2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) 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.
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
2 2H O( ) H O( )l g⎯⎯→←⎯⎯ Fig. 8.2 Dynamic equilibrium between a liquid and its vapor.
http://en.wikipedia.org/wiki/Pressurehttp://en.wikipedia.org/wiki/Vaporhttp://en.wikipedia.org/wiki/Thermodynamic_equilibriumhttp://en.wikipedia.org/wiki/Phase_(matter)
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 8
8.2 Volatility (휘발성,揮發性) and Intermolecular Forces
Volatile liquid: Liquid with weakly bonded molecules High vapor pressure
Hydrogen bondings in liquid lower the vapor pressure
Ex. Dimethyl ether (C2H6O, 3HC-O-CH3) and Ethanol (C2H6O, 3HC-CH2-OH)
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
Fig. 8.3 Vapor pressures of liquids.
Vapor pressure of water near its normal boiling point
At equilibrium, vap m m( ) ( ) 0G G g G l∆ = − =
For a liquid, no pressure dependence o( , ) ( )G l P G l=m m
m m
For an ideal gas,
o( , ) ( ) lnG g P G g RT P= +
Fig. 8.4 Variation of molar Gibbs energy with pressure
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 8
o ovap m m
o o om m vap
( ) ln ( )
( ) ( ) ln ln
G G g RT P G l
G g G l RT P G RT P
⎡ ⎤∆ = + −⎣ ⎦⎡ ⎤= − + = ∆ +⎣ ⎦
ovapG∆ : Standard Gibbs free energy of vaporization
At equilibrium, . vap 0G∆ =o
vap 0 lnG RT∴ = ∆ + P
o ovap vap vapln
G H SP
o
RT RT∆ ∆ ∆
= − = − +R
ovapH∆ , : approximately independent of temperature
ovapS∆
o o o o ovap vap vap vap vap
2 12 1
1 1ln lnH S H S H
P P1 2RT R RT R R T T
⎛ ⎞ ⎛ ⎞∆ ∆ ∆ ∆ ∆ ⎛ ⎞− = − + − − + = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠
◈ Clausius-Clapeyron equation
ovap2
1 1
1 1 ln HP
P R T T∆ ⎛ ⎞
= −⎜ ⎟⎝ ⎠2
2T T> 11 2
1 1 0T T
⎛ ⎞− >⎜ ⎟
⎝ ⎠
ovap 0H∆ >⎯⎯⎯⎯⎯→ 21
ln 0PP
> 2 1P P>
Vapor pressure increases with increasing temperature.
The increase is greatest for substances with high (strong intermolecular interactions) ovapH∆
cf. Ethanol ( o 1vap 43.5 kJ molH−∆ = ⋅ ) vs. benzene ( ) o 1vap 30.8 kJ molH
−∆ = ⋅
8.4 Boiling
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
ovap2
1 1
1 1lnHP
P R T T∆ ⎛ ⎞
= −⎜ ⎟⎝ ⎠2
, o 1vap 43.5 kJ molH−∆ = ⋅
2 13.3 kPaP = , 2 34.9 273.15 K 308.0 KT = + =
1 1 atm 101.325 kPaP = = ∴ 1 350 KT = (~ 77oC)
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 8
8.5 Freezing and Melting
Normal freezing point ( = Normal melting point ) at 1 atm fT
Supercooling liquid phase below its freezing point
Increased pressure prefers denser phase:
▶ Fe : at 1 atm f 1800 KT = f 1805 KT ′ = at 1000 atm
Melting point increases slightly with increasing pressure
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
Unusual low density of ice compared to liquid water
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 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” All four phases are not in equilibrium
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 8
Fig. 8.6 The phase diagram for water. Fig. 8.7 The phase diagram for CO2. Fig. 8.8 The phase diagram for sulfur.
Fig. 8.9 The phase diagram for water. Fig. 8.10 Vapor pressure of water
Fig. 8.11 Changes undergone by a liquid as its pressure is decreased at constant temperature.
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 8
8.7 Critical Properties (임계성질,臨界性質)
Fig. 8.12 Fig. 8.14
Fig. 8.13 The critical phenomenon. “Critical opalescence” Opal
▶ Supercritical fluids : 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
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) 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.
Fig. 8.18 Formation of dilute solution.
