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Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008)
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Chapter 8
CHAPTER 8 PHYSICAL EQUILIBRIA
Ocean:
Largest solution on earth
1.4 x 1021 kg of the Earth’s surface water
How to purify?
◈ PHASE (상,相) AND PHASE TRANSITIONS (상전이,相轉移)
8.1 Vapor Pressure
▶ Vapor pressure: Pressure of a vapor in equilibrium with its liquid phase at a given temperature
Fig. 8.1 Addition of a water drop in the vacuum above the mercury level.
Æ Vapor pressure is kept constant as long as some liquid remains at a fixed T
▷ Volatile liquid : High vapor pressure
ex. Methanol
▶ Dynamic equilibrium of vaporization
In a closed vessel,
Rate of evaporation = Rate of condensation
⎯⎯
→ H 2 O( g )
H 2 O(l ) ←⎯
⎯
Fig. 8.2 Dynamic equilibrium between
a liquid and its vapor.
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Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008)
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Chapter 8
8.2 Volatility (휘발성,揮發性) and Intermolecular Forces
Volatile liquid: Liquid with weakly bonded molecules Æ High vapor pressure
Hydrogen bondings in liquid lower the vapor pressure
Ex. Dimethyl ether (C2H6O, 3HC-O-CH3) and Ethanol (C2H6O, 3HC-CH2-OH)
Å Hydrogen bonds
3982 Torr at 20oC
40 Torr at 19oC
8.3 The Variation of Vapor Pressure with Temperature
Vapor pressure of a liquid ∝ Temperature
Temperature dependence of vapor pressure
vs. Intermolecular interaction
Vapor pressure of water near its
normal boiling point
Fig. 8.3 Vapor pressures of liquids.
At equilibrium,
∆Gvap = Gm ( g ) − Gm (l ) = 0
For a liquid, Gm (l , P ) = Gm (l )
o
Å no pressure dependence
For an ideal gas,
Gm ( g , P ) = Gm ( g ) + RT ln P
o
Fig. 8.4 Variation of molar Gibbs energy with pressure
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Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008)
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Chapter 8
o
o
∆Gvap = ⎡⎣Gm ( g ) + RT ln P ⎤⎦ − Gm (l )
o
o
o
= ⎡⎣Gm ( g ) − Gm (l ) ⎤⎦ + RT ln P = ∆Gvap + RT ln P
∆Gvap : Standard Gibbs free energy of vaporization
o
∴
At equilibrium, ∆Gvap = 0 .
ln P = −
∆Gvap
RT
∆H vap , ∆S vap :
o
o
0 = ∆Gvap + RT ln P
o
∆H vap
o
=−
RT
o
+
∆S vap
o
R
approximately independent of temperature
⎛ ∆H vap o ∆S vap o ⎞ ⎛ ∆H vap o ∆S vap o ⎞ ∆H vap o ⎛ 1 1 ⎞
+
+
ln P2 − ln P1 = ⎜ −
⎟ −⎜−
⎟=
⎜ − ⎟
⎜
RT2
R ⎟⎠ ⎜⎝
RT1
R ⎟⎠
R ⎝ T1 T2 ⎠
⎝
◈
Clausius-Clapeyron equation
P ∆H vap ⎛ 1 1 ⎞
ln 2 =
⎜ − ⎟
P1
R ⎝ T1 T2 ⎠
o
⎛1 1⎞
P
∆H vapo >0
T2 > T1 Æ ⎜ − ⎟ > 0 ⎯⎯⎯⎯⎯
→ ln 2 > 0 Æ P2 > P1
P1
⎝ T1 T2 ⎠
Vapor pressure increases with increasing temperature.
The increase is greatest for substances with high
cf. Ethanol ( ∆H vap
o
∆H vap (strong intermolecular interactions)
o
= 43.5 kJ ⋅ mol −1 ) vs. benzene ( ∆H vap = 30.8 kJ ⋅ mol −1 )
o
8.4 Boiling
Æ Rapid vaporization throughout the liquid forming vapor bubbles at 1 atm
▶ Normal boiling point: Temperature at which the vapor pressure equals to the atmospheric pressure
Ex. 8.2
Estimating the normal boiling point of a liquid.
Vapor pressure of ethanol : 13.3 kPa at 34.9oC
P ∆H vap ⎛ 1 1 ⎞
ln 2 =
⎜ − ⎟,
P1
R ⎝ T1 T2 ⎠
o
P2 = 13.3 kPa ,
∆H vap = 43.5 kJ ⋅ mol −1
o
T2 = 34.9 + 273.15 K = 308.0 K
P1 = 1 atm = 101.325 kPa
Æ ∴
T1 = 350 K (~ 77oC)
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Chapter 8
8.5 Freezing and Melting
Normal freezing point ( = Normal melting point )
Tf
Å at 1 atm
Supercooling Æ liquid phase below its freezing point
Increased pressure prefers denser phase:
▶ Fe :
Tf = 1800 K at 1 atm
Æ
Tf′ = 1805 K at 1000 atm
Æ Melting point increases slightly with increasing pressure
Æ At the center of Earth, very high pressure keeps
Fe as a solid even at very high temperature.
