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Ch. 11: Liquids and Intermolecular Forces
Learning goals and key skills:
Identify the intermolecular attractive interactions (dispersion, dipole-dipole, hydrogen bonding, ion-dipole) that exist between molecules or ions based on their composition and molecular structure and be able to compare the relative strengths of these intermolecular forces Explain the concept of polarizability and how it relates to dispersion forces Explain the concepts of viscosity and surface tension in liquids Know the names of various changes of state for a pure substanceInterpret heating curves and be able to calculate quantities related to temperature and enthalpies of phase changes Define critical pressure, critical temperature, vapor pressure, normal boiling point, normal melting point, critical point, and triple point Be able to interpret and sketch phase diagrams. Explain how water’s phase diagram differs from most other substances, and why.Understand how the molecular arrangements characteristic or nematic, smectic, and choloesteric liquid crystals differ from ordinary liquids and from each other. Be able to recognize the features of molecules that favor formation of liquid crystalline phases.
GasesGases
In gases the particles/molecules are far apart and the intermolecular forces between them are weak.
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... vs liquids and solids... vs liquids and solids
In liquids and solids this is not true. Instead we must consider the intermolecular forces, i.e., the forces that exist between atoms and molecules.
Strength of intermolecular attractionsStrength of intermolecular attractions
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States of MatterStates of MatterThe fundamental difference between states of matter is the distance between particles.
Because in the solid and liquid states particles are closer together, we refer to them as condensed phases.
The States of MatterThe States of Matter
The state a substance is in at a particular temperature and pressure depends on:
– The kinetic energy of the particles– The strength of the attractions between the
particles
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Intermolecular ForcesIntermolecular Forces
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Intermolecular ForcesIntermolecular Forces•ion-dipole•hydrogen bonding
•dipole-dipole van der Waals•dispersion forces (London) forces
The strength of these intermolecular forces is directly related to the melting/boiling points, enthalpy of fusion, enthalpy of vaporization, and solubility of the substances.
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Dispersion forcesDispersion forces
These induced dipoles lead to intermolecular attractions. This is how nonpolar gas molecules (e.g. He, N2, etc.) can liquefy.
The ease with which the electron distribution in a molecule is distorted is called its polarizability.
London dispersion forcesLondon dispersion forcesDispersion forces tend to increase with increasing molecular weight.
MW Boiling Pt.(g/mol) (K)
F2 38 85Cl2 71 239Br2 160 332I2 254 458
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Factors Affecting Dispersion ForcesFactors Affecting Dispersion Forces
• The shape of the molecule affects the dispersion forces: long, thin molecules (like n-pentane tend to have stronger dispersion forces than short, round ones (like neopentane).
• This is due to the increased surface areain n-pentane.
Polar Covalent Bonds and Electronegativity
Polar Covalent Bonds and Electronegativity
Although atoms often form compounds by sharing electrons, the electrons are not always shared equally.
Electronegativity is the ability of atoms in a molecule to attract electrons to itself.
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Dipole-Dipole ForcesDipole-Dipole Forces
Dipole-Dipole Forces
Dipole-Dipole Forces
Influence of dipole-dipole forces is seen in the boiling points of simple molecules.
MW (g/mol) Boiling Pt (K)N2 28 77CO 28 81Br2 160 332ICl 162 370
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Dipole-Dipole ForcesDipole-Dipole ForcesDipole-dipole forces increase with increasing polarity (dipole moments).
MW Dipole Boiling(g/mol) Moment, μ (D) Pt (K)
CH3CH2CH3 44 0.1 231CH3OCH3 46 1.3 248CH3CHO 44 2.7 294CH3CN 41 3.9 355
Hydrogen BondingHydrogen BondingA special form of dipole-dipole attraction, which enhances dipole-dipole attractions.
H-bonding is strongest when X and Y are N, O, or F
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Boiling Points of Simple Hydrogen-Containing Compounds
Base-Pairing through H-BondsBase-Pairing through H-Bonds
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Portion of a DNA chain
Double helix of DNA
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Ion-dipole forcesIon-dipole forces
HH
water dipole••
••
O-
+
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Intermolecular Forces SummaryIntermolecular Forces Summary
Ionic bonding Ions
Ion-dipole Ion charge, dipole moment
Hydrogen bondingVery polar H-X bond and lone pair of e-
(X = N, O, F)Dipole-dipole Dipole moment
Dispersion forces (Induced dipole-
induced dipole) onlypolarizabilityIn
crea
sing
Stre
ngth
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“Like dissolves like”“Like dissolves like”
Polar molecules are more likely to dissolve in a polar solvent, and nonpolar molecules are more likely to dissolve in a nonpolar solvent.
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LiquidsLiquids
• molecules close together → appreciable intermolecular forces
• liquids are almost incompressible• molecules are in constant motion• finite volume, but indefinite shape
ViscosityViscosity
The resistance of liquids to flow
Increases with stronger intermolecular forces and decreases with increasing temperature.
