11

Chapter 5

Phases and Solutions

22

Stability of Phases

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33

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44

Phase Transitions

Spontaneous conversion of one phase into another phase.�• Melting (or fusion) solid liquid�• Boiling (or vaporization) liquid gas�• Sublimation solid gas�• Condensation gas liquid�• Condensation (or deposition) gas solid�• Solidification (or freezing) liquid solid�• Polymorphism �– (allotropes):

different solid forms of a chemical component �–graphite/diamond; calcium carbonate �– argonite/calcite; structurally different solid water.

55

Heats of vaporization and fusion for several common substances

66

Phase changes and their enthalpy changes

77

A cooling curve for the conversion of gaseous water to ice

1/c

88

Within a phase, a change in heat is accompanied by a change in temperature which is associated with a change in average Ek as the most probable speed of the molecules changes.

Quantitative Aspects of Phase Changes

During a phase change, a change in heat occurs at a constant temperature, which is associated with a change in Ep, as the average distance between molecules changes.

q = (amount)(molar heat capacity)( T)

q = (amount)(enthalpy of phase change)

99

Equilibrium Nature of Phase Changes

Vapour pressure = equilibrium vapourpressure

1010

Transition temperature �– temperature at which the 2phases are in equilibrium and the Gibbs energy is minimized at the prevailing pressure.

Sublimation vp

Vapourpressure

1111

Vapour Pressure and Boiling Point

In an open container, at some temperature, the average Ekof the molecules in the liquid is great enough for bubbles of vapour to form in the interior, and the liquid boils. At any lower temperature, the bubbles collapse as soon as they start to form because external pressure is greater than the vapour pressure inside the bubbles. Boiling point is the temperature at which the vapourpressure equals the external pressure (atm pressure).

1212

Boiling temperature at a specific pressure �–temperature at which the vapour pressure of a liquid is equal to the external pressure.Normal boiling point �–external pressure of 1 atm.Standard boiling point �–external pressure of 1 bar (1 bar = 0.987 atm). Normal bp of water = 100 ºC vs 99.6 ºC for standard bp.

Critical points and boiling points

1313

Heating a liquid in a rigid, closed vessel

No boiling occurs. Vapour pressure and density of vapour rise as temperature is raised. Density of liquid decreases slightly because of expansion.

When density of vapour = density of remaining liquid, surface between gas/liquid disappears critical temperature, Tc of the substance.

Vapour pressure at Tc: critical pressure, Pc.

At and above Tc, a single uniform phase �–supercritical phase �– no liquid phase.

1414

Heating a liquid in a rigid, closed vessel

Sc H2O : 374 ºC, Pc 218 atm.

1515

Melting points and triple points

Melting point - temperature at which, under a specified pressure, the liquid and solid phases of a substance coexist in equilibrium.Melting point = freezing point.Normal freezing point �– freezing temperature when the pressure is 1 atm = normal melting point.Standard freezing point �– freezing temperature when the pressure is 1 bar.Triple point �– point at which the three phase boundaries meet �–solid, liquid, vapour coexist simultaneously in equilibrium. Triple point of water: 273.16 K, 611 Pa.

1616

Triple point �– lowest temperature at which the liquid can exist.Critical point - upper limit �– beyond this point, liquid and gas phase become indistinguishable.

Above Tc, no pressurecan force the H2O molecules into a liquid state.

1717

Supercritical FluidsA SCF is defined as a substance above its critical temperature (TC) and critical pressure (PC).

Supercritical carbon dioxide �– Tc: 304 K or 31 ºC, Pc = 72.9 atm.Density at critical point : 0.45 g cm-3. (0.1 �– 1.2 g cm-3)

Advantages of scCO2

�•Cheap, readily accessible, easily recycled.Disadvantages �•Not a very good solvent, need surfactants to dissolve solutes.

1818

Phase Diagram

Carbon dioxide�• Melting point of solid CO2

increases with pressure.�• Triple point above 1 atmno liquid at normal atm pressureswhatever the temperature.�• Cylinders of CO2 contains the liquid or compressed gas.

Supercriticalfluid

1919

Supercritical water

�• Critical temperature : 374 C�• Critical pressure 218 atm�• Properties of the fluid are highly sensitive to pressure.�• As the density of scH2O decreases, solution change

from aqueous non-aqueous gaseous solutions. Dissolve non-polar substances.

