Burning of 1 .00 gal gasoline to produce carbon-dioxide gas and water vapour, value of Δ H ∘ has to be calculated. Concept Introduction: The standard enthalpy of reaction is quantity of energy released or consumed when one mole of a substance is formed under standard conditions from its pure elemental form. It is denoted as ΔH rxn ° . ΔH rxn ° = ΔH p ° − ΔH r ° ΔH p ° = standard enthalpy of product ΔH r ° = standard enthalpy of reactant

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Answer 1

Question

Chapter 21, Problem 21.153P

a)

Interpretation Introduction

Interpretation:

Burning of $1.00 gal$ gasoline to produce carbon-dioxide gas and water vapour, value of $ΔH∘$ has to be calculated.

Concept Introduction:

The standard enthalpy of reaction is quantity of energy released or consumed when one mole of a substance is formed under standard conditions from its pure elemental form. It is denoted as $ΔHrxn°$ .

$ΔHrxn°= ΔHp°− ΔHr°ΔHp° = standard enthalpy of productΔHr° = standard enthalpy of reactant$

a)

Expert Solution

Explanation of Solution

The given chemical equation:

$2C8H8(l) + 25 O2 (g) → 16CO2 (g) +18 H2O (g)$

As known, heat of reaction is calculated using heat of formation as follows,

$ΔHrxn°=∑m ΔHf(poducts)°−∑m ΔHf(reactants)°$

$ΔHrxn°=[(16 mol CO2)(ΔHf° of CO2) + (18 mol H2O)(ΔHf° of H2O) ]- [(2 mol C8H18)(ΔHf° of C8H18) + (25 mol O2)(ΔHf° of O2) ]ΔHrxn°=[(16 mol CO2)(-393.5 kJ/mol) + (18 mol H2O)(-241.826 kJ/mol) ]- [(2 mol C8H18)(-250.1 kJ/mol) + (25 mol O2)(0 kJ/mol) ]ΔHrxn°= -10148.868 = -10148.9 kJ.$

The energy from $1.00 gal$ of gasoline as follows,

$Energy (kJ) =( 1.00 gal )(4 qt1 gal) (1 L1.057 qt) (1 mL10−3L)(0.7028 gmL) (1 mol C8H18114.22 g C8H18)(−10148.9 kJ2 mol C8H18) = -1.18158 × 105 = -1.18 × 105 kJ.$

Hence, the calculated value of $ΔH∘$ is $-1.18 × 105 kJ.$

b)

Interpretation Introduction

Interpretation:

Liters of $H2$ required to burn the given quantity of energy has to be calculated.

Concept introduction:

Ideal gas: A hypothetical gas whose molecules occupy negligible space and have no interactions, and which consequently obeys the gas law exactly.

Ideal gas equation: An equation describes the relationship among the four variables P, V, T, and n.

$PV = nRT$

The unit of each term in the equation is,

Where, R is proportionality constant called gas constant $(atm.L/mol.K)$

V is volume $(L)$

N is number of moles $(moles)$

P is Pressure $(atm)$

T is temperature $(K)$

Standard temperature and pressure: The conditions $0oC$ and $1 atm$ are called standard atmospheric conditions.

b)

Expert Solution

Explanation of Solution

The reaction as follows,

$H2(g) + 12 O2 (g) → H2O (g)$

The heat of formation of water vapour is $ΔHo = -241.826 kJ.$

The moles of hydrogen needed to produce the energy from given part a) is,

$Moles of H2 =( -1.18158 × 105 kJ)(1 mol H2-241.826 kJ) = 488.6 mol.$

In order to determine the volume, the ideal gas equation is used as shown below,

$V = nRTP = ((488.6 mol H2)(0.0821 L.atmmol.K)(298 K)1.00 atm) = 1.195 × 104 = 1.20 × 104 L.$

Therefore, the volume required calculated as $1.20 × 104 L.$

c)

Interpretation Introduction

Interpretation:

For the amount of $H2$ to produce the time required by electrolysis with a current of $1.00 × 103 A at 6.00 V$ has to be calculated.

c)

Expert Solution

Explanation of Solution

The reaction as follows,

$2H2O(l) + 2 e- → H2 (g) + 2OH- (aq)$

Using the known unit conversion; $1A = 1 C/s$

$Time (s) = (488.6 mol H2)(2 mol e-2 mol H2) (96485 C1 mol e-)(s1.00 × 103 C) = 9.43 × 104 seconds.$

Therefore, the required time calculated as $9.43 × 104 seconds.$

d)

Interpretation Introduction

Interpretation:

Power required generating the amount of $H2$ in kilowatt-hours has to be calculated.

d)

Expert Solution

Explanation of Solution

The coulombs involved in electrolysis of $488.6 mol of H2.$

$Coulombs =(488.6 mol H2)(2 mol e−1 mol H2) (96485 C1 mol e−)= 94,286,589 C.$

Conversion of Coulomb into Joules:

$Joules = C × V= 94,286,589 C × 6.00V = 565,719,534 J.$

Conversion of Joules into Power:

$Power (kW.h) = (565,719,534 J) [1 kW.h3.6 ×106J] = 157.144 = 157 kW.h.$

Therefore, the Power required in order of generating the amount of $H2$ in kilowatt-hours calculated as $157 kW.h.$

e)

Interpretation Introduction

Interpretation:

The cost of producing the amount of $H2$ equivalent to $1.00 gal$ of gasoline has to be calculated.

e)

Expert Solution

Explanation of Solution

Given:

Cell has efficiency of $88.0 %$ ; electricity costs of $Per kW.h$ is $$ 0.123.$

Additional electricity is necessary to produce sufficient hydrogen; the purpose of $(100 %80 %)$ factor.

