For the following reaction, the ΔS rxn o value has to be calculated at 298 K , using the given equilibrium K value. Concept introduction: 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.

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Answer 1

Question

Chapter 20, Problem 20.95P

(a)

Interpretation Introduction

Interpretation:

For the following reaction, the $ΔSrxno$ value has to be calculated at $298K$ , using the given equilibrium $K$ value.

Concept introduction:

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.

(a)

Expert Solution

Answer to Problem 20.95P

The standard entropy changes of the given reaction value is $ΔSrxno =121.5 J/K_.$

Explanation of Solution

The given equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Entropy change $ΔS°system$

$ΔSrxno$ Formations of values are,

$NO(g)= 210.65 J/K⋅molBr2(g) = 245.38 J/K⋅molNOBr(g) =272.6 J/K⋅mol$

Calculate the change in entropy for this reaction as follows,

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

Calculate the change in entropy for this reaction as follows

$ΔSrxno = [(2 mol NO) ( So of NO) + (1 mol Br2) ( So of Br2) ] −[ (2 mol of NOBr) ( So of NOBr)]$

Substituting the values of standard entropy $(So)$ of individual species in the above equation

$ΔSrxno = [(2 mol of NO )(210.65 J/mol⋅K ) + (1 mol of Br2 )(245.38 J/mol⋅K)]− [(2 mol of NOBr)(272.6 J/mol⋅K )] ΔSrxno = 121.48 =121.5 J/K.$

The standard entropy of the reaction $ΔSrxno =121.5 J/K_$ and the entropy change is positive.

(b)

Interpretation Introduction

Interpretation:

For the following reaction, the $ΔGrxno$ value has to be calculated at $373 K$ , using the given data’s.

Concept introduction:

Free energy change $ΔG$ : change in the free energy takes place while reactants convert to product where both are in standard state. It depends on the equilibrium constant $K$ ,

$ΔG = ΔGo+ RT ln (K) ΔGo=- RT ln (K)$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔGo$ is standard free energy change.

(b)

Expert Solution

Answer to Problem 20.95P

Given reaction the standard free $ΔG°rxn$ energy value is $2.7×103J/mol.$

Explanation of Solution

The given equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Solving for $ΔGrxno$ using following free energy equation,

$ΔG= ΔGo+RTln(Q)$

Calculate the $ΔGrxno$ of given equilibrium reaction is as follows,

$ΔG= 0=ΔGrxno+RTlnKΔGrxno=− RTlnK= −(8.314 J/mol×K)(373 K) In(0.42)=−(8.314 J/mol×K)(373 K) ×(−0.867500)=2690.223 J/molΔGrxno=2.7×103J/mol.$

Therefore, the given reaction $(ΔGrxno)$ value is $2.7×103J/mol.$

(c)

Interpretation Introduction

Interpretation:

For the following reaction, the $ΔHrxno$ value has to be calculated at $373 K$ , using the given data’s.

Concept introduction:

Free energy change 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.

$ΔGo = ΔΗo- TΔSo$

Where,

$ΔΗo$ is the change in enthalpy of the system

$ΔGo$ is the standard change in free energy of the system

$T$ is the absolute value of the temperature

(c)

Expert Solution

Answer to Problem 20.95P

Given reaction enthalpy $(ΔΗrxno)$ changes is $4.80×104 J/mol$

Explanation of Solution

The given equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Calculation for enthalpy change $ΔHrxno$

Standared Free energy change equation is,

$ΔΗrxno = ΔGrxno- TΔSrxno$

Calcualted entropy $ΔSrxno$ and $ΔGrxno$ values are

$ΔSrxno=121.48 J/K$

$ΔGrxno=2690.225 J/mol$

These values are plugging following above equation,

$ΔΗrxno= 2690.225 J/mol + (373 K) (121.48 J/K)= 4.8002265×104ΔΗrxno=4.80×104 J/mol.$

Therefore, the given reaction $(ΔΗrxno)$ value is $4.80×104 J/mol.$

(d)

Interpretation Introduction

Interpretation:

For the following reaction, the $ΔHfo$ value has to be calculated at $298 K$ .

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,

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

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

(d)

Expert Solution

Answer to Problem 20.95P

Given reaction standard enthalpy changes is $ΔH°rxn =81.7 kJ/mol_.$

Explanation of Solution

Consider the equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Standard enthalpy change is,

The equation for the standard enthalpy of the above reaction is,

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

Calculate the change in entropy for this reaction as follows

$ΔHrxno = [(2 mol NO) ( ΔHfo of NO) + (1 mol Br2) ( ΔHfo of Br2)] −[ (2 mol of NOBr) ( ΔHfo of NOBr)]$

Substituting the values of standard enthalpies of individual species in the above equation

$ΔHrxno = [(2 mol of NO )(90.29 kJ/mol ) + (1 mol of Br2 )(30.91 kJ/mol )]− [(2 mol of NOBr)(4.8002265×104 J )(1 kJ/103 J)] ΔHrxnoof NOBr = 81.7438675= 81.7 kJ/mol.$

Hence, the standard enthalpy of the reaction $ΔH°rxn =81.7 kJ/mol_.$

(e)

Interpretation Introduction

Interpretation:

For the following reaction, the $ΔGrxno$ value has to be calculated at $298 K$ .

Concept introduction:

Free energy (or) entropy change 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.

$ΔGo = ΔΗo- TΔSo$

Where,

$ΔGo$ is the standard change in free energy of the system

$ΔΗo$ is the standard change in enthalpy of the system

$T$ is the absolute value of the temperature

$ΔSo$ is the change in entropy in the system

(e)

Expert Solution

Answer to Problem 20.95P

The $ΔGrxno$ value of the reaction is $1.18×104 J/mol_.$

Explanation of Solution

Consider the equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Calculate the Free energy change $ΔGrxno$

Standared Free energy change equation is,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Calcualted enthalpy and entropy values are

$ΔΗrxno=4.8002265×104 J/mol$

$ΔSrxno=121.48 kJ/K$

These values are plugging following standard free energy equation,

$ΔGrxno =4.8002265×104 J/mol-(298 K)(121.48 J/mol⋅K)= 1.1801225×105 ΔGrxno= 1.18×104 J/mol.$

Therefore, the $ΔGrxno$ value of the reaction is $1.18×104 J/mol_.$

(f)

Interpretation Introduction

Interpretation:

For the following reaction, the $NOBr$ $ΔGfo$ value has to be calculated at $298 K$ .

