For the given set of reactions in problem 20.51 the 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 ° reactants is the standard enthalpy of the reactants ΔH ° 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 20.53P

Interpretation Introduction

Interpretation:

For the given set of reactions in problem 20.51 the 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,

$ΔH°reactants$ is the standard enthalpy of the reactants

$ΔH°produdcts$ 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 20.53P

Reaction-A the standard free energy value is $ΔGrxno= -1138 kJ_$

Reaction-B the standard free energy value is $ΔGrxno= -1379 kJ_$

Reaction-C the standard free energy value is $ΔGrxno= -226 kJ_$

Explanation of Solution

Reaction-A

Given,

Chemical reaction is,

$BaO(s)+ CO2(g) → BaCO3(s)$

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

So, there $ΔGfo$ values are zero the solid is less than gas.

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

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

$ΔH°rxn = [(1 mol MgO) ( ΔH°fof MgO)] −[(2 mol Mg)(ΔH°f of Mg) + (1 mol O2)(ΔH°f of O2)] ΔH°rxn = [(2 mol MgO)(−601.2 kJ/mol ) ]− [(2 mol Mg)(0 kJ/mol ) +(1 mol O2)(0 kJ/mol )] ΔH°rxn = -1202.4 kJ$

The enthalpy change is negative. Hence, the enthalpy $(ΔH°rxn)$ value is $-1202.4 kJ_$

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.

$ΔS°rxn = [(1 mol MgO) (So of MgO)]− [ (2 mol Mg)(So of Mg) + (1mol O2) (So of O2)] ΔS°rxn = [(1 mol MgO)(26.9 J/mol×K) ]− [(2 mol N2)(32.69 J/mol×K) + (1 mol O2)(205.0 J/mol×K) ] ΔS°rxn =-216.58 J/K$

Therefore, the $(ΔS°rxn)$ of the reaction is $-216.58 J/K_$

Next calculate the Free enrgy change $ΔGrxno$

Standared Free energy change equation iss,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Free enrgy change $ΔGfo$

Calcualted enthalpy and entropy values are

$ΔΗrxno=-1202.4 kJ$

$ΔSrxno=-216.58 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= −1202.4 kJ - [(298 K)(−216.58 J/K)(1 kJ/103J)]ΔGrxno= −1137.859 kJ$

Therefore, the standard free energy value is $ΔGrxno= -1138 kJ_$

Reaction-B

Given reaction is,

$2CH3OH (g)+ 3O2(g) → 2CO2(g) + 4H2O(g)$

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

So, there $ΔGfo$ values are zero the solid is less than gas.

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

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

$ΔΗrxno= [(2 mol CO2) ( ΔΗrxnoof CO2) + (4 mol H2O) ( ΔΗrxnoof H2O) ] −[(2 mol CH3OH)(ΔΗrxnoof CH3OH) + (3 mol O2)(ΔΗrxno of O2) ] ΔΗrxno = [(2 mol CO2)(−393.5 kJ/mol ) + (4 mol H2O)(−241.826 kJ/mol ) ]− [(2 mol CH3OH)(−201.2 kJ/mol ) +(3 mol O2)(0 kJ/mol )] ΔΗrxno = -1351.904 kJ$

The enthalpy change is negative.

Hence, the enthalpy $(ΔH°rxn)$ value is $-1351.904 kJ_$

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.

$ΔSrxno= [(2 mol CO2) ( Soof CO2) + (4 mol H2O) ( Soof H2O) ] −[(2 mol CH3OH)( Soof CH3OH) + (3 mol O2)( So of O2) ] ΔSrxno = [(2 mol CO2)(213.7 J/mol⋅K ) + (4 mol H2O)(188.72 J/mol⋅K ) ]− [(2 mol CH3OH)(238 J/mol⋅K ) +(3 mol O2)(205.0 J/mol⋅K )] ΔSrxno = -91.28 J/K$

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

$ΔΗrxno=-1351.904 kJ$

$ΔSrxno=-91.28 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= −1351.904 kJ - [(298 K)(−91.28 J/K)(1 kJ/103J)]ΔGrxno= −1379.105 kJ$

Therefore, the standard free energy value is $ΔGrxno= -1379 kJ_$

Reaction-C

Given reaction is,

$BaO(s)+ CO2(g) → BaCO3(s)$

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

So, there $ΔGfo$ values are zero the solid is less than gas.

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

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

$ΔΗrxno= [(1 mol BaCO3) ( ΔΗrxnoof BaCO3)] −[(1 mol BaO)(ΔΗrxnoof BaO) + (1 mol CO2)(ΔΗrxno of CO2)] ΔΗrxno = [(1 mol BaCO3)(−1219 kJ/mol )]− [(1 mol BaO)(−548.1 kJ/mol ) +(1 mol CO2)(−393.5 kJ/mol )] ΔΗrxno = -277.4 kJ$

The enthalpy change is negative. Hence, the enthalpy $(ΔH°rxn)$ value is $-277.4 kJ_$

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.

