Consider the following reaction: H 2 O ( g ) + CI 2 O ( g ) ⇌ 2 HOCI ( g ) K 298 = 0.090 For Cl 2 O( g ), Δ G f o = 97.9 KJ/mol Δ H f o = 80.3 KJ/mol S ∘ = 266.1 J / K ⋅ mol a. Calculate ∆G° for the reaction using the equation ∆G°=−RT In(K). b. Use bond energy values (Table 3-3) to estimate ∆H° for the reaction. c. Use the results from parts a and b to estimate ∆S° for the reaction . d. Estimate Δ H f o and S ° for HOCI(g). e. Estimate the value of K at 500. K. f. Calculate ∆G at 25°C when P H 2 O = 18 torr , P CI 2 O = 2.0 torr , a n d P HOCI = 0.10 torr .

Question

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Answer 1

Textbook Question

Chapter 17, Problem 96AE

Consider the following reaction:

$H 2 O ( g ) + CI 2 O ( g ) ⇌ 2 HOCI ( g ) K 298 = 0.090$

For Cl₂O(g),

$Δ G f o = 97.9 KJ/mol Δ H f o = 80.3 KJ/mol S ∘ = 266.1 J / K ⋅ mol$

a. Calculate ∆G° for the reaction using the equation ∆G°=−RT In(K).

b. Use bond energy values (Table 3-3) to estimate ∆H° for the reaction.

c. Use the results from parts a and b to estimate ∆S° for the reaction.

d. Estimate $Δ H f o$ and S° for HOCI(g).

e. Estimate the value of K at 500. K.

f. Calculate ∆G at 25°C when $P H 2 O = 18 torr , P CI 2 O = 2.0 torr , a n d P HOCI = 0.10 torr$ .

(a)

Expert Solution

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Explanation of Solution

Explanation

Given

Temperature is $298 K$ .

The value of equilibrium constant is $0.090$ .

The stated reaction is,

$H2O(g)+Cl2O(g)⇌2HOCl(g)$

Formula

The expression for free energy change is,

$ΔG°=−RTln(K)$

Where,

$ΔG°$ is the standard Gibbs free energy change.
$R$ is the gas law constant $(8.3145 J/K⋅mol)$ .
$T$ is the absolute temperature.
$K$ is the equilibrium constant.

Substitute the values of $R, T$ and $K$ in the above expression.

$ΔG°=−RTln(K)=−(8.3145 J/K⋅mol)(298 K)ln(0.090)=6.0×103 J/mol$

The conversion of joule per mole $(J/mol)$ into kilo-joule per mole $(kJ/mol)$ is done as,

$1 J/mol=10−3 kJ/mol$

Hence,

The conversion of $6.0×103 J/mol$ into kilo-joule per mole is,

$6.0×103 J/mol=(6.0×103 ×10−3) kJ/mol=6.0 kJ/mol_$

(b)

Expert Solution

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Explanation of Solution

Explanation

The reaction that takes place is,

$H2O(g)+Cl2O(g)⇌2HOCl(g)$

Refer to Table $3-3$ .

The value of bond energy, $D°(kJ/mol)$ , for the given reactant and product is,

Bonds	$D°(kJ/mol)$
$O−H$	$467$
$Cl−O$	$203$
$O−Cl$	$203$

The formula of $ΔH°$ is,

$ΔH°=∑npD°(bonds broken)−∑nfD°(bonds formed)$

Where,

$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$D°(bonds broken)$ is the bond energy of broken bonds.
$D°(bonds formed)$ is the bond energy of bonds formed.

Substitute all values from the table in the above equation.

$ΔH°=∑npD°(bonds broken)−∑nfD°(bonds formed)=(2×DO−H+2×DCl−O)−(2×DO−H+2×DO−Cl)=[(2×467)+(2×203)−{(2×467)+(2×203)}] kJ/mol$

Simplify the above equation.

$ΔH°=[(2×467)+(2×203)−{(2×467)+(2×203)}] kJ/mol=0 kJ/mol_$

(c)

Expert Solution

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Explanation of Solution

Explanation

The value of $ΔG°$ is $6.0×103 J/mol$ .

The value of $ΔH°$ is $0 J/mol$ .

Temperature is $298 K$ .

The formula of $ΔG°$ is,

$ΔG°=ΔH°−TΔS°$

Where,

$ΔH°$ is the standard enthalpy of reaction.
$ΔG°$ is the free energy change.
$T$ is the given temperature.
$ΔS°$ is the standard entropy of the reaction.

Substitute the values of $ΔG°,ΔH°$ and $T$ in equation (1).

$ΔG°=ΔH°−TΔS°6.0×103 J/mol=0−298(ΔS°)ΔS°=−20.13 J/K⋅mol_$

(d)

Expert Solution

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Explanation of Solution

Explanation

Refer Appendix $4$ .

The value of standard enthalpy of $H2O$ is $−241.82 kJ/mol$ .

The value of standard enthalpy of $Cl2O$ is $80.3 kJ/mol$ .

The formula of $ΔH°$ is,

$ΔH°=∑npΔH°(product)−∑nfΔH°(reactant)$

Where,

$ΔH°$ is the standard enthalpy of reaction.
$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$ΔH°(product)$ is the standard enthalpy of product at a pressure of $1 atm$ .
$ΔH°(reactant)$ is the standard enthalpy of reactant at a pressure of $1 atm$ .

Substitute all values from the table in the above equation.

$ΔH°=∑npΔH°(product)−∑nfΔH°(reactant)=[2(ΔH°HOCl)−(−241.82)−(80.3)] kJ/mol=−80.76 kJ/mol_$

Refer Appendix $4$ .

The value of standard entropy of $H2O$ is $188.83 J/K⋅mol$ .

The value of standard entropy of $Cl2O$ is $266.1 J/K⋅mol$ .

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

Where,

$ΔS°$ is the standard enthalpy of reaction.
$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$ΔS°(product)$ is the standard entropy of the product at a pressure of $1 atm$ .
$ΔS°(reactant)$ is the standard entropy of the reactant at a pressure of $1 atm$ .

Substitute all values from the table in the above equation.

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)−20.13 J/K⋅mol=[2(ΔS°HOCl)−(188.83)−(266.1)] J/K⋅molΔS°HOCl=−217.4 J/K⋅mol_$

(e)

Expert Solution

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Explanation of Solution

Explanation

The value of $ΔS°$ is $−20.3 J/K⋅mol$ .

The value of $ΔH°$ is $0 J/mol$ .

Temperature is $500 K$ .

It is assumed that $ΔH°$ and $ΔS°$ are independent of $T$ .

The formula of $ΔG°$ is,

$ΔG°=ΔH°−TΔS°$

Where,

$ΔH°$ is the standard enthalpy of reaction.
$ΔG°$ is the free energy change.
$T$ is the given temperature.
$ΔS°$ is the standard entropy of the reaction.

Substitute the values of $ΔG°,ΔH°$ and $T$ in equation (1).

$ΔG°=ΔH°−TΔS°=0−500(−20.13 J/K⋅mol)=1.01×104 J/mol_$

Formula

The expression for free energy change is,

$ΔG°=−RTln(K)ln(K)=−ΔG°RT$

Where,

$ΔG$ is the free energy change for a reaction at specified pressure.
$R$ is the gas law constant $(8.3145 J/K⋅mol)$ .
$K$ is the equilibrium constant.
$T$ is the absolute temperature.