Concept introduction:

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

(f)

Expert Solution

Answer to Problem 20.95P

The standard free energy value is $ΔGfo= 82.3 kJ/mol.$

Explanation of Solution

Given equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Free energy changes $ΔG°rxn$

Gibbs free energy equation is,

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

Free energy change for the reaction is calculated as follows,

$ΔG°rxn = [(2 mol of NO) ( ΔGfoof NO) + (1 mol of Br2) ( ΔGfoof Br2)] −[(2 mol of NOBr)(ΔGfo of NOBr)] ΔG°rxn = [(2 mol of NO)(86.60 kJ/mol ) + (1 mol of Br2)(3.13 kJ/mol )]− [(2 mol of NOBr)(1.1801225×104 J )(1 kJ/103 J)] ΔGfo of NOBr = 82.264=82.3 kJ/mol.$

Hence, the standard free energy value is $ΔGfo= 82.3 kJ/mol.$

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Answer 2

Question

Chapter 20, Problem 20.95P

(a)

Interpretation Introduction

Interpretation:

For the following reaction, the $ΔSrxno$ value has to be calculated at $298K$ , using the given equilibrium $K$ value.

Concept introduction:

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.

(a)

Expert Solution

Answer to Problem 20.95P

The standard entropy changes of the given reaction value is $ΔSrxno =121.5 J/K_.$

Explanation of Solution

The given equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Entropy change $ΔS°system$

$ΔSrxno$ Formations of values are,

$NO(g)= 210.65 J/K⋅molBr2(g) = 245.38 J/K⋅molNOBr(g) =272.6 J/K⋅mol$

Calculate the change in entropy for this reaction as follows,

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

Calculate the change in entropy for this reaction as follows

$ΔSrxno = [(2 mol NO) ( So of NO) + (1 mol Br2) ( So of Br2) ] −[ (2 mol of NOBr) ( So of NOBr)]$

Substituting the values of standard entropy $(So)$ of individual species in the above equation

$ΔSrxno = [(2 mol of NO )(210.65 J/mol⋅K ) + (1 mol of Br2 )(245.38 J/mol⋅K)]− [(2 mol of NOBr)(272.6 J/mol⋅K )] ΔSrxno = 121.48 =121.5 J/K.$

The standard entropy of the reaction $ΔSrxno =121.5 J/K_$ and the entropy change is positive.

(b)

Interpretation Introduction

Interpretation:

For the following reaction, the $ΔGrxno$ value has to be calculated at $373 K$ , using the given data’s.

Concept introduction:

Free energy change $ΔG$ : change in the free energy takes place while reactants convert to product where both are in standard state. It depends on the equilibrium constant $K$ ,

$ΔG = ΔGo+ RT ln (K) ΔGo=- RT ln (K)$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔGo$ is standard free energy change.

(b)

Expert Solution

Answer to Problem 20.95P

Given reaction the standard free $ΔG°rxn$ energy value is $2.7×103J/mol.$

Explanation of Solution

The given equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Solving for $ΔGrxno$ using following free energy equation,

$ΔG= ΔGo+RTln(Q)$

Calculate the $ΔGrxno$ of given equilibrium reaction is as follows,

$ΔG= 0=ΔGrxno+RTlnKΔGrxno=− RTlnK= −(8.314 J/mol×K)(373 K) In(0.42)=−(8.314 J/mol×K)(373 K) ×(−0.867500)=2690.223 J/molΔGrxno=2.7×103J/mol.$

Therefore, the given reaction $(ΔGrxno)$ value is $2.7×103J/mol.$

(c)

Interpretation Introduction

Interpretation:

For the following reaction, the $ΔHrxno$ value has to be calculated at $373 K$ , using the given data’s.

Concept introduction:

Free energy change 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.

$ΔGo = ΔΗo- TΔSo$

Where,

$ΔΗo$ is the change in enthalpy of the system

$ΔGo$ is the standard change in free energy of the system

$T$ is the absolute value of the temperature

(c)

Expert Solution

Answer to Problem 20.95P

Given reaction enthalpy $(ΔΗrxno)$ changes is $4.80×104 J/mol$

Explanation of Solution

The given equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Calculation for enthalpy change $ΔHrxno$

Standared Free energy change equation is,

$ΔΗrxno = ΔGrxno- TΔSrxno$

Calcualted entropy $ΔSrxno$ and $ΔGrxno$ values are

$ΔSrxno=121.48 J/K$

$ΔGrxno=2690.225 J/mol$

These values are plugging following above equation,

$ΔΗrxno= 2690.225 J/mol + (373 K) (121.48 J/K)= 4.8002265×104ΔΗrxno=4.80×104 J/mol.$

Therefore, the given reaction $(ΔΗrxno)$ value is $4.80×104 J/mol.$

(d)

Interpretation Introduction

Interpretation:

For the following reaction, the $ΔHfo$ value has to be calculated at $298 K$ .

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,

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

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

(d)

Expert Solution

Answer to Problem 20.95P

Given reaction standard enthalpy changes is $ΔH°rxn =81.7 kJ/mol_.$

Explanation of Solution

Consider the equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Standard enthalpy change is,

The equation for the standard enthalpy of the above reaction is,

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

Calculate the change in entropy for this reaction as follows

$ΔHrxno = [(2 mol NO) ( ΔHfo of NO) + (1 mol Br2) ( ΔHfo of Br2)] −[ (2 mol of NOBr) ( ΔHfo of NOBr)]$

Substituting the values of standard enthalpies of individual species in the above equation

$ΔHrxno = [(2 mol of NO )(90.29 kJ/mol ) + (1 mol of Br2 )(30.91 kJ/mol )]− [(2 mol of NOBr)(4.8002265×104 J )(1 kJ/103 J)] ΔHrxnoof NOBr = 81.7438675= 81.7 kJ/mol.$

Hence, the standard enthalpy of the reaction $ΔH°rxn =81.7 kJ/mol_.$

(e)

Interpretation Introduction

Interpretation:

For the following reaction, the $ΔGrxno$ value has to be calculated at $298 K$ .

Concept introduction:

Free energy (or) entropy change 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.

