**UFBA**

# Físico-Químico - Atkins (8 ed) (Resoluções)

(Parte **1** de 5)

Student Solutions Manual to Accompany

PHYSIC~l CHEMISTRY

Eighth Edition

P. W. Atkins

Professor of Chemistry, University of Oxford and Fellow of Uncoln College

C. A. Trapp

Professor of Chemistry, University of Louisville, Louisville, Kentucky, USA

M. P. Cady

Professor of Chemistry, Indiana University Southeast, New Albany Indiana, USA

C. Giunta

Professor of Chemistry, Le Mayne College, Syracuse, NY, USA

I w. H. Freeman and Company New York

Student's solutions manual to accompany Physical Chemistry, Eighth Edition © Oxford University Press, 2006 All rights reserved

ISBN-13: 978-0-7167-6206-5 ISBN-lO: 0-7167-6206--4

Published in Great Britain by Oxford University Press This edition has been authorized by Oxford University Press for sale in the United States and Canada only and not for export therefrom.

Library of Congress Cataloging in Publication Data Data available

Second printing

W. H. Freeman and Company 41 Madison Avenue New York, NY 10010 w.whfreeman.com

Preface

This manual provides detailed solutions to all the end-of-chapter (b) Exercises, and to the odd-numbered Discussion Questions and Problems. Solutions to Exercises and Problems carried over from previous editions have been reworked, modified, or corrected when needed.

The solutions to the Problems in this edition rely more heavily on the mathematical and molecular modelling software that is now generally accessible to physical chemistry students, and this is particularly true for many of the new Problems that request the use of such software for their solutions. But almost all of the Exercises and many of the Problems can still be solved with a modem hand-held scientific calculator. When a quantum chemical calculation or molecular modelling process has been called for, we have usually provided the solution with PC Spartan pro™ because of its common availability.

In general, we have adhered rigorously to the rules for significant figures in displaying the final answers. However, when intermediate answers are shown, they are often given with one more figure than would be justified by the data. These excess digits are indicated with an overline.

We have carefully cross-checked the solutions for errors and expect that most have been eliminated.

We would be grateful to any readers who bring any remaining errors to our attention.

We warmly thank our publishers for their patience in guiding this complex, detailed project to completion.

P. W. A. e.A.T. M. P.e. e. G.

Contents

PART 1 Equilibrium 1 5 Simple mixtures 91

1 The properties of gases 3 Answers to discussion questions 91 Solutions to exercises 91

Solutions to problems 98

Answers to discussion questions 3 Solutions to numerical problems 98

Solutions to exercises 4 Solutions to theoretical problems 104

Solutions to problems 13 Solutions to applications 107 Solutions to numerical problems 13

Solutions to theoretical problems 18 Solutions to applications 20 6 Phase diagrams 112

Answers to discussion questions 112 2 The First Law 2 Solutions to exercises 113

Solutions to problems 119

Answers to discussion questions 2 Solutions to numerical problems 119 Solutions to exercises 23 Solutions to theoretical problems 124

Solutions to problems 3 Solutions to applications 124 Solutions to numerical problems 3

Solutions to theoretical problems 41 7 Chemical equilibrium 127 Solutions to applications 47

Answers to discussion questions 127 3 The Second Law 50 Solutions to exercises 128

Solutions to problems 137

Answers to discussion questions 50 Solutions to numerical problems 137 Solutions to exercises 51 Solutions to theoretical problems 148 Solutions to problems 58 Solutions to applications 150

Solutions to numerical problems 58 Solutions to theoretical problems 68 Solutions to applications 74 PART 2 Structure 155

4 Physical transformations of pure 8 Quantum theory: introduction and substances 78 prinCiples 157

Answers to discussion questions 78 Answers to discussion questions 157 Solutions to exercises 80 Solutions to exercises 158 Solutions to problems 83 Solutions to problems 162

Solutions to numerical problems 83 Solutions to numerical problems 162 Solutions to theoretical problems 86 Solutions to theoretical problems 165 Solutions to applications 87 Solutions to applications 172 viii Contents

9 Quantum theory: techniques and 14 Molecular spectroscopy 2: applications 176 electronic transitions 280

Answers to discussion questions 176 Answers to discussion questions 280 Solutions to exercises 176 Solutions to exercises 281 Solutions to problems 183 Solutions to problems 284

