1. CH. 11 INTERMOLECULAR FORCES FORCES >types >influences >strength LIQUIDS - SOLIDS >physical properties TRENDS PHASES - PHASE CHANGE >heating curves >  H calculations Equations Clausius-Claperyron

2. RECALL 3 physical states solid -- liquid -- gas condensed phases Phase change related to: intermolecular forces + KE Chemical behavior in diff phases - same Physical behavior in diff phases - diff WHY????? Due to strength of inter- forces PE depends on (Coulomb’s) *charges of particles; dist bet ** KE speed  absol. T

3. FORCES Intra -F : w / i the molecule Inter -F : bet molecules POINTS most all liquids @ room temp are molecules I nt ra -F give rise to covalent bonding influence 1) molecular shape 2) bond E’s 3) chem behavior Ch.8 attraction of gases deviates Ideal Gas Law

4. PHYSICAL PROPERTIES OF LIQUIDS - SOLIDS Due to inter-F show table 11.1 pg 444

5. Characteristic Properties of G - L - S understood in terms of …. 1) E of motion (KE) of 2) particles (atoms, molecules, ions) of 3) g-l-s states compared to E of inter-F bet particle GAS E of attraction bet particles <<< KE ave >allows gas to expand

6. LIQUIDS >Inter-F liquid > in gases >hold particles together >denser, less compressible than gases >partilces move among others, allows “pouring” SOLIDS >Inter-F >>> gases/liquids >“lock”particles in rigid form* E & motion >little free space >Crystalline: orderly structured arrangement

7. Free Space: gas > liquid > solid condensed phase Then how can you  state? heating or cooling;  KE

8. show fig 11.2 pg 445 show how the molecules are arranged among themselves in diff phases

9. EX. NaCl @ 1 atm (incr temp) room: solid 801 o C: melts, liq 1413 o C: boils, gas N 2 O (decr temp) room: gas -88.5 o C: melts, liq -90.8 o C: solid

10. Bonding Ionic --- Covalent --- Metallic Forces vary for diff subst Inter-F < Ionic < Covalent E to vaporize liq E to evaporate liq E to melt solid INTERMOLECULAR FORCES < E break covalent bond

11. show fig 11.3 pg 446 431 kJ required to break H - Cl bond 16 kJ required to separate 2 HCl molecules

12. Notice: as  states, molecules remain intact show fig 11.2 pg 445

13. Forces Influence 1) BP weak inter-F, lowers 2) MP stronger inter-F, raises Van der Waals Forces >bet neutral molecules Viscosity resistancce to flow, inter- attraction slow liq movement incr T -- decr viscosity stronger inter- > higher visc depends on: T & shape of molecule larger molecules higher visc long shape higher visc than small round shape (thnk of contact)

14. 3 Types Intermolecular Forces bet neutral molecules 1) Dipole-Dipole * 2) London Dispersion * 3) Hydrogen Bonding Solutions Ion-Dipole * Van der Waals Forces electrostatic much weaker than covalent / ionic + -

15. Involves cations - anions Ion attracted to polar molecule ex. NaCl in H 2 O attraction incr as 1) ion charge incr 2) dipole moment incr ION - DIPOLE Na +1 O H O H O H O H O H O H O H O H Cl -1 O H

16. similar to ion, but bet neutral charge polar molecules; + / - ends of polar molecule attract D-D < I-D DIPOLE - DIPOLE + - + - + - + - + -

17. diff molecules of approx = size & mass: attraction incr w / incr polarity,  BP incr Polar vs Non higher bp More E to overcome forces -- higher bp show fig 11.7 pg 448

18. Which subst has the strongest dipole-dipole attraction? fig. 11.8 pg.449

19. >only force bet NP molecules - no dipole moment >molecules w / NO permanent polarity caused by momentary movement of e - charge in atoms are present bet all particles DISPERSION (LONDON) FORCES Then how can a nonpolar gas, N 2, be liquified? Must be some type of attraction!

