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This text provides an overview of intermolecular forces and their influences on the physical properties of liquids and solids. It discusses various types of intermolecular forces, their strength, and their effects on boiling point, melting point, and viscosity. The text also explains the concepts of surface tension, capillarity, and viscosity.
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CH. 10 INTERMOLECULARFORCES 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 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
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
PHYSICAL PROPERTIES OF LIQUIDS - SOLIDS Due to inter-F 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
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
Free Space: gas > liquid > solid condensed phase Then how can you state? heating or cooling; KE show how the molecules are arranged among themselves in diff phases
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
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
431 kJ required to break H - Cl bond 16 kJ required to separate 2 HCl molecules Notice: as states, molecules remain intact
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)
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
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
- + - + - - - + + + similar to ion, but bet neutral charge polar molecules; +/- ends of polar molecule attract DIPOLE - DIPOLE D-D < I-D
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
fig. 11.8 pg.449 Which subst has the strongest dipole-dipole attraction?
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!
.. - - + + .. .. 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
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
CH3 CH CH3 O-H CH3 CH CH3 O-H .. H--Br .. .. .. .. H--F .. .. .. .. .. H - BONDING HF - HBr CH3CH(OH)CH3 - CH3CH(OH)CH3 ?
higher b.p. Type molecular bonding CH3Br -- CH3F CH3CH2CH2OH -- CH3CH2OCH3 C2H6 -- C3H8 dipole - dipole H-bonding dipole-dipole London CH3Br CH3CH2CH2OH C3H8
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
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)
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
gas -------> liq condensation (exo) (endo) vaporization <------- liq --------> solid freezing (exo) (endo; fusion) melting <-------
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
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
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
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.
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)
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
COOLING CURVE Shows changes that occur when add/remove heat @ const T fig 11.22, pg 440
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
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
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
SOLID - GAS EQUILIBRA: Sublimation Solids decr vp < liq Solids: high vp Phase Diagrams???
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
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/moloC Cliq = 75.3 J/moloC Cgas = 33.1 J/moloC
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)