If two volatile and miscible liquids are combined to form...

© 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:

98


9.99.9 Real Solutions Exhibit Deviations from Raoult’s LawReal Solutions Exhibit Deviations from Raoult’s Law


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

100


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.


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.


• 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


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

104


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.


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


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


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 Oethanol

ethanol

V n V n V

V n V n V

V n VV

n


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.

109


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 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


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

111



• 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

112



• 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

113


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.

114


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

115


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

116



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

117


• 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:

118

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


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.

119



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.

120



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?

121


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

122


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

..

...

123


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


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


ExerciseExercise

• Find the pressure such that the activity of liquid water is equal to 1.0100 at 298.15 K.


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

127



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.

128

© 2010 Pearson Education South Asia Pte LtdPhysical Chemistry 129


From From AlbertyAlberty, 4e. Fig A.14, 4e. Fig A.14

Fig A.14 Partial pressure of ether‐acetone solution at 30 ℃.

130


From From AlbertyAlberty 4e, Table 4e, Table 6.56.5

131


• 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

132


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

133


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

134


• 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

135


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 ℃

136


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

137

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