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

PHYSICAL PROPERTIES OF LIQUIDS - SOLIDS

Due to inter-F

show table 11.1 pg 444

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

Free Space: gas > liquid > solid

condensed phase

Then how canyou state?

heating or cooling; KE

show fig 11.2 pg 445show how the molecules are arrangedamong 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

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

show fig 11.3 pg 446431 kJ required to break H - Cl bond16 kJ required to separate 2 HCl molecules

Notice: as states, molecules remain intact

show fig 11.2 pg 445

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

Which subst has the strongest dipole-dipole attraction?

fig. 11.8 pg.449

>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

COOLING CURVE

Shows changes that occur when add/remove heat @ const T

fig 11.22, pg 440

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)