Page 1

CHAPTER 10: LIQUIDS + SOLIDS

INTERMOLECULAR FORCES

GENERALITIES

DIPOLE-DIPOLE FORCES

HYDROGEN BONDS

H Cl H Cl HC

H

O HC

H

O

HO

H

HO

H HN

H

H

HN

H

H

H F H F

HO

H

HO

HH

OH

EIadentwithinmolecule

into O'THEV between molecules

gi St

1 fSt 8 St 8 St ain I 8

T dipdip

DipoledipoleForce attraction of dipoles ft 8

made from polar bonds

8

f shi g

St StStJst 8T

Hydrogen bond a strong dipole dipoleforce between 8TH and 8 on O N F

St sO H O H

N H N H

F H F H

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LONDON DISPERSION FORCES (LDF)

Electronic orientations at different times Temporarily induced dipoles

Helium is a liquid at 4 Kelvin Methane (CH4) is a liquid at –160 ˚C

Titan (Saturn’s largest moon) has liquid CH4 rivers, oceans, and rain.

Sample Problems:

Identify the IMF pointed to by an arrow. Use a dashed line to show the strongest IMF possible between these two molecules. Also identify the IMF.

IMF: IMF:

He HeHe

HC

H

H H

HC

H

H H

I Cl I ClC

NH

H

H

HH

OH

H

O O Q O

glopfided LOF

LondonDispersionForce LDF attraction of temporarydipole made from distorted electron clouds

w everything including nonpolar

Iii as

St 8 8T g 8T St

s sSt8T a St

dipoledipole hydrogenbonds

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BOILING POINT

BOILING PROCESS + DHvap

Liquid H2O (water) Which of these represents gaseous H2O (steam)?

→

Boiling (liquid to gas) involves…

Heat of Vaporization (DH˚vap):

Water DHvap = +40.7 kJ/mol

BOILING POINT TRENDS

Molar Mass (g/mol)

DH˚vap (kJ/mol)

Boiling Point (˚C)

O2 32.00 6.8 –183.0

F2 38.00 6.6 –188.1

Cl2 70.90 20.4 –34.0

Br2 159.80 30.0 58.8

Bromine (liquid/gas); Chlorine (gas only)

liquid

0breaking intermolecularforces MFS

energy to vaporize 1 mot liquid

l g endothermic g l exothermic

higherDHvaphigher boilingpoint

to more E input

c

the

needed

Hightop strong IMF

As P molar mass strength of LOF Pmore e clouds to distort t easier to distortas e are further from nucleus

Brz L at room

tempstrongest LOF

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Molar Mass (g/mol)

DH˚vap (kJ/mol)

Boiling Point (˚C)

Methanol

(CH3OH)

IMF:

32.05 38.3 64.7

Formaldehyde

(CH2O) IMF:

30.03 23.3 –19.2

Ethane

(CH3CH3) IMF:

30.08 14.7 –88.6

Sample Problems:

In each pair, determine which should have the higher boiling point, and explain the trend using intermolecular forces.

HCl vs. HF Propane (C3H8) vs. Pentane (C5H12)

HC

O

H H

H

HO

CH

H H

HC

H

O HC

H

O

HC

C

H H

H

H H

HC

C

H H

H

H H

gin 88 St

Hbond8

i St

St gdip dipforce

nonpolar

LDF sameLOFsameMM

IMF strength LDF dip dip L H b ds

Jimi w.no i i.iE s

very polar ft 8large

It is very smallso 8 charges can

get close

gas liquidSts Idipdip Hbonds

1Pentane has higher MM soHF higher op stronger

IMF H bonds dipdie stronger LOF.at Tbp

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LIQUID PROPERTIES

SURFACE TENSION

Water beading on waxy surface Scott Kelly with water droplet in space Water skipper

CAPILLARY ACTION

Concave meniscus with water Convex meniscus with mercury

GlassO

Si

OH

H

H

O HHO

HSurface

Surface tension resistance of liquidto increase its surface area

west surface area sphere

why bailing up maximizes IMFas more IMFs are experienced on

interior than surface

Capillary Action rising of liquid against gravityinto tube or up a papertowel

HgSt 8 not

Adhesive attractedSt S toforce water

Hgbondscohesiveforces strong ballsupstickingtoeachother due to surface

tension

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VISCOSITY

Honey is viscous Structure of sugar (sucrose, C12H22O11)

“Pitch drop experiment” with asphalt / tar General structure of tar / crude oil

Sample Problems:

Which liquid is the most viscous?

