Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
10th Edition
ISBN: 9781337399074
Author: John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher: Cengage Learning
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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
Check Mark

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 1molofN2H4at25oc 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
Check Mark

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/Kmol-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/Kmol.

The standard entropy value for O2(g) is 205.07 J/Kmol.

The standard entropy value for H2O(l) is 69.95 J/Kmol.

The standard entropy value for N2(g) is 191.56 J/Kmol.

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°=fH°(products)fH°(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
Check Mark

Answer to Problem 97SCQ

The temperature change expected in a heating system containing 5.5×104 L of water is 2.7×103 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/gK.

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 12g of 12C.

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

  Molesofsubstance GivenmassofsubstanceMolecularmass

(d)

Expert Solution
Check Mark

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 12g of 12C.

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

  Molesofsubstance GivenmassofsubstanceMolecularmass

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
Check Mark

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 12g of 12C.

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

  Molesofsubstance GivenmassofsubstanceMolecularmass

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
Check Mark

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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Chapter 18 Solutions

Chemistry & Chemical Reactivity

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