2 –component solutionsGeneral: A (solvent) and B (solute) arbitrary if the solutions are completely miscible properties that define solution: P, T, cA, cB, (V)

Examples: benzene & toluene

ethanol & chloroform CH3-CH2-OH CHCl3

CH3

|

Ideal solution?

Add water to a 10 ml graduated cylinder until it reaches the 5 ml mark.

Add denatured ethanol to a 2nd 10 ml graduated cylinder until it reaches the 5 ml mark.

Measure the temperature of the water and ethanol.

Add the ethanol solution into the graduated cylinder with the water.

Mix the solution with the temperature probe.Measure the volume and temperature of the resultant solution.

Table Volume T (before) T (after)

LF 9.75 21.2 27.1

CF 9.61 21.4 26.9

RF 9.59 22.9 28.1

LB 9.8 22.0 26.1

CB 9.58 21.7 28.2

RB 9.46 21.5 27.3

Average

Table Volume T (before) T (after) moles water

LF 9.75 21.2 27.1 0.278CF 9.61 21.4 26.9 c(H2O) 0.76RF 9.59 22.9 28.1 moles ethanol =LB 9.8 22 26.1 0.086CB 9.58 21.7 28.2 c(eth) 0.24RB 9.46 21.5 27.3

Average 9.6 21.8 27.3Stdev = 0.1 0.6 0.8DVmix = -0.4 DTmix = 5.5DSmix = 0.20 J (if ideal)DGmix = -59 J (if ideal)

we could calculate actual DG if we knew Cp for water/ethanol mixture

Raoult’s Law Pi = ci • Pi*Applies to ideal solutions

is the same as....ideal solution

= solvent (A)

= solute (B) (or i)

B & A have same size & shape

B--A interactions same as B--B

Vmix = Hmix = Umix = 0

Smix = - nAR lnA – nBR lnB

Gmix = RT (nAlnA + nBlnB)

CH3

|

Raoult’s Law

1. mi,v (P) = mi,v + RT ln(P/P) ideal gas

Pi = ci • Pi*Vapor pressure solution mole fraction pure liquid VP

From (dG/dP)T = V we can know how the chemical pressure of an IG varies with Pi. If the vapor is ideal that expression also applies to gas mixtures.

2. mi,sln = mi* + RT lnci ideal solution

For a liquid mixture the chemical potential of one component is reduced by mixing relative to the pure liquid form of that component. Follows from DGmix equation.

3. i,sln = i,v equilibrium condition equate 1 & 2

When a solution comes to equilibrium with the vapor phase, the chemical potential of each component is the same in the solution and in the vapor.

Here is what we already know from previous chapters …..

Raoult’s Law

2. mi,sln = mi*(Pi) + RT lnci ideal solution

1. mi,v (Pi) = mi,v + RT ln(Pi/P) ideal gas

3. i,sln = i,v equilibrium condition equate 1 & 2

mi *(Pi) + RT lnci = mi,v + RT ln(Pi/P)

mi*(Pi) - mi*(Pi*) + RT lnci = RT ln(Pi/Pi*) (dDG/dP)T = DV (the red part is ~ 0)

mi*(Pi

*) = mi,v*(Pi

*) = mi,v° + RT ln(Pi*/P°) pure liquid at T, note Pi ≠ Pi*

subtract from above

RT lnci = RT ln(Pi/Pi*) RT then ex (inverse ln)

ci = Pi/Pi*

Pi = ci • Pi*

Pi = ci Pi* = ci,v P

from Dalton’s Law of Partial Pressures

P

0 cB 1

PB*

PA*

PB

P

P = B(PB* - PA

*) + PA*

P = PA + PB = cAPA* + cBPB*

PA

PA = cAPA*

PB = cBPB*

P

0 cB 1

Bubble point line: P = B(PB

* - PA*) + PA

*

liquid

f = c – p + 2

Vapor begins to form

At some lower P all of the liquid will be converted into vapor.This is called the dew point.