$Cost = (157 kW.h)(100 %80 %) [0.123 cents1 kW.h] = 21.96 = 22.0 cents.$

Therefore, the cost of producing the amount of $H2$ equivalent to $1.00 gal$ of gasoline calculated as $22.0 cents.$

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Answer 2

Question

Chapter 21, Problem 21.153P

a)

Interpretation Introduction

Interpretation:

Burning of $1.00 gal$ gasoline to produce carbon-dioxide gas and water vapour, value of $ΔH∘$ has to be calculated.

Concept Introduction:

The standard enthalpy of reaction is quantity of energy released or consumed when one mole of a substance is formed under standard conditions from its pure elemental form. It is denoted as $ΔHrxn°$ .

$ΔHrxn°= ΔHp°− ΔHr°ΔHp° = standard enthalpy of productΔHr° = standard enthalpy of reactant$

a)

Expert Solution

Explanation of Solution

The given chemical equation:

$2C8H8(l) + 25 O2 (g) → 16CO2 (g) +18 H2O (g)$

As known, heat of reaction is calculated using heat of formation as follows,

$ΔHrxn°=∑m ΔHf(poducts)°−∑m ΔHf(reactants)°$

$ΔHrxn°=[(16 mol CO2)(ΔHf° of CO2) + (18 mol H2O)(ΔHf° of H2O) ]- [(2 mol C8H18)(ΔHf° of C8H18) + (25 mol O2)(ΔHf° of O2) ]ΔHrxn°=[(16 mol CO2)(-393.5 kJ/mol) + (18 mol H2O)(-241.826 kJ/mol) ]- [(2 mol C8H18)(-250.1 kJ/mol) + (25 mol O2)(0 kJ/mol) ]ΔHrxn°= -10148.868 = -10148.9 kJ.$

The energy from $1.00 gal$ of gasoline as follows,

$Energy (kJ) =( 1.00 gal )(4 qt1 gal) (1 L1.057 qt) (1 mL10−3L)(0.7028 gmL) (1 mol C8H18114.22 g C8H18)(−10148.9 kJ2 mol C8H18) = -1.18158 × 105 = -1.18 × 105 kJ.$

Hence, the calculated value of $ΔH∘$ is $-1.18 × 105 kJ.$

b)

Interpretation Introduction

Interpretation:

Liters of $H2$ required to burn the given quantity of energy has to be calculated.

Concept introduction:

Ideal gas: A hypothetical gas whose molecules occupy negligible space and have no interactions, and which consequently obeys the gas law exactly.

Ideal gas equation: An equation describes the relationship among the four variables P, V, T, and n.

$PV = nRT$

The unit of each term in the equation is,

Where, R is proportionality constant called gas constant $(atm.L/mol.K)$

V is volume $(L)$

N is number of moles $(moles)$

P is Pressure $(atm)$

T is temperature $(K)$

Standard temperature and pressure: The conditions $0oC$ and $1 atm$ are called standard atmospheric conditions.

b)

Expert Solution

Explanation of Solution

The reaction as follows,

$H2(g) + 12 O2 (g) → H2O (g)$

The heat of formation of water vapour is $ΔHo = -241.826 kJ.$

The moles of hydrogen needed to produce the energy from given part a) is,

$Moles of H2 =( -1.18158 × 105 kJ)(1 mol H2-241.826 kJ) = 488.6 mol.$

In order to determine the volume, the ideal gas equation is used as shown below,

$V = nRTP = ((488.6 mol H2)(0.0821 L.atmmol.K)(298 K)1.00 atm) = 1.195 × 104 = 1.20 × 104 L.$

Therefore, the volume required calculated as $1.20 × 104 L.$

c)

Interpretation Introduction

Interpretation:

For the amount of $H2$ to produce the time required by electrolysis with a current of $1.00 × 103 A at 6.00 V$ has to be calculated.

c)

Expert Solution

Explanation of Solution

The reaction as follows,

$2H2O(l) + 2 e- → H2 (g) + 2OH- (aq)$

Using the known unit conversion; $1A = 1 C/s$

$Time (s) = (488.6 mol H2)(2 mol e-2 mol H2) (96485 C1 mol e-)(s1.00 × 103 C) = 9.43 × 104 seconds.$

Therefore, the required time calculated as $9.43 × 104 seconds.$

d)

Interpretation Introduction

Interpretation:

Power required generating the amount of $H2$ in kilowatt-hours has to be calculated.

d)

Expert Solution

Explanation of Solution

The coulombs involved in electrolysis of $488.6 mol of H2.$

$Coulombs =(488.6 mol H2)(2 mol e−1 mol H2) (96485 C1 mol e−)= 94,286,589 C.$

Conversion of Coulomb into Joules:

$Joules = C × V= 94,286,589 C × 6.00V = 565,719,534 J.$

Conversion of Joules into Power:

$Power (kW.h) = (565,719,534 J) [1 kW.h3.6 ×106J] = 157.144 = 157 kW.h.$

Therefore, the Power required in order of generating the amount of $H2$ in kilowatt-hours calculated as $157 kW.h.$

e)

Interpretation Introduction

Interpretation:

The cost of producing the amount of $H2$ equivalent to $1.00 gal$ of gasoline has to be calculated.

e)

Expert Solution

Explanation of Solution

Given:

Cell has efficiency of $88.0 %$ ; electricity costs of $Per kW.h$ is $$ 0.123.$

Additional electricity is necessary to produce sufficient hydrogen; the purpose of $(100 %80 %)$ factor.

$Cost = (157 kW.h)(100 %80 %) [0.123 cents1 kW.h] = 21.96 = 22.0 cents.$

Therefore, the cost of producing the amount of $H2$ equivalent to $1.00 gal$ of gasoline calculated as $22.0 cents.$

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Answer 3

Question

Chapter 21, Problem 21.153P

a)

Interpretation Introduction

Interpretation:

Burning of $1.00 gal$ gasoline to produce carbon-dioxide gas and water vapour, value of $ΔH∘$ has to be calculated.

Concept Introduction:

The standard enthalpy of reaction is quantity of energy released or consumed when one mole of a substance is formed under standard conditions from its pure elemental form. It is denoted as $ΔHrxn°$ .

$ΔHrxn°= ΔHp°− ΔHr°ΔHp° = standard enthalpy of productΔHr° = standard enthalpy of reactant$

a)

Expert Solution

Explanation of Solution

The given chemical equation:

$2C8H8(l) + 25 O2 (g) → 16CO2 (g) +18 H2O (g)$

As known, heat of reaction is calculated using heat of formation as follows,

$ΔHrxn°=∑m ΔHf(poducts)°−∑m ΔHf(reactants)°$

$ΔHrxn°=[(16 mol CO2)(ΔHf° of CO2) + (18 mol H2O)(ΔHf° of H2O) ]- [(2 mol C8H18)(ΔHf° of C8H18) + (25 mol O2)(ΔHf° of O2) ]ΔHrxn°=[(16 mol CO2)(-393.5 kJ/mol) + (18 mol H2O)(-241.826 kJ/mol) ]- [(2 mol C8H18)(-250.1 kJ/mol) + (25 mol O2)(0 kJ/mol) ]ΔHrxn°= -10148.868 = -10148.9 kJ.$

The energy from $1.00 gal$ of gasoline as follows,

$Energy (kJ) =( 1.00 gal )(4 qt1 gal) (1 L1.057 qt) (1 mL10−3L)(0.7028 gmL) (1 mol C8H18114.22 g C8H18)(−10148.9 kJ2 mol C8H18) = -1.18158 × 105 = -1.18 × 105 kJ.$

Hence, the calculated value of $ΔH∘$ is $-1.18 × 105 kJ.$

b)

Interpretation Introduction

Interpretation:

Liters of $H2$ required to burn the given quantity of energy has to be calculated.

Concept introduction:

Ideal gas: A hypothetical gas whose molecules occupy negligible space and have no interactions, and which consequently obeys the gas law exactly.

Ideal gas equation: An equation describes the relationship among the four variables P, V, T, and n.

$PV = nRT$

The unit of each term in the equation is,

Where, R is proportionality constant called gas constant $(atm.L/mol.K)$

V is volume $(L)$

N is number of moles $(moles)$

P is Pressure $(atm)$

T is temperature $(K)$

Standard temperature and pressure: The conditions $0oC$ and $1 atm$ are called standard atmospheric conditions.

b)

Expert Solution

Explanation of Solution

The reaction as follows,

$H2(g) + 12 O2 (g) → H2O (g)$

The heat of formation of water vapour is $ΔHo = -241.826 kJ.$

The moles of hydrogen needed to produce the energy from given part a) is,

$Moles of H2 =( -1.18158 × 105 kJ)(1 mol H2-241.826 kJ) = 488.6 mol.$

In order to determine the volume, the ideal gas equation is used as shown below,

$V = nRTP = ((488.6 mol H2)(0.0821 L.atmmol.K)(298 K)1.00 atm) = 1.195 × 104 = 1.20 × 104 L.$

Therefore, the volume required calculated as $1.20 × 104 L.$

c)

Interpretation Introduction

Interpretation:

For the amount of $H2$ to produce the time required by electrolysis with a current of $1.00 × 103 A at 6.00 V$ has to be calculated.

c)

Expert Solution

Explanation of Solution

The reaction as follows,

$2H2O(l) + 2 e- → H2 (g) + 2OH- (aq)$

Using the known unit conversion; $1A = 1 C/s$

$Time (s) = (488.6 mol H2)(2 mol e-2 mol H2) (96485 C1 mol e-)(s1.00 × 103 C) = 9.43 × 104 seconds.$

Therefore, the required time calculated as $9.43 × 104 seconds.$

d)