$ΔSrxno= [(2 mol BaCO3) ( Soof BaCO3)] −[(1 mol BaO)( Soof BaO) + (1 mol CO2)( So of CO2) ] ΔSrxno = [(1 mol )(112 J/mol⋅K )]− [(1 mol )(72.07 J/mol⋅K ) +(1 mol)(213.7 J/mol⋅K )] ΔSrxno = -173.77 J/K$

Therefore, the $(ΔS°rxn)$ of the reaction is $-173.77 J/K_$

Finally calculate the Free enrgy change $ΔGrxno$

Standared Free energy change equation iss,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Free enrgy change $ΔGfo$

Calcualted enthalpy and entropy values are

$ΔΗrxno=-277.4 kJ$

$ΔSrxno=-173.77 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= −277.4 kJ - [(298 K)(−173.77 J/K)(1 kJ/103J)]ΔGrxno= −225.6265 kJ$

Therefore, the standard free energy value is $ΔGrxno= -226 kJ_$

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Answer 2

Question

Chapter 20, Problem 20.53P

Interpretation Introduction

Interpretation:

For the given set of reactions in problem 20.51 the 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,

$ΔH°reactants$ is the standard enthalpy of the reactants

$ΔH°produdcts$ 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 20.53P

Reaction-A the standard free energy value is $ΔGrxno= -1138 kJ_$

Reaction-B the standard free energy value is $ΔGrxno= -1379 kJ_$

Reaction-C the standard free energy value is $ΔGrxno= -226 kJ_$

Explanation of Solution

Reaction-A

Given,

Chemical reaction is,

$BaO(s)+ CO2(g) → BaCO3(s)$

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

So, there $ΔGfo$ values are zero the solid is less than gas.

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

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

$ΔH°rxn = [(1 mol MgO) ( ΔH°fof MgO)] −[(2 mol Mg)(ΔH°f of Mg) + (1 mol O2)(ΔH°f of O2)] ΔH°rxn = [(2 mol MgO)(−601.2 kJ/mol ) ]− [(2 mol Mg)(0 kJ/mol ) +(1 mol O2)(0 kJ/mol )] ΔH°rxn = -1202.4 kJ$

The enthalpy change is negative. Hence, the enthalpy $(ΔH°rxn)$ value is $-1202.4 kJ_$

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.

$ΔS°rxn = [(1 mol MgO) (So of MgO)]− [ (2 mol Mg)(So of Mg) + (1mol O2) (So of O2)] ΔS°rxn = [(1 mol MgO)(26.9 J/mol×K) ]− [(2 mol N2)(32.69 J/mol×K) + (1 mol O2)(205.0 J/mol×K) ] ΔS°rxn =-216.58 J/K$

Therefore, the $(ΔS°rxn)$ of the reaction is $-216.58 J/K_$

Next calculate the Free enrgy change $ΔGrxno$

Standared Free energy change equation iss,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Free enrgy change $ΔGfo$

Calcualted enthalpy and entropy values are

$ΔΗrxno=-1202.4 kJ$

$ΔSrxno=-216.58 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= −1202.4 kJ - [(298 K)(−216.58 J/K)(1 kJ/103J)]ΔGrxno= −1137.859 kJ$

Therefore, the standard free energy value is $ΔGrxno= -1138 kJ_$

Reaction-B

Given reaction is,

$2CH3OH (g)+ 3O2(g) → 2CO2(g) + 4H2O(g)$

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

So, there $ΔGfo$ values are zero the solid is less than gas.

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

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

$ΔΗrxno= [(2 mol CO2) ( ΔΗrxnoof CO2) + (4 mol H2O) ( ΔΗrxnoof H2O) ] −[(2 mol CH3OH)(ΔΗrxnoof CH3OH) + (3 mol O2)(ΔΗrxno of O2) ] ΔΗrxno = [(2 mol CO2)(−393.5 kJ/mol ) + (4 mol H2O)(−241.826 kJ/mol ) ]− [(2 mol CH3OH)(−201.2 kJ/mol ) +(3 mol O2)(0 kJ/mol )] ΔΗrxno = -1351.904 kJ$

The enthalpy change is negative.

Hence, the enthalpy $(ΔH°rxn)$ value is $-1351.904 kJ_$

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.

$ΔSrxno= [(2 mol CO2) ( Soof CO2) + (4 mol H2O) ( Soof H2O) ] −[(2 mol CH3OH)( Soof CH3OH) + (3 mol O2)( So of O2) ] ΔSrxno = [(2 mol CO2)(213.7 J/mol⋅K ) + (4 mol H2O)(188.72 J/mol⋅K ) ]− [(2 mol CH3OH)(238 J/mol⋅K ) +(3 mol O2)(205.0 J/mol⋅K )] ΔSrxno = -91.28 J/K$

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

$ΔΗrxno=-1351.904 kJ$

$ΔSrxno=-91.28 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= −1351.904 kJ - [(298 K)(−91.28 J/K)(1 kJ/103J)]ΔGrxno= −1379.105 kJ$

Therefore, the standard free energy value is $ΔGrxno= -1379 kJ_$

Reaction-C

Given reaction is,

$BaO(s)+ CO2(g) → BaCO3(s)$

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

So, there $ΔGfo$ values are zero the solid is less than gas.

