The balanced chemical equation for the given reaction between hydrazine and oxygen also the oxidizing and reducing agents in the reaction should be identified. Concept introduction: Redox reaction : It occurs when oxidation and reduction takes place at the same time in a chemical reaction . Chemical equation is the representation of a chemical reaction, in which the reactants and products of the reactions are represented left and right side of an arrow respectively by using their respective chemical formulas. Reactant of a chemical reaction is the substrate compounds or the compounds which undergo a chemical reaction. Product of a chemical reaction is the produced compounds or the compounds formed after a chemical reaction. Balanced chemical equation of a reaction is written according to law of conservation of mass. Stoichiometry of a chemical reaction is the relation between reactants and products of the reaction and it is represented by the coefficients used for the reactants and products involved in the chemical equation. The hot water system can corrode because of the presence of dissolved oxygen in water. This dissolved oxygen leads to corrosion. However, this oxygen can be removed from water with the use of hydrazine. The formula for hydrazine is N 2 H 4 . It reacts with oxygen to produce water and N 2 .

Question

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Answer 1

Question

Chapter 18, Problem 97SCQ

(a)

Interpretation Introduction

Interpretation:

The balanced chemical equation for the given reaction between hydrazine and oxygen also the oxidizing and reducing agents in the reaction should be identified.

Concept introduction:

Redox reaction: It occurs when oxidation and reduction takes place at the same time in a chemical reaction.

Chemical equation is the representation of a chemical reaction, in which the reactants and products of the reactions are represented left and right side of an arrow respectively by using their respective chemical formulas.

Reactant of a chemical reaction is the substrate compounds or the compounds which undergo a chemical reaction.

Product of a chemical reaction is the produced compounds or the compounds formed after a chemical reaction.

Balanced chemical equation of a reaction is written according to law of conservation of mass.

Stoichiometry of a chemical reaction is the relation between reactants and products of the reaction and it is represented by the coefficients used for the reactants and products involved in the chemical equation.

The hot water system can corrode because of the presence of dissolved oxygen in water. This dissolved oxygen leads to corrosion. However, this oxygen can be removed from water with the use of hydrazine. The formula for hydrazine is $N2H4$ . It reacts with oxygen to produce water and $N2$ .

(a)

Expert Solution

Answer to Problem 97SCQ

The balanced chemical equation for the reaction of hydrazine and oxygen is,

$N2H4(l)+O2(g)→2H2O(l)+N2(g)$

Oxygen is oxidising agent and hydrazine is a reducing agent.

Explanation of Solution

The formula for hydrazine is $N2H4$ . It reacts with oxygen to produce water and $N2$ .

The balanced chemical equation for the reaction of hydrazine and oxygen is,

$N2H4(l)+O2(g)→2H2O(l)+N2(g)$

Oxygen acts as an oxidising agent in the reaction and changes its oxidation number from $0$ to $−2$ . Hydrazine acts as a reducing agent and loses hydrogen.

(b)

Interpretation Introduction

Interpretation:

The value of $ΔrH°$ , $ΔrS°$ and $ΔrGo$ for the given reaction of $1 mol of N2H4 at 25oc$ should be identified.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by $ΔGo$ . It is related to entropy and entropy by the following expression,

$ΔGo= ΔHo- TΔSo$

The sign of $ΔGo$ should be positive for a product-favored reaction. Thus, spontaneous reactions are referred to those that have negative free energy formation.

The sign of $ΔSo$ should be positive for an entropy-favoured reaction and the sign of $ΔH∘$ should be negative for an enthalpy favoured-reaction.

$ΔrHo$ is negative for an exothermic reaction and positive for endothermic reactions. The entropy change is positive whenever number of moles of gases is increasing in any reaction.

(b)

Expert Solution

Answer to Problem 97SCQ

The value of $ΔrH°$ for the reaction of hydrazine and water is $−622.29 kJ/mol-rxn$ .

The value of $ΔrS°$ for the reaction of hydrazine and water is $4.87 J/K⋅mol-rxn$ .

The value of $ΔrGo$ for the reaction of hydrazine and water is $−623.74 kJ/mol-rxn$ .

Explanation of Solution

The value of $ΔrH°$ , $ΔrS°$ and $ΔrGo$ for the reaction of hydrazine and water is calculated below.

Given:

Refer to Appendix L for the values of standard entropies and enthalpies.

The standard entropy value for $N2H4(l)$ is $121.52 J/K⋅mol$ .

The standard entropy value for $O2(g)$ is $205.07 J/K⋅mol$ .

The standard entropy value for $H2O(l)$ is $69.95 J/K⋅mol$ .

The standard entropy value for $N2(g)$ is $191.56 J/K⋅mol$ .

The standard enthalpy value for $N2H4(l)$ is $50.63 kJ/mol$ .

The standard enthalpy value for $O2(g)$ is $0 kJ/mol$ .

The standard enthalpy value for $H2O(l)$ is $−285.83 kJ/mol$ .

The standard enthalpy value for $N2(g)$ is $0 kJ/mol$ .

The given reaction is,

$N2H4(l)+O2(g)→2H2O(l)+N2(g)$

The standard enthalpy change is,

$ΔrH°=∑nΔfH°(products)−∑nΔfH°(reactants)=[[(2 mol H2O(l)/mol-rxn)ΔfH°[H2O(l)]+(1 mol N2(g)/mol-rxn)ΔfH°[N2(g)]]-[(1 mol O2(g)/mol-rxn)ΔfH°[O2(g)]+(1 mol N2H4(l)/mol-rxn)ΔfH°[N2H4(l)]]]$

Substitute the values,

$ΔrH°=[[(2 mol H2O(l)/mol-rxn)(-285.83 kJ/mol)+(1 mol N2(g)/mol-rxn)(0 kJ/mol)]-[(1 mol O2(g)/mol-rxn)(0 kJ/mol)+(1 mol N2H4(l)/mol-rxn)(50.63 kJ/mol)]]=-622.29 kJ/mol-rxn$

The standard entropy change is,

$ΔrS°= ∑nS°(products)-∑nS°(reactants)=[[(2 mol H2O(l)/mol-rxn)S°[H2O(l)]+(1 mol N2(g)/mol-rxn)S°[N2(g)]]-[(1 mol O2(g)/mol-rxn)S°[O2(g)]+(1 mol N2H4(l)/mol-rxn)S°[N2H4(l)]]]$

Substitute the values,

$ΔrS°= [[(2 mol H2O(l)/mol-rxn)(69.95 J/K×mol)+(1 mol N2(g)/mol-rxn)(191.56 J/K×mol)]-[(1 mol O2(g)/mol-rxn)(205.07 J/K×mol)+(1 mol N2H4(l)/mol-rxn)(121.52 J/K×mol)]]= 4.87 J/K×mol-rxn$

Now,

$ΔrGo= ΔrHo- TΔrSo$

Substitute the value of $ΔrHo$ and $ΔrSo$ .

$ΔGo= -622.29 kJ/mol-rxn-[(1000 K)(4.87 J/K×mol-rxn)](1 kJ1000 J)= -623.74 kJ/mol-rxn$

(c)

Interpretation Introduction

Interpretation:

The change in temperature expected in heating system that has $5.5×104L$ of water should be identified.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by $ΔGo$ . It is related to entropy and entropy by the following expression,

$ΔGo= ΔHo- TΔSo$

The sign of $ΔGo$ should be positive for a product-favored reaction. Thus, spontaneous reactions are referred to those that have negative free energy formation.

The sign of $ΔSo$ should be positive for an entropy-favoured reaction and the sign of $ΔHo$ should be negative for an enthalpy favoured-reaction.

$ΔrHo$ is negative for an exothermic reaction and positive for endothermic reactions. The entropy change is positive whenever number of moles of gases is increasing in any reaction.

(c)

Expert Solution

Answer to Problem 97SCQ

The temperature change expected in a heating system containing $5.5×104 L$ of water is $2.7×10−3 K$ .

Explanation of Solution

The temperature change expected in a heating system containing $5.5×104 L$ of water is calculated below.

Given:

The value of $ΔrHo$ for the reaction of hydrazine and water is $−622.29 kJ/mol-rxn$ .

Thus, one mole of hydrazine releases $622.29×103 J$ of heat.

The density of water is $0.996 g/ml$ . Thus, the value of mass $m$ is,

$mass = (density)(volume)= (996 g/L)(5.5×104 L)= 5478×104g$

The specific heat $C$ of water is $4.184 J/g⋅K$ .

The heat of the system is related to specific heat and temperature by the expression,

$q = m c ΔT$

Substitute the values,

$622.29×103 J = (5478×104g) (4.184 J/g×K) ΔTΔT = 2.7×10-3 K$

(d)

Interpretation Introduction

Interpretation:

The amount of $O2$ present in hot water heating system of $5.5×104kg$ should be identified.

Concept introduction:

Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in $12 g$ of $12C$ .

From given mass of substance moles could be calculated by using the following formula,

$Moles of substance = Given mass of substanceMolecular mass$

(d)

Expert Solution

Answer to Problem 97SCQ

The number of moles of $O2$ present in the system is $7.5 moles$ .

Explanation of Solution

The number of moles of $O2$ present in the system is calculated below.

Given:

The solubilty of oxygen is in water at $25 °C$ is $0.000434g/100g$ of water.

Thus, the number of moles $n$ can be calculated by dividing the given mass by the molar mass and multipying it with the solubilty given.