Interpretation Introduction

Interpretation:

Power required generating the amount of $H2$ in kilowatt-hours has to be calculated.

d)

Expert Solution

Explanation of Solution

The coulombs involved in electrolysis of $488.6 mol of H2.$

$Coulombs =(488.6 mol H2)(2 mol e−1 mol H2) (96485 C1 mol e−)= 94,286,589 C.$

Conversion of Coulomb into Joules:

$Joules = C × V= 94,286,589 C × 6.00V = 565,719,534 J.$

Conversion of Joules into Power:

$Power (kW.h) = (565,719,534 J) [1 kW.h3.6 ×106J] = 157.144 = 157 kW.h.$

Therefore, the Power required in order of generating the amount of $H2$ in kilowatt-hours calculated as $157 kW.h.$

e)

Interpretation Introduction

Interpretation:

The cost of producing the amount of $H2$ equivalent to $1.00 gal$ of gasoline has to be calculated.

e)

Expert Solution

Explanation of Solution

Given:

Cell has efficiency of $88.0 %$ ; electricity costs of $Per kW.h$ is $$ 0.123.$

Additional electricity is necessary to produce sufficient hydrogen; the purpose of $(100 %80 %)$ factor.

$Cost = (157 kW.h)(100 %80 %) [0.123 cents1 kW.h] = 21.96 = 22.0 cents.$

Therefore, the cost of producing the amount of $H2$ equivalent to $1.00 gal$ of gasoline calculated as $22.0 cents.$

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Answer 4

Question

Chapter 21, Problem 21.153P

a)

Interpretation Introduction

Interpretation:

Burning of $1.00 gal$ gasoline to produce carbon-dioxide gas and water vapour, value of $ΔH∘$ has to be calculated.

Concept Introduction:

The standard enthalpy of reaction is quantity of energy released or consumed when one mole of a substance is formed under standard conditions from its pure elemental form. It is denoted as $ΔHrxn°$ .

$ΔHrxn°= ΔHp°− ΔHr°ΔHp° = standard enthalpy of productΔHr° = standard enthalpy of reactant$

a)

Expert Solution

Explanation of Solution

The given chemical equation:

$2C8H8(l) + 25 O2 (g) → 16CO2 (g) +18 H2O (g)$

As known, heat of reaction is calculated using heat of formation as follows,

$ΔHrxn°=∑m ΔHf(poducts)°−∑m ΔHf(reactants)°$

$ΔHrxn°=[(16 mol CO2)(ΔHf° of CO2) + (18 mol H2O)(ΔHf° of H2O) ]- [(2 mol C8H18)(ΔHf° of C8H18) + (25 mol O2)(ΔHf° of O2) ]ΔHrxn°=[(16 mol CO2)(-393.5 kJ/mol) + (18 mol H2O)(-241.826 kJ/mol) ]- [(2 mol C8H18)(-250.1 kJ/mol) + (25 mol O2)(0 kJ/mol) ]ΔHrxn°= -10148.868 = -10148.9 kJ.$

The energy from $1.00 gal$ of gasoline as follows,

$Energy (kJ) =( 1.00 gal )(4 qt1 gal) (1 L1.057 qt) (1 mL10−3L)(0.7028 gmL) (1 mol C8H18114.22 g C8H18)(−10148.9 kJ2 mol C8H18) = -1.18158 × 105 = -1.18 × 105 kJ.$

Hence, the calculated value of $ΔH∘$ is $-1.18 × 105 kJ.$

b)

Interpretation Introduction

Interpretation:

Liters of $H2$ required to burn the given quantity of energy has to be calculated.

Concept introduction:

Ideal gas: A hypothetical gas whose molecules occupy negligible space and have no interactions, and which consequently obeys the gas law exactly.

Ideal gas equation: An equation describes the relationship among the four variables P, V, T, and n.

$PV = nRT$

The unit of each term in the equation is,

Where, R is proportionality constant called gas constant $(atm.L/mol.K)$

V is volume $(L)$

N is number of moles $(moles)$

P is Pressure $(atm)$

T is temperature $(K)$

Standard temperature and pressure: The conditions $0oC$ and $1 atm$ are called standard atmospheric conditions.

b)

Expert Solution

Explanation of Solution

The reaction as follows,

$H2(g) + 12 O2 (g) → H2O (g)$

The heat of formation of water vapour is $ΔHo = -241.826 kJ.$

The moles of hydrogen needed to produce the energy from given part a) is,

$Moles of H2 =( -1.18158 × 105 kJ)(1 mol H2-241.826 kJ) = 488.6 mol.$

In order to determine the volume, the ideal gas equation is used as shown below,

$V = nRTP = ((488.6 mol H2)(0.0821 L.atmmol.K)(298 K)1.00 atm) = 1.195 × 104 = 1.20 × 104 L.$

Therefore, the volume required calculated as $1.20 × 104 L.$

c)

Interpretation Introduction

Interpretation:

For the amount of $H2$ to produce the time required by electrolysis with a current of $1.00 × 103 A at 6.00 V$ has to be calculated.

c)