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

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

$ΔΗrxno= [(1 mol BaCO3) ( ΔΗrxnoof BaCO3)] −[(1 mol BaO)(ΔΗrxnoof BaO) + (1 mol CO2)(ΔΗrxno of CO2)] ΔΗrxno = [(1 mol BaCO3)(−1219 kJ/mol )]− [(1 mol BaO)(−548.1 kJ/mol ) +(1 mol CO2)(−393.5 kJ/mol )] ΔΗrxno = -277.4 kJ$

The enthalpy change is negative. Hence, the enthalpy $(ΔH°rxn)$ value is $-277.4 kJ_$

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.

$ΔSrxno= [(2 mol BaCO3) ( Soof BaCO3)] −[(1 mol BaO)( Soof BaO) + (1 mol CO2)( So of CO2) ] ΔSrxno = [(1 mol )(112 J/mol⋅K )]− [(1 mol )(72.07 J/mol⋅K ) +(1 mol)(213.7 J/mol⋅K )] ΔSrxno = -173.77 J/K$

Therefore, the $(ΔS°rxn)$ of the reaction is $-173.77 J/K_$

Finally calculate the Free enrgy change $ΔGrxno$

Standared Free energy change equation iss,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Free enrgy change $ΔGfo$

Calcualted enthalpy and entropy values are

$ΔΗrxno=-277.4 kJ$

$ΔSrxno=-173.77 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= −277.4 kJ - [(298 K)(−173.77 J/K)(1 kJ/103J)]ΔGrxno= −225.6265 kJ$

Therefore, the standard free energy value is $ΔGrxno= -226 kJ_$

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Answer 3

Question

Chapter 20, Problem 20.53P

Interpretation Introduction

Interpretation:

For the given set of reactions in problem 20.51 the 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,

$ΔH°reactants$ is the standard enthalpy of the reactants

$ΔH°produdcts$ 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 20.53P

Reaction-A the standard free energy value is $ΔGrxno= -1138 kJ_$

Reaction-B the standard free energy value is $ΔGrxno= -1379 kJ_$

Reaction-C the standard free energy value is $ΔGrxno= -226 kJ_$

Explanation of Solution

Reaction-A

Given,

Chemical reaction is,

$BaO(s)+ CO2(g) → BaCO3(s)$

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

So, there $ΔGfo$ values are zero the solid is less than gas.

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

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

$ΔH°rxn = [(1 mol MgO) ( ΔH°fof MgO)] −[(2 mol Mg)(ΔH°f of Mg) + (1 mol O2)(ΔH°f of O2)] ΔH°rxn = [(2 mol MgO)(−601.2 kJ/mol ) ]− [(2 mol Mg)(0 kJ/mol ) +(1 mol O2)(0 kJ/mol )] ΔH°rxn = -1202.4 kJ$

The enthalpy change is negative. Hence, the enthalpy $(ΔH°rxn)$ value is $-1202.4 kJ_$

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.

$ΔS°rxn = [(1 mol MgO) (So of MgO)]− [ (2 mol Mg)(So of Mg) + (1mol O2) (So of O2)] ΔS°rxn = [(1 mol MgO)(26.9 J/mol×K) ]− [(2 mol N2)(32.69 J/mol×K) + (1 mol O2)(205.0 J/mol×K) ] ΔS°rxn =-216.58 J/K$

Therefore, the $(ΔS°rxn)$ of the reaction is $-216.58 J/K_$

Next calculate the Free enrgy change $ΔGrxno$

Standared Free energy change equation iss,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Free enrgy change $ΔGfo$

Calcualted enthalpy and entropy values are

$ΔΗrxno=-1202.4 kJ$

$ΔSrxno=-216.58 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= −1202.4 kJ - [(298 K)(−216.58 J/K)(1 kJ/103J)]ΔGrxno= −1137.859 kJ$

Therefore, the standard free energy value is $ΔGrxno= -1138 kJ_$

Reaction-B

Given reaction is,

$2CH3OH (g)+ 3O2(g) → 2CO2(g) + 4H2O(g)$

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

So, there $ΔGfo$ values are zero the solid is less than gas.

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

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

$ΔΗrxno= [(2 mol CO2) ( ΔΗrxnoof CO2) + (4 mol H2O) ( ΔΗrxnoof H2O) ] −[(2 mol CH3OH)(ΔΗrxnoof CH3OH) + (3 mol O2)(ΔΗrxno of O2) ] ΔΗrxno = [(2 mol CO2)(−393.5 kJ/mol ) + (4 mol H2O)(−241.826 kJ/mol ) ]− [(2 mol CH3OH)(−201.2 kJ/mol ) +(3 mol O2)(0 kJ/mol )] ΔΗrxno = -1351.904 kJ$

The enthalpy change is negative.

Hence, the enthalpy $(ΔH°rxn)$ value is $-1351.904 kJ_$

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.

