CHAPTER 8 PHYSICAL EQUILIBRIA - X-Y.netrhbestsh.x-y.net/Studying/SNU_Science/Chemistry/Lecture...8.6 Phase Diagrams Phase diagram: A map showing which phase is most stable at a different

2009년도 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 (相) AND PHASE TRANSITIONS (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)


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


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 is 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 =


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


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.


8.7 Critical Properties (임계성질,臨界性質)

Fig. 8.12 Fig. 8.14

Fig. 8.13 The critical phenomenon. “Critical opalescence”

▶ Supercritical fluids : Dense fluid above Tc and Pc , Used as solvents

▷ Supercritical CO2 : 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


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 ⇓


▶ London forces

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 (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 of a gas is directly proportional to its partial pressure

H s k P= : Henry’s law constant Hk

(2) Vapor pressure of a volatile solute is proportional to its 2P

mole fraction in solution at low mole fractions 2X

William Henry (英,1775-1836)

2 2 2 P k X= : Henry’s law constant k2


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.


8.12 The Enthalpy of Solution

▶ 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.

First step : Ions separating from the solid to form a gas of ions Lattice enthalpy

NaCl(s) Na (g) Cl (g)+ −⎯⎯→ + 1L 787 kJ molH−∆ = ⋅

Second 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−∆ = − ⋅


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∆ > ∆


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

G∆ depends on the concentration of solute

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.

2009년도 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?

11solute

solvent

(mol) (10.5 g)/(58.44 g mol ) 0.719 mol kg 0.719 (kg) 0.250 kg

nm 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, , by

rearranging the equation defining molality, Eq.(3), into

soluten solventm

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

= =


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 solute solventmolality / n m=

16 6 3

0.150 molMolality of C H 1.92 mol kg0.850 92.13 10 kg

−−= =× ×

⋅


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

−= ⋅ × = ×


solute in 1L solution

1 1(1.06 mol L ) (1 L) (342.3 g mol ) 363 gsucrosem− −= ⋅ × × ⋅ =


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 conc.

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.


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


8.16 Boiling-Point Elevation and Freezing-Point Depression

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


, 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 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.


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: solutesolutesolute

mMn

=

2. Osmometry (삼투측정법)

Step 1: ciRTΠ

= where gdhΠ =

Step 2: soluten c= V

Step 3: solutesolutesolute

mMn

=

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

28oC. What is the molar mass and molecular formula of sulfur?

Step 1. Assume that

i = 1 (Sulfur is nonelectrolyte). Then find

the molality of solute.

1

3 1

0.28 KMolality29.8 K kg mol

9.40 10 mol kg

−

− −

=⋅ ⋅

= × ⋅

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 7.9432.1 g mol

x−

−

× ⋅= =

⋅

Ex. 8.10 Using osmometry to determine molar mass

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


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 5(4.49 10 mol L ) (0.100 L) 4.50 10 molPEn 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 A A,liquid A,pureP x P= B B,liquid B,pureP x P=

Dalton’s law: A B A,liquid A,pure B,liquid B,pureP 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,liquid 3x = , 2toluene,liquid 3x = )? benzene,pure 94.6 TorrP = , toluene,pure 29.1 TorrP =


From

A A,liquid A,pureP x P=

B B,liquid B,pureP x P=

( )1benzene 3 94.6 Torr 31.5 TorrP = × =

( )2toluene 3 29.1 Torr 19.4 TorrP = × =

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,vapor

A B

P PxP P P

= =+

,

B BB,vapor

A B

P PxP P P

= =+

From Raoult’s law,

A,liquid A,pureA,vapor

A,liquid A,pure B,liquid B,pure

x Px

x P x P=

+

B,liquid B,pureB,vapor


x Px

x P x P=

+

Fig. 8.36 The composition of the vapor vs. the composition

of the liquid in equilibrium with each other.

If , A,pure B,pureP P>

A,vapor A,pure

A,liquid

1x Px P

= > and B,vapor B,pureB,liquid

1x Px P

= < cf. Fig. 8.35

Vapor of the mixture is richer than the liquid in the more volatile component.

When A,liquid benzene,liquid 0.333x x= = , A,vapor benzene,vapor 0.619x x= = benzene,vapor benzene,liquid x x∴ >

Ex. 8.12 Predicting the composition of the vapor in equilibrium with a binary liquid mixture.

benzene,liquid 0.333x = benzene,vapor ?x = at 25oC


From

B A1x x= −

benzene,liquid 0.333x =

toluene,liquid 1 0.333 0.667x = − =

From

A,liquid A,pureA,vapor


x Px

x P x P=

+

benzene,vapor

0.333 94.6 Torr0.333 94.6 Torr 0.667 29.1 Torr0.619

x×

=× + ×

=

toluene,vapor 0.381x =

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 ( ) is in equilibrium with benzene,liquid 0.45x =

the vapor at B ( benzene,vapor 0.73x = ).

Tie line connects Points A and B.

Fig. 8.38 Fractional distillation steps. Fig. 8.39 Fractional distillation (分別蒸溜) process.


8.20 Azeotropes (불변끓음 혼합물)

◈ Deviations from Raoult’s law

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 Fig. 8.41

▶ 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 Fig. 8.42


IMPACT ON BIOLOGY AND MATERIALS

8.21 Colloids

◈ Colloids : a dispersion 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

▶ 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 Robert Brown (1827)

Theory by Einstein (1905), Smoluchowski (1906) Diffusion


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 made from corn and soybean

▷ Polylactic acid or Polylactide (PLA)

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

▷ Control-release drug delivery systems

made from liquid crystals

Forming liposomes 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


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 trigger explosive gel swelling, releasing insulin

Various implants Implant containing live cells Nano-sizs drug capsule

inserted into spinal column bursts open

MAJOR TECHNIQUE 4 CHROMATOGRAPHY

◆ Solvent extraction

▷ Iodine extraction by CCl4

Partition equilibrium between H2O and CCl4▷ Decaffeinate coffee by “supercritical” CO2


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

Fig. 1. Paper chromatography Fig. 2. Column chromatography

▷ High-performance liquid chromatography (HPLC)

▷ Thin-layer chromatography (TLC)

▶ Gas Chromatography

Fig. 3. A gas chromatogram Fig. 4. Gas chromatograph

▶ Gas Chromatography-Mass Spectrometry (GC-MS)

Documents

CHAPTER 8 PHYSICAL EQUILIBRIA - X-Y.netrhbestsh.x-y.net/Studying/SNU_Science/Chemistry/Lecture...8.6 Phase Diagrams Phase diagram: A map showing which phase is most stable at a different