▶ Water: Melting point decreases with increasing pressure
Æ Unusual low density of ice compared to liquid water
Æ Collapsing of the hydrogen bonds in open structure of ice
Fig. 8.5 Open structure of ice.
8.6 Phase Diagrams
◈ Phase diagram: A map showing which phase is most stable at a different P and T
▶ Phase boundary : A line along which two phases are in equilibrium
▶ Triple point (三重點): A point where three phase boundaries meet (three phases are in equilibrium)
Water: Slight decrease in freezing point with increasing pressure
Steep negative slope of solid/liquid phase boundary (liquid is more dense)
Triple point of water (4.6 Torr and 0.01oC) is used to define the size of kelvin:
There are 273.16 kelvins between absolute zero and the triple point.
Since the normal freezing point of water is 0.01 K below the triple point,
0oC corresponds to 273.15 K.
At very high pressure, several solid phases exist. (Ice-VIII above 20000 atm and 100oC)
CO2:
Vapor at 10oC and 2 atm Æ
condenses to liquid at 10oC and 10 atm
At 1 atm, there is no liquid phase and the solid sublimates to vapor directly.
Steep positive slope of solid/liquid phase boundary (solid is more dense)
Sulfur: Two solid phases (rhombic and monoclinic) and three triple points
No “quadruple point” Å All four phases are not in equilibrium
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Fig. 8.6 The phase diagram for water.
Fig. 8.7 The phase diagram for CO2.
Fig. 8.9 The phase diagram for water.
담당교수: 신국조
Chapter 8
Fig. 8.8 The phase diagram for sulfur.
Fig. 8.10 Vapor pressure of water
Fig. 8.11 Changes undergone by a liquid as its pressure is decreased at constant temperature.
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Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008)
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Chapter 8
8.7 Critical Properties (임계성질,臨界性質)
Fig. 8.12
Fig. 8.14
Fig. 8.13 The critical phenomenon.
▶ Supercritical fluids :
“Critical opalescence”
Dense fluid above Tc and Pc , Used as solvents
▷ Supercritical CO2 (Tc = 31oC, Pc =73 atm) : No worry about contamination of solvents
Removing caffeine from coffee beans, Extracting perfumes from flowers
▷ Supercritical hydrocarbons: Dissolve coals and separate it from ash,
Extracting oils from oil-rich tar sands
◈ SOLUBILITY
8.8 The Limits of Solubility
▶ Molar solubility, s : Molar concentration of a substance in a saturated solution
Opal
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Chapter 8
Fig. 8.16 The unsaturated and saturated aqueous glucose solution.
Fig. 8.15 The events taking place at the interface of a solid ionic solute and a solvent (water).
Fig. 8.17 The solute in a saturated solution is in dynamic equilibrium with the undissolved solute.
8.9 The Like-Dissolves-Like Rule
Polar solvent (water) dissolves polar or ionic compounds
Dry cleaning : hexane and tetrachloroethene (Cl2C=CCl2)
dissolve nonpolar compounds (wax)
▶ Formation of a dilute solution
Step 1. Separation of solute molecules from one another.
Step 2. Some solvent molecules move apart forming cavities.
Step 3. Solutes occupy cavities in solvent releasing energy.
Overall energy change is the sum of the energies involved
in these steps. Æ Red arrow ⇓ in Fig. 8.18
Fig. 8.18 Formation of dilute solution.
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▶ London forces,
EP ∝ −
담당교수: 신국조
Chapter 8
α1α 2 (induced dipole-induced dipole)
r6
Principal cohesive forces between solute molecules
Solvent with a similar London force
ex. S8
ex. CS2
▷ Cleaning with a soap
Polar carboxylate head group: Hydrophilic (親水性)
Fig. 8.19 The solubility of sulfur in
water and carbon disulfide.
Nonpolar hydrocarbon tails: Hydrophobic (疏소水性)
Formation of micelles in soaping action
▷ Detergent (세제,洗劑) : Surfactants (界面活性劑)
Æ Sulfur atoms in polar head group
Fig. 8.20 The soaping action.