PB*

PA*

all liquidP > ‘bubble point’

1st vaporat bubble point

2 phases dew point < PP < bubble point

Last drop at dew point

Table Volume T (before) T (after) moles water

LF 9.75 21.2 27.1 0.278CF 9.61 21.4 26.9 c(H2O) 0.76RF 9.59 22.9 28.1 moles ethanol =LB 9.8 22 26.1 0.086CB 9.58 21.7 28.2 c(eth) 0.24RB 9.46 21.5 27.3

Average 9.6 21.8 27.3Stdev = 0.1 0.6 0.8DVmix = -0.4 DTmix = 5.5DSmix = 0.20 J (if ideal)DGmix = -59 J (if ideal)

we could calculate actual DG if we knew Cp for water/ethanol mixture

Raoult’s Law Pi = ci • Pi*Applies to ideal solutions

P

0 cB 1

PB*

PA*

Bubble point line: P = B(PB

* - PA*) + PA

*

liquid

P = PB*PA

* B,v(PA

* - PB*) + PB

*

Dew Point line vapor

both

f = c – p + 2

The original vapor is enriched in the more volatile component

The last drop of liquid is enriched in the less volatile component

At a P where f = 2 (both phases present), a phase diagram allows you to determine the composition of each phase.

tie line

20

30

40

50

60

70

80

0.0 0.2 0.4 0.6 0.8 1.0

P

cben

P at which vapor forms bubble point

P = ben(Pben* - Ptol

*) + Ptol*

74.7 torr

22.3 torr

Benzene/Toluene (ideal solution) (T ~ 18ºC)

20

30

40

50

60

70

80

0.0 0.2 0.4 0.6 0.8 1.0

P

cben

Benzene/Toluene (ideal solution) (T ~ 18ºC)

Composition of initial vapor

P = Pben*Ptol

* ben,v(Ptol

* - Pben*) + Pben

*

74.7 torr

22.3 torr

20

30

40

50

60

70

80

0.0 0.2 0.4 0.6 0.8 1.0

P

cben

Benzene/Toluene (ideal solution)

3 theoretical plates

Pressure Distillation 74.7 torr

22.3 torr

20

30

40

50

60

70

80

0.0 0.2 0.4 0.6 0.8 1.0

P

cben

Benzene/Toluene 18°C (ideal solution)

Lever Rule nlll = nvlv

vapor

liquid

74.7 torr

22.3 torr

0.00

0.50

1.00

1.50

2.00

2.50

3.00

3.50

0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00

Pb*

Pt*

cb

real ideal

Bubble point line

Dew point line

Benzene-Toluene 120ºC

Pi = ai • Pi* real solutions – ai = activity

Pi = ci • Pi* applies to ideal solutions only

ai = gi • ci gi = activity coefficient

2.98 atm

1.34 atmActivity, ai

The mole fraction a component would have if it behaved ideally.

0.00

0.50

1.00

1.50

2.00

2.50

3.00

3.50

0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00

Pb*

Pt*

cb

real ideal

Bubble point line

Dew point line

Benzene-Toluene 120ºC

Pi = ai • Pi* real solutions – ai = activity

Pi = ci • Pi* applies to ideal solutions only

ai = gi • ci gi = activity coefficient

2.98 atm

1.34 atm

0.00 0.20 0.40 0.60 0.80 1.000

50100150200250300350400450500

P (torr) Peth Pchl

cchl P (torr) Peth Pchl Peth (id) Pchl (id)

0.00 172.8 172.8 0.0 172.8 0.0

0.20 298.2 138.4 159.8 138.2 86.7

0.40 391.0 111.9 279.1 103.7 173.4

0.60 435.2 92.5 342.7 69.1 260.1

0.80 454.5 70.5 384.0 34.6 346.8

1.00 433.5 0.0 433.5 0.0 433.5

Pi = ai • Pi*

ai = gi • ci

Ethanol – chloroform mixture

P (torr)

cchl

0.00 0.20 0.40 0.60 0.80 1.000

50100150200250300350400450500

P (torr) Peth Pchl

Pi = ai • Pi*

ai = gi • ci

Ethanol – chloroform mixture

P (torr)

cchl

What is the activity of ethanol when cchl = 0.8? a) 0.2 b) 0.58 c) 0.42 c) 0.90

cchl P (torr) Peth Pchl Peth (id) Pchl (id)

0.00 172.8 172.8 0.0 172.8 0.0

0.20 298.2 138.4 159.8 138.2 86.7

0.40 391.0 111.9 279.1 103.7 173.4

0.60 435.2 92.5 342.7 69.1 260.1

0.80 454.5 70.5 384.0 34.6 346.8

1.00 433.5 0.0 433.5 0.0 433.5

Pi = ai • Pi*

ai = gi • ci

Calculate aeth and geth when cchl = 0.8 is?

