Intermolecular Forces: Liquids, Solids, and Phase Changes
Chapter 12 Intermolecular Forces: Liquids, Solids, and Phase Changes
12.1 An Overview of Physical States and Phase Changes
12.2 Quantitative Aspects of Phase Changes
12.3 Types of Intermolecular Forces
12.4 Properties of the Liquid State
12.5 The Uniqueness of Water
12.6 The Solid State: Structure, Properties, and Bonding12.7 Advanced Materials
OMIT
Intermolecular forces are forces between molecules while intramolecular forces are forces “within” a molecule.
Intermolecular Force
ForcesBetween Molecules
ForcesWithin AMolecule
Intramolecular ForceIonic or covalent bonding forces within a molecule or ionic substance (i.e. internal chemical bonds).
attractive forces between molecules
Intermolecular forces explain many important physical properties of compounds on the planet.
Wetting or Not Wetting
Surface Tension
ViscosityBoiling and Freezing Point
capillary action
Intermolecular forces play a critical role in life as we understand it.
Aquoues Interior
Embedded Proteins
Non-polar tail
Polar heads in aqueous exterior
Intermolecular forces (hydrogen bonding) hold together the double helix of DNA.
Paper is “dried” network of cellulose fibers held together by hydrogen bonds and Van der waals forces.
~0.7 mm
~0.7
mm
Smook’s Handbook 1992
entangled pulp fibers
Fiber Dimensions:10 nm to 10 mm
GAS
LIQUIDSOLID
A phase is a state of matter that is homogeneous, chemically uniform and has physically distinct properties (density, crystal structure, index of refraction).
We recognize 4-states of matter:
1. Gas2. Liquids3. Solids4. Plasma (hot ionized gas)
-low density-high compressibility-high Kinetic energy-large distance between molecules-no IMF
-very high density-not compressible-low Kinetic energy-small distance between molecules-IMF large
-high density-not compressible-mid Kinetic energy-molecules close but fluid-IMF large
PhaseProperties
Kinetic molecular theory states that the temperature of collection of particles (system) is proportional to the average kinetic energy of the collection.
KE = Ek = 32 RT
Kinetic Energy Temperature
TemperatureKE Intermolecular
Force
Liquid
Vs
The phase of a substance depends on the “competition” between KE, which keeps molecules far apart and moving, and IMF which keep molecules close together and condensed.
Gas
Property Gases Liquid Solids
Molecular Mobility
Expand to occupy container
Assumes shape of container
Fixed Shape molecules fixed
Density Very low density High High
Compressibility High Very Very low Not-compressible
Diffusion high molecular speeds
Molecule slide past one another
--flow
Atoms and molecules are
fixedKinetic Energy High Medium Low
IntermolecularForces
Few and small Many Many
The phase of a substance shows unique and varying different properties.
Think of the different phases as classes of possible molecular motion due to different kinetic energies (caused by temperature differences) and varying degrees of intermolecular forces.
MELTING
FREEZING
DEPO
SITIO
N
CONDENSATIO
N
EVAPORATIO
N
ΔH < 0
ΔH > 0
ΔH < 0ΔH < 0
ΔH > 0ΔH > 0
Gas
LiquidSolid
SUBL
IMAT
ION
Enthalpy is just another word for heat (q) given off or absorbed by a system at constant pressure.
qP = !H = !E + P!V
Internal Energy = !E = qp - P!V Solve for qp
Chemists measure the enthalpies of many reactions, tabulate them in handbooks and give them “specific names”.
The latent heat of fusion and the latent heat of vaporization are measured enthalpies for a pure solid and liquid that tell us something about the IMF between molecules.
!H˚vap
!H˚fus
• Molar Heat of Fusion ("Hfus): The energy required to melt one mole of solid (in kJ).
• Molar Heat of Vaporization("Hvap): The energy (in kJ) required to vaporize one mole of liquid.
• Molar Heat of Sublimation ("Hsub): The energy (in kJ) required to vaporize one mole of solid to gas.
The differences in energy required to vaporize or melt a pure substance tells us something about the IMF’s that try to keep these molecules together in the condensed phase.