2020

Example 1: A liquid is in equilibrium with its vapour in a closed vessel at a fixed temperature. The vessel is connected by a stopcock to an evacuated vessel. When the stopcock is opened, will the final pressure of the vapour be different from its initial value if (a) some liquid remains (b) all the liquid is first removed?

Solution 1a) The final pressure will be the same, since the vapor pressure is constant as long as some liquid is present.b) The final pressure will be lower, according to Boyle�’s Law.

2121

Example 2: The phase diagram for a substance A has a solid-liquid line with a positive slope, and that for substance B has a solid-liquid line with a negative slope. What macroscopic property can distinguish A from B ?Solution 2A: the solid is denser than the liquid. B: the solid is less dense than the liquid. Example 3: Why does water vapour at 100 C cause a more severe burn than liquid water at 100 C ? Heat capacity of liquid water 75.4 J/mol.K; Hvap = -40.7 kJ/molSolution 3When liquid water at 100°C touches skin, the heat released is from the lowering of the temperature of the water. The specific heat of water is approximately 75 J/mol�•K. When steam at 100°C touches skin, the heat released is from the condensation of the gas with a heat of condensation of approximately 41 kJ/mol.

2222

�• Measure of the potential that a substance has for undergoing change in a system.

�• At equilibrium, the chemical potential of a substance is the same throughout a sample, regardless of how many phases are present.

�• The phase with the lowest chemical potential at a specified temperature is the most stable phase.

�• For a one-component system, molar Gibbs energy = chemical potential.

Chemical Potential,

G

2323

Thermodynamic criterion of equilibriumAt equilibrium, the chemical potential of a substance is the same throughout a sample, regardless of how many phases are present.

At low temperatures and if pressure is not too low, the solid phase of a substance has the lowest chemical potential.

Above a certain temperature, another phase may be the lowest spontaneous transition to the second phase.

Consider liquid and solid phases of water at a fixed T, PIf s(T,P) = l(T,P), then liquid water and ice coexist.If s(T,P) > l(T,P), then water is in the liquid phase. If s(T,P) < l(T,P), then water is in the solid phase.

Fundamental equation

24

VP

ST

nGdPVdTSduVdPSdTdG

TP

0T

0Srd Law: 3P

Negative slope

2525

Temperature dependence of phase stability

As temperature is raised, the chemical potential of a pure substance decreases:

> 0 so slope of line is negative.

Slope decreases fromgas > liquid > solid because

As temperature is raised, change in phase.

S

)s(S)l(S)g(S

2626

Response of melting to applied pressureMost substances melt at a higher temperature when subjected to pressure (exception being water).

mT

VP

An increase in pressure raises the chemical potential of any pure substance (because Vm >0). As Vm(l) > Vm(s), this raisesthe melting temperature.

2727

For water, Vm(l) < Vm(s).An increase in pressure increases the chemical potential of the solid more than the liquid.Melting point is slightly lowered.

2828

The location of phase boundariesWhen two phases are in equilibrium, their chemical potentials must be equal.

)T,P()T,P(

In single phase regions, one of the chemical potentials is lower than the other. T and P can be changed independently without changing phases.

Phase rule: F = 3 �– P

F = number of degrees of freedom, P= numberof phases that coexist.

2929

Clapeyron equation

Consider dP/dT. Let P and T be changed infinitesimally so that and are in equilibrium.

dPVdTSdPVdTS

dPV dT S -dGd

dT)SS(dPVV

VS

dTdP

dddd

Exact expression for the slope ofthe phase boundary

On coexistence line

3030

Another way is to use express in terms of H and V.

Substitute into Clapeyron equation,

TdT

VHdP

i

f

TTln

VHP,gIntegratin

VTH

VS

dTdP

TH

SSTH

S - THS - TH

ForSTHG

BA

line, ecoexistenc the on

3131

The solid-liquid boundary

Melting �– accompanied by a molar enthalpy change fusHand occurs at temperature T.

THS fus

fus

VTH

dTdP

fus

fus

Integrating,

*)TT(V*T

H*PPfus

fus As pressure is raised, meltingtemperature rises.

+

+ and small

*TTln

VH*PP

fus

fus

T close to T*

*T*TT

*T*TTln

*TTln 1

3232

The liquid-vapour boundary

�• Entropy of vaporization at T Svap = vapH/T

�• Clapeyron equation for liquid-vapour boundary

VTH

dTdP

vap

vap+

+ and large

Boiling temperature is more responsive to pressure than freezing point (because dT/dP is large).

