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# Physical Chemistry I CHEM 343

Hope College

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This 94 page Class Notes was uploaded by Linnea Wilderman Jr. on Monday October 12, 2015. The Class Notes belongs to CHEM 343 at Hope College taught by Staff in Fall. Since its upload, it has received 15 views. For similar materials see /class/222150/chem-343-hope-college in Chemistry at Hope College.

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Date Created: 10/12/15

Lecture 28 Chapters 16 and 18 Osmotic Pressure and Diffusion Announce 0 Quiz Wed like usual and new HW right away 0 Quiz next week will be ready early Please send email if you would like to take it early There will be a normal Wed 300 offering 0 Exam coming already the Friday after Thanksgiving Then 3 more lectures and nals YIKES Outline osmotic pressure Diffusion Review Activities Remember it is true for all systems that AllRT1npf PA For ideal solutions it 1 u2lt1gt RTInXA For nonideal things we use actiVity MA 1 M210 RTlnaA Where aA YA XA YA is not constant Finally we can write chemical potential in terms of ideal and nonideal components A LL RTlnXA RTlnyA And then expanding this to the total free energy of the solution AGmix motRT XAlnXA XBlnXB XAlnYA XBlnYB ideal mixing excess Gibbs energy of mixing ACTmix A ideal ACTexcess Colligative properties properties that depend only on the number of solute molecules at fixed p I I I I I I TMelt TMelt TVap TVap T Boiling point elevation derivation 2 AT RT XE AWH Three approximations solute has negligible vapor pressure XB ltlt 1 and AT small Freezing point depression Using similar arguments the depression of the freezing point is found to be Since for water AvapH 4505 1d mol391 and Tb 373 K and AfusH 601 1d mol391 and Tf 273 K we can see that for the same amount of solute there will be a greater depression of the freezing point 10 K for the same as above than elevation of the boiling point Osmotic pressure osmosis the passage of a pure solvent into a solution separated from it by a semipermeable membrane impermeable to the solute osmotic pressure the pressure that must be applied to the solution to the stop the transfer of solvent Jp Jp139l HAG t 11 HBO t 11 The pure solvent is thermodynamically driven to the solution side because of the lower chemical potential for A in solution Note that my A and B are opposite Dill because he has strangely changed convention Because we are at equilibrium it must be that the change in the chemical potential for mixing is exactly equal to the change in chemical potential for additional pressure AmixG from our above MA 1 u l RTln XA ApressG from dG Vdp 7 SdT constant T so S term goes away H RTlnXA I dep P If we make two additional assumptions assumption 0 7 constant T l Vm molar volume of pure liquid doesn39t change for the pressure range of interest ie pressure change is modest 2 ln XA ln 1 XB N XB for small XB Hmm de ja vouX RTXB HVm For small X3 X3 nBnAnB Z nBnA And nAVAm V Thus we can rewrite this equation in a very tantalizing form UV nBRT or more commonly H CERT CE is the molarity OH The practical use for this arrangement is the determination of molecular masses of large biochemical molecules ie put a known mass of stuff in solution measure I this gives n Check out Dill s discussion of partition coef cients and desolvation p291 296 Partition coef cient For instance people often look at two immiscible solvents eg hexane sitting above water and look at difference in solubility of some solute K j h B would be water A hydrocarbon s the solute SA not surprisingly this is related to the difference in free energies chemical potentials of s dissolved in each of the solvents 111KB B A NC A kBT kBT Dill uses the lattice model to relate this All to differences in individual interaction energies Figure 1611 5 solvent is cost of opening a cavity in solvent and inserting a single solute so it is the free energy of mixing at infinite dilution chat for a few seconds about desolvation We discussed desolvation a bit before when we talked about hydrophobic effect One of Dill s equation is andjmzr Wm wsB wss WAB lnqi W 61qu So dimerization can be driven by any number of things which may or may not include the attraction for A with B Recall hydrophobic effect is about water really liking itself not so much that water hates oil Partition function term includes possible changes in monomers when they dimerize Chapt 18 stuff Diffusion Remember way back to random walk stuff We ended up with dRMS 1 K612 m1 Diffusion also called Brownian motion We talked then very brie y about diffusion from the standpoint that gas particles can have high velocities but diffusion is still slow because the number of collisions is high So we have covered diffusion at the molecular level 7 let s look at macroscopic level Consider a box partitioned by a door C1 C2 V If C1 gt C2 what happens when door opens concentrations will eventually equalize In terms of diffusion matter is owing in X direction This is simply because there are more particles with the opportunity to move X than 7X wa O V ame p V Flux is a measure of the net ow through the doorway 7 ptclsunit time J Clearly the direction of ow is opposite the concentration gradient It seems sensible that the net ux depends on the disparity in concentrations This is expressed by Fick s first law of diffusion Jx I D BCZ X D is the diffusion coefficient Note that Ohm s law is the same thing for electricity and Fourier s law is the same thing for heat ow Calculating the diffusion constant is a big deal 7 very complicated We can hand wave and say that D cc ltugtl u is velocity mean free path As TT uT so DT as pT M so Di many collisions make it hard to move far as molecule gets smaller IT and uT so DT Also Fick s second law of diffusion also called The Diffusion Equation Ba fl 8151 This says that you have to be near a change in concentration in space in order for the concentration to change in time t0 tslightly later tmuch later t in nity What if we start with all our matter at one spot example 183 0 X The solution to Fick s second law for the initial condition above Cx 0 8 4D What shape is this Gaussian 0 x This is the same as a probability distribution for how far particles might be from zero as a function of time We can solve for the average squared displacement x2 Ix2Cx x2 d2 2Dt or D one dimension Eq 1853 three dimensions 1 For N2 recall that drms 056 cm DN2 7 056 cm2 6 x 1 sec 7 45 x 10391 cmzs39l Let s compare this value to diffusion in liquid media diffusion coefficient 1 2 1 diffusion of N2 in N2 air 45 X 1039 cm s self diffusion of liquid water 226 X 10395 cm2 s391 slower in liquid diffusion of protein myoglobin in water 113 X 10396 cm2 s391 We can also rearrange the above equation to see how long it takes a particle on average 2 to diffuse a certain distance I 61 6D If you were Einstein you could prove such a thing from first principles So it takes 7 02 s to diffuse 10 microns appr0X cell diameter Important questions How do we actually determine D and what does this tell us about the size and shape of compleX molecules Driving force for diffusion is thermal kinetic energy working against friction kBT Again Einstein can show D Dill derives this 7 it is the EinsteinSmoluchowski Eq 7 famous At the end of chapt 18 he points out that this actually comes through use of the Fluctuationdissipation theorem I hope that we might come back to this later in the semester Please read this section as well as the one on Onsager Note that Dill uses where I will use f Usually D is measured allowing calculation of the friction fExp fEXP Theoretically fTHEO 6ir77r 7 is viscosity coefficient r for sphere 47 Comparing these two frictions usually fExp gtfTHEO Why solvent sticks to diffusing particle making it larger Also particle not spherical 7 makes r effectively larger In simplest sense V massdensity mass speci c volume So for bare particle VMv2 N A V2 is speci c volume ofbare particle cm3g same as fTHEO given above If we want to include the solvent sticking effect M V 51V N A V1 speci c volume of the free solvent cm3 g39l inverse of density V2 partial speci c volume of the macromolecule cm3 g39l dependent on pH salt concentration etc shape of macromolecule can respond to many parameters 51 hydration unitless weight of solvent bound per weight of molecule If we solve this for radius 1 3Mv2 51v1 3 r 47cN A Typical values Proteins v2 070 075 cm3 g 1 51 02 06 g of water per g ofprotein DNA v2 040 050 cm3 g 1 51 04 05 g of water per g ofnucleic acid From these parameters we can see that DNA interacts more strongly and is more dense with the solvent than proteins do Typically hydration increases the frictional coef cient by 1020 slows down diffusional transport Frictional coef cients can be calculated for any shape Show Tinoco 68Dill 1813 Notice that the frictional coef cient increases as the object becomes more non spherical fexpftheor Why is diffusion an important thing to know in biochemistry l the measurement of diffusion coef cients can be used to infer the size and shape of macromolecules fexpf 1cm determines shape Lecture 8 7 Chapt 7 first half Connect E to Q Fundamental Eqs Driving Forces Outline agiliQJNV Fundamental equations for S and U Average energy T p and m driving forces Quick review Boltzmann 7 very important man S kan connects microscopic to macroscopic also written as n k pl pl weight e g 6788 139 the normalization factor is Q 7 the partition function Since we know all the probabilities we can calculate the average energy E Ejpj Your book drops the partition function for a little while but I would like to show you one more connection here at this time So if we know the partition function then we can calculate the average energy of our system This is huge and we ll come back to it later Now that brings up a good question What do we mean by our system Whatever we are interested in Very nebulous but convenient Just have to be consistent Text does a nice job of de ning a number of important things at the beginning of Chapt 7 make sure you pay attention to this One thing I ll mention quickly here is extensive vs intensive properties Extensive depends on the amount of stuff U S H V N n Intensive doesn t T P concentration u In last chapter Dill gave examples of how S maximizes with respect to V N and energy U Not surprisingly then he reveals that S is often expressed as SU VN Of course P and Vare related so we could change these variables around if we wanted to but these three work out most naturally Similarly U USVN These are called the fundamental thermodynamic equations for entropy and energy Very grand titles and they are important Of course there are other fundamental equations that we ll come upon later Two is enough for today though Of course in chemistry we are often going to be concerned with how things are behaving when S or U is changing We might like to understand how entropy changes as ice melts for instance How are we going to represent a small change in S with a derivative of course 7 61839 What does this look like gas dUE j dVZ j dN w VyN 8V UyN 139 07vi i UVNJ I We are also going to want to know about energy in the same way dU j d5lt j dVZ j dN 0 s VN 07V SN 1 W1 SVNJ 1 Ok then If we want to know how eg energy changes with respect to volume then we need to know how that coefficient works what it means What do you think If we are looking at some system and we cut the volume in half that has some effect on the energy What should we call that Why not pressure Kind of describes how the system pushes back on you if you are trying to compress it Similarly for the others 3U p 3V SYN TZEan 65 VN an 1 av WNW This leaves us with fundamental equation in differential form dU TdS pdV Z yile Lecture 9 7 Chapt 7 second half PV work Outline System and Surroundings First Law AU q w Expansion Work 1 Free expansion 2 Isothermal constant P 3 Isothermal reversible 4 Adiabatic expansion Quick review We will quickly get through to our fundamental thermodynamic equations for S and U Begin with de nitions S SU KN U Um VN We can then easily define dS and dU Mi Mi chili W W VN W UN 1 I UVNJl 1 9U 9U 9U dag 3 dag j W 1 W 6639 VN 9V 2 WI SYVYNM and we de ne the coefficients as familiar driVing forces 1 d3 dU dV Zampdivl T T l T dU TaS pdVZylle Remember with thermodynamics we are keeping track of energy as it ows around While ago I put up these sketches of heat and work from a microscopic perspective Remember 1 if 5 7 it a 6339 Work llxlnslbl ol39 Htillilllllhlt l wl cm cgy L hngCN energy changes motion ol lloms in motion ul uloms in uniform mumnu L lmnlic mumm We can think of energy as the capacity to do work Heat is energy that is transferred by a temperature gradient Today we will get into several descriptions of energy flow using the rst law One nal point about work and energy before we get rolling is how we de ne what is positive The system is our set of particles of interest In the above example it is the gas inside the piston The surroundings is everything else Positive work then is work done on the system by the surroundings This is positive because work in this direction increases the energy of