if two volatile and miscible liquids are combined to form...
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© 2010 Pearson Education South Asia Pte LtdPhysical Chemistry
9.9 Real Solutions Exhibit Deviations from Raoult’s Law9.9 Real Solutions Exhibit Deviations from Raoult’s Law
• If two volatile and miscible liquids are combined to form a solution, Raoult’s law is not obeyed.
• Use the experimental data in Table 9.3:
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9.99.9 Real Solutions Exhibit Deviations from Raoult’s LawReal Solutions Exhibit Deviations from Raoult’s Law
© 2010 Pearson Education South Asia Pte LtdPhysical Chemistry
9.9 Real Solutions Exhibit Deviations from Raoult’s Law9.9 Real Solutions Exhibit Deviations from Raoult’s Law
• Assuming that A and B are miscible for real solution, we have
• Since ΔVmixing and ΔHmixing can be positive or negative, depending on the nature of the A–B interaction in the solution.
0
0
0
0
mixing
mixing
mixing
mixing
H
V
S
G
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Figure 9.13Figure 9.13
Figure 9.13 The data in Table 9.3 are plotted versus xCS2. The dashed lines showthe expected behavior if Raoult’s law were obeyed.
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Figure 9.14Figure 9.14
Figure 9.14
Deviation in the volume fromthe behavior expected for 1 molof an ideal solution are shownfor the acetone-chloroformsystem as a function of themole fraction of chloroform.
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• The deviation of the volume from ideal behavior can beunderstood by defining the concept of partial molar quantities.
• Partial molar volume
The partial molar volume of a component in a solution is defined as the volume by which the solution changes if 1 mole of the component is added to such a large volume that the solution composition can be assumed constant.
9.9 Real Solutions Exhibit Deviations from 9.9 Real Solutions Exhibit Deviations from Raoult’sRaoult’s LawLaw
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9.9 Real Solutions Exhibit Deviations from 9.9 Real Solutions Exhibit Deviations from Raoult’sRaoult’s LawLaw
• The partial molar volume of solution is defined as the volume by which the solution changes if 1 mol of the component is added to that solution composition.
• Thus the volume of a binary solution is given
2,,1
211 ,,,nTPn
VnnTPV
21222111 ,,,,,, nnTPVnnnTPVnV
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9.99.9 Real Solutions Exhibit Deviations from Real Solutions Exhibit Deviations from Raoult’sRaoult’s LawLaw
Figure 9.15
The partial molar volumes of a chloroform (yellow curve) and acetone (Red curve) in a chloroform-acetone binary solution are shown as a function of xchloroform.
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9.99.9 Real Solutions Exhibit Deviations from Raoult’s LawReal Solutions Exhibit Deviations from Raoult’s Law supportsupport
Vtotal=nAVA+nBVB
mole fraction AA
A B
nxn n
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ProblemProblem
• P9.18) The partial molar volumes of ethanol in a solution with = 0.60 at 25ºC are 17 and 57 cm3 mol–1, respectively. Calculate the volume change upon mixing sufficient ethanol with 2 mol of water to give this concentration. The densities of water and ethanol are 0.997 and 0.7893 g cm–3, respectively, at this temperature.
Solution:
107
22
2
2
2 2
2
3 ? 3 ?17.0 cm mol and 57.0 cm mol
2.00 and 0.600
2 mol 0.600; 1.3332 mol
H O EtH O Et
H O Et
H OH O H O
H O Et
EtEt
V n V n V
V Vn
n xn n
nn
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ProblemProblem
• P9.25) A solution is made up of 184.2 g of ethanol and 108.1 g of H2O. If the volume of the solution is 333.4 cm3 and the partial molar volume of H2O is 17.0 cm3, what is the partial molar volume of ethanol under these conditions?
Solution:
108
2 2
2 2
2 2
3 3 ??
3 ?
?
108.1 g333.4 cm 17.0 cm mol18.02 g mol 57.8 cm mol184.2 g
46.04 g mol
H O H O ethanol ethanol
H O H O ethanol ethanol
H O H Oethanol
ethanol
V n V n V
V n V n V
V n VV
n
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ProblemProblem
• P9.27) The densities of pure water and ethanol are 997 and 789 kg m–3, respectively. The partial molar volumes of ethanol and water in a solution with xethanol = 0.20 are 55.2 and 17.8 × 10–3 L mol–1, respectively. Calculate the change in volume relative to the pure components when 1.00 L of a solution with xethanol = 0.20 is prepared.
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ProblemProblem
Solution:
110
2 2
2 2
2
2 2
3 ? ? ?
?
?
