Physical Chemistry CHEM 3520
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CHEM 3520 Spring 2008 Unit XIII Concepts of Electrochemistry A Electrochemical Cells 1 Redox reactions a 0906 A redox reaction is a reaction in which there is a transfer of electrons from one species to another Oxidation is the process of Reduction is the process of The electron donor is called The electron acceptor is called 2 Types and components of an electrochemical cell a An electrochemical cell consists of two electrodes which are typically metallic conductors in contact with an electrolyte an ionic conductor solution liquid or solid El An electrode and its electrolyte form an electrode compartment El If the electrolytes are different the two compartments of the electrochemical cell may be joined by a salt bridge to complete the electrical circuit The electrode at which oxidation occurs is called the electrode at which reduction occurs is called A galvanic cell is an electrochemical cell that as a result of the spontaneous reaction occurring inside it CI The cathode is the CI The anode is the electrode n Matirm 1 reduction lt electrode 247 CHEM 3520 Spring 2008 An electrolytic cell is an electrochemical cell in which a nonspontaneous reaction is driven by an external source of current CI The cathode is the electrode n Ham 1 reduction CI The anode is the electrode Any redox reaction may be expressed as a difference of two reduction halfreactions which are reactions showing the gain of electrons El Example Cu2aq Zns gt Cus Zn2aq reaction can be expressed as the difference between Cu2aq 2e gt Cus and Zn2aq 2e gt Zns The oxidized labeled Ox and reduced labeled Red species in a half reaction form El In general a redox couple is written as stands for oxidizingreducing agent or oxidantreductant eg Cu2Cu CI The corresponding reduction halfreaction is Ox ve gt Red 3 Types of electrodes a Metalmetalion electrode El Redox couple MlM El Halfreaction Maq e gt Ms Metalinsolublesalt electrode El Redox couple MXMX El Halfreaction MXs e gt Ms X aq Gas electrode PtsX2gXaq or PtsX2gX aq El Redox couple XVXZ or XZX El Halfreaction Xaq e gt 12X2g or 12X2g e gt X aq Redox electrode El Redox couple M2M El Halfreaction M2aq e gt Maq 248 CHEM 3520 Spring 2008 MS Mgtlts s t V R a w M aq X1361 a b amp W X aq M BQYM2 BQ C d 4 Representing electrochemical cells a Notations D Phase boundaries are denoted by a vertical bar l D Liquid junctions are denoted by 3 D Interfaces for which is assumed that the junction potential has been eliminated are denoted by a double bar H Examples MS l H2 g l HClaQl AgClSl AgS Zns l ZnSO4 aq CuSO4aq l Cus Znsl ZnSO4aq H CuSO4aql Cus MS l H2g l HCKaqacr H HClaq962 l H2g l MS The cell reaction is the reaction in the cell sways written on the assumption that the right 7 hand electrode is 3 the spontaneous reaction is one in which reduction is taking place in the anOJM l More J campanmems compartment 249 CHEM 3520 Spring 2008 5 Thermodynamics of electrochemical cells a An electrochemical ie galvanic cell in which the overall cell reaction has not reached chemical equilibrium can do electrical work as the reaction drives electrons through the external circuit The work that a given transfer of electrons can accomplish depends on the potential difference between the two electrodes which is called The maximum nonPVor electrical work that a system can do is AG wemax AG To measure ArG the cell should operate reversibly at a constant composition and this is done by balancing the cell potential with an opposing source of potential so that the cell reaction occurs reversibly The resulting potential difference is called The relationship between the reaction Gibbs energy and the electromotive force of a cell ArG VFE O F is the Faraday constant F 96500 Cmol The reaction Gibbs energy is related to the composition of the reaction mixture ArG ArG RTan VFE gt E ArG an Eo an VF VF VF o This is 0 ArG VF is called As the galvanic cell is a combination of 2 electrodes each one can be considered to make a characteristic contribution to the overall cell potential 250 CHEM 3520 Spring 2008 To measure the contribution of a single electrode de ne the potential of one electrode as zero and then assign values to the others on that basis The specially selected electrode is the s 9 Hydrogen silver chlonde electrode PtsHzgHaq The standard potential for other electrodes is assigned by constructing a cell in which it is the righthand electrode and the SHE is the lefthand electrode The standard emf of a cell in this case and in general is obtained as E0 E0 right EO left E0 cathode EO anode Eedmlyte Similarly for the emf of a cell not in the standard condition E Eright Eleft Ecathode Eanode The relation between the electrode potential and the standard electrode potential OX vei gt Red EE 1n R edsE lncR ed VF aOX 1F COX D Example Cu2 aq 2e7 gt Cus GA acu2 l c cu2 ECu2Cu E Cu2Cu gm E Cu2Cu 1n 2F D Example Fe3 aqe7 gt Fe2 aq 0 RT 0 2 EFe3lFe2E Fe33tFe2 ln Fe F aFe3 The electrochemical series is a list of standard potentials 251 CHEM 3520 Spring 2008 p The temperature dependence of the emf gives AS and AH AIS M P M P ArH ArG TArS nFE nFTEa E M P 252 CHEM 3520 Spring 2008 B Unit Review 1 Important Terminology Redox reaction Oxidation Reduction OXidizing agent Reducing agent Electrochemical cell Galvanic cell Electrolytic cell Cathode Anode Halfreaction Types of electrodes Representation of phase boundariesliquid junctionsinterfaces 253 CHEM 3520 Spring 2004 Thermodynamic Functions and Thermodynamic Equations The Helmholtz energy 7 Criterion for spontaneous processes for systems at constant volume and temperature dU5q w w7 extalV EqSTdS 7 At constant volume SW 0 3 dU S TdS 3 dU TdS S 0 7 For isolated systems dU 0 61 2 0 7 Both T and Vare constant dU TS S 0 7 De ne a new thermodynamic state function called Helmholtz energy A U TS dA S 0 constant T and V where dA 0 at equilibrium 7 If a system is held at constant T and V the Helmholtz energy decreases until all the possible spontaneous processes have occurred at which time the system will be in equilibrium andA will be a minimum 7 Isothermal change from one state to another AA AU TAS S 0 constant T and V 7 If AA gt 0 the process cannot take place spontaneously at constant T and Vand work should be done on the system to make the change process AA represents the compromise between the tendency of a system to decrease its energy and increase its entropy 7 If AU lt 0 and AS gt 0 both contribute to AA lt 0 AU more important at low T AS more important at high T 7 