3333

Sample: Estimate the typical size of the effect of increasing pressure on the boiling point of a liquid.

VTH

dTdP

vap

vap

(g)V(l)V(g)VV mmmvap

Molar volume of an ideal gas at 1 atm and slightly above room temperature ~ 25 dm3 mol-1

113133

1-1

K atm0.034 K Pa 10 x3.4 molm10 x 25

molK J 85dTdP

Trouton�’s constant: 85 J K-1 mol-1

1atm K 29dPdT

Clausius-Clapeyron equation

dTRTH

Plnd

dTRTH

pdP

)P/RT(TH

dTdP

vap

vap

vap

2

2

34

Approximations: 1.Molar volume of gas is

>> molar volume of liquid, therefore neglect volume of liquid. VvapVm(g).

2.Assume ideal gas behavior, so that Vm(g) = RT/P

if

vap

i

f

TTRH

PPln 11

34

Integrating,

Predict how vapour pressure varies with temperature and how boiling point varies with pressure

Assume independent

of temperature

3535

Effect of Temperature on Vapour PressureRaising the temperature of a liquid increases the fraction of molecules moving fast enough to escape into the liquid anddecreases the fraction moving slowly enough to be recaptured. The higher the temperature, the higher the

vapour pressure.

3636

Effect of Intermolecular Forces on Vapour Pressure

The weaker the intermolecularforces are, the higher the vapourpressure.

A linear plot of the relationship between vapor pressure and

temperature .

Hvap/R

3737

SAMPLE PROBLEM Using the Clausius-ClapeyronEquation

SOLUTION:

PROBLEM: The vapor pressure of ethanol is 115 torr at 34.90C. If Hvap of ethanol is 40.5 kJ/mol, calculate the temperature (in 0C) when the vapor pressure is 760 torr.

ln

P2P1

= - Hvap

R1

T2

1T1

34.90C = 308.0K

ln760 torr115 torr

=-40.5 x103 J/mol8.314 J/mol*K

1T2

1308K

-

T2 = 350K = 770C

3838

Solid-Vapour Boundary

Enthalpy of sublimation >Enthalpy of vaporization

subH = fusH + vapH

Steeper slope for sublimation curve than for vaporization curve.

VTH

dTdP

sub

sub

Two-component systems

Total number of variables = 4T, P, xA, yA

Constraints due to coexistence = 2

A(l) = A(g)

B(l) = B(g)Total number of independent variablesF = 4 -2 = 2Know T, P uniquely determines the

compositions in the liquid and gas phase.

39

A(g), yA T, PB(g), yB = 1 - yA

A(liq), xAB(liq), xB = 1 - xA

Gibbs phase rule: F = C �– P + 2F: number of degrees of freedom (independent variables)C: number of componentsP: number of phases

For 1-component system, F = 3 �– PP = 1, F = 2 can vary freely in the T, P region.P = 2, F = 1 can vary along coexistence curveT(P).P = 3, F = 0 Triple point.

40

41

Ideal Solutions

A solution is any homogeneous phase that contains more than one component. Solvent - larger proportion of the solution.Solute �– component lesser proportion.A solution is considered to be ideal when there is a complete uniformity of intermolecular forces, arising from similarity in molecular size and structure.Partial vapour pressures of the individual components within the solution are a good measure of the individual components �– measures the escaping tendency of a molecule from solution.

4242

Raoult�’s Law

Francois Raoult: The vapour pressure PA of solvent A is equal to its mole fraction in the solution multiplied by the vapour pressure PA* of the pure solvent - Raoult�’s Law

Some mixtures obey Raoult�’s law very well �– structurally similar components.

Raoult�’s law assumes a linear dependence.

PA = xA PA*

43

PA*

P

O 1xB

PA* = vapor pressure of pure A at temperature T

0Px

)P - x1(

PxPPP

*AB

*AA

*AA

*AA

*A

A(g), yA T, P

A(liq), xAB(liq), xB = 1 - xA

A: volatile solvent, B: involatile solute. Solvent and solute do not interact, like in mixture of ideal gases.