Notice how IMF’s are weaker in solid-liquid transition (!H˚fus) vs liquid-gas transition (!H˚vap)
IMF’s are greater in the liquid phase.
More energy per mole needed to vaporize than to melt!
Here’s the same curve now applying the conservation of energy (sum of the heats)
Ice
Ice + Water mix
Water
SteamHeating solid ice to 0˚C
Heat Added (kJ/mole)4.12 38 79.3 305 309
Tem
pera
ture
˚C
Melting solid ice to 0˚C water
Heating water to boiling 100˚C
Boiling all water to steam 100˚C
Heating steam past
100˚C
0
-100
100
220
q = sice mice ∆T
q = mice ∆Hfus
q = sH2O mH2O ∆T
q = mH2O ∆Hvap
q = sstm mstm ∆T
Phase transition
Phase transition
A
B C
D E
F
Temperature does not change during a phase transition.
Temperature does not change during a phase transition.
Water + steam mix
Heat Added (Joules)
Tem
pera
ture
˚C
We can graphically show the qualitative and quantitative aspects of a phase change by plotting the temperature of a substance vs heat (q) added to a substance.
-100
0
100
220
Here’s the same curve now applying the conservation of energy (sum of the heats)
Ice
Ice + Water mix
Water
SteamHeating solid ice to 0˚C
Heat Added (kJ/mole)4.12 38 79.3 305 309
Tem
pera
ture
˚C
Melting solid ice to 0˚C water
Heating water to boiling 100˚C
Boiling all water to steam 100˚C
Heating steam past
100˚C
0
-100
100
220
q = sice mice ∆T
q = mice ∆Hfus
q = sH2O mH2O ∆T
q = mH2O ∆Hvap
q = sstm mstm ∆T
Phase transition
Phase transition
A
B C
D E
F
Temperature does not change during a phase transition.
Temperature does not change during a phase transition.
Water + steam mix
The molar heat capacity of a substance is the amount of heat (q) required to raise the temperature of one mole of the substance by one degree Celsius.
q = s m !T
specific heat capacityJ/g °C or J/mol K # of moles
or grams
Change in Temp: ˚C or K
Heat absorbed (Joules)
The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius.
The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius. (How thermally sensitive a substance is to the addition of energy!)
Heat (q) absorbed or released:
q = s m !T
Calculate the amount of heat required to convert 500 grams of ice at -20.0˚C to steam at 120.˚C. The specific heat capacities of water, ice and water vapor are 4.18, 2.06 and 1.84 J/g ˚C respectively, and the latent heat of fusion and vaporization, !Hf and !Hv, are 6.02 and 40.7 kJ/mol respectively.
n�
i=1
qi = 0 sum the q’s baby
q = si mi !T for non-phase transitions
qsolid=>liquid = # moles !H˚fusion
qliquid=>gas = # moles !H˚vaporization for phase transitions
Calculate the amount of heat required to convert 500 grams of ice at -20˚C to steam at 120˚C. The specific heat capacities of water, ice and water vapor are 4.18, 2.06 and 1.84 J/g ˚C respectively, and the latent heat of fusion and vaporization, !Hf and !Hv, are 6.02 and 40.7 kJ/mol respectively.
Calculate the amount of heat required to convert 500 grams of ice at -20˚C to steam at 120˚C. The specific heat capacities of water, ice and water vapor are 4.18, 2.06 and 1.84 J/g ˚C respectively, and the latent heat of fusion and vaporization, !Hf and !Hv, are 6.02 and 40.7 kJ/mol respectively.
1. Heat ice from -20˚C to ice at 0˚C = 500. g x 2.06 J/g ˚C x 20˚C 2. Melt ice at 0˚C to water at 0˚C = 500. g/(18 g/mol) x 6.02 kJ/mol 3. Heat water from 0˚C to water at 100˚C= 500. g x 4.18 J/g ˚C x 100˚C 4. Evap water at 100˚C to vap at 100˚C = 500. g/(18 g/mol) x 40.7 kJ/mol5. Heat vap from 100˚C to vap at 120˚C = 500. g x 1.84 J/g ˚C x 20˚C
1. = 20.6 kJ2. = 167.2 kJ3. = 209.0 kJ4. = 1130.5 kJ5. = 18.4 kJ
Total = 1545.6 kJ
Dynamic chemical equilibrium is condition reached when the rate of evaporation and rate of condensation are equal (no net change).