Solutions to numerical problems 183 Solutions to numerical problems 284 Solutions to theoretical problems 186 Solutions to theoretical problems 289 Solutions to applications 195 Solutions to applications 292

10 Atomic structure and atomic 15 Molecular spectroscopy 3: spectra 199 magnetic resonance 297

Answers to discussion questions 199 Answers to discussion questions 297 Solutions to exercises 200 Solutions to exercises 299 Solutions to problems 207 Solutions to problems 305

Solutions to numerical problems 207 Solutions to numerical problems 305 Solutions to theoretical problems 211 Solutions to theoretical problems 309 Solutions to applications 218 Solutions to applications 311

1 Molecular structure 221 16 Statistical thermodynamics 1 : the concepts 315

Answers to discussion questions 221 Solutions to exercises 223 Answers to discussion questions 315 Solutions to problems 226 Solutions to exercises 315

Solutions to numerical problems 226 Solutions to problems 322 Solutions to theoretical problems 238 Solutions to numerical problems 322

Solutions to applications 241 Solutions to theoretical problems 326 Solutions to applications 329

12 Molecular symmetry 244 17 Statistical thermodynamics 2:

Answers to discussion questions 244 applications 331 Solutions to exercises 245 Solutions to problems 249 Answers to discussion questions 331

Solutions to applications 255 Solutions to exercises 332 Solutions to problems 338

13 Molecular spectroscopy 1 : rotational Solutions to numerical problems 338 and vibrational spectra 259 Solutions to theoretical problems 345 Solutions to applications 353

Answers to discussion questions 259 Solutions to exercises 260 18 Molecular interactions 357 Solutions to problems 269

Solutions to numerical problems 269 Answers to discussion questions 357 Solutions to theoretical problems 275 Solutions to exercises 358 Solutions to applications 276 Solutions to problems 361

Contents ix

Solutions to numerical problems 361 2 The rates of chemical reactions 440 Solutions to theoretical problems 366

Solutions to applications 368 Answers to discussion questions 440 Solutions to exercises 443

19 Materials 1: macromolecules and Solutions to problems 450 aggregates 370 Solutions to numerical problems 450 Solutions to theoretical problems 455

Answers to discussion questions 370 Solutions to applications 458 Solutions to exercises 372 Solutions to problems 375 23 The kinetics of complex reactions 464 Solutions to numerical problems 375

Solutions to theoretical problems 379 Answers to discussion questions 464 Solutions to applications 383 Solutions to exercises 465 Solutions to problems 468

20 Materials 2: the solid state 389 Solutions to numerical problems 468 Solutions to theoretical problems 471

Answers to discussion questions 389 Solutions to applications 478 Solutions to exercises 390

Solutions to problems 398 24 Molecular reaction dynamics 489 Solutions to numerical problems 398

Solutions to theoretical problems 405 Answers to discussion questions 489 Solutions to applications 408 Solutions to exercises 490

Solutions to problems 497

Solutions to numerical problems 497 Solutions to theoretical problems 502 PART 3 Change 411 Solutions to applications 506

21 Molecules in motion 413 25 Processes at solid surfaces 509

Answers to discussion questions 413 Answers to discussion questions 509 Solutions to exercises 414 Solutions to exercises 511 Solutions to problems 424 Solutions to problems 521

Solutions to numerical problems 424 Solutions to numerical problems 521 Solutions to theoretical problems 430 Solutions to theoretical problems 531 Solutions to applications 433 Solutions to applications 534

PART 1 Equilibrium PART 1 Equilibrium

The properties of gases

Answers to discussion questions

01.1 An equation of state is an equation that relates the variables that define the state of a system to each other.

Boyle, Charles, and Avogadro established these relations for gases at low pressures (perfect gases) by appropriate experiments. Boyle determined how volume varies with pressure (V ex lip), Charles how volume varies with temperature (V ex T), and Avogadro how volume varies with amount of gas (V ex n). Combining all of these proportionalities into one we find nT

Vex -. p

Inserting the constant of proportionality, R, yields the perfect gas equation

RnT

V = --or pV = nRT. p

01.3 Consider three temperature regions:

(1) T < TB. At very low pressures, all gases show a compression factor, Z ~ I. At high pressures, all gases have Z > I, signifying that they have a molar volume greater than a perfect gas, which implies that repulsive forces are dominant. At intermediate pressures, most gases show Z < I, indicating that attractive forces reducing the molar volume below the perfect value are dominant.