$ΔGo = ΔΗo- TΔSo$

Where,

$ΔGo$ is the standard change in free energy of the system

$ΔΗo$ is the standard change in enthalpy of the system

$T$ is the absolute value of the temperature

$ΔSo$ is the change in entropy in the system

(e)

Expert Solution

Answer to Problem 20.95P

The $ΔGrxno$ value of the reaction is $1.18×104 J/mol_.$

Explanation of Solution

Consider the equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Calculate the Free energy change $ΔGrxno$

Standared Free energy change equation is,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Calcualted enthalpy and entropy values are

$ΔΗrxno=4.8002265×104 J/mol$

$ΔSrxno=121.48 kJ/K$

These values are plugging following standard free energy equation,

$ΔGrxno =4.8002265×104 J/mol-(298 K)(121.48 J/mol⋅K)= 1.1801225×105 ΔGrxno= 1.18×104 J/mol.$

Therefore, the $ΔGrxno$ value of the reaction is $1.18×104 J/mol_.$

(f)

Interpretation Introduction

Interpretation:

For the following reaction, the $NOBr$ $ΔGfo$ value has to be calculated at $298 K$ .

Concept introduction:

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

(f)

Expert Solution

Answer to Problem 20.95P

The standard free energy value is $ΔGfo= 82.3 kJ/mol.$

Explanation of Solution

Given equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Free energy changes $ΔG°rxn$

Gibbs free energy equation is,

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

Free energy change for the reaction is calculated as follows,

$ΔG°rxn = [(2 mol of NO) ( ΔGfoof NO) + (1 mol of Br2) ( ΔGfoof Br2)] −[(2 mol of NOBr)(ΔGfo of NOBr)] ΔG°rxn = [(2 mol of NO)(86.60 kJ/mol ) + (1 mol of Br2)(3.13 kJ/mol )]− [(2 mol of NOBr)(1.1801225×104 J )(1 kJ/103 J)] ΔGfo of NOBr = 82.264=82.3 kJ/mol.$

Hence, the standard free energy value is $ΔGfo= 82.3 kJ/mol.$

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Answer 3

Question

Chapter 20, Problem 20.95P

(a)

Interpretation Introduction

Interpretation:

For the following reaction, the $ΔSrxno$ value has to be calculated at $298K$ , using the given equilibrium $K$ value.

Concept introduction:

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.

(a)

Expert Solution

Answer to Problem 20.95P

The standard entropy changes of the given reaction value is $ΔSrxno =121.5 J/K_.$

Explanation of Solution

The given equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Entropy change $ΔS°system$

$ΔSrxno$ Formations of values are,

$NO(g)= 210.65 J/K⋅molBr2(g) = 245.38 J/K⋅molNOBr(g) =272.6 J/K⋅mol$

Calculate the change in entropy for this reaction as follows,

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

Calculate the change in entropy for this reaction as follows

$ΔSrxno = [(2 mol NO) ( So of NO) + (1 mol Br2) ( So of Br2) ] −[ (2 mol of NOBr) ( So of NOBr)]$

Substituting the values of standard entropy $(So)$ of individual species in the above equation

$ΔSrxno = [(2 mol of NO )(210.65 J/mol⋅K ) + (1 mol of Br2 )(245.38 J/mol⋅K)]− [(2 mol of NOBr)(272.6 J/mol⋅K )] ΔSrxno = 121.48 =121.5 J/K.$

The standard entropy of the reaction $ΔSrxno =121.5 J/K_$ and the entropy change is positive.

(b)

Interpretation Introduction

Interpretation:

For the following reaction, the $ΔGrxno$ value has to be calculated at $373 K$ , using the given data’s.

Concept introduction:

Free energy change $ΔG$ : change in the free energy takes place while reactants convert to product where both are in standard state. It depends on the equilibrium constant $K$ ,

$ΔG = ΔGo+ RT ln (K) ΔGo=- RT ln (K)$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔGo$ is standard free energy change.

(b)

Expert Solution

Answer to Problem 20.95P

Given reaction the standard free $ΔG°rxn$ energy value is $2.7×103J/mol.$

Explanation of Solution

The given equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Solving for $ΔGrxno$ using following free energy equation,

$ΔG= ΔGo+RTln(Q)$

Calculate the $ΔGrxno$ of given equilibrium reaction is as follows,

$ΔG= 0=ΔGrxno+RTlnKΔGrxno=− RTlnK= −(8.314 J/mol×K)(373 K) In(0.42)=−(8.314 J/mol×K)(373 K) ×(−0.867500)=2690.223 J/molΔGrxno=2.7×103J/mol.$

Therefore, the given reaction $(ΔGrxno)$ value is $2.7×103J/mol.$

(c)

Interpretation Introduction

Interpretation:

For the following reaction, the $ΔHrxno$ value has to be calculated at $373 K$ , using the given data’s.

Concept introduction:

Free energy change 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.

$ΔGo = ΔΗo- TΔSo$

Where,

$ΔΗo$ is the change in enthalpy of the system

$ΔGo$ is the standard change in free energy of the system

$T$ is the absolute value of the temperature

(c)

Expert Solution

Answer to Problem 20.95P

Given reaction enthalpy $(ΔΗrxno)$ changes is $4.80×104 J/mol$

Explanation of Solution

The given equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Calculation for enthalpy change $ΔHrxno$

Standared Free energy change equation is,

$ΔΗrxno = ΔGrxno- TΔSrxno$

Calcualted entropy $ΔSrxno$ and $ΔGrxno$ values are

$ΔSrxno=121.48 J/K$

$ΔGrxno=2690.225 J/mol$

These values are plugging following above equation,

$ΔΗrxno= 2690.225 J/mol + (373 K) (121.48 J/K)= 4.8002265×104ΔΗrxno=4.80×104 J/mol.$

Therefore, the given reaction $(ΔΗrxno)$ value is $4.80×104 J/mol.$

(d)

Interpretation Introduction

Interpretation:

For the following reaction, the $ΔHfo$ value has to be calculated at $298 K$ .

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,

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

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

(d)

Expert Solution

Answer to Problem 20.95P

Given reaction standard enthalpy changes is $ΔH°rxn =81.7 kJ/mol_.$

Explanation of Solution

Consider the equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Standard enthalpy change is,

The equation for the standard enthalpy of the above reaction is,

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

Calculate the change in entropy for this reaction as follows

$ΔHrxno = [(2 mol NO) ( ΔHfo of NO) + (1 mol Br2) ( ΔHfo of Br2)] −[ (2 mol of NOBr) ( ΔHfo of NOBr)]$

Substituting the values of standard enthalpies of individual species in the above equation

$ΔHrxno = [(2 mol of NO )(90.29 kJ/mol ) + (1 mol of Br2 )(30.91 kJ/mol )]− [(2 mol of NOBr)(4.8002265×104 J )(1 kJ/103 J)] ΔHrxnoof NOBr = 81.7438675= 81.7 kJ/mol.$

Hence, the standard enthalpy of the reaction $ΔH°rxn =81.7 kJ/mol_.$

(e)

Interpretation Introduction

Interpretation:

For the following reaction, the $ΔGrxno$ value has to be calculated at $298 K$ .