Substitute the values of $R, ΔG°$ and $T$ in the above expression.

$ln(K)=−ΔG°RT=−1.01×104 J/mol(8.3145 J/K⋅mol)(500 K)K=0.088_$

(f)

Expert Solution

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Explanation of Solution

Explanation

Given

Partial pressure of $H2O$ , $PH2O$ , is $18 torr$ .

Partial pressure of $Cl2O$ , $PCl2O$ , is $2.0 torr$ .

Partial pressure of $HOCl$ , $PHOCl$ , is $0.10 torr$ .

The stated reaction is,

$H2O(g)+Cl2O(g)⇌2HOCl(g)$

If the partial pressure of reactant and product is given, then equilibrium pressure is expressed as $Kp$ and its expression is written as,

$Kp=(Partial pressure of product)a (Partial pressure of reactant)b$

Where,

$a$ is the number of moles of product.
$b$ is the number of moles of reactant.

The equilibrium constant expression for the given reaction is,

$Kp=(PHOCl)2(PH2O)(PCl2O)$

Substitute the values of $PCl2O, PH2O$ and $PHOCl$ in the above expression.

$Kp=(PHOCl)2(PH2O)(PCl2O)=(0.10 torr)2(18 torr)(2.0 torr)=2.7×10−4_$

The value of $Kp$ is $2.7×10−4$ .

Temperature is $298 K$ .

The expression for free energy change is,

$ΔG=ΔG°+RTln(Kp)$

Where,

$ΔG$ is the free energy change for a reaction at specified pressure.
$R$ is the gas law constant $(8.3145 J/K⋅mol)$ .
$T$ is the absolute temperature.
$Kp$ is the equilibrium constant.

Substitute the values of $R, Kp$ and $T$ in the equation.

$ΔG=ΔG°+RTln(Kp)=[6.0×103+(8.3145)(298)ln(2.7×10−4)] J/mol=−14.3×103 J/mol$

The conversion of joule-per mole $(J/mol)$ into kilo-joule per mole $(kJ/mol)$ is done as,

$1 J/mol=10−3 kJ/mol$

Hence,

The conversion of $−14.3×103 J/mol$ into kilo- joule per mole is,

$−14.3×103 J/mol=(−14.3×103×10−3) kJ/mol=−14.3 kJ/mol_$

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Answer 2

Textbook Question

Chapter 17, Problem 96AE

Consider the following reaction:

$H 2 O ( g ) + CI 2 O ( g ) ⇌ 2 HOCI ( g ) K 298 = 0.090$

For Cl₂O(g),

$Δ G f o = 97.9 KJ/mol Δ H f o = 80.3 KJ/mol S ∘ = 266.1 J / K ⋅ mol$

a. Calculate ∆G° for the reaction using the equation ∆G°=−RT In(K).

b. Use bond energy values (Table 3-3) to estimate ∆H° for the reaction.

c. Use the results from parts a and b to estimate ∆S° for the reaction.

d. Estimate $Δ H f o$ and S° for HOCI(g).

e. Estimate the value of K at 500. K.

f. Calculate ∆G at 25°C when $P H 2 O = 18 torr , P CI 2 O = 2.0 torr , a n d P HOCI = 0.10 torr$ .

(a)

Expert Solution

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Explanation of Solution

Explanation

Given

Temperature is $298 K$ .

The value of equilibrium constant is $0.090$ .

The stated reaction is,

$H2O(g)+Cl2O(g)⇌2HOCl(g)$

Formula

The expression for free energy change is,

$ΔG°=−RTln(K)$

Where,

$ΔG°$ is the standard Gibbs free energy change.
$R$ is the gas law constant $(8.3145 J/K⋅mol)$ .
$T$ is the absolute temperature.
$K$ is the equilibrium constant.

Substitute the values of $R, T$ and $K$ in the above expression.

$ΔG°=−RTln(K)=−(8.3145 J/K⋅mol)(298 K)ln(0.090)=6.0×103 J/mol$

The conversion of joule per mole $(J/mol)$ into kilo-joule per mole $(kJ/mol)$ is done as,

$1 J/mol=10−3 kJ/mol$

Hence,

The conversion of $6.0×103 J/mol$ into kilo-joule per mole is,

$6.0×103 J/mol=(6.0×103 ×10−3) kJ/mol=6.0 kJ/mol_$

(b)

Expert Solution

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Explanation of Solution

Explanation

The reaction that takes place is,

$H2O(g)+Cl2O(g)⇌2HOCl(g)$

Refer to Table $3-3$ .

The value of bond energy, $D°(kJ/mol)$ , for the given reactant and product is,

Bonds	$D°(kJ/mol)$
$O−H$	$467$
$Cl−O$	$203$
$O−Cl$	$203$

The formula of $ΔH°$ is,

$ΔH°=∑npD°(bonds broken)−∑nfD°(bonds formed)$

Where,

$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$D°(bonds broken)$ is the bond energy of broken bonds.
$D°(bonds formed)$ is the bond energy of bonds formed.

Substitute all values from the table in the above equation.

$ΔH°=∑npD°(bonds broken)−∑nfD°(bonds formed)=(2×DO−H+2×DCl−O)−(2×DO−H+2×DO−Cl)=[(2×467)+(2×203)−{(2×467)+(2×203)}] kJ/mol$

Simplify the above equation.

$ΔH°=[(2×467)+(2×203)−{(2×467)+(2×203)}] kJ/mol=0 kJ/mol_$

(c)

Expert Solution

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Explanation of Solution

Explanation

The value of $ΔG°$ is $6.0×103 J/mol$ .

The value of $ΔH°$ is $0 J/mol$ .

Temperature is $298 K$ .

The formula of $ΔG°$ is,

$ΔG°=ΔH°−TΔS°$

Where,

$ΔH°$ is the standard enthalpy of reaction.
$ΔG°$ is the free energy change.
$T$ is the given temperature.
$ΔS°$ is the standard entropy of the reaction.

Substitute the values of $ΔG°,ΔH°$ and $T$ in equation (1).

$ΔG°=ΔH°−TΔS°6.0×103 J/mol=0−298(ΔS°)ΔS°=−20.13 J/K⋅mol_$

(d)

Expert Solution

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Explanation of Solution

Explanation

Refer Appendix $4$ .

The value of standard enthalpy of $H2O$ is $−241.82 kJ/mol$ .

The value of standard enthalpy of $Cl2O$ is $80.3 kJ/mol$ .

The formula of $ΔH°$ is,

$ΔH°=∑npΔH°(product)−∑nfΔH°(reactant)$

Where,

$ΔH°$ is the standard enthalpy of reaction.
$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$ΔH°(product)$ is the standard enthalpy of product at a pressure of $1 atm$ .
$ΔH°(reactant)$ is the standard enthalpy of reactant at a pressure of $1 atm$ .

Substitute all values from the table in the above equation.

$ΔH°=∑npΔH°(product)−∑nfΔH°(reactant)=[2(ΔH°HOCl)−(−241.82)−(80.3)] kJ/mol=−80.76 kJ/mol_$

Refer Appendix $4$ .

The value of standard entropy of $H2O$ is $188.83 J/K⋅mol$ .

The value of standard entropy of $Cl2O$ is $266.1 J/K⋅mol$ .

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

Where,

$ΔS°$ is the standard enthalpy of reaction.
$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$ΔS°(product)$ is the standard entropy of the product at a pressure of $1 atm$ .
$ΔS°(reactant)$ is the standard entropy of the reactant at a pressure of $1 atm$ .

