Chemistry: The Molecular Nature of Matter and Change
Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781259631757
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
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Chapter 21, Problem 21.153P

a)

Interpretation Introduction

Interpretation:

Burning of 1.00gal 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
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Explanation of Solution

The given chemical equation:

2C8H8(l)+ 25O2(g)16CO2(g)+18H2O(g)

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

ΔHrxn°=ΔHf(poducts)°ΔHf(reactants)°

ΔHrxn°=[(16mol CO2)(ΔHf°ofCO2)+(18mol H2O)(ΔHf°ofH2O)]-[(2molC8H18)(ΔHf°ofC8H18)+(25mol O2)(ΔHf°ofO2)]ΔHrxn°=[(16mol CO2)(-393.5kJ/mol)+(18mol H2O)(-241.826kJ/mol)]-[(2molC8H18)(-250.1kJ/mol)+(25mol O2)(0kJ/mol)]ΔHrxn°=-10148.868=-10148.9kJ.

The energy from 1.00gal of gasoline as follows,

Energy (kJ)=(1.00gal)(4qt1gal)(1L1.057qt)(1mL103L)(0.7028gmL)(1molC8H18114.22gC8H18)(10148.9kJ2molC8H18)=-1.18158×105=-1.18×105kJ.

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

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 1atm are called standard atmospheric conditions.

b)

Expert Solution
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Explanation of Solution

The reaction as follows,

H2(g)12O2(g)H2O(g)

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

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

Moles of H2=(-1.18158×105kJ)(1molH2-241.826kJ)=488.6mol.

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

V=nRTP=((488.6molH2)(0.0821L.atmmol.K)(298K)1.00atm)=1.195×104=1.20×104L.

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

c)

Interpretation Introduction

Interpretation:

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

c)

Expert Solution
Check Mark

Explanation of Solution

The reaction as follows,

2H2O(l)+ 2e-H2(g)+2OH-(aq)

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

Time (s)=(488.6molH2)(2mole-2molH2)(96485C1mole-)(s1.00×103C)=9.43×104seconds.

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

d)

Interpretation Introduction

Interpretation:

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

d)

Expert Solution
Check Mark

Explanation of Solution

The coulombs involved in electrolysis of 488.6molofH2.

Coulombs=(488.6molH2)(2mole1molH2)(96485C1mole)=94,286,589C.

Conversion of Coulomb into Joules:

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

Conversion of Joules into Power:

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

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

e)

Interpretation Introduction

Interpretation:

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

e)

Expert Solution
Check Mark

Explanation of Solution

Given:

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

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

Cost=(157kW.h)(100%80%)[0.123cents1kW.h]=21.96=22.0cents.

