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CH. 11 INTERMOLECULAR FORCES

CH. 11 INTERMOLECULAR FORCES. FORCES >types >influences >strength LIQUIDS - SOLIDS >physical properties TRENDS PHASES - PHASE CHANGE >heating curves >  H calculations. Equations. Clausius-Claperyron. RECALL. 3 physical states solid -- liquid -- gas.

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CH. 11 INTERMOLECULAR FORCES

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  1. CH. 11 INTERMOLECULARFORCES 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 **KEspeeda absol. T

  3. FORCES Intra-F: w/i the molecule Inter-F: bet molecules POINTS most all liquids @ room temp are molecules Intra-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 show table 11.1 pg 426 Due to inter-F

  5. Characteristic Properties of G - L - Sunderstood in terms of …. 1) E of motion (KE) of 2) particles (atoms, molecules, ions) of 3) g-l-s statescompared to E of inter-F bet particle GAS E of attraction bet particles <<< KEave >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 427 show how the molecules are arranged among themselves in diff phases

  9. EX. NaCl @ 1 atm (incr temp) room: solid 801oC: melts, liq 1413oC: boils, gas N2O (decr temp) room: gas -88.5oC: melts, liq -90.8oC: solid

  10. INTERMOLECULAR FORCES < E break covalent bond 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

  11. show fig 11.3 pg 428 431 kJ required to break H - Cl bond 16 kJ required to separate 2 HCl molecules

  12. show fig 11.2 pg 427 Notice: as  states, molecules remain intact

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

  15. O H H O H H O H H O H H O H H O H H O H H O H H O H H Na+1 Cl-1 Involves cations - anions Ion attracted to polar molecule ex. NaCl in H2O ION - DIPOLE attraction incr as 1) ion charge incr 2) dipole moment incr

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

  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 430

  18. fig. 11.8 pg.431 Which subst has the strongest dipole-dipole attraction?

  19. DISPERSION (LONDON) FORCES >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 Then how can a nonpolar gas, N2, be liquified? Must be some type of attraction!

  20. .. - - + + .. .. Ar .. .. .. .. .. Ar .. .. .. Ar .. 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

  21. O H H O H H .. H N H H .. .. .. .. .. Cl F C H H .. .. .. .. .. H - BONDING 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 H2O - H2O NH3 - CH2FCl

  22. CH3 CH CH3 O-H CH3 CH CH3 O-H .. H--Br .. .. .. .. H--F .. .. .. .. .. H - BONDING HF - HBr CH3CH(OH)CH3 - CH3CH(OH)CH3 ?

  23. higher b.p. Type molecular bonding CH3Br -- CH3F CH3CH2CH2OH -- CH3CH2OCH3 C2H6 -- C3H8 dipole - dipole H-bonding dipole-dipole London CH3Br CH3CH2CH2OH C3H8

  24. Which has a lower boiling pt in each pair NH3 -- PH3 NaBr -- PBr3 H2O -- HBr PH3, dipole-dipole forces weaker, as stronger H-bonding w/ NH3 PBr3, dipole-dipole forces, as NaBr stronger ionic bonding HBr dipole-dipole forces weaker than H-bonding in water

  25. PROPERTIES OF LIQUIDS Surface Tension @ surface molecules attracted only downward (no molecules above), so need KKE 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)

  26. 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

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

  28. Enthalpy Change DHovap DHofus H2O (l) ------> H2O (g) DH = DHovap = 40.7 kJ/mol DH = DHovap = -40.7 kJ/mol <---------- E wise? DHovap >DHofus recall, dist & motion

  29. 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

  30. At 34.10C, vpH2O = 40.1 torr. Find the vp @ 88.50C. DHvap = 40.7*103 N-m 1 N-m = 1 J P1 = 40.1 torr T1 = 273.15 + 34.1 = 307.25 K T2 = 361.65 K P2 = 11.0835*(40.1 torr) = 444 torr Talked about bp - What exactly is bp? -- is the T when ext.P = vp

  31. From the data: bp = 78.5oC DHvap = 40.5 kJ/mol cgas = 1.43 J/g-oC cliq = 2.45 J/g-oC At constant P (1 atm), how much heat needed to convert 0.333 mol of ethanol gas at 300oC to liquidfy at 25.0oC A liquid has a VP of 641 torr at 85.2oC, and bp of 95.6oC at 1 atm. Calculate DHvap.

