°produdcts-∑ΔH°reactants$

Where,

$ΔHro(reactants)$ is the standard enthalpy of the reactants

$ΔHro(produdcts)$ is the standard enthalpy of the products

Entropy is the measure of randomness in the system. Standard entropy change in a reaction is the difference in entropy of the products and reactants. $(ΔS°rxn)$ can be calculated by the following equation.

$ΔS°rxn = ∑m S°Products- ∑n S°reactants$

Where,

$S°reactants$ is the standard entropy of the reactants

$S°Products$ is the standard entropy of the products

Standard entropy change in a reaction and entropy change in the system are same.

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

$ΔG°rxn=∑mΔGf°(Products)-∑nΔGf°(Reactants)$

Where,

$nΔGf°(Reactants)$ is the standard entropy of the reactants

$mΔGf°(products)$ is the standard free energy of the products

Expert Solution & Answer

Answer to Problem 20.54P

Reaction-A the standard free energy value is $ΔGrxno= 2.4 kJ_$

Reaction-B the standard free energy value is $ΔGrxno= -48.4 kJ_$

Reaction-C the standard free energy value is $ΔGrxno= 91.2 kJ_$

Explanation of Solution

Reaction-A

Given chemical reaction is,

$H2 (g)+ I2(g) → 2HI(g)$

The number of particle also decreases, indicating a decrease in entropy.

So, given reaction $ΔGfo$ value is zero.

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

$ΔH°rxn = ∑m ΔH°f(Products)- ∑n ΔH°f(reactants)$

$ΔH°rxn = [(2 mol HI) ( ΔH°fof HI)] −[(1 mol H2)(ΔH°f of H2) + (1 mol I2)(ΔH°f of I2)] ΔH°rxn = [(2 mol HI)(25.9 kJ/mol )]− [(1 mol H2)(0 kJ/mol ) +(1 mol I2)(0 kJ/mol )] ΔH°rxn = -51.8 kJ$

The enthalpy change is negative.

Hence, the enthalpy $(ΔH°rxn)$ value is $-51.8 kJ_$

Entropy change $ΔS°system$

Standard entropy change equation is,

$ΔS°rxn = ∑m S°Products- ∑n S°reactants$

Where, $(m) and (n)$ are the stoichiometric co-efficient.

$ΔS°rxn = [(2 mol HI)(206.33 J/mol×K) ]− [(2 mol H2)(130.6 J/mol×K) + (1 mol I2)(116.14 J/mol×K) ] ΔS°rxn =165.92 J/K$

Therefore, the $(ΔS°rxn)$ of the reaction is $165.92 J/K_$

Next calculate the Free enrgy change $ΔGrxno$

Standared Free energy change equation iss,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Free energy change $ΔGfo$

Calcualted enthalpy and entropy values are

$ΔΗrxno=-51.8 kJ$

$ΔSrxno=165.92 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= 51.8 kJ - [(298 K)(165.92 J/K)(1 kJ/103J)]ΔGrxno= 2.3558 kJ$

Therefore, the standard free energy value is $ΔGrxno= 2.4 kJ_$

Reaction-B

Given reaction is,

$MnO2 (s)+ 2CO(g) → Mn(s) + 4CO2(g)$

The number of particle also increases, indicating a increase in entropy.

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

$ΔH°rxn = ∑m ΔH°f(Products)- ∑n ΔH°f(reactants)$

$ΔΗrxno = [(1 mol Mn)(0 kJ/mol ) + (2 mol CO2)(−393.5 kJ/mol )]− [(1 mol MnO2)(−520.9 kJ/mol ) +(2 mol CO)(−110.5 kJ/mol )] ΔΗrxno = -45.1 kJ$

The enthalpy change is negative.

Hence, the enthalpy $(ΔH°rxn)$ value is $-45.1 kJ_$

Entropy change $ΔS°system$

Standard entropy change equation is,

$ΔS°rxn = ∑m S°Products- ∑n S°reactants$

Where, $(m) and (n)$ are the stoichiometric co-efficient.

$ΔSrxno = [(1 mol Mn)(31.8 J/mol⋅K ) + (2 mol CO2)(213.7 J/mol⋅K ) ]− [(1 mol MnO2)(53.1 J/mol⋅K ) +(2 mol CO)(197.5 J/mol⋅K )] ΔSrxno = 11.1 J/K$

Therefore, the $(ΔS°rxn)$ of the reaction is $-91.28 J/K_$

Next calculate the Free energy change $ΔGrxno$

Standared Free energy change equation iss,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Free enrgy change $ΔGfo$

Calcualted enthalpy and entropy values are

$ΔΗrxno=-45.1 kJ$

$ΔSrxno=-91.28 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= −45.1 kJ - [(298 K)(−11.1 J/K)(1 kJ/103J)]ΔGrxno= −48.4078 kJ$

Therefore, the standard free energy value is $ΔGrxno= -48.4 kJ_$

Reaction-C

Given reaction is,

$NH4Cl (s) → NH3(g) + HCl(g)$

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

$ΔH°rxn = ∑m ΔH°f(Products)- ∑n ΔH°f(reactants)$

$ΔΗrxno = [(1 mol NH3)(−45.9 kJ/mol ) + (1 mol HCl)(−92.3 kJ/mol ) ]− [(1 mol NH4Cl)(−314.4 kJ/mol ) ] ΔΗrxno = 176.2 kJ$

The enthalpy change is positive.