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. Red arrow in Fig. 8.18 ⇓
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 8
▶ London forces, 1 26PE r
α α∝ − (induced dipole-induced dipole)
Principal cohesive forces between solute molecules ex. S8
Solvent with a similar London force ex. CS2
▷ Cleaning with a soap
Polar carboxylate head group: Hydrophilic (親水性)
Nonpolar hydrocarbon tails: Hydrophobic (疏소水性)
Fig. 8.19 The solubility of sulfur in
water and carbon disulfide.
Formation of micelles in soaping action
▷ Detergent (세제,洗劑) : Surfactants (界面活性劑)
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:
H s k P= : Henry’s constant Hk
(2) Vapor pressure of a volatile solute (in a mixture of two volatile liquids) 2P
is proportional to its mole fraction in solution at low mole fractions: 2X
2 2 2 P k X= : Henry’s law constant 2 H )(k k=
William Henry (英,1775-1836)
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) 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
Increase in vapor pressure with increasing temperature
O2 14 mg/L at 0oC 6 mg/L at 40oC
CO2 3.4 g/L at 0oC ~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.”
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 8
8.12 The Enthalpy of Solution (용해溶解엔탈피)
▶ Enthalpy of solution, solH∆
Change in molar enthalpy when a substance dissolves
▶ Limiting enthalpy of solution
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 Lattice enthalpy
NaCl(s) Na (g) Cl (g)+ −⎯⎯→ + 1L 787 kJ molH−∆ = ⋅
2nd step: Gaseous ions plunging into water forming the final solution Hydration enthalpy (< 0)
Na (g) Cl (g) Na (aq) Cl (aq)+ − + −+ ⎯⎯→ + 1hyd 784 kJ molH−∆ = − ⋅
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 8
sol L hydH H H∆ = ∆ + ∆
NaCl(s) Na (aq) Cl (aq)+ −⎯⎯→ +
Lattice Enthalpy of Enthalpy ofenthalpy hydration solution
1 1sol L hyd
1787 kJ mol 784 kJ mol 3 kJ molH H H − − −∆ = ∆ + ∆ = ⋅ − ⋅ = ⋅
2 ↑
S
> 0 (Fig. 8.23b)
Nitrates have big, singly charged anions. Low lattice enthalpy, High hydration enthalpy (H-bonds with water)
Carbonates have big, doubly charged anions. Higher lattice enthalpy, less soluble
Hydrogen carbonates (bicarbonates) have singly charged anions more soluble than carbonates
▶ Hard water :
Rainwater with dissolved CO2 forms a very dilute solution of carbonic acid:
2 2 2 3CO (g) H O(l) H CO (aq)+ ⎯⎯→
As the water flows through the grounds, the carbonic acid reacts with CaCO3 of limestone:
23 2 3 3CaCO (s) H CO (aq) Ca (aq) 2 HCO (aq)
−++ ⎯⎯→ +
These two reactions are reversed when water containing Ca(HCO3)2 is heated in a furnace (Purification):
23 3 2Ca (aq) 2 HCO (aq) CaCO (s) H O(l) CO (g) −+ + ⎯⎯→ ↓ + +
8.13 The Gibbs Free Energy of Solution
G H T∆ = ∆ − ∆
0S∆ > for dissolution of solids as long as 0G∆ < 0H∆ <
0S∆ < for cage formation by solvent molecules 0G∆ < only when 0H∆ < with H T S∆ > ∆
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 8
G H T∆ = ∆ − ∆S
0H∆ > for endothermic dissolving 0G∆ < only when 0T S∆ > and that T S H∆ > ∆
Endothermic dissolving depends on the balance between S∆ of the system and that of the surroundings.
Strong Insoluble ( 0)H∆ >
∵ So much energy leaves the surroundings and enters the solution Decrease in overall disorder (entropy)
Spontaneous dissolving ( ) with increasing T Only when 0G∆ < 0S∆ >
For extensively hydrated substances and gases 0S∆ < for dissolution
Solubility decreases with increasing T
▶ depends on the concentration of solute: a : activity G∆ o lnG G nRT a= +
0G∆ < for low concentration (spontaneous dissolution) and 0G∆ > for high concentration (insoluble)
Saturated solution when 0G∆ =
Fig. 8.24 The dissolution process for a solid. Increase in disorder. Fig. 8.25 Spontaneous dissolution vs. spontaneous precipitation. Saturation at . 0G∆ =
◈ 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 independent of temperature
cf. Molarity, M, depends on temperature
amount of solute (mol)Molality of solutemass of solvent (kg)
= (3)
Fig. 8.26 Preparation of solution in molality.