Substitute all values from the table in the above equation.

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)−20.13 J/K⋅mol=[2(ΔS°HOCl)−(188.83)−(266.1)] J/K⋅molΔS°HOCl=−217.4 J/K⋅mol_$

(e)

Expert Solution

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Explanation of Solution

Explanation

The value of $ΔS°$ is $−20.3 J/K⋅mol$ .

The value of $ΔH°$ is $0 J/mol$ .

Temperature is $500 K$ .

It is assumed that $ΔH°$ and $ΔS°$ are independent of $T$ .

The formula of $ΔG°$ is,

$ΔG°=ΔH°−TΔS°$

Where,

$ΔH°$ is the standard enthalpy of reaction.
$ΔG°$ is the free energy change.
$T$ is the given temperature.
$ΔS°$ is the standard entropy of the reaction.

Substitute the values of $ΔG°,ΔH°$ and $T$ in equation (1).

$ΔG°=ΔH°−TΔS°=0−500(−20.13 J/K⋅mol)=1.01×104 J/mol_$

Formula

The expression for free energy change is,

$ΔG°=−RTln(K)ln(K)=−ΔG°RT$

Where,

$ΔG$ is the free energy change for a reaction at specified pressure.
$R$ is the gas law constant $(8.3145 J/K⋅mol)$ .
$K$ is the equilibrium constant.
$T$ is the absolute temperature.

Substitute the values of $R, ΔG°$ and $T$ in the above expression.

$ln(K)=−ΔG°RT=−1.01×104 J/mol(8.3145 J/K⋅mol)(500 K)K=0.088_$

(f)

Expert Solution

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Explanation of Solution

Explanation

Given

Partial pressure of $H2O$ , $PH2O$ , is $18 torr$ .

Partial pressure of $Cl2O$ , $PCl2O$ , is $2.0 torr$ .

Partial pressure of $HOCl$ , $PHOCl$ , is $0.10 torr$ .

The stated reaction is,

$H2O(g)+Cl2O(g)⇌2HOCl(g)$

If the partial pressure of reactant and product is given, then equilibrium pressure is expressed as $Kp$ and its expression is written as,

$Kp=(Partial pressure of product)a (Partial pressure of reactant)b$

Where,

$a$ is the number of moles of product.
$b$ is the number of moles of reactant.

The equilibrium constant expression for the given reaction is,

$Kp=(PHOCl)2(PH2O)(PCl2O)$

Substitute the values of $PCl2O, PH2O$ and $PHOCl$ in the above expression.

$Kp=(PHOCl)2(PH2O)(PCl2O)=(0.10 torr)2(18 torr)(2.0 torr)=2.7×10−4_$

The value of $Kp$ is $2.7×10−4$ .

Temperature is $298 K$ .

The expression for free energy change is,

$ΔG=ΔG°+RTln(Kp)$

Where,

$ΔG$ is the free energy change for a reaction at specified pressure.
$R$ is the gas law constant $(8.3145 J/K⋅mol)$ .
$T$ is the absolute temperature.
$Kp$ is the equilibrium constant.

Substitute the values of $R, Kp$ and $T$ in the equation.

$ΔG=ΔG°+RTln(Kp)=[6.0×103+(8.3145)(298)ln(2.7×10−4)] J/mol=−14.3×103 J/mol$

The conversion of joule-per mole $(J/mol)$ into kilo-joule per mole $(kJ/mol)$ is done as,

$1 J/mol=10−3 kJ/mol$

Hence,

The conversion of $−14.3×103 J/mol$ into kilo- joule per mole is,

$−14.3×103 J/mol=(−14.3×103×10−3) kJ/mol=−14.3 kJ/mol_$

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Answer 3

Textbook Question

Chapter 17, Problem 96AE

Consider the following reaction:

$H 2 O ( g ) + CI 2 O ( g ) ⇌ 2 HOCI ( g ) K 298 = 0.090$

For Cl₂O(g),

$Δ G f o = 97.9 KJ/mol Δ H f o = 80.3 KJ/mol S ∘ = 266.1 J / K ⋅ mol$

a. Calculate ∆G° for the reaction using the equation ∆G°=−RT In(K).

b. Use bond energy values (Table 3-3) to estimate ∆H° for the reaction.

c. Use the results from parts a and b to estimate ∆S° for the reaction.

d. Estimate $Δ H f o$ and S° for HOCI(g).

e. Estimate the value of K at 500. K.

f. Calculate ∆G at 25°C when $P H 2 O = 18 torr , P CI 2 O = 2.0 torr , a n d P HOCI = 0.10 torr$ .

(a)

Expert Solution

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Explanation of Solution

Explanation

Given

Temperature is $298 K$ .

The value of equilibrium constant is $0.090$ .

The stated reaction is,

$H2O(g)+Cl2O(g)⇌2HOCl(g)$

Formula

The expression for free energy change is,

$ΔG°=−RTln(K)$

Where,

$ΔG°$ is the standard Gibbs free energy change.
$R$ is the gas law constant $(8.3145 J/K⋅mol)$ .
$T$ is the absolute temperature.
$K$ is the equilibrium constant.

Substitute the values of $R, T$ and $K$ in the above expression.

$ΔG°=−RTln(K)=−(8.3145 J/K⋅mol)(298 K)ln(0.090)=6.0×103 J/mol$

The conversion of joule per mole $(J/mol)$ into kilo-joule per mole $(kJ/mol)$ is done as,

$1 J/mol=10−3 kJ/mol$

Hence,

The conversion of $6.0×103 J/mol$ into kilo-joule per mole is,

$6.0×103 J/mol=(6.0×103 ×10−3) kJ/mol=6.0 kJ/mol_$

(b)

Expert Solution

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Explanation of Solution

Explanation

The reaction that takes place is,

$H2O(g)+Cl2O(g)⇌2HOCl(g)$

Refer to Table $3-3$ .

The value of bond energy, $D°(kJ/mol)$ , for the given reactant and product is,

Bonds	$D°(kJ/mol)$
$O−H$	$467$
$Cl−O$	$203$
$O−Cl$	$203$

The formula of $ΔH°$ is,

$ΔH°=∑npD°(bonds broken)−∑nfD°(bonds formed)$

Where,

$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$D°(bonds broken)$ is the bond energy of broken bonds.
$D°(bonds formed)$ is the bond energy of bonds formed.

Substitute all values from the table in the above equation.

$ΔH°=∑npD°(bonds broken)−∑nfD°(bonds formed)=(2×DO−H+2×DCl−O)−(2×DO−H+2×DO−Cl)=[(2×467)+(2×203)−{(2×467)+(2×203)}] kJ/mol$

Simplify the above equation.

$ΔH°=[(2×467)+(2×203)−{(2×467)+(2×203)}] kJ/mol=0 kJ/mol_$

(c)

Expert Solution

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Explanation of Solution

Explanation

The value of $ΔG°$ is $6.0×103 J/mol$ .

The value of $ΔH°$ is $0 J/mol$ .

Temperature is $298 K$ .

The formula of $ΔG°$ is,

$ΔG°=ΔH°−TΔS°$

Where,

$ΔH°$ is the standard enthalpy of reaction.
$ΔG°$ is the free energy change.
$T$ is the given temperature.
$ΔS°$ is the standard entropy of the reaction.