Concept introduction:

Free energy (or) entropy change 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.

$ΔGo = ΔΗo- TΔSo$

Where,

$ΔGo$ is the standard change in free energy of the system

$ΔΗo$ is the standard change in enthalpy of the system

$T$ is the absolute value of the temperature

$ΔSo$ is the change in entropy in the system

(e)

Expert Solution

Answer to Problem 20.95P

The $ΔGrxno$ value of the reaction is $1.18×104 J/mol_.$

Explanation of Solution

Consider the equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Calculate the Free energy change $ΔGrxno$

Standared Free energy change equation is,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Calcualted enthalpy and entropy values are

$ΔΗrxno=4.8002265×104 J/mol$

$ΔSrxno=121.48 kJ/K$

These values are plugging following standard free energy equation,

$ΔGrxno =4.8002265×104 J/mol-(298 K)(121.48 J/mol⋅K)= 1.1801225×105 ΔGrxno= 1.18×104 J/mol.$

Therefore, the $ΔGrxno$ value of the reaction is $1.18×104 J/mol_.$

(f)

Interpretation Introduction

Interpretation:

For the following reaction, the $NOBr$ $ΔGfo$ value has to be calculated at $298 K$ .

Concept introduction:

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

(f)

Expert Solution

Answer to Problem 20.95P

The standard free energy value is $ΔGfo= 82.3 kJ/mol.$

Explanation of Solution

Given equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Free energy changes $ΔG°rxn$

Gibbs free energy equation is,

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

Free energy change for the reaction is calculated as follows,

$ΔG°rxn = [(2 mol of NO) ( ΔGfoof NO) + (1 mol of Br2) ( ΔGfoof Br2)] −[(2 mol of NOBr)(ΔGfo of NOBr)] ΔG°rxn = [(2 mol of NO)(86.60 kJ/mol ) + (1 mol of Br2)(3.13 kJ/mol )]− [(2 mol of NOBr)(1.1801225×104 J )(1 kJ/103 J)] ΔGfo of NOBr = 82.264=82.3 kJ/mol.$

Hence, the standard free energy value is $ΔGfo= 82.3 kJ/mol.$

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Answer 4

Question

Chapter 20, Problem 20.95P

(a)

Interpretation Introduction

Interpretation:

For the following reaction, the $ΔSrxno$ value has to be calculated at $298K$ , using the given equilibrium $K$ value.

Concept introduction:

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.

(a)

Expert Solution

Answer to Problem 20.95P

The standard entropy changes of the given reaction value is $ΔSrxno =121.5 J/K_.$

Explanation of Solution

The given equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Entropy change $ΔS°system$

$ΔSrxno$ Formations of values are,

$NO(g)= 210.65 J/K⋅molBr2(g) = 245.38 J/K⋅molNOBr(g) =272.6 J/K⋅mol$

Calculate the change in entropy for this reaction as follows,

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

Calculate the change in entropy for this reaction as follows

$ΔSrxno = [(2 mol NO) ( So of NO) + (1 mol Br2) ( So of Br2) ] −[ (2 mol of NOBr) ( So of NOBr)]$

Substituting the values of standard entropy $(So)$ of individual species in the above equation

$ΔSrxno = [(2 mol of NO )(210.65 J/mol⋅K ) + (1 mol of Br2 )(245.38 J/mol⋅K)]− [(2 mol of NOBr)(272.6 J/mol⋅K )] ΔSrxno = 121.48 =121.5 J/K.$

The standard entropy of the reaction $ΔSrxno =121.5 J/K_$ and the entropy change is positive.

(b)

Interpretation Introduction

Interpretation:

For the following reaction, the $ΔGrxno$ value has to be calculated at $373 K$ , using the given data’s.

Concept introduction:

Free energy change $ΔG$ : change in the free energy takes place while reactants convert to product where both are in standard state. It depends on the equilibrium constant $K$ ,

$ΔG = ΔGo+ RT ln (K) ΔGo=- RT ln (K)$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔGo$ is standard free energy change.

(b)

Expert Solution

Answer to Problem 20.95P

Given reaction the standard free $ΔG°rxn$ energy value is $2.7×103J/mol.$

Explanation of Solution

The given equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Solving for $ΔGrxno$ using following free energy equation,

$ΔG= ΔGo+RTln(Q)$

Calculate the $ΔGrxno$ of given equilibrium reaction is as follows,

$ΔG= 0=ΔGrxno+RTlnKΔGrxno=− RTlnK= −(8.314 J/mol×K)(373 K) In(0.42)=−(8.314 J/mol×K)(373 K) ×(−0.867500)=2690.223 J/molΔGrxno=2.7×103J/mol.$

Therefore, the given reaction $(ΔGrxno)$ value is $2.7×103J/mol.$

(c)

Interpretation Introduction

Interpretation:

For the following reaction, the $ΔHrxno$ value has to be calculated at $373 K$ , using the given data’s.

Concept introduction:

Free energy change 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.

$ΔGo = ΔΗo- TΔSo$

Where,

$ΔΗo$ is the change in enthalpy of the system

$ΔGo$ is the standard change in free energy of the system

$T$ is the absolute value of the temperature

(c)

Expert Solution

Answer to Problem 20.95P

Given reaction enthalpy $(ΔΗrxno)$ changes is $4.80×104 J/mol$

Explanation of Solution

The given equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Calculation for enthalpy change $ΔHrxno$

Standared Free energy change equation is,

$ΔΗrxno = ΔGrxno- TΔSrxno$

Calcualted entropy $ΔSrxno$ and $ΔGrxno$ values are

$ΔSrxno=121.48 J/K$

$ΔGrxno=2690.225 J/mol$

These values are plugging following above equation,

$ΔΗrxno= 2690.225 J/mol + (373 K) (121.48 J/K)= 4.8002265×104ΔΗrxno=4.80×104 J/mol.$

Therefore, the given reaction $(ΔΗrxno)$ value is $4.80×104 J/mol.$

(d)

Interpretation Introduction

Interpretation:

For the following reaction, the $ΔHfo$ value has to be calculated at $298 K$ .