$n = (5.5×107g)1 mol O2(32.00 g)(0.00434 g100 g)= 7.5 moles O2$

(e)

Interpretation Introduction

Interpretation:

The mass of solution that has $5%$ dissolved hydrazine in water required to totally consume the dissolved amount of $O2$ .

Concept introduction:

Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in $12 g$ of $12C$ .

From given mass of substance moles could be calculated by using the following formula,

$Moles of substance = Given mass of substanceMolecular mass$

Mass: It is the quantitative measure of a substance. The amount of matter present in substance is expressed as mass. The $S.I.$ unit of mass is $kg$ .

(e)

Expert Solution

Answer to Problem 97SCQ

The mass of hydrazine solution that should be added to totally consume the dissolved oxygen is $4.8×103 g$ .

Explanation of Solution

The mass of hydrazine solution that should be added to totally consume the dissolved oxygen is calculated below.

Given:

Hydrazine is available as $5 %$ solution in water. If the total solution is $100 g$ then the mass of hydrazine is $5 g$ . The molar mass of hydrazine is $32.05 g$ .

The number of moles of oxygen present is $7.5 moles$ .

Thus the mass $m$ of hydrazine solution that should be added to totally consume the dissolved oxygen is,

$m = (7.5 mol O21 mol N2H4 )(100 g solution5 g)(32.05 g N2H4)= 4.8×103 g solution$

(f)

Interpretation Introduction

Interpretation:

The volume of $N2$ produced under given conditions should be calculated.

Concept introduction:

Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in $12 g$ of $12C$ .

From given mass of substance moles could be calculated by using the following formula,

$Moles of substance = Given mass of substanceMolecular mass$

Mass: It is the quantitative measure of a substance. The amount of matter present in substance is expressed as mass. The $S.I.$ unit of mass is $kg$ .

(f)

Expert Solution

Answer to Problem 97SCQ

The volume of $N2(g)$ that will be produced is $168 L$ .

Explanation of Solution

The volume of $N2(g)$ that will be produced is calculated below.

Given:

The given reaction is,

$N2H4(l)+O2(g)→2H2O(l)+N2(g)$

Thus, $7.5 moles$ of $O2(g)$ will release $7.5 moles$ of $N2(g)$ . At standard temperature and pressure, the volume of nitrogen gas $V$ is,

$V=(7.5 moles)(22.5 L/mol)=168 L$

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Answer 2

Question

Chapter 18, Problem 97SCQ

(a)

Interpretation Introduction

Interpretation:

The balanced chemical equation for the given reaction between hydrazine and oxygen also the oxidizing and reducing agents in the reaction should be identified.

Concept introduction:

Redox reaction: It occurs when oxidation and reduction takes place at the same time in a chemical reaction.

Chemical equation is the representation of a chemical reaction, in which the reactants and products of the reactions are represented left and right side of an arrow respectively by using their respective chemical formulas.

Reactant of a chemical reaction is the substrate compounds or the compounds which undergo a chemical reaction.

Product of a chemical reaction is the produced compounds or the compounds formed after a chemical reaction.

Balanced chemical equation of a reaction is written according to law of conservation of mass.

Stoichiometry of a chemical reaction is the relation between reactants and products of the reaction and it is represented by the coefficients used for the reactants and products involved in the chemical equation.

The hot water system can corrode because of the presence of dissolved oxygen in water. This dissolved oxygen leads to corrosion. However, this oxygen can be removed from water with the use of hydrazine. The formula for hydrazine is $N2H4$ . It reacts with oxygen to produce water and $N2$ .

(a)

Expert Solution

Answer to Problem 97SCQ

The balanced chemical equation for the reaction of hydrazine and oxygen is,

$N2H4(l)+O2(g)→2H2O(l)+N2(g)$

Oxygen is oxidising agent and hydrazine is a reducing agent.

Explanation of Solution

The formula for hydrazine is $N2H4$ . It reacts with oxygen to produce water and $N2$ .

The balanced chemical equation for the reaction of hydrazine and oxygen is,

$N2H4(l)+O2(g)→2H2O(l)+N2(g)$

Oxygen acts as an oxidising agent in the reaction and changes its oxidation number from $0$ to $−2$ . Hydrazine acts as a reducing agent and loses hydrogen.

(b)

Interpretation Introduction

Interpretation:

The value of $ΔrH°$ , $ΔrS°$ and $ΔrGo$ for the given reaction of $1 mol of N2H4 at 25oc$ should be identified.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by $ΔGo$ . It is related to entropy and entropy by the following expression,

$ΔGo= ΔHo- TΔSo$

The sign of $ΔGo$ should be positive for a product-favored reaction. Thus, spontaneous reactions are referred to those that have negative free energy formation.

The sign of $ΔSo$ should be positive for an entropy-favoured reaction and the sign of $ΔH∘$ should be negative for an enthalpy favoured-reaction.

$ΔrHo$ is negative for an exothermic reaction and positive for endothermic reactions. The entropy change is positive whenever number of moles of gases is increasing in any reaction.

(b)

Expert Solution

Answer to Problem 97SCQ

The value of $ΔrH°$ for the reaction of hydrazine and water is $−622.29 kJ/mol-rxn$ .

The value of $ΔrS°$ for the reaction of hydrazine and water is $4.87 J/K⋅mol-rxn$ .

The value of $ΔrGo$ for the reaction of hydrazine and water is $−623.74 kJ/mol-rxn$ .

Explanation of Solution

The value of $ΔrH°$ , $ΔrS°$ and $ΔrGo$ for the reaction of hydrazine and water is calculated below.

Given:

Refer to Appendix L for the values of standard entropies and enthalpies.

The standard entropy value for $N2H4(l)$ is $121.52 J/K⋅mol$ .

The standard entropy value for $O2(g)$ is $205.07 J/K⋅mol$ .

The standard entropy value for $H2O(l)$ is $69.95 J/K⋅mol$ .

The standard entropy value for $N2(g)$ is $191.56 J/K⋅mol$ .

The standard enthalpy value for $N2H4(l)$ is $50.63 kJ/mol$ .

The standard enthalpy value for $O2(g)$ is $0 kJ/mol$ .

The standard enthalpy value for $H2O(l)$ is $−285.83 kJ/mol$ .

The standard enthalpy value for $N2(g)$ is $0 kJ/mol$ .

The given reaction is,

$N2H4(l)+O2(g)→2H2O(l)+N2(g)$

The standard enthalpy change is,

$ΔrH°=∑nΔfH°(products)−∑nΔfH°(reactants)=[[(2 mol H2O(l)/mol-rxn)ΔfH°[H2O(l)]+(1 mol N2(g)/mol-rxn)ΔfH°[N2(g)]]-[(1 mol O2(g)/mol-rxn)ΔfH°[O2(g)]+(1 mol N2H4(l)/mol-rxn)ΔfH°[N2H4(l)]]]$

Substitute the values,

$ΔrH°=[[(2 mol H2O(l)/mol-rxn)(-285.83 kJ/mol)+(1 mol N2(g)/mol-rxn)(0 kJ/mol)]-[(1 mol O2(g)/mol-rxn)(0 kJ/mol)+(1 mol N2H4(l)/mol-rxn)(50.63 kJ/mol)]]=-622.29 kJ/mol-rxn$

The standard entropy change is,

$ΔrS°= ∑nS°(products)-∑nS°(reactants)=[[(2 mol H2O(l)/mol-rxn)S°[H2O(l)]+(1 mol N2(g)/mol-rxn)S°[N2(g)]]-[(1 mol O2(g)/mol-rxn)S°[O2(g)]+(1 mol N2H4(l)/mol-rxn)S°[N2H4(l)]]]$

Substitute the values,

$ΔrS°= [[(2 mol H2O(l)/mol-rxn)(69.95 J/K×mol)+(1 mol N2(g)/mol-rxn)(191.56 J/K×mol)]-[(1 mol O2(g)/mol-rxn)(205.07 J/K×mol)+(1 mol N2H4(l)/mol-rxn)(121.52 J/K×mol)]]= 4.87 J/K×mol-rxn$

Now,

$ΔrGo= ΔrHo- TΔrSo$

Substitute the value of $ΔrHo$ and $ΔrSo$ .

$ΔGo= -622.29 kJ/mol-rxn-[(1000 K)(4.87 J/K×mol-rxn)](1 kJ1000 J)= -623.74 kJ/mol-rxn$

(c)

Interpretation Introduction

Interpretation:

The change in temperature expected in heating system that has $5.5×104L$ of water should be identified.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by $ΔGo$ . It is related to entropy and entropy by the following expression,

$ΔGo= ΔHo- TΔSo$

The sign of $ΔGo$ should be positive for a product-favored reaction. Thus, spontaneous reactions are referred to those that have negative free energy formation.

The sign of $ΔSo$ should be positive for an entropy-favoured reaction and the sign of $ΔHo$ should be negative for an enthalpy favoured-reaction.

$ΔrHo$ is negative for an exothermic reaction and positive for endothermic reactions. The entropy change is positive whenever number of moles of gases is increasing in any reaction.

(c)

Expert Solution

Answer to Problem 97SCQ

The temperature change expected in a heating system containing $5.5×104 L$ of water is $2.7×10−3 K$ .

Explanation of Solution

The temperature change expected in a heating system containing $5.5×104 L$ of water is calculated below.

Given:

The value of $ΔrHo$ for the reaction of hydrazine and water is $−622.29 kJ/mol-rxn$ .

Thus, one mole of hydrazine releases $622.29×103 J$ of heat.

The density of water is $0.996 g/ml$ . Thus, the value of mass $m$ is,

$mass = (density)(volume)= (996 g/L)(5.5×104 L)= 5478×104g$

The specific heat $C$ of water is $4.184 J/g⋅K$ .