8.10 Pressure and Gas Solubility: Henry’s Law
◈ Henry’s law
(1) Solubility, s, of a gas is directly proportional to its partial pressure P:
s = kH P
(2)
Vapor pressure
k H : Henry’s constant
P2 of a volatile solute (in a mixture of two volatile liquids)
is proportional to its mole fraction
P2 = k2 X 2
X 2 in solution at low mole fractions:
k2 ( = k H ) : Henry’s law constant
William Henry
(英,1775-1836)
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Chapter 8
Fig. 6.15 from Oxtoby
Vapor pressure above a mixture of two
volatile liquids. Henry’s law is shown as dilute
solution limit of nonideal mixture.
Fig. 8.21 The variation of molar solubilities
with partial pressure.
8.11 Temperature and Solubility
▶ Solubility of a gas in water decreases with increasing temperature
Æ Increase in vapor pressure with increasing temperature
O2
14 mg/L at 0oC Æ 6 mg/L at 40oC
CO2
3.4 g/L
at 0oC Æ ~1 g/L
at 40oC
▶ Solubility of most solid compounds in water increases with increasing temperature
Exception: Na2SO4
Fig. 8.22 The variation of solubilities of solid compounds in water with temperature.
“Most gases are less soluble in warm water than in cold water; solids show a more varied behavior.”
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Chapter 8
8.12 The Enthalpy of Solution (용해溶解엔탈피)
▶ Enthalpy of solution,
∆H sol
Æ Change in molar enthalpy when a substance dissolves
▶ Limiting enthalpy of solution
Æ Enthalpy of solution in a dilute solution where solute-solute interactions are negligible
Exothermic dissolution: LiCl, AlBr3
Endothermic dissolution: NH4NO3, AgI
◆ Hypothetical two step dissolving process
Fig. 8.23 The enthalpies of solution in (a) an exothermic and (b) an endothermic cases.
1st step : Ions separating from the solid to form a gas of ions Å Lattice enthalpy
NaCl(s) ⎯⎯
→ Na + (g) + Cl − (g)
∆H L = 787 kJ ⋅ mol−1
2nd step: Gaseous ions plunging into water forming the final solution Å Hydration enthalpy (< 0)
Na + (g) + Cl − (g) ⎯⎯
→ Na + (aq) + Cl − (aq)
∆H hyd = −784 kJ ⋅ mol−1
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Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008)
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Chapter 8
∆H sol = ∆H L + ∆H hyd
NaCl(s) ⎯⎯
→ Na + (aq) + Cl − (aq)
∆H sol = ∆H L + ∆H hyd = 787 kJ ⋅ mol−1 − 784 kJ ⋅ mol −1 = 3 kJ ⋅ mol−1 > 0
Lattice
enthalpy
Enthalpy of
hydration
(Fig. 8.23b)
Enthalpy of
solution
Nitrates have big, singly charged anions. Æ Low lattice enthalpy, High hydration enthalpy (H-bonds with water)
Carbonates have big, doubly charged anions. Æ Higher lattice enthalpy, less soluble
Hydrogen carbonates (bicarbonates) have singly charged anions Æ more soluble than carbonates
▶ Hard water :
Rainwater with dissolved CO2 forms a very dilute solution of carbonic acid:
CO 2 (g) + H 2 O(l) ⎯⎯
→ H 2 CO3 (aq)
As the water flows through the grounds, the carbonic acid reacts with CaCO3 of limestone:
−
CaCO3 (s) + H 2 CO3 (aq) ⎯⎯
→ Ca 2+ (aq) + 2 HCO3 (aq)
These two reactions are reversed when water containing Ca(HCO3)2 is heated in a furnace (Purification):
−
Ca 2+ (aq) + 2 HCO3 (aq) ⎯⎯
→ CaCO3 (s) ↓ + H 2 O(l) + CO2 (g) ↑
8.13 The Gibbs Free Energy of Solution
∆G = ∆H − T ∆S
∆S > 0 for dissolution of solids Æ ∆G < 0 as long as ∆H < 0
∆S < 0 for cage formation by solvent molecules Æ ∆G < 0 only when ∆H < 0 with ∆H > T ∆S
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Chapter 8
∆G = ∆H − T ∆S
∆H > 0 for endothermic dissolving Æ ∆G < 0 only when T ∆S > 0 and that T ∆S > ∆H
Endothermic dissolving depends on the balance between
Strong ∆H ( > 0) Æ
∆S of the system and that of the surroundings.
Insoluble
∵ So much energy leaves the surroundings and enters the solution Æ Decrease in overall disorder (entropy)
Spontaneous dissolving ( ∆G < 0 ) with increasing T Æ Only when
For extensively hydrated substances and gases Æ
∆S > 0
∆S < 0 for dissolution
Æ Solubility decreases with increasing T
▶
∆G depends on the concentration of solute:
G = G o + nRT ln a
a : activity
∆G < 0 for low concentration (spontaneous dissolution) and ∆G > 0 for high concentration (insoluble)
Saturated solution when
∆G = 0
Fig. 8.24 The dissolution process for a solid. Increase in disorder.