Expert Solution

Explanation of Solution

The reaction as follows,

$2H2O(l) + 2 e- → H2 (g) + 2OH- (aq)$

Using the known unit conversion; $1A = 1 C/s$

$Time (s) = (488.6 mol H2)(2 mol e-2 mol H2) (96485 C1 mol e-)(s1.00 × 103 C) = 9.43 × 104 seconds.$

Therefore, the required time calculated as $9.43 × 104 seconds.$

d)

Interpretation Introduction

Interpretation:

Power required generating the amount of $H2$ in kilowatt-hours has to be calculated.

d)

Expert Solution

Explanation of Solution

The coulombs involved in electrolysis of $488.6 mol of H2.$

$Coulombs =(488.6 mol H2)(2 mol e−1 mol H2) (96485 C1 mol e−)= 94,286,589 C.$

Conversion of Coulomb into Joules:

$Joules = C × V= 94,286,589 C × 6.00V = 565,719,534 J.$

Conversion of Joules into Power:

$Power (kW.h) = (565,719,534 J) [1 kW.h3.6 ×106J] = 157.144 = 157 kW.h.$

Therefore, the Power required in order of generating the amount of $H2$ in kilowatt-hours calculated as $157 kW.h.$

e)

Interpretation Introduction

Interpretation:

The cost of producing the amount of $H2$ equivalent to $1.00 gal$ of gasoline has to be calculated.

e)

Expert Solution

Explanation of Solution

Given:

Cell has efficiency of $88.0 %$ ; electricity costs of $Per kW.h$ is $$ 0.123.$

Additional electricity is necessary to produce sufficient hydrogen; the purpose of $(100 %80 %)$ factor.

$Cost = (157 kW.h)(100 %80 %) [0.123 cents1 kW.h] = 21.96 = 22.0 cents.$

Therefore, the cost of producing the amount of $H2$ equivalent to $1.00 gal$ of gasoline calculated as $22.0 cents.$

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Answer 5

Chapter 21, Problem 21.153P

a)

Interpretation Introduction

Interpretation:

Burning of $1.00 gal$ gasoline to produce carbon-dioxide gas and water vapour, value of $ΔH∘$ has to be calculated.

Concept Introduction:

The standard enthalpy of reaction is quantity of energy released or consumed when one mole of a substance is formed under standard conditions from its pure elemental form. It is denoted as $ΔHrxn°$ .

$ΔHrxn°= ΔHp°− ΔHr°ΔHp° = standard enthalpy of productΔHr° = standard enthalpy of reactant$

a)

Expert Solution

Explanation of Solution

The given chemical equation:

$2C8H8(l) + 25 O2 (g) → 16CO2 (g) +18 H2O (g)$

As known, heat of reaction is calculated using heat of formation as follows,

$ΔHrxn°=∑m ΔHf(poducts)°−∑m ΔHf(reactants)°$

$ΔHrxn°=[(16 mol CO2)(ΔHf° of CO2) + (18 mol H2O)(ΔHf° of H2O) ]- [(2 mol C8H18)(ΔHf° of C8H18) + (25 mol O2)(ΔHf° of O2) ]ΔHrxn°=[(16 mol CO2)(-393.5 kJ/mol) + (18 mol H2O)(-241.826 kJ/mol) ]- [(2 mol C8H18)(-250.1 kJ/mol) + (25 mol O2)(0 kJ/mol) ]ΔHrxn°= -10148.868 = -10148.9 kJ.$

The energy from $1.00 gal$ of gasoline as follows,

$Energy (kJ) =( 1.00 gal )(4 qt1 gal) (1 L1.057 qt) (1 mL10−3L)(0.7028 gmL) (1 mol C8H18114.22 g C8H18)(−10148.9 kJ2 mol C8H18) = -1.18158 × 105 = -1.18 × 105 kJ.$

Hence, the calculated value of $ΔH∘$ is $-1.18 × 105 kJ.$

b)

Interpretation Introduction

Interpretation:

Liters of $H2$ required to burn the given quantity of energy has to be calculated.

Concept introduction:

Ideal gas: A hypothetical gas whose molecules occupy negligible space and have no interactions, and which consequently obeys the gas law exactly.

Ideal gas equation: An equation describes the relationship among the four variables P, V, T, and n.

$PV = nRT$

The unit of each term in the equation is,

Where, R is proportionality constant called gas constant $(atm.L/mol.K)$

V is volume $(L)$

N is number of moles $(moles)$

P is Pressure $(atm)$

T is temperature $(K)$

Standard temperature and pressure: The conditions $0oC$ and $1 atm$ are called standard atmospheric conditions.

b)

Expert Solution

Explanation of Solution

The reaction as follows,

$H2(g) + 12 O2 (g) → H2O (g)$

The heat of formation of water vapour is $ΔHo = -241.826 kJ.$

The moles of hydrogen needed to produce the energy from given part a) is,

$Moles of H2 =( -1.18158 × 105 kJ)(1 mol H2-241.826 kJ) = 488.6 mol.$

In order to determine the volume, the ideal gas equation is used as shown below,

$V = nRTP = ((488.6 mol H2)(0.0821 L.atmmol.K)(298 K)1.00 atm) = 1.195 × 104 = 1.20 × 104 L.$

Therefore, the volume required calculated as $1.20 × 104 L.$

c)