The heat of the system is related to specific heat and temperature by the expression,

$q = m c ΔT$

Substitute the values,

$622.29×103 J = (5478×104g) (4.184 J/g×K) ΔTΔT = 2.7×10-3 K$

(d)

Interpretation Introduction

Interpretation:

The amount of $O2$ present in hot water heating system of $5.5×104kg$ should be identified.

Concept introduction:

Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in $12 g$ of $12C$ .

From given mass of substance moles could be calculated by using the following formula,

$Moles of substance = Given mass of substanceMolecular mass$

(d)

Expert Solution

Answer to Problem 97SCQ

The number of moles of $O2$ present in the system is $7.5 moles$ .

Explanation of Solution

The number of moles of $O2$ present in the system is calculated below.

Given:

The solubilty of oxygen is in water at $25 °C$ is $0.000434g/100g$ of water.

Thus, the number of moles $n$ can be calculated by dividing the given mass by the molar mass and multipying it with the solubilty given.

$n = (5.5×107g)1 mol O2(32.00 g)(0.00434 g100 g)= 7.5 moles O2$

(e)

Interpretation Introduction

Interpretation:

The mass of solution that has $5%$ dissolved hydrazine in water required to totally consume the dissolved amount of $O2$ .

Concept introduction:

Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in $12 g$ of $12C$ .

From given mass of substance moles could be calculated by using the following formula,

$Moles of substance = Given mass of substanceMolecular mass$

Mass: It is the quantitative measure of a substance. The amount of matter present in substance is expressed as mass. The $S.I.$ unit of mass is $kg$ .

(e)

Expert Solution

Answer to Problem 97SCQ

The mass of hydrazine solution that should be added to totally consume the dissolved oxygen is $4.8×103 g$ .

Explanation of Solution

The mass of hydrazine solution that should be added to totally consume the dissolved oxygen is calculated below.

Given:

Hydrazine is available as $5 %$ solution in water. If the total solution is $100 g$ then the mass of hydrazine is $5 g$ . The molar mass of hydrazine is $32.05 g$ .

The number of moles of oxygen present is $7.5 moles$ .

Thus the mass $m$ of hydrazine solution that should be added to totally consume the dissolved oxygen is,

$m = (7.5 mol O21 mol N2H4 )(100 g solution5 g)(32.05 g N2H4)= 4.8×103 g solution$

(f)

Interpretation Introduction

Interpretation:

The volume of $N2$ produced under given conditions should be calculated.

Concept introduction:

Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in $12 g$ of $12C$ .

From given mass of substance moles could be calculated by using the following formula,

$Moles of substance = Given mass of substanceMolecular mass$

Mass: It is the quantitative measure of a substance. The amount of matter present in substance is expressed as mass. The $S.I.$ unit of mass is $kg$ .

(f)

Expert Solution

Answer to Problem 97SCQ

The volume of $N2(g)$ that will be produced is $168 L$ .

Explanation of Solution

The volume of $N2(g)$ that will be produced is calculated below.

Given:

The given reaction is,

$N2H4(l)+O2(g)→2H2O(l)+N2(g)$

Thus, $7.5 moles$ of $O2(g)$ will release $7.5 moles$ of $N2(g)$ . At standard temperature and pressure, the volume of nitrogen gas $V$ is,

$V=(7.5 moles)(22.5 L/mol)=168 L$

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Answer 3

Question

Chapter 18, Problem 97SCQ

(a)

Interpretation Introduction

Interpretation:

The balanced chemical equation for the given reaction between hydrazine and oxygen also the oxidizing and reducing agents in the reaction should be identified.

Concept introduction:

Redox reaction: It occurs when oxidation and reduction takes place at the same time in a chemical reaction.

Chemical equation is the representation of a chemical reaction, in which the reactants and products of the reactions are represented left and right side of an arrow respectively by using their respective chemical formulas.

Reactant of a chemical reaction is the substrate compounds or the compounds which undergo a chemical reaction.

Product of a chemical reaction is the produced compounds or the compounds formed after a chemical reaction.

Balanced chemical equation of a reaction is written according to law of conservation of mass.

Stoichiometry of a chemical reaction is the relation between reactants and products of the reaction and it is represented by the coefficients used for the reactants and products involved in the chemical equation.

The hot water system can corrode because of the presence of dissolved oxygen in water. This dissolved oxygen leads to corrosion. However, this oxygen can be removed from water with the use of hydrazine. The formula for hydrazine is $N2H4$ . It reacts with oxygen to produce water and $N2$ .

(a)

Expert Solution

Answer to Problem 97SCQ

The balanced chemical equation for the reaction of hydrazine and oxygen is,

$N2H4(l)+O2(g)→2H2O(l)+N2(g)$

Oxygen is oxidising agent and hydrazine is a reducing agent.

Explanation of Solution

The formula for hydrazine is $N2H4$ . It reacts with oxygen to produce water and $N2$ .

The balanced chemical equation for the reaction of hydrazine and oxygen is,

$N2H4(l)+O2(g)→2H2O(l)+N2(g)$

Oxygen acts as an oxidising agent in the reaction and changes its oxidation number from $0$ to $−2$ . Hydrazine acts as a reducing agent and loses hydrogen.

(b)

Interpretation Introduction

Interpretation:

The value of $ΔrH°$ , $ΔrS°$ and $ΔrGo$ for the given reaction of $1 mol of N2H4 at 25oc$ should be identified.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by $ΔGo$ . It is related to entropy and entropy by the following expression,

$ΔGo= ΔHo- TΔSo$

The sign of $ΔGo$ should be positive for a product-favored reaction. Thus, spontaneous reactions are referred to those that have negative free energy formation.

The sign of $ΔSo$ should be positive for an entropy-favoured reaction and the sign of $ΔH∘$ should be negative for an enthalpy favoured-reaction.

$ΔrHo$ is negative for an exothermic reaction and positive for endothermic reactions. The entropy change is positive whenever number of moles of gases is increasing in any reaction.

(b)

Expert Solution

Answer to Problem 97SCQ

The value of $ΔrH°$ for the reaction of hydrazine and water is $−622.29 kJ/mol-rxn$ .

The value of $ΔrS°$ for the reaction of hydrazine and water is $4.87 J/K⋅mol-rxn$ .

The value of $ΔrGo$ for the reaction of hydrazine and water is $−623.74 kJ/mol-rxn$ .

Explanation of Solution

The value of $ΔrH°$ , $ΔrS°$ and $ΔrGo$ for the reaction of hydrazine and water is calculated below.

Given:

Refer to Appendix L for the values of standard entropies and enthalpies.

The standard entropy value for $N2H4(l)$ is $121.52 J/K⋅mol$ .

The standard entropy value for $O2(g)$ is $205.07 J/K⋅mol$ .

The standard entropy value for $H2O(l)$ is $69.95 J/K⋅mol$ .

The standard entropy value for $N2(g)$ is $191.56 J/K⋅mol$ .

The standard enthalpy value for $N2H4(l)$ is $50.63 kJ/mol$ .

The standard enthalpy value for $O2(g)$ is $0 kJ/mol$ .

The standard enthalpy value for $H2O(l)$ is $−285.83 kJ/mol$ .

The standard enthalpy value for $N2(g)$ is $0 kJ/mol$ .

The given reaction is,

$N2H4(l)+O2(g)→2H2O(l)+N2(g)$

The standard enthalpy change is,

$ΔrH°=∑nΔfH°(products)−∑nΔfH°(reactants)=[[(2 mol H2O(l)/mol-rxn)ΔfH°[H2O(l)]+(1 mol N2(g)/mol-rxn)ΔfH°[N2(g)]]-[(1 mol O2(g)/mol-rxn)ΔfH°[O2(g)]+(1 mol N2H4(l)/mol-rxn)ΔfH°[N2H4(l)]]]$

Substitute the values,

$ΔrH°=[[(2 mol H2O(l)/mol-rxn)(-285.83 kJ/mol)+(1 mol N2(g)/mol-rxn)(0 kJ/mol)]-[(1 mol O2(g)/mol-rxn)(0 kJ/mol)+(1 mol N2H4(l)/mol-rxn)(50.63 kJ/mol)]]=-622.29 kJ/mol-rxn$

The standard entropy change is,

$ΔrS°= ∑nS°(products)-∑nS°(reactants)=[[(2 mol H2O(l)/mol-rxn)S°[H2O(l)]+(1 mol N2(g)/mol-rxn)S°[N2(g)]]-[(1 mol O2(g)/mol-rxn)S°[O2(g)]+(1 mol N2H4(l)/mol-rxn)S°[N2H4(l)]]]$

Substitute the values,

$ΔrS°= [[(2 mol H2O(l)/mol-rxn)(69.95 J/K×mol)+(1 mol N2(g)/mol-rxn)(191.56 J/K×mol)]-[(1 mol O2(g)/mol-rxn)(205.07 J/K×mol)+(1 mol N2H4(l)/mol-rxn)(121.52 J/K×mol)]]= 4.87 J/K×mol-rxn$

Now,

$ΔrGo= ΔrHo- TΔrSo$

Substitute the value of $ΔrHo$ and $ΔrSo$ .

$ΔGo= -622.29 kJ/mol-rxn-[(1000 K)(4.87 J/K×mol-rxn)](1 kJ1000 J)= -623.74 kJ/mol-rxn$

(c)

Interpretation Introduction

Interpretation:

The change in temperature expected in heating system that has $5.5×104L$ of water should be identified.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by $ΔGo$ . It is related to entropy and entropy by the following expression,

$ΔGo= ΔHo- TΔSo$

The sign of $ΔGo$ should be positive for a product-favored reaction. Thus, spontaneous reactions are referred to those that have negative free energy formation.