Hence, the enthalpy $(ΔH°rxn)$ value is $176.2 kJ_$

Entropy change $ΔS°system$

Standard entropy change equation is,

$ΔS°rxn = ∑m S°Products- ∑n S°reactants$

Where, $(m) and (n)$ are the stoichiometric co-efficient.

$ΔSrxno = [(1 mol NH3 )(193 J/mol⋅K ) + (1 mol HCl)(186.79 J/mol⋅K )] − [(1 mol NH4Cl )(94.6 J/mol⋅K ) ] ΔSrxno = 285.19 J/K$

Therefore, the $(ΔS°rxn)$ of the reaction is $285.19 J/K_$

Finally calculate the Free enrgy change $ΔGrxno$

Standared Free energy change equation iss,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Free energy change $ΔGfo$

Calcualted enthalpy and entropy values are

$ΔΗrxno=176.2 kJ$

$ΔSrxno=285.19 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= 176.2 kJ - [(298 K)(285.19 J/K)(1 kJ/103J)]ΔGrxno= 91.213 kJ$

Therefore, the standard free energy value is $ΔGrxno= 91.2 kJ_$

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Answer 3

Question

Chapter 20, Problem 20.54P

Interpretation Introduction

Interpretation:

For the given set of reactions in problem 20.52 the standard free energy $ΔGo$ value have to be calculated using the $ΔHfo and So$ values.

Concept introduction:

Enthalpy is the amount energy absorbed or released in a process.

The enthalpy change in a system $(ΔΗsys)$ can be calculated by the following equation.

$ΔHrxn = ∑ΔH°produdcts-∑ΔH°reactants$

Where,

$ΔHro(reactants)$ is the standard enthalpy of the reactants

$ΔHro(produdcts)$ is the standard enthalpy of the products

Entropy is the measure of randomness in the system. Standard entropy change in a reaction is the difference in entropy of the products and reactants. $(ΔS°rxn)$ can be calculated by the following equation.

$ΔS°rxn = ∑m S°Products- ∑n S°reactants$

Where,

$S°reactants$ is the standard entropy of the reactants

$S°Products$ is the standard entropy of the products

Standard entropy change in a reaction and entropy change in the system are same.

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

$ΔG°rxn=∑mΔGf°(Products)-∑nΔGf°(Reactants)$

Where,

$nΔGf°(Reactants)$ is the standard entropy of the reactants

$mΔGf°(products)$ is the standard free energy of the products

Expert Solution & Answer

Answer to Problem 20.54P

Reaction-A the standard free energy value is $ΔGrxno= 2.4 kJ_$

Reaction-B the standard free energy value is $ΔGrxno= -48.4 kJ_$

Reaction-C the standard free energy value is $ΔGrxno= 91.2 kJ_$

Explanation of Solution

Reaction-A

Given chemical reaction is,

$H2 (g)+ I2(g) → 2HI(g)$

The number of particle also decreases, indicating a decrease in entropy.

So, given reaction $ΔGfo$ value is zero.

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

$ΔH°rxn = ∑m ΔH°f(Products)- ∑n ΔH°f(reactants)$

$ΔH°rxn = [(2 mol HI) ( ΔH°fof HI)] −[(1 mol H2)(ΔH°f of H2) + (1 mol I2)(ΔH°f of I2)] ΔH°rxn = [(2 mol HI)(25.9 kJ/mol )]− [(1 mol H2)(0 kJ/mol ) +(1 mol I2)(0 kJ/mol )] ΔH°rxn = -51.8 kJ$

The enthalpy change is negative.

Hence, the enthalpy $(ΔH°rxn)$ value is $-51.8 kJ_$

Entropy change $ΔS°system$

Standard entropy change equation is,

$ΔS°rxn = ∑m S°Products- ∑n S°reactants$

Where, $(m) and (n)$ are the stoichiometric co-efficient.

$ΔS°rxn = [(2 mol HI)(206.33 J/mol×K) ]− [(2 mol H2)(130.6 J/mol×K) + (1 mol I2)(116.14 J/mol×K) ] ΔS°rxn =165.92 J/K$

Therefore, the $(ΔS°rxn)$ of the reaction is $165.92 J/K_$

Next calculate the Free enrgy change $ΔGrxno$

Standared Free energy change equation iss,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Free energy change $ΔGfo$

Calcualted enthalpy and entropy values are

$ΔΗrxno=-51.8 kJ$

$ΔSrxno=165.92 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= 51.8 kJ - [(298 K)(165.92 J/K)(1 kJ/103J)]ΔGrxno= 2.3558 kJ$

Therefore, the standard free energy value is $ΔGrxno= 2.4 kJ_$

Reaction-B

Given reaction is,

$MnO2 (s)+ 2CO(g) → Mn(s) + 4CO2(g)$

The number of particle also increases, indicating a increase in entropy.

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

$ΔH°rxn = ∑m ΔH°f(Products)- ∑n ΔH°f(reactants)$

$ΔΗrxno = [(1 mol Mn)(0 kJ/mol ) + (2 mol CO2)(−393.5 kJ/mol )]− [(1 mol MnO2)(−520.9 kJ/mol ) +(2 mol CO)(−110.5 kJ/mol )] ΔΗrxno = -45.1 kJ$

The enthalpy change is negative.