0.80 17.8 10 L mol 0.20 55.2 10 L mol
0.0253 L mol
1.00 L = 39.6 mol0.0253 L mol
H O H O ethanol ethanol
H O H O ethanol ethanoltotal
total
total H O ethanol
ethanol ethanol
H O H O
V n V n V
V x V x VnV
n
n n n
x nx n
2
2
2
2
? ? ? ?
? ?
1; 31.7 mol 7.90 mol
4
46.07 10 kg mol 18.02 10 kg mol7.90 mol + 31.7 mol = 1.034 L789 kg m 998 kg m0.034 L
H O ethanol
H Oethanolideal ethanol H O
ethanol H O
ideal
n n
MMV n n
V V V
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9.10 The Ideal Dilute Solution9.10 The Ideal Dilute Solution
• We define the dimensionless activity, asolvent, of the solvent by
• For a nonideal solution, the activity coefficient defined by
• The activity coefficient quantifies the degree to which the solution is non-ideal.
*solvent
solventsolvent P
P
solvent
solventsolvent x
a
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9.10 The Ideal Dilute Solution9.10 The Ideal Dilute Solution
• For real solution the chemical potential of a component related to its activity:
• For ideal dilute solution, solute and solvent are defined by the conditions, xsolute→0 and xsolvent→1.
1 aln* RTisolventi
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9.10 The Ideal Dilute Solution9.10 The Ideal Dilute Solution
• Henry’s law states that
where i = ideal dilute solutionkH = Henry’s law constant
• Ideal dilute solution is a solution in which the solvent is described using Raoult’s law and the solute is described using Henry’s law.
0 as iiHii xkxP
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9.11 Activities Are Defined with Respect to Standard States9.11 Activities Are Defined with Respect to Standard States
• The Henry’s law standard state is a state in which the pure solute has a vapor pressure kH,solute rather than its actual value P*solute.
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9.11 Activities Are Defined with Respect to Standard States9.11 Activities Are Defined with Respect to Standard States
• Henry’s law standard state chemical potential is given by
• The activity and activity coefficient based on Henry’s law are defined, respectively, by
*** kln
solute
soluteH
soluteH
solute PRT
i
iiH
i
ii x
akPa and
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Example 9.8Example 9.8
Calculate the activity and activity coefficient for CS2 at xCS2=0.3502. Assume a Raoult’s law standard state.
Solution:
99713502069940
6994035123358
2
22
2
22
...
xa
...a *
CS
RCSR
CS
CS
CSRCS P
P
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Example 9.9Example 9.9
Calculate the activity and activity coefficient for CS2 at xCS2=0.3502. Assume a Henry’s law standard state.
Solution:
509003502017830
178302010
3358
2
22
2
2
2
...
xa
..a,
CS
HCSH
CS
CSH
CSHCS k
P
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• P9.29) Calculate the activity and activity coefficient for CS2 at = 0.7220 using the data in Table 9.3 for both a Raoult’s law and a Henry’s law standard state.
Solution:
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2
2
2
2
2
2
*446.9 Torr 0.8723512.3 Torr
0.8723 1.2080.7220
CSRCS
CS
RCSR
CSCS
Pa
P
ax
2
2
2
2
2
2
,
446.9 Torr 0.22232010 Torr
0.2223 0.30790.7220
CSHCS
H CS
HCSH
CSCS
Pa
k
ax
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9.12 Henry’s Law and the Solubility of Gases in a Solvent9.12 Henry’s Law and the Solubility of Gases in a Solvent
• The ideal dilute solution model can be applied to the solubility of gases in liquid solvents.
• We will see how to model the dissolution of a gas in a liquid.
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Example 9.12Example 9.12
The average human with a body weight of 70 kg has a blood volume of 5.00 L. The Henry’s law constant for the solubility of N2 in H2O is 9.04 × 104 bar at 298 K. Assume that this is also the value of the Henry’s law constant for blood and that the density of blood is 1.00 kg L-1.
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Example 9.12Example 9.12
a. Calculate the number of moles of nitrogen absorbed in this amount of blood in air of composition 80% N2 at sea level, where the pressure is 1 bar, and at a pressure of 50 bar.
b. Assume that a diver accustomed to breathing compressed air at a pressure of 50 bar is suddenly brought to sea level. What volume of N2 gas is released as bubbles in the diver’s bloodstream?
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SolutionSolution
a. We use the symbol csolute to designate the solute molarity, and co to indicate a 1 molar concentration.
mol..barmol.
bar.bar.
molkg.kg.
130105250 bar, 50 Atpressure total 1 at 1052
1004980
100218 1.00 05
3
3
413
1
2
2
222
N
NH
NOHN
n
LLkPnn
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SolutionSolution
b.
The symptoms induced by the release of air into the bloodstream are known to divers as the bends. The volume of N2 just calculated is far more than is needed to cause the formation of arterial blocks due to gas-bubble embolisms.
LbarP
nRTV 13 01
30010314810521250 23
..