the system OK then so let s recall the First law of Thermodynamics The internal energy of an isolated system is constant AU q w In other words energy is neither produced or consumed In yet another way q and w are the only ways to change energy As pointed out by Abby last week note that q and w are what kinds of functions path fnctns But their sum is a state fnctn One common kind of work Expansion work work arising from a change in volume This takes place in internal combustion engines So why not picture this as a gas expanding against a piston several specific types of expansion work A cross 7 sectional area Work force x distance dw Fdz multiply by A A dw FemA A dz FAp dw pextA dz Adz dV dw pexth as this is we have the piston doing work on the system So we need to reverse this And if p isn t constant we need to inte grate ldw 39lpext dV Armed with this equation let s look at 1 Free Expansion Pm 0 There is no force opposing the expansion Therefore dw 0 for each stage of expansion and w 0 overall j0dV 0 This makes sense because it takes no effort to push against nothing Also if there is nothing to push against then we aren t transferring energy to any particles outside the piston 2 Isothermal expansion against constant pressure Pat is constant throughout the expansion so our integral is easy to evaluate w Pext dV Pext de PextAV V We ll show later that A U I CVdT 1 this is really just the definition of heat capacity CV dUdT and if CV is constant AU CV AT so since this is also isothermal what does the rst law say AUqw isothermal AU 0 therefore Notice that the work is just the area under the PV curve We will use this a lot il cxl X l inul I o chl 7111111111 I sz15 Q Thcrmoslalcr39i to keep 1 constant 3 Isothermal reversible expansion Reversible change in thermodynamics is a change that can be reversed by an in nitesimal modi cation of a variable Essentially we are always at equilibrium AS 0 Thus PM must be equal to Pgas throughout the expansion We can put together a hypothetical example of a piston with a pile of sand on it Slowly removing sand from piston LPCXI chl P HMLM Mi Thcrmnslalcd In keep 139 constant Our work integral is the same but Pm which Pgasis not constant So we can use any of our favorite gas equations to describe P We ll use ideal gas right now but could be Van der Waals or whatever w Pm dV Pm nRT V So w f nRT dVV nRTf dV V why can we pull out T isothermal nRTln Vf nRTln V nRT 1nVf V Note that reversible path is the maximum amount of expansion work minimum compression Why Because reversible by de nition utilizes the smallest possible pressure to induce change in volume If we were irreversible then Pext would be at the final value for the entire expansion 9 P lt P at all points except the very end extirreversible extreversible Notice also that doubling the volume at low temperature or with few molecules present is less work than doubling the volume at high temperature or with many molecules present Remember that these items came from Pext 7 so higher T or n means that Pext has to be greater to ensure reversibility 4 Reversible adiabatic expansion Remember isothermal was AT 0 For adiabatic what is zero heat q 0 Adiabatic gives less work than isothermal We will show in a second that T drops as the volume increases 7 thus P drops even more than in the isothermal case What does the rst law let us say about an adiabatic expansion AU w T2 So then w AU fCVdT T1 If Cv is constant for what gas Ideal must have no vibration and no distancedependent interactions 9 Ideal monatomic gas Then w C V AT During expansion work is done on the surroundings 9 w lt 0 AT lt 0 This is our proof that adiabatic reversible is less work than isothermal reversible Lecture 25 Chapter 15 Thermodynamics of Mixing Announce 0 HW due next Monday 0 We ll go over exams on Wednesday Outline 8V Partlal molar volumes VJ I Vtot 11A VA IIB VB J p7n39 Mixing of ideal solutions AHmix 0 ASmix quot101R XAIUXA XBlnXB ACTmix quottotRTXA1nXA Bl j g Mixing of nonideal solutions AHmix i 0 AGmix could be anything Liquidgas equilibria A 1 LL 1 RT1np J PA Review Remember last lecture when Brian played ba1tender and mixed up our giant Scotch and Water Rule 1 Scotch should be served neat Observations 1 Volumes don t add up What does this say about molecular interactions ethanol and water must be more attracted to each other than they are to themselves 2 Temperature increases What does this say about AmixH We measured a negative AmixH Internal energy of mixture vs individual components Lower We ll get to these in a minute Recall that for gases we used Dalton s law pi Xi pm to calculate the partial pressures This is valid for mixtures that behave ideally Clearly molecules are obviously interacting with each other in the liquid phase can we use something similar to Dalton s law for liquid mixtures To deal with the rst observation let s introduce the concept of the partial molar volume which is similar to Dalton s law of partial pressures Show diagram for water ethanol at 298K We started with 500 ml or 500 g 277 moles of water and 500 ml or 395 g 86 moles of ethanol ethanol 1 0 Pure water Pure EtOH Note that pure water normally 18 mLmol and ethanol 58 mLmol Let s operate in terms of ethanol Methanol 86 molV101 depends on T because V depends on T 86095 905 M or 86090 956 M mammal 86 mol EtOH05 kg solvent H20 172 mol kg391 172 m independent of T What happens to M and m for dilute aqueous solutions M E m Xema ol 86 mol363 mol 025 wemml 395 g895 g 044 Look on diagram for Xemml 025 gt Vemml 58 7 25 555 ml mol391 and vver 18 7 05 175 ml mol39l More generally we would like to know how V depends on individual components For a binary mixture with components A and B we can de ne a total derivative for V dV dnA dnB ail1 pm 8 13 NM Not surprisingly we will define molar volume as I W pTn39 9 This gives us dV 126171 17361713 Finally we can integrate this expression to get the total volume V V101 nA V A 113 V3 of course you knew this from unit conversion Finally then for our gigantic Scotch and water V 86 mol x 555 ml mol391 277 mol x 175 ml mol391 962 ml As we experimentally determined the total volume for the mixture is less than the summed volume of the two components Thermodynamics of Mixing things lattice Let s think of the lattice model in the book 7 simulates a liquid of two different There are N total sites each filled with either an A or a B molecule sketch N NA N 3 Let s think about the entropy of the situation What are the number of arrangements What about A and B before mixing WA NA NA so we can ignore these contributions W N NANB W3 NBUNB ASMW kan kanA kanB zkan kN1nN NA1nNA NB lnNB kNA 1nNNB lnN NA lnNA NB lnNB kNAlnN lnNANBlnN lnNB kNA1nXANBlnXB kNXA1nXA XB1nXB or ASmix 39ntotR XAlnXA XB1UXB Now it turns out this is true for gasses or liquids or whatever What about enthalpy if we had ideal gasses AHmix zero because there are no interactions between the gases So then it is easy to get AGmix AGmiX AHmiX TASmix n0RT XAlnXA XBlnXB AGmix is ALWAYS negative if no interactions called an ideal mixture In other words ideal mixing is always spontaneous But back to our big Scotch and water AT i 0 so AmixH i 0 Obviously not an ideal mixture How do we deal with this This is great stuff by the way we are learning about molecular interactions 9 chemistry We ll do this a bit differently than the textbook so please read the book and think about this from both points of View Dependence of chemical potential on composition for liquids Let s pretend we have two things in a closed system at equilibrium Noting that uliquid ugas because we are at equilibrium Ag Bg Mg MAO Mg HBO Al Bl For now de ne substance A as the solvent and substance B as the solute Let s divert for a minute to think about just pure substance mg Ma Mg Note notation A0 for pure substance Thinking back to chapt 11 the chemical potential for a gas is P u HQ RT ln we gave this before using G P Because we have a liquid and vapor at equilibrium us must be equal and p must be the vapor pressure M210 M lug 2 If we now allow for the presence of another substance solute we replace the pure quantities in the above expression MA 1 M lug 2 By equating these two relations through the standard chemical potential Hf we can eliminate the dependence on the standard chemical potential m1 u1 mug if pA lt pA then MA 1 lt u l mixing lowers the chemical potential Therefore the chemical potential for the liquid substance A in a mixture is dependent on the chemical potential for the pure substance plus a factor that is dependent on the ratio of the vapor pressures of p in mixture to p of pure substance Raoult s Law Let s simplify this last relationship that connects chemical potential with ratio of vapor pressures We can use the experimental result that for similar liquids the mole fraction of the substance A in the mixture is approximately p Ap A or the ratio of the partial vapor pressures of each component in a mixture to its vapor pressure as a pure liquid pA XAPA Raoult s law Binary mixture obeying Raoult s Law Fig 103 from MT N O O O O Vapor Pressure torr 0 I I i I 025 05 075 10 Pure toluene Pure benzene X be nze ne We can see since the mole fraction is always less than 1 that the vapor pressure is always reduced for a substance when additional substances are added to the mixture The reason for this can be understood simply by considering that only molecules near the surface of the liquid may easily escape So it makes sense that the mole fraction in the mixture should directly affect the vapor pressure Raoult s law is valid for similar substances 7 this is because then interactions with other substance will closely mimic those with the pure substance We can use Raoult s law to write the equation for the chemical potential of the liquid A MA 1 M20 RTlnXA This equation is valid only for ideal solutions similar interactions AmixH 0 Remember that AmixH 0 does not mean that there are no interactions it means that the interactions in the mixture are the same as those in either pure substance Even for nonideal solutions under what conditions might Raoult s law be obeyed When the mole fraction of a substance approaches one In this case the solvent obeys Raoult s law The molecular explanation for this limiting behavior is that each of the molecules will have only molecules identical to itself in its nextneighbor sphere and the solute obeys Henry s law pB XBKB Henry s law Notice here that while the vapor pressure is proportional to the mole fraction as is the case for Raoult s law the constant of proportionality is not the vapor pressure of the pure substance This constant K3 is chosen to be tangent to the experimentally determined curve at XB 0 Show diagram for transition from Henry s law regime to Raoult s law regime Positive Deviation from Raoult s Law 1 Slope KB I I I I I Actual Vapor Pressure Raoult s Law Ideal dilute solutions are those that behave as above 7 nonideal mixtures for which the solvent obeys Raoult s law and the solute obeys Henry s law Henry s law constants are inversely proportional to the solubility vapor pressure is inversely proportional to solubility Right If something is very soluble then we would expect it to not escape into the vapor phase Sample Henry s law constants for aqueous solutions 298 K from Tinoco Solute KB atm He 131 x 103 N2 86 x 103 Lecture 6 7 Chapters 4 and 5 More Random Walk amp Kinetic Molecular Theory State Functions Exact Differentials Outline More Random Walk Kinetic Molecular Theory State Functions Exact Differentials Review 7 First Law and Second Law First U q w Heat and work are different expressions of energy Second systems tend to want to maximize their multiplicity Contrast heat and work 1 When heat is transferred we change how many particles are at various energy levels but the levels themselves stay the same 2 When work is done then the particles stay at the same energy levels but the levels themselves change their energy Both heat and work change the energy but in different ways Random Walk N Whatifp q 12 Then m 7 text proves this using Stirling39s approx very important lnN NlnN N Stirling s approx is valid when N is large At end we talked about displacement d m N ml 2m N d M N d 2 NJ 0 Not surprising 7 on average nothing should move Okay then what about the RMS deviation d2 lt2m 1W2 d2 4ltm2gt NW to get here we substituted ltmgt N2 I won t derive this but qu l 4 This leads us to N2N ltd2gt4 4 N212Nl2 amQ N What do these results for d and aims mean If we have many particles their average position will be zero but each of the particles is moving Any individual particle on average will move by the square root of N 2 particles starting at zero have moved but avg still zero 0 o For example steps sequences result d d2 1 L 1 0 2 2 R 1 i 2 2 2 2 LL 2 0 8 LR RL 0 Z 2 RR 2 4 3 LLL 3 0 LLRLRLRLL 1 g a RRLRLRLRR 1 8 RRR 3 To connect these random walk results to real life 1 Replace the number of steps per second by the collision rate zA 2 Replace average step size by the mean free path 7M Intermolecular collisions Consider the volume swept out by some hard sphere particle of diameter d in among a bunch of other particles of diameter d in the time dt i miss V V dt hit Kr Volume of