Hence, the enthalpy $(ΔH°rxn)$ value is $-45.1 kJ_$

Entropy change $ΔS°system$

Standard entropy change equation is,

$ΔS°rxn = ∑m S°Products- ∑n S°reactants$

Where, $(m) and (n)$ are the stoichiometric co-efficient.

$ΔSrxno = [(1 mol Mn)(31.8 J/mol⋅K ) + (2 mol CO2)(213.7 J/mol⋅K ) ]− [(1 mol MnO2)(53.1 J/mol⋅K ) +(2 mol CO)(197.5 J/mol⋅K )] ΔSrxno = 11.1 J/K$

Therefore, the $(ΔS°rxn)$ of the reaction is $-91.28 J/K_$

Next calculate the Free energy change $ΔGrxno$

Standared Free energy change equation iss,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Free enrgy change $ΔGfo$

Calcualted enthalpy and entropy values are

$ΔΗrxno=-45.1 kJ$

$ΔSrxno=-91.28 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= −45.1 kJ - [(298 K)(−11.1 J/K)(1 kJ/103J)]ΔGrxno= −48.4078 kJ$

Therefore, the standard free energy value is $ΔGrxno= -48.4 kJ_$

Reaction-C

Given reaction is,

$NH4Cl (s) → NH3(g) + HCl(g)$

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

$ΔH°rxn = ∑m ΔH°f(Products)- ∑n ΔH°f(reactants)$

$ΔΗrxno = [(1 mol NH3)(−45.9 kJ/mol ) + (1 mol HCl)(−92.3 kJ/mol ) ]− [(1 mol NH4Cl)(−314.4 kJ/mol ) ] ΔΗrxno = 176.2 kJ$

The enthalpy change is positive.

Hence, the enthalpy $(ΔH°rxn)$ value is $176.2 kJ_$

Entropy change $ΔS°system$

Standard entropy change equation is,

$ΔS°rxn = ∑m S°Products- ∑n S°reactants$

Where, $(m) and (n)$ are the stoichiometric co-efficient.

$ΔSrxno = [(1 mol NH3 )(193 J/mol⋅K ) + (1 mol HCl)(186.79 J/mol⋅K )] − [(1 mol NH4Cl )(94.6 J/mol⋅K ) ] ΔSrxno = 285.19 J/K$

Therefore, the $(ΔS°rxn)$ of the reaction is $285.19 J/K_$

Finally calculate the Free enrgy change $ΔGrxno$

Standared Free energy change equation iss,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Free energy change $ΔGfo$

Calcualted enthalpy and entropy values are

$ΔΗrxno=176.2 kJ$

$ΔSrxno=285.19 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= 176.2 kJ - [(298 K)(285.19 J/K)(1 kJ/103J)]ΔGrxno= 91.213 kJ$

Therefore, the standard free energy value is $ΔGrxno= 91.2 kJ_$

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Answer 4

Question

Chapter 20, Problem 20.54P

Interpretation Introduction

Interpretation:

For the given set of reactions in problem 20.52 the standard free energy $ΔGo$ value have to be calculated using the $ΔHfo and So$ values.

Concept introduction:

Enthalpy is the amount energy absorbed or released in a process.

The enthalpy change in a system $(ΔΗsys)$ can be calculated by the following equation.

$ΔHrxn = ∑ΔH°produdcts-∑ΔH°reactants$

Where,

$ΔHro(reactants)$ is the standard enthalpy of the reactants

$ΔHro(produdcts)$ is the standard enthalpy of the products

Entropy is the measure of randomness in the system. Standard entropy change in a reaction is the difference in entropy of the products and reactants. $(ΔS°rxn)$ can be calculated by the following equation.

$ΔS°rxn = ∑m S°Products- ∑n S°reactants$

Where,

$S°reactants$ is the standard entropy of the reactants

$S°Products$ is the standard entropy of the products

Standard entropy change in a reaction and entropy change in the system are same.

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

$ΔG°rxn=∑mΔGf°(Products)-∑nΔGf°(Reactants)$

Where,

$nΔGf°(Reactants)$ is the standard entropy of the reactants

$mΔGf°(products)$ is the standard free energy of the products

Expert Solution & Answer

Answer to Problem 20.54P

Reaction-A the standard free energy value is $ΔGrxno= 2.4 kJ_$

Reaction-B the standard free energy value is $ΔGrxno= -48.4 kJ_$

Reaction-C the standard free energy value is $ΔGrxno= 91.2 kJ_$

Explanation of Solution

Reaction-A

Given chemical reaction is,

$H2 (g)+ I2(g) → 2HI(g)$

The number of particle also decreases, indicating a decrease in entropy.

So, given reaction $ΔGfo$ value is zero.

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

$ΔH°rxn = ∑m ΔH°f(Products)- ∑n ΔH°f(reactants)$

$ΔH°rxn = [(2 mol HI) ( ΔH°fof HI)] −[(1 mol H2)(ΔH°f of H2) + (1 mol I2)(ΔH°f of I2)] ΔH°rxn = [(2 mol HI)(25.9 kJ/mol )]− [(1 mol H2)(0 kJ/mol ) +(1 mol I2)(0 kJ/mol )] ΔH°rxn = -51.8 kJ$

The enthalpy change is negative.

