https://1drv.ms/b/s!AmkCsf2WlV7n3g6VtJuK3DHt7GQo?e=QSK0rH
Student Solutions Manual to Accompany Atkins’ Physical Chemistry ELEVENTH EDITION Peter Bolgar Haydn Lloyd Aimee North Vladimiras Oleinikovas Stephanie Smith and James Keeler Department of Chemistry University of Cambridge UK 1 Great Clarendon Street, Oxford, OX2 6DP, United Kingdom Oxford University Press is a department of the University of Oxford It furthers the University’s objective of excellence in research, scholarship, and education by publishing worldwide Oxford is a registered trade mark of Oxford University Press in the UK and in certain other countries © Oxford University Press 2018 The moral rights of the authors have been asserted Eighth edition 2006 Ninth edition 2010 Tenth edition 2014 Impression: All rights reserved No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, without the prior permission in writing of Oxford University Press, or as expressly permitted by law, by licence or under terms agreed with the appropriate reprographics rights organization Enquiries concerning reproduction outside the scope of the above should be sent to the Rights Department, Oxford University Press, at the address above You must not circulate this work in any other form and you must impose this same condition on any acquirer Published in the United States of America by Oxford University Press 198 Madison Avenue, New York, NY 10016, United States of America British Library Cataloguing in Publication Data Data available ISBN 978–0–19–255086–6 Printed in Great Britain by Bell & Bain Ltd., Glasgow Links to third party websites are provided by Oxford in good faith and for information only Oxford disclaims any responsibility for the materials contained in any third party website referenced in this work Table of contents Preface vii The properties of gases 1A The perfect gas 1B The kinetic model 12 1C Real gases 24 Internal energy 41 2A Internal energy 41 2B Enthalpy 47 2C Thermochemistry 50 2D State functions and exact differentials 58 2E Adiabatic changes 66 The second and third laws 73 3A Entropy 73 3B Entropy changes accompanying specific processes 79 3C The measurement of entropy 91 3D Concentrating on the system 101 3E Combining the First and Second Laws 107 Physical transformations of pure substances 119 4A Phase diagrams of pure substances 119 4B Thermodynamic aspects of phase transitions 121 Simple mixtures 137 5A The thermodynamic description of mixtures 137 5B The properties of solutions 149 5C Phase diagrams of binary systems: liquids 165 5D Phase diagrams of binary systems: solids 173 5E Phase diagrams of ternary systems 179 5F 184 Activities Chemical equilibrium 199 6A The equilibrium constant 199 iv TABLE OF CONTENTS 6B The response of equilibria to the conditions 208 6C Electrochemical cells 221 6D Electrode potentials 228 Quantum theory 243 7A The origins of quantum mechanics 243 7B Wavefunctions 250 7C Operators and observables 254 7D Translational motion 263 7E Vibrational motion 277 7F 288 Rotational motion Atomic structure and spectra 299 8A Hydrogenic Atoms 299 8B Many-electron atoms 308 8C Atomic spectra 311 Molecular Structure 321 9A Valence-bond theory 321 9B Molecular orbital theory: the hydrogen molecule-ion 324 9C Molecular orbital theory: homonuclear diatomic molecules 329 9D Molecular orbital theory: heteronuclear diatomic molecules 333 9E Molecular orbital theory: polyatomic molecules 339 10 Molecular symmetry 353 10A Shape and symmetry 353 10B Group theory 363 10C Applications of symmetry 374 11 Molecular Spectroscopy 385 11A General features of molecular spectroscopy 385 11B Rotational spectroscopy 394 11C Vibrational spectroscopy of diatomic molecules 408 11D Vibrational spectroscopy of polyatomic molecules 421 11E Symmetry analysis of vibrational spectroscopy 424 11F Electronic spectra 426 11G Decay of excited states 437 TABLE OF CONTENTS 12 Magnetic resonance 445 12A General principles 445 12B Features of NMR spectra 449 12C Pulse techniques in NMR 458 12D Electron paramagnetic resonance 467 13 Statistical thermodynamics 473 13A The Boltzmann distribution 473 13B Partition functions 477 13C Molecular energies 487 13D The canonical ensemble 496 13E The internal energy and entropy 497 13F Derived functions 