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,

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

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

(d)

Expert Solution

Answer to Problem 20.95P

Given reaction standard enthalpy changes is $ΔH°rxn =81.7 kJ/mol_.$

Explanation of Solution

Consider the equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Standard enthalpy change is,

The equation for the standard enthalpy of the above reaction is,

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

Calculate the change in entropy for this reaction as follows

$ΔHrxno = [(2 mol NO) ( ΔHfo of NO) + (1 mol Br2) ( ΔHfo of Br2)] −[ (2 mol of NOBr) ( ΔHfo of NOBr)]$

Substituting the values of standard enthalpies of individual species in the above equation

$ΔHrxno = [(2 mol of NO )(90.29 kJ/mol ) + (1 mol of Br2 )(30.91 kJ/mol )]− [(2 mol of NOBr)(4.8002265×104 J )(1 kJ/103 J)] ΔHrxnoof NOBr = 81.7438675= 81.7 kJ/mol.$

Hence, the standard enthalpy of the reaction $ΔH°rxn =81.7 kJ/mol_.$

(e)

Interpretation Introduction

Interpretation:

For the following reaction, the $ΔGrxno$ value has to be calculated at $298 K$ .

Concept introduction:

Free energy (or) entropy change 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.

$ΔGo = ΔΗo- TΔSo$

Where,

$ΔGo$ is the standard change in free energy of the system

$ΔΗo$ is the standard change in enthalpy of the system

$T$ is the absolute value of the temperature

$ΔSo$ is the change in entropy in the system

(e)

Expert Solution

Answer to Problem 20.95P

The $ΔGrxno$ value of the reaction is $1.18×104 J/mol_.$

Explanation of Solution

Consider the equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Calculate the Free energy change $ΔGrxno$

Standared Free energy change equation is,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Calcualted enthalpy and entropy values are

$ΔΗrxno=4.8002265×104 J/mol$

$ΔSrxno=121.48 kJ/K$

These values are plugging following standard free energy equation,

$ΔGrxno =4.8002265×104 J/mol-(298 K)(121.48 J/mol⋅K)= 1.1801225×105 ΔGrxno= 1.18×104 J/mol.$

Therefore, the $ΔGrxno$ value of the reaction is $1.18×104 J/mol_.$

(f)

Interpretation Introduction

Interpretation:

For the following reaction, the $NOBr$ $ΔGfo$ value has to be calculated at $298 K$ .

Concept introduction:

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

(f)

Expert Solution

Answer to Problem 20.95P

The standard free energy value is $ΔGfo= 82.3 kJ/mol.$

Explanation of Solution

Given equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Free energy changes $ΔG°rxn$

Gibbs free energy equation is,

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

Free energy change for the reaction is calculated as follows,

$ΔG°rxn = [(2 mol of NO) ( ΔGfoof NO) + (1 mol of Br2) ( ΔGfoof Br2)] −[(2 mol of NOBr)(ΔGfo of NOBr)] ΔG°rxn = [(2 mol of NO)(86.60 kJ/mol ) + (1 mol of Br2)(3.13 kJ/mol )]− [(2 mol of NOBr)(1.1801225×104 J )(1 kJ/103 J)] ΔGfo of NOBr = 82.264=82.3 kJ/mol.$

Hence, the standard free energy value is $ΔGfo= 82.3 kJ/mol.$

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Answer 5

Chapter 20, Problem 20.95P

(a)

Interpretation Introduction

Interpretation:

For the following reaction, the $ΔSrxno$ value has to be calculated at $298K$ , using the given equilibrium $K$ value.

Concept introduction:

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.

(a)

Expert Solution

Answer to Problem 20.95P

The standard entropy changes of the given reaction value is $ΔSrxno =121.5 J/K_.$

Explanation of Solution

The given equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Entropy change $ΔS°system$

$ΔSrxno$ Formations of values are,

$NO(g)= 210.65 J/K⋅molBr2(g) = 245.38 J/K⋅molNOBr(g) =272.6 J/K⋅mol$

Calculate the change in entropy for this reaction as follows,

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

Calculate the change in entropy for this reaction as follows

$ΔSrxno = [(2 mol NO) ( So of NO) + (1 mol Br2) ( So of Br2) ] −[ (2 mol of NOBr) ( So of NOBr)]$

Substituting the values of standard entropy $(So)$ of individual species in the above equation

$ΔSrxno = [(2 mol of NO )(210.65 J/mol⋅K ) + (1 mol of Br2 )(245.38 J/mol⋅K)]− [(2 mol of NOBr)(272.6 J/mol⋅K )] ΔSrxno = 121.48 =121.5 J/K.$

The standard entropy of the reaction $ΔSrxno =121.5 J/K_$ and the entropy change is positive.

(b)

Interpretation Introduction

Interpretation:

For the following reaction, the $ΔGrxno$ value has to be calculated at $373 K$ , using the given data’s.

Concept introduction:

Free energy change $ΔG$ : change in the free energy takes place while reactants convert to product where both are in standard state. It depends on the equilibrium constant $K$ ,

$ΔG = ΔGo+ RT ln (K) ΔGo=- RT ln (K)$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔGo$ is standard free energy change.

(b)

Expert Solution

Answer to Problem 20.95P

Given reaction the standard free $ΔG°rxn$ energy value is $2.7×103J/mol.$

Explanation of Solution

The given equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Solving for $ΔGrxno$ using following free energy equation,

$ΔG= ΔGo+RTln(Q)$

Calculate the $ΔGrxno$ of given equilibrium reaction is as follows,

$ΔG= 0=ΔGrxno+RTlnKΔGrxno=− RTlnK= −(8.314 J/mol×K)(373 K) In(0.42)=−(8.314 J/mol×K)(373 K) ×(−0.867500)=2690.223 J/molΔGrxno=2.7×103J/mol.$

Therefore, the given reaction $(ΔGrxno)$ value is $2.7×103J/mol.$

(c)

Interpretation Introduction

Interpretation:

For the following reaction, the $ΔHrxno$ value has to be calculated at $373 K$ , using the given data’s.

Concept introduction:

Free energy change 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.