Hence, the enthalpy $(ΔH°rxn)$ value is $-51.8 kJ_$

Entropy change $ΔS°system$

Standard entropy change equation is,

$ΔS°rxn = ∑m S°Products- ∑n S°reactants$

Where, $(m) and (n)$ are the stoichiometric co-efficient.

$ΔS°rxn = [(2 mol HI)(206.33 J/mol×K) ]− [(2 mol H2)(130.6 J/mol×K) + (1 mol I2)(116.14 J/mol×K) ] ΔS°rxn =165.92 J/K$

Therefore, the $(ΔS°rxn)$ of the reaction is $165.92 J/K_$

Next calculate the Free enrgy change $ΔGrxno$

Standared Free energy change equation iss,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Free energy change $ΔGfo$

Calcualted enthalpy and entropy values are

$ΔΗrxno=-51.8 kJ$

$ΔSrxno=165.92 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= 51.8 kJ - [(298 K)(165.92 J/K)(1 kJ/103J)]ΔGrxno= 2.3558 kJ$

Therefore, the standard free energy value is $ΔGrxno= 2.4 kJ_$

Reaction-B

Given reaction is,

$MnO2 (s)+ 2CO(g) → Mn(s) + 4CO2(g)$

The number of particle also increases, indicating a increase in entropy.

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

$ΔH°rxn = ∑m ΔH°f(Products)- ∑n ΔH°f(reactants)$

$ΔΗrxno = [(1 mol Mn)(0 kJ/mol ) + (2 mol CO2)(−393.5 kJ/mol )]− [(1 mol MnO2)(−520.9 kJ/mol ) +(2 mol CO)(−110.5 kJ/mol )] ΔΗrxno = -45.1 kJ$

The enthalpy change is negative.

Hence, the enthalpy $(ΔH°rxn)$ value is $-45.1 kJ_$

Entropy change $ΔS°system$

Standard entropy change equation is,

$ΔS°rxn = ∑m S°Products- ∑n S°reactants$

Where, $(m) and (n)$ are the stoichiometric co-efficient.

$ΔSrxno = [(1 mol Mn)(31.8 J/mol⋅K ) + (2 mol CO2)(213.7 J/mol⋅K ) ]− [(1 mol MnO2)(53.1 J/mol⋅K ) +(2 mol CO)(197.5 J/mol⋅K )] ΔSrxno = 11.1 J/K$

Therefore, the $(ΔS°rxn)$ of the reaction is $-91.28 J/K_$

Next calculate the Free energy change $ΔGrxno$

Standared Free energy change equation iss,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Free enrgy change $ΔGfo$

Calcualted enthalpy and entropy values are

$ΔΗrxno=-45.1 kJ$

$ΔSrxno=-91.28 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= −45.1 kJ - [(298 K)(−11.1 J/K)(1 kJ/103J)]ΔGrxno= −48.4078 kJ$

Therefore, the standard free energy value is $ΔGrxno= -48.4 kJ_$

Reaction-C

Given reaction is,

$NH4Cl (s) → NH3(g) + HCl(g)$

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

$ΔH°rxn = ∑m ΔH°f(Products)- ∑n ΔH°f(reactants)$

$ΔΗrxno = [(1 mol NH3)(−45.9 kJ/mol ) + (1 mol HCl)(−92.3 kJ/mol ) ]− [(1 mol NH4Cl)(−314.4 kJ/mol ) ] ΔΗrxno = 176.2 kJ$

The enthalpy change is positive.

Hence, the enthalpy $(ΔH°rxn)$ value is $176.2 kJ_$

Entropy change $ΔS°system$

Standard entropy change equation is,

$ΔS°rxn = ∑m S°Products- ∑n S°reactants$

Where, $(m) and (n)$ are the stoichiometric co-efficient.

$ΔSrxno = [(1 mol NH3 )(193 J/mol⋅K ) + (1 mol HCl)(186.79 J/mol⋅K )] − [(1 mol NH4Cl )(94.6 J/mol⋅K ) ] ΔSrxno = 285.19 J/K$

Therefore, the $(ΔS°rxn)$ of the reaction is $285.19 J/K_$

Finally calculate the Free enrgy change $ΔGrxno$

Standared Free energy change equation iss,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Free energy change $ΔGfo$

Calcualted enthalpy and entropy values are

$ΔΗrxno=176.2 kJ$

$ΔSrxno=285.19 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= 176.2 kJ - [(298 K)(285.19 J/K)(1 kJ/103J)]ΔGrxno= 91.213 kJ$

Therefore, the standard free energy value is $ΔGrxno= 91.2 kJ_$

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Answer 5

Chapter 20, Problem 20.54P

Interpretation Introduction

Interpretation:

For the given set of reactions in problem 20.52 the standard free energy $ΔGo$ value have to be calculated using the $ΔHfo and So$ values.

Concept introduction:

Enthalpy is the amount energy absorbed or released in a process.

The enthalpy change in a system $(ΔΗsys)$ can be calculated by the following equation.