$ΔSrxno= [(2 mol CO2) ( Soof CO2) + (4 mol H2O) ( Soof H2O) ] −[(2 mol CH3OH)( Soof CH3OH) + (3 mol O2)( So of O2) ] ΔSrxno = [(2 mol CO2)(213.7 J/mol⋅K ) + (4 mol H2O)(188.72 J/mol⋅K ) ]− [(2 mol CH3OH)(238 J/mol⋅K ) +(3 mol O2)(205.0 J/mol⋅K )] ΔSrxno = -91.28 J/K$

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

$ΔΗrxno=-1351.904 kJ$

$ΔSrxno=-91.28 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= −1351.904 kJ - [(298 K)(−91.28 J/K)(1 kJ/103J)]ΔGrxno= −1379.105 kJ$

Therefore, the standard free energy value is $ΔGrxno= -1379 kJ_$

Reaction-C

Given reaction is,

$BaO(s)+ CO2(g) → BaCO3(s)$

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

So, there $ΔGfo$ values are zero the solid is less than gas.

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

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

$ΔΗrxno= [(1 mol BaCO3) ( ΔΗrxnoof BaCO3)] −[(1 mol BaO)(ΔΗrxnoof BaO) + (1 mol CO2)(ΔΗrxno of CO2)] ΔΗrxno = [(1 mol BaCO3)(−1219 kJ/mol )]− [(1 mol BaO)(−548.1 kJ/mol ) +(1 mol CO2)(−393.5 kJ/mol )] ΔΗrxno = -277.4 kJ$

The enthalpy change is negative. Hence, the enthalpy $(ΔH°rxn)$ value is $-277.4 kJ_$

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.

$ΔSrxno= [(2 mol BaCO3) ( Soof BaCO3)] −[(1 mol BaO)( Soof BaO) + (1 mol CO2)( So of CO2) ] ΔSrxno = [(1 mol )(112 J/mol⋅K )]− [(1 mol )(72.07 J/mol⋅K ) +(1 mol)(213.7 J/mol⋅K )] ΔSrxno = -173.77 J/K$

Therefore, the $(ΔS°rxn)$ of the reaction is $-173.77 J/K_$

Finally calculate the Free enrgy change $ΔGrxno$

Standared Free energy change equation iss,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Free enrgy change $ΔGfo$

Calcualted enthalpy and entropy values are

$ΔΗrxno=-277.4 kJ$

$ΔSrxno=-173.77 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= −277.4 kJ - [(298 K)(−173.77 J/K)(1 kJ/103J)]ΔGrxno= −225.6265 kJ$

Therefore, the standard free energy value is $ΔGrxno= -226 kJ_$

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Answer 4

Question

Chapter 20, Problem 20.53P

Interpretation Introduction

Interpretation:

For the given set of reactions in problem 20.51 the 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,

$ΔH°reactants$ is the standard enthalpy of the reactants

$ΔH°produdcts$ 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 20.53P

Reaction-A the standard free energy value is $ΔGrxno= -1138 kJ_$

Reaction-B the standard free energy value is $ΔGrxno= -1379 kJ_$

Reaction-C the standard free energy value is $ΔGrxno= -226 kJ_$

Explanation of Solution

Reaction-A

Given,

Chemical reaction is,

$BaO(s)+ CO2(g) → BaCO3(s)$

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

So, there $ΔGfo$ values are zero the solid is less than gas.

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

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

$ΔH°rxn = [(1 mol MgO) ( ΔH°fof MgO)] −[(2 mol Mg)(ΔH°f of Mg) + (1 mol O2)(ΔH°f of O2)] ΔH°rxn = [(2 mol MgO)(−601.2 kJ/mol ) ]− [(2 mol Mg)(0 kJ/mol ) +(1 mol O2)(0 kJ/mol )] ΔH°rxn = -1202.4 kJ$

The enthalpy change is negative. Hence, the enthalpy $(ΔH°rxn)$ value is $-1202.4 kJ_$

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.

$ΔS°rxn = [(1 mol MgO) (So of MgO)]− [ (2 mol Mg)(So of Mg) + (1mol O2) (So of O2)] ΔS°rxn = [(1 mol MgO)(26.9 J/mol×K) ]− [(2 mol N2)(32.69 J/mol×K) + (1 mol O2)(205.0 J/mol×K) ] ΔS°rxn =-216.58 J/K$

Therefore, the $(ΔS°rxn)$ of the reaction is $-216.58 J/K_$

Next calculate the Free enrgy change $ΔGrxno$

Standared Free energy change equation iss,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Free enrgy change $ΔGfo$

Calcualted enthalpy and entropy values are

$ΔΗrxno=-1202.4 kJ$

$ΔSrxno=-216.58 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= −1202.4 kJ - [(298 K)(−216.58 J/K)(1 kJ/103J)]ΔGrxno= −1137.859 kJ$

Therefore, the standard free energy value is $ΔGrxno= -1138 kJ_$

Reaction-B

Given reaction is,

$2CH3OH (g)+ 3O2(g) → 2CO2(g) + 4H2O(g)$

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

So, there $ΔGfo$ values are zero the solid is less than gas.