$ΔGo = ΔΗo- TΔSo$

Where,

$ΔΗo$ is the change in enthalpy of the system

$ΔGo$ is the standard change in free energy of the system

$T$ is the absolute value of the temperature

(c)

Expert Solution

Answer to Problem 20.95P

Given reaction enthalpy $(ΔΗrxno)$ changes is $4.80×104 J/mol$

Explanation of Solution

The given equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Calculation for enthalpy change $ΔHrxno$

Standared Free energy change equation is,

$ΔΗrxno = ΔGrxno- TΔSrxno$

Calcualted entropy $ΔSrxno$ and $ΔGrxno$ values are

$ΔSrxno=121.48 J/K$

$ΔGrxno=2690.225 J/mol$

These values are plugging following above equation,

$ΔΗrxno= 2690.225 J/mol + (373 K) (121.48 J/K)= 4.8002265×104ΔΗrxno=4.80×104 J/mol.$

Therefore, the given reaction $(ΔΗrxno)$ value is $4.80×104 J/mol.$

(d)

Interpretation Introduction

Interpretation:

For the following reaction, the $ΔHfo$ value has to be calculated at $298 K$ .

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,

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

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

(d)

Expert Solution

Answer to Problem 20.95P

Given reaction standard enthalpy changes is $ΔH°rxn =81.7 kJ/mol_.$

Explanation of Solution

Consider the equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Standard enthalpy change is,

The equation for the standard enthalpy of the above reaction is,

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

Calculate the change in entropy for this reaction as follows

$ΔHrxno = [(2 mol NO) ( ΔHfo of NO) + (1 mol Br2) ( ΔHfo of Br2)] −[ (2 mol of NOBr) ( ΔHfo of NOBr)]$

Substituting the values of standard enthalpies of individual species in the above equation

$ΔHrxno = [(2 mol of NO )(90.29 kJ/mol ) + (1 mol of Br2 )(30.91 kJ/mol )]− [(2 mol of NOBr)(4.8002265×104 J )(1 kJ/103 J)] ΔHrxnoof NOBr = 81.7438675= 81.7 kJ/mol.$

Hence, the standard enthalpy of the reaction $ΔH°rxn =81.7 kJ/mol_.$

(e)

Interpretation Introduction

Interpretation:

For the following reaction, the $ΔGrxno$ value has to be calculated at $298 K$ .

Concept introduction:

Free energy (or) entropy change 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.

$ΔGo = ΔΗo- TΔSo$

Where,

$ΔGo$ is the standard change in free energy of the system

$ΔΗo$ is the standard change in enthalpy of the system

$T$ is the absolute value of the temperature

$ΔSo$ is the change in entropy in the system

(e)

Expert Solution

Answer to Problem 20.95P

The $ΔGrxno$ value of the reaction is $1.18×104 J/mol_.$

Explanation of Solution

Consider the equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Calculate the Free energy change $ΔGrxno$

Standared Free energy change equation is,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Calcualted enthalpy and entropy values are

$ΔΗrxno=4.8002265×104 J/mol$

$ΔSrxno=121.48 kJ/K$

These values are plugging following standard free energy equation,

$ΔGrxno =4.8002265×104 J/mol-(298 K)(121.48 J/mol⋅K)= 1.1801225×105 ΔGrxno= 1.18×104 J/mol.$

Therefore, the $ΔGrxno$ value of the reaction is $1.18×104 J/mol_.$

(f)

Interpretation Introduction

Interpretation:

For the following reaction, the $NOBr$ $ΔGfo$ value has to be calculated at $298 K$ .

Concept introduction:

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

(f)

Expert Solution

Answer to Problem 20.95P

The standard free energy value is $ΔGfo= 82.3 kJ/mol.$

Explanation of Solution

Given equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Free energy changes $ΔG°rxn$

Gibbs free energy equation is,

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

Free energy change for the reaction is calculated as follows,

$ΔG°rxn = [(2 mol of NO) ( ΔGfoof NO) + (1 mol of Br2) ( ΔGfoof Br2)] −[(2 mol of NOBr)(ΔGfo of NOBr)] ΔG°rxn = [(2 mol of NO)(86.60 kJ/mol ) + (1 mol of Br2)(3.13 kJ/mol )]− [(2 mol of NOBr)(1.1801225×104 J )(1 kJ/103 J)] ΔGfo of NOBr = 82.264=82.3 kJ/mol.$

Hence, the standard free energy value is $ΔGfo= 82.3 kJ/mol.$

Answer 6

Chapter 20, Problem 20.95P

(a)

Interpretation Introduction

Interpretation:

For the following reaction, the $ΔSrxno$ value has to be calculated at $298K$ , using the given equilibrium $K$ value.

Concept introduction:

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.

(a)

Expert Solution

Answer to Problem 20.95P

The standard entropy changes of the given reaction value is $ΔSrxno =121.5 J/K_.$

Explanation of Solution

The given equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Entropy change $ΔS°system$

$ΔSrxno$ Formations of values are,

$NO(g)= 210.65 J/K⋅molBr2(g) = 245.38 J/K⋅molNOBr(g) =272.6 J/K⋅mol$

Calculate the change in entropy for this reaction as follows,

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

Calculate the change in entropy for this reaction as follows

$ΔSrxno = [(2 mol NO) ( So of NO) + (1 mol Br2) ( So of Br2) ] −[ (2 mol of NOBr) ( So of NOBr)]$

Substituting the values of standard entropy $(So)$ of individual species in the above equation

$ΔSrxno = [(2 mol of NO )(210.65 J/mol⋅K ) + (1 mol of Br2 )(245.38 J/mol⋅K)]− [(2 mol of NOBr)(272.6 J/mol⋅K )] ΔSrxno = 121.48 =121.5 J/K.$

The standard entropy of the reaction $ΔSrxno =121.5 J/K_$ and the entropy change is positive.

Answer 7

Chapter 20, Problem 20.95P

(a)

Interpretation Introduction

Interpretation:

For the following reaction, the $ΔSrxno$ value has to be calculated at $298K$ , using the given equilibrium $K$ value.

Concept introduction:

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.