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 8 Ex. 8.5 Calculating the molality of a solute.
10.5 g of NaCl dissolved in 250 g water Molality?
1
1solute
solvent
(mol) (10.5 g)/(58.44 g mol )molality 0.719 mol kg 0.719 (kg) 0.250 kg
n mm
−−⋅= = = ⋅ =
TOOLBOX 8.1 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, , present in a given mass of solvent, , soluten solventm
by rearranging the equation defining molality, Eq.(3), into
solute solventmolality n m= ×
Step 1.2 Use the molar mass of solute, , to find the mass of the solute from its amount: soluteM
solute solute solutem n M=
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 solutex , the amount of solute molecules in a total of 1 mol of molecules is
solute solute moln x=
Step 2.2 If there is only one solute, the mole fraction of solvent molecules is solute1 x− . The amount of
solvent molecules in a total of 1 mol of molecules is then solvent solute(1 )n x= − mol. Convert
this amount into mass in grams by using the molar mass of the solvent, : solventM
( ){ }solvent solvent solvent solute solvent1 molm n M x M= = −
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, , by using the molar mass, , of the solvent: solventn solventM3
solventsolvent solvent
1 kg 10 gnM M
= =
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 8
We already know the amount of solute molecules, , in the solution: soluten
solute molality (1 kg)n = ×
Step 3.2 Calculate the mole fraction from the amounts of solvent and solute molecules:
solutesolute
solute solvent
nxn n
=+
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,
of the solution: 3solution (10 mL)m d= ×
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:
solute molarity (1 L)n cV= = ×
solute solute solute solute molarity (1 L)m m M M= = × ×
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, solvent solution solutem m m= −
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. Molality ?
Step 2.1 Calculate the amount of solute in a total 1 mol of solution molecules:
benzene benzenemol 0.150 1 mol 0.150 moln x= = × =
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 olute solventmolality / n m= s
16 6
0.150 molMolality of C H 1.92 mol kg0.0783 kg
−= = ⋅
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: cal Principles, 4th ed., Freeman (2008) Chapter 8 P. Atkins / L. Jones, Chemi
Ex. 8.7 Converting molarity into molality
Density of 1.06 M sucrose, C12H22O11(aq), solution is 1.14 g·mL–1 Molality?
Step 4.1 Find the mass of
1 L of solution from
3(10 mL)m d= ×
1 3 3solution (1.14 g mL ) (10 mL) 1.14 10 gm
−= ⋅ × = ×
Step 4.2 Find the mass of
solute in 1L solution
1 1sucrose (1.06 mol L ) (1 L) (342.3 g mol ) 363 gm
− −= ⋅ × × ⋅ =
Step 4.3 Find the mass of
water in 1 L solution
water solution solutem m m= −
31.14 10 363 g 0.78 kg= × − =
Step 4.4 From
solute solventmolality /n m=
Molality of C12H22O11 11.06 mol 1.4 mol kg
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.
solvent p ru eP x P=o
1( )x P=
▶ Ideal solution ~ obeys Raoult’s law at all concentrations.
Interactions between [solute-solvent]solution
= [solute-solute]pure state
= [solvent-solvent]pure state
solution 0H∆ =
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.
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 8
Ex 8.8 Using Raoult’s law
10.00 g of nonelectrolyte sucrose, C12H22O11, dissolved in 100.0 g of water at 20oC. in solution water ?P =
[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 solventsolventsolute solvent
0.555 mol 0.995(0.0292 0.555) mol
nxn n
= = =+ +
From and , solvent pureP x P=o
pure (water,20 ) 17.54 TorrP C =
water 0.995 (17.54 Torr) 17.45 TorrP = × =
◈ Vapor pressure lowering (from Oxtoby)
o o o o o1 1 1 1 1 1 1 1 2 1 ( 1) P P P x P P x P x P∆ = − = − = − = − 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, m m(pure liquid solvent) (vapor from pure solvent)G G=
In an ideal solution, solutes increases the entropy but the enthalpy remains constant.