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

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

$ΔΗrxno= [(2 mol CO2) ( ΔΗrxnoof CO2) + (4 mol H2O) ( ΔΗrxnoof H2O) ] −[(2 mol CH3OH)(ΔΗrxnoof CH3OH) + (3 mol O2)(ΔΗrxno of O2) ] ΔΗrxno = [(2 mol CO2)(−393.5 kJ/mol ) + (4 mol H2O)(−241.826 kJ/mol ) ]− [(2 mol CH3OH)(−201.2 kJ/mol ) +(3 mol O2)(0 kJ/mol )] ΔΗrxno = -1351.904 kJ$

The enthalpy change is negative.

Hence, the enthalpy $(ΔH°rxn)$ value is $-1351.904 kJ_$

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.

$ΔSrxno= [(2 mol CO2) ( Soof CO2) + (4 mol H2O) ( Soof H2O) ] −[(2 mol CH3OH)( Soof CH3OH) + (3 mol O2)( So of O2) ] ΔSrxno = [(2 mol CO2)(213.7 J/mol⋅K ) + (4 mol H2O)(188.72 J/mol⋅K ) ]− [(2 mol CH3OH)(238 J/mol⋅K ) +(3 mol O2)(205.0 J/mol⋅K )] ΔSrxno = -91.28 J/K$

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

$ΔΗrxno=-1351.904 kJ$

$ΔSrxno=-91.28 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= −1351.904 kJ - [(298 K)(−91.28 J/K)(1 kJ/103J)]ΔGrxno= −1379.105 kJ$

Therefore, the standard free energy value is $ΔGrxno= -1379 kJ_$

Reaction-C

Given reaction is,

$BaO(s)+ CO2(g) → BaCO3(s)$

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

So, there $ΔGfo$ values are zero the solid is less than gas.

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

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

$ΔΗrxno= [(1 mol BaCO3) ( ΔΗrxnoof BaCO3)] −[(1 mol BaO)(ΔΗrxnoof BaO) + (1 mol CO2)(ΔΗrxno of CO2)] ΔΗrxno = [(1 mol BaCO3)(−1219 kJ/mol )]− [(1 mol BaO)(−548.1 kJ/mol ) +(1 mol CO2)(−393.5 kJ/mol )] ΔΗrxno = -277.4 kJ$

The enthalpy change is negative. Hence, the enthalpy $(ΔH°rxn)$ value is $-277.4 kJ_$

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.

$ΔSrxno= [(2 mol BaCO3) ( Soof BaCO3)] −[(1 mol BaO)( Soof BaO) + (1 mol CO2)( So of CO2) ] ΔSrxno = [(1 mol )(112 J/mol⋅K )]− [(1 mol )(72.07 J/mol⋅K ) +(1 mol)(213.7 J/mol⋅K )] ΔSrxno = -173.77 J/K$

Therefore, the $(ΔS°rxn)$ of the reaction is $-173.77 J/K_$

Finally calculate the Free enrgy change $ΔGrxno$

Standared Free energy change equation iss,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Free enrgy change $ΔGfo$

Calcualted enthalpy and entropy values are

$ΔΗrxno=-277.4 kJ$

$ΔSrxno=-173.77 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= −277.4 kJ - [(298 K)(−173.77 J/K)(1 kJ/103J)]ΔGrxno= −225.6265 kJ$

Therefore, the standard free energy value is $ΔGrxno= -226 kJ_$

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Answer 5

Chapter 20, Problem 20.53P

Interpretation Introduction

Interpretation:

For the given set of reactions in problem 20.51 the 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,

$ΔH°reactants$ is the standard enthalpy of the reactants

$ΔH°produdcts$ 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 20.53P

Reaction-A the standard free energy value is $ΔGrxno= -1138 kJ_$

Reaction-B the standard free energy value is $ΔGrxno= -1379 kJ_$

Reaction-C the standard free energy value is $ΔGrxno= -226 kJ_$

Explanation of Solution

Reaction-A

Given,

Chemical reaction is,

$BaO(s)+ CO2(g) → BaCO3(s)$

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

So, there $ΔGfo$ values are zero the solid is less than gas.

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

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

$ΔH°rxn = [(1 mol MgO) ( ΔH°fof MgO)] −[(2 mol Mg)(ΔH°f of Mg) + (1 mol O2)(ΔH°f of O2)] ΔH°rxn = [(2 mol MgO)(−601.2 kJ/mol ) ]− [(2 mol Mg)(0 kJ/mol ) +(1 mol O2)(0 kJ/mol )] ΔH°rxn = -1202.4 kJ$

The enthalpy change is negative. Hence, the enthalpy $(ΔH°rxn)$ value is $-1202.4 kJ_$

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.