$ΔHrxn = ∑ΔH°produdcts-∑ΔH°reactants$

Where,

$ΔHro(reactants)$ is the standard enthalpy of the reactants

$ΔHro(produdcts)$ is the standard enthalpy of the products

Entropy is the measure of randomness in the system. Standard entropy change in a reaction is the difference in entropy of the products and reactants. $(ΔS°rxn)$ can be calculated by the following equation.

$ΔS°rxn = ∑m S°Products- ∑n S°reactants$

Where,

$S°reactants$ is the standard entropy of the reactants

$S°Products$ is the standard entropy of the products

Standard entropy change in a reaction and entropy change in the system are same.

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

$ΔG°rxn=∑mΔGf°(Products)-∑nΔGf°(Reactants)$

Where,

$nΔGf°(Reactants)$ is the standard entropy of the reactants

$mΔGf°(products)$ is the standard free energy of the products

Expert Solution & Answer

Answer to Problem 20.54P

Reaction-A the standard free energy value is $ΔGrxno= 2.4 kJ_$

Reaction-B the standard free energy value is $ΔGrxno= -48.4 kJ_$

Reaction-C the standard free energy value is $ΔGrxno= 91.2 kJ_$

Explanation of Solution

Reaction-A

Given chemical reaction is,

$H2 (g)+ I2(g) → 2HI(g)$

The number of particle also decreases, indicating a decrease in entropy.

So, given reaction $ΔGfo$ value is zero.

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

$ΔH°rxn = ∑m ΔH°f(Products)- ∑n ΔH°f(reactants)$

$ΔH°rxn = [(2 mol HI) ( ΔH°fof HI)] −[(1 mol H2)(ΔH°f of H2) + (1 mol I2)(ΔH°f of I2)] ΔH°rxn = [(2 mol HI)(25.9 kJ/mol )]− [(1 mol H2)(0 kJ/mol ) +(1 mol I2)(0 kJ/mol )] ΔH°rxn = -51.8 kJ$

The enthalpy change is negative.

Hence, the enthalpy $(ΔH°rxn)$ value is $-51.8 kJ_$

Entropy change $ΔS°system$

Standard entropy change equation is,

$ΔS°rxn = ∑m S°Products- ∑n S°reactants$

Where, $(m) and (n)$ are the stoichiometric co-efficient.

$ΔS°rxn = [(2 mol HI)(206.33 J/mol×K) ]− [(2 mol H2)(130.6 J/mol×K) + (1 mol I2)(116.14 J/mol×K) ] ΔS°rxn =165.92 J/K$

Therefore, the $(ΔS°rxn)$ of the reaction is $165.92 J/K_$

Next calculate the Free enrgy change $ΔGrxno$

Standared Free energy change equation iss,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Free energy change $ΔGfo$

Calcualted enthalpy and entropy values are

$ΔΗrxno=-51.8 kJ$

$ΔSrxno=165.92 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= 51.8 kJ - [(298 K)(165.92 J/K)(1 kJ/103J)]ΔGrxno= 2.3558 kJ$

Therefore, the standard free energy value is $ΔGrxno= 2.4 kJ_$

Reaction-B

Given reaction is,

$MnO2 (s)+ 2CO(g) → Mn(s) + 4CO2(g)$

The number of particle also increases, indicating a increase in entropy.

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

$ΔH°rxn = ∑m ΔH°f(Products)- ∑n ΔH°f(reactants)$

$ΔΗrxno = [(1 mol Mn)(0 kJ/mol ) + (2 mol CO2)(−393.5 kJ/mol )]− [(1 mol MnO2)(−520.9 kJ/mol ) +(2 mol CO)(−110.5 kJ/mol )] ΔΗrxno = -45.1 kJ$

The enthalpy change is negative.

Hence, the enthalpy $(ΔH°rxn)$ value is $-45.1 kJ_$

Entropy change $ΔS°system$

Standard entropy change equation is,

$ΔS°rxn = ∑m S°Products- ∑n S°reactants$

Where, $(m) and (n)$ are the stoichiometric co-efficient.

$ΔSrxno = [(1 mol Mn)(31.8 J/mol⋅K ) + (2 mol CO2)(213.7 J/mol⋅K ) ]− [(1 mol MnO2)(53.1 J/mol⋅K ) +(2 mol CO)(197.5 J/mol⋅K )] ΔSrxno = 11.1 J/K$

Therefore, the $(ΔS°rxn)$ of the reaction is $-91.28 J/K_$

Next calculate the Free energy change $ΔGrxno$

Standared Free energy change equation iss,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Free enrgy change $ΔGfo$

Calcualted enthalpy and entropy values are

$ΔΗrxno=-45.1 kJ$

$ΔSrxno=-91.28 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= −45.1 kJ - [(298 K)(−11.1 J/K)(1 kJ/103J)]ΔGrxno= −48.4078 kJ$

Therefore, the standard free energy value is $ΔGrxno= -48.4 kJ_$

Reaction-C

Given reaction is,

$NH4Cl (s) → NH3(g) + HCl(g)$

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

$ΔH°rxn = ∑m ΔH°f(Products)- ∑n ΔH°f(reactants)$

$ΔΗrxno = [(1 mol NH3)(−45.9 kJ/mol ) + (1 mol HCl)(−92.3 kJ/mol ) ]− [(1 mol NH4Cl)(−314.4 kJ/mol ) ] ΔΗrxno = 176.2 kJ$

The enthalpy change is positive.