200

220

240

260

280

300

320

340

360

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

acetone - chloroform

ca

P

ideal

real

Azeotrope = Constant boiling mixture liquid and vapor phases have same concentration

Vapor becomes enriched in Chl

Vapor becomes enriched in acetone

330

335

340

345

350

355

360

365

370

375

380

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Benzene - Toluene: theoretical

ben

T

liquid

vapor

A,l = {P - PB*(T)}/{PA

*(T) - PB*(T)}

A,v = {PA*(T)/P} A

-100

-50

0

50

100

150

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

HCl - water azeotrope

T

HCl

7.12 & 7.16A solution contains a 1:1 molar ratio of hexane and cyclohexane.Phex* = 151.4 torr and Pcyc* = 97.6 torr.What is P?What is cv,hex and cv,cyc ?

7.14A solution of ethanol and methanol has P = 350.0 torr at 50°C. Pmeth* = 413.5 torr and Peth* = 221.6 torr. What is the composition of the solution? Assume ideal solution

Set up expression for Pmeth and Peth using Raoult’s law

Dalton’s Law of partial pressures: P = Pmeth + Peth — This was used to derive bubble point line

Reduce one variable by letting ceth = 1- cmeth. Solve for cmeth

Find Phex and Pcyc using Raoult’s law

Use Dalton’s Law to find P

Since P n, use P ratios to find cv,hex and cv,cyc

0.00 0.20 0.40 0.60 0.80 1.000

50100150200250300350400450500

Pchl

Pi = ai • Pi* ai = gi • ciEthanol – chloroform mixture

P (torr)

cchl

Use the graph to estimate achl and gchl when cchl = 0.8

achl ~ 0.89 and gchl = 1.11

Ideally dilute solutionideal solution

= solvent (A)

= solute (B) (or i)

B & A have same size & shape

B--A interactions same as B--B

B and A are very different

Standard state ≠ pure B but rather ideally dilute solution extrapolated to fictitious ‘pure’ B state where each B behaves as if it were surrounded by A molecules.

Raoult’s Law: Pi = ci • Pi*

Real soln: Pi = ai • Pi*

Henry’s Law: Pi = ci • Ki

Applied to extremely non-ideal solutions particularly when solute is not miscible in solvent.Particularly useful for gases dissolved in water.

Raoult’s law is still used (and very accurate) for the solvent.

0.00 0.20 0.40 0.60 0.80 1.000

100

200

300

400

500

600

700

800

900

1000

P (torr) Peth Pchl

Ethanol – chloroform mixture

cchl P (torr) Peth Pchl Peth (id) Pchl (id)0.00 172.8 172.8 0.0 172.8 0.00.20 298.2 138.4 159.8 138.2 86.70.40 391.0 111.9 279.1 103.7 173.40.60 435.2 92.5 342.7 69.1 260.10.80 454.5 70.5 384.0 34.6 346.81.00 433.5 0.0 433.5 0.0 433.5

Raoult’s Law: Pi = ci • Pi*

Real soln: Pi = ai • Pi*

Henry’s Law: Pi = ci • Ki

cchl

P (torr)

Kchl ~ 850 torr

Keth ~ 400 torr

What is Peth when ceth = 0.1? Peth ~ 0.1 • 400 ~ 40 torr

P*chl = 433.5 torr

P*eth =

172.8 torr

Kchl = Pchl/ci = 159.8/0.2 ~ 800

What is the concentration of oxygen dissolved in a lake at 25ºC? KO2 = 4.34 x 109 Pa

What will happen to KO2 and the concentration of oxygen dissolved if global warming causes an increase in T to 30ºC?Discuss this with a partner and then explain your answer.