Molecules in liquid begin to vaporize Molecules vaporizing and
condensing at such a rate that no net change in numbers
occure
Equilibrium
Time
At the melting point a solid begins to change into a liquid as heat is added. As long no heat is added or removed melting (red arrows) and freezing (black arrows) occur at the same rate an the number of particles in the solid remains constant.
Dynamic equilibria also occurs with melting and sublimation and also in most chemical reactions.
Reaction Rate of the forward reaction =
aA + bB cC + dDReactants Products
= Reaction Rate of Reverse reaction
Solid
Liquid
The equilibrium vapor pressure is the pressure exerted by a vapor over its liquid phase (measured under vacuum) when a dynamic equilibrium exists between condensation and evaporation.
Evaporation
Liquid
Vapor Pressure
Liquid
In open containers molecules that have enough KE can overcome IMF’s at the surface and “evaporate” into the atmosphere.
In closed containers, molecules vaporize and condense until there is no further change in concentration. The trapped molecules in the gas phase form a equilibrium “vapor pressure” above the liquid’s surface.
no equilibrium equilibrium
The vapor pressure of a liquid is the partial pressure exerted by the liquid’s vapor when it is in dynamic equilibrium with a liquid at a constant temperature.
As T " vapor pressure "
Vapor Pressure
As a liquid’s temperature increases, so does its KE. The number of molecules with enough energy to overcome IMF’s increase allowing vaporization.
KE = Ek = 32
RT
Kinetic Energy Temperature
KE needed to overcome IMF in liquid phase
Speed
Molecules with enough speed to vaporize
T1 T2
T1
T2 > T1
The temperature at which the vapor pressure equals the external pressure over the liquid is the boiling point.
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm.
Evaporation Boiling
The vapor pressure of a pure liquid depends on the IMF’s between molecules. The stronger the attractive IMF forces in the liquid phase => the lower the vapor pressure => the less volatile the liquid it is.
Vapo
r Pre
ssur
e (to
rr)
Temperature (˚C)
Which of the following has the highest vapor pressure at 1 atm?
Which is the least volatile at 1 atm?.66 atm
2 atm
Shape of the curve is exponential
1.00 atm
Vapor pressure plotted as a function of temperature
ln (vapor pressure) plotted as a function of 1/Temp
If we plot ln (vapor pressure) vs 1/Temp we observe a linear relationship.
1/T
ln P
Vap
or p
ress
ure
(tor
r)
Temperature ˚C
• The Clausius-Claperyron equation relates the vapor pressure (P) of a pure liquid to the liquid’s temperature (T) and the liquids molar heat of vaporization (!Hvap).
• By taking measurements at two temps, we get:
slope = !Hvap/R
1/T
ln P
ln
P2P1
= -!HvapR
1T2
"1T1
#
$ %
&
' (
y = m x + b
ln P = -!Hvap
R1T"
# $
%
& ' + C
R = 8.31 J/K mol
Vapor Pressure of Some LiquidsVapor Pressure (torr) of Some Liquids
Heat of Vaporization, Boiling Pts of Liquids
Vapor pressure of pure etOH is 115 torr at 34.9˚C. If !Hvap = 40.5 kJ/mol calculate the temperature when the vapor pressure of etOH is 760 torr. R is the gas constant given at 8.314 J/mol K
P1=115 torrT1= 34.9 + 273.15 = 308.0K
P2=760 torr
Vapor pressure of pure etOH is 115 torr at 34.9˚C. If !Hvap = 40.5 kJ/mol calculate the temperature when the vapor pressure of etOH is 760 torr. R is the gas constant given at 8.314 J/mol K
!Hvap =40.5 kJ/mol
ln
P2P1
= -!HvapR
1T2
"1T1
#
$ %
&
' (
ln 760 torr115 torr =
-40.5 x103 J/mol
8.314 J/mol*K
1T2
1308K-
T2 = 350K = 77˚C
The following diagram shows a close-up view of part of the vapor-pressure curves for a solvent (red curve) and a solution of the solvent with a second liquid (green curve). Which solvent is more volatile?