Hence, the enthalpy $(ΔH°rxn)$ value is $176.2 kJ_$

Entropy change $ΔS°system$

Standard entropy change equation is,

$ΔS°rxn = ∑m S°Products- ∑n S°reactants$

Where, $(m) and (n)$ are the stoichiometric co-efficient.

$ΔSrxno = [(1 mol NH3 )(193 J/mol⋅K ) + (1 mol HCl)(186.79 J/mol⋅K )] − [(1 mol NH4Cl )(94.6 J/mol⋅K ) ] ΔSrxno = 285.19 J/K$

Therefore, the $(ΔS°rxn)$ of the reaction is $285.19 J/K_$

Finally calculate the Free enrgy change $ΔGrxno$

Standared Free energy change equation iss,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Free energy change $ΔGfo$

Calcualted enthalpy and entropy values are

$ΔΗrxno=176.2 kJ$

$ΔSrxno=285.19 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= 176.2 kJ - [(298 K)(285.19 J/K)(1 kJ/103J)]ΔGrxno= 91.213 kJ$

Therefore, the standard free energy value is $ΔGrxno= 91.2 kJ_$

Answer 6

Chapter 20, Problem 20.54P

Interpretation Introduction

Interpretation:

For the given set of reactions in problem 20.52 the standard free energy $ΔGo$ value have to be calculated using the $ΔHfo and So$ values.

Concept introduction:

Enthalpy is the amount energy absorbed or released in a process.

The enthalpy change in a system $(ΔΗsys)$ can be calculated by the following equation.

$ΔHrxn = ∑ΔH°produdcts-∑ΔH°reactants$

Where,

$ΔHro(reactants)$ is the standard enthalpy of the reactants

$ΔHro(produdcts)$ is the standard enthalpy of the products

Entropy is the measure of randomness in the system. Standard entropy change in a reaction is the difference in entropy of the products and reactants. $(ΔS°rxn)$ can be calculated by the following equation.

$ΔS°rxn = ∑m S°Products- ∑n S°reactants$

Where,

$S°reactants$ is the standard entropy of the reactants

$S°Products$ is the standard entropy of the products

Standard entropy change in a reaction and entropy change in the system are same.

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

$ΔG°rxn=∑mΔGf°(Products)-∑nΔGf°(Reactants)$

Where,

$nΔGf°(Reactants)$ is the standard entropy of the reactants

$mΔGf°(products)$ is the standard free energy of the products

Expert Solution & Answer

Answer to Problem 20.54P

Reaction-A the standard free energy value is $ΔGrxno= 2.4 kJ_$

Reaction-B the standard free energy value is $ΔGrxno= -48.4 kJ_$

Reaction-C the standard free energy value is $ΔGrxno= 91.2 kJ_$

Explanation of Solution

Reaction-A

Given chemical reaction is,

$H2 (g)+ I2(g) → 2HI(g)$

The number of particle also decreases, indicating a decrease in entropy.

So, given reaction $ΔGfo$ value is zero.

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

$ΔH°rxn = ∑m ΔH°f(Products)- ∑n ΔH°f(reactants)$

$ΔH°rxn = [(2 mol HI) ( ΔH°fof HI)] −[(1 mol H2)(ΔH°f of H2) + (1 mol I2)(ΔH°f of I2)] ΔH°rxn = [(2 mol HI)(25.9 kJ/mol )]− [(1 mol H2)(0 kJ/mol ) +(1 mol I2)(0 kJ/mol )] ΔH°rxn = -51.8 kJ$

The enthalpy change is negative.

Hence, the enthalpy $(ΔH°rxn)$ value is $-51.8 kJ_$

Entropy change $ΔS°system$

Standard entropy change equation is,

$ΔS°rxn = ∑m S°Products- ∑n S°reactants$

Where, $(m) and (n)$ are the stoichiometric co-efficient.

$ΔS°rxn = [(2 mol HI)(206.33 J/mol×K) ]− [(2 mol H2)(130.6 J/mol×K) + (1 mol I2)(116.14 J/mol×K) ] ΔS°rxn =165.92 J/K$

Therefore, the $(ΔS°rxn)$ of the reaction is $165.92 J/K_$

Next calculate the Free enrgy change $ΔGrxno$

Standared Free energy change equation iss,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Free energy change $ΔGfo$

Calcualted enthalpy and entropy values are

$ΔΗrxno=-51.8 kJ$

$ΔSrxno=165.92 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= 51.8 kJ - [(298 K)(165.92 J/K)(1 kJ/103J)]ΔGrxno= 2.3558 kJ$

Therefore, the standard free energy value is $ΔGrxno= 2.4 kJ_$

Reaction-B

Given reaction is,

$MnO2 (s)+ 2CO(g) → Mn(s) + 4CO2(g)$

The number of particle also increases, indicating a increase in entropy.

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

$ΔH°rxn = ∑m ΔH°f(Products)- ∑n ΔH°f(reactants)$

$ΔΗrxno = [(1 mol Mn)(0 kJ/mol ) + (2 mol CO2)(−393.5 kJ/mol )]− [(1 mol MnO2)(−520.9 kJ/mol ) +(2 mol CO)(−110.5 kJ/mol )] ΔΗrxno = -45.1 kJ$

The enthalpy change is negative.