  32. 3 steps: gas; gas-liq; liq CH3CH2OH: 46.0 g/mol 1st: find mass: 0.333 mol*(46.0 g/mol) = 15.3 g Cooling vapor to bp: q = Cgas*mass*DT = (1.43 J/g-oC)*(15.3 g)*(78.5 - 300) = -4846 J Condensation (*direction) q = n*(-DHcond) = (0.333 mol)*(-40.5 kJ/mol) = -13.4865 kJ*1000 = -13487 J Cooling liquid to 25.0oC q = Cliq*mass*DT = (2.45 J/g-oC)*(15.3 g)*(25.0 - 78.5) = -2055 J Total q = qvapor+qcond+qliq = (-4846 J)+(-13487 J)+(-2005 J) = -20,338 J (-2.03*104 J)

  33. At bp: ext P = VP use Clasius-Clapeyron eqn 1st: convert 641 torr to atm 641 torr/760 = 0.8434 atm 2 points: P1 = 1 atm T1 = 273.15 + 95.6 = 368.75 K P2 = 0.8434 atm T2 = 358.35 K

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

  35. TEMP DHovap DHofus Heat Flow Out removed ----------> GAS GAS-LIQ LIQUID SOLID LIQ-SOLID GAS-LIQ: const T & EKE ave speed is same at given T decr ave EPE but not D EKE H2O (g) & H2O (l) same EKE liq EPE < gas EPE @ same T; heat released = moles * (-vap) q = n*(-DHovap) GAS: q = n*Cgas*DT results in largest amt of heat released WHY??? decr PE due from condensing dist. bet molecules

  36. LIQUIDq = n*Cliq*DT * loss of heat results in decr T decr molecular speed, this decr EKE LIQ-SOLID * inter- attraction > motion of molecules * loss EPE form crystalline solid * const T & EKE * H2O (l) & H2O (soln) same EKE solid EPE < liq EPE @ same T; heat released = moles * (-fusion) q = n*(-DHofus) SOLID q = n*Csol*DT motion restricted; decr T reduced ave speed TOTAL HEAT RELEASED Use Hess’ Law sum of 5 steps 2 pts @ const P w/i phase: Dq is DT (DEKE) depends on: amt subst (n), C for phase, DT during phase D: Dq (@T)(DEPE), dist bet molecules changes

  37. LIQ-GAS EQUILIBRA @ const T open closed vaporize condense Sys reaches pt of dynamic balance @ equilibria & vp const 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 Weaker bonds = higher vp,lower bp

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

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

  40. COOLING - HEATING CURVES #4) Calculate the enthalpy change (H) when 18.0 g of ice at -25oC is converted to vapor at 125oC. Hfus = 6.02 kJ Hvap = 40.7 kJ Csolid = 37.6 J/moloC Cliq = 75.3 J/moloC Cgas = 33.1 J/moloC

  41. 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 = Csolid*mol*DT = (37.6 J/mol-oC)*(1.00 mol)*(0 --25) = 940 J Fusion (*direction) q = n*(DHfus) = (1.00 mol)*(6.02 kJ/mol) = 6.02 kJ*1000 = 6020 J Heating liquid to bp q = Cliq*mol*DT = (75.2 J/mol-oC)*(1.00 mol)*(100.0 - 0.0) = 7520 J Vaporization(*direction) q = n*(DHvap) = (1.00 mol)*(40.7 kJ/mol) = 40.7 kJ*1000 = 40700 J Heat gas to 125oC q = Cgas*mol*DT = (33.1 J/mol-oC)*(1.00 mol)*(125.0 - 100.0) = 830 J Total q = qsolid + qfus + qliq+ qvap + qgas = (940 J)+(6020 J)+(7520 J)+(40700 J)+(830 J) = 56,000 J (56 kJ)

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