511 14 Molecular Interactions 521 14A Electric properties of molecules 521 14B Interactions between molecules 533 14C Liquids 539 14D Macromolecules 542 14E Self-assembly 554 15 Solids 561 15A Crystal structure 561 15B Diffraction techniques 564 15C Bonding in solids 571 15D The mechanical properties of solids 576 15E The electrical properties of solids 578 15F The magnetic properties of solids 580 15G The optical properties of solids 583 16 Molecules in motion 589 16A Transport properties of a perfect gas 589 16B Motion in liquids 595 16C Diffusion 601 17 Chemical kinetics 17A The rates of chemical reactions 611 611 v vi TABLE OF CONTENTS 17B Integrated rate laws 617 17C Reactions approaching equilibrium 634 17D The Arrhenius equation 638 17E Reaction mechanisms 642 17F Examples of reaction mechanisms 648 17G Photochemistry 652 18 Reaction dynamics 671 18A Collision theory 671 18B Diffusion-controlled reactions 676 18C Transition-state theory 679 18D The dynamics of molecular collisions 691 18E Electron transfer in homogeneous systems 693 19 Processes at solid surfaces 699 19A An introduction to solid surfaces 699 19B Adsorption and desorption 704 19C Heterogeneous catalysis 717 19D Processes at electrodes 719 Preface This manual provides detailed solutions to the (a) Exercises and the odd-numbered Discussion questions and Problems from the 11th edition of Atkins’ Physical Chemistry Conventions used is presenting the solutions We have included page-specific references to equations, sections, figures and other features of the main text Equation references are denoted [14B.3b–595], meaning eqn 14B.3b located on page 595 (the page number is given in italics) Other features are referred to by name, with a page number also given Generally speaking, the values of physical constants (from the first page of the main text) are used to significant figures except in a few cases where higher precision is required In line with the practice in the main text, intermediate results are simply truncated (not rounded) to three figures, with such truncation indicated by an ellipsis, as in 0.123 ; the value is used in subsequent calculations to its full precision The final results of calculations, generally to be found in a box , are given to the precision warranted by the data provided We have been rigorous in including units for all quantities so that the units of the final result can be tracked carefully The relationships given on the back of the front cover are useful in resolving the units of more complex expressions, especially where electrical quantities are involved Some of the problems either require the use of mathematical software or are much easier with the aid of such a tool In such cases we have used Mathematica (Wolfram Research, Inc.) in preparing these solutions, but there are no doubt other options available Some of the Discussion questions relate directly to specific section of the main text in which case we have simply given a reference rather than repeating the material from the text Acknowledgements In preparing this manual we have drawn on the equivalent volume prepared for the 10th edition of Atkins’ Physical Chemistry by Charles Trapp, Marshall Cady, and Carmen Giunta In particular, the solutions which use quantum chemical calculations or molecular modelling software, and some of the solutions to the Discussion questions, have been quoted directly from the solutions manual for the 10th edition, without significant modification More generally, we have benefited from the ability to refer to the earlier volume and acknowledge, with thanks, the influence that its authors have had on the present work This manual has been prepared by the authors using the LATEX typesetting system, in the implementation provided by MiKTEX (miktex.org); the vast majority of the figures and graphs have been generated using PGFPlots We are grateful to the community who maintain and develop these outstanding resources Finally, we are grateful to the editorial team at OUP, Jonathan Crowe and Roseanna Levermore, for their invaluable support in bringing this project to a conclusion viii PREFACE Errors and omissions In such a complex undertaking some errors will no doubt have crept in, despite the authors’ best efforts Readers who identify any errors or omissions are invited to pass them on to us by email to pchem@ch.cam.ac.uk The properties of gases 1A The perfect gas Answers to discussion questions D1A.