Hence, the enthalpy $(ΔH°rxn)$ value is $-45.1 kJ_$

Entropy change $ΔS°system$

Standard entropy change equation is,

$ΔS°rxn = ∑m S°Products- ∑n S°reactants$

Where, $(m) and (n)$ are the stoichiometric co-efficient.

$ΔSrxno = [(1 mol Mn)(31.8 J/mol⋅K ) + (2 mol CO2)(213.7 J/mol⋅K ) ]− [(1 mol MnO2)(53.1 J/mol⋅K ) +(2 mol CO)(197.5 J/mol⋅K )] ΔSrxno = 11.1 J/K$

Therefore, the $(ΔS°rxn)$ of the reaction is $-91.28 J/K_$

Next calculate the Free energy change $ΔGrxno$

Standared Free energy change equation iss,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Free enrgy change $ΔGfo$

Calcualted enthalpy and entropy values are

$ΔΗrxno=-45.1 kJ$

$ΔSrxno=-91.28 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= −45.1 kJ - [(298 K)(−11.1 J/K)(1 kJ/103J)]ΔGrxno= −48.4078 kJ$

Therefore, the standard free energy value is $ΔGrxno= -48.4 kJ_$

Reaction-C

Given reaction is,

$NH4Cl (s) → NH3(g) + HCl(g)$

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

$ΔH°rxn = ∑m ΔH°f(Products)- ∑n ΔH°f(reactants)$

$ΔΗrxno = [(1 mol NH3)(−45.9 kJ/mol ) + (1 mol HCl)(−92.3 kJ/mol ) ]− [(1 mol NH4Cl)(−314.4 kJ/mol ) ] ΔΗrxno = 176.2 kJ$

The enthalpy change is positive.

Hence, the enthalpy $(ΔH°rxn)$ value is $176.2 kJ_$

Entropy change $ΔS°system$

Standard entropy change equation is,

$ΔS°rxn = ∑m S°Products- ∑n S°reactants$

Where, $(m) and (n)$ are the stoichiometric co-efficient.

$ΔSrxno = [(1 mol NH3 )(193 J/mol⋅K ) + (1 mol HCl)(186.79 J/mol⋅K )] − [(1 mol NH4Cl )(94.6 J/mol⋅K ) ] ΔSrxno = 285.19 J/K$

Therefore, the $(ΔS°rxn)$ of the reaction is $285.19 J/K_$

Finally calculate the Free enrgy change $ΔGrxno$

Standared Free energy change equation iss,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Free energy change $ΔGfo$

Calcualted enthalpy and entropy values are

$ΔΗrxno=176.2 kJ$

$ΔSrxno=285.19 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= 176.2 kJ - [(298 K)(285.19 J/K)(1 kJ/103J)]ΔGrxno= 91.213 kJ$

Therefore, the standard free energy value is $ΔGrxno= 91.2 kJ_$

Answer 7

Chapter 20, Problem 20.54P

Interpretation Introduction

Interpretation:

For the given set of reactions in problem 20.52 the standard free energy $ΔGo$ value have to be calculated using the $ΔHfo and So$ values.

Concept introduction:

Enthalpy is the amount energy absorbed or released in a process.

The enthalpy change in a system $(ΔΗsys)$ can be calculated by the following equation.

$ΔHrxn = ∑ΔH°produdcts-∑ΔH°reactants$

Where,

$ΔHro(reactants)$ is the standard enthalpy of the reactants

$ΔHro(produdcts)$ is the standard enthalpy of the products

Entropy is the measure of randomness in the system. Standard entropy change in a reaction is the difference in entropy of the products and reactants. $(ΔS°rxn)$ can be calculated by the following equation.

$ΔS°rxn = ∑m S°Products- ∑n S°reactants$

Where,

$S°reactants$ is the standard entropy of the reactants

$S°Products$ is the standard entropy of the products

Standard entropy change in a reaction and entropy change in the system are same.

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

$ΔG°rxn=∑mΔGf°(Products)-∑nΔGf°(Reactants)$

Where,

$nΔGf°(Reactants)$ is the standard entropy of the reactants

$mΔGf°(products)$ is the standard free energy of the products

Answer 8

Expert Solution & Answer

Answer to Problem 20.54P

Reaction-A the standard free energy value is $ΔGrxno= 2.4 kJ_$

Reaction-B the standard free energy value is $ΔGrxno= -48.4 kJ_$

Reaction-C the standard free energy value is $ΔGrxno= 91.2 kJ_$

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Reaction-A

Given chemical reaction is,

$H2 (g)+ I2(g) → 2HI(g)$

The number of particle also decreases, indicating a decrease in entropy.

So, given reaction $ΔGfo$ value is zero.

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

$ΔH°rxn = ∑m ΔH°f(Products)- ∑n ΔH°f(reactants)$

$ΔH°rxn = [(2 mol HI) ( ΔH°fof HI)] −[(1 mol H2)(ΔH°f of H2) + (1 mol I2)(ΔH°f of I2)] ΔH°rxn = [(2 mol HI)(25.9 kJ/mol )]− [(1 mol H2)(0 kJ/mol ) +(1 mol I2)(0 kJ/mol )] ΔH°rxn = -51.8 kJ$

The enthalpy change is negative.