PA

Vapor pressure is lowered in the mixture

PA < PA*

4444

B

Both A and B volatile

Benzene-toluene,Ethylene bromide-ethylene chlorideCarbon tetrachloride �– trichloroethylene

A(g), yA T, PB(g), yB = 1 - yA

A(liq), xAB(liq), xB = 1 - xA

A

*BB

*AABA

*BBB

*AAA

PxPxPPP

PxPandPxP

Ideal solutions =Both components obey Raoult�’s Law.

45

If the vapour phase is treated as an ideal gas, then according to Dalton�’s law of partial pressuresPtotal = P1 + P2

From Raoult�’s law, Ptotal = x1P1* + x2P2*

Ptotal = x1P1* + (1 �– x1)P2*

Ptotal = P2* + (P1* �– P2*)x1 Straight line with slopeP1* �– P2*

46

Slope = (P1* �– P2*)Intercept = P2*

Liquid solutions will boil at different temperatures depending on their composition and vapour pressures of the pure components.

47

Example problemAn ideal solution can be approximated using the liquid hydrocarbons hexane and heptane. At 25 ºC, hexane has an equilibrium vapor pressure of 151.4 Torr and heptane has an equilibrium vapor pressure of 45.70 Torr. What is the equilibrium vapor pressure of a 50:50 molar hexane and heptane solution in a closed system ?

Using Raoult�’s law,P1 = (0.50)(151.4 Torr) = 75.70 TorrP2 = (0.50)(45.70 Torr) = 22.85 TorrBy Dalton�’s law, Ptotal = 75.70 + 22.85 = 98.55 Torr

48

Example problemA hexane/heptane solution is used to establish a constant 65 ºC temperature in a closed system that has a pressure of 500 Torr. At 65 º C, the vapor pressures of hexane and heptane are 674.9 and 253.5 Torr. What is the composition of the solution ?Ptotal = P2* + (P1* �– P2*)x1

Mole fraction of hexane : heptane = 0.5850 : 0.4150

**

*tot

PPPPx

21

21

5850044215246

525396745253500 .

..

...

49

At 25 C, hexane has an equilibrium vapor pressure of 151.4 Torr and heptane has an equilibrium vapor pressure of 45.70 Torr. P1 = (0.50)(151.4 Torr) = 75.70 Torr (hexane)P2 = (0.50)(45.70 Torr) = 22.85 Torr (heptane)By Dalton�’s law, Ptotal = 75.70 + 22.85 = 98.55 Torr

In the gas phase, y1 = 75.70/98.55 = 0.77y2 = 22.85/98.55 = 0.23

Vapour always contains relatively more of the more volatile component than does the liquid.

50

Example problemAt temperature T, the vapor pressure of pure benzene, C6H6, is 0.256 bar and the vapor pressure of pure toluene is 0.0925 bar. If the mole fraction of toluene in the solution is 0.600 and there is some empty space in the system, what is the total vapor pressure in equilibrium with the liquid, and what is the composition of the vapor in terms of mole fraction ?

Using Raoult�’s law, pbenzene = (0.4) x (0.256 bar) = 0.102 barPtoluene = (0.6) x (0.0925 bar) = 0.0555 barPtotal = 0.102 + 0.0555 = 0.158 barytoluene = 0.0555/0.158 = 0.351ybenzene = 0.102/0.158 = 0.646

Vapor phase is enriched in benzene over original solution

51

Composition of the vapour

What are the mole fractions of the two components in the vapour phase ?Total pressure 1212211121 1 x)PP(PP)x(PxPPP *****

total

x)PP(P

PxPPy ***

*

total 1212

1111

y)PP(P

Pyx ***

*

1121

211

y)PP(P

PPP ***

**

total1121

21

Let y1 and y2 denote the gas-phase mole fractions,

Rearranging,

Substituting into Ptotal:

52

Total pressure of the vapour phase vs mole fraction of one component is not a straight line !

liquid

vapour

, liquid curve

, vapour curve

Constant temperatureL + V

53

Tie-line �– connect the liquid-phase composition with the vapour-phase composition.

Coexistence curve or bubble line

Coexistence curve or dew line

54

1

4

xB(1)

xB, yB

yB(2) xB(1)

P1 yB(2)P1

yB(4) = xB(1)xB(4)

xB(4)At B, liquid phase (xB(1)) with a very small amount of gas phase (yB(2)). On decreasing the pressure, composition of the liquid and gas phases change along the coexistence lines until yB(4) = xB(1). Then, there is no more liquid left.

yB(4)

Lever RuleHow much is in each phase ?