(2) T ~ TB. Z ~ I at low pressures, slightly greater than I at intermediate pressures, and significantly greater than I only at high pressures. There is a balance between the attractive and repulsive forces at low to intermediate pressures, but the repulsive forces predominate at high pressures where the molecules are very close to each other.

(3) T > TB. Z > I at all pressures because the frequency of collisions between molecules increases with temperature.

01.5 The van der Waals equation 'corrects' the perfect gas equation for both attractive and repulsive interactions between the molecules in a real gas. See Justification 1.1 for a fuller explanation.

The Bertholet equation accounts for the volume of the molecules in a manner similar to the van der Waals equation but the term representing molecular attractions is modified to account for the effect of temperature. Experimentally one finds that the van der Waals a decreases with increasing temperature. Theory (see Chapter 18) also suggests that intermolecular attractions can decrease with temperature.

E1.1(b)

E1.2(b) E1.3(b)

4 STUDENT'S SOLUTIONS MANUAL

This variation of the attractive interaction with temperature can be accounted for in the equation of state by replacing the van der Waals a with a/To

Solutions to exercises

(a) The perfect gas law is pV = nRT implying that the pressure would be nRT

All quantities on the right are given to us except n, which can be computed from the given mass of Ar.

n = 25 g = 0.626 mol 39.95 g mol-1

(0.626 mol) x (8.31 x 10-2 dm3 bar K-1 mol-I) x (30 + 273 K) so P = 3 = . 10.5 bar . 1.5dm not 2.0 bar.

(b) The van der Waals equation is

P = V -b -V2 m m

(8.31 X 10-2 dm3 bar K-1 mol-I) x (30 + 273) K sop = (1.53 dm3 /0.626 mol) -3.20 x 10-2 dm3 mol- 1

(1.337dm6atmmol-2) x (1.013baratm-1) _I -1 -3 -10.4 bar . (1.5 dm /0.626 mo1)2

(a) Boyle's law applies: P V = constant so pf Vf = Pi Vi and prVr (1.97 bar) x (2.14dm3) 1 1 Pi = --= = 1.07 bar Vi (2.14 + 1.80) dm3 (b) The original pressure in bar is

( 1 atm) (760 TOrr) I I Pi = (1.07 bar) x x = 803 Torr 1.013 bar I atm

The relation between pressure and temperature at constant volume can be derived from the perfect gas law pV = nRT Pi so P ex: T and Ti Pr Tr

E1.4(b)

E1.S(b) E1.6(b)

E1.7(b)

THE PROPERTIES OF GASES 5

The final pressure, then, ought to be

= pjTr = (125 kPa) x (I + 273) K = 1120 kPa I Pr Tj (23 + 273) K

According to the perfect gas law, one can compute the amount of gas from pressure, temperature, and volume. Once this is done, the mass of the gas can be computed from the amount and the molar mass using pV = nRT pV (1.0atm) x (1.013 x 105 Paatm-I) x (4.0 x 103m3) 5 so n ---= 1.6 x 10 mol . -RT -(8.3145J K-Imol-I) x (20+273) K and m = (1.6 x lOS mol) x (16.04 g mol-I) = 2.67 x 106g = 12.67 X 103 kg 1

Identifying Pex in the equation P = Pex + pgh [1.3] as the pressure at the top of the straw and P as the atmospheric pressure on the liquid, the pressure difference is

P -Pex = pgh = (1.0 x 103 kg m-3) x (9.81 m s-2) x (0.15 m)

= 1.5 x 103 Pa 1(= 1.5 X 10-2 atm)

The pressure in the apparatus is given by P = Palm + pgh [1.3]

Palm = 760 Torr = I atm = 1.013 x 105 Pa pgh = 13.5 g cm-3 x x (1O:~m3) x 0.100 m x 9.806 m s-2 = 1.3 X 104 Pa

P = 1.013 X 105 Pa + 1.3 x 104 Pa = 1.146 x 105 Pa = 1115 kPa I

All gases are perfect in the limit of zero pressure. Therefore the extrapolated value of pVm/T will give the best value of R.

