2 –component solutions general: a (solvent) and b (solute) arbitrary if the solutions are...
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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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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