cylinder height times area Area6Tld2 what is height vAdt what is area il39dA2 we call this the cross section 01 note use of d rather than radius accounts for collision with other particles of same size VA qvAdt This is nice but we have left out one tidbit All particles are moving not just our particle The physics trick for this is to use the reduced mass to get a relative velocity rather than just the velocity of the one particle mlm2 m m1m2 2 Since in energy m and v are related by square root it works out that ltVReIgt 5ltquotAgt So our effective volume then is VA 2 GAVAdt Why do we care about this volume We can use it to gure out of collisions How can we get the total number of collisions our happy molecule sees multiply this Vtimes the number of particles per unit volume the number density that our molecule will bump into N quotNA pNA I7 V V RT kBT at STP 71 p 269 X 1019 molecules cm393 note that sometimes p is used instead of 71 collisions VAn A JEUAltVAgtHA dt MZltVAgtnA dt d 1139 39 collisionrate 2A 2 W JEUAltVAgtHA MZltVAgtnA For N2 at STP d 0370 nm so 2A 74 X 109 s391 or one collision every 0135 nanoseconds Note that the collision frequency also determines the gas kinetic rate of chemical reactions no gas phase chemical reaction can proceed faster than the gas kinetic rate Mean free path 1 or Z the average distance a particle travels between collisions V This must be the mean speed divided by the collision frequency 1 l Q ZA b 39 39 39 1d 7 1 kBT su st1tutlon y1e s JEOAn A Jag1p 1 Note that X cc so increasing pressure decreases distance between collisions However 1 does not depend directly on the speed For N2 at STP 7 61 nm Because N2 has a molecular diameter of about 04 nm d ltlt 7 so the kinetic theory of gases holds since the molecules spend much more time apart than in collisions this implies that the ideal gas should be a very good approximation for N2 at STP For N2 at 10398 bar a pressure achievable with the diffusion pump in the pchem lab 7 6 m Molecules collide with the walls of system much more often than they collide with each other Getting back to our random walk results for displacement M For N2 at STP 160nm ZA7X1098391 dWS W1 lsecx 7 x109 sec 160x10 9m 050cm in one second Note that for N2 at STP V 474 ms So dRMS is small not because the particles aren39t moving quickly but because all the collisions keep reversing direction roughly remember 3 dimensions For all of these kinds of things we always want to try to think about what things depend on 6 MM ltVAgt ltVAgt 2m ZA J Jami mz 39dzp 211sz Now then what do we have from here that is important dRMS x J 1 61 DC T The Tdependence is approximate since it is also in v P l d cc 1am J Anything important that we don t have here How about mass contained in v Lecture 18 Chapter 11 7 Tdependence of heat capacity No seminar today but things get rolling again next week Outline vibration electronic heat capacity Tdependence Review Vibration 7 Harmonic Oscillator o Vx kx2 h M o QM solutionis swbvv2 K vhvv V 012 75 1 Note stupid use of V vee and V nu O vibrational temperature 8m hV k 0 solve for A8hv how many Vibrational lines 1 2L F vib freq light freq 7 1 0 D0 De 12hV D0 is dissociation energy DC is ground state electronic energy where zero is the energy of the atoms in nitely far apart DC is the bottom of the well Do is the lowest state 0 large force constant strong bond or small reduced mass gt large spacing hVZkT 827 e e i UKT 787 1 6 V 1 6 0 9m Electronic energies all atoms and molecules are generally measured by spectroscopy 7 measures transition energies 7 differences between two levels you all know this from GenChem o Schrodinger equation can only be solved exactly for hydrogen other atoms and molecules require extensive approximation use experiments 0 electronic energies not a big concern for thermo 7 usually only ground state 0 So we won t really get into Be sure to note the degeneracy gn Several common units for transition energy from spectroscopists 7 relate to light 0 frequency 8 hvunits of s391 also called Hz 0 wavenumbers cm391 170 1 clear from dimensional analysis 0 wavelength length N lenergy c 11 so A note that A All rightee newstuff for today which you have already seen through Katie s and Julie s HW problems Degrees of freedom 0 always total must be 3atoms DOF are not lost or gained by bonds always 3 translational DOF total for molecule either 2 or 3 rotational DOF depending on shape remainder are vibrational OOO Now let s remember that the partition function is what we re really after and look at this some more HCl Let s consider a volume of 1 mL a small glass bulb size at 298K V 10396 m3 and m MNA where M 0036 kg mol39l 32 Recall that gum 27Wij V hi m 2X7rx60gtlt1026kggtlt138gtlt10Z3JK1gtlt298K gm x10 6m3 22x10 6626gtlt10 3 1s2 J At room temperature the translational energy is spread out over an incredible number of energy levels E U NkT2 m 3Nkr or 32 RT permole trans trans 2 Oman lt5U5Tgtv 32 R on a molar basis How many DOF in monatomic ideal gas so 12 R of heat capacity per DOF for translation The rotational partition function depends on the moment of inertia which is related to the structure of the molecule and the mass of the associated atoms and the temperature HCl 298 K 8m 1502 K 039 1 recall qRUt 00R 298K 15 20 1X 1502K We can similarly calculate ltEmgt ltEmgt NkT2 8T Em NkT2 817M NkT2 XNkTZ NkT V 3 quot771 1 I 2 A Note nonlinear molecule has q a bc M so we would get E 2NkT ml 0 112 ml 2 On a molar basis Umt RT vamt is a triVial result vamt R Keep in mind that this is for HCl 7 2 rotational DOF Something that is nonlinear would have 3 rotational DOF so vamt would be 32R Vibrational the one with zero point correction 1 Ifwe use qvib W HCl 298 K 8W3 4227 K If 0T is gtgt 1 then what is q N l qlviid z b 1e422 What does this mean At room temperature HCl like most molecules is almost solely in its ground 10000007 Vibrational state eihVZkT eieZT How does this translate to energy and heat capacity using qwb 1W 1W e e dlnq 8 8 E I Vlb Vlb Vlb lt V bgt dT 2 eewbT l 8 8V Uvib RTb eew Tb1 molar bas1s dltE b 8 I 2 eiembT C V39 R Vquot molar basis vvib dT T 1 eiemh T 2 just like Einstein Does anyone see something new For the rst time we have T dependence per vibrational DOF 7U Heat Capacity 05 10 TGVib This was one of the big early results that helped establish quantum mechanics Cm HCl 298 K 00012 J K391 mcr1 very small compared to 32 R Cm HCl 2000 K 581 J K391 mcr1 signi cant compared to 32 R vavib H1 298 K 00170 J K391 mol391 smaller theta than HCl due to larger mass Cm HI 2000 K 668 J K391 mcr1 vavib HI in nite temp 831 J K391 mol391 the maximum contribution to the heat capacity for each Vibrational mode is R High temperature limit or classical limit is T gt 9 indicate on gure Book miss uses this in Katie s HW We can see here that the vibrational contribution to the heat capacity depends on the temperature and bond strength of the molecule frequencies of its vibrational modes This vibrational contribution is the primary reason why heat capacity does in general depend on temperature for polyatomic molecules Finally for electronic the energy spacings are huge so the ground state is the only one that contributes even at very high temperatures 7B9 qelec Z gje levels HCl 298 K qgeccl 1 energy of lowest excited state is much higher than kT at 298 K degeneracy of ground state is typically 1 For most atoms and molecules the electronic degree of freedom does not signi cantly contribute to the average energy or to the heat capacity ltEelecgt 0 0r Uelec 0 and Cvelec 0 Lecture 7 7 Chapters 5 and 6 Exact Differentials Boltzmann Outline Exact Differentials Boltzmann Second law revisited Review 7 dRMS ltd2gt 1W collisions VAn A 0AltvAgtnA dt 39d v n dt d 1139 39 collisionrate 2A 2 W 0AltvAgtnA MZltVAgtnA For N2 at STP d 0370 nm and ltVAgt 474ms so 2A 74 X 109 s391 or one collision every 0135 nanoseconds ki JEO39AnA 450117 1 Note that X cc so increasing pressure decreases distance between collisions I7 Also 2 0c T at constant p because ptcls are farther apart However 1 does not depend directly on the speed For N2 at STP 7 61 nm For N2 at 10398 bar a pressure achievable with the diffusion pump in the pchem lab 7 6 m For N2 at STP 160nm ZA7X1098391 dRMS 4N lsecx 74gtlt109 sec 161X10 9m 056cm in one second Recall that for N2 at STP V 474 ms So dRMS is small not because the particles aren39t moving quickly but because all the collisions keep reversing direction roughly remember 3 dimensions Derivatives and Differentials OK let39s switch gears a bit to chapter 5 Like I said we ll have these things again and again so I m too worried that you have all of this down pat right now However I do want to make sure everyone is 100 clear on a few things Total derivative of some function f that depends on only x and y df dx dy what ifwe had z just add another term y x Exact differential df is an exact differential if 31 5 notice how this relates the coefficients we had above If f does have an exact differential then we call it a state function What does this mean It means that the change in f on going from point A to point B only depends on the conditions at A and B not on the conditions of any intermediate point Thus de dy xy fxByB fxA yA This is true for things like temperature energy pressure location Not true for things like work heat work and heat and others are called Path functions because their value depends on path For example if its 2 d and one at mid field Culpepper can do a little QB sneak ahead for two yards and pick of the first down The ball is now on the 48 Or the Vikes know they have two downs left so they can be a little crazy and run a reverse to Moss who goes the whole width of the field but only gains two yards Same result right First down ball on the 48 In this case the down and position of the ball are state functions However the amount of work done by the ball carrier is a path function Duante ran a few yards Randy ran about 40 completely different So for work and heat the above equation is not enough You must also specify the path that you are taking We39ll get to that more in later chapters Chapt 6 I don t know if you guys really appreciate it or not but this Chapter 6 is pretty awesome Dill has really done some neat stuff Hopefully it is not all lost in the math Toward that end I ll try to do some of the same things but again I ll deviate a bit from the book I ll also not dive into much math here That isn t to say that the calculus is not important Quite the contrary I just want to have a venue where the focus is on physics Dill talks about entropy in terms of probability distributions Let s think about that in terms of some of what we have done for the last week If you think back to the second day when we were ipping coins I put up the excel spreadsheet with several Gaussian distributions from different N values What was the point That as N gets huge the distribution gets narrow And the effect is dramatic Remember that for 100000 ips 49000 51000 heads contains 9999999993 of the distribution VERY IMPORTANT 7 as N becomes huge the distribution gets very narrow So if we have something like 1023 particles then even though we can t know the energy of any particular particle at all we can know the most probable energy and the average energy very well How did we involve entropy in here We said that the system will tend to maximize its multiplicity 12 heads and 12 tails gives maximum multiplicity 7 that is why it is most likely Let s now shift gears to a chemical kind of system Now what we are concerned with are lots of energy states and how they are populated Boltzmann did two huge things for us here 1 S k In W This tells us how to relate the microscopic world of energy states distributions and multiplicity to the macroscopic world of entropy 2 the following probability equation tells us what the probability of occupying each state is based on its energy Note that your book states S in two ways one the an thing that we have above the other in terms of probabilities S I Z 1 lnpi So if we know the energy levels that exist in our system then we can know the entropy Now here is where I really thought the book was doing some nice stuff The principle of maximum entropy can be applied to any situation In particular when your knowledge is incomplete the book uses example of rolling dice and knowing only the sum of the rolls you can use max entropy to have a good guess at the probability distribution That is if I just tell you the sum of a bunch of rolls of dice you can use max entropy to calculate the most likely probability of each face of the dye coming up In part this is because of what we have mentioned several times 7 that entropy is not some magical thing it just says that the most likely thing is what occurs It is just probabilities Anyway think about what I just said You know that the average of many rolls of a sixsided dye should be 35 Max entropy says that if you nd that a particular dye has an average of 28 then we can make the fairest prediction about what the probability of each dye face is ie probability of rolling a l 2 3 etc Of course it might not be right but it is the most likely Ok then let s shift away from the book a