PHYSICAL CHEMISTRY CHEM 113C
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This 30 page Class Notes was uploaded by Ashley Kunze on Thursday October 22, 2015. The Class Notes belongs to CHEM 113C at University of California Santa Barbara taught by Staff in Fall. Since its upload, it has received 22 views. For similar materials see /class/226953/chem-113c-university-of-california-santa-barbara in Chemistry and Biochemistry at University of California Santa Barbara.
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Date Created: 10/22/15
Chem 113C Homework Assignment 5 Solutions Problem 1 Consider the schematic reaction P If the reaction is onehalf order with respect to A what is the integrated rate law expression for this reaction What plot would you construct to determine the rate constant k for the reaction What would be the halflife for this reaction Will it depend on initial concentration of the reactant 3 83 a 4M Aquot 1 I A kdr if A 1 11Ak m A o 2A1 2 4A 1 lrr m At AAt quot l b A plot of AT1 st thll be linear and the rate constant is equal to two times the slope of the line c At rm A 05An lAl krm 4 i 2 7 IrrxvgA I 2 frat kr2 1 ANAL The halflife depends on the squarei39oot of the initial concentration Problem 2 The halflife of 238U is 45 X 109 years How many disintegrations occur in 1 minute for a lOmg sample of this element The rate constant can be found from the halflife n n I i ki quot Ir quotl39hus 0693 lyr lday 1hr 7 7 7 4gtlt109yr536425day 24hrs 60min 94gtlt10 16 111in l C unverung l0 Lugs into the number of 23U atoms Nagy 1x104 g23s gumr 6022gtlt1013 mor 143gtlt10 Employing the rate law the number of disintegrations in 1 minute is N 24 L43X1024 849410 mmquot x mm l 43 X l OH The rate constant is so small that a negligible number of disintegrations occur in one minute Problem 3 Derive the rate law expression for the production of product P for the following reaction set ABgt P The rate expression for the formation of the product is d P 7 1 A B d r 1 The equilibrium constant can be expressed as K Em All All therefore Substituting this into the rate expression for P yields dle Ir 393 m 1 4 A B d r 1 H m 3 I 3 7191 Alll B Where If 2k Problem 4 For a type II secondorder reaction the reaction is 60 complete in 60 seconds when A0 01 M and Bo 05 M a What is the rate constant for this reaction b Will the time for the reaction to reach 60 completion change if the initial reactant concentrations are decreased by a factor of two a The integrated ratelaw expression for a secondorder reaction of type ll is 7 1 131410 A l l at t 60 s A 004 M with 11 stoichiometry so that B 044 M Substituting these values into the above expression and using I 60 s yie ls BL A 0 05 M 004 M x 01 M 044 M k 1 605 05 MilllM J 004l7 Mquot squot0788 00329 M391s391 b The tune will double assiulu39ug the k Value is the same Numerically checking this Bl expectauou B l 1 In 0 A ll l h I 00329 Mquot squot 025 Me 005 M L002 M 1 t rllBlrlAlu 022 M 023 M 005 M 152 s0788 120 s chem 113C Homework Asslgqmenwz Saturnquot Problem 1 For a sel ofnondegmerale levels erh energy ak o 100 and 200 K calculale the r on and 000K h lrmmn ahl valuev The probability ofoccupyiug a speci c energy 1e el a gixen by 4 2 97 q 2 e 4 94 News 20 r YVI e Me r r r r scales 7 l and 2 respectively is T v I 39 39 39 39 39 e r pl obabillty of belllg populated or 13 m ml example for rllree nondegenerate energy In 615 Problem 2 Consth state What ts the temperature ofthe collecttom mmm 4 39 w t t levels is given by wt to detennine the temperature The 139 t o 0th D n 1 39 r a l a is 8 U at a he separation between energy 12 13 he emperamre is Problem 3 J the rst exctted state be halfthat of the gound state7 Problem4 ofmonawmxc Jdeal gases 1s othe form N INZIV D 11123 adlel a whae 3 2m kT 2 11041 V i12 Show that a 20 NZ Problem 5 Derive an expression for the partition function for a molecule with two states separated by energy hv Le 21 0 and 2 hv Ifwe consider a canonical ensemble ofN such molecules derive expressions for the total internal energy and the heat