Fig. 8.25 Spontaneous dissolution vs. spontaneous
precipitation. Saturation at ∆G = 0 .
◈ COLLIGATIVE PROPERTIES (총괄성질,總括性質)
▶ Properties that depend on the relative numbers of solute and solvent molecules and not on the chemical
identity of the solute
▶ Used to determine the molar mass of a solute
▶ Vapor pressure lowering, Elevation of boiling point,
Depression of freezing point, Osmosis
8.14 Molality, m Æ independent of temperature
cf. Molarity, M, depends on temperature
amount of solute (mol)
Molality of solute =
mass of solvent (kg)
Fig. 8.26 Preparation of solution in molality.
(3)
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Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008)
Ex. 8.5
담당교수: 신국조
Chapter 8
Calculating the molality of a solute.
10.5 g of NaCl dissolved in 250 g water Æ Molality?
molality =
TOOLBOX 8.1
nsolute (mol) (10.5 g)/(58.44 g ⋅ mol −1 )
=
= 0.719 mol ⋅ kg −1 = 0.719 m
0.250 kg
msolvent (kg)
HOW TO USE THE MOLALITY
1. Calculating the mass of solute in a given mass of solvent from the molality.
Step 1.1 Calculate the amount of solute molecules,
nsolute , present in a given mass of solvent, msolvent ,
by rearranging the equation defining molality, Eq.(3), into
nsolute = molality × msolvent
Step 1.2 Use the molar mass of solute,
M solute , to find the mass of the solute from its amount:
msolute = nsolute M solute
2. Calculating the molality from a mole fraction.
Step 2.1 Consider a solution comprised of a total of 1 mol of molecules. If the mole fraction of the solute
is
xsolute , the amount of solute molecules in a total of 1 mol of molecules is
nsolute = xsolute mol
Step 2.2 If there is only one solute, the mole fraction of solvent molecules is
solvent molecules in a total of 1 mol of molecules is then
1 − xsolute . The amount of
nsolvent = (1 − xsolute ) mol. Convert
this amount into mass in grams by using the molar mass of the solvent,
M solvent :
msolvent = nsolvent M solvent = {(1 − xsolute ) mol} M solvent
and then convert grams into kilograms.
Step 2.3 Calculate the molality of the solute by dividing the amount of solute molecules (step 1)
by the mass of solvent (step 2).
3. Calculating the mole fraction from the molality.
Step 3.1 Consider a solution containing exactly 1 kg of solvent. Convert that mass of solvent
into an amount of solvent molecules, nsolvent , by using the molar mass,
nsolvent
1 kg
103 g
=
=
M solvent M solvent
M solvent , of the solvent:
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We already know the amount of solute molecules,
담당교수: 신국조
Chapter 8
nsolute , in the solution:
nsolute = molality × (1 kg)
Step 3.2 Calculate the mole fraction from the amounts of solvent and solute molecules:
xsolute =
nsolute
nsolute + nsolvent
4. Calculating the molality, given the molarity
Step 4.1 Calculate the total mass of exactly 1 L (103 mL) of solution by using the density, d,
msolution = d × (103 mL)
of the solution:
Step 4.2 The molarity gives the amount of solute in 1 L of solution. Use the molar mass of the
solute to convert that amount into the mass of solute present in 1 L of solution:
nsolute = cV = molarity × (1 L)
msolute = msolute M solute = molarity × (1 L) × M solute
Step 4.3 Subtract the mass of solute (step 2) from the total mass (step 1) to find the mass of solvent
in 1 L of solution,
m solvent = msolution − msolute
and convert the mass of solvent into kilograms.
Step 4.4 The molality is the amount of solute (given by the molarity) divided by the mass of the solvent
in kilograms (step 3).
Ex. 8.6
Calculating a molality from a mole fraction.
Mole fraction of C6H6 (benzene) dissolved in toluene (C6H5CH3) is 0.150. Æ Molality ?
Step 2.1 Calculate the amount of solute in a total 1 mol of solution molecules:
nbenzene = xbenzene mol = 0.150 × 1 mol = 0.150 mol
Step 2.2 Find the mass of solvent (in kg) present:
Mass of toluene (kg) = (1 – 0.150)mol x (92.13 g·mol–1) x (1 kg / 103 g)
= 0.850 x 92.13 x 10–3 kg = 0.0783 kg
Step 2.3 From
molality = nsolute / msolvent
Molality of C6 H 6 =
0.150 mol
= 1.92 mol ⋅ kg −1
0.0783 kg
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Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008)
Ex. 8.7
담당교수: 신국조
Chapter 8
Converting molarity into molality
Density of 1.06 M sucrose, C12H22O11(aq), solution is 1.14 g·mL–1 Æ Molality?
