physical chemistry i for biochemists chem340 lecture 32 (4...
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Physical Chemistry I for BiochemistsChem340
Lecture 32 (4/4/11)
Yoshitaka IshiiYoshitaka IshiiCh8.8-8.12
If you have a note 33, skip printing p2-3.
8.5 The Gibbs-Duhem Equation• In Ch 6, we learned
dG = -SdT + VdP + idni
For a binary system at constant T and PFor a binary system at constant T and P .
dG = 1dn1 + 2dn2 (1)On the other hand,
G = 1n1 + 2n2,
dG = d1n1 + d2n2+ 1dn1 + 2dn2 (2)(2) – (1) yields
0 = (d1)n1 + (d2)n2
d2 = -(x1/x2)d(1* + RTlnx1) = -(x1/x2)RT(dx1/x1)
d2 = -(n1/n2)d1 = -(x1/x2)d1 Changes in 1 and 2
are NOT independent
2
Relation of molar fractions in gas phase (y) and solution phase (x)
• Ptotal = P1 + P2 = x1P1* + (1-x1)P2
*
=P * + x (P *-P *)
[Q1]
y1 = P1/Ptotal = x1P1*/{P2
* +x1(P1*-P2*)}
x1 = y1P2*/{P1
* +y1(P2*-P1*)}
=P2 + x1(P1 -P2 )[Q1]
[Q2]
[Q3]
• y1 = (P1*Ptotal – P1
* P2* )/{Ptotal(P1* - P2*)}
• Ptotal = P2* + x1(P1*-P2*)
= P1*P2
*/{P1* +y1(P2*-P1*)}
[Q4A] [Q4B]
• P8.4) A and B form an ideal solution. At a total pressure of 0.900 bar, yA = 0.450 and xA
= 0.650. Using this information, calculate the vapor pressure of pure A and of pure Bvapor pressure of pure A and of pure B.
• yB = 1 – yA & xB = 1 -xA
• So there are two unknown. PA* and PB*
We need two equations:• We need two equations:(1) Ptotal = xAPA* + xBPA*
(2) yA = xAPA* /Ptotal[Q2]
[Q1]
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• P8.5) A and B form an ideal solution at 298 K, with xA = 0.600, and
• a. Calculate the partial pressures of A and B in the gas phase.
PA* 105 Torr, PB
* 63.5 Torr.
•PA = xAPA*[Q1] •PB = (1-xA)PB
*[Q1]
• b. A portion of the gas phase is removed and condensed in a separate container. Calculate the partial pressures of A and B in equilibrium with this liquid sample at 298 K.
PA xAPA[Q1] PB = (1 xA)PB[Q1]
• Assume that xA (new solution) = yA (old gas) and xB(new solution) = yB(old gas).
PA(new) = xA(new)PA* = yA(old)PA*
yA(old) = x1(old)P1*/{P2
* +x1(old)(P1*-P2*)}[Q2]
How pressure Ptotal depends on molar fractions
in gas phase (y) and solution phase (x)?
ybenzene
•Ptotal = P1*P2
*/{P1* +y1(P2*-P1*)}
•Ptotal =P2* + x1(P1*-P2*)
Xbenzene[Q1 ] Ybenzene[Q2 ]
Data fromBenzene + Toluene
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P-Z (average composition) diagramZ1 = (n1
liquid+ n1gas)/(n1
liquid+ n1gas + n2
liquid+ n2gas)
P-XP
ress
ure
P-Y
Q2. How much is Xbenezen and Ybenzene at the point b?
Q. How is the phase changed along the line from a to d?
Q3. Is nliquid > ngas correct at the point b?
lb = ZB – XB
= nBtotal/ntotal - nB
liq/nB+Tliq
b = YB – ZB
= nBgas/nB+T
gas -nBtotal/ntotal
LiqGas
Tie Line
LiqGas
XB YBZB
(ZB-XB) / (YB-ZB) = ntotalGas / ntotal
Liq
Lever RuleQ. What is the molar ratio between liquid and vapor phases (ngas/nliq) at point b (B & T all mixed) ?
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Physical Chemistry Fundamentals: Figure 5.33Q. How is the phase changed along the line from a to e?
X1b bXAb
XAc
cXAd
d
pb
pc
pd
e
yAb
yAc
© 2009 W.H. Freeman
Q2. How much is ngas/nliquid at the point c?