55

12 3

Liquid

gas

yA(1)yA(2) xA(3)

)2(y)1(y)1(y)3(x

liquidgas

AA

AA

56

Fractional distillation

Liquid phase

Gas phase

57

Temperature-Composition Diagrams

Constant pressure

Liquid

Both bubble point and dew point lines are curved !

Dew line

Bubble line

58

Theoretical plates �– number ofeffective vaporization and condensation steps that are required to achieve a condensateof a given composition from agiven distillate.

5959

The process of fractional distillation.

Gas

Gasoline 380CKerosene 1500C

Heating oil 2600C

Lubricating oil 3150C-3700C

Crude oil vapors from heater

Steam

Residue (asphalt, tar)

Condenser

Gasoline vapors

C

E

Bubble caps

Bubble trays hold the bubble caps. Overflow weir to allow some liquid to fall back down to be evaporated again

A bubble cap forces the vapours and liquid to mix.

61

Fractionating column �– provides the contacting surface for mass transfer between the liquid and vapour at a desired rate. At each level in the column, vapour from the level or plate below bubbles through a thin film of liquid C, at a temperature slightly lower than that of the vapour coming through the bubble cap. Partial condensation of the vapor occurs. The lower boiling mixture remains as vapour and move to the next plate. The vapour leaving each plate is enriched in the more volatile component compared to the entering vapour from the plate below. Excess liquid at each plate is returned to the plates below via overflow tubes, E.http://www.youtube.com/watch?v=26AN1LfbUPc

62http://www.metacafe.com/watch/yt-0x2-8dedmE4/fractional_distillation/

Chemical potentials of ideal solutions

63

)P,T,g()P,T,l( AAA

Assuming ideal gas,

AoAo

AAAA PlnRT)T,g(

PPlnRT)bar1,T,g()P,T,g(

Substituting into 1st equation,

AoAA PlnRT)T,g()P,T,l(

For pure A, *A

oA

*A PlnRT)T,g()P,T,l(

At coexistence,

For mixture,

*A

A*A

A*A

*AA

PPlnRT)P,T,l(

PlnRTPlnRT)P,T,l()P,T,l(

64

Using Raoult�’s Law: PA = xAPA*

A*AA xlnRT)P,T,l()P,T,l( for an ideal solution

Chemical potential of A in mixture

Chemical potential of A in pure liquid

Mole fraction of A in mixture

)pure)(P,T,l()mixture)(P,T,l( *AA

Chemical potential for a mixture is always lower than pureliquid at same T,P.

Free energy change in ideal solutions

Before mixing, BBAAi nnG

BBBAAAf xlnRTnxlnRTnG

BABBAA

BBAABBAA

BBAABBBAAAmix

n n n where xlnxxlnxnRT RTlnxnx RTlnxnx xlnRTnxlnRTn

nnxlnRTnxlnRTnG

After mixing,

Gibbs energy of mixing

BBAAmix

P

mixmix

xlnxxlnxnRST

GS

SdTVdPG

0STGH mixmixmix

No enthalpy change on mixing for ideal solutions.

0T

mixmix P

GV

No volume change on mixing.

Non-ideal solutionsIn ideal solutions, non-interacting molecules. Gmix comes all from entropy of mixing. No change in volume upon mixing, just like ideal gas.Non-ideal solutions �– molecules interact differently with liquid molecules of another species giving rise to positive or negative deviation from Raoult�’s Law

65

A

A

B

B

A

B

A

B

uAA < 0 uBB < 0 uAB uABInteraction energy u = 2uAB �– (uAA �– uBB)

Departure from ideality

66

Positive Deviation u > 0

Ethanol/benzene, ethanol/chloroform,ethanol/water

Positive deviation:If strength of interaction between like molecules A-A or B-B > A-B, the tendency will be to force both components into the vapour phase. Observed for dissimilar liquids.

67

68

Negative deviation u < 0

Acetone/chloroform

Occurs when the attractions between components A and B are strong. Mixing is energetically favorable in liquid phase.Holding back of molecules that would otherwise go into the vapour state.

3HCC O

H3CC

Cl

Cl

Cl

HH-bonding

Azeotropic composition �– composition of the liquid and vapour in equilibrium have the same mole fraction. E.g. water and ethanol have a minimum boiling point at 78.2 ºC and is 96 % ethanol and 4 % water.