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 ∝ 1/p), Charles how volume varies with temperature (V ∝ T), and Avogadro how volume varies with amount of gas (V ∝ n) Combining all of these proportionalities into one gives nT V∝ p Inserting the constant of proportionality, R, yields the perfect gas equation V =R nT p or pV = nRT Solutions to exercises E1A.1(a) From the inside the front cover the conversion between pressure units is: atm ≡ 101.325 kPa ≡ 760 Torr; bar is 105 Pa exactly (i) A pressure of 108 kPa is converted to Torr as follows 108 kPa × E1A.2(a) atm 760 Torr × = 810 Torr 101.325 kPa atm (ii) A pressure of 0.975 bar is 0.975 × 105 Pa, which is converted to atm as follows atm 0.975 × 105 Pa × = 0.962 atm 101.325 kPa The perfect gas law [1A.4–8], pV = nRT, is rearranged to give the pressure, p = nRT/V The amount n is found by dividing the mass by the molar mass of Xe, 131.29 g mol−1 n (8.2057 × 10−2 dm3 atm K−1 mol−1 ) × (298.15 K) (131 g) (131.29 g mol−1 ) 1.0 dm3 = 24.4 atm p= SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY (b) Again assuming V T 3/2 = constant for the adiabatic stage, the volume after Stage can be related to the initial volume VD = VA × ( Th 3/2 373 K 3/2 ) = (1.00 dm3 ) × ( ) Tc 273 K = 1.59 dm3 = 1.60 dm3 (c) As shown in Section 3A.3(a) on page 82 the heat transferred reversibly during an isothermal gas expansion is q rev = nRT ln (Vf /Vi ), thus the heats for Stage and Stage are, respectively VB ) VA = (0.100 mol) × (8.3145 J K−1 mol−1 ) × (373 K) × ln (2) q = q h = nRTh ln ( = +2.14 × 102 J = +215 J q = q c = nRTc ln ( VD ) VC = (0.100 mol) × (8.3145 J K−1 mol−1 ) × (273 K) × ln ( = −1.57 × 102 J = −157 J 1.59 dm3 ) 3.19 dm3 Because there is no heat exchange during adiabatic processes, rhe heat transfer for Stages and are q = and q = , respectively (d) At the beginning and end of the cycle the temperature is the same Because the working substance is a perfect gas, ∆U = over the cycle The First Law [2A.2–38], ∆U = w + q, therefore implies that w = −q, that is, the net heat over the cycle is converted to work This net heat is the difference between that extracted from the hot source and deposited into the cold sink (e) The efficiency is defined in [3A.7–84], η = ∣w∣/∣q h ∣ As has been explained, ∣w∣ is the net heat ∣w∣ = ∣q h ∣ − ∣q c ∣ = ∣ + 2.14 × 102 J∣ − ∣ − 1.57 × 102 J∣ hence = +5.7 × 101 J = +58 J η= ∣w∣ ∣ + 5.7 × 101 J∣ = = 0.268 = 27% ∣q h ∣ ∣ + 2.14 × 102 J∣ (f) The Carnot efficiency is given by [3A.9–84], η =1− 273 K Tc =1− = 0.268 = 26.8% Th 373 K the result is the same as the above (the difference is due to the use of fewer significant figures in the previous calculation) 77 78 THE SECOND AND THIRD LAWS Using the values of the heat transfer calculated above in equation [3A.6– 84] gives q c q c 214 J −157 J + = + Th Tc 373 K 273 K = 0.0 J the result is zero, as expected from a Carnot cycle P3A.5 (a) Consider a process in which heat dq c is extracted from the cold source at temperature Tc , and heat dq h is discarded into the hot sink at temperature Th The overall entropy change of such process is dS = dq c dq h + Tc Th Assume that dq c = −dq and dq h = +dq, where dq is a positive quantity It follows that dS = +dq −dq 1 + = dq × ( − ) Th Tc Th Tc Because Th > Tc , the term in parentheses is negative, therefore dS is negative The process is therefore not spontaneous and not allowed by the Second Law If work is done on the engine, ∣dq h ∣ will become greater than ∣dq c ∣ and eventually dS will be greater than zero (b) Assuming q c = −∣q∣ and q h = ∣q∣ + ∣w∣ the overall change in entropy is ∆S = −∣q∣ ∣q∣ + ∣w∣ + Tc Th For the process to be permissible by the Second Law the Clausius inequality defined in [3A.12–86], dS ≥ 0, must be satisfied Therefore which implies −∣q∣ ∣q∣ + ∣w∣ + ≥0 Tc Th ∣w∣ ≥ ∣q∣ × ( P3A.7 Th Th − 1) = ∣q∣ × ( − 1) Tc Tc Suppose two adiabatic paths intersect at point A as shown in the figure Two remote points corresponding to the same temperature on each adiabat, A and B, are then connected by an isothermal path forming a