The following diagram shows a close-up view of part of the vapor-pressure curves for a solvent (red curve) and a solution of the solvent with a second liquid (green curve). Which solvent is more volatile?
The more volatile solvent will have a higher vapor pressure (more gas molecules in the gas phase above its liquid). The green solvent has a higher vapor pressure at all temperatures. It has weaker IMF’s.
AB is the solid-liquid (sublimation) curve for ice; BD the vapor pressure curve for water-steam; BC the solid-liquid (melting point line); point B the triple point point D labels the critical point.
Solid
Liquid
Gas
374
D3
Pres
sure
(Atm
) 1.0
218
0.006
NormalBoiling Point
Triple Point
Critical PointFreezingPoint
A
.43
Melting
Freezing
Evaporation
Condensation
Sublimation
Deposition
A phase diagram summarizes the conditions at which a substance exists as a solid, liquid, or gas phase of matter.
Gas
SupercriticalFluid
Line AB is the vapor-pressure curve for solid-vapor (sublimation curve); BD the vapor pressure curve for liquid-vapor water; BC the melting point line (notice how it has a negative slope = solid is lower density); point B the triple point: only point when three phases are in equilibrium; point D is the critical point--the temperature at which no pressure can liquefy a gas (new phase of matter).
A phase diagram summarizes the conditions at which a substance exists as a solid, liquid, or gas phase of matter.
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy, no matter how great the applied pressure.
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature.
The triple point is the single P,T point at which all three phases are in equilibrium with one another.
The normal boiling point is the temperature at which the vapor pressure of a liquid equals 1 atm.
The normal freezing point is the temperature at which the liquid is in equilibrium with its solids phase at 1 atm.
Terminology in a Phase Diagram Water is one of the few substances on earth for which the liquid phase is more dense than its solid phase. We see this in its phase diagram as the negative slope of the solid-liquid line (higher pressure decreases the melting point)
more pressure at a fixed temperature leads to a liquid not a solid
more pressure at a fixed temperature leads to a solid not a liquid
CO2H2O
Water CO2
K1000
760
From Line J to E what is occurring?Between what two points is the vapor pressure curve.Which two points define the melting point curveWhat is the normal boiling point of this substance?What is the triple point of this substance?
A
C
B
Temp ˚C
Pres
sure
(tor
r)
From Line E to G what is occurring?
D
GAS SOLIDLIQUID
Generally speaking, which phase of a substance should have the highest density (mass/volume)? Name a compound that does not follow this trend?
Electronegativity is an element’s inherent property to draw electrons to itself when chemically bonded to another atom in a molecule.
F is the most electronegative element
Cs the least
Lookout for bonds having: F, O, Cl, N, Br, I
Polar with net dipolemoment
!+!-
!- !-
!-
!-
!-
!-
!+!+!+
non-polar with no net dipole moment
non-polar with no net dipolemoment
Polar with net dipolemoment
Polar with net dipolemoment
Connecting Dots: large electronegativity differences in a bond => polar molecules => large dipole moments => gives rise to large IMF => decreased vapor pressure => increased boiling and melting points, higher viscosity, higher surface tension.
We must consider molecular and electron geometry to determine a molecule is polar!
2 EG
3 EG 4 EG
5 EG 6 EG
2. Dipole-dipole: occurs between neutral polar molecules (they can be different or the same interacting molecules).
1. Ion-dipole: occur between neighboring an ion solution and a polar molecule (dipole) also in solution.
Na+
4. London Dispersion Forces are attractive IMF’s that occur between all molecules. Spontaneous dipoles are formed randomly or induced by other charged species in neutral polarizable molecules.
3. Induced-dipoles: occur when a ion or a dipole induces a spontaneous dipole in a neutral polarizable molecule. +- +-
Ion or Dipole Induced Dipole
Intermolecular forces result from electrostatic forces between molecules. There are 4-types of IMF’s.
Summary of 4 types of IMF’s
--an ion interacts with a polar molecule (dipole).
--two polar molecules (dipoles) interact electrostatically.
--ion or a dipole induces a dipole in a polarizable non-polar molecule.