The sign of $ΔSo$ should be positive for an entropy-favoured reaction and the sign of $ΔHo$ should be negative for an enthalpy favoured-reaction.

$ΔrHo$ is negative for an exothermic reaction and positive for endothermic reactions. The entropy change is positive whenever number of moles of gases is increasing in any reaction.

(c)

Expert Solution

Answer to Problem 97SCQ

The temperature change expected in a heating system containing $5.5×104 L$ of water is $2.7×10−3 K$ .

Explanation of Solution

The temperature change expected in a heating system containing $5.5×104 L$ of water is calculated below.

Given:

The value of $ΔrHo$ for the reaction of hydrazine and water is $−622.29 kJ/mol-rxn$ .

Thus, one mole of hydrazine releases $622.29×103 J$ of heat.

The density of water is $0.996 g/ml$ . Thus, the value of mass $m$ is,

$mass = (density)(volume)= (996 g/L)(5.5×104 L)= 5478×104g$

The specific heat $C$ of water is $4.184 J/g⋅K$ .

The heat of the system is related to specific heat and temperature by the expression,

$q = m c ΔT$

Substitute the values,

$622.29×103 J = (5478×104g) (4.184 J/g×K) ΔTΔT = 2.7×10-3 K$

(d)

Interpretation Introduction

Interpretation:

The amount of $O2$ present in hot water heating system of $5.5×104kg$ should be identified.

Concept introduction:

Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in $12 g$ of $12C$ .

From given mass of substance moles could be calculated by using the following formula,

$Moles of substance = Given mass of substanceMolecular mass$

(d)

Expert Solution

Answer to Problem 97SCQ

The number of moles of $O2$ present in the system is $7.5 moles$ .

Explanation of Solution

The number of moles of $O2$ present in the system is calculated below.

Given:

The solubilty of oxygen is in water at $25 °C$ is $0.000434g/100g$ of water.

Thus, the number of moles $n$ can be calculated by dividing the given mass by the molar mass and multipying it with the solubilty given.

$n = (5.5×107g)1 mol O2(32.00 g)(0.00434 g100 g)= 7.5 moles O2$

(e)

Interpretation Introduction

Interpretation:

The mass of solution that has $5%$ dissolved hydrazine in water required to totally consume the dissolved amount of $O2$ .

Concept introduction:

Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in $12 g$ of $12C$ .

From given mass of substance moles could be calculated by using the following formula,

$Moles of substance = Given mass of substanceMolecular mass$

Mass: It is the quantitative measure of a substance. The amount of matter present in substance is expressed as mass. The $S.I.$ unit of mass is $kg$ .

(e)

Expert Solution

Answer to Problem 97SCQ

The mass of hydrazine solution that should be added to totally consume the dissolved oxygen is $4.8×103 g$ .

Explanation of Solution

The mass of hydrazine solution that should be added to totally consume the dissolved oxygen is calculated below.

Given:

Hydrazine is available as $5 %$ solution in water. If the total solution is $100 g$ then the mass of hydrazine is $5 g$ . The molar mass of hydrazine is $32.05 g$ .

The number of moles of oxygen present is $7.5 moles$ .

Thus the mass $m$ of hydrazine solution that should be added to totally consume the dissolved oxygen is,

$m = (7.5 mol O21 mol N2H4 )(100 g solution5 g)(32.05 g N2H4)= 4.8×103 g solution$

(f)

Interpretation Introduction

Interpretation:

The volume of $N2$ produced under given conditions should be calculated.

Concept introduction:

Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in $12 g$ of $12C$ .

From given mass of substance moles could be calculated by using the following formula,

$Moles of substance = Given mass of substanceMolecular mass$

Mass: It is the quantitative measure of a substance. The amount of matter present in substance is expressed as mass. The $S.I.$ unit of mass is $kg$ .

(f)

Expert Solution

Answer to Problem 97SCQ

The volume of $N2(g)$ that will be produced is $168 L$ .

Explanation of Solution

The volume of $N2(g)$ that will be produced is calculated below.

Given:

The given reaction is,

$N2H4(l)+O2(g)→2H2O(l)+N2(g)$

Thus, $7.5 moles$ of $O2(g)$ will release $7.5 moles$ of $N2(g)$ . At standard temperature and pressure, the volume of nitrogen gas $V$ is,

$V=(7.5 moles)(22.5 L/mol)=168 L$

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Answer 4

Question

Chapter 18, Problem 97SCQ

(a)

Interpretation Introduction

Interpretation:

The balanced chemical equation for the given reaction between hydrazine and oxygen also the oxidizing and reducing agents in the reaction should be identified.

Concept introduction:

Redox reaction: It occurs when oxidation and reduction takes place at the same time in a chemical reaction.

Chemical equation is the representation of a chemical reaction, in which the reactants and products of the reactions are represented left and right side of an arrow respectively by using their respective chemical formulas.

Reactant of a chemical reaction is the substrate compounds or the compounds which undergo a chemical reaction.

Product of a chemical reaction is the produced compounds or the compounds formed after a chemical reaction.

Balanced chemical equation of a reaction is written according to law of conservation of mass.

Stoichiometry of a chemical reaction is the relation between reactants and products of the reaction and it is represented by the coefficients used for the reactants and products involved in the chemical equation.

The hot water system can corrode because of the presence of dissolved oxygen in water. This dissolved oxygen leads to corrosion. However, this oxygen can be removed from water with the use of hydrazine. The formula for hydrazine is $N2H4$ . It reacts with oxygen to produce water and $N2$ .

(a)

Expert Solution

Answer to Problem 97SCQ

The balanced chemical equation for the reaction of hydrazine and oxygen is,

$N2H4(l)+O2(g)→2H2O(l)+N2(g)$

Oxygen is oxidising agent and hydrazine is a reducing agent.

Explanation of Solution

The formula for hydrazine is $N2H4$ . It reacts with oxygen to produce water and $N2$ .

The balanced chemical equation for the reaction of hydrazine and oxygen is,

$N2H4(l)+O2(g)→2H2O(l)+N2(g)$

Oxygen acts as an oxidising agent in the reaction and changes its oxidation number from $0$ to $−2$ . Hydrazine acts as a reducing agent and loses hydrogen.

(b)

Interpretation Introduction

Interpretation:

The value of $ΔrH°$ , $ΔrS°$ and $ΔrGo$ for the given reaction of $1 mol of N2H4 at 25oc$ should be identified.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by $ΔGo$ . It is related to entropy and entropy by the following expression,

$ΔGo= ΔHo- TΔSo$

The sign of $ΔGo$ should be positive for a product-favored reaction. Thus, spontaneous reactions are referred to those that have negative free energy formation.

The sign of $ΔSo$ should be positive for an entropy-favoured reaction and the sign of $ΔH∘$ should be negative for an enthalpy favoured-reaction.

$ΔrHo$ is negative for an exothermic reaction and positive for endothermic reactions. The entropy change is positive whenever number of moles of gases is increasing in any reaction.

(b)

Expert Solution

Answer to Problem 97SCQ

The value of $ΔrH°$ for the reaction of hydrazine and water is $−622.29 kJ/mol-rxn$ .

The value of $ΔrS°$ for the reaction of hydrazine and water is $4.87 J/K⋅mol-rxn$ .

The value of $ΔrGo$ for the reaction of hydrazine and water is $−623.74 kJ/mol-rxn$ .

Explanation of Solution

The value of $ΔrH°$ , $ΔrS°$ and $ΔrGo$ for the reaction of hydrazine and water is calculated below.

Given:

Refer to Appendix L for the values of standard entropies and enthalpies.

The standard entropy value for $N2H4(l)$ is $121.52 J/K⋅mol$ .

The standard entropy value for $O2(g)$ is $205.07 J/K⋅mol$ .

The standard entropy value for $H2O(l)$ is $69.95 J/K⋅mol$ .

The standard entropy value for $N2(g)$ is $191.56 J/K⋅mol$ .

The standard enthalpy value for $N2H4(l)$ is $50.63 kJ/mol$ .

The standard enthalpy value for $O2(g)$ is $0 kJ/mol$ .

The standard enthalpy value for $H2O(l)$ is $−285.83 kJ/mol$ .

The standard enthalpy value for $N2(g)$ is $0 kJ/mol$ .

The given reaction is,

$N2H4(l)+O2(g)→2H2O(l)+N2(g)$

The standard enthalpy change is,

$ΔrH°=∑nΔfH°(products)−∑nΔfH°(reactants)=[[(2 mol H2O(l)/mol-rxn)ΔfH°[H2O(l)]+(1 mol N2(g)/mol-rxn)ΔfH°[N2(g)]]-[(1 mol O2(g)/mol-rxn)ΔfH°[O2(g)]+(1 mol N2H4(l)/mol-rxn)ΔfH°[N2H4(l)]]]$

Substitute the values,

$ΔrH°=[[(2 mol H2O(l)/mol-rxn)(-285.83 kJ/mol)+(1 mol N2(g)/mol-rxn)(0 kJ/mol)]-[(1 mol O2(g)/mol-rxn)(0 kJ/mol)+(1 mol N2H4(l)/mol-rxn)(50.63 kJ/mol)]]=-622.29 kJ/mol-rxn$

The standard entropy change is,

$ΔrS°= ∑nS°(products)-∑nS°(reactants)=[[(2 mol H2O(l)/mol-rxn)S°[H2O(l)]+(1 mol N2(g)/mol-rxn)S°[N2(g)]]-[(1 mol O2(g)/mol-rxn)S°[O2(g)]+(1 mol N2H4(l)/mol-rxn)S°[N2H4(l)]]]$

Substitute the values,

$ΔrS°= [[(2 mol H2O(l)/mol-rxn)(69.95 J/K×mol)+(1 mol N2(g)/mol-rxn)(191.56 J/K×mol)]-[(1 mol O2(g)/mol-rxn)(205.07 J/K×mol)+(1 mol N2H4(l)/mol-rxn)(121.52 J/K×mol)]]= 4.87 J/K×mol-rxn$

Now,

$ΔrGo= ΔrHo- TΔrSo$

Substitute the value of $ΔrHo$ and $ΔrSo$ .