Interpretation Introduction

Interpretation:

For the amount of $H2$ to produce the time required by electrolysis with a current of $1.00 × 103 A at 6.00 V$ has to be calculated.

c)

Expert Solution

Explanation of Solution

The reaction as follows,

$2H2O(l) + 2 e- → H2 (g) + 2OH- (aq)$

Using the known unit conversion; $1A = 1 C/s$

$Time (s) = (488.6 mol H2)(2 mol e-2 mol H2) (96485 C1 mol e-)(s1.00 × 103 C) = 9.43 × 104 seconds.$

Therefore, the required time calculated as $9.43 × 104 seconds.$

d)

Interpretation Introduction

Interpretation:

Power required generating the amount of $H2$ in kilowatt-hours has to be calculated.

d)

Expert Solution

Explanation of Solution

The coulombs involved in electrolysis of $488.6 mol of H2.$

$Coulombs =(488.6 mol H2)(2 mol e−1 mol H2) (96485 C1 mol e−)= 94,286,589 C.$

Conversion of Coulomb into Joules:

$Joules = C × V= 94,286,589 C × 6.00V = 565,719,534 J.$

Conversion of Joules into Power:

$Power (kW.h) = (565,719,534 J) [1 kW.h3.6 ×106J] = 157.144 = 157 kW.h.$

Therefore, the Power required in order of generating the amount of $H2$ in kilowatt-hours calculated as $157 kW.h.$

e)

Interpretation Introduction

Interpretation:

The cost of producing the amount of $H2$ equivalent to $1.00 gal$ of gasoline has to be calculated.

e)

Expert Solution

Explanation of Solution

Given:

Cell has efficiency of $88.0 %$ ; electricity costs of $Per kW.h$ is $$ 0.123.$

Additional electricity is necessary to produce sufficient hydrogen; the purpose of $(100 %80 %)$ factor.

$Cost = (157 kW.h)(100 %80 %) [0.123 cents1 kW.h] = 21.96 = 22.0 cents.$

Therefore, the cost of producing the amount of $H2$ equivalent to $1.00 gal$ of gasoline calculated as $22.0 cents.$

Answer 6

Chapter 21, Problem 21.153P

a)

Interpretation Introduction

Interpretation:

Burning of $1.00 gal$ gasoline to produce carbon-dioxide gas and water vapour, value of $ΔH∘$ has to be calculated.

Concept Introduction:

The standard enthalpy of reaction is quantity of energy released or consumed when one mole of a substance is formed under standard conditions from its pure elemental form. It is denoted as $ΔHrxn°$ .

$ΔHrxn°= ΔHp°− ΔHr°ΔHp° = standard enthalpy of productΔHr° = standard enthalpy of reactant$

a)

Expert Solution

Explanation of Solution

The given chemical equation:

$2C8H8(l) + 25 O2 (g) → 16CO2 (g) +18 H2O (g)$

As known, heat of reaction is calculated using heat of formation as follows,

$ΔHrxn°=∑m ΔHf(poducts)°−∑m ΔHf(reactants)°$

$ΔHrxn°=[(16 mol CO2)(ΔHf° of CO2) + (18 mol H2O)(ΔHf° of H2O) ]- [(2 mol C8H18)(ΔHf° of C8H18) + (25 mol O2)(ΔHf° of O2) ]ΔHrxn°=[(16 mol CO2)(-393.5 kJ/mol) + (18 mol H2O)(-241.826 kJ/mol) ]- [(2 mol C8H18)(-250.1 kJ/mol) + (25 mol O2)(0 kJ/mol) ]ΔHrxn°= -10148.868 = -10148.9 kJ.$

The energy from $1.00 gal$ of gasoline as follows,

$Energy (kJ) =( 1.00 gal )(4 qt1 gal) (1 L1.057 qt) (1 mL10−3L)(0.7028 gmL) (1 mol C8H18114.22 g C8H18)(−10148.9 kJ2 mol C8H18) = -1.18158 × 105 = -1.18 × 105 kJ.$

Hence, the calculated value of $ΔH∘$ is $-1.18 × 105 kJ.$

Answer 7

Chapter 21, Problem 21.153P

a)

Interpretation Introduction

Interpretation:

Burning of $1.00 gal$ gasoline to produce carbon-dioxide gas and water vapour, value of $ΔH∘$ has to be calculated.

Concept Introduction:

The standard enthalpy of reaction is quantity of energy released or consumed when one mole of a substance is formed under standard conditions from its pure elemental form. It is denoted as $ΔHrxn°$ .