Decrease in [ ]m m(solvent in solution) (pure liquid solvent)G G<
In order to reach an equilibrium,
m m(solvent in solution) (vapor from solution)G G=
Decrease in [ ]m m(vapor from solution) (vapor from pure solvent)G G<
Decrease in vapor pressure cf. o( , ) ( ) lnG g P G g RT P= +m m
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) 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
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 8
, b bT k m∆ = × bk : boiling point elevation constant (ebullioscopic constant)
, : freezing point depression constant (cryoscopic constant) f fT k∆ = − ×m fk
1b (water) 0.51 K kg molk
−= ⋅ ⋅ , 1f (water) 1.86 K kg molk−= ⋅ ⋅ b fT T∆ < ∆
m m
In an electrolyte solution,
b biT k∆ = × and i, van’t Hoff i-factor f fiT k∆ = − ×
i = 1 for dilute nonelectrolyte solution
i = 2 for MX salts, NaCl Na Cl+ −⎯⎯→ +
i = 3 for MX2 salts, 2CaCl Ca 2Cl+ −⎯⎯→ +
In dilute solution of HCl, i = 1 in toluene, i = 2 in water.
molecular form in toluene but fully deprotonated in water
In an aqueous solution of a weak acid. 5% deprotonated, i = 0.95 + (0.05x2) = 1.05
8.17 Osmosis (삼투현상,渗透現象)
Flow of solvent through a semipermeable (반투과성,半透過性) membrane into a more concentrated solution.
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 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) 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
iRTcΠ = i : van’t Hoff i -factor c : molarity of solute in solution
TOOLBOX 8.2 HOW TO USE COLLIGATIVE PROPERTIES TO DETERMINE MOLAR MASS
1. Cryoscopy (어는점 측정법)
Step 1: f
freezing-point depressionMolality i k
= −
Step 2: solute solvent = molality n m×
Step 3: Molar mass: solutesolutesolute
mMn
= for a given mass of solute, . solutem
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 8
2. Osmometry (삼투측정법)
Step 1: ciRTΠ
= where hg dΠ = ⋅ ⋅
Step 2: soluten c= V
Step 3: Molar mass: solutesolutesolute
mMn
= for a given mass of solute, solutem
Ex. 8.9 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 non- electrolyte). Then find
the molality of solute.
f
1f
3 1
0.28 KMolality29.8 K kg mol
9.40 10 mol kg
Tk −
− −
∆ −= − = −
⋅ ⋅
= × ⋅
Step 2. Calculate the
amount of solute
present.
3 1S
4
(0.100 kg) (9.40 10 mol kg )
9.40 10 molx
n − −
−
= × × ⋅
= ×
Step 3. Determine the
molar mass of the
solute.
2 14
0.24 g 2.55 10 g mol9.40 10 molxS
M −−= = × ⋅×
Step 4. Use the molar
mass of atomic sulfur
to find the value of x in
Sx.
2 1
1
2.55 10 g mol 7.94 832.1 g mol
x−
−
× ⋅= = ≈
⋅
Ex. 8.10 Using osmometry to determine molar mass Most accurate method!
21.10 10 atm−Π = × 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 PE is a nonelectrolyte
Molar mass of polymer is very high Order of kg per mole
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 8
Step 1.
24 1
1 1
1.10 10 atm 4.49 10 mol L(0.0821 L atm K mol ) (298 K)
ciRT
−− −
− −
Π ×= = = × ⋅
⋅ ⋅ ⋅ ×
Step 2. Find the amount of solute in solution.
4 1 5PE (4.49 10 mol L ) (0.100 L) 4.50 10 moln cV
− − −= = × ⋅ × = ×
Step 3. Find the molar mass of solute.
4 15
2.20 gMolar mass of PE 4.89 10 g mol4.50 10 mol
−−= = ××
⋅
▶ Reverse osmosis (역삼투현상)
A pressure greater than the osmotic pressure is applied to the solution side of the membrane.
Solvent molecules leave the solution
Purification of seawater
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: and oP x P=A A A B B B
A B A A B B
oP x P=
Dalton’s law: o oP P P x P x P= + = + 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 ( 1benzene 3x = , 2toluene 3x = )? , o
benzene 94.6 TorrP =o
toluene 29.1 TorrP =
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 8
From Raoult’s law, o
A A AP x P= o
B B BP x P=
( )1benzene 3 94.6 31.5 TorrP = × = Torr
( )2toluene 3 29.1 19.4 TorrP = × = Torr
From Dalton’s law,
A BP P P= +
total benzene toluene (31.5 19.4) Torr 50.9 TorrP P P= + = + =
◆ Express the composition of the vapor in equilibrium with the liquid phase of a binary liquid mixture.
From Dalton’s law,
A AA
A B
P PyP P P
= =+
,
B BB
A B
P PyP P P
= =+
From Raoult’s law,
oA A
A o oA A B B
x Pyx P x P
=+
oB B
B o oA A B B
x Pyx P x P
=+
Fig. 8.36 The composition of the vapor vs. the composition
of the liquid in equilibrium with each other.