Answer 8

(a)

Expert Solution

Answer to Problem 20.95P

The standard entropy changes of the given reaction value is $ΔSrxno =121.5 J/K_.$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The given equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Entropy change $ΔS°system$

$ΔSrxno$ Formations of values are,

$NO(g)= 210.65 J/K⋅molBr2(g) = 245.38 J/K⋅molNOBr(g) =272.6 J/K⋅mol$

Calculate the change in entropy for this reaction as follows,

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

Calculate the change in entropy for this reaction as follows

$ΔSrxno = [(2 mol NO) ( So of NO) + (1 mol Br2) ( So of Br2) ] −[ (2 mol of NOBr) ( So of NOBr)]$

Substituting the values of standard entropy $(So)$ of individual species in the above equation

$ΔSrxno = [(2 mol of NO )(210.65 J/mol⋅K ) + (1 mol of Br2 )(245.38 J/mol⋅K)]− [(2 mol of NOBr)(272.6 J/mol⋅K )] ΔSrxno = 121.48 =121.5 J/K.$

The standard entropy of the reaction $ΔSrxno =121.5 J/K_$ and the entropy change is positive.

Answer 13

(b)

Interpretation Introduction

Interpretation:

For the following reaction, the $ΔGrxno$ value has to be calculated at $373 K$ , using the given data’s.

Concept introduction:

Free energy change $ΔG$ : change in the free energy takes place while reactants convert to product where both are in standard state. It depends on the equilibrium constant $K$ ,

$ΔG = ΔGo+ RT ln (K) ΔGo=- RT ln (K)$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔGo$ is standard free energy change.

(b)

Expert Solution

Answer to Problem 20.95P

Given reaction the standard free $ΔG°rxn$ energy value is $2.7×103J/mol.$

Explanation of Solution

The given equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Solving for $ΔGrxno$ using following free energy equation,

$ΔG= ΔGo+RTln(Q)$

Calculate the $ΔGrxno$ of given equilibrium reaction is as follows,

$ΔG= 0=ΔGrxno+RTlnKΔGrxno=− RTlnK= −(8.314 J/mol×K)(373 K) In(0.42)=−(8.314 J/mol×K)(373 K) ×(−0.867500)=2690.223 J/molΔGrxno=2.7×103J/mol.$

Therefore, the given reaction $(ΔGrxno)$ value is $2.7×103J/mol.$

Answer 14

(b)

Interpretation Introduction

Interpretation:

For the following reaction, the $ΔGrxno$ value has to be calculated at $373 K$ , using the given data’s.

Concept introduction:

Free energy change $ΔG$ : change in the free energy takes place while reactants convert to product where both are in standard state. It depends on the equilibrium constant $K$ ,

$ΔG = ΔGo+ RT ln (K) ΔGo=- RT ln (K)$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔGo$ is standard free energy change.

Answer 15

(b)

Expert Solution

Answer to Problem 20.95P

Given reaction the standard free $ΔG°rxn$ energy value is $2.7×103J/mol.$

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

The given equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Solving for $ΔGrxno$ using following free energy equation,

$ΔG= ΔGo+RTln(Q)$

Calculate the $ΔGrxno$ of given equilibrium reaction is as follows,

$ΔG= 0=ΔGrxno+RTlnKΔGrxno=− RTlnK= −(8.314 J/mol×K)(373 K) In(0.42)=−(8.314 J/mol×K)(373 K) ×(−0.867500)=2690.223 J/molΔGrxno=2.7×103J/mol.$

Therefore, the given reaction $(ΔGrxno)$ value is $2.7×103J/mol.$

Answer 20

(c)

Interpretation Introduction

Interpretation:

For the following reaction, the $ΔHrxno$ value has to be calculated at $373 K$ , using the given data’s.

Concept introduction:

Free energy change 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.

$ΔGo = ΔΗo- TΔSo$

Where,

$ΔΗo$ is the change in enthalpy of the system

$ΔGo$ is the standard change in free energy of the system

$T$ is the absolute value of the temperature

(c)

Expert Solution

Answer to Problem 20.95P

Given reaction enthalpy $(ΔΗrxno)$ changes is $4.80×104 J/mol$

Explanation of Solution

The given equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Calculation for enthalpy change $ΔHrxno$

Standared Free energy change equation is,

$ΔΗrxno = ΔGrxno- TΔSrxno$

Calcualted entropy $ΔSrxno$ and $ΔGrxno$ values are

$ΔSrxno=121.48 J/K$

$ΔGrxno=2690.225 J/mol$

These values are plugging following above equation,

$ΔΗrxno= 2690.225 J/mol + (373 K) (121.48 J/K)= 4.8002265×104ΔΗrxno=4.80×104 J/mol.$

Therefore, the given reaction $(ΔΗrxno)$ value is $4.80×104 J/mol.$

Answer 21

(c)

Interpretation Introduction

Interpretation:

For the following reaction, the $ΔHrxno$ value has to be calculated at $373 K$ , using the given data’s.

Concept introduction:

Free energy change 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.

$ΔGo = ΔΗo- TΔSo$

Where,

$ΔΗo$ is the change in enthalpy of the system

$ΔGo$ is the standard change in free energy of the system

$T$ is the absolute value of the temperature

Answer 22

(c)

Expert Solution

Answer to Problem 20.95P

Given reaction enthalpy $(ΔΗrxno)$ changes is $4.80×104 J/mol$

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

The given equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Calculation for enthalpy change $ΔHrxno$

Standared Free energy change equation is,

$ΔΗrxno = ΔGrxno- TΔSrxno$

Calcualted entropy $ΔSrxno$ and $ΔGrxno$ values are

$ΔSrxno=121.48 J/K$

$ΔGrxno=2690.225 J/mol$

These values are plugging following above equation,

$ΔΗrxno= 2690.225 J/mol + (373 K) (121.48 J/K)= 4.8002265×104ΔΗrxno=4.80×104 J/mol.$

Therefore, the given reaction $(ΔΗrxno)$ value is $4.80×104 J/mol.$

Answer 27

(d)

Interpretation Introduction

Interpretation:

For the following reaction, the $ΔHfo$ value has to be calculated at $298 K$ .

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,

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

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

(d)

Expert Solution

Answer to Problem 20.95P

Given reaction standard enthalpy changes is $ΔH°rxn =81.7 kJ/mol_.$

Explanation of Solution

Consider the equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Standard enthalpy change is,

The equation for the standard enthalpy of the above reaction is,

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

Calculate the change in entropy for this reaction as follows

$ΔHrxno = [(2 mol NO) ( ΔHfo of NO) + (1 mol Br2) ( ΔHfo of Br2)] −[ (2 mol of NOBr) ( ΔHfo of NOBr)]$

Substituting the values of standard enthalpies of individual species in the above equation

$ΔHrxno = [(2 mol of NO )(90.29 kJ/mol ) + (1 mol of Br2 )(30.91 kJ/mol )]− [(2 mol of NOBr)(4.8002265×104 J )(1 kJ/103 J)] ΔHrxnoof NOBr = 81.7438675= 81.7 kJ/mol.$

Hence, the standard enthalpy of the reaction $ΔH°rxn =81.7 kJ/mol_.$

Answer 28

(d)

Interpretation Introduction

Interpretation:

For the following reaction, the $ΔHfo$ value has to be calculated at $298 K$ .