0.0 0.2 0.4 0.6 0.8 1.00.0

10.0

20.0

30.0

40.0

50.0

60.0

70.0

80.0

90.0

100.0

bubble dew

d(ln P)/d(1/T) = - DHm/R & P(atm) = exp{-DHm/R •(1/293 – 1/Tnbp)}

DHvap Tnbp

Water 40,660 373.15Methanol 39,400 337.8

P*meth = 89 torr

P*H2O = 21.1 torr

cmeth bubble dew0.0 21.1 21.10.1 27.9 22.80.2 34.7 24.90.3 41.5 27.40.4 48.3 30.40.5 55.1 34.10.6 61.8 38.90.7 68.6 45.30.8 75.4 54.10.9 82.2 67.31.0 89.0 89.0

P = Pmeth*Pw

* meth,v(Pw

* - Pmeth*) + Pmeth

*P = meth(Pmeth

* - Pw*) + Pw

*

0.0 0.2 0.4 0.6 0.8 1.00.0

10.0

20.0

30.0

40.0

50.0

60.0

70.0

80.0

90.0

100.0

bubble dew

d(ln P)/d(1/T) = - DHm/R & P(atm) = exp{-DHm/R •(1/293 – 1/Tnbp)}

P*meth = 89 torr

P*H2O = 21.1 torr

1st vapor: cmeth ~ 0.8 & cH2O ~ 0.2 last drop: cmeth ~ 0.19 & cH2O ~ 0.81 (2 plates)

F = c – p + 2 = 2 – 2 + 2 = 2 ll = .12, lv = .21 nl = 0.64, nv = 0.36

Vapor: cmeth ~ 0.61 & cH2O ~ 0.39 Liquid: cmeth ~ 0.28 & cH2O ~ 0.72

nmeth ~ 0.22 & nH2O ~ 0.14 nmeth ~ 0..18 & nH2O ~ 0.46

Continuing to assume this is an ideal solution, calculate DHmix, DSmix, and DGmix for the initial liquid when cmeth = 0.4, T = 298K and ntotal = 1 mole. DHmix = 0 DSmix = -nH2O • ln (cH2O) - nmeth • ln (cmeth) = -0.6 • ln 0.6 - 0.4 • ln 0.4 = 0.67 J

DGmix = DHmix – TDSmix = -200. J

If the dissolution process is exothermic, comment on the actual value of DGmix relative to the ideal value. DGmix, real < -200 J since the reduction in DHmix makes the process more favorable.

T

cB

p = 1 miscible

p = 2

Tc

partially miscible liquids

cB,2 Asat in BcB,1 Bsat in A

T

cB

B & C solution

solid B & C

solution + Cssoln. + Bs

2c solid-liquid phase diagram

nfp B

nfp C

Colligative Properties

solute added to solvent changes A such that A,soln < A*

This affects the equilibrium between phases so that component A will ‘escape’

from the phase with the higher A.

Vapor Pressure Lowering

Boiling Point Elevation

Freezing Point Depression

Osmotic Pressure

Vapor Pressure Lowering Derive ..... P = -BPA*

assume nonvolatile solute in IDS

Define P as P - PA*

Raoult’s Law: P = PA = APA*

P = PA = APA* sub into above ....

P = P - PA*

P = APA* - PA

* factor .....

P = (A – 1)PA* = {(1 - B) – 1} PA

*

P = -B PA*

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AA AA

AAAAAAAAAAAAAAA

AA AA

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

A

A

Boiling Point Elevation

Tb = Tb - Tb*

A,v(Tb,P) = A(sln) (Tb ,P) = A* + RT ln(A)

Concept:vapor pressure lowering means that a solution will exert less pressure than the pure solvent, therefore it requiresa higher T before the vapor P = atm P (i.e. boiling point).

Tb = kb mB where kb = MARTb*2/(Hvap,A)

7.49

Freezing Point Depression Tf = -kf mB

Tf = Tf - Tf*

A(s)*(Tf ,P) = A(sln) (Tf ,P)

A(sln) = A* + RT ln(A)

kf = MARTf*2/(Hfus,A)

A

Al

As

Tf*

Asln

Tf

PR = Patm + PL = Patm

P = nB RT/V = cBRT

Osmotic Pressure P = cBRT

,A L > ,A R not at equilibrium

mA,sln = mA* + RT ln cA

mA,L = mA,R or mA* (P) = mA,sln (P + P)

7.55

7.55 Osmotic Pressure P = cBRT

Freezing Point Depression Tf = -kf mB

Boiling Point Elevation Tb = -kb mB

Assume 1 L of solution so cB = nB

55 Osmotic P MW T mb P (Pa) P (atm) g solute DTb DTf

185000 310 1.16E-05 30 0.000296 2.15 5.94E-06 -2.16E-05

Assume mB ~ cB = nB = (30/101325)/(0.08206 • 310) =

For Wednesday 7.49