capacity at 1 Zia ii i 6 0 433 and LL75 UL Fvs 39 A V 3mm Am 4amp3 LTV AM 1 lt3 NWELBV z 1 weM Eh vinyll 7 All 5 1 s t fquot Cv dTV it a V pluv I hve hquot Cv N kJ ZG VSIJiv Lv 7 Problem 6 em calculate the probablllty of occupylng the rst exalted vlbrat onal level at 300 and 1000 K For Cl at 300 K39 1087 WWquot ylsmllw me HM 7 mm 51640quot I am mle 7 me JKquotUmDKl 1585 Problem7 h the ht ht h tt t alamce gas tam tt anngas parttetes Assume thatboth M and N are large and Lhatthe dehstty ofpameles 15 small 1 e17quot ltlt 1 D a mmeLhemulhphcltyW forplacmg the Npameles m the M sttes assumthg that the pamcles are thdtstthgutshabte b Wnte an aipresslon for entropy as a funetton ofM and Nbased on the Boltzmann equatton We use Ihe Bollununn cqumion tt 5 klnH kln LlMi J and SLirling s approximal ion x1 a 1quot In gel MM 5 klquot mew 39 Inn dolbili x V M so 7 39 Chem 113C Homework Assignment 1 Solutions Problem 1 a What is the possible number of microstates associated With tossing a coin N times and having it come up Htimes heads and T times tails b For a series of 1000 tosses What is the total number of microstates associated With 50 heads and 50 tails c How less probable is the outcome that the coin Will land 40 heads and 60 tails a In this case the number of coin tosses is equal to the number of units N and ch unit ca exist in one of two states heads H or tails T Since the number ofmicrostates is equal to the weight W N 1177 HT For this case N 1000 H 500 and T 500 Substituting into the above expression for W W7 N 7 1000 Hr 5002 The factorials that require evaluation are generally too large to determine on a alculator therefore evaluation of Wis performed using Sterling s approximation an 7 m100072m500 1000m100071000 7 25001n5007 500 1000m100071000m5001000112 693 Therefore W expifsa3 c Proceeding as in 1aiTllmt with H 400 and T 60039 111 W 7111000007 11140037 1600 7 iooomuooo7 1000 7 400111900 400 7 6001n600 600 1000h110007 400111400 7 6001111600 7 673 Therefore W Cxp73 Comparing the answers to parts 1 and c the H 500 T 500 outcome is approxrniately exp20 01485 X 108 times more likely than the H 400 T600 outcome Problem 2 a Realizing that the most probable outcome from a series of N coin tosses is N2 heads and N2 tails What is the expression for WW corresponding to this outcome b Given your answer for part a derive the following relationship between the Weight for an outcome other than the most probable an 39mm H log c We can define the deviation of a given outcome from the most probable outcome using a deviation index a Show that the number of heads or tails can be expressed as H la and T lia d Finally demonstrate that W NE a Wp p N 7 N w w 7 2 H 6 j 13 ml quotCV hil39l397hernlnlyiln l w LuN71nHh1T7lnNlh1J mHmm2m4 rHlnHHTlnTTN1nA7N 7HlnH7T1uTNlnA 7HluH7Th1TtHTh1L 7Hln H r 7 7Tb 7 N N t 2 x Z c Substituting the de nition of part a into the expressions forH and T 2 7 2 2 tit t t tf 2 t 2 1 Substituting in the result of part c into the nal equation ofpart b m71111n1a7g17Dr1nl0 m If MltltL then 1110 or 1 1 therefore lah1laili00111041 1aa7liaiaiNa3 Problem 3 Consider the case of 10 oscillators and eight quanta of energy Determine the dominant con guration of energy for this system by identifying energy con gumtions and calculating the corresponding Weights What is the probability of observing the dominant con guration Th configuration with the maxinnun weight 15 I3 1 I2 l 11 3 no 5 with the corresponding weig i l 0 I W 1135Y Vriting down all possible con gurations and sluinning the weight results in WWW 20170 Therefore the probability of observing the dominant configuration is 5040 A W 5040 W 20170 my 3 025 Problem 4 A box of nails contains 90 straight nails and 10 nails too bent to be used You randomly select 10 nails from the box What is the probability that you removed 8 straight nails and 2 bent nails To nd the probability rst derive a general expression using the following notation Nb n RS 4 T11 where N is the total number of nails in the box N5 and Nb are the number of straight and bent nails respectively 1 