----also called London Dispersion Forces ----occurs between non-polar molecules via polarizability
Highest
Lowest
Type of Interaction
+-+Ion Dipole
Strength Energy (kJ/mol)
50-600
5-25
2-15
0.05-40
Interaction
DipoleDipole
+-+-
Induced Dipole
+-+Ion
Induced Dipole
+-Induced Dipole
+-
Dipole
+- Energy of Attraction is:
Coulomb’s Law: The electric force acting on a point charge q1 as a result of the presence of a second point charge q2 some r meters away is given by:
Intermolecular forces arise from electrostatic forces and depend on distance between molecules.
F =k q1 q2
r2
E =k q1 q2
r
Because Energy = Force x Distance
+ -
k depends on medium (kair = 9.0 x 109 N•m2/C2)
units of coulombs
units of metersr
+-
If we can predict charge separation or polarity in molecules we can identify IMF’s in play for that molecule.
Comparing the strengths of Intermolecular Forces
Force Strength Energy-Distance Interactions
London Forces 1-10 kJ/mole 1/r6 All-molecules
Dipole-Dipole 3-40 kJ/mole 1/r3 Polar
molecules
Hydrogen Bonding 10-40 kJ/mole 1/r3 O-H
N-H H-F
Ion-Dipole 10-50 kJ/mole 1/r2 Ions-Polar molecules
Ion-Ion >>200 kJ/mole 1/r Cation-Anion
1. Ion-dipole forces are attractive forces between an ion in solution and a neighboring polar molecule.
An anionin solution
Molecules with a dipole moment in solution
A cationin solution
- end of dipole attracted to + ions
+ end of dipole attracted to -ions
+-
2. Dipole-Dipole intermolecular forces are attractive forces that occur between polar molecules (same or not the same).
Notice the very large difference in boiling point caused by IMF’s present in acetone (lacking in butane)!
VS
Polar Molecule (dipole moment)
Non-polar Molecule (no dipole moment)
3. Hydrogen bonding is a special case of dipole-dipole intermolecular force that occurs between a hydrogen atom and an unshared pair of electrons in a polar N-H, O-H, or H-F bond.
Water is a highly polar molecule due to oxygen inherent ability to attract electrons more so than hydrogen.
There is more electron density near the oxygen atom and less around the hydrogen (arrows).
This polarity property gives water its dissolving properties of other polar substances.
The boiling points of covalent binary hydrides increase with increasing molecular mass down a Group but the hydrides of NH3, H2O and HF have abnormally high BP because of hydrogen bonding.
?Increased London Forces
Period
Tem
pera
ture
˚C
100
0
-100
As dipole moments increase in polar substances of the same mass, IMF’s increase as do boiling points (lower vapor pressure) and melting points.
Non-polar
Very polar
Polar
4. London Dispersion Forces (or Van deer waal forces) are attractive intermolecular forces when temporary dipoles are formed due to random electron motions in all polarizable molecules. Two non-polar
polarizable molecules
Induced-Dipole Momentand IMF between molecules
spontaneous movement of electrons forming an instantaneousdipole moment
Polarizability is the ease with which an electron distribution (cloud) in the atom or molecule can be distorted by an outside ion or dipole.
non-polar molecule (electron cloud)
induced dipoledispersion
London Dispersion Force
spontaneousinduced dipole that depends on polarizability of substance
All molecules have at least this IMF
Polarizability and London dispersion trends mirror atomic size trends in the periodic table.
POLARIZABILITY
•Size and polarizability increases down a group
•Polarizability increases right to left (bigger size more distortable.
•Cations are less polarizable than their parent atom because they are smaller and more compact.•Anions are more polarizable than their parent atom because they are larger.
•Polarizability increases with molar mass (# e-) of a molecule
London dispersion forces increase with size and polarizability.
Increasing Size &
Polarizability
Increasing Size/Polarizability
•Cations are less polarizable than their parent ground state atoms while anions have a larger polarizability than their ground state atoms.
Rank the following in order of increasing polarizability:
Na vs Na+ Br vs Br- Mg2+ vs Al Cl- vs Cl
Rank the following in order of increasing polarizability:
Na > Na+ Br < Br- Mg2+ < Al Cl- > Cl
Larger unbranched molecules are more polarizable than compact branched molecules.
neopentane bp=10 oC tetrahedral
normal pentanebp=36 oC
extended structure
Larger polarizability in unbranched molecules explains boiling and melting points trends in isomers.