If , o oA BP P>
oA A
A
1y Px P
= > and o
B B
B
1y Px P
= < cf. Fig. 8.35
Vapor of the mixture is richer than the liquid in the more volatile component.
When , A benzene 0.333x x= = A benzene 0.619y y= = benzene benzene y x∴ >
Ex. 8.12 Predicting the composition of the vapor in equilibrium with a binary liquid mixture.
benzene 0.333x = benzene ?y = at 25oC
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 8 From
B A1x x= − benzene 0.333x =
toluene 1 0.333 0.667x = − =
From o
A AA o o
A A B B
x Pyx P x P
=+
benzene
0.333 94.6 Torr0.333 94.6 Torr 0.667 29.1 Torr0.619
y×
=× + ×
=
toluene 0.381y =
8.19 Distillation (증류,蒸溜)
◈ Temperature-Composition diagram
Fig. 8.37 A temperature-composition diagram for benzene and toluene.
▶ , ob (pure toluene) 110.6 CT =o
b (pure benzene) 80.1 CT =
▶ 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 ( benzene 0.45x = ) is in equilibrium with
the vapor at B ( benzene 0.73y = ). 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 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 8
8.20 Azeotropes (불변끓음 혼합물) or Azeotropic mixtures
◈ Deviations from Raoult’s law non-ideal solution
Enthalpy of mixing, : enthalpy difference between mixture and pure components mixH∆
Impossible to separate by distillation
▶ Positive deviation, mix 0H∆ >
Endothermic : Higher vapor pressure than predicted by Raoult’s law
Solute-solute attraction > Solute-solvent attraction
Minimum-boiling azeotrope (obtained as the initial distillate)
Fig. 8.40a Positive deviation Fig. 8.41 Minimum-boiling azeotrope
▶ Negative deviation, mix 0H∆ <
Exothermic : Lower vapor pressure than predicted by Raoult’s law
Solute-solute attraction < Solute-solvent attraction
Maximum-boiling azeotrope (left in the flask)
Fig. 8.40b Negative deviation Fig. 8.42 Maximum-boiling azeotrope
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) 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
Intermediate between a solution and a heterogeneous mixture
Homogenous appearance but scatters light
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 Fig. 8.44 Cross section of a bilayered Fig. 8.45 Colloid of metallic
from particles suspended in the air. 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
Discovered by a botanist Robert Brown (1827)
Theory by Einstein (1905), Smoluchowski (1906) Diffusion process
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 8
8.22 Bio-Based Materials and Biomimetic Materials
◈ Bio-based material (생체바탕 재료)
Materials that are taken from or made from natural materials in living things
▷ Packing pellets (포장입상체,包裝粒狀體) made from corn and soybean
▷ Polylactic acid or Polylactide (PLA) : 플라스틱포장재
a biodegradable, thermoplastic, aliphatic polyester
derived from corn starch or sugarcanes
▷ Hyaluronic acid (or Hyaluronan)
hydrogen bonds by –OH groups
lubricating fluid for joints
repairing skin tissues
sports medicine
Fig. 8.46 Hyaluronic acid
◈ Biomimetic materials (생체모방 재료)
▷ Gels or flexible polymers mimic lifelike movements (기어가는 벌레 흉내, nanometer worm)
▷ Control-release drug delivery systems
made from liquid crystals
Forming liposomes (artificial vesicle) from artificial membranes made from phospholipids
encapsulating drug molecules
http://en.wikipedia.org/wiki/Biodegradablehttp://en.wikipedia.org/wiki/Thermoplastichttp://en.wikipedia.org/wiki/Aliphatichttp://en.wikipedia.org/wiki/Polyesterhttp://en.wikipedia.org/wiki/Corn_starchhttp://en.wikipedia.org/wiki/Sugarcane
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 8
BOX 8.1 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 Implant containing live cells Nano-sizs drug capsule
inserted into spinal column bursting open
MAJOR TECHNIQUE 4 CHROMATOGRAPHY
◆ Solvent extraction
▷ Iodine extraction by CCl4
Partition equilibrium between H2O and CCl4▷ Decaffeinate coffee by “supercritical” CO2 fluid
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 8
Left: Upper water layer dissolving iodine on top of CCl4 layer
Right: After shaking, iodine dissolves preferentially in CCl4 layer
◈ Chrmatography (“color writing”)
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)