little bit Get back to the Boltzmann probability distribution We ll get to this in more detail in your book later but I want to give you a little peak ahead 7 into the physics Boltzmann tells us how energy affects probability 7 we will use this a million times VERY IMPORTANT We encounter the denominator in that Boltzmann expression often and it actually tells us all we need to know about a system We call it the PARTITION FUNCTION QNV5 Ee39W It turns out that B is closely related to temperature 5 l kBT So P and Q can also be written in terms of temperature but 5 is often easier to deal with Back to the spreadsheet to show probability vs E at several temps What are the rami cations of this to NMR Low Bf1eld means 2 spin states are close in energy so populations are very similar High Bf1eld means larger energy difference so significantly fewer protons in upper state Q is a weighted sum of the total of con gurations in the system In other words it represents the number of con gurations that are accessible at the current T V and N Of course what we are really interested in is the average energy not just the probability of any particular individual energy What is the average of any quantity in terms of probability X 2i P iXi So average energy is note that our P from above is already normalized E 1 PiEi I ll relate this average energy to Q but do it in reverse because it will make more sense recall dlnx ldfx BB NV x i Q lt95 i BZe pEj derivative of sums is sum of derivatives l p39a 111 N Em II M So then we have the second most important equation we have gotten to so far second to the probability equation that we just did E NV Quickly now before you go 7 details in book For N noninteracting particles Q 3 li q is the partition function for an individual molecule For noninteracting particles consider only translation Just trust me 3 q 21IMJZV th Lecture 22 Chapter 13 amp 14 7 Extent of reaction and equilibrium Remember next exam is 1 week from Friday Today will be last quiz before exam GRE practice today Outline van t Hoff partition function phase diagrams Review Ame nRTXA ln xA xB ln xB Grem AGRX1 witho ut mixing Gpro d AC TMix E Recall we looked at N204 9 2N0 N204 g ltgt 2 N02 g Initial amounts 1 0 Equilibrium amounts 1 2E Total amount 1 E Equilibrium mole fractions 1 1 E 2E1 E K prOdUCtS PNOZ2 XNOZPTUI 2 F reactants PD204 X NZOAPM l SolV1ng for E E 1 1 4 KP7 251 xP2 4 2P 1 1 xP 1 OK then to clarify from last time The equilibrium constant is constant ONLY with respect to concentration In the case above we are using pressure to measure concentration so K is constant wrt pressure but not Temperature So when we see ArGO RT In K the AG must be constant T and K both move to balance each other Response of equilibria to temperature We can use Le Chatelier s principle to guide what to expect for temperature dependence Exothermic reactions generate heat 9 heat is one of the products So if we increase the temperature The equilibrium will shift to make less heat less products Endothermic reactions absorb heat 9 heat is a reactant If we increase T we can expect that the equilibrium will shift to absorb more heat more products OK so now we are going to do a little derivation in that direction FAST Recall that ArGO RT In K ArGO T so Ran Recall also that AG AH ms or AGOT AHOT As0 If we take a derivative wrt T a a A f J A i Ho and So do not depend on T T Z substitute for G0 dR1nK AH dT T2 A little cleanup gives us the van t Hoff a Dutch guy equation RT2 dan AH dT J or in a different form that is sometimes more convenient d In K A H Q J d1 T R We can see from the first form of the van t Hoff equation that if a reaction is exothermic AH lt 0 d In K dT will be negative Therefore as the temperature rises K will become smaller Similarly if the reaction is endothermic AH gt 0 K will increase as the temperature is raised In fact AngZ is often determined by measuring the equilibrium constant as a function of temperature To help these eXperimentalists we need to recast the van t Hoff equation in a way to get a linear dependence of some function of K on some function of T How do we integrate the second form of the van t Hoff rearrange to get common variables on the same side of the equation and do definite integration K2 1 1 1 2 Q 1 d1nK AH d K1 R 111 T What assumption should we make Ang3 does not vary significantly with T o anZ an12 A H gtlt i L R T2 T1 an2 A H ian1 i R T T Therefore a plot of In K vs lT gives a slope with ArH R Recall that K 9 ARG through ARG RT In K KT gt AH Therefore we also can get what AS AH AGT So measuring K as a functn of T gives us all the important thermodynamic quantities ltPAUSEgt Okay finally the moment you all have been waiting for let s relate our equilibrium constant to the partition function Let s go back and make our general reaction look slightly more complicated aA bB gt yY zZ What should our equilibrium constant look like in terms of concentration y 2 K T CA CB We ll get back to this in a minute First Dill showed back in chapter 11 that 1 RT Inf q is of course the molecular partition functn Putting this together with GibbsDuhem and dG 0 gives NW2 2 q q MW 61261 Stop for a second and think about what this means Draw version of Figure 131 Population will distribute itself among all energy states of the system according to Boltzmann irrespective of the molecular identities Okay then let s get back to K Recognizing that concentration is NV y z LY 12 V V K T qA qB V V Of course q quansqmtqvibqelec for each species 139 in the chemical reaction For the H I gt 2 HI reaction we have 2 e 2 V qHI qi qquIz V V the expression is similar to example 132 in Dill KT 32 H17 H27 12 min 4838 l e el 1 e 9 mzb E D KT T We mHZmIZ 8m l e 3 translational rotational Vibrational electronic Where is our T dependence Only in the Vibrational and electronic parts of the expression and the biggest contributor is Vibrational part again With our usual approximation that T gtgt 81 Let s look at how K changes as temp changes As T increases the exponential term goes toward 1 and 1 exponential goes toward zero A big 8W means that the 1exp term decreases more slowly So if reactant 8m gt product 8m then numerator reactant decreases more slowly than denominator product therefore K increases as T increases Because evil hvkB and v 127c1du12 gt GM is proportional to k force constant Lecture 4 7 Chapter 2 Outline nish some probability potential energy surfaces minimize energy maximize multiplicity lattice models Review Coulomb s Law U qlqz 47r8R12 Magnitudes 0f Intermolecular interactions HH 500 kJ mol391 chemical bond H20 H20 20 1d mol391 intermolecular electrostatic bond Ar Ar l kJ mol391 intermolecular dispersion bond can be larger in bigger molecules It is the magnitude of these energies that determines the condensed phase properties of materials Ether is a lowboiling compound because it is held together in the liquid phase by weak dispersion interactions and water is a high boiling compound because it is held together by strong electrostatic interactions hydrogen bonds We also put together some formulas to describe coin ipping We ended up with N PW h1lh This formula works when the probability of heads tails More generally the 12N needs to be diVided up into the two different probabilities PNh ph 1 ph h So then the above general form further generalized for t different possible events rather than 2 for coins or 6 for dice is PNh pf pz pa upfquot quot1712713 quott remember this is for N dice of t sides each 5 n1n2n3n4n5n6 What if we were rolling ve dice like Yahtzeel P 5 6 OK there are some important statistical things to notice Let s go back to all our ipping coins Let s notice a couple more things about coins What ifwe ipped 2 coins 7gt w I 2 heads probability of getting exactly half heads Phalf 2 4 50 but notice ratio of Phalf heads Pzer0 heads 2 l 2 What ifwe ipped 4 coins H 3 heads probability of getting exactly half heads Phalf 6 l6 375 But again ratio of Phalf Pzer0 6 l 6 Go back to spreadsheet and show for ips going up to 150 Probability of the dominant contribution falls off but ratio of most likely to least likely increases dramatically For 100000 ips 49000 51000 heads contains 9999999993 ofthe distribution VERY IMPORTANT 7 as N becomes huge the distribution gets very narrow So if we have something like 1023 particles then even though we can t know the energy of any particular particle at all we can know the most probable energy and the average energy very well OK let s look at how to calculate averages and std deviations and the like what if we have several discrete things like test scores 67 83 95 100 what is the average add them up and divide by 71 67 83 95 100 x 4 1 1 1 1 2 6797 8397 95Z100Z Z xpx jxpxdx other important property is the standard deviation 039 but we usually use the variance 0392 oz x W x2 ltxgt2 Note from the text that the average is more formally known as the rst moment and the variance is the second moment Today we get rolling on some fundamental physics If this seems simple good it is vital that everyone is absolutely clear on these things Probably no one has used lattice models Dill has made a living off of them so we ll run into them a lot this semester Therefore we ll so them thoroughly today First define a few critical things V potential energy For instance the energy stored in the bond of a diatomic that V vs R plot that I put up during the math review dV f Force It drives the system toward the lowest potential energy f d x Please remember that force is the negative gradient of the energy Several kinds of equilibria wrt to energy Stable A global minimum see Fig 22 Metastable A local not global minumum see Fig 24 Unstable A local or global maximum see Fig 25 Clearly derivatives help us distinguish these d V Any of these equilibria will have 5 O d Stable and metastable have d gt 0 hard to tell apart 7 this is a huge problem for 2 x computational work especially for complex systems like protein folding 2 d V Unstable has lt 0 dx2 The physics here is that things tend to want to minimize their energy We ll do this over and over again this semester Second bit of physics is that things tend to maximize their multiplicity We saw this last time with the coins Recall that each speci c sequence of ips eg HTTH HHTT HTHT was equally likely to happen But if the ips are indistinguishable then all we can tell is the composition In this case what is most likely the composition with equal numbers H and T Why because it has the largest number of combinations 6 of 16 ie the largest multiplicity The important nomenclature is the multiplicity This is the number of occurrences of the composition in question e g in this case the 6 is the multiplicity Recall that it comes from the combinatorics term nllnzln3 n Can anyone see or did they read what key quantity this is related to Entropy In fact 4 Lecture 1 7 introduction to the course bring copy of homework rubric 3x5 cards 7 make sure series are in textbook first two math review problems Yes this is PChem and I have to say I am more excited about this course this semester than ever before I like the new textbook and now that we are in the second year things should run smoothly Homework After class on this free 3x5 card please give me your name major hometown something interesting to help me 39 what vou hopeplan to learn concerns about the course Bring to my office in person sometime this week When you go home for Tgiving and aunt asks for your favorite class you say just checking First day syllabus amp course mechanics we ll get to this at end Complex course 7 easy to loose big picture 7 paint this today PChem 7 what is it to you 0 For me show sizeline ranging from atom to bulk it is how atomic properties and interactions yield molecular properties and how in turn how molecular properties and interactions yield bulk physical properties Prompt for intermediate line labels molecule and polymernanocrystal Maybe a good time for superballLN2 demo The energy ow is completely different as ball changes temp because of different ratio of intermolecular interactions to thermal energy Second semester 7 QM first semester 7 thermodynamics and kinetics Thermo is the study of energy ow and how simple rules about energy allow us to understand the physical properties of matter 7 independent of atomic model in book Read Einstein quote While thermo deals with systems at equilibrium kinetics deals with processes of change Kinetics uses simple models for molecules and interactions to calculate the rate of a process yielding information about mechanisms We will utilize many mathematical tools to help us connect physical properties with molecular properties interactions and energy ow As much as the math might dominate what we are doing sometimes it is not chemistry amp chemistry is the focus When aunt asks what good is PChem I dug up some recent Natures 7 show articles ask Also mention SH3 domains Anyone heard of SH3 domains kinases 0 Important signalingregulatory proteins 0 What chemical industry might care 0 Pharma like Pharmacia cares a lot spends lots of 0 Show Wendell s work 7 all thermodynamically controlled Show DNARNA polymerase movies 7 more thermodynamics at work So thermo is at