$ΔGo= -622.29 kJ/mol-rxn-[(1000 K)(4.87 J/K×mol-rxn)](1 kJ1000 J)= -623.74 kJ/mol-rxn$

(c)

Interpretation Introduction

Interpretation:

The change in temperature expected in heating system that has $5.5×104L$ of water should be identified.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by $ΔGo$ . It is related to entropy and entropy by the following expression,

$ΔGo= ΔHo- TΔSo$

The sign of $ΔGo$ should be positive for a product-favored reaction. Thus, spontaneous reactions are referred to those that have negative free energy formation.

The sign of $ΔSo$ should be positive for an entropy-favoured reaction and the sign of $ΔHo$ should be negative for an enthalpy favoured-reaction.

$ΔrHo$ is negative for an exothermic reaction and positive for endothermic reactions. The entropy change is positive whenever number of moles of gases is increasing in any reaction.

(c)

Expert Solution

Answer to Problem 97SCQ

The temperature change expected in a heating system containing $5.5×104 L$ of water is $2.7×10−3 K$ .

Explanation of Solution

The temperature change expected in a heating system containing $5.5×104 L$ of water is calculated below.

Given:

The value of $ΔrHo$ for the reaction of hydrazine and water is $−622.29 kJ/mol-rxn$ .

Thus, one mole of hydrazine releases $622.29×103 J$ of heat.

The density of water is $0.996 g/ml$ . Thus, the value of mass $m$ is,

$mass = (density)(volume)= (996 g/L)(5.5×104 L)= 5478×104g$

The specific heat $C$ of water is $4.184 J/g⋅K$ .

The heat of the system is related to specific heat and temperature by the expression,

$q = m c ΔT$

Substitute the values,

$622.29×103 J = (5478×104g) (4.184 J/g×K) ΔTΔT = 2.7×10-3 K$

(d)

Interpretation Introduction

Interpretation:

The amount of $O2$ present in hot water heating system of $5.5×104kg$ should be identified.

Concept introduction:

Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in $12 g$ of $12C$ .

From given mass of substance moles could be calculated by using the following formula,

$Moles of substance = Given mass of substanceMolecular mass$

(d)

Expert Solution

Answer to Problem 97SCQ

The number of moles of $O2$ present in the system is $7.5 moles$ .

Explanation of Solution

The number of moles of $O2$ present in the system is calculated below.

Given:

The solubilty of oxygen is in water at $25 °C$ is $0.000434g/100g$ of water.

Thus, the number of moles $n$ can be calculated by dividing the given mass by the molar mass and multipying it with the solubilty given.

$n = (5.5×107g)1 mol O2(32.00 g)(0.00434 g100 g)= 7.5 moles O2$

(e)

Interpretation Introduction

Interpretation:

The mass of solution that has $5%$ dissolved hydrazine in water required to totally consume the dissolved amount of $O2$ .

Concept introduction:

Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in $12 g$ of $12C$ .

From given mass of substance moles could be calculated by using the following formula,

$Moles of substance = Given mass of substanceMolecular mass$

Mass: It is the quantitative measure of a substance. The amount of matter present in substance is expressed as mass. The $S.I.$ unit of mass is $kg$ .

(e)

Expert Solution

Answer to Problem 97SCQ

The mass of hydrazine solution that should be added to totally consume the dissolved oxygen is $4.8×103 g$ .

Explanation of Solution

The mass of hydrazine solution that should be added to totally consume the dissolved oxygen is calculated below.

Given:

Hydrazine is available as $5 %$ solution in water. If the total solution is $100 g$ then the mass of hydrazine is $5 g$ . The molar mass of hydrazine is $32.05 g$ .

The number of moles of oxygen present is $7.5 moles$ .

Thus the mass $m$ of hydrazine solution that should be added to totally consume the dissolved oxygen is,

$m = (7.5 mol O21 mol N2H4 )(100 g solution5 g)(32.05 g N2H4)= 4.8×103 g solution$

(f)

Interpretation Introduction

Interpretation:

The volume of $N2$ produced under given conditions should be calculated.

Concept introduction:

Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in $12 g$ of $12C$ .

From given mass of substance moles could be calculated by using the following formula,

$Moles of substance = Given mass of substanceMolecular mass$

Mass: It is the quantitative measure of a substance. The amount of matter present in substance is expressed as mass. The $S.I.$ unit of mass is $kg$ .

(f)

Expert Solution

Answer to Problem 97SCQ

The volume of $N2(g)$ that will be produced is $168 L$ .

Explanation of Solution

The volume of $N2(g)$ that will be produced is calculated below.

Given:

The given reaction is,

$N2H4(l)+O2(g)→2H2O(l)+N2(g)$

Thus, $7.5 moles$ of $O2(g)$ will release $7.5 moles$ of $N2(g)$ . At standard temperature and pressure, the volume of nitrogen gas $V$ is,

$V=(7.5 moles)(22.5 L/mol)=168 L$

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Answer 5

Chapter 18, Problem 97SCQ

(a)

Interpretation Introduction

Interpretation:

The balanced chemical equation for the given reaction between hydrazine and oxygen also the oxidizing and reducing agents in the reaction should be identified.

Concept introduction:

Redox reaction: It occurs when oxidation and reduction takes place at the same time in a chemical reaction.

Chemical equation is the representation of a chemical reaction, in which the reactants and products of the reactions are represented left and right side of an arrow respectively by using their respective chemical formulas.

Reactant of a chemical reaction is the substrate compounds or the compounds which undergo a chemical reaction.

Product of a chemical reaction is the produced compounds or the compounds formed after a chemical reaction.

Balanced chemical equation of a reaction is written according to law of conservation of mass.

Stoichiometry of a chemical reaction is the relation between reactants and products of the reaction and it is represented by the coefficients used for the reactants and products involved in the chemical equation.

The hot water system can corrode because of the presence of dissolved oxygen in water. This dissolved oxygen leads to corrosion. However, this oxygen can be removed from water with the use of hydrazine. The formula for hydrazine is $N2H4$ . It reacts with oxygen to produce water and $N2$ .

(a)

Expert Solution

Answer to Problem 97SCQ

The balanced chemical equation for the reaction of hydrazine and oxygen is,

$N2H4(l)+O2(g)→2H2O(l)+N2(g)$

Oxygen is oxidising agent and hydrazine is a reducing agent.

Explanation of Solution

The formula for hydrazine is $N2H4$ . It reacts with oxygen to produce water and $N2$ .

The balanced chemical equation for the reaction of hydrazine and oxygen is,

$N2H4(l)+O2(g)→2H2O(l)+N2(g)$

Oxygen acts as an oxidising agent in the reaction and changes its oxidation number from $0$ to $−2$ . Hydrazine acts as a reducing agent and loses hydrogen.

(b)

Interpretation Introduction

Interpretation:

The value of $ΔrH°$ , $ΔrS°$ and $ΔrGo$ for the given reaction of $1 mol of N2H4 at 25oc$ should be identified.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by $ΔGo$ . It is related to entropy and entropy by the following expression,

$ΔGo= ΔHo- TΔSo$

The sign of $ΔGo$ should be positive for a product-favored reaction. Thus, spontaneous reactions are referred to those that have negative free energy formation.

The sign of $ΔSo$ should be positive for an entropy-favoured reaction and the sign of $ΔH∘$ should be negative for an enthalpy favoured-reaction.

$ΔrHo$ is negative for an exothermic reaction and positive for endothermic reactions. The entropy change is positive whenever number of moles of gases is increasing in any reaction.

(b)

Expert Solution

Answer to Problem 97SCQ

The value of $ΔrH°$ for the reaction of hydrazine and water is $−622.29 kJ/mol-rxn$ .

The value of $ΔrS°$ for the reaction of hydrazine and water is $4.87 J/K⋅mol-rxn$ .

The value of $ΔrGo$ for the reaction of hydrazine and water is $−623.74 kJ/mol-rxn$ .

Explanation of Solution

The value of $ΔrH°$ , $ΔrS°$ and $ΔrGo$ for the reaction of hydrazine and water is calculated below.

Given:

Refer to Appendix L for the values of standard entropies and enthalpies.

The standard entropy value for $N2H4(l)$ is $121.52 J/K⋅mol$ .

The standard entropy value for $O2(g)$ is $205.07 J/K⋅mol$ .

The standard entropy value for $H2O(l)$ is $69.95 J/K⋅mol$ .

The standard entropy value for $N2(g)$ is $191.56 J/K⋅mol$ .

The standard enthalpy value for $N2H4(l)$ is $50.63 kJ/mol$ .

The standard enthalpy value for $O2(g)$ is $0 kJ/mol$ .