Step 4.1 Find the mass of
msolution = (1.14 g ⋅ mL−1 ) × (103 mL) = 1.14 × 103 g
1 L of solution from
m = d × (103 mL)
msucrose = (1.06 mol ⋅ L−1 ) × (1 L) × (342.3 g ⋅ mol −1 ) = 363 g
Step 4.2 Find the mass of
solute in 1L solution
m water = m solution − m solute
Step 4.3 Find the mass of
= 1.14 × 103 − 363 g = 0.78 kg
water in 1 L solution
Step 4.4 From
Molality of C12H22O11
molality = nsolute / msolvent
=
1.06 mol
= 1.4 mol ⋅ kg −1
0.78 kg
8.15 Vapor-Pressure Lowering
◈ Raoult’s law:
The vapor pressure of a solvent is proportional
to its mole fraction in a solution.
P = xsolvent Ppure ( = x1P o )
▶ Ideal solution ~ obeys Raoult’s law at all concentrations.
Interactions between [solute-solvent]solution
= [solute-solute]pure state
= [solvent-solvent]pure state
Æ
∆H solution = 0
ex. Benzene/Toluene
▶ Nonideal solutions behave as ideal solutions
at low concentrations.
☺ Raoult’s law is a limiting law.
Fig. 8.27
Raoult’s law predicts that the
vapor pressure of a solvent in a solution
should be proportional to the mole fraction
of the solvent.
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Ex 8.8
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Chapter 8
Using Raoult’s law
10.00 g of nonelectrolyte sucrose, C12H22O11, dissolved in 100.0 g of water at 20oC. Æ
Pwater = ? in solution
[Solution] Amount of sucrose = 10.00 g / 342.3 g·mol–1 = 0.0292 mol
Amount of water = 100.0 g / 18.02 g·mol–1 = 0.555 mol
From
xsolvent =
nsolvent
0.555 mol
=
= 0.995
nsolute + nsolvent (0.0292 + 0.555) mol
From P = xsolvent Ppure and
Ppure (water, 20o C ) = 17.54 Torr ,
Pwater = 0.995 × (17.54 Torr) = 17.45 Torr
◈ Vapor pressure lowering (from Oxtoby)
∆P1 = P1 − P1o = x1P1o − P1o = ( x1 − 1) P1o = − x2 P1o
subscript 1 (solvent), subscript 2 (solute)
Fig. 8.28 Vapor pressure lowering of a solvent
by a nonvolatile solute in solution in a
barometer tube.
Left column: Small volume of pure water
floating on the mercury
Right column: Small volume of 10 m NaCl(aq)
solution floating on the mercury
◆ Thermodynamics of vapor pressure lowering of solvent in solution
At equil,
Gm (pure liquid solvent) = Gm (vapor from pure solvent)
In an ideal solution, solutes increases the entropy but the enthalpy remains constant.
Æ Decrease in
Gm (solvent in solution)
[ < Gm (pure liquid solvent)]
In order to reach an equilibrium,
Gm (solvent in solution) = Gm (vapor from solution)
Æ Decrease in
Gm (vapor from solution)
Æ Decrease in vapor pressure
[ < Gm (vapor from pure solvent)]
cf. Gm ( g , P ) = Gm ( g ) + RT ln P
o
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Chapter 8
8.16 Boiling-Point Elevation (비점상승沸點上昇) and Freezing-Point Depression (융점강하融點降下)
Fig. 8.29 (a) Stability of phases (b) Boiling point elevation (lowering of G due to entropy increase for solution).
Fig. 8.30 (a) Stability of phases (b) Freezing point depression
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Chapter 8
∆Tb = kb × m ,
kb : boiling point elevation constant (ebullioscopic constant)
∆Tf = −kf × m ,
kf : freezing point depression constant (cryoscopic constant)
kb (water) = 0.51 K ⋅ kg ⋅ mol −1 , kf (water) = 1.86 K ⋅ kg ⋅ mol−1
Æ
∆Tb < ∆Tf
In an electrolyte solution,
∆Tb = ikb × m and ∆Tf = −ikf × m
i, van’t Hoff i-factor
i = 1 for dilute nonelectrolyte solution
i = 2 for MX salts,
NaCl ⎯⎯
→ Na + + Cl−
i = 3 for MX2 salts,
CaCl2 ⎯⎯
→ Ca + + 2Cl−
In dilute solution of HCl, i = 1 in toluene, i = 2 in water.