the heart of most current bio and lots of other stuff I ll try to bring in lots of examples like this throughout the term Help you connect the PChem foundation we will build with current research If anyone nds something they think is interesting NYTimes or whatever bring it in to me and maybe we ll use a few minutes of lecture time here to discuss it This is also the primary reason for our textbook change The current book deals a lot with bio 7 primary author is one of the main protein structurefolding guys On to the syllabus 7 hand out and go over 0 o o o o O Lecture 27 Chapter 16 Activity and Colligative Properties Announce 0 HW due next Monday 0 Remember Seminars today 7 Hledin at 300 Buchwalter at 400 Outline Activity Non ideal ity colligative properties boil pt elevation freeze pt depression osmotic pressure Review Raoult s Law PA XApA Raoult s law Idealdilute solutions Solvent obeys Raoult s Law and the solute obeys Henry s law pB XBKB Henry s law Positive Deviation from Raoult s Law Vapor Pressure Raoult s Law What should 4 be for an ideal solution iiil BiB AiB 2 Alternatively we can de ne w as the nonideality w zuArB uArA uBrB u w 0 for ideal w lt 0 for stronger attraction w gt 0 for weaker attractionrepulsion Activities For an ideal solution Raoult s law is obeyed and we can write the chemical potential MA 1 M210 RTlnXA When the solution does not obey Raoult s law the form of the above expression can be retained by substituting the activity for the mole fraction MA 1 M210 RTlnaA The activity is an effective mole fraction for nonideal solutions just as fugacity is an effective pressure for nonideal gases We didn t do fugacity in lecture but Chapt 8 talks about it By comparison to our general equation for the chemical potential of a mixture we can see that aA pA p AR Therefore the activity can simply measured as the ratio of the vapor pressure of component in A in a mixture and its vapor pressure as a pure substance Solvent activity Since all substances obey Raoult s law as the XA gt 1 we can say as XA gt 1 then pA gt p and therefore aA gt XA We can de ne a new term called the activity coef cient for substance A YA aA YA XA YA is not constant It follows then that XA gt 1 VA gt 1 and therefore aA gt XA as it should The chemical potential for a solvent in a non ideal solution is therefore MA u RTlnXA RTlnyA Notice that the second two terms are zero for XA 1 so asXA gt 1uA gtuA Can we quantify non ideality Remember that AGmix 710RT X AlnXA XBInXB for ideal solutions Keeping in mind that MA u RTln XA RTlnyA then what might AGmix be for our non ideal case AGmix n0RT XAlnXA XBlnXB XAlnyA XBlnYB ideal mixing excess Gibbs energy of mixing ACTmix ACTideal Aquotexcess This is great stuff allows us to quantify the intermolecular interactions Colligative properties properties that depend only on the number of solute molecules at fixed p I I I I I I TMelt TMelt TVap TVap T Boiling point elevation derivation Notation solvent is component A and solute is component B MA 1 M20 RTlnXA At equilibrium MA g MA 1 and solute does not contribute to gas so WA g HA 1 mquot g In 1 RT 1n XA This equation can be rearranged if A a ma X8 mg ma 2 RT RT AvapG AvapH T AMPS For a general solution with XB nonzero vapH AvapS RT R Ina XE For the pure liquid A XB 0 at its boiling point T AvapH AVEPS lnl F RT R The last two equations can be solved for AMPSR and equated to give 1 1 X 4 A n B R T Tquot If we further consider the case of a very dilute solution of component B lnl X3 5 XB for X13 ltlt 1 For instance ln 1 001 001005 ln 1 01 0105 AvpH 1 1 X gtlt B R T T And if T Z T are not too different from each other 1 1 1 1 T T AT 1X 1gtlt 1 z T T T T TT T For instance 1375 71373 1430 X 10395 375 7 373 3732 1438 x10395 Thus finally the elevation in the boiling point is AWH AT R W 2 AT RT XE AWH Thus the change in solvent boiling point depends only on the amount of solute not any of the solute properties TV and AWPH are solvent properties B This is supposedly one reason why salt is added to boiling water used for cooking pasta But if enough salt is added to make XNacl 01 this is a lot of salt 12 pound to a quart of water AT is only 28 K Any other ideas Freezing point depression Using similar arguments the depression of the freezing point is found to be Since for water AvapH 4505 1d mol391 and Th 373 K and AfusH 601 1d mol391 and Tf 273 K we can see that for the same amount of solute there will be a greater depression of the freezing point 10 K for the same as above than elevation of the boiling point Practical uses of the freezing point depression 1 making ice cream adding salt lowers the temperature of the ice water solution so as to make the ice cream harden 2 treating slippery roads adding salt lowers the freezing point so that the ice melts at the ambient temperature Osmotic pressure osmosis the passage of a pure solvent into a solution separated from it by a semipermeable membrane impermeable to the solute osmotic pressure the pressure that must be applied to the solution to the stop the transfer of solvent Jp Jpl39l HAG 11 HBP 11 The pure solvent is thermodynamically driven to the solution side because of the lower chemical potential for A in solution Note that my A and B are opposite Dill because he has strangely changed convention Because we are at equilibrium it must be that the change in the chemical potential for mixing is exactly equal to the change in chemical potential for additional pressure AmixG from our above MA 1 u l RTlnXA ApressG from dG Vdp 7 SdT constant T so S term goes away pH RTlnXA j dep P If we make two additional assumptions assumption 0 7 constant T l Vm molar volume of pure liquid doesn t change for the pressure range of interest 2 ln XA ln 1 XB N XB for small XB RTXB HVm Lecture 17 Chapter 11 7 QM energy levels and Tdependence of heat capacity No seminar today but things get rolling again next week Outline vibration electronic heat capacity Tdependence Review Translational partition function 2ka 3 I so gnarlst 112 V Rotational energy only molecules 7 can t discern atom rotation Rigid rotator model assumes two masses at a fixed distance note that L should really be I but I m using caps for clarity 2 o 8quotthan LL 1 L 012 degeneracy gL 2L 1 note hbar instead of h hi 0 Again we have a rotational temperature 0R0 21k Note error in text on p204 8 w math T ZIkT o 2L1e m d12L1 T q lt gt g lt gt2 0ng 0 where 039 is a symmetry factor that accounts for overcounting rotationally symmetric arrangements 039 l heteronuclear diatomics HBr 039 2 homonuclear diatomics Oz 039 2 water 039 l2 methane 039 l2 benzene 0 Vibrational energy only molecules 7 vibration requires a bond 0 assume a bond looks like masses on a spring 7 Hooke s law I f kx eq for force k is force constant big stiff leads to Xt Acos2TEvt draw and de ne A V T Vx J f xdx indefinite integral so add constant gives Vx kx2 usually define constant 0 7gt draw parabola V2 h k 0 QM solution is SW l l v hvv V 012 add states to parab Z k j Note stupid use of V vee and V nu 0 Reduced mass 1 M m1 m2 OK let s break into 4 groups I would like each group to answer the following questions just a few minutes What is nu what is vibrational temp what are the A8 observed by spectroscopy what is min Evib V Lf 27 u j 0 Again de ne vibrational temperature everything except quantum number 8m hVk l k 0 solve for A8hv how many Vibrational lines 1 2 vib freq light freq 7 N l k 0 also often use wavenumbers for convenience V 27w 1 0 minimum vibrational energy is called the zero point energy 12hV Allright back to normal 0 harmonic oscillator has equal level spacing real molecules do not sketch Morse vs parabola 0 Show harmonic oscillator levels and zero point energy 7 always some vib energy 0 D0 De 12hV D0 is dissociation energy DC is ground state electronic energy where zero is the energy of the atoms in nitely far apart DC is the bottom of the well Do is the lowest state 0 large force constant strong bond or small reduced mass gt large spacing m 7 H1 kT i i i 0 Looking at the partition function qwb Ze H1 V 6 NWT 16 WW e W e 3mm 0 V 0 Can we use our same trick of making this an integral No spacing is too large 7hv2kT 7827 e e 0 But we can use a series approx1mation to end up With qwb W W N h h d l o ote t at it is common to ignore t e groun state energy to give qvib W Electronic energies all atoms and molecules are generally measured by spectroscopy 7 measures transition energies 7 differences between two levels you all know this from GenChem o Selecyhydmgenyn 2l7869 x 103918 n2 Joules show H atom levels 0 n is the electronic quantum number 123 derived through Schrodinger equation 0 Schrodinger equation can only be solved exactly for hydrogen other atoms and molecules require extensive approximation use experiments electronic energies not a big concern for thermo 7 usually only ground state So we won t really get into Be sure to note the degeneracy gn 00 Several common units for transition energy from spectroscopists 7 relate to light 0 frequency 8 hvunits of s391 also called Hz 0 wavenumbers cm391 170 1 clear from dimensional analysis 0 wavelength length N lenergy c 11 so A note that A Several common units for transition energy from spectroscopists 7 relate to light 0 frequency 8 hvunits of s391 also called Hz 0 wavenumbers cm391 170 1 clear from dimensional analysis 0 wavelength length N lenergy c 11 so A note that A 0 Degrees of freedom 0 always total must be 3atoms DOF are not lost or gained by bonds always 3 translational DOF total for molecule either 2 or 3 rotational DOF depending on shape remainder are Vibrational 000 Now let s remember that the partition function is what we re really after and look at this some more HCl Let s consider a volume ofl mL a small glass bulb size at 298K V 10396 m3 and m MNA where M 0036 kg mol39l 32 Recall that qnmw 27ij gtlt gtltgtlt 26 x x 23 lx gigs 2 7 60 10 kg 138 102 JK 298K xlowmszziz oz 6626X10 34Js At room temperature the translational energy is spread out over an incredible number of energy levels E U NkT2 m 3NkT or 32 RT permole trans trans 2 vamms lt5U5Tgtv 32 R on a molar basis Remember my question about of DOF in monatomic ideal gas so 12 R ofheat capacity per DOF in this case The rotational partition function depends on the moment of inertia which is related to the structure of the molecule and the mass of the associated atoms and the temperature HCl 298 K 8m 1502 K recall qRat 00R HO 2 298K 2 1X1502K We can similarly calculate ltEmgt Bl ltErmgt NkT2 amp 3T V EmNkT2x 39 NkT2gtlt 39 gtlt 39 NkT2x meNkT 3T T 3T T om V On a molar basis Umt RT vamt is a trivial result vamt R Keep in mind that this is for HCl 7 2 rotational DOF Something that is nonlinear would have 3 rotational DOF so vamt would be 32R Vibrational l Ifwe use qvib W HCl 298 K 8W3 4227 K If 0T is gtgt 1 then what is q N l qIViiCl z b helf What does this mean At room temperature HCl like most molecules is almost solely in its ground 10000007 Vibrational state hVZkT 927 e e How does this translate to energy and heat capa01ty us1ng qwb 1W 1W e e dlnqvi 8w 8w Em NkT2 dT b Nk 2b eemTb1 8 8 Uvib R 2v e mb VTquot 1 molar basis d E I I 2 femT C vnbgtR8vnb 6 molar basis Wh dT T 1 e 9mbT2 Does anyone see something new For the rst time we have T dependence Heat Capacity 05 10 TGVib Cm HCl 298 K 00012 J K391 mcr1 very small compared to 32 R CVvib HCl 2000 K 581 J K391 mcr1 signi cant compared to 32 R Cm H1 298 K 00170 J K391 mcl391 Cm HI 2000 K 668 J K391 mcr1 vavib HI in nite temp 831 J K391 mol391 the maximum contribution to the heat capacity for each Vibrational mode is R High temperature limit is T gt 9 indicate on gure We can see here that the vibrational contribution to the heat capacity depends on the temperature and bond strength of the molecule frequencies of its vibrational modes This vibrational contribution is the primary reason why heat capacity does in general depend on temperature for polyatomic molecules Finally for electronic the energy spacings are huge so the ground state is the only one that contributes even at very high temperatures 7B9 qelec Z gje levels HCl 298 K qgeccl 1 energy of lowest excited state is much higher than kT at 298 K degeneracy of ground state is typically 1 For most atoms and molecules the electronic degree of freedom does not signi cantly contribute to the average energy or to the heat capacity ltEelecgt 0 0r Uelec 0 and Cvelec 0 Summary OK so let s go back and summarize For HCl the partition function is ask for values huge for translational 1029 a handful for rotational 20 just over 1 for vibrational 10000007 and exactly 1 for electronic 1 We can now sum up the total heat capacity for a molecule like HCl at 298 K Cv298 K Cvtrans 1 Cvr0t 1 Cvvib 1 Cvelec 32 R R 1 Cvvib 1 Cvelec 1246 831 00012 0 J K391 mcl391 2078 J K391 mcl391 The experimental value at 298 K is 2081 J K39lmol39l Lecture 13 7 Chapt 10 Announcements Friday HUGE Science division dedication thing HW is posted 7 short but nonzero Back on normal schedule next week The week after is fall break Outline Boltzmann distribution partition function review We have introduced almost all of the tools to start doing serious things Today and Monday are good stuff We ll get back to