1. Ion-Ion > Ion-Dipole > Dipole-Dipole > London Forces
4. For non-polar molecules of the same molecular mass, longer less compact molecules are more polarizable than compact molecules and show London forces.
2. For polar molecules of approximately the same mass and shape and volume (i.e. polarizability the same), dipole-dipole forces dictate the difference in physical properties.
3. Hydrogen bonding (special dipole-dipole) occurs with specific polar bonds: H-F, O-H, H-N C=O and an unshared pair of electrons on a nearby electronegative atom usually F, O, or N.
5. For non-polar molecules of widely varying molecular mass, those with more mass are typically more polarizable and experience greater London dispersion forces and exhibit higher boiling points and melting points.
Key Generalizations Good To Know About IMF’s
Flowchart for classifying intermolecular forces ! Arrange the following substances in order of increasing boiling points.
C2H6, NH3, Ar, NaCl
Ar < C2H6 < NH3 < NaCl nonpolar nonpolar polar ionic London London dipole-dipole H-bonding ion-ion
! Arrange the following substances in order of increasing boiling points.
C2H6, NH3, Ar, NaCl, AsH3
Which of the following substances exhibits H bonding? For those that do, draw two molecules of the substance with the H bonds between them.
C2H6(a) CH3OH(b) CH3C NH2
O(c)
PLAN: Find molecules in which H is bonded to N, O or F. Draw H bonds in the format -B: H-A-.
SOLUTION:
Which of the following substances exhibits H bonding? For those that do, draw two molecules of the substance with the H bonds between them.
C2H6(a) CH3OH(b) CH3C NH2
O(c)
PLAN: Find molecules in which H is bonded to N, O or F. Draw H bonds in the format -B: H-A-.
(a) C2H6 has no H bonding sites.
(c)(b)C O HH
HH
COHH
HH
CH3C N
O
H
H
CH3CN
O
H
H
CH3CN
O
H
H
CH3CN
O
H
H
Which substances experience dipole-dipole intermolecular forces? SiF4,CHBr3, CO2, SO2
SiF4, geometry tetrahedral, Si-F bonds are polar, but no molecular dipole; bond dipoles cancel. No dipole-dipole interactions.
CO2, linear geometry, C-O bonds are polar but symmetry cancels net diple, but no molecular dipole; bond dipoles cancel and no dipole IMF’s
SO2, bent geometry, S-O bonds are polar and do not cancel. Sulfur lone pair dipole only partially offsets net bond dipole. Has dipole-dipole forces
CHBr3, tetrahedral geometry, C-H and C-Cl bonds are polar and do not cancel
SO2 and CHCl3 experience dipole-dipole intermolecular forces.
Which substances experience dipole-dipole intermolecular forces? SiF4,CO2, SO2, CHBr3,
Arrange the following non-polar molecules in order of increasing melting point.
SiF4, CS2, CI4, GeCl4
Arrange the following non-polar molecules in order of increasing melting point.
SiF4, CS2, CI4, GeCl4Solution. None of these molecules poses a net dipole moment. Only dispersion forces exist and these are expected to increase with increasing molecular mass (more polarizable as a molecule gets larger). The molar masses of these substances follow:
SubstanceMolar MassCS2 76.131SiF4 104.077CI4 519.631GeCl4 214.402
The intermolecular forces, and the melting points, should increase in the following order: CS2 < SiF4 < GeCl4 < CI4 The experimentally determined melting points are -110.8, -90, -49.5, and 171 oC, respectively.
For each pair of substances, identify the dominant intermolecular forces in each substance, and select the substance with the higher boiling point.
(a) MgCl2 or PCl3(b) CH3NH2 or CH3F(c) CH3OH or CH3CH2OH(d) Hexane
(CH3CH2CH2CH2CH2CH3) or 2,2-dimethylbutane
CH3CCH2CH3
CH3
CH3
PLAN: •Bonding forces are stronger than nonbonding(intermolecular) forces.
•Hydrogen bonding is a strong type of dipole-dipole force.