Example mixing of two gases at constant T and V AU 0 and A y1Rlny1 y2Rlny2 3 AA RTy11ny1 y21ny2lt 0 3 spontaneous process 7 Physical interpretation of the Helmholtz energy A AA AU TAS lt 0 for an irreversible process but consider the reversible path connecting the two states TAS qrev 3 AA AU qrev wrev isothermal reversible work CHEM 3520 Spring 2004 AA lt 0 wrev maximum work that can be obtained from the system work done by system in a reversible process 7 less if irreversible due to friction AA gt 0 wrev the work needed to be done on the system to produce the change reversibly 7 irreversibly the work required is even bigger The Gibbs energy 7 Criterion for spontaneous processes for systems at constant pressure and temperature dU S T dS PdV dU TdS PdV S 0 dU TS PV S 0 at constant Tand P 7 De ne a new thermodynamic state function called Helmholtz energy GU TSPVH TSAPV AG S 0 constant T and P where dG 0 at equilibrium 7 If a system is held at constant T and P the Gibbs energy decreases until all the possible spontaneous processes have occurred at which time the system will be in equilibrium and G will be a minimum 7 The Gibbs energy G is analog of the Helmholtz energyA for processes that take place at constant T and P 7 Isothermal change from one state to another AG AH TAS S 0 constant T and P 7 If AG gt 0 the process cannot take place spontaneously at constant T and P and work should be done on the system to make the change process AG represents the compromise between the tendency of a system to decrease its energy and increase its entropy 7 If AH lt 0 and AS gt 0 both contribute to AG lt 0 AU more important at low T AS more important at high T 7 Example NH3g HClg gt NH4Cls at 298 K AIH 71762kJ 3 ArG 9l21kJ 3 spontaneous process AIS 70285kJK 7 Example H20l gt H20g CHEM 3520 Spring 2004 AvapG AvapH TA S vap A H 4065kJmol vap 3Avap G Oldmol at 373 K and 1 atm AvapS 10891dKmol At 363 K AvapG 110kJmol 1 g not spontaneous At 383 K AvapG l081dmol 1 g spontaneous 7 Physical interpretation of the Gibbs energy G G U TS PV 3 dGdU TdS SdTPdVVdP But dU TdS w 3 dG 7SdT VdP wrev PdV rev But wrev PdV EWHOHPV work other than PVwork electrical work 3 dG 7SdT VdP wnoan 3 dG wnonPV reversible constant T and P 3 AG Wman reversible constant T and P AG lt 0 Wman maximum work exclusive of PVwork that can be done by system AG gt 0 Wman the work exclusive of PVthat can be done on the system 7 Example H2 1 02 g gt H20l AG 237 l kJmol AG is the maximum nonPVwork that can be given by system 7 Example H20l gt 02 g H2 g AG 237l kJmol AG is the minimum nonPVwork that can be put for the reaction to occur Maxwell relations and dependences of state functions 7 Not all thermodynamic functions can be measured directly so some quantities are expressed in terms of others that can be experimentally determined 7 The temperature and volume dependence of the Helmholtz energy AU TS 3 dAdU TdS SdT For a reversible process dU TdS PdV 3 dA PdV SdT 6A 6A But AAVT 3 dA dV dT 6V T 6T V 3 l nuoi spnng 2004 CHEM3520 43 EJ 4 8V T 37 VT AA7J39PdV at constant 7 K V dV erRTln VI 7 TAS V V K rIdeal gas AA RT also because AU 0 at constant Tforideal gas 7The Volume dependence ofthe entropy 82A 7 82A aVaT 7 876V 15ng But i E density at constant temperature a Maxwell relation 87 t n t t we V AS jdV at constant 7 37 Vl T choose V as a Volume when gas behave ideally STVr S d 1 ij Em 24545 Jmolquot Kquot for ethane at 400 K V Ideal gas E AS nRId V nlean 2 lsothennal process 37 V V V V l P77b RT gas hi Rln 1 V Tb TIhe Volume dependence ofthe energy TFohldeal gas Udepmds only on Tand not on V What about real gases7 J 4 r r r AUTS 7 8V 8V 8V E and 1 1 8V T 8V T 37 T CHEM 3520 Spnng 2004 epwf av T aT V emtegrate between avolume where there rs an xdeal behamor andthe eurrerrt volume V aP UTvrU d 77 7P v ateorrstantn w 3quotquot V rExamples Idealgas ePT ePP0 av T v 51775 RT gas Redhcherong gas p r t j 3A m rebate m 37 T 2T127 ray p g av 7 a7 P ar V aT P we we Idl CrC T R eagas P V VP 1 2 specny E a7 av T K 1 v where 527 L rstheeoemerentomermalexpansror v aT P g the rsothermal eompressrbrlrty v aP T rThe temperature and volume dependenee ofthe Gbbs energy G Ur 79 Pv m we SdTerSVdPPdV but 411 mu PdV m exam WP 8G ButGGTP dG d7 j dp aT P aP T CHEM 3520 Spnng 2004 E V and E S T 8T P e e thbs energy oleereases wth lncreaslng Tandlncreases wth lnereasln P eThe pressure olepenolenee ofthe entropy But aPaT BTBP 2 e E 2 a Maxwell relatron 3P 7 8739 P 7 Useful relataonshrp beeause allow deterrnrnatron ofthe ehange tn Swth P at eonstantternperature AS2r dP attonstantD 5 8T P 24 7 Choose P as apressure where gas behave ldeally p m sages I j dp tn pm 57 p F m id 246 45 J rnolquot Kquot for ethane at400 K o no mu mquot P Ideal gas 2 AS 21Pan lsotherrnal proeess 8T P P 7 Same as the olepenolenee wth respeet to V eThe pressure olepenolenee ofthe enthalpy eForroleal gas Holepenols only on Tand not on P What aboutreal gases7 Genem f r 2 yand 7 8P 7 31quot p 8P 8H av 2 i y 7 me 3P T 37 P T p a n 2 AH ver jdP at eonstant 7 8T e F to not no F mo 0 Mutual l on rum CHEM 3520 Spring 2004 Natural independent variables for various thermodynamic functions 7 U H S A G depends upon natural sets of variables that allow simple equations to connect these quantities 7 Energy Consider S and Vto be independent variables of U UUSV3dU a U dS6 U dV 6S V 6V S 7 Compare with dU TdS PdV 3 6 U Tand6 U P 6SV BVS 3 Natural variables of U are S and V 7 Criterion for spontaneous processes AU lt 0 dU lt 0 for systems at constant S and V Consider T and Vto be independent variables of U 3 dU 46 13 7P dVCVdT 6T V 7 Enthalpy Consider S and P to be independent variables of H HHSP3dHai dS6 dP 6S P 6P S 7 Compare with dH TdS VdP 3 32 Tand 31 V 6S 6PS 3 Natural variables of H are S and P 7 Criterion for spontaneous processes AH lt 0 dH lt 0 for systems at constant S and P Consider T and P to be independent variables of H 3dH V T a V dPCPdT 6T P l P 7 Entropy Rewrlte the above equatlon as dS dU dV 6S 1 6S P 3 and 6U V T 6V U T 3 Natural variables of S are U and V CHEM 3520 Spring 2004 7 Criterion for spontaneous processes AS gt 0 dS gt 0 for systems at constant U and V Rewrite the above equation as dS dH gdP 6S 1 6S V 3 and 6H P T 6P H T 3 Natural variables of S are H and P 7 Criterion for spontaneous processes AS gt 0 dS gt 0 for systems at constant H and P 7 Helmholtz energy dA SdT PdV 3 Sand P 6T V 6V T 3 Natural variables of A are T and V 7 Criterion for spontaneous processes AA lt 0 61A lt 0 for systems at constant T and V 7 Gibbs energy dG SdT VdP 3 E Sand E V 6TP 6PT 3 