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Surface TensionSurface Tension
Surface tension results from the net inward force experienced by the molecules on the surface of a liquid.
Cohesive and adhesive forcesCohesive and adhesive forcesCohesive forces
- between the liquid molecules
Adhesive forces- between the liquid molecules and another substance
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Properties of liquidsProperties of liquids
Viscosity – the resistance of liquids to flow; “the thickness”
Surface tension – energy required to break through the surface or to disrupt a liquid drop and make the drop spread out like a film.
Cohesive forces – forces between liquid molecules
Adhesive forces – forces between liquid molecules and another substance
Capillary action – the rise of liquids up narrow tubes and other surfaces
Phase changesPhase changes
Changes of state involve energy (at constant T)
Ice + (heat of fusion) → waterWater + (heat of vaporization) → steam
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Phase changesPhase changes
Energy changes associated with changes of state
Energy changes associated with changes of state
For water DHfus = 6.01 kJ/mol or 334 J/gDHvap = 40.67 kJ/mol or 2258 J/g
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Heating/Cooling Curve for WaterHeating/Cooling Curve for Water
Heating/Cooling Curve for WaterHeating/Cooling Curve for Water
The total heat is the sum of each step qtotal = ∑ qiThe total heat is the sum of each step qtotal = ∑ qi
The ice is heated qice = sicemΔT
The ice melts to water qfus = ΔHfusm
The water is heated qwater = swatermΔT
The water evaporates to steam qvap = ΔHvapm
The steam is heated qsteam = ssteammΔT
The ice is heated qice = sicemΔT
The ice melts to water qfus = ΔHfusm
The water is heated qwater = swatermΔT
The water evaporates to steam qvap = ΔHvapm
The steam is heated qsteam = ssteammΔT
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Example: Heating curve problemExample: Heating curve problemDetermine the amount of heat (in kJ) required to heat 500. g of ice from -50.0 °C to steam at 200. °C.
sice = 2.09 J/gKDHfus = 334 J/gswater = 4.184 J/gKDHvap = 2258 J/gssteam=1.84 J/gK
qice, -50.0 to 0 °C = +52 250 J
qfus, ice to water = +167 000 J
qwater, 0 to 100 °C = +209 200 J
qvap, water to steam = +1 129 000 J
qsteam, 100 to 200. °C = +92 000 J
qtotal = 1 650 000 J = 1650 kJ
Note that most of the heat is used to convert water into steam.
Example: Heating curve problemExample: Heating curve problemDetermine the amount of heat (in kJ) required to convert 42.0 g of ethanol from -155 °C to 78 °C. ethanol (C2H5OH) melts at -114 C and boils at 78 °C. The enthalpy of fusion of ethanol is 5.02 kJ/mol and the enthalpy of vaporization is 38.56 kJ/mol. The specific heats of solid and liquid ethanol are 0.97 J/gKand 2.3 J/gK, respectively.
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Critical Temperature: The highest temperature at which a distinct liquid phase can form
Critical Pressure: Pressure required to bring about liquefaction at this critical temperature.
Supercritical Fluid: Above this critical point, we have a supercritical fluid where the density is similar to a liquid and the viscosity is similar to a gas.
Vapor PressureVapor Pressure
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Temperature effects on the distribution of energy in a liquid
Temperature effects on the distribution of energy in a liquid
E needed to evaporate liquid
Liquids that evaporate rapidly are volatile
Vapor pressure curvesVapor pressure curves
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Equilibrium Vapor Pressure & the Clausius-Clapeyron Equation
Equilibrium Vapor Pressure & the Clausius-Clapeyron Equation
• Clausius-Clapeyron equation — used to find heat of vaporization, ∆Hvap.
• The logarithm of the vapor pressure P is proportional to ∆Hvap and to 1/T.
• ln P = –(∆Hvap/RT) + C
D
121
2 11lnTTR
HPP vap
Clausius-Clapeyron EquationClausius-Clapeyron Equation
Liquid bromine has a vapor pressure of 400. torr at 41.0 °C and a normal boiling point of 58.2 °C. Calculate the heat of vaporization of bromine (in kJ/mol).
D
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2 11lnTTR
HPP vap
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Phase DiagramsPhase DiagramsPhase diagrams display the state of a substance at various pressures and temperatures and the places where equilibria exist between phases.
Normal Phase Diagram: CO2Normal Phase Diagram: CO2
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Phase Diagram of WaterPhase Diagram of Water
Note the high critical temperature and critical pressure.
–These are due to the strong van der Waals forces between water molecules.
Info from P-T phase diagramsInfo from P-T phase diagramsFrom a (pressure-temperature) phase diagram we can find:
• the normal melting point• the normal boiling point• the triple point• whether the liquid is more or less dense than the solid
• the vapor pressure curve• the sublimation pressure curve
• the critical temperature (Tc) and critical pressure (Pc)
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Liquid CrystalsLiquid Crystals
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Liquid Crystals• Rod-shaped and contain double or triple bond near the middle• Polar groups create a dipole moment and promote alignment
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