Physical Chemistry Fundamentals: Figure 5.34
© 2009 W.H. Freeman
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8.7 Freezing Point Depression& Boiling Point Elevation
solvent(T) = solvent*(T0) + Sliq (T-T0) + RTln(xsolvent)
Q1. How much is RTln(xsolvent) for pure liquid?
0
Q2. For a mixture, what is the signf RTl ( )?of RTln(xsolvent)?
Negative
At a freezing point,
• liq(T) = solid*(T)
solvent*(T) + RTln(xsolvent) = solid*(T)
Freezing Point Depression (Derivation)
ln(xsolv) = - {solv*(T) - solid*(T)}/RT
= -(Gfusion, m)/RT Let’s find dT vs d(lnx)
dTH
dTTG
xd mfusionmfusionsolv 2
1 )()/()(ln ,,
Gibbs-Helmholtz eq.
RTTRsolv 2)(
purefusionfusion
mfusionTf
pureTf
mfusionsolv
x
TTR
HdT
TR
Hxd
solv
,
,
,
, )()()(ln
1112
1
mfusion
solv
purefusionfusion H
xR
TT ,,
ln
11
7
• Using
mfusion
solv
pureff H
xR
TT ,,
ln
11
2
11
purefpureffpureff T
T
TT
T
TT,,,
~
Q. Does this depend
and lnxsolvent = ln(1-xsolute) ~ -xsolute
pp
mfusion
solutepureff H
xTRT
,
,
Mnnn
Molality of Solute(mol/kg of solvent)
depend on solute?
solventsolutesolvent
solventsolute
solvent
solute
solutesolvent
solutesolute Mm
w
Mn
n
n
nn
nx
~
solutefsolutemfusion
solventpureff mKm
H
MTRT
,
,
Boiling Point Elevation
At a freezing point,
• (T) = *(T)• liq(T) = gas (T)
solvent*(T) + RTln(xsolvent) = gas*(T)
ln(xsolv) = {gas*(T) - solvent*(T)}/RT
= (Gvap, m)/RTIt does NOTdepend
solutebsolutemvap
solventpurebb mKm
H
MTRT
,
,
solutefsolutemfusion
solventpureff mKm
H
MTRT
,
,
pon solute!
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8.8 The Osmotic PressureWater and Sugar Water are separated by a membrane that permits only water go through.
H2O + Sugar
H2O
go through.
Q. Which way is H2O likely to move to?Q2. How can we make the two sections in equilibrium?
d(P, T) = VmdP – SmdT
solvent (P+P, T)* + RTln(xsolvent) = solvent*(P,T)
solvent (P, T)* + VmP + RTln(xsolvent) = solvent*(P, T)
Vm = - RTln(xsolvent) = - (RT/Vm)ln(xsolvent)
Elucidation of the Osmotic Pressure =P
ln(xsolvent) =ln(1 - xsolute) ~ -xsolute ~ -nsolute/nsolvent
= (RT/Vm)(nsolute/nsolvent)
= n RT/(V n )= nsoluteRT/(Vmnsolvent)
= nsoluteRT/V
or
= csoluteRT, where csolute = nsolute/V (molarity).
(Vant’ Hoff Equation)
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Osmotic Pressures-Molecular weight determination
• =nsoluteRT/V • = c RT• = csoluteRT
• If we use Csolute = w/V = nsoluteRT/V
= (wsolute/Msolute)(RT/V) {( /V)/M }RT= {(wsolute/V)/Msolute}RT
= {Csolute/Msolute}RT Msolute = RTCsolute/
Pturgor = cell - medium
At isotonicccell RT = cmedium RT
Q. A medium is a mixture. How do you define Cmedium?Q2. What is Ccell?
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Osmotic Pressure for a poly-disperse or mixture system
• Total = i = ci RT = (Ci /Mi)RT,
where ci and Ci are molar and mass concentration of i, respectively.
So cmedium or ccell ci
Define as =(C /M)RT and C = CDefine as Total =(CT/M)RT and CT = Ci
M = (Ci)/(Ci/Mi) = (niMi/V)/(ni/V)
= (niMi)/(ni)
= (niMi)/(nTotal)
• P8.36) Assume sucrose and water form an ideal solution. What is the equilibrium vapor pressure of a solution of 2.0 grams of sucrose
1(molecular weight 342g mol–1) at T = 293 K if the vapor pressure of pure water at 293 K is 17.54 Torr. Assume 100.0 g of water. (Note density 1.00 mL/g.)