Substitute the values of $ΔG°,ΔH°$ and $T$ in equation (1).

$ΔG°=ΔH°−TΔS°6.0×103 J/mol=0−298(ΔS°)ΔS°=−20.13 J/K⋅mol_$

(d)

Expert Solution

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Explanation of Solution

Explanation

Refer Appendix $4$ .

The value of standard enthalpy of $H2O$ is $−241.82 kJ/mol$ .

The value of standard enthalpy of $Cl2O$ is $80.3 kJ/mol$ .

The formula of $ΔH°$ is,

$ΔH°=∑npΔH°(product)−∑nfΔH°(reactant)$

Where,

$ΔH°$ is the standard enthalpy of reaction.
$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$ΔH°(product)$ is the standard enthalpy of product at a pressure of $1 atm$ .
$ΔH°(reactant)$ is the standard enthalpy of reactant at a pressure of $1 atm$ .

Substitute all values from the table in the above equation.

$ΔH°=∑npΔH°(product)−∑nfΔH°(reactant)=[2(ΔH°HOCl)−(−241.82)−(80.3)] kJ/mol=−80.76 kJ/mol_$

Refer Appendix $4$ .

The value of standard entropy of $H2O$ is $188.83 J/K⋅mol$ .

The value of standard entropy of $Cl2O$ is $266.1 J/K⋅mol$ .

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

Where,

$ΔS°$ is the standard enthalpy of reaction.
$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$ΔS°(product)$ is the standard entropy of the product at a pressure of $1 atm$ .
$ΔS°(reactant)$ is the standard entropy of the reactant at a pressure of $1 atm$ .

Substitute all values from the table in the above equation.

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)−20.13 J/K⋅mol=[2(ΔS°HOCl)−(188.83)−(266.1)] J/K⋅molΔS°HOCl=−217.4 J/K⋅mol_$

(e)

Expert Solution

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Explanation of Solution

Explanation

The value of $ΔS°$ is $−20.3 J/K⋅mol$ .

The value of $ΔH°$ is $0 J/mol$ .

Temperature is $500 K$ .

It is assumed that $ΔH°$ and $ΔS°$ are independent of $T$ .

The formula of $ΔG°$ is,

$ΔG°=ΔH°−TΔS°$

Where,

$ΔH°$ is the standard enthalpy of reaction.
$ΔG°$ is the free energy change.
$T$ is the given temperature.
$ΔS°$ is the standard entropy of the reaction.

Substitute the values of $ΔG°,ΔH°$ and $T$ in equation (1).

$ΔG°=ΔH°−TΔS°=0−500(−20.13 J/K⋅mol)=1.01×104 J/mol_$

Formula

The expression for free energy change is,

$ΔG°=−RTln(K)ln(K)=−ΔG°RT$

Where,

$ΔG$ is the free energy change for a reaction at specified pressure.
$R$ is the gas law constant $(8.3145 J/K⋅mol)$ .
$K$ is the equilibrium constant.
$T$ is the absolute temperature.

Substitute the values of $R, ΔG°$ and $T$ in the above expression.

$ln(K)=−ΔG°RT=−1.01×104 J/mol(8.3145 J/K⋅mol)(500 K)K=0.088_$

(f)

Expert Solution

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Explanation of Solution

Explanation

Given

Partial pressure of $H2O$ , $PH2O$ , is $18 torr$ .

Partial pressure of $Cl2O$ , $PCl2O$ , is $2.0 torr$ .

Partial pressure of $HOCl$ , $PHOCl$ , is $0.10 torr$ .

The stated reaction is,

$H2O(g)+Cl2O(g)⇌2HOCl(g)$

If the partial pressure of reactant and product is given, then equilibrium pressure is expressed as $Kp$ and its expression is written as,

$Kp=(Partial pressure of product)a (Partial pressure of reactant)b$

Where,

$a$ is the number of moles of product.
$b$ is the number of moles of reactant.

The equilibrium constant expression for the given reaction is,

$Kp=(PHOCl)2(PH2O)(PCl2O)$

Substitute the values of $PCl2O, PH2O$ and $PHOCl$ in the above expression.

$Kp=(PHOCl)2(PH2O)(PCl2O)=(0.10 torr)2(18 torr)(2.0 torr)=2.7×10−4_$

The value of $Kp$ is $2.7×10−4$ .

Temperature is $298 K$ .

The expression for free energy change is,

$ΔG=ΔG°+RTln(Kp)$

Where,

$ΔG$ is the free energy change for a reaction at specified pressure.
$R$ is the gas law constant $(8.3145 J/K⋅mol)$ .
$T$ is the absolute temperature.
$Kp$ is the equilibrium constant.

Substitute the values of $R, Kp$ and $T$ in the equation.

$ΔG=ΔG°+RTln(Kp)=[6.0×103+(8.3145)(298)ln(2.7×10−4)] J/mol=−14.3×103 J/mol$

The conversion of joule-per mole $(J/mol)$ into kilo-joule per mole $(kJ/mol)$ is done as,

$1 J/mol=10−3 kJ/mol$

Hence,

The conversion of $−14.3×103 J/mol$ into kilo- joule per mole is,

$−14.3×103 J/mol=(−14.3×103×10−3) kJ/mol=−14.3 kJ/mol_$

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Answer 4

Textbook Question

Chapter 17, Problem 96AE

Consider the following reaction:

$H 2 O ( g ) + CI 2 O ( g ) ⇌ 2 HOCI ( g ) K 298 = 0.090$

For Cl₂O(g),

$Δ G f o = 97.9 KJ/mol Δ H f o = 80.3 KJ/mol S ∘ = 266.1 J / K ⋅ mol$

a. Calculate ∆G° for the reaction using the equation ∆G°=−RT In(K).

b. Use bond energy values (Table 3-3) to estimate ∆H° for the reaction.

c. Use the results from parts a and b to estimate ∆S° for the reaction.

d. Estimate $Δ H f o$ and S° for HOCI(g).

e. Estimate the value of K at 500. K.

f. Calculate ∆G at 25°C when $P H 2 O = 18 torr , P CI 2 O = 2.0 torr , a n d P HOCI = 0.10 torr$ .

(a)

Expert Solution

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Explanation of Solution

Explanation

Given

Temperature is $298 K$ .

The value of equilibrium constant is $0.090$ .

The stated reaction is,

$H2O(g)+Cl2O(g)⇌2HOCl(g)$

Formula

The expression for free energy change is,

$ΔG°=−RTln(K)$

Where,

$ΔG°$ is the standard Gibbs free energy change.
$R$ is the gas law constant $(8.3145 J/K⋅mol)$ .
$T$ is the absolute temperature.
$K$ is the equilibrium constant.

Substitute the values of $R, T$ and $K$ in the above expression.

$ΔG°=−RTln(K)=−(8.3145 J/K⋅mol)(298 K)ln(0.090)=6.0×103 J/mol$

The conversion of joule per mole $(J/mol)$ into kilo-joule per mole $(kJ/mol)$ is done as,

$1 J/mol=10−3 kJ/mol$

Hence,

The conversion of $6.0×103 J/mol$ into kilo-joule per mole is,

$6.0×103 J/mol=(6.0×103 ×10−3) kJ/mol=6.0 kJ/mol_$

(b)

Expert Solution

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Explanation of Solution

Explanation

The reaction that takes place is,

$H2O(g)+Cl2O(g)⇌2HOCl(g)$

Refer to Table $3-3$ .