Minimum boiling azeotrope

At this point,x1 = y1

Maximum-boiling azeotrope

7171

Henry�’s Law

In ideal solutions, the solute and the solvent obey Raoult�’s law.For real solutions at low concentrations, although the vapour pressure of the solute is proportional to its mole fraction, the constant of proportionality is not the vapour pressure of the pure substance.

PB = xBKB

xB: mole fraction of the solute; KB: empirical constant (with dimensions of pressure) chosen so that the plot of the vapour pressure of B against its mole fraction is tangent to the experimental curve at xB = 0.

7272

Mixtures for which the solute obeys Henry�’s law and the solvent obeys Raoult�’s law �– ideal-dilute solutions.

1. xB 1 �• B is the solvent.�• Raoult�’s law applies

PB = xBPB*�• B molecules sees mostly

other B molecules.

2. xB 0 �• B is the solute.�• Henry�’s law applies

PB = xBKBKB depends on A.Positive deviation: KB > PB*Negative deviation: KB < PB*

7373

In dilute solution, the solventmolecules are in an environment that differs only slightly from that of the pure solvent. Solute particles are in an environment totally unlike that of the pure solute.

7474

7575

Liquid-Gas SystemLiquid/gas solutions are not ideal, therefore Raoult�’s law does not apply. Examples of liquid/gas solutions, fizzy drinks, oxygen in water, carbon dioxide in water.HCl �– high solubility in water.Oxygen �– low solubility, 0.0013 M.

William Henry (1774 �– 1836) �– found that the mass of gas mB dissolved by a given volume of solvent at constant temperature is proportional to the pressure of the gas in equilibrium with the solution.

mB = kBPB B: solute, kB is the Henry�’s law constant.

Gas solubility in water decreases with rising temperature. Gases have weak intermolecular forces and as temperature rises, more gas particles re-enters the gas phase.

Thermal pollution

Water taken from lakes, rivers for cooling ofliquids, gases, equipment. This waterhas to be cooled beforeit exits the plant.

7777

For practical reason, Henry�’s law is expressed in terms of the molality, b, of the solute: PB = bBKB

K = P/b

7878

Hyperbaric oxygen chamber �– treatment of certain diseases �– CO poisoning, diseases caused by anaerobic bacteria �– gas gangrene and tetanus.

Scuba diving �– air is supplied at a higher pressure so that pressure within diver�’s chest matches pressure exerted by the surrounding water (1 atm for every 10 m descent). Nitrogen is much more soluble in fatty tissues than in water, so it tends to dissolve in the central nervous system, bone marrow, fat reserves nitrogen narcosis, bends, arterial embolisms.

7979

Example: Estimate the molar solubility of oxygen in water at 25 C and a partial pressure of 21 kPa,

PB = bBKB where KB = 7.9 x 104 kPa kg mol-1

1414

O

OO kg mol10 x 2.9

mol kg kPa 10 x 7.9kPa 21

KP

b2

2

2

3-3-1-OHO2

-3

dm mmol 0.29 dm kg 0.99709 x kg mmol 0.29 x b ][O

dm kg 0.99709 water of density Taking

22

1415

N

NN kg mol10 x 1.5

mol kg kPa 10 x1.56 kPa 79

KP

b2

2

2

1415

N

NN kg mol10 x 3.1

mol kg kPa 10 x1.56 kPa 21

KP

b2

2

2

8080

Example problem

The Henry�’s law constant Ki for CO2 in water is 1.67 x 108

Pa at some temperature T. If the pressure of CO2 in equilibrium with water were 1.00 x 106 Pa, what is the mole fraction of CO2 in the solution ? Can you estimate the molarity of the CO2 solution ?Solution:PCO2 = KCO2xCO2 where KCO2 = 1.67 x 108 Pa

1.00 x 106 Pa = 1.67 x 108 Pa xCO2

xCO2 = 0.00599

xCO2 = nCO2/(nCO2 + nH2O) nCO2/nH2O = 0.00599

1 mole water = 18 g = 18 cm3 = 0.018 L

Molarity of CO2 solution = 0.00599/0.018 L = 0.333 M

Mole fraction

Example problemThe partial pressure of carbon dioxide gas into a bottle of coca cola is 4 atm at 25 C. What is the solubility of CO2 ? The Henry�’s law constant for CO2 dissolved in water is 1.67 x 108 Pa.PCO2 = KCO2xCO2 where KCO2 = 1.67 x 108 Pa