Æ molecular form in toluene but fully deprotonated in water
In an aqueous solution of a weak acid. Æ 5% deprotonated, i = 0.95 + (0.05x2) = 1.05
8.17 Osmosis (삼투현상,渗透現象)
Æ Flow of solvent through a semipermeable (반투과성,半透過性) membrane into a more concentrated solution.
Æ The pressure needed to stop the flow of solvent is the osmotic pressure,
Π
Fig. 8.31 Osmosis: Original heights of the water levels of the beaker (water) and the tube (solution separated by a
membrane) are the same. As solvent molecules pass into the tube by osmosis, the water level of the tube rises above
that of the beaker.
2010년도 제1학기
화 학 1
Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008)
담당교수: 신국조
Chapter 8
Fig. 8.32 Red blood cells in solutions of (a) suitable, (b) dilute (bursting), and (c) dense (shrivel up) solute concentrations.
Fig. 8.33 Decrease in Gm for the solution.
Fig. 8.34 The pressure at the bottom of a fluid column.
◈ van’t Hoff equation for the osmotic pressure,
Π = iRTc
TOOLBOX 8.2
i : van’t Hoff i -factor
Π
c : molarity of solute in solution
HOW TO USE COLLIGATIVE PROPERTIES TO DETERMINE MOLAR MASS
1. Cryoscopy (어는점 측정법)
freezing-point depression
i kf
Step 1:
Molality = −
Step 2:
nsolute = molality × msolvent
Step 3:
Molar mass: M solute =
msolute
nsolute
for a given mass of solute,
msolute .
2010년도 제1학기
화 학 1
Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008)
담당교수: 신국조
Chapter 8
2. Osmometry (삼투측정법)
Π
iRT
Step 1:
c=
Step 2:
nsolute = cV
Step 3:
Molar mass:
Ex. 8.9
where
Π = g ⋅d ⋅h
M solute =
msolute
nsolute
for a given mass of solute,
msolute
Determining molar mass cryoscopically. (assuming i = 1)
Addition of 0.24 g of sulfur to 100.g of the solvent carbon tetrachloride, CCl4, lowers the freezing point by
0.28oC. What is the molar mass and molecular formula of sulfur?
Step 1. Assume that
i = 1 (Sulfur is nonelectrolyte). Then find
Molality = −
∆Tf
−0.28 K
=−
kf
29.8 K ⋅ kg ⋅ mol −1
= 9.40 × 10−3 mol ⋅ kg −1
the molality of solute.
Step 2. Calculate the
amount of solute
nSx = (0.100 kg) × (9.40 ×10−3 mol ⋅ kg −1 )
present.
= 9.40 × 10−4 mol
Step 3. Determine the
molar mass of the
M Sx =
0.24 g
= 2.55 × 102 g ⋅ mol −1
−4
9.40 × 10 mol
solute.
Step 4. Use the molar
mass of atomic sulfur
x=
2.55 × 102 g ⋅ mol −1
= 7.94 ≈ 8
32.1 g ⋅ mol −1
to find the value of x in
Sx.
Ex. 8.10 Using osmometry to determine molar mass Å Most accurate method!
Π = 1.10 × 10 −2 atm due to 2.20 g of polyethylene (PE) dissolved in enough benzene to produce 100.0 mL
of solution at 25oC. Average molar mass of the polymer?
i = 1 Å PE is a nonelectrolyte
Molar mass of polymer is very high Æ Order of kg per mole
2010년도 제1학기
화 학 1
Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008)
담당교수: 신국조
Chapter 8
Step 1.
c=
Π
1.10 ×10−2 atm
=
= 4.49 × 10−4 mol ⋅ L−1
−1
−1
iRT (0.0821 L ⋅ atm ⋅ K ⋅ mol ) × (298 K)
Step 2. Find the amount of solute in solution.
nPE = cV = (4.49 × 10−4 mol ⋅ L−1 ) × (0.100 L) = 4.50 × 10−5 mol
Step 3. Find the molar mass of solute.
Molar mass of PE =
2.20 g
= 4.89 × 104 g ⋅ mol −1
4.50 × 10 −5 mol
▶ Reverse osmosis (역삼투현상)
A pressure greater than the osmotic pressure is applied to the solution side of the membrane.
Æ Solvent molecules leave the solution
Æ Purification of seawater
Æ Cellulose acetate membrane used at 70 atm
◈ BINARY LIQUID MIXTURES
Volatile solute in solution
Separation by distillation
8.18 The Vapor Pressure of a Binary Liquid Mixture
Ideal binary mixture of the volatile liquids A and B
(A: benzene, B: toluene)
Raoult’s law:
PA = xA PA and PB = xB PB
Dalton’s law:
P = PA + PB = xA PA + xB PB
o
o
o
o
Fig. 8.35 The vapor pressures of the two
components of an ideal binary mixture.