Natural variables of G are T and P 7 Criterion for spontaneous processes AG lt 0 dG lt 0 for systems at constant T and P 7 All the equation can be deduced quite easily dU TdS PdV Add dPV PdV VdP to both sides dU PV TdS PdVPdVVdP 3 dH TdS VdP Subtract dTS TdS SdT from both sides dH TSTdSVdP TdS SdT 3 dG 7SdT VdP Subtract dPV PdV VdP from both sides dG PV 7SdT VdP PdV VdP 3 511 7SdT PdV CHEM 3520 Spring 2004 Thermodynamic Energy Natural Variable Differential Expression Maxwell Relation U SandV dUTdS7PdV 6 T 6 13 6V S as V H SandP dHTdSVdP 6 T 6 V 6P S as P A T and V dA 7SdT 7PdV 6 13 6V 6T G TandP dG7SdTVdP g 6 V 6P 6T 7 Application the correction to the entropy due to nonideality behavior at 1 bar 7 The standard state for a gas is hypothetical ideal gas at one bar and if the gas does not behave ideally at 1 bar some corrections should be made Real gas Correction for Hypothetical ideal gas gt at 1 bar non39ideahty at 1 bar Use real gas Use ideal gas equation of state equation of state Real Ideal gas at very low pressure Pid lbar lbar 39 6V 6V SP1dS1bar J dP J dP So1barSP1d J39 56113 lbar 6T P Pid 6T P Pid P lbar 3 SO at 1 bar at 1 bar J Bid P dL M P P J where S O at 1 bar is the standard molar entropy of the gas hypothetical ideal gas and at 1 bar is the molar entropy from heat capacity data and heat of transition B T 7 Example l 2V P equation of state RT 2 6 V 5dBZV 6T P P dT dB 3 S atlbar Sat lbar 6172 X l bar P1d is neglected 7 For N2 the correction is 002 Jmol 1 K 1 CHEM 3520 Spring 2004 Temperature and pressure dependence of the Gibbs energy 7 The Gibbs 7Helmholtz equation 7 give the temperature dependence of G GH7TS 39 7S T T BGT H 1 6H 6S atconstantP 6T P T2 T 6T P 6TP Wig 0 T 6T P 6T P BGT H 3 2 6T P T2 The Gibbs 7Helmholtz equation BAG T g 6T P T 2 7 The temperature dependence of G can be determined directly 7 The case ofa gas that has only one solid phase Tfus Tvap T HT H0 ICISTdTAquH ICPTdTAvapH IC TdT 0 Tfus Tvap Tfus s Tvap 1 A H T Cg T sT j CPTdT AfusH j CPTdT Vap j P dT 0 T T Tm T T Twp T 2 Tm 170 17T 170 T T T7I70 G T is a continuous function of T AH because Ttrs AHSS 3 AHSG 0 E S so the slope is smaller for solid state 6T P than for liquid state than for gas state and is discontinuous in going from one phase to another T 7 The H T H 0 S T and G T H 0 are tabulated quantities CHEM 3520 Spring 2004 7 The pressure dependence of G P2 V 3 AG GP2T GPIT IVdP at constant T 6P T Pl Similarly lgl 17 WP Jr PZdP P P 7The ideal gas case AG RTJ RTln 2 AH 0 and AS Rln 2 Pl P P1 Pl 7 Usually P1 1 bar 3 C7TP G TRTlnP1 bar where GO T is the standard molar Gibbs energy the Gibbs energy of one mole of ideal gas at 1 bar and depends only on temperature 3 C7TP GO T increases logarithmically with P 7 For liquids and solids Vis almost constant 2 ACT 7P2T CP1T Ilzl7dP 17P2 7P1 P1 7 The nonideal gas case 1BZPTPB3PTP2 P P 6113 P P 2 j 516 RT j RTBZPT deRTB3PT deP Pid Pid P Pid Pid P RTB TP2 RTBZPTPLN Pld 2 Pid Po RTB TP2 PRTBZPPW Po 7TPCTPidRT1n But TPid G TRT1n 2 7TPG TRT1n 7 It is more convenient to rewrite this equation by de ning a new thermodynamic function called fugacity and denoted f P T 2 6TPG Tern where fPT P o exp32PTP B3PTP2 CHEM 3520 Spring 2004 GasPhase Reaction Dynamics 7 Present some of the current models that are used to describe the molecular aspects of bimolecular gasphase reactions 7 Introduce few new concepts such as reaction cross section impact parameter and potential energy surface The rate constant of a bimolecular gasphase reaction using hardsphere collision theory k 7 Consider a general bimolecular elementary gasphase reaction Ag Bg3 Products d A 7 The rate ofthis reaction v kAB 7 As a rst approximation use hardsphere collision theory to estimate the rate constant k V ZAB O39AB lt r gt PAPB where 039 AB is the hardsphere collision cross section of A and B molecules lt MI gt is the average relative speed ofa colliding pair ofA and B p A pB are the number densities of A and B molecules ZAB has the units of collisions m 3 s 1 or n f3 s 1 Z AB 3 k O39AB lt MI gt k has the units of that is molecules 1 m3 sil 7 To transform to more common units of mol l dm3 s 1 multiply by N A and 1000 dm3 m 3 2 k 1000 dm3 m 3NAo AB lt ur gt 7 Example Calculate k in mol 1 dm3 sil using hardsphere collision theory for H2g C2H4g 3 C2H6g reaction 270 pm 430 pm 2 J 38510 19 m2 2 O39AB diB n39 12 723 1 8kBT 8xl38110 JK x298K ltur gt m 72y 7rgtlt31210 kg 2 k 1000 dm3 m 360221023 moI 138510 19 1112183103 ms 1 12 J 183103ms 1 3 k 4241011 dm3molilsi1 experimental k 34910 26 dm3molilsil The rate constant of a bimolecular gasphase reaction and the reaction cross section 7 Hardsphere collision theory predicts a temperature dependence of k as TlZ CHEM 3520 Spring 2004 7 Hardsphere collision theory predicts that all the molecules collide while having a relative speed of lt MI gt 7 Improve the theory by assuming that reaction occur only the relative speed is sufficient to overcome the repulsive forces 7 Replace the collision cross section 039 AB by a reaction cross section 0r ur that depends on the relative speed of the reactants 3 kur urar ur 7 Calculate the observed rate constant by averaging over all possible collision speeds 2 k Ikurfurdur lure urwadu 0 0 7 The f ur is the distribution of relative speeds in the gas ur f urdur is given by kinetic theory of gases 3 2 2 12 2 ur uadu i uSe W quotBTdu kBT 7239 7 Transform the dependent variable from relative speed ur to the relative kinetic energy Er 12 12 2E Erlmr2ur r 3dur dEr 2 u 2uEr 2 32 1 12 3 urfur dur EreiEikBTdEr B 3 2 12 00 7 Substituting in the expression for k k L L IEIe EMkBToKEr dEr kBT 7 0 7 A simple model for 0Er is to assume that only collisions with higher relative kinetic 0 Er lt E 0 energy than a threshold energy E0 are reactive 0r Er dz E gt E AB r 0 32 12 00 E 3 k i L IEre EMkBTmgBaEr lt MI gt aABe EOkBT 1 0 kBT Lm39 E kBT 0 7 Same result as obtained using collision of hard spheres whose kinetic energy exceed a threshold energy but here this dependence is included into the expression through the reaction cross section 3 explore different models of Ur Er that will give different values for the rate constant CHEM 3520 Spring 2004 7 Example For the H2 g C2H4g 3 C2H6 g reaction using the hardsphere collision theory k 4241011 dm3molils71 while experimental value is k 349 10 26 dm3molils71 What is the value of E0 that give the experimental value of k e EokBT 1E 0 3394910726 3 E0 223 kJmol kBT 4241011 Experimental activation energy E a 180 ldmol The lineof centers model and the impact parameter 7 Just a simple