4.00 x 105 Pa = 1.67 x 108 Pa xCO2

xCO2 = 0.0024

xCO2 = nCO2/(nCO2 + nH2O) nCO2/nH2O = 0.0024

1 mole water = 18 g = 18 cm3 = 0.018 L

Molarity of CO2 solution = 0.0024/0.018 L = 0.13 M

8282

Colligative Properties of SolutionsProperties of solutions in the dilute limit.Property of a solution with a non-volatile solute may be different from that of the pure solvent. Properties are independent of the identity of the solute molecules but related only to the number of solute molecules colligativeproperties.Concentrations in (a)Mole fraction xB = nB/(nA + nB) nB/nA

(b)Molality: mB = moles solute/kg solvent = nB/(nAMA) where MA is the mass in kg of 1 mole of solvent.

8383

All colligative properties stem from reduction of the chemical potential of the liquid solvent as a result of the presence of the solute.

Lowering of the liquid�’s chemical potential has a greater effect on the freezing point than on the boiling point.

)P,T,l()P,T,l( pureA

mixA

Four Colligative Properties

Vapour pressure depression:

Boiling point elevation:

Freezing point depression

Osmotic pressure:

84

*AB

*AAA PxPPP

H)T(RK

xKTTT

vap

2*b

b

Bb*bbb

H)T(RK

xKTTT

fus

2*fus

f

Bf*fff

VnRT B

8585

Vapour pressure of pure liquid reflects the tendency of the solution towards greater entropy. When a solute is present, the disorder of the condensed phase is higher than that of the pure liquid, and there is a decreased tendency to form the vapour lower vapour pressure higher boiling point.

8686

The effect of a solute on the vapor pressure of a solution.

8787

How does the amount of solute affect the magnitude of the vapour pressure ?

0PxPPP

P)x1(PxP*AB

*AAA

*AB

*AAA

1. Vapour pressure lowering �– just Raoult�’s law.

8888

Using Raoult�’s Law to Find the Vapor Pressure Lowering

SOLUTION:

PROBLEM: Calculate the vapor pressure lowering, P, when 10.0 mL of glycerol (C3H8O3) is added to 500. mL of water at 50.0C. At this temperature, the vapor pressure of pure water is 92.5 torr and its density is 0.988 g/mL. The density of glycerol is 1.26 g/mL.

10.0 mL C3H8O31.26 g C3H8O3

mL C3H8O3

mol C3H8O3

92.09 g C3H8O3= 0.137 mol C3H8O3

500.0 mL H2O0.988 g H2O

mL H2O

mol H2O18.02 g H2O

= 27.4 mol H2O

P = 0.137 mol C3H8O3

0.137 mol C3H8O3 + 27.4 mol H2O92.5 torrx

x

x

= 0.461 torr

x

x

8989

Heterogeneous equilibrium for boiling point elevation

xlnRT)l()g( A*A

*A

9090

Elevation of boiling point

In the presence of a solute at a mole fraction, xB, the boiling point increases from T* to T*+ T where

T = KbxB

For practical applications, express mole fraction of B in its molality, mb, in the solution

T = K�’bmb

HRTK

vap

*b

b

2

where K�’b is the empirical boiling-point constant of the solvent.

H)T(RMK

vap

2*bA'

b

91

RTG

RT)l()g(

lnx

g,Rearrangin

xlnRT)l()g(

vap*A

*

A

A*A

*

A

A

2vapvapA

RTH

dT)T/G(d

R1

dTxlnd

dTRT

Hxlnd 2

vapA

T

*T

vapxln

A dTT

HR

xlnd

,gIntegratinA

20

1

*T1

T1

RH

)x-(1 ln lnx

constant, H Assuming

vapBA

vap

T1

*T1

...)x31x 32

RH

x

21- x- x)-ln(1 (from

x )x-ln(1 then 1, x If

vapB

BBB

2*TT

*TT*TT

T1

*T1

T*, T Also

BBvap

2

2vap

B

KxxH*TRT

*RTTH

x

Derivation

9292

Heterogeneous equilibrium for lowering of freezing point

At the freezing point, the chemical potentials of A in the 2 phases are equal:

xlnRT)l()s( A*A

*A

Nonvolatile solutes will make it harder forsolvent molecules to crystallize at theirnormal melting point.

9393

Depression of freezing point

T = -KfxB HRT

Kfus

*f

f

2

where T is the freezing point depression and fusH is the enthalpy of fusion of the solvent.