20. NONPOLAR overall: equal distr of e - charge - polarity cancels (average) actual: e - movement at any moment causes e - density to be conen at one end creating instantaneous dipole. Not permanent will change Ar.. Ar.. Ar.. + + - -

22. special Dip-Dip when H bonded to small size, high EN atom w/ unbonded e - pair plays imprt role in biological sys Criteria molecule 1: H bonded to O, N, or F molecule 2: unbonded e - on O, N, or F result: H on molecule 1 interact w/ unbonded e - on molecule 2 H 2 O - H 2 O NH 3 - CH 2 FCl H - BONDING O H O H.. Cl F C H H.. H N H H..

23. HF - HBr CH 3 CH(OH)CH 3 - CH 3 CH(OH)CH 3 H - BONDING.. H--F.. H--Br.. ? CH 3 CH CH 3 O-H.. CH 3 CH CH 3 O-H..

24. CH 3 Br -- CH 3 F CH 3 CH 2 CH 2 OH -- CH 3 CH 2 OCH 3 C 2 H 6 -- C 3 H 8 dipole - dipole H-bonding dipole-dipole London CH 3 Br CH 3 CH 2 CH 2 OH C 3 H 8 Type molecular bonding higher b.p.

25. NH 3 -- PH 3 NaBr -- PBr 3 H 2 O -- HBr PH 3, dipole-dipole forces weaker, as stronger H-bonding w/ NH 3 PBr 3, dipole-dipole forces, as NaBr stronger ionic bonding HBr dipole-dipole forces weaker than H-bonding in water Which has a lower boiling pt in each pair

26. PROPERTIES OF LIQUIDS Surface Tension @ surface molecules attracted only downward (no molecules above), so need K KE to break thru surface stronger forces > surface tension Capillarity liq rises in small space against pull of gravity; forces acting bet cohesive (w/i liq) & adhesive forces Viscosity resistancce to flow, inter- attraction slow liq movement incr T -- decr viscosity stronger inter- > higher visc depends on: T & shape of molecule larger molecules higher visc long shape higher visc than small round shape (thnk of contact)

27. GAS - no attraction (far apart); random; highly compress; flow/diffuse ezly LIQUID - some attraction (contact); random; not compress; flow/diffuse slower SOLID - strongest attraction (fixed position); not compress; flow/diffuse not Phase Changes

28. gas -------> liq condensation (exo) (endo) vaporization <------- liq --------> solid freezing (exo) (endo; fusion) melting <-------

29. Enthalpy Change  H o vap  H o fus H 2 O (l) ------> H 2 O (g)  H =  H o vap = 40.7 kJ/mol  H =  H o vap = -40.7 kJ/mol <---------- E wise?  H o vap >  H o fus recall, dist & motion

31. VP - depends on T, inter- forces effects of incr T: incr n to vaporize, decr amt condense higher T -- higher vp SOLID - strongest attraction (fixed position); not compress; flow/diffuse not Universal Gas Const R = 8.31 J/mol-K Hold 3 variables const, vary 1, can find 4th Clausius - Clapeyron Eqn

32. At 34.1 0 C, vp H2O = 40.1 torr. Find the vp @ 88.5 0 C.  H vap = 40.7*10 3 N-m P 1 = 40.1 torr T 1 = 273.15 + 34.1 = 307.25 K T 2 = 361.65 K 1 N-m = 1 J P 2 = 11.0835*(40.1 torr) = 444 torr Talked about bp - What exactly is bp? -- is the T when ext.P = vp

33. From the data: bp = 78.5 o C  H vap = 40.5 kJ/mol c gas = 1.43 J/g- o C c liq = 2.45 J/g- o C At constant P (1 atm), how much heat needed to convert 0.333 mol of ethanol gas at 300 o C to liquidfy at 25.0 o C A liquid has a VP of 641 torr at 85.2 o C, and bp of 95.6 o C at 1 atm. Calculate  H vap.