is the total number of nails removed from the box and us and 11 are the number of straight and bent nails removed from the box respectively Solution Since the probability of pulling a bent nail from the box is independent of pulling a straight nail from the box we can easily write a general expression for the probability of pulling T15 straight nails and rib bent nails from a box of N N5 Nb straight and bent nails We rst write an expression for the number of ways of pulling 713 straight nails from a box of NS straight nails W NS N5 5 n5 NS nsjins and a similar expression is obtained for the bent nails W Na 712 nb Nb 71b 71b The expression for the number of ways of pulling T1 nails either straight or bent from N total nails is My N N n N njlnl The probability we require is then given by W smlb N 7131711 h N5 l Nb 1 3 squot w f L N ll ENS nsjlnslll L Nb nbynblll Now ifwe let NS 90 Nb 10 8 and m 2 then N 100 and n 10 and the resulting probability is 9010 Hum 3ml 2 U520 10 15 N 20 Problem 5 A set of 13 particles occupies states with energies of 0 100 and 200 cm391 Calculate the total energy and number of microstates for the following con gurations of energy aaa8a5 anda20 Do any of these con gumtions correspond to the Boltzmann distribution The total energy is equal to the sum of energy associated with a given level times the number ofparucles in that level For the occupation numbers in a E Z Emu 30110 5101 511 I 0 curl81oo emquot5200 cur 0 500 cmquot Repeating the calculation for the occupation niunbers in b and c yields the 5 me enerr of 500 cnfl The Lumber of microstates associated with each distribution is given by the weight N v 739 771339 1287 Hiram 8V510l Wb 7 2860 90800 W i 858 10121 The ratio of any two occupation numbers for a set ofnondegenerate energy levels is given by The above expression suggests that the ratio of occupation numbers can be used to determine the temperature For set b comparing the occupation numbers for level 2 and level 0 results in 9 Tl3lK The distribution of energy in b is in accord with the Boltzmann distribution Problem 6 Consider 25 players on a professional baseball team At any point 9 players are in the batting order How many 9player batting orders are there if the allstar center elder must bat inthe 43911 spot in the order 7291 7 l i39zzmxio 16E P7IjP248 mblem7 wuplayaslatmachmemLasVegas Fax every cmnwumsenthexe mum auwames 1 yaulase a wuwm 5a wmpm msn a wuwmss 5mm pm mu suppase Lhatwu nd Lhatymx avenge expectedpw t We manymals 5 sn tanbad Fmd39hemaxxmu39n quotmay dnsmbuuanfmthz mmmmsnp andPg fax absuwng each mm mm amcames 1 wan4 h m 3st 1 1 40 d a g v i 7 mm k V1 X w Dirn M F A 4 Law S M x WM 39 iv ed 1 n1 3 waadw a z Lu 4 kinztmmwy it IIWA mama 7 5 mm u 2M rum 5 Kim 3 M km a mu lt9 13 A1 9 z A 239 1 r m AL quotq u 7 c A Problem 8 Consider the following alternate de nition of entropy for some integer n gt 1 N 5 1nP i1 Note also that in this ensemble the subsystems all have the same energy ie E E a Show that this de nition is consistent with entropy being an extensive property b Using the above formula for entropy determine the set Pi that maximizes 5 subject to the constraint Pi 1 Solution a We partition the ensemble into two identical independent parts A and B with N NA l N E subsystems The probability of being in a particular state of the system for the total ensemble is given by p515 pf pf and the entropy is then NA M3 NA N3 5 ln 22ng quot ln 22951415 i1 j1 i1 j1 NA N3 n Zeairp i1 j1 NA N3 i ln 2011 ln 2915 5A 58 i1 i1 b We now need to optimize S with respect to each of the Pi s subject to the constraint Pi 1 to obtain a set Pf that maximizes 5 Since this same set also maximizes Piquot it is easiest to optimize the argument of the logarithm rather than the logarithm itself For each of the Pi s we solve the equation where a is the Lagrange multiplier for the constraint The solution to this equation for each of the Pi s is the easily found anf 1 a 0 Solving for Pf yields Chem 113C Homework Assignment 4 Solutions Problem 1 The escape velocity from the Earth s