$ΔS°rxn = [(1 mol MgO) (So of MgO)]− [ (2 mol Mg)(So of Mg) + (1mol O2) (So of O2)] ΔS°rxn = [(1 mol MgO)(26.9 J/mol×K) ]− [(2 mol N2)(32.69 J/mol×K) + (1 mol O2)(205.0 J/mol×K) ] ΔS°rxn =-216.58 J/K$

Therefore, the $(ΔS°rxn)$ of the reaction is $-216.58 J/K_$

Next calculate the Free enrgy change $ΔGrxno$

Standared Free energy change equation iss,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Free enrgy change $ΔGfo$

Calcualted enthalpy and entropy values are

$ΔΗrxno=-1202.4 kJ$

$ΔSrxno=-216.58 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= −1202.4 kJ - [(298 K)(−216.58 J/K)(1 kJ/103J)]ΔGrxno= −1137.859 kJ$

Therefore, the standard free energy value is $ΔGrxno= -1138 kJ_$

Reaction-B

Given reaction is,

$2CH3OH (g)+ 3O2(g) → 2CO2(g) + 4H2O(g)$

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

So, there $ΔGfo$ values are zero the solid is less than gas.

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

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

$ΔΗrxno= [(2 mol CO2) ( ΔΗrxnoof CO2) + (4 mol H2O) ( ΔΗrxnoof H2O) ] −[(2 mol CH3OH)(ΔΗrxnoof CH3OH) + (3 mol O2)(ΔΗrxno of O2) ] ΔΗrxno = [(2 mol CO2)(−393.5 kJ/mol ) + (4 mol H2O)(−241.826 kJ/mol ) ]− [(2 mol CH3OH)(−201.2 kJ/mol ) +(3 mol O2)(0 kJ/mol )] ΔΗrxno = -1351.904 kJ$

The enthalpy change is negative.

Hence, the enthalpy $(ΔH°rxn)$ value is $-1351.904 kJ_$

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.

$ΔSrxno= [(2 mol CO2) ( Soof CO2) + (4 mol H2O) ( Soof H2O) ] −[(2 mol CH3OH)( Soof CH3OH) + (3 mol O2)( So of O2) ] ΔSrxno = [(2 mol CO2)(213.7 J/mol⋅K ) + (4 mol H2O)(188.72 J/mol⋅K ) ]− [(2 mol CH3OH)(238 J/mol⋅K ) +(3 mol O2)(205.0 J/mol⋅K )] ΔSrxno = -91.28 J/K$

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

$ΔΗrxno=-1351.904 kJ$

$ΔSrxno=-91.28 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= −1351.904 kJ - [(298 K)(−91.28 J/K)(1 kJ/103J)]ΔGrxno= −1379.105 kJ$

Therefore, the standard free energy value is $ΔGrxno= -1379 kJ_$

Reaction-C

Given reaction is,

$BaO(s)+ CO2(g) → BaCO3(s)$

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

So, there $ΔGfo$ values are zero the solid is less than gas.

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

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

$ΔΗrxno= [(1 mol BaCO3) ( ΔΗrxnoof BaCO3)] −[(1 mol BaO)(ΔΗrxnoof BaO) + (1 mol CO2)(ΔΗrxno of CO2)] ΔΗrxno = [(1 mol BaCO3)(−1219 kJ/mol )]− [(1 mol BaO)(−548.1 kJ/mol ) +(1 mol CO2)(−393.5 kJ/mol )] ΔΗrxno = -277.4 kJ$

The enthalpy change is negative. Hence, the enthalpy $(ΔH°rxn)$ value is $-277.4 kJ_$

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.

$ΔSrxno= [(2 mol BaCO3) ( Soof BaCO3)] −[(1 mol BaO)( Soof BaO) + (1 mol CO2)( So of CO2) ] ΔSrxno = [(1 mol )(112 J/mol⋅K )]− [(1 mol )(72.07 J/mol⋅K ) +(1 mol)(213.7 J/mol⋅K )] ΔSrxno = -173.77 J/K$

Therefore, the $(ΔS°rxn)$ of the reaction is $-173.77 J/K_$

Finally calculate the Free enrgy change $ΔGrxno$

Standared Free energy change equation iss,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Free enrgy change $ΔGfo$

Calcualted enthalpy and entropy values are

$ΔΗrxno=-277.4 kJ$

$ΔSrxno=-173.77 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= −277.4 kJ - [(298 K)(−173.77 J/K)(1 kJ/103J)]ΔGrxno= −225.6265 kJ$

Therefore, the standard free energy value is $ΔGrxno= -226 kJ_$

Answer 6

Chapter 20, Problem 20.53P

Interpretation Introduction

Interpretation:

For the given set of reactions in problem 20.51 the 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,

$ΔH°reactants$ is the standard enthalpy of the reactants

$ΔH°produdcts$ 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 20.53P

Reaction-A the standard free energy value is $ΔGrxno= -1138 kJ_$

Reaction-B the standard free energy value is $ΔGrxno= -1379 kJ_$

Reaction-C the standard free energy value is $ΔGrxno= -226 kJ_$

Explanation of Solution

Reaction-A

Given,

Chemical reaction is,

$BaO(s)+ CO2(g) → BaCO3(s)$

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

So, there $ΔGfo$ values are zero the solid is less than gas.

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

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

$ΔH°rxn = [(1 mol MgO) ( ΔH°fof MgO)] −[(2 mol Mg)(ΔH°f of Mg) + (1 mol O2)(ΔH°f of O2)] ΔH°rxn = [(2 mol MgO)(−601.2 kJ/mol ) ]− [(2 mol Mg)(0 kJ/mol ) +(1 mol O2)(0 kJ/mol )] ΔH°rxn = -1202.4 kJ$

The enthalpy change is negative. Hence, the enthalpy $(ΔH°rxn)$ value is $-1202.4 kJ_$

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.