cycle Consider energy changes for each Stage of the cycle Stage (A → B) is adiabatic and, thus, no heat exchange takes place q = Therefore, the total change in internal energy is ∆U = w + q = w Stage (B → C) is an isothermal change and assuming that the system energy is a function of temperature only (e.g SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY Pressure, p A Stage Stage C Stage B Volume, V Figure 3.1 ideal gas): ∆U = w + q = Stage (C → A) is again adiabatic, q = 0, with ∆U = w + q = w Because the system energy is a function of temperature only, U B = U C and, thus ∆U = U A − U C = U A − U B = −∆U This implies that w = −w Because internal energy is a state function and the cycle is closed: U cycle = w cycle + q cycle = = ∆U + ∆U + ∆U Finally, analyse the net work done, w cycle = w + w + w = w , and the net heat absorbed, q cycle = q + q + q = q , over the cycle It is apparent that the sole result of the process is the absorption of heat q and its convertion to work w , which directly contradicts the statement of the Second Law by Kelvin, unless the q = w = 0, i.e points B and C are the same and correspond to the same path Therefore, no two such adiabatic paths exist 3B Entropy changes accompanying specific processes Answer to discussion question D3B.1 The explanation of Trouton’s rule is that a comparable change in volume is expected whenever any unstructured liquid forms a vapour; accompanying this will be a comparable change in the number of accessible microstates Hence, all unstructured liquids can be expected to have similar entropies of vaporization Liquids that show significant deviations from Trouton’s rule so on account of strong molecular interactions that restrict molecular motion As a result there is a greater dispersal of matter and energy when such liquids vaporize 79 80 THE SECOND AND THIRD LAWS Water is an example of a liquid with strong intermolecular interactions (hydrogen bonding) which tend to organize the molecules in the liquid, hence its entropy of vaporization is expected to be greater than the value predicted by Trouton’s rule The same is true for ethanol, which is also hydrogen bonded in the liquid Mercury has quite strong interactions between the atoms, as evidenced by its cohesiveness, and so its entropy of vaporization is expected to be greater than that predicted by Trouton’s rule Solutions to exercises E3B.1(a) The entropy change of a phase transition is given by [3B.4–89], ∆ trs S = ∆ trs H/Ttrs As discussed in Section 3B.2 on page 89 because there is no hydrogen bonding in liquid benzene it is safe to apply Trouton’s rule That is ∆ vap S −○ = +85 J K−1 mol−1 It follows that ∆ vap H −○ = Tb × ∆ vap S −○ = (273.15 K + 80.1 K) × (+85 J K−1 mol−1 ) = 3.00 × 104 J mol−1 = +30 kJ mol−1 E3B.2(a) (i) The entropy change of a phase transition is given by [3B.4–89], ∆ trs S = ∆ trs H/Ttrs For vaporisation this becomes ∆ vap S −○ = ∆ vap H −○ + 29.4 × 103 J mol−1 = Tb 334.88 K = +87.8 J K−1 mol−1 E3B.3(a) (ii) Because the system at the transition temperature is at equilibrium, ∆S tot = 0, thus ∆S sur = −∆ vap S −○ = −87.8 J K−1 mol−1 The change in entropy with temperature is given by [3B.6–90], ∆S = S(Tf ) − S(Ti ) = ∫ Tf Ti Cp dT T Assuming that C p is constant in the temperature range Ti to Tf , this becomes ∆S = C p ln (Tf /Ti ) as detailed in Section 3B.3 on page 90 Thus, the increase in the molar entropy of oxygen gas is ∆S m = S m (348 K) − S m (298 K) = (29.355 J K−1 mol−1 ) × ln ( = +4.55 J K−1 mol−1 348 K ) 298 K SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY E3B.4(a) As explained in Section 3B.3 on page 90 the temperature variation of the entropy at constant volume is given by ∆S = S(Tf ) − S(Ti ) = ∫ Tf Ti CV dT T Assuming that C V = 32 R, the ideal gas limit, for the temperature range of interest, the molar entropy at 500 K is given by S m (500 K) = S m (298 K) + ∫ 500 K dT R T 500 K = S m (298 K) + × ln ( ) 298 K = (146.22 J K−1 mol−1 ) 298 K R + ( 32 × 8.3145 J K−1 mol−1 ) × ln ( = 153 J K−1 mol−1 E3B.5(a) 500 K ) 298 K Two identical blocks must come to their average temperature Therefore the final temperature is Tf = 12 (T1 + T2 ) = × (50 ○ C + ○ C) = 25 ○ C = 298 K Although the above result may seem self-evident, the more detailed