•Dispersion forces are decisive when the difference is molar mass or molecular shape.
SOLUTION:(a) Mg2+ and Cl- are held together by ionic bonds while PCl3 is covalently bonded and the molecules are held together by dipole-dipole interactions. Ionic bonds are stronger than dipole interactions and so MgCl2 has the higher boiling point. (b) CH3NH2 and CH3F are both covalent compounds and have bonds which are polar. The dipole in CH3NH2 can H bond while that in CH3F cannot. Therefore CH3NH2 has the stronger interactions and the higher boiling point.(c) Both CH3OH and CH3CH2OH can H bond but CH3CH2OH has more CH for more dispersion force interaction. Therefore CH3CH2OH has the higher boiling point.(d) Hexane and 2,2-dimethylbutane are both nonpolar with only dispersion forces to hold the molecules together. Hexane has the larger surface area, thereby the greater dispersion forces and the higher boiling point.
For each pair of substances, identify the dominant intermolecular forces in each substance, and select the substance with the higher boiling point.
Any questions on homework from Chapter 11?Name the types of intermolecular forces in the followingcompounds.
CH3CH2CH2CH3CH3OH CH3CONH2 CH3COOH CH3CH2CH2CH2OH CH3CH2COCH3
C2H6, HCHCl3, CH3CH2OH, HNO3, PH3
In which substances would hydrogen bonding forces occur between molecules?
C2H6, HCHCl3, CH3CH2OH, HNO3, PH3
Solution. Hydrogen is bonded to one of the very electronegative atoms in CH3CH2OH and HNO3. Hydrogen bonding should occur in both of these substances.
In which substances would hydrogen bonding forces occur between molecules?
Water Is A Unique Substance
•great solvent due to high polarity and hydrogen bonding ability
•exceptional high specific heat capacity and "Hvaporization
•high surface tension and capillarity
•large density differences of liquid and solid states
Ice is less dense than water
Maximum Density40C
Density of Water
Temperature ˚C
Den
sity
(g/c
m3 )
--High H-bonding = high cohesive forces responsible for the transport of water in roots and xylem of trees and plants.
--Base-pairing in double-stranded DNA
--Heat capacity, high heat of vaporization, inverted density of water and ice.
Water Is Wild and Unique and Taken For Granted
Three common physical properties of liquids that depend on the magnitude of IMF’s include surface tension, capillarity and viscosity.
surface tension
capillarity
ViscosityScientists say that the leaf has a “low surface energy” as water will not spread out (it takes energy to spread the water out)
molecules at the surface feel
a net force downward
molecules in the interior experience
equal force in 3-D
Surface tension is the amount of energy required to stretch or increase the surface area of a liquid by a unit area (J/m2).
Substance Formula Surface Tension (J/m2) at 200C
Major IMF’s
diethyl ether
ethanol
butanol
water
mercury
dipole-dipole; dispersion
H bonding
H bonding; dispersion
H bonding
metallic bonding
1.7x10-2
2.3x10-2
2.5x10-2
7.3x10-2
48x10-2
CH3CH2OCH2CH3
CH3CH2OH
CH3CH2CH2CH2OH
H2O
Hg
Surface Tension of Some LiquidsThe physical properties of a pure liquid are related to its intermolecular forces.
Capillary effect occurs when the adhesive IMFʼs between a liquid and a substance are stronger than the cohesive IMFʼs inside the liquid.
Cohesion is the intermolecular attraction between like molecules.
Adhesion is an attraction between unlike molecules
AdhesionCohesion
– Capillary rise implies that the:• Adhesive forces > cohesive forces
– Capillary fall implies that the:• Cohesive forces > adhesive forces
Si-O glass
Water will rise to different heights in tubes depending on the diameter of the capillary. Can you surmise why and how?
Water will rise to different heights depending on the diameter of the capillary. Why?
In general, viscosity decreases as temperature increases and visa versa.
• Viscosity is the measure of a liquid’s resistance to flow relative to one another, and is thus related to the intermolecular forces.
! Oil for your car is bought based on this property: 10W30 or 5W30 describes the viscosity of the oil at high and low temperatures (W means winter dudes higher the number the more viscous).
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Figure 12.49 The viscosity of a polymer in solution.