$ΔS°rxn = [(1 mol MgO) (So of MgO)]− [ (2 mol Mg)(So of Mg) + (1mol O2) (So of O2)] ΔS°rxn = [(1 mol MgO)(26.9 J/mol×K) ]− [(2 mol N2)(32.69 J/mol×K) + (1 mol O2)(205.0 J/mol×K) ] ΔS°rxn =-216.58 J/K$

Therefore, the $(ΔS°rxn)$ of the reaction is $-216.58 J/K_$

Next calculate the Free enrgy change $ΔGrxno$

Standared Free energy change equation iss,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Free enrgy change $ΔGfo$

Calcualted enthalpy and entropy values are

$ΔΗrxno=-1202.4 kJ$

$ΔSrxno=-216.58 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= −1202.4 kJ - [(298 K)(−216.58 J/K)(1 kJ/103J)]ΔGrxno= −1137.859 kJ$

Therefore, the standard free energy value is $ΔGrxno= -1138 kJ_$

Reaction-B

Given reaction is,

$2CH3OH (g)+ 3O2(g) → 2CO2(g) + 4H2O(g)$

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

So, there $ΔGfo$ values are zero the solid is less than gas.

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

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

$ΔΗrxno= [(2 mol CO2) ( ΔΗrxnoof CO2) + (4 mol H2O) ( ΔΗrxnoof H2O) ] −[(2 mol CH3OH)(ΔΗrxnoof CH3OH) + (3 mol O2)(ΔΗrxno of O2) ] ΔΗrxno = [(2 mol CO2)(−393.5 kJ/mol ) + (4 mol H2O)(−241.826 kJ/mol ) ]− [(2 mol CH3OH)(−201.2 kJ/mol ) +(3 mol O2)(0 kJ/mol )] ΔΗrxno = -1351.904 kJ$

The enthalpy change is negative.

Hence, the enthalpy $(ΔH°rxn)$ value is $-1351.904 kJ_$

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.

$ΔSrxno= [(2 mol CO2) ( Soof CO2) + (4 mol H2O) ( Soof H2O) ] −[(2 mol CH3OH)( Soof CH3OH) + (3 mol O2)( So of O2) ] ΔSrxno = [(2 mol CO2)(213.7 J/mol⋅K ) + (4 mol H2O)(188.72 J/mol⋅K ) ]− [(2 mol CH3OH)(238 J/mol⋅K ) +(3 mol O2)(205.0 J/mol⋅K )] ΔSrxno = -91.28 J/K$

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

$ΔΗrxno=-1351.904 kJ$

$ΔSrxno=-91.28 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= −1351.904 kJ - [(298 K)(−91.28 J/K)(1 kJ/103J)]ΔGrxno= −1379.105 kJ$

Therefore, the standard free energy value is $ΔGrxno= -1379 kJ_$

Reaction-C

Given reaction is,

$BaO(s)+ CO2(g) → BaCO3(s)$

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

So, there $ΔGfo$ values are zero the solid is less than gas.

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

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

$ΔΗrxno= [(1 mol BaCO3) ( ΔΗrxnoof BaCO3)] −[(1 mol BaO)(ΔΗrxnoof BaO) + (1 mol CO2)(ΔΗrxno of CO2)] ΔΗrxno = [(1 mol BaCO3)(−1219 kJ/mol )]− [(1 mol BaO)(−548.1 kJ/mol ) +(1 mol CO2)(−393.5 kJ/mol )] ΔΗrxno = -277.4 kJ$

The enthalpy change is negative. Hence, the enthalpy $(ΔH°rxn)$ value is $-277.4 kJ_$

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.

$ΔSrxno= [(2 mol BaCO3) ( Soof BaCO3)] −[(1 mol BaO)( Soof BaO) + (1 mol CO2)( So of CO2) ] ΔSrxno = [(1 mol )(112 J/mol⋅K )]− [(1 mol )(72.07 J/mol⋅K ) +(1 mol)(213.7 J/mol⋅K )] ΔSrxno = -173.77 J/K$

Therefore, the $(ΔS°rxn)$ of the reaction is $-173.77 J/K_$

Finally calculate the Free enrgy change $ΔGrxno$

Standared Free energy change equation iss,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Free enrgy change $ΔGfo$

Calcualted enthalpy and entropy values are

$ΔΗrxno=-277.4 kJ$

$ΔSrxno=-173.77 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= −277.4 kJ - [(298 K)(−173.77 J/K)(1 kJ/103J)]ΔGrxno= −225.6265 kJ$

Therefore, the standard free energy value is $ΔGrxno= -226 kJ_$

Answer 7

Chapter 20, Problem 20.53P

Interpretation Introduction

Interpretation:

For the given set of reactions in problem 20.51 the 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,

$ΔH°reactants$ is the standard enthalpy of the reactants

$ΔH°produdcts$ 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 20.53P

Reaction-A the standard free energy value is $ΔGrxno= -1138 kJ_$

Reaction-B the standard free energy value is $ΔGrxno= -1379 kJ_$

Reaction-C the standard free energy value is $ΔGrxno= -226 kJ_$

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Reaction-A

Given,

Chemical reaction is,

$BaO(s)+ CO2(g) → BaCO3(s)$

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

So, there $ΔGfo$ values are zero the solid is less than gas.