energydependent reaction cross section is unrealistic the two collision below occur at same relative collision energy but the one depicted in the right side is less likely to generate products gt 7007 lt versus 7 A more realistic model for the reaction cross section is one that considers the component of the relative kinetic energy along the line connecting the centers of colliding molecules This model is called line of centers model of 039I Er m l M quotEv 7 Denote the kinetic energy along the line of centers by E10c and assume that the reaction occurs when E10c gt E0 7 Look for the expression for Ur Er in this model 7 Relative velocity 11r uA uB relative kinetic energy Er 12Wr2 7 De ne the impact parameter b as the perpendicular distance between the lines of the trajectories before collision 7 Collision occur if b lt rA rB dAB molecules will miss each other if b gt dAB Cross section 7 The kinetic energy along the line of centers depends on the impact parameter when b 0 3 E100 Er when b gt dAB then O39rEr 0 7 Finally after the derivation of the expression of the reaction cross section 0 Er lt E0 E afar nd B 0 Er 2 E0 Er Collision energy CHEM 3520 Spring 2004 7 Substituting in the expression for k gt k Sk BTjngBe EnW lt ur gt UABe E k r 71 7 Correlation With Arrhenius parameters Ea kBTZ E0 kBT A lt ur gt UABelZ 7 The values ofA calculated using this model are bigger than the experimental ones the reaction cross sections determined experimentally present athreshold energy but a different shape compared to the calculated values base on the above model gt the simple hard sphere picture is not accumte Other factors in uencing the reactivity of molecular collisions 7 Depending on the shape of the reactants the rate constant for a gasphase reaction depends on the orientations of the colliding molecules 7 Example Rbg CH 3Ig gt Rblg CH 3 g reaction occurs only When the rubidium atom collides With iodomethane in the vicinity of the iodine atom collisions With the methyl end are nonreactive gt cone of nonreactfviry 7 The reaction cross section depends on the internal energy of the reactants 7 Example H g Heg gt HeHg Hg The reaction cross section at the same total energy depends on the vibmtional state in Which HE is For v 2 4 v is the vibrational quantum number there is no threshold energy Em gt E0 so no additional translational energy is needed gt Chemical reactivity depends not only on the total energy but also on its distribution among the internal energy levels Reactive collisions in a centerof mass coordinate system 7 Consider the collision and the scattering process for a bimolecular reaction Ag Bg gt Cg Dg 7 Look at the collision from the center of mass of the tWo colliding molecules CHEM 3520 Spring 2004 7 The center of mass lies along the vector 1 rA rB that connects the centers of the two colliding molecules and the location of the center of mass R depends on the masses of the two molecules R mArA mBrB mArA mBrB m A mB 7 The velocity of the center of mass is the time derivative of the position m All A mBuB T 110m 7 The total kinetic energy of the reactants is the sum of the kinetic energy of the reactants and can be rewritten in terms of the reduced mass u mAmB M the relative speed of the two molecules ur lurl luA uB and the speed ofthe center ofmass ucm KEreact mAui mBu Mu02m Wr2 Also uA 11cm nI Bur and HR 11cm rXI Aur 7 The kinetic energy is a sum of two contributions one due to the center of mass and one due to the relative motion of colliding molecules 7 The kinetic energy of the center of mass is constant if there are no external forces acting on reactant molecules 3 molecules remain in the plane traveling at the speed of the center of mass during the collision Only the relative component of the velocity is important in determining the energy available for the reaction 7 After the collision the centerofmass position and the centerofmass velocity are given by mCI C mD1 D I u mCuC lelD cm R 7 Similar to the reactants the total kinetic energy of the products u39 and u39 are the reduced mass and the relative speed of product molecules 1 2 1 2 1 2 1 I 2 KEprOd 7mCuC 7mDuD 7Mucm 72 r m m Also uC ucm Du and HD ucm Cu39r M M 7 During the collision the mass the linear momentum the total energy must be conserved mA ME mC mD mAuA mBuB mCuC mDuD a 11cm constant 1 2 1 I 2 Ereactint 7 WI Eprodint 7 I r where Emath and Eprodint are the total internal energies of the reactants and products includes all the degrees of freedom except translation CHEM 3520 Spring 2004 7 The conservation of the momentum and energy does not give information about the angles between the vectors 11r and 11 Many reactions show anisotropic scattering angles 7 Example Fg D2 g 3 DFg Dg reaction where the relative kinetic energy of reactants is KE react 762 kJmol Treat reactants and products as hard spheres 3 Eprodim Ereacum De D2 De DF l40 kJmol Determine relative speed ofthe products and the speed of each product relative to the center of mass m m u L 55210 27 kg and u39 m 30510 27 kg mD2 m1 mm mD 12 2KE ur reaaj 214103 ms u 39 r 127104 ms u 12 2E E 3 u iuz prod1nt 39 reactant J mD I mD 39 3 3 luDF 11ch Vlurl Vur 11110 ms 3 luD ucmlulu 116104 ms Reactions can produce vibrationally excited product molecules 7 Investigate Fg D2 g 3 DFg Dg 50 v 5 reaction and look at the potential energy A V 0 along reaction coordinate potential energy E T diagram Also shown are the harmonic in 50 energy levels of D2 and DF E 7 Considering also the vibrational levels 100 Fg D2 V 0 3 DFV Dg 150 DF D 7 Conservation of energy condition Reaction coordinate I 39 Etot Etrans Eint Etrans Eint Etot Etrans Erot Evib Eelec Etrans Eiot Evib Eelec Eelec DeD2 7 Numerical example 39 elec De DF and De D2 De DF l40 ldmol Consider Emms 762 kJmol Erot Egot 0 Evib vhuD2 179 kJmol Etrans Etrans Erot Evib Eelec iot Evib Eelec 166 kJm01 vib Em v 9h uDF v 9038 kJmol lt 166 kJmol 2 v s 4 CHEM 3520 Spring 2004 Reactive collisions are investigated using crossed molecular beam method 7 The molecular beams are used to produce the velocities ofmolecules in the reactant beams The collimated beam ofmolecules has a very narrow distribution in molecular speeds With loW rotational and vibrational energies 7 The product molecules are detected using a mass spectrometer that can also rotate in the plane ofthe two molecular beams 7 The energydependent reaction cross section 17 EY can be determined by measuring the product yield as a function ofcollision energy 7 Measuring the number ofproduct molecules as a function of time one can resolve the