The value of bond energy, $D°(kJ/mol)$ , for the given reactant and product is,

Bonds	$D°(kJ/mol)$
$O−H$	$467$
$Cl−O$	$203$
$O−Cl$	$203$

The formula of $ΔH°$ is,

$ΔH°=∑npD°(bonds broken)−∑nfD°(bonds formed)$

Where,

$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$D°(bonds broken)$ is the bond energy of broken bonds.
$D°(bonds formed)$ is the bond energy of bonds formed.

Substitute all values from the table in the above equation.

$ΔH°=∑npD°(bonds broken)−∑nfD°(bonds formed)=(2×DO−H+2×DCl−O)−(2×DO−H+2×DO−Cl)=[(2×467)+(2×203)−{(2×467)+(2×203)}] kJ/mol$

Simplify the above equation.

$ΔH°=[(2×467)+(2×203)−{(2×467)+(2×203)}] kJ/mol=0 kJ/mol_$

(c)

Expert Solution

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Explanation of Solution

Explanation

The value of $ΔG°$ is $6.0×103 J/mol$ .

The value of $ΔH°$ is $0 J/mol$ .

Temperature is $298 K$ .

The formula of $ΔG°$ is,

$ΔG°=ΔH°−TΔS°$

Where,

$ΔH°$ is the standard enthalpy of reaction.
$ΔG°$ is the free energy change.
$T$ is the given temperature.
$ΔS°$ is the standard entropy of the reaction.

Substitute the values of $ΔG°,ΔH°$ and $T$ in equation (1).

$ΔG°=ΔH°−TΔS°6.0×103 J/mol=0−298(ΔS°)ΔS°=−20.13 J/K⋅mol_$

(d)

Expert Solution

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Explanation of Solution

Explanation

Refer Appendix $4$ .

The value of standard enthalpy of $H2O$ is $−241.82 kJ/mol$ .

The value of standard enthalpy of $Cl2O$ is $80.3 kJ/mol$ .

The formula of $ΔH°$ is,

$ΔH°=∑npΔH°(product)−∑nfΔH°(reactant)$

Where,

$ΔH°$ is the standard enthalpy of reaction.
$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$ΔH°(product)$ is the standard enthalpy of product at a pressure of $1 atm$ .
$ΔH°(reactant)$ is the standard enthalpy of reactant at a pressure of $1 atm$ .

Substitute all values from the table in the above equation.

$ΔH°=∑npΔH°(product)−∑nfΔH°(reactant)=[2(ΔH°HOCl)−(−241.82)−(80.3)] kJ/mol=−80.76 kJ/mol_$

Refer Appendix $4$ .

The value of standard entropy of $H2O$ is $188.83 J/K⋅mol$ .

The value of standard entropy of $Cl2O$ is $266.1 J/K⋅mol$ .

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

Where,

$ΔS°$ is the standard enthalpy of reaction.
$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$ΔS°(product)$ is the standard entropy of the product at a pressure of $1 atm$ .
$ΔS°(reactant)$ is the standard entropy of the reactant at a pressure of $1 atm$ .

Substitute all values from the table in the above equation.

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)−20.13 J/K⋅mol=[2(ΔS°HOCl)−(188.83)−(266.1)] J/K⋅molΔS°HOCl=−217.4 J/K⋅mol_$

(e)

Expert Solution

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Explanation of Solution

Explanation

The value of $ΔS°$ is $−20.3 J/K⋅mol$ .

The value of $ΔH°$ is $0 J/mol$ .

Temperature is $500 K$ .

It is assumed that $ΔH°$ and $ΔS°$ are independent of $T$ .

The formula of $ΔG°$ is,

$ΔG°=ΔH°−TΔS°$

Where,

$ΔH°$ is the standard enthalpy of reaction.
$ΔG°$ is the free energy change.
$T$ is the given temperature.
$ΔS°$ is the standard entropy of the reaction.

Substitute the values of $ΔG°,ΔH°$ and $T$ in equation (1).

$ΔG°=ΔH°−TΔS°=0−500(−20.13 J/K⋅mol)=1.01×104 J/mol_$

Formula

The expression for free energy change is,

$ΔG°=−RTln(K)ln(K)=−ΔG°RT$

Where,

$ΔG$ is the free energy change for a reaction at specified pressure.
$R$ is the gas law constant $(8.3145 J/K⋅mol)$ .
$K$ is the equilibrium constant.
$T$ is the absolute temperature.

Substitute the values of $R, ΔG°$ and $T$ in the above expression.

$ln(K)=−ΔG°RT=−1.01×104 J/mol(8.3145 J/K⋅mol)(500 K)K=0.088_$

(f)

Expert Solution

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Explanation of Solution

Explanation

Given

Partial pressure of $H2O$ , $PH2O$ , is $18 torr$ .

Partial pressure of $Cl2O$ , $PCl2O$ , is $2.0 torr$ .

Partial pressure of $HOCl$ , $PHOCl$ , is $0.10 torr$ .

The stated reaction is,

$H2O(g)+Cl2O(g)⇌2HOCl(g)$

If the partial pressure of reactant and product is given, then equilibrium pressure is expressed as $Kp$ and its expression is written as,

$Kp=(Partial pressure of product)a (Partial pressure of reactant)b$

Where,

$a$ is the number of moles of product.
$b$ is the number of moles of reactant.

The equilibrium constant expression for the given reaction is,

$Kp=(PHOCl)2(PH2O)(PCl2O)$

Substitute the values of $PCl2O, PH2O$ and $PHOCl$ in the above expression.

$Kp=(PHOCl)2(PH2O)(PCl2O)=(0.10 torr)2(18 torr)(2.0 torr)=2.7×10−4_$

The value of $Kp$ is $2.7×10−4$ .

Temperature is $298 K$ .

The expression for free energy change is,

$ΔG=ΔG°+RTln(Kp)$

Where,

$ΔG$ is the free energy change for a reaction at specified pressure.
$R$ is the gas law constant $(8.3145 J/K⋅mol)$ .
$T$ is the absolute temperature.
$Kp$ is the equilibrium constant.

Substitute the values of $R, Kp$ and $T$ in the equation.

$ΔG=ΔG°+RTln(Kp)=[6.0×103+(8.3145)(298)ln(2.7×10−4)] J/mol=−14.3×103 J/mol$

The conversion of joule-per mole $(J/mol)$ into kilo-joule per mole $(kJ/mol)$ is done as,

$1 J/mol=10−3 kJ/mol$

Hence,

The conversion of $−14.3×103 J/mol$ into kilo- joule per mole is,

$−14.3×103 J/mol=(−14.3×103×10−3) kJ/mol=−14.3 kJ/mol_$

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Explanation of Solution

Explanation

Given

Temperature is $298 K$ .

The value of equilibrium constant is $0.090$ .

The stated reaction is,

$H2O(g)+Cl2O(g)⇌2HOCl(g)$

Formula

The expression for free energy change is,

$ΔG°=−RTln(K)$

Where,

$ΔG°$ is the standard Gibbs free energy change.
$R$ is the gas law constant $(8.3145 J/K⋅mol)$ .
$T$ is the absolute temperature.
$K$ is the equilibrium constant.

Substitute the values of $R, T$ and $K$ in the above expression.