Ex. 8.11
Predicting the vapor pressure of a mixture of two liquids.
What is the vapor pressure of each component at 25oC and the total vapor pressure of a binary mixture of
benzene and toluene ( xbenzene
= 13 , xtoluene = 23 )? Pbenzene = 94.6 Torr , Ptoluene = 29.1 Torr
o
o
2010년도 제1학기
화 학 1
Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008)
Pbenzene = ( 13 ) × 94.6 Torr = 31.5 Torr
From Raoult’s law,
PA = xA PA
PB = xB PB
담당교수: 신국조
Chapter 8
o
Ptoluene = ( 23 ) × 29.1 Torr = 19.4 Torr
o
From Dalton’s law,
Ptotal = Pbenzene + Ptoluene = (31.5 + 19.4) Torr = 50.9 Torr
P = PA + PB
◆ Express the composition of the vapor in equilibrium with the liquid phase of a binary liquid mixture.
From Dalton’s law,
yA =
PA
PA
,
=
P PA + PB
yB =
PB
PB
=
P PA + PB
From Raoult’s law,
o
xA PA
yA =
o
o
xA PA + xB PB
o
yB =
xB PB
o
o
xA PA + xB PB
Fig. 8.36 The composition of the vapor vs. the composition
of the liquid in equilibrium with each other.
If PA > PB ,
o
o
o
yA PA
=
> 1 and
xA
P
o
yB PB
=
<1
xB
P
cf. Fig. 8.35
Æ Vapor of the mixture is richer than the liquid in the more volatile component.
Æ When
xA = xbenzene = 0.333 ,
Ex. 8.12
yA = ybenzene = 0.619
∴ ybenzene > xbenzene
Predicting the composition of the vapor in equilibrium with a binary liquid mixture.
xbenzene = 0.333 Æ ybenzene = ? at 25oC
2010년도 제1학기
화 학 1
Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008)
From
x
= 0.333
담당교수: 신국조
Chapter 8
benzene
x B = 1 − xA
xtoluene = 1 − 0.333 = 0.667
ybenzene
From
o
xA PA
yA =
o
o
xA PA + xB PB
0.333 × 94.6 Torr
0.333 × 94.6 Torr + 0.667 × 29.1 Torr
= 0.619
=
ytoluene = 0.381
8.19 Distillation (증류,蒸溜)
◈ Temperature-Composition diagram
Fig. 8.37 A temperature-composition diagram for benzene and toluene.
▶
Tb (pure toluene) = 110.6o C , Tb (pure benzene) = 80.1o C
▶ Lower curve: Boiling point of the mixture as a function of composition
▶ Upper curve: Composition of the vapor in equilibrium with the liquid
at each boiling point.
▶ Point B shows the vapor composition for a mixture that boils at Point A.
Boiling liquid mixture at A ( xbenzene
the vapor at B (
= 0.45 ) is in equilibrium with
ybenzene = 0.73 ).
Tie line connects Points A and B.
Fig. 8.38 Fractional distillation(분별증류,分別蒸溜) steps.
Fig. 8.39 Fractional distillation process.
2010년도 제1학기
화 학 1
Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008)
담당교수: 신국조
Chapter 8
8.20 Azeotropes (불변끓음 혼합물) or Azeotropic mixtures
◈ Deviations from Raoult’s law Å non-ideal solution
Æ Enthalpy of mixing,
∆H mix : enthalpy difference between mixture and pure components
Æ Impossible to separate by distillation
▶ Positive deviation,
∆H mix > 0
Æ Endothermic : Higher vapor pressure than predicted by Raoult’s law
Æ Solute-solute attraction > Solute-solvent attraction
Æ Minimum-boiling azeotrope (obtained as the initial distillate)
Fig. 8.40a Positive deviation
▶ Negative deviation,
Fig. 8.41 Minimum-boiling azeotrope
∆H mix < 0
Æ Exothermic : Lower vapor pressure than predicted by Raoult’s law
Æ Solute-solute attraction < Solute-solvent attraction
Æ Maximum-boiling azeotrope (left in the flask)
Fig. 8.40b Negative deviation
Fig. 8.42 Maximum-boiling azeotrope
2010년도 제1학기
화 학 1
Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008)
담당교수: 신국조
Chapter 8
◈ IMPACT ON BIOLOGY AND MATERIALS
8.21 Colloids
◈ Colloids : a dispersion (분산,分散) of large particles (1 nm ~ 1 μm) in a solvent
Æ Intermediate between a solution and a heterogeneous mixture
Æ Homogenous appearance but scatters light
Æ Milk, Smoke, Clay particles
▶ Sol (솔,졸) : A suspension (현탁액,懸濁液) of solids in a liquid ex. Muddy water
▶ Emulsion (에멀션: 유탁액,乳濁液) : A suspension of one liquid in another
ex. Milk (suspension of fat in water), Mayonnaise (water droplets suspended in vegetable oil)
▶ Foam : A suspension of a gas in a liquid or in a solid
ex. Styrofoam, Aerogels, Zeolites (boiling stone: microporous, aluminosilicate minerals commonly
used as commercial adsorbents)
▶ Solid emulsion : A suspension of a liquid or solid phase in a solid
ex. Opal (partly hydrate silica fills the interstices between close-packed microspheres of silica aggregates),
Stained galsses (colloidal clusters of Cu, Ag, Au in glass)
▷ Gel (젤,겔) : a type of solid emulsion
ex. Gelatin desserts, Photographic emulsions (AgBr)
Fig. 8.43 Invisible laser beams scatter
from particles suspended in the air.