velocity distribution of the product molecules Measuring the total number of product molecules of product as a function of scattering angle one can resolve the angular distribution of the product molecules The velocity and the angular distribution of the producw of a reactive collision 7 Investigate F D2 v 0 3 DFv D reaction at a relative translational energy of Eums 762 kJmol 7 Look at a constant value ofimpact parameter B is spherical so the scattering center is cylindrically symmetric to molecule A No of The angle takes all possible values With equal probability but molecules MAW not same for 9 damsel Time to reach detector 7 Translational kinetic energy the time to get to the detector depends on the internal vibrational energy of DF The translational energies of DF and D are not independent and are related by conservation laWs 7 The dependence of reaction product formation on the scattering angle 0 is represented in a twodimensional polar contour plot 7 The center of mass is at the center of contour plot 7 The distance from the origin to any point is the speed ofDF relative to the center ofmass uDF 7 ucm W7 The horizontal axis lies along the relative velocity vector of the reactants 9 0 When DF travels in the same direction as incident F atom B 180 When DF travels in the opposite direction to incident F atom 2m r kf CHEM 3520 Spring 2004 7 The arrows below the plot indicate the directions with which reactants approach each other 7 The contours represent a constant number of DF molecules 7 The dashed circles correspond to the maximum relative speed allowed for a product molecule in a given vibrational state 7 The region between dashed circles corresponds to rotationally excited products rotational excited state can be calculated based on point location and the rotational constant of DF 7 The relative population on the vibrational states is not described by Boltzmann distribution 7 Because the product preferentially scatters backwards t9 180 this type of reaction is called rebound reaction The velocity and the angular distribution of the products of a reactive collision 7 Example Kg 12 g gt KIg Ig reaction with a relative translational energy of Etrans 1513 kJmol 7 For this reaction the product K1 is preferentially scattered in the forward direction same direction as reactant K atom This type of reaction is called stripping reaction 17 V 7 The reaction cross section for this reaction is 125 X106pm2 m 7 The hardsphere collision cross section is 65 gtlt105 pm2 y W 7 The trajectories of reactants are drawn together by the longrange potential stronger than van der Waals interactions 7 The rst step takes place when reactants are still separated and produces a pair Jiqu 7quot of ions that attract each other strongly Kg 12g gt K4r g I g This K 7 i l mechanism is called the harpoon mechanism 7 Example Og Brz g 3 BrOg Brg reaction with a relative translational energy of Etrans 1255 kJmol 7 The contour indicates the relative number of Br0 molecules observed 7 The product molecule BrO is forward and backward scatter with equal intensity 7 The reacting molecule should forget the original collision geometry m39 7 The collision form a atommolecule complex whose lifetime is long compared with its rotational period 7 The long lifetime allow the complex to rotate many times before generating u a k 5 products so the angular distribution of the products became independent of their initial collision geometry CHEM 3520 Spring 2004 Potential energy surfaces 7 An important property in discussing beam results and calculating reaction cross sections 7 According to the BomOppenheimer approximation the wavefunction of a system composed of N nuclei and n electrons is written as a product of nuclear wavefunction depending on the positions of the nuclei and an electronic wavefunction depending on the positions of the electrons within a xed nuclear con guration This allows solving the Schrodinger equation for the wavefunction of the electrons alone at a speci c nuclear con guration Modifying the nuclear con guration and solving again the Schrodinger equation one get a different energy Energy 7 The case of diatomic molecules representing the electronic energy versus the interatomic distance one obtains a potential energy curve 7 The case of triatomic molecules there are three geometric parameters that de ne the molecular geometry and representing all three of them plus the energy requires a 4dimensional representation 7 Example the geometry of water is de ned by bond lengths and one angle or A 3 bond lengths or 1 bond length and 2 bond angles or 3 bond angles HA 0 HB 7 The minimum number of geometric parameters necessary to de ne the geometry of a system ofN atoms N Z 3 is 3N 6 7 When the potential energy depends on more than one single geometric parameter then use the phrase potential energy surface 7 The entire potential energy function cannot be plotted because the plotting is limited to 3 dimensions To overcome this one usually represents the energy function of only 2 geometric parameters keeping the otherothers at a xed value Such a plot is a cross sectional cut of the full potential energy surface 7 Example water 7 a 3dimensional plot VrOHA rOHB a constant versus rOHA and rOHB gives information about how the potential energy of water molecule changes when the bond lengths are varied at a constant angle a A series of crosssectional plots at different values of a give information of how the potential energy depends on the angle a 7 The representation of the potential energy surface for the molecules with more than 3 atoms is much more complicated CHEM 252 55 2mm 7 5 5u55555 m cm r m mm 5mm mum 29m 51m rmwmw 515 55 pm 5 wwwm mm m a m smebemenDAandDB m1 mm 31 mm m5 m mtnmldear msmes 51M calhsmn anglz r r 7 5 l5 5 7 upquot 5 555 55mmquot r5 5 quot5 m r r 5 any 5 5 315 Mm 555 5 5xp5nm5uya55m5 7 55 1 555555 m 555qu 5m 55 afanexgyxs 25my5555 5 5 mm m 555 7w 55 5 mm 555 m 55 55mmquot 5 5 5 M 92 m5 m 5 my a 5 5 mm mm 5 m 555 maleculz whznthz mmmm emxgypnth mam path mm pmths Th2 mm pm 15 1 7 Th Hananm 52m 5 hypxsnrface 555 1 reachm xegmn mm m mam xegmn andxtcamnms m 5 pm 1854 CHRISTOPHER A PARR AND DONALD G TRUHLAR 1 l quotI l m l IIIIIJIIIIIIMMII mm N a L l quotiiilllllllll l ill l39lgllmk III I quot Miii lli quotlilliiilllllll liviiiiie39lilii 39 liiilllllllll ng 59 may Figure 10 Stereo View of Persky and Klein s potential surface II for collinear H H01 gt 110 H 0 64 Left dz 136 right qb 140 I I I I I I W V in s x h W 39l N W Na mil u I ll I 3 amp 1 quotI l I 47 M il amp v I I 39 