probabilities today We ll visit our friend Boltzmann and his pet the partition function We ll also see a few new tricks the partition function can do Recall U E ZEipi dU dE Z Eidpi Z pidEi We take it as one of the postulates of quantum mechanics that the microscopic energy levels El do not depend on temperature 7 only the populations pl Thus E depends only on V and N When V and N are constant dE 0 and also note that work 0 Thus dU Sq and dU the Edp term 2 Eidpi 2 heat 2 pidEi 2 work I had described microscopic heat and work before now we have shown in more detail that heat deals with changes in populations and work with changes in the energy levels themselves Now in a HW set a bit back you all derived some probabilities based on weighted dice Not surprisingly there are some applications to molecules as well Recall with the dice that we knew the average value of many rolls and the value of each side of the die We then were able to calculate the most likely set of probabilities to explain that average Let s shift gears a bit and say that we know the average energy over many microstates and the energy of each microstate El Shouldn t we then be able to calculate the most likely set of probabilities to explain that average How did we nd the answer for the dice Lagrange multipliers What did the answer look like exponential So if we assume we know and all the El we just use maximum entropy to solve for all the pl What do you think we get Boltzmann E E pf 2 6765 Q and of course we all know that 6 turns out to be JkT Dill does a nice job of explaining how this relates to our familiar notion that things tend toward minimum energy If we look at the individual microstates any one molecule is likely to be at any of several energy levels It is the fact that our system ensemble of molecules are restricted constrained to some energy that makes them tend toward lower energy Lower energy per particle allows more arrangements of the particles given the same total or average energy Again let s take a minute to think about the partition function Q is the number of effectively accessible states IfT is high then Q is N of states If Tis low then Q N 1 just ground state contributes Q tells you how energy is partitioned in the system The state s energy compared to kT tells you how accessible the state is Clarify microstates states vs macrostates levels microstates are all the possible distinguishable arrangements of the system macrostates are all the possible energetically different arrangements of the system So there can be many microstates per macrostate our normal definition of Q is in terms of microstates Q always counts microstates However it is often convenient to deal with macrostates 7 energy levels 7 instead Something like 7317 E Wy l rquot Wr m Can someone think of a good think to call W degeneracy Note that many people use g for this If we do want to do things in terms of energy level then it might be good to recast Q in terms of energy level E Q22648080808182823golgl2 2 139 does anyone see how to recast this in general summing over levels and using W l E 7 J Q Z Wje 4T where the E are now energy levels rather than individual states 139 We have done everything so far in terms of Q 7 the partition function for the system What if we are concerned with the partition function of an individual particle Note that we will get to individual particle energies on Monday Let s call this q Let s assume that the total energy of the system is just the sum of all the molecular energies a b 9 E1 6 ekel 139 is over states of the system a b c indicate different particles and j k l are different states of each of those particles Is this valid only if particles are noninteracting ie ideal gas But cross terms are small in most systems so this is still reasonable in most real systems Q 2e MK 1 1 a i39 z e enepem jkl 2e 532 4 5 2e we If we define tlie molecular partition function q q 12 4 then Q qaqchu But usually all the particles we have in our system if its homogeneous have essentially the same energy levels available to them Then what can we say qa qb and so on QWVT9VTN The above is true for independent distinguishable particles different atoms or molecules Is this true in general No If particles are indistinguishable e g ideal gas then we have over counted a bit because we can t tell the difference between particle at with 827 and b with PM from the opposite arrangement a b c a b c 6162E3 2El63m How does this change things N 1 N It is our old overcounting friend from binomial days Please take time to look over Table 101 7 everything in terms of Q connecting microscopic to macrosopic 0 an Uni j J VyN U S kan F U TS kT1nQ and so on Review for Exam note that we actually began the day talking about DNA which is back in the Lecture 12 notes Pchem is all about intermolecular interactions 8 F VU lh lwluxiw 1 Km lHlL lMlUH U R 2 Mlmmu LennardJones potential nonpolar things a 12 a 5 W a H Coulomb s Law polarcharged things U qlqz 39 4mm Ur olt can be positive or negative mn1 1 monopolemonopole qlq2 r dipole 7 dipole uluz r3 quadquad 9192 r5 dipole 7 quad 192 r4 Note that electrostatic interactions have orientation dependence They can be either attractive or repulsive Note that for real gasses such as described by the Van der Waals eqtn Note differences between Ideal and Van der Waals equations PV nRT P 17 h RT P Approximate Magnitudes 0f the Chemical Bonding for Representative Systems HH 500 kJ mol391 chemical bond H20 H20 20 1d mol391 intermolecular electrostatic bond Ar Ar l kJ mol391 intermolecular dispersion bond can be larger in bigger molecules Shift gears to Statistical Thermodynamics The general form for t different possible events rather than 2 for coins or 6 for dice is PNh p1 p2quot2paquot3pquot L n1n2n3 71 remember this is for N dice of t sides each As N becomes huge the distribution of average outcomes gets very narrow So if we have something like 1023 particles then even though we can t know the energy of any particular particle at all we can know the most probable energy and the average energy very well General formula for average ltxgt Z W J xpltxgtdx standard deviation 039 or the variance 0392 02 x W x2gt ltxgt2 Key extremum principles Energy tends to minimize Multiplicity tends to maximize 2quotd law While we re on laws First law thermodynamics U q w Remember microscopic pictures of heat and work heat 7 changes how particles are distributed among levels work 7 changes the energies of the levels themselves We can think of energy as the capacity to do work Heat is energy that is transferred by a temperature gradient Random Walk N mN m p is prob of forward step and q is prob of backward step PNm pmq We worked our way around to d 0 d2 4N24 N N2 N12 dmml If we have many particles their average position will be zero but each of the particles is moving Any individual particle on average will move by the square root of N These results can be related to real gasses by substituting the mean free path 7 in place of l and the collision rate 2 in place ofN kBT Jig1n A Jig1P d 1139 39 collisionirate 2A 2 W JEGAltVAgtH A wiltvAgtn A A quick aside on some calculusphysics stuff If f is a state function then it must be true that df is an exact differential or E33143 OK now back to chemistry Let s now shift gears to a chemical kind of system Now what we are concerned with are lots of energy states and how they are populated Boltzmann did two huge things for us here l S k In W This tells us how to relate the microscopic world of energy states distributions and multiplicity to the macroscopic world of entropy This is also S I Z 1 lnpi 2 the following probability equation tells us what the probability of occupying each state is based on its energy 39 the 1 J 39 of state r 39 quot vs energy and temperature r The denominator normalization is the partition function Q 7 hugely important Q is a weighted sum of the total of con gurations in the system In other words it represents the number of con gurations that are accessible at the current T V and N E NV Onto Energy entropy stuff SSUVM U U8 VN dS dU dV ZE j dN w VN 07V UN i 07 UyVme i dU as dV 2 dN S VN W SN 1 WI SVN l Remembering our favorite energy force relationship dU TdS pdV 2 Mom And similarly for entropy l dS dU 5W Z div T T T 39 Work 5W 39lpext dV Energy TZ AU In CVJT So isothermal means that AU 0 H U pV dH ms Vdp MN r2 AH der isothermal also means that AH 0 Helmholtz Free Energy F U7 TS other books use A dF SdTpdVZudN Fm KN If we restrict ourselves to constant Vand constant T dF 5g 7 T dS which means AF 0 for spontaneous processes Gibbs Free Energy G H 7 TS dG SdT Vdp 2 de 6mm Also AG AH7 TAS at constant T Adiabatic processes q 0 thus AU w And it works out that for an ideal gas and constant C V VZT V1le P 1 Vly P 2V21 in general CF CV R 3R for monatomic ideal gas 7 2 2 3 3 R 2 for all gasses 7gt l Ru L IlllL imthcrmul l V 7 C lellllll Ru crxihlc adiabatic PR5 7 constant Lecture 2 Chapt 1 Outline 13180110513th ulomblc forces mnlhpoles Intro to Slat Themo Three 00m in a fountain Note Overa what is m the text I39m gmng to try to present some faction ofwhat is m the chapter In a shgh y diffeiml waythan the ext highlighhnglhlngs Ithlnk are Imp crtant Review F These lnlemchons lead to more Ie hshc equation ofslale Van der Waals Equation m 17h RT p E v b 7 Didn39t get to Elechoslahcs last lime Electrostatic Forces important or polar molecules amp inns Coulomb s law charges iepel or attract U cm 5RD electric dipole consists oftwo electric chargesq and q separated by a distance land The magmtude cfthe vector 15 qquotl and 15 called the electnc dipole rumnail DebyeumlOD 1D 3335xlu393 Cm dismbu on HVF nonzero NVN zero symmemc Thus we refer Io HF as apolar molecule What about cop CCL cmcl Larger molecules are more compllcaled sums of charges and posmons Usually reasonably 11m lm p L L L dlpoles polnllng ln same dnecuon moment alone Expand wuh fan1y sample terms charge dlsulbullon monopole charge dlpole moment quadrupole momeru ocmpole momem monopole n 0 one palm charge Na39 dipole n 1 two palm charges um has no monopole 11 Fquot quadrupole n 2 four palm charges Ihm as no dlpole 0 2 c 0quot naupnle n 3 mg palm charges um has no dlpole or quadrupole cm and so on an expans lon Distance Dependence of Elemostau39c Forces multlpole lndlces for the mmactmg molecules 1 hunt I u m can be posmve ox negatwe monopolermonopole m x dlpole dlpole mu m quad u 99 dipole e quad39J 9114 long range or short rangew xepulswe Draw ahgned and antlrallgled and ask which is amac ve ms 1s a big dlffemnce between electzostabc and dispersion foxces Dupelslon have no duecbonahty lsotzoplc and always are attzacbve forces are most ofmn Imponanhn non polar systems such as benzene HVH 500 u mol chemcal bond 110 7 H10 20 u mol mtermoleculax 6160010502110 bond A 7 An 1 u mol mmmolecular dlspmsmn bond can be large m blggex molecules electrostath lntemctlons hydrogen bonds Now we get backto the beginning 7 Chapter 1 Classmal lhermodynarnlcs o Predams the acceptance of awnno theon 0 mpirical39 theo o Smple Based on 3 fundamental laws and ver geneml 0 Combmanm ofsn nphmty and genemhty make It extzemely powexful th mom dlffemntkmds ofthmgs 1t xelams and the mom extended ms 105 area of apphcablhty Therefoxe the deep Impression Whlch 013551031 thambdynamms made upon me It s the only physmal theory ofumvexsal Corlan concennng nt 0 Clearly properties depend on materials so should be able to reconcile the detail of atomic theory with generality of thermodynamics O This is Statistical Thermodynamics our text is a synthesis of classical and statistical thermodynamics 0 Properties of Statistical Thermodynamics 0 Use a model for molecular behavior average over an ensemble of molecules to get the bulk properties of thermo like temperature and pressure 0 Molecular model of choice is quantum mechanics 0 All we need om quantum are the energy levels of the molecules this is incredible from a chemists perspective because it means that detailed molecular properties like atomic masses electron con gurations bond strengthslengths etc lead directly to bulk properties This is like saying that if we just know how big and heavy a car is and how fast it goes then we can understand exactly how traf c ows Of course the human drivers make this much more complicated but you get the idea 0 So all we need to know are values of energy levels We ll get to this later be waiting for them 0 In the meantime we are going to develop some ways to deal with statistics Clearly this is important for statistical thermodynamics This is really great stuff I am also doing things quite differently than the book so it would be good to read these sections carefully to give you two different perspectives Everybody get a coin while I get spreadsheet up and running You are all going to ip 8 times and keep track on paper eg HHTHTTHT Go through spreadsheet hope illy gets close to 50 heads as more trials are done Go back and get exact order of rst 4 tosses using the book s terminology 4 elementary events combine to form 1 composite event Point out that all 16 possible con gurations are equally likely But if all we care about is total heads how does this change things Each speci c con guration is