$ΔG°=−RTln(K)=−(8.3145 J/K⋅mol)(298 K)ln(0.090)=6.0×103 J/mol$

The conversion of joule per mole $(J/mol)$ into kilo-joule per mole $(kJ/mol)$ is done as,

$1 J/mol=10−3 kJ/mol$

Hence,

The conversion of $6.0×103 J/mol$ into kilo-joule per mole is,

$6.0×103 J/mol=(6.0×103 ×10−3) kJ/mol=6.0 kJ/mol_$

(b)

Expert Solution

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Explanation of Solution

Explanation

The reaction that takes place is,

$H2O(g)+Cl2O(g)⇌2HOCl(g)$

Refer to Table $3-3$ .

The value of bond energy, $D°(kJ/mol)$ , for the given reactant and product is,

Bonds	$D°(kJ/mol)$
$O−H$	$467$
$Cl−O$	$203$
$O−Cl$	$203$

The formula of $ΔH°$ is,

$ΔH°=∑npD°(bonds broken)−∑nfD°(bonds formed)$

Where,

$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$D°(bonds broken)$ is the bond energy of broken bonds.
$D°(bonds formed)$ is the bond energy of bonds formed.

Substitute all values from the table in the above equation.

$ΔH°=∑npD°(bonds broken)−∑nfD°(bonds formed)=(2×DO−H+2×DCl−O)−(2×DO−H+2×DO−Cl)=[(2×467)+(2×203)−{(2×467)+(2×203)}] kJ/mol$

Simplify the above equation.

$ΔH°=[(2×467)+(2×203)−{(2×467)+(2×203)}] kJ/mol=0 kJ/mol_$

(c)

Expert Solution

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Explanation of Solution

Explanation

The value of $ΔG°$ is $6.0×103 J/mol$ .

The value of $ΔH°$ is $0 J/mol$ .

Temperature is $298 K$ .

The formula of $ΔG°$ is,

$ΔG°=ΔH°−TΔS°$

Where,

$ΔH°$ is the standard enthalpy of reaction.
$ΔG°$ is the free energy change.
$T$ is the given temperature.
$ΔS°$ is the standard entropy of the reaction.

Substitute the values of $ΔG°,ΔH°$ and $T$ in equation (1).

$ΔG°=ΔH°−TΔS°6.0×103 J/mol=0−298(ΔS°)ΔS°=−20.13 J/K⋅mol_$

(d)

Expert Solution

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Explanation of Solution

Explanation

Refer Appendix $4$ .

The value of standard enthalpy of $H2O$ is $−241.82 kJ/mol$ .

The value of standard enthalpy of $Cl2O$ is $80.3 kJ/mol$ .

The formula of $ΔH°$ is,

$ΔH°=∑npΔH°(product)−∑nfΔH°(reactant)$

Where,

$ΔH°$ is the standard enthalpy of reaction.
$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$ΔH°(product)$ is the standard enthalpy of product at a pressure of $1 atm$ .
$ΔH°(reactant)$ is the standard enthalpy of reactant at a pressure of $1 atm$ .

Substitute all values from the table in the above equation.

$ΔH°=∑npΔH°(product)−∑nfΔH°(reactant)=[2(ΔH°HOCl)−(−241.82)−(80.3)] kJ/mol=−80.76 kJ/mol_$

Refer Appendix $4$ .

The value of standard entropy of $H2O$ is $188.83 J/K⋅mol$ .

The value of standard entropy of $Cl2O$ is $266.1 J/K⋅mol$ .

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

Where,

$ΔS°$ is the standard enthalpy of reaction.
$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$ΔS°(product)$ is the standard entropy of the product at a pressure of $1 atm$ .
$ΔS°(reactant)$ is the standard entropy of the reactant at a pressure of $1 atm$ .

Substitute all values from the table in the above equation.

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)−20.13 J/K⋅mol=[2(ΔS°HOCl)−(188.83)−(266.1)] J/K⋅molΔS°HOCl=−217.4 J/K⋅mol_$

(e)

Expert Solution

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Explanation of Solution

Explanation

The value of $ΔS°$ is $−20.3 J/K⋅mol$ .

The value of $ΔH°$ is $0 J/mol$ .

Temperature is $500 K$ .

It is assumed that $ΔH°$ and $ΔS°$ are independent of $T$ .

The formula of $ΔG°$ is,

$ΔG°=ΔH°−TΔS°$

Where,

$ΔH°$ is the standard enthalpy of reaction.
$ΔG°$ is the free energy change.
$T$ is the given temperature.
$ΔS°$ is the standard entropy of the reaction.

Substitute the values of $ΔG°,ΔH°$ and $T$ in equation (1).

$ΔG°=ΔH°−TΔS°=0−500(−20.13 J/K⋅mol)=1.01×104 J/mol_$

Formula

The expression for free energy change is,

$ΔG°=−RTln(K)ln(K)=−ΔG°RT$

Where,

$ΔG$ is the free energy change for a reaction at specified pressure.
$R$ is the gas law constant $(8.3145 J/K⋅mol)$ .
$K$ is the equilibrium constant.
$T$ is the absolute temperature.

Substitute the values of $R, ΔG°$ and $T$ in the above expression.

$ln(K)=−ΔG°RT=−1.01×104 J/mol(8.3145 J/K⋅mol)(500 K)K=0.088_$

(f)

Expert Solution

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Explanation of Solution

Explanation

Given

Partial pressure of $H2O$ , $PH2O$ , is $18 torr$ .

Partial pressure of $Cl2O$ , $PCl2O$ , is $2.0 torr$ .

Partial pressure of $HOCl$ , $PHOCl$ , is $0.10 torr$ .

The stated reaction is,

$H2O(g)+Cl2O(g)⇌2HOCl(g)$

If the partial pressure of reactant and product is given, then equilibrium pressure is expressed as $Kp$ and its expression is written as,

$Kp=(Partial pressure of product)a (Partial pressure of reactant)b$

Where,

$a$ is the number of moles of product.
$b$ is the number of moles of reactant.

The equilibrium constant expression for the given reaction is,

$Kp=(PHOCl)2(PH2O)(PCl2O)$

Substitute the values of $PCl2O, PH2O$ and $PHOCl$ in the above expression.

$Kp=(PHOCl)2(PH2O)(PCl2O)=(0.10 torr)2(18 torr)(2.0 torr)=2.7×10−4_$

The value of $Kp$ is $2.7×10−4$ .

Temperature is $298 K$ .

The expression for free energy change is,

$ΔG=ΔG°+RTln(Kp)$

Where,

$ΔG$ is the free energy change for a reaction at specified pressure.
$R$ is the gas law constant $(8.3145 J/K⋅mol)$ .
$T$ is the absolute temperature.
$Kp$ is the equilibrium constant.

Substitute the values of $R, Kp$ and $T$ in the equation.

$ΔG=ΔG°+RTln(Kp)=[6.0×103+(8.3145)(298)ln(2.7×10−4)] J/mol=−14.3×103 J/mol$

The conversion of joule-per mole $(J/mol)$ into kilo-joule per mole $(kJ/mol)$ is done as,

$1 J/mol=10−3 kJ/mol$

Hence,

The conversion of $−14.3×103 J/mol$ into kilo- joule per mole is,

$−14.3×103 J/mol=(−14.3×103×10−3) kJ/mol=−14.3 kJ/mol_$

Answer 8

(a)

Expert Solution

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Explanation of Solution

Explanation

Given

Temperature is $298 K$ .

The value of equilibrium constant is $0.090$ .

