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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]

3

• 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

4

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) ?

5

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).