$ΔHrxn°= ΔHp°− ΔHr°ΔHp° = standard enthalpy of productΔHr° = standard enthalpy of reactant$

Answer 8

a)

Expert Solution

Explanation of Solution

The given chemical equation:

$2C8H8(l) + 25 O2 (g) → 16CO2 (g) +18 H2O (g)$

As known, heat of reaction is calculated using heat of formation as follows,

$ΔHrxn°=∑m ΔHf(poducts)°−∑m ΔHf(reactants)°$

$ΔHrxn°=[(16 mol CO2)(ΔHf° of CO2) + (18 mol H2O)(ΔHf° of H2O) ]- [(2 mol C8H18)(ΔHf° of C8H18) + (25 mol O2)(ΔHf° of O2) ]ΔHrxn°=[(16 mol CO2)(-393.5 kJ/mol) + (18 mol H2O)(-241.826 kJ/mol) ]- [(2 mol C8H18)(-250.1 kJ/mol) + (25 mol O2)(0 kJ/mol) ]ΔHrxn°= -10148.868 = -10148.9 kJ.$

The energy from $1.00 gal$ of gasoline as follows,

$Energy (kJ) =( 1.00 gal )(4 qt1 gal) (1 L1.057 qt) (1 mL10−3L)(0.7028 gmL) (1 mol C8H18114.22 g C8H18)(−10148.9 kJ2 mol C8H18) = -1.18158 × 105 = -1.18 × 105 kJ.$

Hence, the calculated value of $ΔH∘$ is $-1.18 × 105 kJ.$

Answer 9

a)

Expert Solution

Answer 10

a)

Expert Solution

Answer 11

Expert Solution

Answer 12

b)

Interpretation Introduction

Interpretation:

Liters of $H2$ required to burn the given quantity of energy has to be calculated.

Concept introduction:

Ideal gas: A hypothetical gas whose molecules occupy negligible space and have no interactions, and which consequently obeys the gas law exactly.

Ideal gas equation: An equation describes the relationship among the four variables P, V, T, and n.

$PV = nRT$

The unit of each term in the equation is,

Where, R is proportionality constant called gas constant $(atm.L/mol.K)$

V is volume $(L)$

N is number of moles $(moles)$

P is Pressure $(atm)$

T is temperature $(K)$

Standard temperature and pressure: The conditions $0oC$ and $1 atm$ are called standard atmospheric conditions.

b)

Expert Solution

Explanation of Solution

The reaction as follows,

$H2(g) + 12 O2 (g) → H2O (g)$

The heat of formation of water vapour is $ΔHo = -241.826 kJ.$

The moles of hydrogen needed to produce the energy from given part a) is,

$Moles of H2 =( -1.18158 × 105 kJ)(1 mol H2-241.826 kJ) = 488.6 mol.$

In order to determine the volume, the ideal gas equation is used as shown below,

$V = nRTP = ((488.6 mol H2)(0.0821 L.atmmol.K)(298 K)1.00 atm) = 1.195 × 104 = 1.20 × 104 L.$

Therefore, the volume required calculated as $1.20 × 104 L.$

Answer 13

b)

Interpretation Introduction

Interpretation:

Liters of $H2$ required to burn the given quantity of energy has to be calculated.

Concept introduction:

Ideal gas: A hypothetical gas whose molecules occupy negligible space and have no interactions, and which consequently obeys the gas law exactly.

Ideal gas equation: An equation describes the relationship among the four variables P, V, T, and n.

$PV = nRT$

The unit of each term in the equation is,

Where, R is proportionality constant called gas constant $(atm.L/mol.K)$

V is volume $(L)$

N is number of moles $(moles)$

P is Pressure $(atm)$

T is temperature $(K)$

Standard temperature and pressure: The conditions $0oC$ and $1 atm$ are called standard atmospheric conditions.

Answer 14

b)

Expert Solution

Explanation of Solution

The reaction as follows,

$H2(g) + 12 O2 (g) → H2O (g)$

The heat of formation of water vapour is $ΔHo = -241.826 kJ.$

The moles of hydrogen needed to produce the energy from given part a) is,

$Moles of H2 =( -1.18158 × 105 kJ)(1 mol H2-241.826 kJ) = 488.6 mol.$

In order to determine the volume, the ideal gas equation is used as shown below,

$V = nRTP = ((488.6 mol H2)(0.0821 L.atmmol.K)(298 K)1.00 atm) = 1.195 × 104 = 1.20 × 104 L.$

Therefore, the volume required calculated as $1.20 × 104 L.$

Answer 15

b)

Expert Solution

Answer 16

b)

Expert Solution

Answer 17

Expert Solution

Answer 18

c)

Interpretation Introduction

Interpretation:

For the amount of $H2$ to produce the time required by electrolysis with a current of $1.00 × 103 A at 6.00 V$ has to be calculated.

c)

Expert Solution

Explanation of Solution

The reaction as follows,

$2H2O(l) + 2 e- → H2 (g) + 2OH- (aq)$

Using the known unit conversion; $1A = 1 C/s$

$Time (s) = (488.6 mol H2)(2 mol e-2 mol H2) (96485 C1 mol e-)(s1.00 × 103 C) = 9.43 × 104 seconds.$

Therefore, the required time calculated as $9.43 × 104 seconds.$

Answer 19

c)

Interpretation Introduction

Interpretation:

For the amount of $H2$ to produce the time required by electrolysis with a current of $1.00 × 103 A at 6.00 V$ has to be calculated.