surface is given by vE 2gR 2 where g is gravitational acceleration 980 m s and R is the radius ofthe Earth 637 X 106 m a At what temperature will vmp for N2 be equal to the escape velocity b How does the answer for part a change if the gas of interest is He 1 The escape velocity of ea11h is given as vE 2980 uls l637gtlt106 m vE 111x104 ms 1 For N 4 1 28314Jum1quotKquotr llZgtltlO ms 71 0028 kg uiol 112x104 m s 3 0023 kg mor 1 1 21ogtlt10i K 28314Jmol39 K39 b The temperature should decrease as the mass decreases therefore we would expect the coirespoudiug temperature for He to be lower than that osz RTN 2RTire My MHe 111 T H 139 MM I 0004 kg Luolquot THC rl 0028 kg IDOl TH 300x10 K Jzioxlo K Problem 2 Starting with the Maxwell speed distribution demonstrate that the probability distribution for translational energy for 577 gtgt kT is given by 1 3 z f5Trd5Tr 2 eiETTkTgiizdgTr nkT The translational energy of a particle can be related to the Velocity of the particle by the expressions 1 1 5 71mquot 2 V 25 111 2 l l l 2 y 77 i dg 7 7 day in 2 gm l 1119 Substituting this result into the Maxwell Speed distribution w m rzs r e 1in m i 2 dgfr l 2 Int J m H 28 g d 475 7 7 f 7 b 27rij m J 32 1 7 2 e I d n Problem 3 Using the distribution of particle translational energy provided in Problem 2 derive expressions for the avemge and most probable translational energies for a collection of aseous particles 32m Z J s e n 157 iTkT o z 73 2 1 J 3 J II T 4 kT ikT 2 The most probable energy is determined by taking the derivative of the probability distribution function setting the derivative equal to zero then solving for energy 1 a l 39 B i f 027r7 7 age IZ39kT Bar 1 31 1 5 12 k 027J 75 e i cL e H ITkT 2 kT Ll 02r Irer The equality will be true when the term in square brackets equals zero 1 l agn 1 4w Pm 073 77 2 quotW kT l 712 1 12 55mg Egmp l g 7 k1quot 11 2 Problem 4 Imagine a cubic container with sides 1 cm in length that contains 1 atm of AI at 298 K HoW many gasiwall collisions are there per second The collisional rate is given by INC 7 PAVN 7 PA N Av mg It 4 k2quot 4R1quot With 1 0040 kg Limb1 and T 298 K the average speed is 8R 88314 Jnlol39 Kquot298 K v tg 7770040 kg 11101quot I39M 459 m 5 1 Substituting into the expression for the collisional rate 75 3 2 7 1 MMW lcm 1 J 6022X10Jtnolquot459ins391 lOOcin dt 7 48314Jnm1quotKquot298K dNC 2823gtlt1033 coll per sec per wall dt Taking into account the six Walls that compiise the container the total collisional rate is 15 6 walls 2823gtlt1023 cell 5quot wall391 1N 170x101 coll 5quot dt Problem 5 You are a NASA engineer faced With the task of ensuring that the material on the hull of a spacecra can Withstand puncturing by space debris The initial cabin air pressure in the cra ofl atm can drop to 07 atm before the safety ofthe crew is jeopardized The Volume ofthe cabin is 100 m3 and the tempemture in the cabin is 285 K Assuming it takes the space shuttle about 8 hours from entry into orbit until landing What is the largest circular aperture created by a hull puncture that can be safely tolerated assuming that the ow of gas out of the spaceship is effusive Can the escaping gas from the spaceship be considered as an effusive process You can assume that the air is adequately represented by N2 1Pi4 IT 1 71 V Zmu PRje 74 K7 J 07 annl atme V 25 A288gtlt104 s 138x1039 JK 1285 K 100 m3 7 f 0028 kg marl r l 6022x10 3 moiquot 334x10 inquot 107x10 i m1 A If the process can be considered effusive then the diameter of the hole should be smaller than the mean free path The mean free path 1 for N at 1 mm and 285 K is L 39 PNVNA Jim 7 8314Jmol 1Kquot298 K 1 7 101325Pa6022x1023mor1 J543x10quot9m9 639X1039xn1 C 0111pmquot son of this length scale to the apertiue dimensions demonstrates that this process is not et lsive Problem 6 The diffusion coef cient for CO2 at 273 K and 1 atm is 100 X 10 5m2 s71 Estimate the collisional cross section of CO2 given this diffusion coef cient D l VIMl 3 J21 1SIJRT1 1 3 1M PM Rearranging the above equation to isolate the collisional cross section U71 312T