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CH. 11 INTERMOLECULARFORCES
FORCES >types >influences >strength
LIQUIDS - SOLIDS >physical properties
TRENDS
PHASES - PHASE CHANGE >heating curves >H calculations
EquationsEquations
Clausius-Claperyron
12
vap
1
2
vap
T
1 -
T
1
R
H-
P
Pln
RT
H- PLn
RECALL
3 physical states solid -- liquid -- gas
condensed phases
Phase change related to: intermolecular forces + KE
Chemical behavior in diff phases - samePhysical behavior in diff phases - diffWHY????? Due to strength of inter- forces
PE depends on (Coulomb’s) *charges of particles; dist bet
**KEspeed absol. T
FORCESIntra-F: w/i the molecule
Inter-F: bet molecules
POINTSmost 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.8attraction of gases deviates Ideal Gas Law
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 statescompared to E of inter-F bet particle
GASE 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
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
Bonding Ionic --- Covalent --- Metallic
Forces vary for diff subst Inter-F < Ionic < Covalent
E to vaporize liqE to evaporate liq E to melt solid
INTERMOLECULAR FORCES
< E break covalent bond
Forces Influence
1) BP weak inter-F, lowers2) MP stronger inter-F, raises
Van der Waals Forces >bet neutral molecules
Viscosityresistancce to flow, inter- attraction slow liq movement incr T -- decr viscosity stronger inter- > higher viscdepends on: T & shape of molecule larger molecules higher visc long shape higher visc than small round shape (thnk of contact)
3 Types Intermolecular Forcesbet neutral molecules1) Dipole-Dipole*2) London Dispersion*3) Hydrogen BondingSolutionsIon-Dipole *Van der Waals Forces
electrostaticmuch weaker than covalent/ionic
+ -+ -
Involves cations - anionsIon attracted to polar moleculeex. NaCl in H2O
attraction incr as 1) ion charge incr 2) dipole moment incr
ION - DIPOLE
Na+1
OH
H OH H
O
H H
OH
H
OH H
O
H H
O
H H O H
H
Cl-1
O
H H
similar to ion, but bet neutral charge polar molecules;+/- ends of polar molecule attract
D-D < I-D
DIPOLE - DIPOLE
+ -
+ -+ - + -
+ -
diff molecules of approx = size & mass: attraction incr w/ incr polarity, BP incr
Polar vs Non higher bp
More E to overcomeforces -- higher bp
show fig 11.7 pg 448
>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 anonpolar gas, N2,
be liquified?
Must be sometype of attraction!
NONPOLARoverall: 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 ....
....+
+-
-
special Dip-Dip when H bonded to small size, high EN atom w/ unbonded e- pairplays imprt role in biological sys
Criteriamolecule 1: H bonded to O, N, or Fmolecule 2: unbonded e- on O, N, or F
result: H on molecule 1 interact w/ unbonded e- on molecule 2
H2O - H2O NH3 - CH2FCl
H - BONDINGH - BONDING
O
H H
O
H H..
..
....
ClF C H H
......
......
H N H H
..
HF - HBr CH3CH(OH)CH3 - CH3CH(OH)CH3
H - BONDINGH - BONDING
..H--F .... H--Br
..
..
..? CH3 CH CH3
O-H
.... CH3 CH CH3
O-H
....
CH3Br -- CH3F
CH3CH2CH2OH -- CH3CH2OCH3
C2H6 -- C3H8
dipole - dipole
H-bondingdipole-dipole
London
CH3Br
CH3CH2CH2OH
C3H8
Type molecular bondingType molecular bonding higher b.p.higher b.p.