Answer 20

c)

Expert Solution

Explanation of Solution

The reaction as follows,

$2H2O(l) + 2 e- → H2 (g) + 2OH- (aq)$

Using the known unit conversion; $1A = 1 C/s$

$Time (s) = (488.6 mol H2)(2 mol e-2 mol H2) (96485 C1 mol e-)(s1.00 × 103 C) = 9.43 × 104 seconds.$

Therefore, the required time calculated as $9.43 × 104 seconds.$

Answer 21

c)

Expert Solution

Answer 22

c)

Expert Solution

Answer 23

Expert Solution

Answer 24

d)

Interpretation Introduction

Interpretation:

Power required generating the amount of $H2$ in kilowatt-hours has to be calculated.

d)

Expert Solution

Explanation of Solution

The coulombs involved in electrolysis of $488.6 mol of H2.$

$Coulombs =(488.6 mol H2)(2 mol e−1 mol H2) (96485 C1 mol e−)= 94,286,589 C.$

Conversion of Coulomb into Joules:

$Joules = C × V= 94,286,589 C × 6.00V = 565,719,534 J.$

Conversion of Joules into Power:

$Power (kW.h) = (565,719,534 J) [1 kW.h3.6 ×106J] = 157.144 = 157 kW.h.$

Therefore, the Power required in order of generating the amount of $H2$ in kilowatt-hours calculated as $157 kW.h.$

Answer 25

d)

Interpretation Introduction

Interpretation:

Power required generating the amount of $H2$ in kilowatt-hours has to be calculated.

Answer 26

d)

Expert Solution

Explanation of Solution

The coulombs involved in electrolysis of $488.6 mol of H2.$

$Coulombs =(488.6 mol H2)(2 mol e−1 mol H2) (96485 C1 mol e−)= 94,286,589 C.$

Conversion of Coulomb into Joules:

$Joules = C × V= 94,286,589 C × 6.00V = 565,719,534 J.$

Conversion of Joules into Power:

$Power (kW.h) = (565,719,534 J) [1 kW.h3.6 ×106J] = 157.144 = 157 kW.h.$

Therefore, the Power required in order of generating the amount of $H2$ in kilowatt-hours calculated as $157 kW.h.$

Answer 27

d)

Expert Solution

Answer 28

d)

Expert Solution

Answer 29

Expert Solution

Answer 30

e)

Interpretation Introduction

Interpretation:

The cost of producing the amount of $H2$ equivalent to $1.00 gal$ of gasoline has to be calculated.

e)

Expert Solution

Explanation of Solution

Given:

Cell has efficiency of $88.0 %$ ; electricity costs of $Per kW.h$ is $$ 0.123.$

Additional electricity is necessary to produce sufficient hydrogen; the purpose of $(100 %80 %)$ factor.

$Cost = (157 kW.h)(100 %80 %) [0.123 cents1 kW.h] = 21.96 = 22.0 cents.$

Therefore, the cost of producing the amount of $H2$ equivalent to $1.00 gal$ of gasoline calculated as $22.0 cents.$

Answer 31

e)

Interpretation Introduction

Interpretation:

The cost of producing the amount of $H2$ equivalent to $1.00 gal$ of gasoline has to be calculated.

Answer 32

e)

Expert Solution

Explanation of Solution

Given:

Cell has efficiency of $88.0 %$ ; electricity costs of $Per kW.h$ is $$ 0.123.$

Additional electricity is necessary to produce sufficient hydrogen; the purpose of $(100 %80 %)$ factor.

$Cost = (157 kW.h)(100 %80 %) [0.123 cents1 kW.h] = 21.96 = 22.0 cents.$

Therefore, the cost of producing the amount of $H2$ equivalent to $1.00 gal$ of gasoline calculated as $22.0 cents.$

Answer 33

e)

Expert Solution

Answer 34

e)

Expert Solution

Answer 35

Expert Solution

Answer 36

Ch. 21.1 - Write a balanced molecular equation for the...Ch. 21.1 - Prob. 21.1BFP Ch. 21.2 - In one half-cell of a voltaic cell, a graphite rod...Ch. 21.2 - In one half-cell of a voltaic cell, a graphite rod...Ch. 21.3 - Prob. 21.3AFP Ch. 21.3 - Prob. 21.3BFP Ch. 21.3 - Prob. 21.4AFP Ch. 21.3 - Prob. 21.4BFP Ch. 21.3 - Combine pairs of the balanced half-reactions (1),...Ch. 21.3 - Prob. 21.5BFP

Videos

Explanation of Solution

Explanation of Solution

Interpretation:

For the amount of $H2$ to produce the time required by electrolysis with a current of $1.00 × 103 A at 6.00 V$ has to be calculated.

Explanation of Solution

Interpretation:

Power required generating the amount of $H2$ in kilowatt-hours has to be calculated.

Explanation of Solution

Interpretation:

The cost of producing the amount of $H2$ equivalent to $1.00 gal$ of gasoline has to be calculated.

Explanation of Solution

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Chapter 21 Solutions

Videos

Explanation of Solution

Explanation of Solution

Explanation of Solution

Interpretation: Power required generating the amount of H2<m:math><m:mrow><m:msub><m:mtext>H</m:mtext><m:mtext>2</m:mtext></m:msub></m:mrow></m:math> in kilowatt-hours has to be calculated.

Explanation of Solution

Explanation of Solution

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Chapter 21 Solutions

Interpretation:

Power required generating the amount of $H2$ in kilowatt-hours has to be calculated.