• Use P l /P l * = x lUse Psolvent/Psolvent xsolvent
• What is the osmotic pressure of the sucrose solution versus pure water? Assume 100.0 g of water. Just use = nRT/V
11
r
Pre
ssur
e (T
orr)
TB
P @
750
Tor
Xchloroform Xchloroform
8.9 Real Solutions Exhibit Deviation from Raoult’s Law
Q.Which is Raoult’s Law? (a) Pi = xi Pi*(b) Pi P(b) Pi = xi Ptotal
(c) Pi = niRT/V
Q2. Which is for a realsolution? Dotted or solidline?
Q3. How does Tb-Xcs2 curvelook like for the solid curve?
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8.10 The Ideal Dilute Solution• i
solution = i* + RTln(Pi/Pi*),
where i* = i
*(vapor) = i*(liquid) i
0(pure liquid, T)
F id l l ti P P * F dil t• For a non-ideal solution Pi xiPi*. For a dilute binary solution, we define activity a as
asolvent = Psolvent/P*solvent.
Naturally, for an ideal solution, asolvent = xsolvent
In general, ai Pi/P*i.
solvent
For a general case, a and x are related by activity constant
solvent = asolvent/xsolvent
isolution = i
* + RTln(ai ).
So use ai for a non-ideal solution in place of xi!
Henry’s law (useful when xi ~ 0)
Pacetone = xacetonekHacetone as xacetone 0 (or xCS2 1)
I l P k i 0In general, Pi = xikHi as xi 0
To find Pi when xi ~ 0 look up the Henry’s law constant for SOLUTEAn ideal dilute solution in this text means a system where Henry’s law is valid for a solute and Raoult’s law is valid for solvent
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Ex. Gas in solution xCO2 = PCO2/kH
CO2
In a soda ~3 g of CO2 (0.05 mol) in 450 mL (H2O ~ 20 mol )xCO2 << 1
8.11 Activities Are Defined with Respect to Standard States
• In 8.10 a and were defined for a dilute solution by relationsy
ai Pi/P*i. Activity a
i = ai/xi
In a standard state for dilute solution (xsolvent 1) (Raoult’s law standard state)a xsolvent ~ 1
solventsolution = *solvent + RTln(asolvent)
In Raoult’s law standard state, the standardchemical potential for solvent (xsolvent 1)is
solventsolution *solvent
ai = xi
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Activity in Henry’s law Standard State Standard µ for a dilute solute
isolution = i
* + RTln(kHixi/Pi*).
* RTl (k i/P *) RTl ( )
Pi = xikHi
= i* + RTln(kH
i/Pi*) + RTln(xi)
= i*H + RTln(xi)
The above equation is correct as xi 0. (In general, use ai
in place of xi). The standard-state (xi ~1; i.e. hypothetical pure solute) chemical potential for a solute i is given by
*H * + RTln(k i/P *)
In Henry’s law standard state, (xi ~ 1)ai = Pi/ki
H (~ xi) & i = ai/xi (~1)
solute iH Solute i
+ RTln(kHi/Pi ).
solute i = Solute i*H + RTln(ai).
In general, Pi = aikHi
From Atkins Physical Chemistry 9th Ed
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8.12 Henry’s Law & Solubility of Gas in a Solvent
• N2(aq, cN2) ↔ N2(gas, PN2)
In this case (xN2~0), we use µ following Henry’s law
• µN2solution= µN2
*H(vapor) + RTln(asolute N2)
The mole fraction in solution is
xN2 aq= nN2 aq/(nN2 aq + nH2O) ~ nN2 aq/nH2O.
The amount of dissolved gas is given by
nN2 = nH2OxN2 ac ~ nH2O PN2/kN2H.
8.13 Chemical Equilibrium in Solution• The concept of activity can be used to express the
thermodynamics equilibrium constant using activities. When a reaction in solution
A + 2B 3C + D is in equilibriumA + 2B ↔ 3C + D is in equilibrium• 3µC + µD - µA - 2µB = 0
For a general equilibrated reaction in a solution,• 0 = ∑iµi(solution)
= ∑iµi*H(solution) +RT∑ln(ai
eq)i
G0reaction = - RT∑ln(ai
eq)i = - RT∑lnK mBeqmAeq
mDeqmceq
aa
aaK DC
For a dilute solution, ieq ~ 1. K ~ (ci
eq/c0)i
K = (aieq)i = (i
eqxieq)i
= (ieqci
eq/c0)i
qq aa BA
By the way, for a dilute solution C0 ~ Csolvent (~constant).