34. 3 steps: gas; gas-liq; liq CH 3 CH 2 OH: 46.0 g / mol 1 st : find mass: 0.333 mol * (46.0 g / mol) = 15.3 g Cooling vapor to bp: q = C gas * mass *  T = (1.43 J / g- o C) * (15.3 g) * (78.5 - 300) = -4846 J Condensation (*direction) q = n * (-  H cond ) = (0.333 mol) * (-40.5 kJ / mol) = -13.4865 kJ * 1000 = -13487 J Cooling liquid to 25.0 o C q = C liq * mass *  T = (2.45 J / g- o C) * (15.3 g) * (25.0 - 78.5) = -2055 J Total q = q vapor + q cond + q liq = (-4846 J) + (-13487 J) + (-2005 J) = -20,338 J (-2.03*10 4 J)

35. At bp: ext P = VP use Clasius-Clapeyron eqn 1st: convert 641 torr to atm 641 torr / 760 = 0.8434 atm 2 points: P 1 = 1 atm T 1 = 273.15 + 95.6 = 368.75 K P 2 = 0.8434 atm T 2 = 358.35 K

36. COOLING CURVE Shows changes that occur when add/remove heat @ const T fig 11.22, pg 440

37. TEMP Heat Flow Out removed ----------> GAS GAS-LIQ LIQUID LIQ-SOLID SOLID  H o vap  H o fus GAS : q = m*C gas *  T GAS-LIQ : const T & E KE ave speed is same at given T decr ave E PE but not  E KE H 2 O (g) & H 2 O (l) same E KE liq E PE < gas E PE @ same T; heat released = moles * (-vap) q = n*(-  H o vap ) results in largest amt of heat released WHY??? decr PE due from condensing dist. bet molecules

38. LIQUID LIQUID q = m*C liq *  T * loss of heat results in decr T decr molecular speed, this decr E K E LIQ-SOLID * inter- attraction > motion of molecules * loss E PE form crystalline solid * const T & E KE * H 2 O (l) & H 2 O (soln) same E KE solid E PE < liq E PE @ same T; heat released = moles * (-fusion) q = n*(-  H o fus ) SOLID q = m*C sol *  T motion restricted; decr T reduced ave speed TOTAL HEAT RELEASED Use Hess’ Law sum of 5 steps 2 pts @ const P w/i phase:  q is  T (  E KE ) depends on: amt subst (n), C for phase,  T during phase  :  q (@T)(  E PE ), dist bet molecules changes

39. LIQ-GAS EQUILIBRA @ const T open closed vaporize condense Weaker bonds = higher vp,lower bp Keep in Mind: when a sys at equil is distr, will react in way to counteract said disturb to regain a state at a new equilib Sys reaches pt of dynamic balance @ equilibria & vp const

40. Solids decr vp < liq Solids: high vp SOLID - GAS EQUILIBRA: Sublimation Phase Diagrams???

41. SUMMARY CH. 11 Clausius-Claperyron EQUILIBRA gas -------> liq condensation (exo) (endo) vaporization <------- liq --------> solid freezing (exo) (endo; fusion) melting <------- COOLING CURVE INTERMOLECULAR FORCES bond type; bp/mp higher/lower q = n*C PHASE *  T

42. COOLING - HEATING CURVES #4) Calculate the enthalpy change (  H) when 18.0 g of ice at -25 o C is converted to vapor at 125 o C.  H fus = 6.02 kJ  H vap = 40.7 kJ C solid = 37.6 J / mol  o C C liq = 75.3 J / mol  o C C gas = 33.1 J / mol  o C

43. 4) 5 steps: solid; solid-liq; liquid; liq-gas; gas given 18.0 g = 1 mol *** look at labels on “ C ” Heat solid to mp: q = C solid * mol *  = (37.6 J / mol- o C) * (1.00 mol) * (0 - - 25) = 940 J Fusion (*direction) q = n * (  H fus ) = (1.00 mol) * (6.02 kJ / mol) = 6.02 kJ * 1000 = 6020 J Heating liquid to bp q = C liq * mol *  T = (75.2 J / mol- o C) * (1.00 mol) * (100.0 - 0.0) = 7520 J Vaporization (*direction) q = n * (  H vap ) = (1.00 mol) * (40.7 kJ / mol) = 40.7 kJ * 1000 = 40700 J Heat gas to 125 o C q = C gas * mol *  T = (33.1 J / mol- o C) * (1.00 mol) * (125.0 - 100.0) = 830 J Total q = q solid + q fus + q liq + q vap + q gas = (940 J) + (6020 J) + (7520 J) + (40700 J) + (830 J) = 56,000 J (56 kJ)