chapter 10: iquids olids intermolecular...
TRANSCRIPT
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
Page 2
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
Page 3
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
Page 4
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
Page 5
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
Page 6
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
Page 7
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
Page 8
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
Page 9
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