42quot s l 391 ill 3 5 739 l Figure 12 Two Views of Timmons and Weston s potential surface for collinear H HBr gt HBr H Left 0 81 45 81 right 0 64 140 g 3 iqIIInIIII 11 It I N 39o C llllliii I u QR gliau 39I II I i1 49 II II Q quot 4 i ll39 V Figure 14 Two Views of Sullivan s potential surface for collinear H H1 gt HI H Left 0 67 1 100 Tight 0 64 140 The Journal of Physical Chemistry Vol 7395 No 19 1971 CHEM 3520 Spring 2004 The Third Law of Thermodynamics The entropy dependence on temperature pressure and volume dU EqTev wTev EqTev TdS wTev PdV 3 dU TdS PdV 3 dSidU dV T T UVT dUVT6U dT6U dV 6T V 6V T dSi a U alTi P a U dV T 6T V T 6V T SVT 2 dSVT dT j dV V T 6T 6V 3 2i 12 LU 6V T T 6V T as 1 6U CV 3 aT V T 6T V T T2 7 At constant volume AS ST2 ST1 wag T1 Similarly dH dU PdVVdP TdSVdP HPT 2 dHPT6 dT j 6113 P T 6T 6P 3 g CPltTgt 6T P T 7 JT JTVl H C T 7 At constant pressure AS ST2 ST1 63 T1 T 7If T1 0 2 ST S0KjdT 0 SmOK 7 Determine S at any temperature from C P T from 0 to temperature of 1nterest CHEM 3520 Spring 2004 The Third Law of Thermodynamics 7 Gives the entropy at 0 K The entropy of a perfect crystal is zero at 0 K 7 History 7 Nernst postulated that AIS gt 0 as T gt 0 3 all pure crystalline substances have the same entropy at 0 K 7 Planck postulated that the entropy of a pure substance approaches zero at 0 K 7 The statement of the Third Law of Thermodynamics Every substance has a nite positive entropy but at zero Kelvin the entropy may become zero and does so in the case of a perfectly crystalline substance 7 the Third Law provides a numerical scale for entropy 7 The Third Law of Thermodynamics also states that C P gt 0 as T gt 0 7 The Third Law was formulated before quantum mechanics can be understood better in terms of molecular quantum states S kB an where W is the number of ways the total energy of a system can be distributed over its various 7 At 0 K the system will be in its lowest state 3 W l 3 S 0 S kBZpJlnpj J where pj is the probability of nding the system in the jLh quantum state with energy E J 7 At 0 K system will be in ground state 3 p0 l pJio 0 3 S 0 But if ground state has a degeneracy of n 3 n quantum states with energy E0 each with a probability of l n quot l l 3 S0K kBZ ln kB lnn F1 n n The Determination of Absolute Entropies T C T 7 The entropy at one temperature ST S0K IPTdT 0 T But S00 2 ST aw 0 CHEM 3520 Spring 2004 7 For a phase transition AnsS 4 quot T115 7 A phase transition is a good example of reversible process at constant T 3 1er 4p AH A H 5 A ms quot 5 T115 7 Example Fusion and vaporization of H20 601kJmol 5 K A H601kJmolgtA S 22JK39lmol39l f f 2731 Ava F 407 kJmol Ava s W109JK39lmol39l P P 37315 K 7 To calculate the entropy of a gaseous substance one add entropy increase during heating and during phase transition Tfuss Tvapl A H TCgTdT Sm IMM IL 0 T Tfus Tm T Tvap TWP T where ngl395 T are heat capacities for solid liquid gas 7 Heat capacities at low temperatures 7 Experimental determination C15D T 0c T3 as T 7gt 0 for nonmetallic CRT 0c aT bT3 as T 7gt 0 for metallic 7 The T3 temperature dependence is valid from 0 to 715 K and is called Debye T 3 laW SoHd l 7 3 1274 T s m 7 Debye theory CPT 5 REE 0 lt T g Tlow Where 8D is called Debye temperature and is characteristic ms T 7 Relationship between entropy and heat capacity T Icpg CP T 0 5130410 339 JIC1 mol391 for c12 113 JK39lmol39l atl4K CHEM 3520 Spnng 2004 e Deterrnrnatron of Praetreal Absolute Entropres tabulated values 7 Calonmemcally from ealonrnetne data 7 glyen heat eapaerty data enthalpres oftranslnon and transrtron ternperatures one ean deterrnrne the absolute entropy eBeeause sueh entroples are ealeulated based on the fact that no K 0 they are calledthlrdrlaw entropres or praetreal absolute entropres eErarnple standard rnolar entropy of N2 at 298 15 K J Kquot rnolquot m 10 K 2 0 V 10 gt 35 61 K 25 79 s transrtron o 43 f lnoe 3561a6315 2341 lon ll 20 W3 6315a77 36 1146 o yaponzatron 72 00 0 mu m m w 77 3o gt 29815 39 25 1 nonrdealrty eorreetron 0 02 Total 191 o e The lncrease m entropy for yaponzatron ls rnueh blgger than for fuslon e Uslng partrtron funetrons S kEankETaan NV 8T where QNVT ze ii 39m y est ZEJE 1 r 75ka ly Skalnzss I 47 J 2 s eThe T gt 0 hrnrt Assurnlng n states wrth same energy E 52 5 ground state ls nrfold degenerate then m states wrth hlgher energy Em E Zaram quotSeam mee5lkpr gritmoi Weed mm CHEM 3520 Spring 2004 7E k T 3Ze B aneElkBTasT gt0 j 1nEe SkBlnne ElkBT 1 3 T quot67131 kBT 7E1 kBT E E k lnn 1 1k lnn B T T B 3 as T gt 0 S is proportional to the logarithm of the degeneracy of ground state This means it is negligible even if n gt N A N 7 But QNVT W for ideal gas Assumption indistinguishable particles for which the available number of states is much greater than the number of molecules Blnq 6T 3 S NkB ln q kB lnNNkBT using Stirling formula lnN NlnN N 2 SNkB NkBln 9701 NkBT m N 6T V 7 Example diatomic ideal gas example N2 g at 29815 K 32 7 2T 272ilk T b qVT ZB Vie leDekBT h 1n 27szBTJ32 1752 In Te R h2 NA 219rot 79 T t9vibT ln e b gemT lngel 19m 288K vib 3374K gel 1 go gtians groot ib gglec 3 15044113115gtlt10 30 1915 JK 1 mol 1 7 very close to 1916 J K71 mol 1 based on calorimetric data 7 The accepted literature values are often a combination of statistical thermodynamics and calorimetric data CHEM 3520 Spring 2004 The dependence of the entropy upon physical state molecular mass and molecular structure 7 the standard molar entropy is higher for gases than liquids than solids because solids are more ordered than liquids and gases 7 within a series of similar molecules or atoms the standard molar entropy increases with the increasing of molecular mass 7 Examples noble gases hydrogen halides halogen molecules 7 Explanation An increase in mass leads to an increase in thermal disorder more translational energy levels are available and a greater entropy h2n2 7 Particle in a box onedimensional energy En 2 8Ma 7 the more atoms of a given type in a molecule the greater is the capacity of the molecule to take up energy more different ways in which the molecule can vibrate the greater the entropy 7 Example C2H2lt C2H4lt C2H6 7 for molecules with the same geometry and number of atoms the standard molar entropy increases with increasing molecular mass 7 Example CH3C1 lt CH2C12 lt CHC13 CH4lt CCI4 7 for molecules with approximately the same molecular masses the more compact the molecule is the smaller its entropy 7 Example acetone vs trimethyleneoxide O O 7 free rotation along CC bond more movement 3 more entropy 7 ring is rigid less movement 3 