The stated reaction is,

$H2O(g)+Cl2O(g)⇌2HOCl(g)$

Formula

The expression for free energy change is,

$ΔG°=−RTln(K)$

Where,

$ΔG°$ is the standard Gibbs free energy change.
$R$ is the gas law constant $(8.3145 J/K⋅mol)$ .
$T$ is the absolute temperature.
$K$ is the equilibrium constant.

Substitute the values of $R, T$ and $K$ in the above expression.

$ΔG°=−RTln(K)=−(8.3145 J/K⋅mol)(298 K)ln(0.090)=6.0×103 J/mol$

The conversion of joule per mole $(J/mol)$ into kilo-joule per mole $(kJ/mol)$ is done as,

$1 J/mol=10−3 kJ/mol$

Hence,

The conversion of $6.0×103 J/mol$ into kilo-joule per mole is,

$6.0×103 J/mol=(6.0×103 ×10−3) kJ/mol=6.0 kJ/mol_$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Answer 13

Explanation of Solution

Explanation

Given

Temperature is $298 K$ .

The value of equilibrium constant is $0.090$ .

The stated reaction is,

$H2O(g)+Cl2O(g)⇌2HOCl(g)$

Formula

The expression for free energy change is,

$ΔG°=−RTln(K)$

Where,

$ΔG°$ is the standard Gibbs free energy change.
$R$ is the gas law constant $(8.3145 J/K⋅mol)$ .
$T$ is the absolute temperature.
$K$ is the equilibrium constant.

Substitute the values of $R, T$ and $K$ in the above expression.

$ΔG°=−RTln(K)=−(8.3145 J/K⋅mol)(298 K)ln(0.090)=6.0×103 J/mol$

The conversion of joule per mole $(J/mol)$ into kilo-joule per mole $(kJ/mol)$ is done as,

$1 J/mol=10−3 kJ/mol$

Hence,

The conversion of $6.0×103 J/mol$ into kilo-joule per mole is,

$6.0×103 J/mol=(6.0×103 ×10−3) kJ/mol=6.0 kJ/mol_$

Answer 14

(b)

Expert Solution

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Explanation of Solution

Explanation

The reaction that takes place is,

$H2O(g)+Cl2O(g)⇌2HOCl(g)$

Refer to Table $3-3$ .

The value of bond energy, $D°(kJ/mol)$ , for the given reactant and product is,

Bonds	$D°(kJ/mol)$
$O−H$	$467$
$Cl−O$	$203$
$O−Cl$	$203$

The formula of $ΔH°$ is,

$ΔH°=∑npD°(bonds broken)−∑nfD°(bonds formed)$

Where,

$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$D°(bonds broken)$ is the bond energy of broken bonds.
$D°(bonds formed)$ is the bond energy of bonds formed.

Substitute all values from the table in the above equation.

$ΔH°=∑npD°(bonds broken)−∑nfD°(bonds formed)=(2×DO−H+2×DCl−O)−(2×DO−H+2×DO−Cl)=[(2×467)+(2×203)−{(2×467)+(2×203)}] kJ/mol$

Simplify the above equation.

$ΔH°=[(2×467)+(2×203)−{(2×467)+(2×203)}] kJ/mol=0 kJ/mol_$

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Answer 19

Explanation of Solution

Explanation

The reaction that takes place is,

$H2O(g)+Cl2O(g)⇌2HOCl(g)$

Refer to Table $3-3$ .

The value of bond energy, $D°(kJ/mol)$ , for the given reactant and product is,

Bonds	$D°(kJ/mol)$
$O−H$	$467$
$Cl−O$	$203$
$O−Cl$	$203$

The formula of $ΔH°$ is,

$ΔH°=∑npD°(bonds broken)−∑nfD°(bonds formed)$

Where,

$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$D°(bonds broken)$ is the bond energy of broken bonds.
$D°(bonds formed)$ is the bond energy of bonds formed.

Substitute all values from the table in the above equation.

$ΔH°=∑npD°(bonds broken)−∑nfD°(bonds formed)=(2×DO−H+2×DCl−O)−(2×DO−H+2×DO−Cl)=[(2×467)+(2×203)−{(2×467)+(2×203)}] kJ/mol$

Simplify the above equation.

$ΔH°=[(2×467)+(2×203)−{(2×467)+(2×203)}] kJ/mol=0 kJ/mol_$

Answer 20

(c)

Expert Solution

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Explanation of Solution

Explanation

The value of $ΔG°$ is $6.0×103 J/mol$ .

The value of $ΔH°$ is $0 J/mol$ .

Temperature is $298 K$ .

The formula of $ΔG°$ is,

$ΔG°=ΔH°−TΔS°$

Where,

$ΔH°$ is the standard enthalpy of reaction.
$ΔG°$ is the free energy change.
$T$ is the given temperature.
$ΔS°$ is the standard entropy of the reaction.

Substitute the values of $ΔG°,ΔH°$ and $T$ in equation (1).

$ΔG°=ΔH°−TΔS°6.0×103 J/mol=0−298(ΔS°)ΔS°=−20.13 J/K⋅mol_$

Answer 21

(c)

Expert Solution

Answer 22

(c)

Expert Solution

Answer 23

Expert Solution

Answer 24

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Answer 25

Explanation of Solution

Explanation

The value of $ΔG°$ is $6.0×103 J/mol$ .

The value of $ΔH°$ is $0 J/mol$ .

Temperature is $298 K$ .

The formula of $ΔG°$ is,

$ΔG°=ΔH°−TΔS°$

Where,

$ΔH°$ is the standard enthalpy of reaction.
$ΔG°$ is the free energy change.
$T$ is the given temperature.
$ΔS°$ is the standard entropy of the reaction.

Substitute the values of $ΔG°,ΔH°$ and $T$ in equation (1).

$ΔG°=ΔH°−TΔS°6.0×103 J/mol=0−298(ΔS°)ΔS°=−20.13 J/K⋅mol_$

Answer 26

(d)

Expert Solution

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Explanation of Solution

Explanation

Refer Appendix $4$ .

The value of standard enthalpy of $H2O$ is $−241.82 kJ/mol$ .

The value of standard enthalpy of $Cl2O$ is $80.3 kJ/mol$ .

The formula of $ΔH°$ is,

$ΔH°=∑npΔH°(product)−∑nfΔH°(reactant)$

Where,

$ΔH°$ is the standard enthalpy of reaction.
$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$ΔH°(product)$ is the standard enthalpy of product at a pressure of $1 atm$ .
$ΔH°(reactant)$ is the standard enthalpy of reactant at a pressure of $1 atm$ .

Substitute all values from the table in the above equation.

$ΔH°=∑npΔH°(product)−∑nfΔH°(reactant)=[2(ΔH°HOCl)−(−241.82)−(80.3)] kJ/mol=−80.76 kJ/mol_$

Refer Appendix $4$ .

The value of standard entropy of $H2O$ is $188.83 J/K⋅mol$ .

The value of standard entropy of $Cl2O$ is $266.1 J/K⋅mol$ .

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

Where,

$ΔS°$ is the standard enthalpy of reaction.
$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$ΔS°(product)$ is the standard entropy of the product at a pressure of $1 atm$ .
$ΔS°(reactant)$ is the standard entropy of the reactant at a pressure of $1 atm$ .

Substitute all values from the table in the above equation.