explaination is as follows The heat capacity at constant volume is defined in [2A.14– 43], C V = (∂U/∂T)V As shown in Section 2A.4(b) on page 43, if the heat capacity is constant, the internal energy changes linearly with the change in temperature That is ∆U = C V ∆T = C V (Tf − Ti ) For the two blocks at the initial temperatures of T1 and T2 , the change in internal energy to reach the final temperature Tf is ∆U = C V ,1 (Tf − T1 ) and ∆U = C V ,2 (Tf − T2 ), respectively The blocks of metal are made of the same substance and are of the same size, therefore C V ,1 = C V ,2 = C V Because the system is isolated the total change in internal energy is ∆U = ∆U + ∆U = This means that ∆U = C V ((Tf − T1 ) − (Tf − T2 )) = C V × (2Tf − (T1 + T2 )) = 0, which implies that the final temperature is Tf = 12 (T1 + T2 ), as stated above The temperature variation of the entropy at constant volume is given by [3B.7– 90], ∆S = C V ln (Tf /Ti ), with C p replaced by C V Expressed with the specific heat C V ,s = C V /m it becomes ∆S = mC V ,s ln ( Tf ) Ti Note that for a solid the internal energy does not change significantly with the volume or pressure, thus it can be assumed that C V = C p = C The entropy 81 82 THE SECOND AND THIRD LAWS change for each block is found using this expression ∆S = mC V ,s ln ( Tf ) T1 = (1.00 × 103 g) × (0.385 J K−1 g−1 ) × ln ( 298 K ) 50 K + 273.15 K = (1.00 × 103 g) × (0.385 J K−1 g−1 ) × ln ( 298 K ) 273.15 K = −31.0 J K−1 = −31.0 J K−1 Tf ∆S = mC V ,s ln ( ) T2 = 33.7 J K−1 = +33.7 J K−1 The total change in entropy is ∆S tot = ∆S + ∆S = (−31.0 J K−1 ) + (33.7 J K−1 ) = 27.2 J K−1 = +2.7 J K−1 E3B.6(a) Because ∆S tot > the process is spontaneous, in accord with experience Because entropy is a state function, ∆S between the initial and final states is the same irrespective of the path taken Thus the overall process can be broken down into steps that are easier to evaluate First consider heating the initial system at constant pressure to the final temperature The variation of entropy with temperature at constant pressure is given by [3B.7–90], S(Tf ) = S(Ti ) + C p ln (Tf /Ti ) Thus the change in entropy, ∆S = S(Tf ) − S(Ti ), of this step is ∆S = C p ln ( Tf Tf ) = nC p,m ln ( ) Ti Ti Next consider an isothermal change in pressure As explained in Section 3A.2(a) on page 80 the change in entropy of an isothermal expansion of an ideal gas is given by ∆S = nR ln (Vf /Vi ) Because for a fixed amount of gas at fixed temperature p ∝ (1/V ) an equivalent expression for this entropy change is ∆S = nR ln ( pi ) pf Therefore the overall entropy change for the system is ∆S = ∆S + ∆S = nC p,m ln ( Tf pi ) + nR ln ( ) Ti pf 273.15 K + 125 K ) 273.15 K + 25 K 1.00 atm ) + (3.00 mol) × (8.3145 J K−1 mol−1 ) × ln ( 5.00 atm = (3.00 mol) × ( 52 × 8.3145 J K−1 mol−1 ) × ln ( = (+18.0 J K−1 ) + (−40.1 J K−1 ) = −22.1 J K−1 SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY E3B.7(a) Because entropy is a state function, ∆S between the initial and final states is the same irrespective of the path taken Thus the overall process can be broken down into steps that are easier to evaluate First consider heating the ice at constant pressure from the initial temperature to the melting point, Tm The variation of entropy with temperature at constant pressure is given by [3B.7– 90], S(Tf ) = S(Ti ) + C p ln (Tf /Ti ) Thus the change in entropy, ∆S = S(Tf ) − S(Ti ), for this step is ∆S = C p ln ( Tm Tm ) = nC p,m (H2 O(s)) ln ( ) Ti Ti Next consider the phase transition from solid to liquid at the melting temperature The entropy change of a phase transition is given by [3B.4–89], ∆ trs S = ∆ trs H/Ttrs , thus ∆S = n − ○ ∆ fus H m Tm Then the liquid is heated to the boiling temperature, Tb In analogy to the first step ∆S = nC p,m (H2 O(l)) ln ( Tb ) Tm The next phase transition is from liquid to gas ∆S = n − ○ ∆ vap H m Tb Finally, the vapour is heated from Tb to Tf ∆S = nC p,m (H2 O(g)) ln ( Tf ) Tb 83 84 THE SECOND AND THIRD LAWS Therefore the overall entropy change for the system is ∆S/n = ∆S + ∆S + ∆S + ∆S + ∆S − ○ Tm ∆ fus H m Tb )+ + C p,m (H2 O(l)) ln ( ) Ti Tm Tm − ○ ∆ vap H m Tf + + C p,m (H2 O(g)) ln ( ) Tb Tb 273.15 K = (37.6 J K−1 mol−1 ) × ln ( ) 273.15 K − 10.0 K 6.01 × 103 J mol−1 + 273.15 