Larger depressions are observed in solvents with low enthalpies of fusion and high melting points.

9494

For dilute solutions, mole fraction is proportional to the molality of the solute, mb,

T = -K�’f mb

where Kf is the empirical freezing-point constant for the solvent. Also called the cryoscopic constant.

Once the freezing-point constant of a solvent is known, the depression of freezing point may be used to measure the molar mass of a solute.

fus

2*fusA'

f H)T(RMK

9595

Example: A solution contains 1.50 g of solute in 30.0 g of benzene and its freezing point is 3.74 ºC. The freezing point of pure benzene is 5.48 ºC. Calculate the molar mass of the solute.Kf = 4.90 K kg mol-1

Solution: K74.148.574.3T

1-1

fkg mol

mol kgK 4.90K 1.74

KT 355.0mmKT bbf

g0.30xmolkg355.0g50.1M

WM/Wm 1solute

solvent

solutesoluteb

-1mol kg 0.141 -1mol g 141

9696

Osmotic Pressure

Assume a system consisting of two compartments which are separated by a semi-permeable membrane. Solvent molecules A can pass freely through this membrane, but solute molecules B cannot pass the membrane.

Osmosis is the spontaneous passage of solvent molecules from the pure solvent (higher A concentration) into the solution (lower concentration of A).

9797

pure solvent

solution

net movement of solvent

semipermeable membrane

solvent molecules

solute molecules

osmotic pressure

Applied pressure needed to prevent volume increase

9898

The osmotic pressure, , is the pressure that must be applied to the solution to stop the influx of solvent. At equilibrium, the chemical potential of the solvent must be the same on each side of the membrane.

At equilibrium:

)T,P,l()T,P,l( *AA

0xlnRT)T,P,l()T,P,l(

)T,P,l(xlnRT)T,P,l(

A?

*A

*A

*AA

*A

dG = -SdT + VdPAt constant T,

dPVddG*A

*A

99

Integrating from P P + ,

*A

P

P

*A

*A

*A VdPV)T,P,l()T,P(,

Assuming volumeconstant of P

0)n/V()n/n(RTn/nx)x1ln(xlnBut

0VxlnRT

AAAB

ABBB

A

*AASo

But VA VA + VB = V as VB << VA

B

B

RTnV0V)n(RT

Van�’t Hoff equation

100100

Van�’t Hoff equation

For dilute solutions, the osmotic pressure is given by van�’t Hoff equation:

Vsoln = nsoluteRT

Used for determination of molar masses of macromolecules such as proteins and synthetic polymers. Important for life �– cell wall membranes �– isotonic solution is 8.9 g/l of NaCl.

101101

Example problemWhat is the osmotic pressure of a 0.01 m solution of sucrose in water? How high would a column of diluted sucrose be at equilibrium? Assume 25 ºC and the density of the solution is 1.01 g/ml.

Solution 0.01 m solution contains 0.01 mole of sucrose in 1000 g of water.Volume of solution = 1000 g /1.01 g/ml = 990.1 ml = 0.9901 L

(0.9901 L) = (0.01 mol) x (0.08314 L.bar mol-1 K-1) x 298 K= 0.250 bar

Substantial osmotic pressure for such a dilute solution

102102

0.250 bar = 0.250 x 105 Pa = 0.250 x 105 Nm-2

P = h g0.250 x 105 Nm-2 = h x 1010 kg m-3 x 9.81 m.s-2

h = 2.52 m

103103

Determining Molar Mass from Osmotic Pressure

SOLUTION:

PROBLEM: Biochemists have discovered more than 400 mutant varieties of hemoglobin, the blood protein that carries oxygen throughout the body. A physician studying a variety associated with a fatal disease first finds its molar mass (M). She dissolves 21.5 mg of the protein in water at 5.00C to make 1.50 mL of solution and measures an osmotic pressure of 3.61 torr. What is the molar mass of this variety of hemoglobin?

mol/L 10 x2.08 K))(278.1KL.atm.mol(0.0821

Torr.atmTorr/7603.61RT

Vn 4-

11

1

mol 10 x3.12 L10 x1.50 x mol/L 10 x2.08 ml1.50 inmoles of Number

7-

-3 -4

147-

-3

g.mol 10 x 6.89mol 10 x3.12 g 10 x 21.5 weight Molecular