Glycerin (glycerol) 1-propanol 2,4-pentanedione

O

CC

C

CC

C

CC

C

O

O

C HO

H HH

HO

H

OH O

H

H

H

HC

H

OH H

H

OH

O

H

H

C

OH

HH

H

OC

CC

O

O

H

H

H

H H H H

HH

CC

CO

H

H H H H

H HC

CC

CC

H

O O

H H

H

HH

HH

CH3 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH3

60-70 of these!

g

t bondsite

coViscosity resistance of liquid to flow thickness

Thick liquids either

many H bonding sites which told molecules

to each other stronglyVery long molecules which entangle in each other

causing sliding difficult

s

C C C st

cmost H bond site

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EVAPORATION

EVAPORATION

VAPOR PRESSURE

Sample Problem:

Equal volumes of liquids A and B are placed in separate beakers on the countertop.

a. After some time, half of “A” has evaporated. After the same amount of time, should there be more or less of “B” remaining?

b. Which has a higher vapor pressure, A or B?

CC

C

O

H

HH

H

HH C

CC

C

H

HH

H

HH

H H

A B

Evaporationl g at tempbelowboiling point

Endothermic takes Ebreaking IMF These

evapEven below the bp a certain of Etomolecules have enough energy to break break1MFS

MFS as theres a range of energies

vapor pressure P caused byevaporation in a closed container

Paco

8

ddiir

Lot

1Weaker IMF LOFso evaporates morequicklyso I of B remains

PVP weaker IMF dbp PReap

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TEMPERATURE + VAPOR PRESSURE

Temp H2O (˚C)

Vapor Pressure H2O (torr)

0 4.6 20 17.5 40 55.3 60 149.4 90 525.8 100 760.0

LINEAR RELATIONSHIP

Inverse Temp (K–1)

Natural log (ln) of Vapor Pressure

H2O 0.00366 1.52 0.00341 2.864 0.00319 4.013 0.00300 5.0066 0.00275 6.2649 0.00268 6.6333

Logarithm Review:

log (100) means 10? = 100 log (100) = 2 because 102 = 100

ln (7.39) means e? = 7.39 ln (7.39) = 2 because e2 = 7.39

m = –5204 b = 20.6

Water, bp = 100 ˚C

00As temp P vapor pressure 9At highertemp a greaterof molecules have enough E Listto break 1Mmore modes of gas

Liquid boils when Pvap appliedPPath

The normal boiling point at 1atm

itis when Puap 760 ton iii ether

bp 55 C

Linear

In Pvap vs YTTink

1natural 105represents anexponentunitless

2.718

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Clausius-Claperyon Equation

ln$%&'() = −∆-&'(. /112 + 4 ln 567689 = ∆:;<=> 5 ?@8 −

?@79

ln = natural log (base “e”, not base 10) Pvap = vapor pressure DHvap = heat of vaporization (J/mol) R = gas constant, 8.3145 J/mol·K T = temperature in Kelvin

P1 = vapor pressure at temperature 1 P2 = vapor pressure at temperature 2 T1 = Kelvin temperature 1 T2 = Kelvin temperature 2

Sample Problems:

Give the graphical data for water, calculate water’s heat of vaporization in kJ/mol. (Note: actual DHvap water = +40.7 kJ/mol.)

The vapor pressure of alcohol (ethanol) at 34.7 ˚C is 100.0 mmHg, and the heat of vaporization of alcohol is 38.6 kJ/mol. Calculate the vapor pressure of alcohol at 65.0 ˚C.

m = –5204 b = 20.6

w 2graph situation

Units on slope

kk

enPyar Ifstope 0Hf

UHrap slope R

5204 4 8.3113

43

In OHLI F 34.7 C 275 K Tz 65.0 C zP toooommHg 338.2 k

en en Pz

e

the 38.6kW

Pz100 0mmHg

eOHLI ta f

eqz j F 8Ik 3074k

14SF1450.00295710.0032481

0.00029113511

In Ppt Lnp In Pz