K 273.15 K + 100.0 K ) + (75.3 J K−1 mol−1 ) × ln ( 273.15 K 40.7 × 103 J mol−1 + 273.15 K + 100.0 K 273.15 K + 115.0 K ) + (33.6 J K−1 mol−1 ) × ln ( 273.15 K + 100.0 K = (+1.40 J K−1 mol−1 ) + (+22.0 J K−1 mol−1 ) = C p,m (H2 O(s)) ln ( + (+23.4 J K−1 mol−1 ) + (+1.09 × 102 J K−1 mol−1 ) + (+1.32 J K−1 mol−1 ) Hence = +1.57 × 102 J K−1 mol−1 ∆S = 10.0 g × (+1.57 × 102 J K−1 ) = +87.3 J K−1 18.02 g mol−1 Solutions to problems P3B.1 Because entropy is a state function, ∆S between the initial and final states is the same irrespective of the path taken Thus the overall process can be broken down into steps that are easier to evaluate First consider heating the water at constant pressure from the initial temperature T to the melting point The variation of the entropy with temperature at constant pressure is given by [3B.7–90], S(Tf ) = S(Ti ) + C p ln (Tf /Ti ) Thus the change in entropy for this step is ∆S = C p ln ( Tm Tm ) = nC p,m (H2 O(l)) ln ( ) T T Next consider the phase transition from liquid to solid at the melting temper− ○ ature; note that freezing is just the opposite of fusion, thus ∆H = n(−∆ fus H m ) The entropy change of a phase transition is given by [3B.4–89], ∆ trs S = ∆ trs H/Ttrs , thus − ○ −∆ fus H m ∆H ∆S = =n Tm Tm SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY The ice is then cooled to the final temperature, T Similarly to the first step ∆S = nC p,m (H2 O(s)) ln ( T ) Tm Therefore the overall entropy change for the system is ∆S = ∆S + ∆S + ∆S ○ Tm T −∆ fus H −m + nC p,m (H2 O(s)) ln ( ) )+n T Tm Tm 273.15 K = (1.00 mol) × (75.3 J K−1 mol−1 ) × ln ( ) 273.15 K − 5.00 K −6.01 × 103 J mol−1 + (1.00 mol) × 273.15 K 273.15 K − 5.00 K ) + (1.00 mol) × (37.6 J K−1 mol−1 ) × ln ( 273.15 K = (+1.39 J K−1 ) + (−22.0 J K−1 ) + (−0.694 J K−1 ) = nC p,m (H2 O(l)) ln ( = −21.3 J K−1 = −21.3 J K−1 Consider enthalphy change for the same path The variation of enthalpy with temperature at constant pressure is given by [2B.6b–49], ∆H = C p ∆T Thus for the first and third steps, respectively ∆H = nC p,m (H2 O(l))(Tm − T) and ∆H = nC p,m (H2 O(s))(T − Tm ) Therefore the overall enthalpy change for the system is ∆H = ∆H + ∆H + ∆H − ○ = nC p,m (H2 O(l))(Tm − T) + n(−∆ fus H m ) + nC p,m (H2 O(s))(T − Tm ) = (1.00 mol) × (75.3 J K−1 mol−1 ) × (+5.00 K) + (1.00 mol) × (−6.01 × 103 J mol−1 ) + (1.00 mol) × (37.6 J K−1 mol−1 ) × (−5.00 K) = (+3.76 × 102 J) + (−6.01 × 103 J) + (−1.88 × 102 J) = −5.82 × 103 J At constant pressure the heat released by the system is the enthalpy change of the system, q = ∆H Because q sur = −q, the entropy change of the surroundings is −q −(−5.82 × 103 J) = T 273.15 K − 5.00 K = +21.7 J K−1 = +21.7 J K−1 ∆S sur = Therefore the total entropy change is ∆S tot = ∆S + ∆S sur = (−21.3 J K−1 ) + (+21.7 J K−1 ) = +0.403 J K−1 = +0.4 J K−1 85 86 THE SECOND AND THIRD LAWS Because the total entropy change is positive, the Second Law implies that the process is spontaneous A similar method is used to find the entropy change when the liquid evaporates at T2 Consider heating the liquid to the boiling temperature Tb , then the phase transition taking place, followed by cooling of the gas back to the temperature T2 The entropy changes are calculated in an analagous way ∆S = ∆S + ∆S + ∆S − ○ ∆ vap H m Tb T2 )+n + nC p,m (H2 O(g)) ln ( ) T2 Tb Tb 273.15 K + 100 K −1 −1 = (1.00 mol) × (75.3 J K mol ) ln ( ) 273.15 K + 95.0 K 4.07 × 104 J mol−1 + (1.00 mol) × 273.15 K + 100 K 273.15 K + 95.0 K ) + (1.00 mol) × (33.6 J K−1 mol−1 ) ln ( 273.15 K + 100 K = (+1.01 J K−1 ) + (+1.09 × 102 J K−1 ) + (−0.453 J K−1 ) = nC p,m (H2 O(l)) ln ( = +1.09 × 102 J K−1 = +110 J K−1 −∆H = − × (∆H + ∆H + ∆H ) T2 T2 − ○ −∆ vap H m Tb − T2 T2 − Tb = − (nC p,m (H2 O(l)) +n + nC p,m (H2 O(g)) ) T2 T2 T2 5.00 K = −(1.00 mol) × (75.3 J K−1 mol−1 ) 273.15 K + 95.0 K 4.07 × 104 J mol−1 − (1.00 mol) × 273.15 K + 95.0 K −5.00 K − (1.00 mol) × (33.6 J K−1 mol−1 ) × 273.15 K + 95.0 K = −(+1.02 J K−1 ) − (+1.10 × 102 J K−1 ) − (−0.456 J K−1 ) ∆S sur = = −1.11 × 102 J K−1 = −111 J K−1 Therefore the total entropy change is ∆S tot = ∆S + ∆S sur = (+1.09 × 102 J K−1 ) + (−1.11 × 102 J K−1 ) = −1.48 J K−1 = −1.5 J K−1 Because the change in the entropy is negative, the Second Law implies that the process is not