The molar mass is obtained from P V = nRT = -RT M

. mRT RT which upon rearrangement gives M = --= p-

The best value of M is obtained from an extrapolation of p / P versus P to P = 0; the intercept is M / RT.

Draw up the following table

0.7500 0.082 0 14 1.428 59 0.500000 0.082 027 1.428 2 0.250000 0.082 0414 1.427 90

From Figure l.l(a), (PVm) = I 0.082 0615 dm3 atm K-I mol-I I T p=O

E1.8(b)

6 STUDENT'S SOLUTIONS MANUAL

::: ··i | ·!·!·i .. ·!· .. !· .. · .. . |

"'M ",,,.,,,.! | ·'·I·I·,·j·j·· . 'i"'!"'!"'!'" .. ,." ..... . |

~ ,,;8.202, ";"';";"';";"';"';" ... i ... i ... i ... ! . ... ·-!-·!·-!-·-!-·!·,·,·!·.·! .. ·i .. ·! | |

! ... ! ... j "!"'! "!"-!-". 'i"+" | |

"M ,,!"''''! | ,',,'!',,··i .. ·!· .. i .. ·!·!· .. |

". :2 ";8.200' ·; | ·;·;·; .. ·; .. ·: .. Y·i ... : ... : ... : .. : ... ; ... '1.0 :::i::: Figure 1.I(a) |

[1 :I:,.!.: ·rr=.·:,i••• ! •••••.••.••••

From Figure 1.l(b), (~) -1.42755 g dm-3 atm-I p p=o-

i .4288 | |

"': | |

"': ·': | ·1.4286' .. |

. ,,? |

. ,: | ': |

.. , | , .. . ,! .. ... ... |

; .. ! .. ; .... : i .... : |

Figure 1.I(b)

M = RT (~) (0.0820615 dm3 atm mol-I K-1) x (273.15 K) x (1.42755 g dm-3atm-l) p p=o

= 131.9987 g mol-I

The value obtained for R deviates from the accepted value by 0.005 percent. The error results from the fact that only three data points are available and that a linear extrapolation was employed. The molar mass, however, agrees exactly with the accepted value, probably because of compensating plotting errors.

The mass density p is related to the molar volume Vrn by

M p where M is the molar mass. Putting this relation into the perfect gas law yields pVrn = RT so pM =RT p

E1.9(b)

THE PROPERTIES OF GASES 7

Rearranging this result gives an expression for M; once we know the molar mass, we can divide by the molar mass of phosphorus atoms to determine the number of atoms per gas molecule

RTp (8.314 Pa m3 mol-I) x [(100 + 273) K] x (0.6388kgm-3) M- --- - ----------------------~----------~----- P - 1.60xl()4Pa

= 0.124 kg mol-I = 124 g mol-I

The number of atoms per molecule is

124g mol- I

----'------,-= 4.0 31.0g mol-I suggesting a formula of ~

Use the perfect gas equation to compute the amount; then convert to mass.

pV P V = nRT so n = RT

We need the partial pressure of water, which is 53 percent of the equilibrium vapor pressure at the given temperature and standard pressure.

p = (0.53) x (2.69 x 103 Pa) = 1.43 x 103 Pa

(1.43 x 103 Pa) x (250 m3) -2 so n = = 1.45 x 10 mol (8.3145 J K-I mol-I) x (23 + 273) K

or m = (l.45 x 102 mol) x (l8.0g mol-I) = 2.61 x 103 g = 12.61 kg 1

E1.10(b) (a) The volume occupied by each gas is the same, since each completely fills the container. Thus solving for V we have (assuming a perfect gas) nJRT 0.225 g V = --nNe = --------=-----,-

PJ 20.18 g mol-I

= 1.115x 10-2 mol, PNe= 8.87kPa, T=300K

(!.l15 x 10-2 mol) x (8.314 dm3 kPa K-I mol-I) x 300 K) -3

V= =3.137dm 8.87 kPa

(b) The total pressure is determined from the total amount of gas, n = nCH4 + nAr + nNe.

(Parte **1** de 5)