The standard enthalpy value for $H2O(l)$ is $−285.83 kJ/mol$ .

The standard enthalpy value for $N2(g)$ is $0 kJ/mol$ .

The given reaction is,

$N2H4(l)+O2(g)→2H2O(l)+N2(g)$

The standard enthalpy change is,

$ΔrH°=∑nΔfH°(products)−∑nΔfH°(reactants)=[[(2 mol H2O(l)/mol-rxn)ΔfH°[H2O(l)]+(1 mol N2(g)/mol-rxn)ΔfH°[N2(g)]]-[(1 mol O2(g)/mol-rxn)ΔfH°[O2(g)]+(1 mol N2H4(l)/mol-rxn)ΔfH°[N2H4(l)]]]$

Substitute the values,

$ΔrH°=[[(2 mol H2O(l)/mol-rxn)(-285.83 kJ/mol)+(1 mol N2(g)/mol-rxn)(0 kJ/mol)]-[(1 mol O2(g)/mol-rxn)(0 kJ/mol)+(1 mol N2H4(l)/mol-rxn)(50.63 kJ/mol)]]=-622.29 kJ/mol-rxn$

The standard entropy change is,

$ΔrS°= ∑nS°(products)-∑nS°(reactants)=[[(2 mol H2O(l)/mol-rxn)S°[H2O(l)]+(1 mol N2(g)/mol-rxn)S°[N2(g)]]-[(1 mol O2(g)/mol-rxn)S°[O2(g)]+(1 mol N2H4(l)/mol-rxn)S°[N2H4(l)]]]$

Substitute the values,

$ΔrS°= [[(2 mol H2O(l)/mol-rxn)(69.95 J/K×mol)+(1 mol N2(g)/mol-rxn)(191.56 J/K×mol)]-[(1 mol O2(g)/mol-rxn)(205.07 J/K×mol)+(1 mol N2H4(l)/mol-rxn)(121.52 J/K×mol)]]= 4.87 J/K×mol-rxn$

Now,

$ΔrGo= ΔrHo- TΔrSo$

Substitute the value of $ΔrHo$ and $ΔrSo$ .

$ΔGo= -622.29 kJ/mol-rxn-[(1000 K)(4.87 J/K×mol-rxn)](1 kJ1000 J)= -623.74 kJ/mol-rxn$

(c)

Interpretation Introduction

Interpretation:

The change in temperature expected in heating system that has $5.5×104L$ of water should be identified.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by $ΔGo$ . It is related to entropy and entropy by the following expression,

$ΔGo= ΔHo- TΔSo$

The sign of $ΔGo$ should be positive for a product-favored reaction. Thus, spontaneous reactions are referred to those that have negative free energy formation.

The sign of $ΔSo$ should be positive for an entropy-favoured reaction and the sign of $ΔHo$ should be negative for an enthalpy favoured-reaction.

$ΔrHo$ is negative for an exothermic reaction and positive for endothermic reactions. The entropy change is positive whenever number of moles of gases is increasing in any reaction.

(c)

Expert Solution

Answer to Problem 97SCQ

The temperature change expected in a heating system containing $5.5×104 L$ of water is $2.7×10−3 K$ .

Explanation of Solution

The temperature change expected in a heating system containing $5.5×104 L$ of water is calculated below.

Given:

The value of $ΔrHo$ for the reaction of hydrazine and water is $−622.29 kJ/mol-rxn$ .

Thus, one mole of hydrazine releases $622.29×103 J$ of heat.

The density of water is $0.996 g/ml$ . Thus, the value of mass $m$ is,

$mass = (density)(volume)= (996 g/L)(5.5×104 L)= 5478×104g$

The specific heat $C$ of water is $4.184 J/g⋅K$ .

The heat of the system is related to specific heat and temperature by the expression,

$q = m c ΔT$

Substitute the values,

$622.29×103 J = (5478×104g) (4.184 J/g×K) ΔTΔT = 2.7×10-3 K$

(d)

Interpretation Introduction

Interpretation:

The amount of $O2$ present in hot water heating system of $5.5×104kg$ should be identified.

Concept introduction:

Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in $12 g$ of $12C$ .

From given mass of substance moles could be calculated by using the following formula,

$Moles of substance = Given mass of substanceMolecular mass$

(d)

Expert Solution

Answer to Problem 97SCQ

The number of moles of $O2$ present in the system is $7.5 moles$ .

Explanation of Solution

The number of moles of $O2$ present in the system is calculated below.

Given:

The solubilty of oxygen is in water at $25 °C$ is $0.000434g/100g$ of water.

Thus, the number of moles $n$ can be calculated by dividing the given mass by the molar mass and multipying it with the solubilty given.

$n = (5.5×107g)1 mol O2(32.00 g)(0.00434 g100 g)= 7.5 moles O2$

(e)

Interpretation Introduction

Interpretation:

The mass of solution that has $5%$ dissolved hydrazine in water required to totally consume the dissolved amount of $O2$ .

Concept introduction:

Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in $12 g$ of $12C$ .

From given mass of substance moles could be calculated by using the following formula,

$Moles of substance = Given mass of substanceMolecular mass$

Mass: It is the quantitative measure of a substance. The amount of matter present in substance is expressed as mass. The $S.I.$ unit of mass is $kg$ .

(e)

Expert Solution

Answer to Problem 97SCQ

The mass of hydrazine solution that should be added to totally consume the dissolved oxygen is $4.8×103 g$ .

Explanation of Solution

The mass of hydrazine solution that should be added to totally consume the dissolved oxygen is calculated below.

Given:

Hydrazine is available as $5 %$ solution in water. If the total solution is $100 g$ then the mass of hydrazine is $5 g$ . The molar mass of hydrazine is $32.05 g$ .

The number of moles of oxygen present is $7.5 moles$ .

Thus the mass $m$ of hydrazine solution that should be added to totally consume the dissolved oxygen is,

$m = (7.5 mol O21 mol N2H4 )(100 g solution5 g)(32.05 g N2H4)= 4.8×103 g solution$

(f)

Interpretation Introduction

Interpretation:

The volume of $N2$ produced under given conditions should be calculated.

Concept introduction:

Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in $12 g$ of $12C$ .

From given mass of substance moles could be calculated by using the following formula,

$Moles of substance = Given mass of substanceMolecular mass$

Mass: It is the quantitative measure of a substance. The amount of matter present in substance is expressed as mass. The $S.I.$ unit of mass is $kg$ .

(f)

Expert Solution

Answer to Problem 97SCQ

The volume of $N2(g)$ that will be produced is $168 L$ .

Explanation of Solution

The volume of $N2(g)$ that will be produced is calculated below.

Given:

The given reaction is,

$N2H4(l)+O2(g)→2H2O(l)+N2(g)$

Thus, $7.5 moles$ of $O2(g)$ will release $7.5 moles$ of $N2(g)$ . At standard temperature and pressure, the volume of nitrogen gas $V$ is,

$V=(7.5 moles)(22.5 L/mol)=168 L$

Answer 6

Chapter 18, Problem 97SCQ

(a)

Interpretation Introduction

Interpretation:

The balanced chemical equation for the given reaction between hydrazine and oxygen also the oxidizing and reducing agents in the reaction should be identified.

Concept introduction:

Redox reaction: It occurs when oxidation and reduction takes place at the same time in a chemical reaction.

Chemical equation is the representation of a chemical reaction, in which the reactants and products of the reactions are represented left and right side of an arrow respectively by using their respective chemical formulas.

Reactant of a chemical reaction is the substrate compounds or the compounds which undergo a chemical reaction.

Product of a chemical reaction is the produced compounds or the compounds formed after a chemical reaction.

Balanced chemical equation of a reaction is written according to law of conservation of mass.

Stoichiometry of a chemical reaction is the relation between reactants and products of the reaction and it is represented by the coefficients used for the reactants and products involved in the chemical equation.

The hot water system can corrode because of the presence of dissolved oxygen in water. This dissolved oxygen leads to corrosion. However, this oxygen can be removed from water with the use of hydrazine. The formula for hydrazine is $N2H4$ . It reacts with oxygen to produce water and $N2$ .

(a)

Expert Solution

Answer to Problem 97SCQ

The balanced chemical equation for the reaction of hydrazine and oxygen is,

$N2H4(l)+O2(g)→2H2O(l)+N2(g)$

Oxygen is oxidising agent and hydrazine is a reducing agent.

Explanation of Solution

The formula for hydrazine is $N2H4$ . It reacts with oxygen to produce water and $N2$ .

The balanced chemical equation for the reaction of hydrazine and oxygen is,

$N2H4(l)+O2(g)→2H2O(l)+N2(g)$

Oxygen acts as an oxidising agent in the reaction and changes its oxidation number from $0$ to $−2$ . Hydrazine acts as a reducing agent and loses hydrogen.

Answer 7

Chapter 18, Problem 97SCQ

(a)

Interpretation Introduction

Interpretation:

The balanced chemical equation for the given reaction between hydrazine and oxygen also the oxidizing and reducing agents in the reaction should be identified.

Concept introduction:

Redox reaction: It occurs when oxidation and reduction takes place at the same time in a chemical reaction.

Chemical equation is the representation of a chemical reaction, in which the reactants and products of the reactions are represented left and right side of an arrow respectively by using their respective chemical formulas.

Reactant of a chemical reaction is the substrate compounds or the compounds which undergo a chemical reaction.

Product of a chemical reaction is the produced compounds or the compounds formed after a chemical reaction.

Balanced chemical equation of a reaction is written according to law of conservation of mass.