spontaneous P3B.3 Consider heating trichloromethane at constant pressure from the initial to final temperatures The variation of the entropy with temperature is given by SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY [3B.6–90], S(Tf ) = S(Ti ) + ∫Ti f (C p /T)dT The contant-pressure molar heat capacity is given as a function of temperature of a form C p,m = a + bT, with a = +91.47 J K−1 mol−1 and b = +7.5 × 10−2 J K−2 mol−1 Thus the change in molar entropy, ∆S m = S m (Tf ) − S m (Ti ), of this process is T ∆S m = ∫ Tf Ti (C p,m /T)dT = ∫ = a×∫ Tf Ti = a × ln ( Tf Ti a + bT dT T Tf dT dT + b × ∫ T Ti Tf ) + b × (Tf − Ti ) Ti 300 K ) 273 K + (+7.5 × 10−2 J K−2 mol−1 ) × (300 K − 273 K) = (+91.47 J K−1 mol−1 ) × ln ( = (+8.62 J K−1 mol−1 ) + (+2.02 J K−1 mol−1 ) = +10.7 J K−1 mol−1 P3B.5 Two identical blocks must come to their average temperature Therefore the final temperature is T = 12 (Tc + Th ) Although the above result may seem self-evident, the more detailed explaination is as follows The heat capacity at constant volume is defined in [2A.14– 43], C V = (∂U/∂T)V As shown in Section 2A.4(b) on page 43, if the heat capacity is constant, the internal energy changes linearly with the change in temperature That is ∆U = C V ∆T = C V (Tf − Ti ) For the two blocks at the initial temperatures of Tc and Th , the change in internal energy to reach the final temperature T is ∆U c = C V ,c (T − Tc ) and ∆U h = C V ,h (T − Th ), respectively The blocks of metal are made of the same substance and are of the same size, therefore C V ,c = C V ,h = C V Note that for a given solid the internal energy does not change significantly on the volume or pressure, thus it can be assumed that C V = C p Assuming the system is isolated the total change in internal energy is ∆U = ∆U c + ∆U h = This means that ∆U = C p ((T − Tc ) − (T − Th )) = C p × (2T − (T1 + T2 )) = 0, which implies that the final temperature is T = 12 (Tc + Th ), as stated above At constant pressure the temperature dependence of the entropy is given by [3B.7–90], Tf ∆S = nC p,m ln ( ) Ti Therefore for the two blocks ∆S c = nC p,m ln ( T ) Tc and ∆S h = nC p,m ln ( T ) Th 87 88 THE SECOND AND THIRD LAWS The total change in entropy is ∆S tot = ∆S c + ∆S h T T ) + nC p,m ln ( ) Tc Th T2 = nC p,m × ln ( ) Tc × Th = nC p,m ln ( = = ⎛ [ (Tc + Th )] ⎞ m C p,m × ln M ⎝ Tc × Th ⎠ (Tc + Th )2 m C p,m ln ( ) M 4(Tc × Th ) where m is the mass of the block and M is the molar mass In the case given ∆S tot = 500 g (250 K + 500 K)2 −1 −1 × (24.4 J K mol ) × ln ( ) × (250 K × 500 K) 63.55 g mol−1 = +22.6 J K−1 P3B.7 The heat produced by the resistor over a time period ∆t is q = power × ∆t = IV ∆t = I R∆t, where the last expression was obtained using Ohm’s law, V = IR Note that care is needed handling the units From the inside of the front cover of the textbook use (1 A) ≡ (1 Cs−1 ) and (1 V) ≡ (1 JC−1 ), so that (1 Ω) ≡ (1 JsC−2 ) Therefore the units of the final expression for the heat are as expected A2 × Ω × s ≡ (C2 s−2 ) × (JsC−2 ) × (s) ≡ J Assuming that all the heat is absorbed by the large metal block at constant pressure, this heat is the change of enthalpy of the system, ∆H = q The enthalpy change on heating is given by [2B.6b–49], ∆H = C p ∆T This is rearranged to give an expression for a temperature change ∆T = ∆H q I R∆t I R∆t = = = Cp Cp Cp (m/M)C p,m where m is the mass, M the molar mass and C p,m the molar heat capacity Thus the final temperature of the metal block is Tf = Ti + ∆T = Ti + = (293 K) + I R∆t (m/M)C p,m (1.00 A)2 × (1.00 × 103 Ω) × (15.0 s) [(500 g)/(63.55 g mol−1 )] × (24.4 J K−1 mol−1 ) = (293 K) + 78.1 K = 3.71 × 102 K SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY The variation of entropy with temperature at constant pressure is given by [3B.7– 90], S(Tf ) = S(Ti ) + C p ln (Tf /Ti ) Therefore the change in entropy is Tf m Tf ) = ( ) C p,m ln ( ) Ti M Ti 500 g 3.71 × 102 K −1 −1 =( ) × (24.4 J K mol ) × ln ( ) 293 K 63.55 g mol−1 ∆S = S(Tf ) − S(Ti ) = C p ln ( = +45.4 J K−1 For the second experiment, the initial and final states of the metal block is the same, therefore ∆S = All the heat is released into surroundings, that is water bath, which can be assumed to be large enough to retain constant temperature Thus