Hence, the enthalpy $(ΔH°rxn)$ value is $-51.8 kJ_$

Entropy change $ΔS°system$

Standard entropy change equation is,

$ΔS°rxn = ∑m S°Products- ∑n S°reactants$

Where, $(m) and (n)$ are the stoichiometric co-efficient.

$ΔS°rxn = [(2 mol HI)(206.33 J/mol×K) ]− [(2 mol H2)(130.6 J/mol×K) + (1 mol I2)(116.14 J/mol×K) ] ΔS°rxn =165.92 J/K$

Therefore, the $(ΔS°rxn)$ of the reaction is $165.92 J/K_$

Next calculate the Free enrgy change $ΔGrxno$

Standared Free energy change equation iss,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Free energy change $ΔGfo$

Calcualted enthalpy and entropy values are

$ΔΗrxno=-51.8 kJ$

$ΔSrxno=165.92 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= 51.8 kJ - [(298 K)(165.92 J/K)(1 kJ/103J)]ΔGrxno= 2.3558 kJ$

Therefore, the standard free energy value is $ΔGrxno= 2.4 kJ_$

Reaction-B

Given reaction is,

$MnO2 (s)+ 2CO(g) → Mn(s) + 4CO2(g)$

The number of particle also increases, indicating a increase in entropy.

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

$ΔH°rxn = ∑m ΔH°f(Products)- ∑n ΔH°f(reactants)$

$ΔΗrxno = [(1 mol Mn)(0 kJ/mol ) + (2 mol CO2)(−393.5 kJ/mol )]− [(1 mol MnO2)(−520.9 kJ/mol ) +(2 mol CO)(−110.5 kJ/mol )] ΔΗrxno = -45.1 kJ$

The enthalpy change is negative.

Hence, the enthalpy $(ΔH°rxn)$ value is $-45.1 kJ_$

Entropy change $ΔS°system$

Standard entropy change equation is,

$ΔS°rxn = ∑m S°Products- ∑n S°reactants$

Where, $(m) and (n)$ are the stoichiometric co-efficient.

$ΔSrxno = [(1 mol Mn)(31.8 J/mol⋅K ) + (2 mol CO2)(213.7 J/mol⋅K ) ]− [(1 mol MnO2)(53.1 J/mol⋅K ) +(2 mol CO)(197.5 J/mol⋅K )] ΔSrxno = 11.1 J/K$

Therefore, the $(ΔS°rxn)$ of the reaction is $-91.28 J/K_$

Next calculate the Free energy change $ΔGrxno$

Standared Free energy change equation iss,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Free enrgy change $ΔGfo$

Calcualted enthalpy and entropy values are

$ΔΗrxno=-45.1 kJ$

$ΔSrxno=-91.28 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= −45.1 kJ - [(298 K)(−11.1 J/K)(1 kJ/103J)]ΔGrxno= −48.4078 kJ$

Therefore, the standard free energy value is $ΔGrxno= -48.4 kJ_$

Reaction-C

Given reaction is,

$NH4Cl (s) → NH3(g) + HCl(g)$

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

$ΔH°rxn = ∑m ΔH°f(Products)- ∑n ΔH°f(reactants)$

$ΔΗrxno = [(1 mol NH3)(−45.9 kJ/mol ) + (1 mol HCl)(−92.3 kJ/mol ) ]− [(1 mol NH4Cl)(−314.4 kJ/mol ) ] ΔΗrxno = 176.2 kJ$

The enthalpy change is positive.

Hence, the enthalpy $(ΔH°rxn)$ value is $176.2 kJ_$

Entropy change $ΔS°system$

Standard entropy change equation is,

$ΔS°rxn = ∑m S°Products- ∑n S°reactants$

Where, $(m) and (n)$ are the stoichiometric co-efficient.

$ΔSrxno = [(1 mol NH3 )(193 J/mol⋅K ) + (1 mol HCl)(186.79 J/mol⋅K )] − [(1 mol NH4Cl )(94.6 J/mol⋅K ) ] ΔSrxno = 285.19 J/K$

Therefore, the $(ΔS°rxn)$ of the reaction is $285.19 J/K_$

Finally calculate the Free enrgy change $ΔGrxno$

Standared Free energy change equation iss,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Free energy change $ΔGfo$

Calcualted enthalpy and entropy values are

$ΔΗrxno=176.2 kJ$

$ΔSrxno=285.19 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= 176.2 kJ - [(298 K)(285.19 J/K)(1 kJ/103J)]ΔGrxno= 91.213 kJ$

Therefore, the standard free energy value is $ΔGrxno= 91.2 kJ_$

Answer 12

Ch. 20.1 - Select the substance with the higher entropy in...Ch. 20.1 - Select the substance with the lower entropy in...Ch. 20.2 - Balance each equation, predict the sign of if...Ch. 20.2 - Balance each equation, predict the sign of if...Ch. 20.2 - Gaseous phosphorus trichloride forms from the...Ch. 20.2 - Prob. 20.3BFP Ch. 20.3 - Use and to calculate at 298 K for this...Ch. 20.3 - Prob. 20.4BFP Ch. 20.3 - Use values to calculate at 298 K: 2NO(g) +...Ch. 20.3 - Prob. 20.5BFP

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