Phases and solutions
-
Upload
molina-thirumal -
Category
Documents
-
view
54 -
download
0
description
Transcript of Phases and solutions
11
Chapter 5
Phases and Solutions
22
Stability of Phases
3KDVH�ದ DQ\�SK\VLFDOO\�GLVWLQFW��KRPRJHQHRXV�SDUW�RI�D�V\VWHP��8QLIRUP�WKURXJKRXW�LQ�FKHPLFDO�FRPSRVLWLRQ�DQG�SK\VLFDO�VWDWH�
&RPSRQHQW ದ XQLTXH�FKHPLFDO�VXEVWDQFH�WKDW�KDV�GHILQLWH�SURSHUWLHV�
3KDVH�WUDQVLWLRQ ದ VSRQWDQHRXV�FRQYHUVLRQ�RI�RQH�SKDVH�LQWR�DQRWKHU�SKDVH��RFFXUV�DW�D�FKDUDFWHULVWLF�WHPSHUDWXUH�IRU�D�JLYHQ�SUHVVXUH�
7UDQVLWLRQ�WHPSHUDWXUH ದ WHPSHUDWXUH DW�ZKLFK�WKH�WZR�SKDVHV�DUH�LQ�HTXLOLEULXP�DQG�WKH�*LEEV�HQHUJ\�LV�PLQLPL]HG�DW�WKH�SUHYDLOLQJ�SUHVVXUH��
33
ExampleD� 6\VWHP�FRQWDLQLQJ�LFH�DQG�ZDWHU6LQJOH�FRPSRQHQW�DQG���SKDVHV
E� $�������VROXWLRQ�RI�ZDWHU�DQG�HWKDQRO��FRPSRQHQWV�DQG���SKDVH
F� $�SUHVVXUL]HG�WDQN�RI�&2��WKDW�FRQWDLQV�OLTXLG�DQG�JDV6LQJOH�FRPSRQHQW�DQG���SKDVHV
G� 3XUH�VROLG�&D26LQJOH�FRPSRQHQW�DQG���SKDVH
H� &D2 DQG�6L2���FRPSRQHQWV�DQG���SKDVHV
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