less entropy How to get standard molar entropy for Br2 g at 29815 K and 1 bar even though Br2 is liquid under these conditions Calculate Brz g using the circle 98 Br2 098 BIZ 1 11 111 Brza 332139 Brag T ASI S13320 S129815C 111T 2 l 111 T AS111 Sg29815 Sg3320 C 111T 1 2 29815 K A 17 Bug ASH gap Sg 3320 S13320 v p Brz CHEM 3520 Spring 2004 The Properties of Gases Ideal Gases 7 the case of any gas in nitely suf ciently dilute so one can ignore the interactions between molecules Sum 7 v lsmm mum 7 Ideal perfect gas equation of state PVnRT or P17RT quot V 1 mm 7 The ideal gas constant R 83145 JmolK R 0083145 LbarmolK R 0082058 L atmmol K Pressma a Pressure 9 0 7 Perfect gas law is obtained by combining quot191 7 Boyle s Law PV ct 7 GayLussac Law V ctx t 27315 or V cth 7 Charles Law P cth c gt a g u E 3 g 3 g u g lncreaslng g 3 temperature T 9 a Increasmg temperamm T E rrauarau un E lvapolanun E Lrapolaimn 0 a u u 0 Volume v 0 IN 7273 Temperature we 0 Temperature WK 7 Ifa gas obeys the ideal gas equation it is called an ideal gas or the gas is said to behave ideally 7 Combined gas equation T1 T 2 7 Dalton s law ofpartial pressures each gas in a mixture ofideal gases acts as ifthe gases were not present anT n Ptotal ZPJ Where P V total ythotal quotJ CHEM 3520 Spring 2004 Real Gases 7 To determine if a gas behaves ideally one could plot denoted Z and called compressibility factor versus P Z 1 for ideal gas Z 1 for real gases at low pressures Z 1 for real gases at high pressure Z gt 1 repulsion forces 17ml gt ideal Z lt 1 attraction forces 17ml lt ideal 2m x 7 The ideal gas equation does not account for the intermolecular interactions m 7 As the temperature increases the real gas I 5 behave more like the ideal gas M 0 200 011 00 800 INN PI Iul 7 Introduce another very popular equation of state to account for intermolecular interactions This is the van der Waals equation a P 7 21V 7b RT Where a and b are called van der Waals constants b Lmol re ects the size ofthe molecules a L2 barmol2 re ects how strongly the molecules interact Pressure a L2 barmolz b Lmol H2 024646 0026665 CH4 23026 0043067 N2 13661 0038577 C6H6 18876 011974 7 Rewrite the equation as P 77i2 I7 a 7 The compressibility factor Z 7 7 RT V 7b RT V CHEM 3520 Spnng 2004 eMolarvolurne rs obtatned by solvrng the eubre equatron W 7b 72 1771 0 P P P e Other equataon ofstates rnore accurate1ncludng a1so only two pararneters dhcherong equatron e PengrRobmson equataon RT p 4 7 V78 V7B 78 5 to at mal Lquot elnvestagate sorne enpenrnenta1 plots of Pversus V at eonstant Tea11ed rsotherrns eThe ease of co2 e Cnncal pornt e Cntreal ternperature Tc rs the ternperature above whreh agas eannot be Tc 304 14 K 30 99 quotc hque edregardless ofthe pressure Z 0 094 Lrno1 eThe conespondmg pressure and molarvolume at the entrea1 pornt are the entrea1 pressure PE and the errtaeal volume Z 372 CHEM 3520 Sprlng 2004 Wth eaeh other The olasheol euwe eonneeung the enols of honzontal hnes ls ealleolthe coexlstence eurve o the lelt there ls only llquld present on the nght there ls only gas on the honzontal hne there are both gas andllquld e The hne beeornes a polnt an ln eeuon polnt at 7C There the rnenlseus apor dlsappears and there ls no dlstlnetlon between ul39face tenslon dlsappears and the gas both have the sarne olenslty ealleol enueal olenslty between llquld andlts v gas andllquld phases eEquataons ofstate that ean be wntten as euble equataons m 7 van olerWaals Redllcherong ean olesenbe both the gaseous and the llquldreglons ofa substanee eThese equatlons show spunous loops at T lt7c eThe DA hne ls olrawn so that the DCE area and EBA area are equal ealleol Maxwellrequalrarea eonstruetlon eThe Gas 7A eornpresslon othe gas eThe AeEeD hne llquld and gas are ln equlllbnum e The Dequulol ehange ofvapor of h quld wlth lnereaslng pressure eThe ArB segrnent rnetastable reglon correspondlng to superheated vapor eThe C7D segrnent rnetastable reglon correspondlng to supercooledllquld eThe BeEec segrnent T gt 0 unstable reglon not observedfor equlllbnum systems eAeEeD hne has 3 solutaons hlgher one polnt A comespondmg to gas lower one pomtD correspondlng to hqulol and the mlddle one pomt E spurlous CHEM 3520 Spring 2004 7 Densities molL of CO 2 experimental vs calculated at 142 69 K and 3 50 atm Experimental van der Walls RedlichKwong PengRobinson p 22491 1414 2013 2361 pg 5241 4615 5147 5564 7 At critical point in ection point 2 0 and 0 T W T 7 Use these conditions to nd the critical constants in terms of van der Waals constants a and b 7 van der Waals equation 73 b 172 2711 0 P P P 7 AtT To there is only one solution of the equation 7 V 3 0 173 4170172 3I7CZI7 I7c3 0 31703bP a2 8quot 27 27bR c 7 Similar but mathematically more complicated one obtains for Redlich Kwong equation 23 13 23 70 384733 Pc 0029894A R T0 034504 1 353 BR 7 Law of Corresponding States PCVL RT 0 7 The values of are independent of the gas P I7 7van der Waals iL 27bR 2 0375 RTc R 27132 8a 8 P I7 7RedlichKwong c c 03333 RTc P075 7 PengRoblnson 03074 RTc P VC 026 030 RTc 7 Experimentally Spring 2004 CHEM 3520 5 The properties of all gases are the same if we compare them under the same conditions relative to their critical point Law of Corresponding States 7 van der Waals equation in terms of reduced parameters 27 RT 2 RT 7 Substituting a and b 64Pc 8P 7 Divide by Pc and 7 use PJ7c RTC and rearrange 5373 Z RT RT AL P V2 7 3 PJZ m 3r L T R To ll VI 7 Introduce reduced quantities PR 1 7R c gt PR VR 7lj TR VR2 3 3 7 This equation does not have any parameters that are gas dependent it is a universal equation for all gases 7 Different gases at the same values ofreduced parameters same PR7RTR behave the same are and are said to be at corresponding states This is an example of the law of corresponding states All gases have the same properties if they are compared at corresponding conditions same values of PR7RTR 7Example CO2 at P 143atm 719 Lmol T 456 K and N2 at P 658 atm 7 l 8 Lmol T 189 K are equivalent because they have the same reducedparameters PR 0196 7R 20 TR 15 7 Expression of compressibility factor in terms of reduced parameters 5 Z VR 7 9 universal equation VR 7 VRTR 7 Z can be expressed as a function of any two reduced parameters VR andTR PR andTR PR and VR H 7 Book has an example of the law of corresponding we r states Z vs PR at various TR should look same quota for any gas M 7 Everything is relative to critical point M CHEM 3520 Spnng 2004 eThe vlnal equauoh of state elt ls the rhostfuholarhehtal equatroh ofstate a ap l u ruralul are ealleol vlnal expanslons