q I R∆t = Tsur Tsur (1.00 A)2 × (1.00 × 103 Ω) × (15.0 s) = = +51.2 J K−1 293 K ∆S sur = P3B.9 As suggested in the hint, first consider heating the folded protein at constant pressure to from the initial temperature T to that of the transition, Ttrs The variation of entropy with temperature at constant pressure is given by [3B.7– 90], S(Tf ) = S(Ti ) + C p ln (Tf /Ti ) Thus the change in molar entropy, ∆S m = S m (Tf ) − S m (Ti ), of this step is ∆S 1,m = C p,m (folded) ln ( Ttrs ) T Next consider the unfolding step The entropy change of such a transition is given by [3B.4–89], ∆ trs S = ∆ trs H/Ttrs , thus ∆S 2,m = ○ ∆ trs H −m Ttrs The final step is cooling the unfolded protein to the initial temperature ∆S 3,m = C p,m (unfolded) ln ( T Ttrs ) = −C p,m (unfolded) ln ( ) Ttrs T The overall entropy change is the sum of above steps ∆S m = ∆S 1,m + ∆S 2,m + ∆S 3,m ○ Ttrs Ttrs ∆ trs H −m − C p,m (unfolded) ln ( )+ ) T Ttrs T Ttrs + [C p,m (folded) − C p,m (unfolded)] × ln ( ) T = C p,m (folded) ln ( = ○ ∆ trs H −m Ttrs 89 90 THE SECOND AND THIRD LAWS Given that C p,m (unfolded) − C p,m (folded) = 6.28 × 103 J K−1 mol−1 , the molar entropy of unfolding at 25.0 ○ C is thus ∆S m = 5.09 × 105 J mol−1 273.15 K + 75.5 K 273.15 K + 75.5 K ) 273.15 K + 25.0 K = (1.45 × 103 J K−1 mol−1 ) + (−9.82 × 102 J K−1 mol−1 ) + (−6.28 × 103 J K−1 mol−1 ) × ln ( = +4.77 × 102 J K−1 mol−1 = +477 J K−1 mol−1 P3B.11 (a) Consider a process in which heat ∣dq∣ is extracted from the cold source at temperature Tc , and heat q h = ∣dq∣ + ∣dw∣ is discarded into the hot sink at temperature Th The overall entropy change of such process is dS = −∣dq∣ ∣dq∣ + ∣dw∣ + Tc Th For the process to be permissible by the Second Law, the Clausius inequality defined in [3A.12–86], dS ≥ 0, must be satisfied Therefore −∣dq∣ ∣dq∣ + ∣dw∣ + ≥0 Tc Th the equality implies the minimum amount of work for which the process is permissible Hence it follows that ∣dq∣ ∣dq∣ + ∣dw∣ = Tc Th (b) The expression in (a) is rearranged to find ∣dw∣ and the given relation, dq = CdTc , is used to give ∣dq∣ − ∣dq∣ Tc dTc ∣dw∣ = CTh ∣ ∣ − C∣dTc ∣ Tc ∣dw∣ = Th Integration of both sides between the appropriate limits gives ∫ w which evaluates to ∣dw ′ ∣ = CTh ∫ Tf Ti ∣w∣ = CTh ∣ln ( ∣ Tf dTc ∣ − C ∫ ∣dTc ∣ Tc Ti Tf )∣ − C∣Tf − Ti ∣ Ti SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY (c) Using C = (m/M)C p,m , the work needed is ∣w∣ = 250 g 273 K −1 −1 )∣ −1 × (75.3 J K mol ) × (293 K) × ∣ln ( 293 K 18.02 g mol 250 g × (75.3 J K−1 mol−1 ) × ∣273 K − 293 K∣ − 18.02 g mol−1 = ∣ − 2.16 × 104 ∣ J − ∣ − 2.08 × 104 ∣ J = +7.47 × 102 J = +7.5 × 102 J (d) Assuming constant temperature, for finite amounts of heat and work, the expression derrived in (a) becomes ∣q∣ ∣q∣ + ∣w∣ = Tc Th This is rearranged to give the work as ∣w∣ = ( Th − 1) × ∣q∣ Tc The heat transferred during freezing is equal to the enthalpy of the transition, which is the opposite of fusion, q = ∆ trs H = (m/M)(−∆ fus H −○ ) Therefore the work needed is ∣w∣ = ( 293 K 250 g × (−6.01 × 103 J K−1 mol−1 )∣ − 1) × ∣ 273 K 18.02 g mol−1 = 6.10 × 103 J = 6.11 × 103 J (e) The total work is the sum of the two steps described in (c) and (d) Therefore w tot = (+7.47 × 102 J) + (6.10 × 103 J) = +6.85 × 103 J = +6.86 kJ (f) Assuming no energy losses, power is the total work divided by the time interval over which the work is done, P = w tot /∆t, hence ∆t = w tot 6.85 × 103 J = = 68.6 s P 100 W 3C The measurement of entropy Answer to discussion question D3C.1 Because solutions of cations cannot be prepared in the absence of anions, the standard molar entropies of ions in solution are reported on a scale in which, 91 .. .Student Solutions Manual to Accompany Atkins? ?? Physical Chemistry ELEVENTH EDITION Peter Bolgar Haydn Lloyd Aimee North Vladimiras Oleinikovas Stephanie Smith and James Keeler Department of Chemistry. .. tot = 0.780 × (0.987 bar) = 0.770 bar p O2 = x O2 p tot = 0.210 × (0.987 bar) = 0.207 bar SOLUTIONS MANUAL TO ACCOMPANY ATKINS? ?? PHYSICAL CHEMISTRY Note: the final values are quite sensitive to. .. rearranged to give the pressure p= = nRT V n 255 × 10−3 g (8.3145 × 10−2 dm3 bar K−1 mol−1 ) × (122 K) × 20.18 g mol−1 3.00 dm3 = 0.0427 bar SOLUTIONS MANUAL TO ACCOMPANY ATKINS? ?? PHYSICAL CHEMISTRY