RT 1 3 nM PNJ r 821X10 l L mm mot1 Kquot 1 273 K X 1000 L 1 1 axm6022x10B marl 100x10 5 mls 1 a 0318 mm1 Problem 7 A thermopane Window consists of two sheew of glass separated by a Volume lled with air which We will model as N2 Where K 00240 J K71 m71 s71 For a thermopane Window that is 1 m2 in areaWith a separation between glass sheets of3 cm What is the loss of energy When the exterior ofthe Window is at a temperature of 10 C and the interior of the Window is at a temperature of 22 C AE7KL JA A39 a AE700240JK39111139l EmiliKMLlo 0 m CH1 DJ 7960 Js39l Chem 113C Homework Assignment 3 Salaam Problem 39 uulmlp39 1 um JUIaCOllectionofone dimensional harmonic oscillators HTkf VZ39LM DT W 7 gt gt N T kr Oh1q Vk luq B n 3V T W70 rlnIV 0139 V iMrTTBqIJ N 1 9T y 1139 M 7 7N 1 7 rpm I39CVe hw slimy N V V Hm e inijilljm 7 Nhr e mv NhCV Problem 2 Detennine the equilibrium constant for the dissociation ofsodium at 298 K NaZCg 2Nag crnquot v 159 cmquot the dissociation energy is 704 kJrnol and the ForNaz B 0155 groundstate e1ectronic degeneracy for Na is 2 q NA Mi 75s K7e I V 2L 002451333 J2522x103 1 a 4 1M 15 LA 938X10733 m3 1 00245 mi 0695 cmquot K39 29815 K I V r 4xhqz134139qz 7 GB X to 526x141 mp 00x10 sin ti 3mm1 4093 15 K lie 922gtlt1033 Problem 3 For a system of energy leVels 5m m2 a Where on is a constant With unis of energy and i 0 1 2 00 What is the internal energy and heat capacity ofthis system in the hightemperature limit Wliting out the expression for the partition function and recognizing that in the high temperature limit the partition nuictiou can be evaluated by mtegrntion q 2 0111 Z gD mn39 E J eHmu39 1m m In 39 u V 04 39V 1fquot E L4 U f f q rl 39 q I did1p r 3 1 wing2 LV gm q 2 a 2 1 2 95 1 a 39 91 2 Problem 4 The standard molar entropy of 02 is 20514 J mol391 K39l Using this information determine the bond length of 02 For this molecule v 1580 cm39l and the ground electronic state degeneracy is three f sL vaL r quotUr111 NU quotith1qmmiklnN r N J r RRhiq7qkqqgiRhil molNAR 760mquQRQNYEh3qrh1qzzhl9rh1515 V 1 hiFh1qxln178751 Jung 7 00245 1nqKm1m3 3 7 572gtlt10 m3 7607051nqx110 44lnqx 8151 Lg G H 0695 cmquot Kquot29815 K B 1z7cmquot o 815 2815 B127 ourl 8711 87139 anquot 392 11 6626x10 Is 7 l 8nz127 Cm 8723300gtlt101 cms391l33gtlt10quot6 kg127 cmquot r2 l66gtlt10 2 m2 r128x10 11 Problem 5 DeteImine limp 17mm and 17m for the following species at 298 K a Ne b Kr 0 CH4 d CsHs 3 C60 Note I kg 1 2 In2 5392 a Ne M 0020 mol ZRT um 7 p M 28314Jmol 1 Kquot298K 4 495 111 5 1 0020 kg 11101 S 8314111101quot 1quotquot 298 K 1 W 559ms1 111M 7r0020 kg 11mlquot 3 831411110141l 298K vnns LRT 4 z 607 m 3 1 M 0020 kg 11101 b Kr M 0084 kg uml39l z a quot1 2 lquot 2M 514 11101 Is 981243u 4 1 M 0084kgmor XJim x IZL39O I 863M131 410m r HE39SZ ks m 633 1410mm IZL390W 109 e S m m7 JOm 831 0500 N m 4 4 X 8624IE410 W HE39S Hi Jim 551 090390 1 W1 145 H A gt1 soz r IJorrIr Hum 138 mm 351 OEO39O V Am I45 II 9017 A 1 862 1 JON I HE39SZ 13 1qu 5x 0 0390W 9HCJ p IJOY 3 9100 A 141 A m 1 8691 Flam I 171539s N WWW JWLNA 1 IS6ZPIIJOTUIHE 88 1218 41m 5319101 N am gt1 862M131 JOIHIHE39Sk RI Vs m 089 Is m SSS Jom x 910390W m3 0 S m 86 4 I410th BX 82800 A N A m 4 186131 VIUNIVIE39SK HE S m tr JOIN 531 mow m m 4 4 L 1 SEMI 41mm VIEW 1918 88314111101391 K 293 K r0721 kg moi 3R7 38514 J moiquot Kquotz9s K 1 Pm 71102ms 1W 0721kginol Consider a collection ofgas paIticles con ned to translate in two dimensions for example a gas molecule on a surface DeIiVe the Maxwell speed distribution for such a gas SRT quotm m V 936 in 3391 Beginning With the MaxwellBoltzmann Velocity distribution in onedimension Ill fV172T em and the de nition 0f speed in 2 dimensions l l v vj The speed distiibutian in 2 dimensions is given by 1th fvx va dvde where ltle is the differential of Velocity in the jth direction Thus m 11 m 1 HIV 1 e 3 21239 cf 1 m x kf quot V 7 e d xd Fdl m 1eWPdvdvv 21 r The differential area is de ned as vxdvV Zm39d v Substituting this into the expression for F dl39 Fdl39 m 4e 27139 Mquot
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