Stoichiometry of a chemical reaction is the relation between reactants and products of the reaction and it is represented by the coefficients used for the reactants and products involved in the chemical equation.

The hot water system can corrode because of the presence of dissolved oxygen in water. This dissolved oxygen leads to corrosion. However, this oxygen can be removed from water with the use of hydrazine. The formula for hydrazine is $N2H4$ . It reacts with oxygen to produce water and $N2$ .

Answer 8

(a)

Expert Solution

Answer to Problem 97SCQ

The balanced chemical equation for the reaction of hydrazine and oxygen is,

$N2H4(l)+O2(g)→2H2O(l)+N2(g)$

Oxygen is oxidising agent and hydrazine is a reducing agent.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The formula for hydrazine is $N2H4$ . It reacts with oxygen to produce water and $N2$ .

The balanced chemical equation for the reaction of hydrazine and oxygen is,

$N2H4(l)+O2(g)→2H2O(l)+N2(g)$

Oxygen acts as an oxidising agent in the reaction and changes its oxidation number from $0$ to $−2$ . Hydrazine acts as a reducing agent and loses hydrogen.

Answer 13

(b)

Interpretation Introduction

Interpretation:

The value of $ΔrH°$ , $ΔrS°$ and $ΔrGo$ for the given reaction of $1 mol of N2H4 at 25oc$ should be identified.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by $ΔGo$ . It is related to entropy and entropy by the following expression,

$ΔGo= ΔHo- TΔSo$

The sign of $ΔGo$ should be positive for a product-favored reaction. Thus, spontaneous reactions are referred to those that have negative free energy formation.

The sign of $ΔSo$ should be positive for an entropy-favoured reaction and the sign of $ΔH∘$ should be negative for an enthalpy favoured-reaction.

$ΔrHo$ is negative for an exothermic reaction and positive for endothermic reactions. The entropy change is positive whenever number of moles of gases is increasing in any reaction.

(b)

Expert Solution

Answer to Problem 97SCQ

The value of $ΔrH°$ for the reaction of hydrazine and water is $−622.29 kJ/mol-rxn$ .

The value of $ΔrS°$ for the reaction of hydrazine and water is $4.87 J/K⋅mol-rxn$ .

The value of $ΔrGo$ for the reaction of hydrazine and water is $−623.74 kJ/mol-rxn$ .

Explanation of Solution

The value of $ΔrH°$ , $ΔrS°$ and $ΔrGo$ for the reaction of hydrazine and water is calculated below.

Given:

Refer to Appendix L for the values of standard entropies and enthalpies.

The standard entropy value for $N2H4(l)$ is $121.52 J/K⋅mol$ .

The standard entropy value for $O2(g)$ is $205.07 J/K⋅mol$ .

The standard entropy value for $H2O(l)$ is $69.95 J/K⋅mol$ .

The standard entropy value for $N2(g)$ is $191.56 J/K⋅mol$ .

The standard enthalpy value for $N2H4(l)$ is $50.63 kJ/mol$ .

The standard enthalpy value for $O2(g)$ is $0 kJ/mol$ .

The standard enthalpy value for $H2O(l)$ is $−285.83 kJ/mol$ .

The standard enthalpy value for $N2(g)$ is $0 kJ/mol$ .

The given reaction is,

$N2H4(l)+O2(g)→2H2O(l)+N2(g)$

The standard enthalpy change is,

$ΔrH°=∑nΔfH°(products)−∑nΔfH°(reactants)=[[(2 mol H2O(l)/mol-rxn)ΔfH°[H2O(l)]+(1 mol N2(g)/mol-rxn)ΔfH°[N2(g)]]-[(1 mol O2(g)/mol-rxn)ΔfH°[O2(g)]+(1 mol N2H4(l)/mol-rxn)ΔfH°[N2H4(l)]]]$

Substitute the values,

$ΔrH°=[[(2 mol H2O(l)/mol-rxn)(-285.83 kJ/mol)+(1 mol N2(g)/mol-rxn)(0 kJ/mol)]-[(1 mol O2(g)/mol-rxn)(0 kJ/mol)+(1 mol N2H4(l)/mol-rxn)(50.63 kJ/mol)]]=-622.29 kJ/mol-rxn$

The standard entropy change is,

$ΔrS°= ∑nS°(products)-∑nS°(reactants)=[[(2 mol H2O(l)/mol-rxn)S°[H2O(l)]+(1 mol N2(g)/mol-rxn)S°[N2(g)]]-[(1 mol O2(g)/mol-rxn)S°[O2(g)]+(1 mol N2H4(l)/mol-rxn)S°[N2H4(l)]]]$

Substitute the values,

$ΔrS°= [[(2 mol H2O(l)/mol-rxn)(69.95 J/K×mol)+(1 mol N2(g)/mol-rxn)(191.56 J/K×mol)]-[(1 mol O2(g)/mol-rxn)(205.07 J/K×mol)+(1 mol N2H4(l)/mol-rxn)(121.52 J/K×mol)]]= 4.87 J/K×mol-rxn$

Now,

$ΔrGo= ΔrHo- TΔrSo$

Substitute the value of $ΔrHo$ and $ΔrSo$ .

$ΔGo= -622.29 kJ/mol-rxn-[(1000 K)(4.87 J/K×mol-rxn)](1 kJ1000 J)= -623.74 kJ/mol-rxn$

Answer 14

(b)

Interpretation Introduction

Interpretation:

The value of $ΔrH°$ , $ΔrS°$ and $ΔrGo$ for the given reaction of $1 mol of N2H4 at 25oc$ should be identified.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by $ΔGo$ . It is related to entropy and entropy by the following expression,

$ΔGo= ΔHo- TΔSo$

The sign of $ΔGo$ should be positive for a product-favored reaction. Thus, spontaneous reactions are referred to those that have negative free energy formation.

The sign of $ΔSo$ should be positive for an entropy-favoured reaction and the sign of $ΔH∘$ should be negative for an enthalpy favoured-reaction.

$ΔrHo$ is negative for an exothermic reaction and positive for endothermic reactions. The entropy change is positive whenever number of moles of gases is increasing in any reaction.

Answer 15

(b)

Expert Solution

Answer to Problem 97SCQ

The value of $ΔrH°$ for the reaction of hydrazine and water is $−622.29 kJ/mol-rxn$ .

The value of $ΔrS°$ for the reaction of hydrazine and water is $4.87 J/K⋅mol-rxn$ .

The value of $ΔrGo$ for the reaction of hydrazine and water is $−623.74 kJ/mol-rxn$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

The value of $ΔrH°$ , $ΔrS°$ and $ΔrGo$ for the reaction of hydrazine and water is calculated below.

Given:

Refer to Appendix L for the values of standard entropies and enthalpies.

The standard entropy value for $N2H4(l)$ is $121.52 J/K⋅mol$ .

The standard entropy value for $O2(g)$ is $205.07 J/K⋅mol$ .

The standard entropy value for $H2O(l)$ is $69.95 J/K⋅mol$ .

The standard entropy value for $N2(g)$ is $191.56 J/K⋅mol$ .

The standard enthalpy value for $N2H4(l)$ is $50.63 kJ/mol$ .

The standard enthalpy value for $O2(g)$ is $0 kJ/mol$ .

The standard enthalpy value for $H2O(l)$ is $−285.83 kJ/mol$ .

The standard enthalpy value for $N2(g)$ is $0 kJ/mol$ .

The given reaction is,

$N2H4(l)+O2(g)→2H2O(l)+N2(g)$

The standard enthalpy change is,

$ΔrH°=∑nΔfH°(products)−∑nΔfH°(reactants)=[[(2 mol H2O(l)/mol-rxn)ΔfH°[H2O(l)]+(1 mol N2(g)/mol-rxn)ΔfH°[N2(g)]]-[(1 mol O2(g)/mol-rxn)ΔfH°[O2(g)]+(1 mol N2H4(l)/mol-rxn)ΔfH°[N2H4(l)]]]$

Substitute the values,

$ΔrH°=[[(2 mol H2O(l)/mol-rxn)(-285.83 kJ/mol)+(1 mol N2(g)/mol-rxn)(0 kJ/mol)]-[(1 mol O2(g)/mol-rxn)(0 kJ/mol)+(1 mol N2H4(l)/mol-rxn)(50.63 kJ/mol)]]=-622.29 kJ/mol-rxn$

The standard entropy change is,

$ΔrS°= ∑nS°(products)-∑nS°(reactants)=[[(2 mol H2O(l)/mol-rxn)S°[H2O(l)]+(1 mol N2(g)/mol-rxn)S°[N2(g)]]-[(1 mol O2(g)/mol-rxn)S°[O2(g)]+(1 mol N2H4(l)/mol-rxn)S°[N2H4(l)]]]$

Substitute the values,

$ΔrS°= [[(2 mol H2O(l)/mol-rxn)(69.95 J/K×mol)+(1 mol N2(g)/mol-rxn)(191.56 J/K×mol)]-[(1 mol O2(g)/mol-rxn)(205.07 J/K×mol)+(1 mol N2H4(l)/mol-rxn)(121.52 J/K×mol)]]= 4.87 J/K×mol-rxn$

Now,

$ΔrGo= ΔrHo- TΔrSo$

Substitute the value of $ΔrHo$ and $ΔrSo$ .