equally likely but since coins are indistinguishable all of the 22 con gurations are equivalent to us that makes it more likely Thus for 4 coins we don t have 16 con gurations there are really only 5 distinguishable con gurations weighted 116 416 616 416116 Super but what if we had a mole of coins we need some kind of formula We want something like weight of con guration total con gurations Can anyone see what the denominator is 2N the total number of con gurations What about numerator Let s say we have 4 coins A B C and D how many ways can we arrange them 4 choices for rst 3 for second 2 for third 1 for fourth so 4321 4 24 This is total arrangements so we need to divide out the number of equivalent ones This is the same type of thing so we get heads tails N so the nal formula is PNh 2 Tryitout 4312A4 24616 416 gtperfect This formula works when the probability of heads tails More generally the 12N needs to be divided up into the two different probabilities PNh ph 1 ph So then the above general form further generalized fort different possible events rather than 2 for coins or 6 for dice is 1 n2 3 N PNh p1quotp2 173317 n1n2n3nt remember this is for N dice oft sides each 5 nln2n3n4n5n6 What if we were rolling five dice like Yahtzeel P 5 6 OK there are some important statistical things to notice Let s go back to all our ipping coins Let s notice a couple more things about coins What if we ipped 2 coins gt 2 1 heads probability of getting exactly half heads Phalf 2 4 50 but notice ratio of Phalf heads Pzero heads 2 1 2 What if we ipped 4 coins 2 heads Lecture 5 7 Chapter 3 and 4 Second Law First Law Random Walk Outline Second Law work and heat First Law of Thermodynamics Random Walk Molecular collisions mean free path total collisions Review 7 The above general form further generalized for tdifferent possible events rather than 2 for coins or 6 for dice is 1 n2 3 39 PNh 121quot 12 p3quot quotPtquot n1n2n3 nt remember this is for N dice of t sides each We also talked about minima and maxima local and global and that systems tend to minimize energy while maximizing entropy Let me put my little 5pa1ticle lattice model up again 11 00000 Wrot WLWR 5005 1 I 5 5 O000E 0 WmWLWRmm25 I I I 00 OOO WWLWR 5 5 100 2 22323222 So evenly mixed is the most likely to occur Again each possible con guration is equally likely it is just the fact that there are more con gurations for the equally mixed compositions that makes them most likely Remember 7 this is entropy drive from ordered top to disordered bottom This is the second law of thermodynamics Your book drives this point home by looking at heat flow pay attention to this OK now to new stuff Your book does a fair bit de ning force work and various energies so I won t do much of that here I do want to really quickly mention conservative forces Can someone name a conservative force gravity force from spring force from electric eld etc With all these things you ght against the force eld but then you can get the energy back Like you work to ride your bike up a hill building up potential energy but then you can coast back down for free converting potential to kinetic Can someone name a non conservative force friction When you ght against friction all your energy dissipates as heat kinetic E You don t store any up as potential energy Note also that total energy is always conserved 7 shouldn t be news to anyone Now onto heat and work the text gives a really nice description of some of the history behind heat The key nal experiments were done by Joule What did they show That heat could be created by doing work therefore the amount of heat was not constant conserved This leads to understanding that heat is simply another form of energy and that heat and work can be interconverted giving us the First Law of Thermodynamics Anyone U q w probably a good thing to remember In other words energy is neither produced or consumed simply converted between heat and work In yet another way q and w are the only ways to change energy Let s contrast heat and work from a microscopic perspective Think about a piston and some gas molecules If we move the piston we do work 7 an organized transfer of energy from the piston to the gas If we just heat the system up then we also increase the energy but now we ve done it in a random unorganized way 3 it jfx 1 K i i d Work ll LlelL l ul llc111mngtlbrul cnl39cg changes energy changes motion nl nlnmx In motion wlhlumx In nnilln ln HMHHCI chaotic Innnncl Another way to view the two is with particles occupying various energy levels When heat is transferred we change how many particles are at various energy levels but the levels themselves stay the same When work is done then the particles stay at the same energy levels but the levels themselves change their energy Both heat and work change the energy but in different ways OK let s sum up quickly We talked about the First Law and just barely mentioned the Second Law First U q w Heat and work are different expressions of energy Second systems tend to want to maximize their multiplicity Chapter 4 The text spends quite a bit of time talking about series approximations Recall we did a bit on this in the math review I am not going to do this here but make sure you work through what he does in the text don t just look at it and say quotThat makes sensequot Actually work out the details Your book then gets into the Random walk We39ll do it a bit differently here so don t get confused Random walk is really quite important used to describe diffusion and polymer behavior among others We won39t get into molecular transport diffusion in earnest until chapt 18 But if we have time today I ll try to connect things just a bit with some of the kinetic molecular theory that you seewill see in PChem lab Random Walk Consider someone walking on a sidewalk who is unable to decide at each step whether to step forward or back say a President s daughter or someone Each step is therefore random and we would consider this a onedimensional random walk Consider the following six random steps In this case the random walker returned to the a starting position 6 Obviously if we take N steps and the probability of forward p and backward q steps are the same gt then on average the total distance traveled should be zero But if N is very large then we don t expect the number of forward steps m to exactly match the number of backward steps N m So individual molecules will drift away from the starting point Keep in mind that we are doing this because we can consider the movement of molecules as a three dimensional random walk Each random step is the direction and distance traveled between collisions this can be either in a gaseous or liquid medium So what we want to know is the probability that afterN total steps we will end up some number of steps m N m 2m N away from where we started What39s the probability we have the sequence bbbf ppqqqp pm That is the probability of that sequence But of course any sequence with m forward and N m backward steps has the same probability Of course this is our old friend the binomial distribution N mN m pquim What if we wanted to know the average m Remember x J xfxdx m JmPNmdm but m is discreet so m ZNZMPNm and quot10 N W Z m2PNm quotF0 N Whatifpq12 Then m 7 text proves this using Stirling39s approx very important lnN NlnN N Stirling s approx is valid when N is large OK we really care about the displacement How do we connect these d m N ml 2m N d M N N d 27 NJ 0 Not surprising 7 on average nothing should move What about RMS deviation d2 lt2m 1W2 d2 4ltm2gt N212 to get here we substituted ltmgt N2 I won t derive this but 2 MN 1 m gt 4 This leads us to 2 d2 4W szlz N12 dMMNW What do these results for d and didIS mean If we have many particles their average position will be zero but each of the particles is moving Any individual particle on average will move by the square root of N 2 particles starting at zero 0 0 have moved but avg still zero For example Lecture 24 Chapt 14 Clausius Clapeyron Equation Announcements We ll review for exam this afternoon 7 no quiz this week Outline Clapeyron Equation dP ASm A H dT AVm my Clausius 7 Clapeyron Equation dp AmH p dT nRT2 lnp AmsH constant Review Phase Diagrams Summarize SolidLiquidGas Behavior of a Substance 3G 3G a V and E S equatmg coef clents 1n eXpress1ons for dG P T p Therefore if we plot G vs T what is the slope negative of the molar entropy at xed p I I I T TM elt TV av If we repeat for many pressures then we can build up a full phase diagram 7 a P vs T plot showing regions of thermodynamic stability We have a generic one below 1 Let s look at phase transitions a little more quantitatively Equilibrium AG must be 0 so between two phases 0L and B umpT u pT Therefore there is no thermodynamic driving force for change We can use this equality to get an equation for the phase boundary If p and T are changed in nitesimally in a way that the two phases 0L and 5 remain in equilibrium Then um u before the change um duet M dug after the change and therefore du01 dug or equivalently for two phases of a pure substance dGmm dG ym plug in fundamental equation dGm Sde dep for each phase Smde medp S yde V ymdp and this rearranges Vum Vumdp sum SocmdT For measurable changes this expression becomes the Clapeyron Equation dp ASm dT AVm Notice that this is a total derivative and applies for any phase change for any pure substance Let s see a quick example How does the vapor pressure for water change as its temperature is changed H20 at 25 0C has a vapor pressure of 24 Torr B is gas 0c is liquid Asm Amsm sm 5 sm 0L 1888 J K391 mol39l 699 J K391 mol39l 1189 J K391 mol39l AVn Ammst Vm B Vm 0L N Vm B RTp 831 J K391 mol391 X 298 K 24 Torr 103 J mol39 torr391 dp ASm 1189 JK lmol 1 1 71115TorrK dT AVm 103 Jmol Torr So at 10 0C the vapor pressure is 24 Torr 115 Torr K391 X 15 K 675 Torr Molecular explanation for why dpdT is always positive higher T means more energy available for molecules to escape attractive forces in liquid What is vapor pressure at 1 C 25 torr 115 torrK 24 K 36 torr What is wrong You can t use this expression very far from the reference temperature 298 K in this case because the ASAV ratio changes with temp For two phases at equilibrium we can change the Clapeyron eq by noting Gm G Hm TS01 Hg TS TS Sm Hg Ho TAM S Amms H Amms S Atrms H T This should look familiar when Aman q We then can rewrite the Clapeyron equation trans dpAH dT TAmv If we make two assumptions 1 AthV Vgas Vliquid or Vsolid Z Vgas volume of liquid or solid is negligible 2 VIgas nRTp ideal gas dp pAmH dT nRT2 same side of the equation we know how to solve this Rearranging to get common variables on the Lecture 34 Chapt 27 sort of Enzyme Kinetics and Intro to Electrostatics Anouncements remember seminar today at VanAndel Exam Friday over all the Kinetics stuff 8llam and l4 pm note seminar at 300 Outline Enzymes MichaelisMenten Electrostatics Review Enzymes Should have pointed out last time that enzymes are normally proteins So whenl was making the point about the ribosome RNA as an enzyme is very surprising Remember our 3 important experimental observations about enzyme kinetics 1 Rate of reaction usually linear with enzyme concentration see plot at right 2 At fixed E rate is linear in S for low S prev plot 7 low S 3 but plateaus saturates at high S prev plot 7 high S amp saturation Michaelis and Menten put these together in 1913 and came up with a mechanism that involves a binding event preceding the reaction This is why there is saturation if all the E is already bound to S then more S won t speed up the reaction L L E 5 E5 E P k1 72 This mechanism leads to what rates for E P and ES dl fl k2 k1ES k1ES k2EP d P k2ES k2EP k1ES k1ES Note that since no enzyme is ever used up E0 E ES We can rewrite the second equation using this E0 and also by using the steadystate approximation on ES This is reasonable if we have excess S because then we expect a basically fixed amount of the E will be bound So setting dESdt 0 and plugging in that E0 relation to get rid of E 1S k2P k1 k2 Plugging this back into the dSdt equation gives us something that doesn t look too bad We do a bunch of algebra and end up with something that looks pretty nasty but contains dESdt inside of it So we can set this to zero and things clean up a lot We can then make a second approximation by saying that we will only care about the early stages of the reaction So the rxn velocity is v0 and S S S0 and P z 0 w Key equation V0 dt Km S0 w 1 represents the apparent dissociation constant of E and S for the reaction written ES gt E S When is Km the actual dissociation constant When k1 gtgt k2 then Km approaches k 1k1 the dissociation constant In other words when equilibrium is reached much faster than the subsequent reaction Notice that k2 is gone from the v0 equation Why PSO where K is the Michaelis constant It m Remember this important equation applies to the initial velocity when there is excess substrate Now let s break into groups and look at a couple questions 1 What happens when S is small 2 when S is big Relate these to our three observations Notice that when S is small Km gtgt S0 and we get m SlolElo and we get something linear in S just