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)−20.13 J/K⋅mol=[2(ΔS°HOCl)−(188.83)−(266.1)] J/K⋅molΔS°HOCl=−217.4 J/K⋅mol_$

Answer 27

(d)

Expert Solution

Answer 28

(d)

Expert Solution

Answer 29

Expert Solution

Answer 30

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Answer 31

Explanation of Solution

Explanation

Refer Appendix $4$ .

The value of standard enthalpy of $H2O$ is $−241.82 kJ/mol$ .

The value of standard enthalpy of $Cl2O$ is $80.3 kJ/mol$ .

The formula of $ΔH°$ is,

$ΔH°=∑npΔH°(product)−∑nfΔH°(reactant)$

Where,

$ΔH°$ is the standard enthalpy of reaction.
$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$ΔH°(product)$ is the standard enthalpy of product at a pressure of $1 atm$ .
$ΔH°(reactant)$ is the standard enthalpy of reactant at a pressure of $1 atm$ .

Substitute all values from the table in the above equation.

$ΔH°=∑npΔH°(product)−∑nfΔH°(reactant)=[2(ΔH°HOCl)−(−241.82)−(80.3)] kJ/mol=−80.76 kJ/mol_$

Refer Appendix $4$ .

The value of standard entropy of $H2O$ is $188.83 J/K⋅mol$ .

The value of standard entropy of $Cl2O$ is $266.1 J/K⋅mol$ .

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

Where,

$ΔS°$ is the standard enthalpy of reaction.
$np$ is the number of moles of each product.
$nr$ is the number of moles each reactant.
$ΔS°(product)$ is the standard entropy of the product at a pressure of $1 atm$ .
$ΔS°(reactant)$ is the standard entropy of the reactant at a pressure of $1 atm$ .

Substitute all values from the table in the above equation.

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)−20.13 J/K⋅mol=[2(ΔS°HOCl)−(188.83)−(266.1)] J/K⋅molΔS°HOCl=−217.4 J/K⋅mol_$

Answer 32

(e)

Expert Solution

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Explanation of Solution

Explanation

The value of $ΔS°$ is $−20.3 J/K⋅mol$ .

The value of $ΔH°$ is $0 J/mol$ .

Temperature is $500 K$ .

It is assumed that $ΔH°$ and $ΔS°$ are independent of $T$ .

The formula of $ΔG°$ is,

$ΔG°=ΔH°−TΔS°$

Where,

$ΔH°$ is the standard enthalpy of reaction.
$ΔG°$ is the free energy change.
$T$ is the given temperature.
$ΔS°$ is the standard entropy of the reaction.

Substitute the values of $ΔG°,ΔH°$ and $T$ in equation (1).

$ΔG°=ΔH°−TΔS°=0−500(−20.13 J/K⋅mol)=1.01×104 J/mol_$

Formula

The expression for free energy change is,

$ΔG°=−RTln(K)ln(K)=−ΔG°RT$

Where,

$ΔG$ is the free energy change for a reaction at specified pressure.
$R$ is the gas law constant $(8.3145 J/K⋅mol)$ .
$K$ is the equilibrium constant.
$T$ is the absolute temperature.

Substitute the values of $R, ΔG°$ and $T$ in the above expression.

$ln(K)=−ΔG°RT=−1.01×104 J/mol(8.3145 J/K⋅mol)(500 K)K=0.088_$

Answer 33

(e)

Expert Solution

Answer 34

(e)

Expert Solution

Answer 35

Expert Solution

Answer 36

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Answer 37

Explanation of Solution

Explanation

The value of $ΔS°$ is $−20.3 J/K⋅mol$ .

The value of $ΔH°$ is $0 J/mol$ .

Temperature is $500 K$ .

It is assumed that $ΔH°$ and $ΔS°$ are independent of $T$ .

The formula of $ΔG°$ is,

$ΔG°=ΔH°−TΔS°$

Where,

$ΔH°$ is the standard enthalpy of reaction.
$ΔG°$ is the free energy change.
$T$ is the given temperature.
$ΔS°$ is the standard entropy of the reaction.

Substitute the values of $ΔG°,ΔH°$ and $T$ in equation (1).

$ΔG°=ΔH°−TΔS°=0−500(−20.13 J/K⋅mol)=1.01×104 J/mol_$

Formula

The expression for free energy change is,

$ΔG°=−RTln(K)ln(K)=−ΔG°RT$

Where,

$ΔG$ is the free energy change for a reaction at specified pressure.
$R$ is the gas law constant $(8.3145 J/K⋅mol)$ .
$K$ is the equilibrium constant.
$T$ is the absolute temperature.

Substitute the values of $R, ΔG°$ and $T$ in the above expression.

$ln(K)=−ΔG°RT=−1.01×104 J/mol(8.3145 J/K⋅mol)(500 K)K=0.088_$

Answer 38

(f)

Expert Solution

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$

To determine: The value of $ΔG°$ for the given reaction, using the equation $ΔG°=−RTln(K)$ .

Explanation of Solution

Explanation

Given

Partial pressure of $H2O$ , $PH2O$ , is $18 torr$ .

Partial pressure of $Cl2O$ , $PCl2O$ , is $2.0 torr$ .

Partial pressure of $HOCl$ , $PHOCl$ , is $0.10 torr$ .

The stated reaction is,

$H2O(g)+Cl2O(g)⇌2HOCl(g)$

If the partial pressure of reactant and product is given, then equilibrium pressure is expressed as $Kp$ and its expression is written as,

$Kp=(Partial pressure of product)a (Partial pressure of reactant)b$

Where,

$a$ is the number of moles of product.
$b$ is the number of moles of reactant.

The equilibrium constant expression for the given reaction is,

$Kp=(PHOCl)2(PH2O)(PCl2O)$

Substitute the values of $PCl2O, PH2O$ and $PHOCl$ in the above expression.

$Kp=(PHOCl)2(PH2O)(PCl2O)=(0.10 torr)2(18 torr)(2.0 torr)=2.7×10−4_$

The value of $Kp$ is $2.7×10−4$ .

Temperature is $298 K$ .

The expression for free energy change is,

$ΔG=ΔG°+RTln(Kp)$

Where,

$ΔG$ is the free energy change for a reaction at specified pressure.
$R$ is the gas law constant $(8.3145 J/K⋅mol)$ .
$T$ is the absolute temperature.
$Kp$ is the equilibrium constant.

Substitute the values of $R, Kp$ and $T$ in the equation.

$ΔG=ΔG°+RTln(Kp)=[6.0×103+(8.3145)(298)ln(2.7×10−4)] J/mol=−14.3×103 J/mol$

The conversion of joule-per mole $(J/mol)$ into kilo-joule per mole $(kJ/mol)$ is done as,

$1 J/mol=10−3 kJ/mol$

Hence,

The conversion of $−14.3×103 J/mol$ into kilo- joule per mole is,

$−14.3×103 J/mol=(−14.3×103×10−3) kJ/mol=−14.3 kJ/mol_$

Answer 39

(f)

Expert Solution

Answer 40

(f)

Expert Solution

Answer 41

Expert Solution

Answer 42

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $HOCl(g)$ , its equilibrium constant at $298 K$ and values of $ΔGf°, ΔHf°, S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$ , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$ΔG=0Q=K$

The expression for free energy change is,

$ΔG°=−RTln(K)$

The formula of $ΔS°$ is,

$ΔS°=∑npΔS°(product)−∑nfΔS°(reactant)$