$ΔGo= -622.29 kJ/mol-rxn-[(1000 K)(4.87 J/K×mol-rxn)](1 kJ1000 J)= -623.74 kJ/mol-rxn$

Answer 20

(c)

Interpretation Introduction

Interpretation:

The change in temperature expected in heating system that has $5.5×104L$ of water should be identified.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by $ΔGo$ . It is related to entropy and entropy by the following expression,

$ΔGo= ΔHo- TΔSo$

The sign of $ΔGo$ should be positive for a product-favored reaction. Thus, spontaneous reactions are referred to those that have negative free energy formation.

The sign of $ΔSo$ should be positive for an entropy-favoured reaction and the sign of $ΔHo$ should be negative for an enthalpy favoured-reaction.

$ΔrHo$ is negative for an exothermic reaction and positive for endothermic reactions. The entropy change is positive whenever number of moles of gases is increasing in any reaction.

(c)

Expert Solution

Answer to Problem 97SCQ

The temperature change expected in a heating system containing $5.5×104 L$ of water is $2.7×10−3 K$ .

Explanation of Solution

The temperature change expected in a heating system containing $5.5×104 L$ of water is calculated below.

Given:

The value of $ΔrHo$ for the reaction of hydrazine and water is $−622.29 kJ/mol-rxn$ .

Thus, one mole of hydrazine releases $622.29×103 J$ of heat.

The density of water is $0.996 g/ml$ . Thus, the value of mass $m$ is,

$mass = (density)(volume)= (996 g/L)(5.5×104 L)= 5478×104g$

The specific heat $C$ of water is $4.184 J/g⋅K$ .

The heat of the system is related to specific heat and temperature by the expression,

$q = m c ΔT$

Substitute the values,

$622.29×103 J = (5478×104g) (4.184 J/g×K) ΔTΔT = 2.7×10-3 K$

Answer 21

(c)

Interpretation Introduction

Interpretation:

The change in temperature expected in heating system that has $5.5×104L$ of water should be identified.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by $ΔGo$ . It is related to entropy and entropy by the following expression,

$ΔGo= ΔHo- TΔSo$

The sign of $ΔGo$ should be positive for a product-favored reaction. Thus, spontaneous reactions are referred to those that have negative free energy formation.

The sign of $ΔSo$ should be positive for an entropy-favoured reaction and the sign of $ΔHo$ should be negative for an enthalpy favoured-reaction.

$ΔrHo$ is negative for an exothermic reaction and positive for endothermic reactions. The entropy change is positive whenever number of moles of gases is increasing in any reaction.

Answer 22

(c)

Expert Solution

Answer to Problem 97SCQ

The temperature change expected in a heating system containing $5.5×104 L$ of water is $2.7×10−3 K$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

The temperature change expected in a heating system containing $5.5×104 L$ of water is calculated below.

Given:

The value of $ΔrHo$ for the reaction of hydrazine and water is $−622.29 kJ/mol-rxn$ .

Thus, one mole of hydrazine releases $622.29×103 J$ of heat.

The density of water is $0.996 g/ml$ . Thus, the value of mass $m$ is,

$mass = (density)(volume)= (996 g/L)(5.5×104 L)= 5478×104g$

The specific heat $C$ of water is $4.184 J/g⋅K$ .

The heat of the system is related to specific heat and temperature by the expression,

$q = m c ΔT$

Substitute the values,

$622.29×103 J = (5478×104g) (4.184 J/g×K) ΔTΔT = 2.7×10-3 K$

Answer 27

(d)

Interpretation Introduction

Interpretation:

The amount of $O2$ present in hot water heating system of $5.5×104kg$ should be identified.

Concept introduction:

Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in $12 g$ of $12C$ .

From given mass of substance moles could be calculated by using the following formula,

$Moles of substance = Given mass of substanceMolecular mass$

(d)

Expert Solution

Answer to Problem 97SCQ

The number of moles of $O2$ present in the system is $7.5 moles$ .

Explanation of Solution

The number of moles of $O2$ present in the system is calculated below.

Given:

The solubilty of oxygen is in water at $25 °C$ is $0.000434g/100g$ of water.

Thus, the number of moles $n$ can be calculated by dividing the given mass by the molar mass and multipying it with the solubilty given.

$n = (5.5×107g)1 mol O2(32.00 g)(0.00434 g100 g)= 7.5 moles O2$

Answer 28

(d)

Interpretation Introduction

Interpretation:

The amount of $O2$ present in hot water heating system of $5.5×104kg$ should be identified.

Concept introduction:

Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in $12 g$ of $12C$ .

From given mass of substance moles could be calculated by using the following formula,

$Moles of substance = Given mass of substanceMolecular mass$

Answer 29

(d)

Expert Solution

Answer to Problem 97SCQ

The number of moles of $O2$ present in the system is $7.5 moles$ .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

The number of moles of $O2$ present in the system is calculated below.

Given:

The solubilty of oxygen is in water at $25 °C$ is $0.000434g/100g$ of water.

Thus, the number of moles $n$ can be calculated by dividing the given mass by the molar mass and multipying it with the solubilty given.

$n = (5.5×107g)1 mol O2(32.00 g)(0.00434 g100 g)= 7.5 moles O2$

Answer 34

(e)

Interpretation Introduction

Interpretation:

The mass of solution that has $5%$ dissolved hydrazine in water required to totally consume the dissolved amount of $O2$ .

Concept introduction:

Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in $12 g$ of $12C$ .

From given mass of substance moles could be calculated by using the following formula,

$Moles of substance = Given mass of substanceMolecular mass$

Mass: It is the quantitative measure of a substance. The amount of matter present in substance is expressed as mass. The $S.I.$ unit of mass is $kg$ .

(e)

Expert Solution

Answer to Problem 97SCQ

The mass of hydrazine solution that should be added to totally consume the dissolved oxygen is $4.8×103 g$ .

Explanation of Solution

The mass of hydrazine solution that should be added to totally consume the dissolved oxygen is calculated below.

Given:

Hydrazine is available as $5 %$ solution in water. If the total solution is $100 g$ then the mass of hydrazine is $5 g$ . The molar mass of hydrazine is $32.05 g$ .

The number of moles of oxygen present is $7.5 moles$ .

Thus the mass $m$ of hydrazine solution that should be added to totally consume the dissolved oxygen is,

$m = (7.5 mol O21 mol N2H4 )(100 g solution5 g)(32.05 g N2H4)= 4.8×103 g solution$

Answer 35

(e)

Interpretation Introduction

Interpretation:

The mass of solution that has $5%$ dissolved hydrazine in water required to totally consume the dissolved amount of $O2$ .

Concept introduction:

Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in $12 g$ of $12C$ .

From given mass of substance moles could be calculated by using the following formula,

$Moles of substance = Given mass of substanceMolecular mass$

Mass: It is the quantitative measure of a substance. The amount of matter present in substance is expressed as mass. The $S.I.$ unit of mass is $kg$ .

Answer 36

(e)

Expert Solution

Answer to Problem 97SCQ

The mass of hydrazine solution that should be added to totally consume the dissolved oxygen is $4.8×103 g$ .

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

The mass of hydrazine solution that should be added to totally consume the dissolved oxygen is calculated below.

Given:

Hydrazine is available as $5 %$ solution in water. If the total solution is $100 g$ then the mass of hydrazine is $5 g$ . The molar mass of hydrazine is $32.05 g$ .

The number of moles of oxygen present is $7.5 moles$ .

Thus the mass $m$ of hydrazine solution that should be added to totally consume the dissolved oxygen is,

$m = (7.5 mol O21 mol N2H4 )(100 g solution5 g)(32.05 g N2H4)= 4.8×103 g solution$

Answer 41

(f)

Interpretation Introduction

Interpretation:

The volume of $N2$ produced under given conditions should be calculated.

Concept introduction:

Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in $12 g$ of $12C$ .

From given mass of substance moles could be calculated by using the following formula,

$Moles of substance = Given mass of substanceMolecular mass$

Mass: It is the quantitative measure of a substance. The amount of matter present in substance is expressed as mass. The $S.I.$ unit of mass is $kg$ .

(f)

Expert Solution

Answer to Problem 97SCQ

The volume of $N2(g)$ that will be produced is $168 L$ .

Explanation of Solution

The volume of $N2(g)$ that will be produced is calculated below.

Given:

The given reaction is,

$N2H4(l)+O2(g)→2H2O(l)+N2(g)$

Thus, $7.5 moles$ of $O2(g)$ will release $7.5 moles$ of $N2(g)$ . At standard temperature and pressure, the volume of nitrogen gas $V$ is,

$V=(7.5 moles)(22.5 L/mol)=168 L$

Answer 42

(f)

Interpretation Introduction

Interpretation:

The volume of $N2$ produced under given conditions should be calculated.

Concept introduction:

Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in $12 g$ of $12C$ .

From given mass of substance moles could be calculated by using the following formula,

$Moles of substance = Given mass of substanceMolecular mass$

Mass: It is the quantitative measure of a substance. The amount of matter present in substance is expressed as mass. The $S.I.$ unit of mass is $kg$ .

Answer 43

(f)

Expert Solution

Answer to Problem 97SCQ

The volume of $N2(g)$ that will be produced is $168 L$ .

Answer 44

(f)

Expert Solution

Answer 45

(f)

Expert Solution

Answer 46

Expert Solution

Answer 47

Explanation of Solution

The volume of $N2(g)$ that will be produced is calculated below.

Given:

The given reaction is,

$N2H4(l)+O2(g)→2H2O(l)+N2(g)$

Thus, $7.5 moles$ of $O2(g)$ will release $7.5 moles$ of $N2(g)$ . At standard temperature and pressure, the volume of nitrogen gas $V$ is,

$V=(7.5 moles)(22.5 L/mol)=168 L$

Answer 48

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