P7 1 l 2 z 2 1 3 3 RT arm V 3mm where the eoemerehts are ealleol vlnal eoemerehts 32V ls the seeohol vlnal eoemereht 33V ls the thrd vlnal eoemereht eAhothervrhal equatroh of state ls wrrtteh by expandmg wrth respeetto P 2 l32pmp33pmpz RT eThe vlnal eoemerehts m the two equauohs are relateol 3W0 Rmzpm re eet the rst devrauoh from rolealrty as the pressure thereases eThe 32pm ear be oleterrruheol from the slope Zversus P rThe gym ls hegatrve at low temperatures theh beeorhes L posmve wth a shallow maxrrhurh eThe 3mm 0 atBoyle temperature the repulslve and x m attraeuve parts othe rhterrholeeularrhteraetrohs eaueel eaeh other 2 gas appears to behave ldeally E 1 rhteraeuohs e Conslderrto be the dlstance between the eehters oftwo rholeeules eAssurhe that all posslble ohehtauohs average out 7 Conslder ho to be the potehtral ehergy rhteraetroh energy oftwo rholeeules at dlstance t 2 3mm 227 e 37 2112 u U0 0 2 32V 73 0 ldeal behavlor ho rhteraetroh Spnng 2004 CHZEM 3 520 r Lemmerones potential 7 An amacuve component 512 r A repulsrve componmt 7 7 512 Comlnmng me two no 7 4r l2 5 l2 c 450 r Rearrangmg no 4s 3 r 3 where 2 6 0 r r 6 40 3 quot a Paramztcrs Ens a measure ofhow strongly me molecules amact a 15 me depth omne well 7 Is a measure omne srze omne molecules a where no so r Subsumung Lemeroms potential m me expressron for 32 B T 2rNTex 3 Ef gy 71 2dr 2quot 7 A0 P kET r r Reorgamze use Tquot M x L s a r Drvrde by 2ra3NA 3 7 Introduce reduced second vmal coefficient depends only on T 8230 82VT 2zra3NA3 82quot 7quot 73J39exprixquot2 r x 571x2 lt T o i solved nurnencally exlsts m tables rPlol of 83 vs Tquot for dclrerenl gases grves only one curve uner ammple omne law ofcorrespondcng states tally and from merr values 7 The 32 are determined expenmen me Lemmerones parametns are deterrnmzd CHEM 3520 Spring 2004 7 Hardsphere potential 7 The attractive part of the potential is missing 00 rlt039 ur 0 rgt039 3 3 BZV T 27m NA 4 times the volume of NA spheres 7 This potential predicts BZV to be independent of T but experimentally the molecules are not really hard 7 Squarewell potential ur 8 0 rlt039 altrltla rgt ta 3 2 BZVT 1 13 AXEW 4 7 Give good agreement with experiment for BZV vs T 7 Combination of hardsphere and LennardJones potentials 00 r lt 039 0 Cs VgtO39 r 27ra3N 27rN 6 32V T A Ag 3 3kBT039 7 Interpretation of BZV T 7 neglecting the higher order terms of the virial expansion 1 2 1 w RT RT 3 BZVT V l7ideal P 7 Van der Waals constants in terms of molecular parameters 7 Rewrite van der Waals equation as RT 1 a RT 1 b2 a 1 T l V 1 bV V2 V 1 2234 b L RT RTV b2 m CHEM 3520 Spring 2004 7 Comparing with virial equation of state a B T b 2V RT 7 Comparing with the expression of BZV obtained for a combination of hard sphere and LennardJones potentials 27ra3N 27rN 6 32V T A 3 3kBTo 3 2 27z39N 3 a 1 c6 3039 a is proportional to 06 is a measure of the interaction between molecules 3 272039 N 3 b 2 TA b is four times the volume of N A spheres 7 From molecular point of view the van der Waals equation is based on intermolecular potential that is hardspheres potential at small distances and a weak attraction potential at larger distances Intermolecular Interactions 7 are the source of the deviations from the ideal gas behavior 7 The form of the r712 repulsion term is not well established 7 Intermolecular forces responsible for the attractive term r76 term in LennardJones potential 7 Interaction between the dipole moments 1 and 2 of polar molecules dipole dipole interaction 7 depends on orientation 7 overall average interaction potential 151510 212 i 4723980 2 3kBT r6 7 Interaction between a dipole moment and an induced dipole dipoleinduced dipole interaction CHEM 3520 Spring 2007 Unit 11 Statistical Thermodynamics A Concepts of Statistical Thermodynamics 1 Introduction a Statistical thermodynamics is the link between molecular properties ie molecular energy levels and the bulk thermodynamic properties ie properties of matter in bulk which deals with the average behavior of a large number of molecules 2 The Boltzmann distribution a Consider a macroscopic system that can be described by specifying El the number of particles El the volume of the system El the forces between particles In principle the energy of a Nbody system could be obtained by solving the Schrodinger equation CI The energy will be written as EJNV El It depends on the number of particles and the volume El j is a indeX of the various states of the system For the case of an ideal gas the total energy is a sum of energies of individual molecules EjNV 8182 8N CI The energy of individual molecules is then a sum of electronic vibrational rotational and translational El For a more general case when the molecules of the system interact with each other the energy EJNV cannot be written as a sum of individual particles energy CHEM 3520 Spring 2007 El We will though still consider for the purpose of discussion the energy of the system as a sum of individual molecular energies d Deducing the Boltzmann distribution El Consider a collection of systems in thermal equilibrium with each other 0 This collection of macroscopic systems in thermal equilibrium with a heat reservoir is called 0 More specifically if N V and T are common the ensemble is called El Determine the probability that a system will be in the state j having energy EJN V 7 The number of systems in state j is a and the total number of systems is A 7 Determine the relative number of systems found in each state 7 Consider 2 states with energies E1 and E2 fE1E2fE1 E2 where fE1Ez is a functional that should be determined and the difference E1 7 E2 is used so that any arbitrary zero of energy will be canceled 7 Consider a third state fE2 E1 a3 a3 3EfE2 E1fE1 E3 f E1 E3 a 3 fE2 E3 612 7 This is similar to exy exey 3 fE e E 3 an e Em En am 3 611 Cei gEj 7 Two constants C and should be determined 7 Determine C by summing over all the number of systems 26 Cze EJ A 3C L Ej J J 2 6 J CHEM 3520 Spring 2007 a 7 De ne 7 as the fraction of systems in the ensemble that will be found in the state j with energy a T j J e A ee Ej pf cm CI The ajA fraction becomes if number of systems is very large the probability of a system to be in the state j with energy EJNV denoted pj gt p e Ej A Z e Ej J E gtpoce 1 o This is O e Ej is called CI The denominator is denoted by Q and is a very important quantity called QNV 8 Ze ENV J El It can be shown that 8 Where ICE is the Boltzmann constant B 2 QNVT ze EfN VkBT J e EjNVkBT 3 pf QNVT
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