like we should for fixed E When S is big S0 gtgt Km and we get VMax 2 k2 Elo we get something linear in E0 and independent of S like we should Because this corresponds to the saturated region we call it vmax k2 then is the turnover or kcat we talked about before Note again E0 is the concentration of active sites not just of enzymes Also in your groups what about when Km S Finally when Km S0 we get kziSloiElo 1 V0 vMax 2S0 2 So the Michaelis constant is the substrate concentration that gives half vmax Note that we can rearrange the important equation above substitute vmax k2E0 then invert the rest VMax 1KM S v0 and then invert to give i1KM V0 VMax VMaxS So plotting lv0 vs 1 S gives KM and vMaX This is called a Lineweaver Burke plot Several other plotting techniques eXist Slope K f m n Finally note that k KM is the specificity constant If two substrates are m VA mohm competing for one enzyme VB quotMBBl So the specificity constant tells you the selectiVity of an enzyme for a particular substrate If A has a large kegKm then its velocity is really fast and A outcompetes B for enzyme Brief intro to Electrostatics Intermolecular Interactions Okay for this week we are changing our focus to talk about various kinds of intermolecular interactions I m a little irritated with the book that this central topic comes so late but we ll do our best We begin with the strongest interactions between molecules electrostatics nuclear forces are stronger but that s about it Electrostatics includes chargecharge interactions chargedipole dipole dipole dipolequadruple etc often described as monopolemultipole interactions Coulomb s law vA Let s start with Coulomb s law which everyone remembers from physics 0 1 quB 4723980 r u is the interaction energy units J VC note W Js VCs 80 is the permittivity of vacuum units Fm CVm CzJm So we can calculate the electrostatic interaction between two charged particles In practice this is a difficult calculation Why Because the interaction falls off very slowly with distance r1 Text does an example of NaCl crystal If you just look at one line of charges that surround a central Na we get something that converges VERY slowly In this case there is a regular arrangments of charges so we can find a series approximation that helps us out defines the Madelung constant e g This is great for cyrstalline solids but for more general charge distributions like dissolved salts amorphous materials or proteins you have to try to just add them all up Fortunately in most media charges are shielded by the surroundings This is especially important in water We can write Coulombs law the way it is really defined where 8 80D where D is the dielectric constant D E l for vacuum D 100059 zl for air D z 2 for hydrocarbons D z 14 often used for proteins 6 Lecture 35 Chapt 20 Coulomb s Law electrostatics Announce 0 Give exam distribution Outline Coulomb s Law Electric Field Electrostatic Potential Internu ecularInterac ons Okay for this week we are changing our focus to talk about various kinds of intermolecular interactions Let s look at the strongest interactions between molecules electrostatics nuclear forces are stronger but that s about it Electrostatics includes chargecharge interactions chargedipole dipoledipole dipolequadruple etc often described as monopolemultipole interactions Coulomb s law Let s start with Coulomb s law which everyone remembers from physics r 1 94 0 Iquot u is the interaction energy units J VC note W Js VCs VA 80 is the permittivity of vacuum units Fm CVm C2Jm So we can calculate the electrostatic interaction between two charged particles In practice this is a difficult calculation Why Because the interaction falls off very slowly with distance r71 Text does an example of NaCl crystal If you just look at one line of charges that surround a central Na 1 6 22 2 2 2 2 we get something that converges VERY slowly In this case there is a regular arrangement of charges so we can find a series approximation that helps us out 2 defines the Madelung constant e g U 17474NA e 75 a This is great for cyrstalline solids but for more general charge distributions like dissolved salts amorphous materials or proteins you have to try to just add them all up Fortunately in most media charges are shielded by the surroundings This is especially important in water We can write Coulombs law the way it is really defined ur 4 where 8 80D where D is the dielectric constant 729 r D E l for vacuum D 100059 z 1 for air D z 2 for hydrocarbons D z 14 often used for proteins D 33 MeOH D 7854 for water Book mentions Bj errum length which is a handy tool Remember our golden tool for connecting microscopic properties to macroscopic behavior Partition function Energy always appears as EkT or ERT depending on units The Bj errum length is just the distance two charged bodies are apart that corresponds to U RT Book points out that Bjerrum length is 80 times shorter tie a factor of D 560 A versus 713 A for NaCl dissolved in water than for two bare Na Cl particles Their interaction is severely weakened by the presence of the water If we had many charges we would just add up the u for each of them until we had accounted for everybody However there are easier ways which we will work up to now Force Of course everyone remembers that force is Bu 1 q q A V A E f u Br 471 Dr2 V So forces add as vectors This can be really a pain to keep track of But force gives us the gateway to talk about the electrostatic field Field If we have some charged particle sitting around the field is force that particle would apply to some other particle if the second particle were present We call the second particle the test charge it probes the electric field Said another way the field represents what would happen to a particle of ql if it were present at any point in space L IAZ f qB 4780 Dr remember that force is units of N so field is NC or Vm Field is still a vector However an important part of the field is the field flux the total amount of field that is perpendicular to some arbitrary surface lt1 DE d surfacz s is the surface normal vector If we have a single particle and we do this integral in spherical coordinates then at some radius r E is constant and always parallel to s so 2 1A lt1 Dngsm DESASFW 4 r2 4m 8 0 0 The distance dependence is gone because the surface area grows at exactly the same rate that the field decreases Or no matter how big one draws their sphere the total flux contained is always the same just depends on total charge It isn t a big leap to realize that the surface could be any shape because the dot product ensures that extra surface gets less impact from the nonperpendicular field The total charge contained will be the same The one next step is a bit harder but you should have all seen it before the flux is independent of the charge s position We could have the charge anywhere inside of the Lecture 20 Chapter 13 7 Extent of reaction and equilibrium Remember next exam is 1 week from Friday This week will be last quiz before exam Outline Extent of reaction Reaction equilibrium Aran at nonstandard state Relating G and E and K and p K and E as fctn ofp Review Partition function values for typical molecule HCl huge for translational 1026 a handful for rotational 20 just over 1 for Vibrational 10000007 and exactly 1 for electronic 1 I m not going to write it all down but remember that we had that nice table that summarized the heat capacities of molecules Where is the temperature dependence all of them technically but usually just vibration And Chapt 12 1 Density of States 2 Fluctuations are critical 3 Negative temp 4 Graphical connection between energy functions Check out p226 carefully Now on to Chapt 13 We have done a lot recently figuring out how to relate microscopic information energy levels and such to macroscopic quantities H T S U through the partition function Today we are going to start really applying this to find out some useful chemical information We are getting into some real freshman chemistry 7 equilibrium constants Of course we ll do it in just a bit more detail than you saw in CHEMl l 1 Let s define the simplest possible reaction 9 zZ If we started with nAo and nzo moles of A and Z then let s de ne the amounts of A and Z at any time I1A nA0 3E 112 1120 ZE E Xi or squiggle is the extent of reaction It measures the progress of the reaction and has units of moles Subtraction is because A is a reactant For instance H 1202 lt gt H20 At start nHz n02 Ileo Later nHz E 1102 111120 E Of course we are really interested in the change in the amount of the substances dnJ j dE negative if reactant We can substitute this expression into what is called the GibbsDuhem equation GibbsDuhem is for constant T and p recall dG SdT Vdp AdNA uZdNZ d6 Z qunJ J So rewriting GibbsDuhem in terms of E gives d6 Z w dig J where products have positive and reactants negative coef cients This can be rewritten as Zj j ArG Kagij J where AG is the reaction Gibbs energy This is de ned for a balanced chemical equation and the reactants and products at their standard states add products subtract reactants Let s look at a simple case Isomerization Reaction llcis retinal lt gt all trans retinal Everyone know what this does vision 5G l l it as w 4 K7 JN Both of these chemical potentials change as the rxn proceeds so this differential depends on the extent of reaction 3G a EJPYT lt 0 Hcis gt trans 36 gt 0 ram 33 lids i l ans ram 0 Heis Htmns V E O 1 100 cis A 100 trans Z When MA gt Hz or lt 0 the reaction A gt Z is spontaneous E must increase to minimize G E a When MA lt Hz or g gt 0the reaction Z gt A is spontaneous E must decrease to minimize G 3 l l K N The minimum in occurs when its value is zero or MA uz really a A 2112 This is when pT the reaction is at equilibrium The reaction stops All right then we can see that equilibrium is determined by the chemical potential Another way of saying this is that the chemical potential is the molar G for each productreactant Note that chemical potential depends on concentration p T etc Note that the standard molar reaction Gibbs energy is different than the reaction Gibbs energy because for the standard energy the composition of reactants and products are fixed at their standards states 1 M or 1 atm rather than an arbitrary composition ArGO ZHZO auAO Calculating the composition of reactions at equilibrium what is the yield Remember that we are talking about equilibrium constants so we want to try and get back to this If A and Z are assumed to be perfect gases we can calculate ArG just as we have done before constant p and T ArG ZMZ all We now need to make use of another expression that shows the pressure concentration dependence of the chemical potential 2 RTlnp comes from Chapt ll 2 kBTln Z p Z AG 3 RT lnp a f RT1npD P P ArG 211 61112 RTzlnp i RTalnp g p p a AS ArGO RT lnpz where ArGO zuzo aqu p We will nd it convenient for more complicated reactions to denote the argument of the logarithm as the reaction quotient Q Terrible notation of course Q is not the partition function AG AG RT 1nQ At equilibrium ArG 0 and Q is equal to the equilibrium constant K by de nition ArGO RT In K This is one of the most useful equations in chemical thermodynamics because it provides a direct relationship between all that thermodynamic data and information on how actual chemical reactions proceed Obviously when K lt l the reactant is favored ArG gt 0 and the reaction is not spontaneous and when K gt 1 the product is favored ArG lt 0 and the reaction is spontaneous in the direction A 9 Z If we have equal concentrations of products and reactants ltPAUSEgt Exergonic and endergonic reactions ArG lt 0 reaction is spontaneous and exergonic workproducing could drive other processes e g ATP 9 ADP or glucose combustion ArG gt 0 reaction is not spontaneous it is endergonic workconsuming eg ADP 9 ATP or electron transfer Discussion of the general form of G vs g If ArGO for a reaction is negative the products are more thermodynamically stable than the reactants Everyone agree Do most reactions go all the way to products or stop somewhere in equilibrium Most stop short of 100 yield Why The standard molar reaction Gibbs energy ArGO does not account for the additional entropy introduced when substances are mixed together In other words ArGO does not include contributions from AmixG AmixG nRTXA 1n X A XB 1n XB Think back to chapt 6 homework X is mole fraction ignore enthalpy So the minimum in AmixG occurs when our A gt B reaction is 50 complete After that point our devious friend entropy ghts the continued conversion of reactants to products Greact Gpro d AC IMix Depending on the relative magnitudes of AG and AmixG the extent of reaction can vary dramatically Now that we have a relationship between equilibrium constants and thermodynamic data we can calculate the extent of reactions 5 for real systems Use this format N204 g ltgt 2 N02 g Initial amounts 1 0 Equilibrium amounts 1 2E Total amount 1 E Equilibrium mole fractions 1 1 E 2E1 E Note that in this case E is now unitless and must be between 0 and 1 because of how we have set up our initial conditions in relation to the stoichiometry 5

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