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,

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

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

Answer 29

(d)

Expert Solution

Answer to Problem 20.95P

Given reaction standard enthalpy changes is $ΔH°rxn =81.7 kJ/mol_.$

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Consider the equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Standard enthalpy change is,

The equation for the standard enthalpy of the above reaction is,

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

Calculate the change in entropy for this reaction as follows

$ΔHrxno = [(2 mol NO) ( ΔHfo of NO) + (1 mol Br2) ( ΔHfo of Br2)] −[ (2 mol of NOBr) ( ΔHfo of NOBr)]$

Substituting the values of standard enthalpies of individual species in the above equation

$ΔHrxno = [(2 mol of NO )(90.29 kJ/mol ) + (1 mol of Br2 )(30.91 kJ/mol )]− [(2 mol of NOBr)(4.8002265×104 J )(1 kJ/103 J)] ΔHrxnoof NOBr = 81.7438675= 81.7 kJ/mol.$

Hence, the standard enthalpy of the reaction $ΔH°rxn =81.7 kJ/mol_.$

Answer 34

(e)

Interpretation Introduction

Interpretation:

For the following reaction, the $ΔGrxno$ value has to be calculated at $298 K$ .

Concept introduction:

Free energy (or) entropy change 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.

$ΔGo = ΔΗo- TΔSo$

Where,

$ΔGo$ is the standard change in free energy of the system

$ΔΗo$ is the standard change in enthalpy of the system

$T$ is the absolute value of the temperature

$ΔSo$ is the change in entropy in the system

(e)

Expert Solution

Answer to Problem 20.95P

The $ΔGrxno$ value of the reaction is $1.18×104 J/mol_.$

Explanation of Solution

Consider the equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Calculate the Free energy change $ΔGrxno$

Standared Free energy change equation is,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Calcualted enthalpy and entropy values are

$ΔΗrxno=4.8002265×104 J/mol$

$ΔSrxno=121.48 kJ/K$

These values are plugging following standard free energy equation,

$ΔGrxno =4.8002265×104 J/mol-(298 K)(121.48 J/mol⋅K)= 1.1801225×105 ΔGrxno= 1.18×104 J/mol.$

Therefore, the $ΔGrxno$ value of the reaction is $1.18×104 J/mol_.$

Answer 35

(e)

Interpretation Introduction

Interpretation:

For the following reaction, the $ΔGrxno$ value has to be calculated at $298 K$ .

Concept introduction:

Free energy (or) entropy change 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.

$ΔGo = ΔΗo- TΔSo$

Where,

$ΔGo$ is the standard change in free energy of the system

$ΔΗo$ is the standard change in enthalpy of the system

$T$ is the absolute value of the temperature

$ΔSo$ is the change in entropy in the system

Answer 36

(e)

Expert Solution

Answer to Problem 20.95P

The $ΔGrxno$ value of the reaction is $1.18×104 J/mol_.$

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Consider the equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Calculate the Free energy change $ΔGrxno$

Standared Free energy change equation is,

$ΔGrxno = ΔΗrxno- TΔSrxno$

Calcualted enthalpy and entropy values are

$ΔΗrxno=4.8002265×104 J/mol$

$ΔSrxno=121.48 kJ/K$

These values are plugging following standard free energy equation,

$ΔGrxno =4.8002265×104 J/mol-(298 K)(121.48 J/mol⋅K)= 1.1801225×105 ΔGrxno= 1.18×104 J/mol.$

Therefore, the $ΔGrxno$ value of the reaction is $1.18×104 J/mol_.$

Answer 41

(f)

Interpretation Introduction

Interpretation:

For the following reaction, the $NOBr$ $ΔGfo$ value has to be calculated at $298 K$ .

Concept introduction:

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

(f)

Expert Solution

Answer to Problem 20.95P

The standard free energy value is $ΔGfo= 82.3 kJ/mol.$

Explanation of Solution

Given equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Free energy changes $ΔG°rxn$

Gibbs free energy equation is,

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

Free energy change for the reaction is calculated as follows,

$ΔG°rxn = [(2 mol of NO) ( ΔGfoof NO) + (1 mol of Br2) ( ΔGfoof Br2)] −[(2 mol of NOBr)(ΔGfo of NOBr)] ΔG°rxn = [(2 mol of NO)(86.60 kJ/mol ) + (1 mol of Br2)(3.13 kJ/mol )]− [(2 mol of NOBr)(1.1801225×104 J )(1 kJ/103 J)] ΔGfo of NOBr = 82.264=82.3 kJ/mol.$

Hence, the standard free energy value is $ΔGfo= 82.3 kJ/mol.$

Answer 42

(f)

Interpretation Introduction

Interpretation:

For the following reaction, the $NOBr$ $ΔGfo$ value has to be calculated at $298 K$ .

Concept introduction:

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 43

(f)

Expert Solution

Answer to Problem 20.95P

The standard free energy value is $ΔGfo= 82.3 kJ/mol.$

Answer 44

(f)

Expert Solution

Answer 45

(f)

Expert Solution

Answer 46

Expert Solution

Answer 47

Explanation of Solution

Given equilibrium reaction is,

$2NOBr(g) ⇌ 2NO(g) + Br2(g) K=0.42 at 373 K$

Free energy changes $ΔG°rxn$

Gibbs free energy equation is,

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

Free energy change for the reaction is calculated as follows,

$ΔG°rxn = [(2 mol of NO) ( ΔGfoof NO) + (1 mol of Br2) ( ΔGfoof Br2)] −[(2 mol of NOBr)(ΔGfo of NOBr)] ΔG°rxn = [(2 mol of NO)(86.60 kJ/mol ) + (1 mol of Br2)(3.13 kJ/mol )]− [(2 mol of NOBr)(1.1801225×104 J )(1 kJ/103 J)] ΔGfo of NOBr = 82.264=82.3 kJ/mol.$