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

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Chapter 21 Solutions

Chemistry: The Molecular Nature of Matter and Change

Ch. 21.4 - Prob. 21.6AFPCh. 21.4 - Prob. 21.6BFPCh. 21.4 - Prob. 21.7AFPCh. 21.4 - Prob. 21.7BFPCh. 21.4 - Prob. 21.8AFPCh. 21.4 - Prob. 21.8BFPCh. 21.7 - The most ionic and least ionic of the common...Ch. 21.7 - Prob. 21.9BFPCh. 21.7 - Prob. 21.10AFPCh. 21.7 - Prob. 21.10BFPCh. 21.7 - Prob. 21.11AFPCh. 21.7 - Prob. 21.11BFPCh. 21.7 - In the final steps of the ETC, iron and copper...Ch. 21.7 - Prob. B21.2PCh. 21 - Prob. 21.1PCh. 21 - Prob. 21.2PCh. 21 - Prob. 21.3PCh. 21 - Water is used to balance O atoms in the...Ch. 21 - Prob. 21.5PCh. 21 - Prob. 21.6PCh. 21 - Prob. 21.7PCh. 21 - Prob. 21.8PCh. 21 - Prob. 21.9PCh. 21 - Prob. 21.10PCh. 21 - Prob. 21.11PCh. 21 - Prob. 21.12PCh. 21 - Prob. 21.13PCh. 21 - Prob. 21.14PCh. 21 - Prob. 21.15PCh. 21 - Prob. 21.16PCh. 21 - Prob. 21.17PCh. 21 - Prob. 21.18PCh. 21 - Prob. 21.19PCh. 21 - Prob. 21.20PCh. 21 - Aqua regia, a mixture of concentrated HNO3 and...Ch. 21 - Consider the following general voltaic...Ch. 21 - Why does a voltaic cell not operate unless the two...Ch. 21 - Prob. 21.24PCh. 21 - Prob. 21.25PCh. 21 - Prob. 21.26PCh. 21 - Consider the following voltaic cell: In which...Ch. 21 - Consider the following voltaic cell: In which...Ch. 21 - Prob. 21.29PCh. 21 - Prob. 21.30PCh. 21 - A voltaic cell is constructed with an Fe/Fe2+...Ch. 21 - Prob. 21.32PCh. 21 - Prob. 21.33PCh. 21 - Prob. 21.34PCh. 21 - Prob. 21.35PCh. 21 - What does a negative indicate about a redox...Ch. 21 - Prob. 21.37PCh. 21 - In basic solution, Se2− and ions react...Ch. 21 - Prob. 21.39PCh. 21 - Prob. 21.40PCh. 21 - Use the emf series (Appendix D) to arrange each...Ch. 21 - Prob. 21.42PCh. 21 - Prob. 21.43PCh. 21 - Prob. 21.44PCh. 21 - Prob. 21.45PCh. 21 - Prob. 21.46PCh. 21 - Prob. 21.47PCh. 21 - Prob. 21.48PCh. 21 - Prob. 21.49PCh. 21 - Prob. 21.50PCh. 21 - Prob. 21.51PCh. 21 - Prob. 21.52PCh. 21 - Prob. 21.53PCh. 21 - Prob. 21.54PCh. 21 - Prob. 21.55PCh. 21 - Prob. 21.56PCh. 21 - Prob. 21.57PCh. 21 - Prob. 21.58PCh. 21 - Prob. 21.59PCh. 21 - Prob. 21.60PCh. 21 - Prob. 21.61PCh. 21 - Prob. 21.62PCh. 21 - Prob. 21.63PCh. 21 - Prob. 21.64PCh. 21 - Prob. 21.65PCh. 21 - Prob. 21.66PCh. 21 - Prob. 21.67PCh. 21 - Prob. 21.68PCh. 21 - Prob. 21.69PCh. 21 - Prob. 21.70PCh. 21 - Prob. 21.71PCh. 21 - Prob. 21.72PCh. 21 - Prob. 21.73PCh. 21 - Prob. 21.74PCh. 21 - Prob. 21.75PCh. 21 - Prob. 21.76PCh. 21 - Prob. 21.77PCh. 21 - Prob. 21.78PCh. 21 - Prob. 21.79PCh. 21 - Prob. 21.80PCh. 21 - Prob. 21.81PCh. 21 - Consider the following general electrolytic...Ch. 21 - Prob. 21.83PCh. 21 - Prob. 21.84PCh. 21 - Prob. 21.85PCh. 21 - Prob. 21.86PCh. 21 - In the electrolysis of molten NaBr: What product...Ch. 21 - Prob. 21.88PCh. 21 - Prob. 21.89PCh. 21 - Prob. 21.90PCh. 21 - Prob. 21.91PCh. 21 - Prob. 21.92PCh. 21 - Prob. 21.93PCh. 21 - Prob. 21.94PCh. 21 - Prob. 21.95PCh. 21 - Prob. 21.96PCh. 21 - Prob. 21.97PCh. 21 - Write a balanced half-reaction for the product...Ch. 21 - Prob. 21.99PCh. 21 - Prob. 21.100PCh. 21 - Prob. 21.101PCh. 21 - Prob. 21.102PCh. 21 - Prob. 21.103PCh. 21 - Prob. 21.104PCh. 21 - Prob. 21.105PCh. 21 - Prob. 21.106PCh. 21 - Prob. 21.107PCh. 21 - Prob. 21.108PCh. 21 - Prob. 21.109PCh. 21 - Prob. 21.110PCh. 21 - Prob. 21.111PCh. 21 - Prob. 21.112PCh. 21 - Prob. 21.113PCh. 21 - Prob. 21.114PCh. 21 - Prob. 21.115PCh. 21 - Prob. 21.116PCh. 21 - Prob. 21.117PCh. 21 - Prob. 21.118PCh. 21 - Prob. 21.119PCh. 21 - Prob. 21.120PCh. 21 - To examine the effect of ion removal on cell...Ch. 21 - Prob. 21.122PCh. 21 - Prob. 21.123PCh. 21 - Prob. 21.124PCh. 21 - Prob. 21.125PCh. 21 - Prob. 21.126PCh. 21 - Commercial electrolytic cells for producing...Ch. 21 - Prob. 21.129PCh. 21 - Prob. 21.130PCh. 21 - The following reactions are used in...Ch. 21 - Prob. 21.132PCh. 21 - Prob. 21.133PCh. 21 - Prob. 21.134PCh. 21 - Prob. 21.135PCh. 21 - If the Ecell of the following cell is 0.915 V,...Ch. 21 - Prob. 21.137PCh. 21 - Prob. 21.138PCh. 21 - Prob. 21.139PCh. 21 - Prob. 21.140PCh. 21 - Prob. 21.141PCh. 21 - Prob. 21.142PCh. 21 - Prob. 21.143PCh. 21 - Prob. 21.144PCh. 21 - Prob. 21.145PCh. 21 - Prob. 21.146PCh. 21 - Prob. 21.147PCh. 21 - Both Ti and V are reactive enough to displace H2...Ch. 21 - For the reaction ∆G° = 87.8 kJ/mol Identity the...Ch. 21 - Two voltaic cells are to be joined so that one...Ch. 21 - Prob. 21.152PCh. 21 - Prob. 21.153P
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