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
Which has a lower boiling pt in each pairWhich has a lower boiling pt in each pair
PROPERTIES OF LIQUIDS
Surface Tension@ surface molecules attracted only downward (no molecules above), so need KKE to break thru surface stronger forces > surface tension
Capillarityliq rises in small space against pull of gravity; forces acting bet cohesive (w/i liq) & adhesive forces
Viscosityresistancce to flow, inter- attraction slow liq movement incr T -- decr viscosity stronger inter- > higher viscdepends 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 Hovap Ho
fus
H2O (l) ------> H2O (g)
H = Hovap = 40.7 kJ/mol
H = Hovap = -40.7 kJ/mol <----------
E wise? Hovap > Ho
fus
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 ConstR = 8.31 J/mol-K
Hold 3 variables const,vary 1, can find 4th
Clausius - Clapeyron Eqn
12
vap
1
2
vap
T
1 -
T
1
R
H-
P
Pln
:pt version-2
b x m y
)T
1(
R
H- Pln k
At 34.10C, vpH2O = 40.1 torr. Find the vp @ 88.50C.Hvap = 40.7*103 N-m
P1 = 40.1 torr T1 = 273.15 + 34.1 = 307.25 K T2 = 361.65 K
11.0835 1.40
P
1.2845 Pln
PlnLn inv
K307.25
1 -
361.65
1
KJ/mol 8.31
J/mol) 10*(40.7-
P
Pln
2
1
2
3
1
2
1 N-m = 1 J
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 Hvap = 40.5 kJ/molcgas = 1.43 J/g-oC cliq = 2.45 J/g-oC
At constant P (1 atm), how much heat needed to convert0.333 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 Hvap.
3 steps: gas; gas-liq; liq
CH3CH2OH: 46.0 g/mol1st: find mass: 0.333 mol*(46.0 g/mol) = 15.3 g
Cooling vapor to bp: q = Cgas*mass*T = (1.43 J/g-oC)*(15.3 g)*(78.5 - 300) = -4846 J
Condensation (*direction)
q = n*(-Hcond) = (0.333 mol)*(-40.5 kJ/mol) = -13.4865 kJ*1000 = -13487 J
Cooling liquid to 25.0oC q = Cliq*mass*T = (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 eqn1st: 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
K75.368
1
35.358
1
KJ/mol 8.314
H-
atm 1
atm 8434.0ln vap
)10*87.7(mol/J314.8
H 17029.0 5vap
J/mol 17990- 10*87.7
J/mol) (8.314(-0.17029) H
5vap
)10*(1.80 J/mol 17990 H 4vap
TE
MP
Heat Flow Out removed ---------->
GAS
GAS-LIQ
LIQUID
LIQ-SOLID
SOLID
Hovap
Hofus
GAS: q = m*Cgas*T GAS-LIQ: const T & EKE
ave speed is same at given T decr ave EPE but not EKE
H2O (g) & H2O (l) same EKE
liq EPE < gas EPE
@ same T; heat released = moles * (-vap)
q = n*(-Hovap)
results in largest amt of heat released WHY??? decr PE due from condensing dist. bet molecules
LIQUIDLIQUID q = m*Cliq*T * 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*(-Hofus)
SOLID q = m*Csol*T motion restricted; decr T reduced ave speed
TOTAL HEAT RELEASEDUse Hess’ Law sum of 5 steps
2 pts @ const Pw/i phase: q is T (EKE)depends on: amt subst (n), C for phase, T during phase : q (@T)(EPE), dist bet molecules changes
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 disturbto regain a state at a new equilib
gas liq
onvaporizati
oncondensati
Sys reaches pt of dynamic balance@ equilibria & vp const
Solids decr vp < liq
Solids: high vp
SOLID - GAS EQUILIBRA: Sublimation
gas solid sublime
Phase Diagrams???
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*CPHASE*T
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 kJCsolid = 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* = (37.6 J/mol-oC)*(1.00 mol)*(0 - -25) = 940 J
Fusion (*direction) q = n*(Hfus) = (1.00 mol)*(6.02 kJ/mol) = 6.02 kJ*1000 = 6020 J
Heating liquid to bp q = Cliq*mol*T = (75.2 J/mol-oC)*(1.00 mol)*(100.0 - 0.0) = 7520 J
Vaporization (*direction) q = n*(Hvap) = (1.00 mol)*(40.7 kJ/mol) = 40.7 kJ*1000 = 40700 J
Heat gas to 125oC q = Cgas*mol*T = (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)