...
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The activity of a Pure Solid or liquidThe activity of a Pure Solid or liquid
• The standard state of a pure solid or liquid is the pure substance at pressure p o.
• At another pressure p’ , then
• Assumed that the solid or liquid has a nearly constant volume.• Now• The activity of a pure liquid or solid at pressure p is given by
oim
oi
p
p imo
imimi ppVpVGpGp o ''' *,
' *,,
*, d
oimi ppVRT *,aln
RTppV o
imi
*,expa
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ExampleExample
• Find the activity of pure liquid water at a pressure of 2,000 bar and a temperature of 298.15 K
Solution:
a = e 1.456 =4.287
45611529831458
120001010805111
125135
...
.
aln*
,
KmolKJbarbarmNmolmx
RTppV o
imi
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ExerciseExercise
• Find the pressure such that the activity of liquid water is equal to 1.0100 at 298.15 K.
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9.13 Chemical Equilibrium in Solutions9.13 Chemical Equilibrium in Solutions
• The concept of activity can be used to express the thermodynamic equilibrium constant in terms of activities for real solutions.
• Using Henry’s law standard state for each solute with Gibbs energy and the chemical potential,
• The equilibrium constant can be derived as
KRTaRTG jveqi
jreaction lnln
j
jj
veqiv
i
eqi
v
i
eqi c
caK
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Example 9.13Example 9.13
a. Write the equilibrium constant for the reactionN2(aq, m) ⇄ N2(g, P)
in terms of activities at 25°C, where m is the molarity of N2(aq).
b. By making suitable approximations, convert the equilibrium constant of part (a) into one in terms of pressure and molarity only.
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From From AlbertyAlberty, 4e. Fig A.14, 4e. Fig A.14
Fig A.14 Partial pressure of ether‐acetone solution at 30 ℃.
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From From AlbertyAlberty 4e, Table 4e, Table 6.56.5
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• Activity coefficient γi apply to nonideal solution with
• If there are positive derivations from Raoult’s law, γi is greater than unity; and if there are negative derivations from Raoult’s law, γi is less than unity.
• As activity ai = Pi / Pi *,• To calculate the activity coefficient of i from experimental data:
Chap 9Chap 9
i i i i xγlnRTlμlμ i i i xγa here
1 1 ii x as γ
*PP xγai
ii i i
*PxPy
*PxP γ
ii
i
ii
ii
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Example A.10 Calculate activity coefficients in ether-acetone solutionCalculate the activity coefficients for ether(1) and acetone(2) in ether-acetone solution at 30℃. The experimental data are given in Table A.5 and are plotted in Figure A.14.
Ans: At 0.5 mole fraction acetone, the activity coefficients of the two components are given by
The activity coefficients of both components, calculated in this way at other mole fractions, are summarized in Table A.5.
Note: as the mole fraction of either component approaches unity, its activity coefficient approaches unity, since the vapor pressure asymptotically approaches that given by Raoult’s law.
kPa 86.1 0.5kPa 52.1
*Px
Pγ1 1
1 1
kPa 37.7 0.5kPa 22.4
*Px
P γ2 2
2 2
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Chap 9Chap 9
0.572kPa 78.40.5
kPa 22.4y
i i
ii xK
P'γ
*Px' Kγ
*PPa
i
iii
i
i i
• For 0.5 mole fraction of acetone, the activity coefficient of acetone based on the derivation from Henry’s law is
• The two types of activity coefficients are related by:
Hi
HHi
H
k*Pγγ
*Pk γγ ii
i i
ii or
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• Since
i i
ii ii i xK
Pyx/KPγ H
0 as 1 ii xγ H
2
2i x
P'K :constant senry'Apparent H
0 tongextraploti and versus ofplot A 2 2 xxxP'K
2
2i
Chap 9Chap 9
i ii i xKγ P H
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Fig A.15Fig A.15
• Evaluation of Henry’s law constant K2 as K2’ at x2 = 0 for acetone in ether‐acetone solutions at 30 ℃
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Example A.9 Proof that if Henry’s law holds for the solute, Raoult’s law holds for the solvent.Show that if Henry’s law holds for the solute (component 2), Raoult’s law holds for the solvent (component 1).
Ans: The Gibbs-Duhem equation provides a relationship between the differentials of the chemical potentials of components 1 and 2 at constant temperature and pressure. If
Since x1 + x2 = 1, dx2 = – dx1 then
222 aln RT μμ *P
Kxln RT μ 2
22
22
2 x xRTμ dd
21
21
21 x
xRTμ
xxμ d-d-d
11
11 xln RT
xxRTμ ddd
*PPlnRT*μxlnRT*μμ
.*μμ ,x If constant xln RTμ
1
11111
111
11
Thereforethen1
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