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

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

$ΔH°rxn = [(1 mol MgO) ( ΔH°fof MgO)] −[(2 mol Mg)(ΔH°f of Mg) + (1 mol O2)(ΔH°f of O2)] ΔH°rxn = [(2 mol MgO)(−601.2 kJ/mol ) ]− [(2 mol Mg)(0 kJ/mol ) +(1 mol O2)(0 kJ/mol )] ΔH°rxn = -1202.4 kJ$

The enthalpy change is negative. Hence, the enthalpy $(ΔH°rxn)$ value is $-1202.4 kJ_$

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.

$ΔS°rxn = [(1 mol MgO) (So of MgO)]− [ (2 mol Mg)(So of Mg) + (1mol O2) (So of O2)] ΔS°rxn = [(1 mol MgO)(26.9 J/mol×K) ]− [(2 mol N2)(32.69 J/mol×K) + (1 mol O2)(205.0 J/mol×K) ] ΔS°rxn =-216.58 J/K$

Therefore, the $(ΔS°rxn)$ of the reaction is $-216.58 J/K_$

Next calculate the Free enrgy change $ΔGrxno$

Standared Free energy change equation iss,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Free enrgy change $ΔGfo$

Calcualted enthalpy and entropy values are

$ΔΗrxno=-1202.4 kJ$

$ΔSrxno=-216.58 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= −1202.4 kJ - [(298 K)(−216.58 J/K)(1 kJ/103J)]ΔGrxno= −1137.859 kJ$

Therefore, the standard free energy value is $ΔGrxno= -1138 kJ_$

Reaction-B

Given reaction is,

$2CH3OH (g)+ 3O2(g) → 2CO2(g) + 4H2O(g)$

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

So, there $ΔGfo$ values are zero the solid is less than gas.

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

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

$ΔΗrxno= [(2 mol CO2) ( ΔΗrxnoof CO2) + (4 mol H2O) ( ΔΗrxnoof H2O) ] −[(2 mol CH3OH)(ΔΗrxnoof CH3OH) + (3 mol O2)(ΔΗrxno of O2) ] ΔΗrxno = [(2 mol CO2)(−393.5 kJ/mol ) + (4 mol H2O)(−241.826 kJ/mol ) ]− [(2 mol CH3OH)(−201.2 kJ/mol ) +(3 mol O2)(0 kJ/mol )] ΔΗrxno = -1351.904 kJ$

The enthalpy change is negative.

Hence, the enthalpy $(ΔH°rxn)$ value is $-1351.904 kJ_$

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.

$ΔSrxno= [(2 mol CO2) ( Soof CO2) + (4 mol H2O) ( Soof H2O) ] −[(2 mol CH3OH)( Soof CH3OH) + (3 mol O2)( So of O2) ] ΔSrxno = [(2 mol CO2)(213.7 J/mol⋅K ) + (4 mol H2O)(188.72 J/mol⋅K ) ]− [(2 mol CH3OH)(238 J/mol⋅K ) +(3 mol O2)(205.0 J/mol⋅K )] ΔSrxno = -91.28 J/K$

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

$ΔΗrxno=-1351.904 kJ$

$ΔSrxno=-91.28 J/K$

These values are plugging above standard free energy equation,

$ΔGrxno= −1351.904 kJ - [(298 K)(−91.28 J/K)(1 kJ/103J)]ΔGrxno= −1379.105 kJ$

Therefore, the standard free energy value is $ΔGrxno= -1379 kJ_$

Reaction-C

Given reaction is,

$BaO(s)+ CO2(g) → BaCO3(s)$

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

So, there $ΔGfo$ values are zero the solid is less than gas.

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

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

$ΔΗrxno= [(1 mol BaCO3) ( ΔΗrxnoof BaCO3)] −[(1 mol BaO)(ΔΗrxnoof BaO) + (1 mol CO2)(ΔΗrxno of CO2)] ΔΗrxno = [(1 mol BaCO3)(−1219 kJ/mol )]− [(1 mol BaO)(−548.1 kJ/mol ) +(1 mol CO2)(−393.5 kJ/mol )] ΔΗrxno = -277.4 kJ$

The enthalpy change is negative. Hence, the enthalpy $(ΔH°rxn)$ value is $-277.4 kJ_$

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.

$ΔSrxno= [(2 mol BaCO3) ( Soof BaCO3)] −[(1 mol BaO)( Soof BaO) + (1 mol CO2)( So of CO2) ] ΔSrxno = [(1 mol )(112 J/mol⋅K )]− [(1 mol )(72.07 J/mol⋅K ) +(1 mol)(213.7 J/mol⋅K )] ΔSrxno = -173.77 J/K$