Fig. 8.44 Cross section of a bilayered
Fig. 8.45 Colloid of metallic
cell membrane formed from surfactant-like
gold clusters (violet liquid)
phospholipids molecules.
prepared by Faraday in 1857.
◆ Brownian motion ~ motion of a small particles resulting from constant bombardment by solvent molecules
Æ Discovered by a botanist Robert Brown (1827)
Æ Theory by Einstein (1905), Smoluchowski (1906)
Å Diffusion process
2010년도 제1학기
화 학 1
Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008)
담당교수: 신국조
Chapter 8
8.22 Bio-Based Materials and Biomimetic Materials
◈ Bio-based material (생체바탕 재료)
Æ Materials that are taken from or made from natural materials in living things
▷ Packing pellets (포장입상체,包裝粒狀體) Æ made from corn and soybean
▷ Polylactic acid or Polylactide (PLA) : 플라스틱포장재
Æ a biodegradable, thermoplastic, aliphatic polyester
derived from corn starch or sugarcanes
▷ Hyaluronic acid (or Hyaluronan)
Æ hydrogen bonds by –OH groups
Æ lubricating fluid for joints
Æ repairing skin tissues
Æ sports medicine
Fig. 8.46 Hyaluronic acid
◈
Biomimetic materials (생체모방 재료)
▷ Gels or flexible polymers Æ mimic lifelike movements (기어가는 벌레 흉내, nanometer worm)
▷ Control-release drug delivery systems
Æ made from liquid crystals
Æ Forming liposomes (artificial vesicle) from artificial membranes made from phospholipids
encapsulating drug molecules
2010년도 제1학기
화 학 1
Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008)
BOX 8.1
담당교수: 신국조
Chapter 8
FRONTIERS OF CHEMISTRY: DRUG DELIVERY
◈ Drug delivery
▶ Transdermal patch (피부에 붙이는 헝겊)
Nitroglycerin (heart disease), Morphine (pain killer), estrogen (female hormone), nicotine
▶ Implants (이식조직,移植組織)
~ Contained in a cylinder of porous foam
▶ Control-release drug delivery system
~ Drug-encapsulated liposomes injected into the body to stick only to cancer cells
~ Nano-size hollow spheres
▶ Smart gels
~ Adjustable dosage of drug delivery (Insulin treatment for diabetes)
~ Adsorption of enough glucose molecules triggers explosive gel swelling, releasing insulin
Various implants
MAJOR TECHNIQUE 4
Implant containing live cells
Nano-sizs drug capsule
inserted into spinal column
bursting open
CHROMATOGRAPHY
◆ Solvent extraction
▷ Iodine extraction by CCl4
Æ Partition equilibrium between H2O and CCl4
▷ Decaffeinate coffee by “supercritical” CO2 fluid
2010년도 제1학기
화 학 1
Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008)
Left:
Right:
담당교수: 신국조
Chapter 8
Upper water layer dissolving iodine on top of CCl4 layer
After shaking, iodine dissolves preferentially in CCl4 layer
◈ Chrmatography (“color writing”)
Æ Stationary and mobile phases
▶ Liquid Chromatography
▷ Paper chromatography
▷ Column chromatography ~ SiO2 or Al2O3 packing (stationary phase)
Fig. 1. Paper chromatography
Fig. 2.
Column chromatography
▷ High-performance liquid chromatography (HPLC)
~ forced elution(용출,溶出) through a long narrow column under pressure (eluate, eluent)
▷ Thin-layer chromatography (TLC)
~ SiO2 or Al2O3 coated on a glass
▶ Gas Chromatography
Fig. 3.
A gas chromatogram
Fig. 4. Gas chromatograph
▶ Gas Chromatography-Mass Spectrometry (GC-MS)