Lect36 42 Gibbs

1

Lecture 36. The Phase Rule

P = number of phasesC = number of components (chemically independentconstituents)F = number of degrees of freedom

xC,P = the mole fraction of component C in phase P

The variables used to describe a system in equilibrium:

1,1312111 ,...,,, −Cxxxx phase 1

2,1322212 ,...,,, −Cxxxx phase 2

PCPPP xxxx ,1321 ,...,,, − phase PT,P

Total number of variables = P(C-1) + 2

Constraints on the system:

m11 =m12 =m13 =…=m1,P P - 1 relationsm21 =m22 =m23 =…=m2,P P - 1 relations

mC,1 =mC,2 =mC,3 =…=mC,P P - 1 relations

2

Total number of constraints = C(P - 1)

Degrees of freedom = variables - constraints

F = P(C - 1) + 2 - C(P - 1)

F = C - P + 2

Single Component Systems: F = 3 - P

In single phase regions, F = 2. Both T and P may vary.

At the equilibrium between two phases, F = 1. ChangingT requires a change in P, and vice versa.

At the triple point, F = 0. Tt and Pt are unique.

3

Four phases cannot be in equilibrium (for a singlecomponent.)

Two Component Systems: F = 4 - P

The possible phases are the vapor, two immiscible (orpartially miscible) liquid phases, and two solid phases.(Of course, they don’t have to all exist. The liquidsmight turn out to be miscible for all compositions.)

4

Liquid-Vapor Equilibrium

Possible degrees of freedom: T, P, mole fraction of A

xA = mole fraction of A in the liquidyA = mole fraction of A in the vaporzA = overall mole fraction of A (for the entire system)

We can plot either T vs zA holding P constant, or P vs zA

holding T constant.

Let A be the more volatile substance:

PA* > PB

* and Tb,A < Tb,B

Pressure-composition diagrams

Fix the temperature at some value, T.

Assume Raoult’s Law:

P = PA*xA + PB

*xB

P = PA*xA + PB

*(1 - xA) = PB*+ (PA

*- PB

*)xA

5

Composition of the vapor

( ) AABAB

AAAA x

xPPP

Px

P

Py >

−+==

***

*

( ) BABAB

BBBB x

xPPP

Px

P

Py <

−+==

***

*

Point a: One phase, F = 3 (T, P, xA)

Point b: Liquid starts to vaporize, F = 2 (T,P; xA not free.)xA = zb, yA = yb” Vapor is rich in A.

Point c: Liquid has lost so much A that its composition isxA = xc’. The vapor is now poorer in A, yA = yc”

Ratio of moles in the two phases is given by the leverrule:

cc

cc

n

n

vap

liq

′′′

=

6

Point d: Liquid is almost all gone, xA = xd’, yA = yd = zA

For points below d, only the vapor is present (F=3).

Numerical example: benzene-toluene at 20 C

Exercise 8.4b. Let B = toluene and A = benzene.

Given: PA* = 74 Torr, PB

* = 22 Torr, zA = 0.5

Q. At what pressure does the mixture begin to boil?A. At point b, P = 0.5x22 + 0.5x74 = 48 Torr.

Q. What is the composition of the vapor at this point?A. PB = 0.5x22 = 11, PA = 0.5x74 = 37, yA = 37/48 = 0.77

Q. What is the composition and vapor pressure of theliquid when the last few drops of liquid boil?A. Point d.

)1(2274

740.5zy AA

AA

A

xx

x

−+===

xA = 0.229, xB = 0.771

P = 0.229x74 + 0.771x22= 33.9 Torr

7

Lecture 37. Construction of the Temperature-Composition Diagram

Fix the total pressure at some chosen value, P. Boilingdoes not occur at a unique temperature, but rather over arange of temperatures. (Contrast with a pure substance.)

The vapor pressure curve is determined by the liquidcomposition:

P = PA(T) + PB(T)

PA(T) is the vapor pressure of pure A at temperature T.

8

Assume Raoult’s Law:

)()( ** TPxTPxP BBAA +=)()1()( ** TPxTPx BAAA −+=))()(()( *** TPTPxTP BAAB −+=

The vapor pressure pure liquid A at any temperature T isgiven by the Clapeyron-Clausius equation:

−

∆−=

0

,

0*

* 11

)(

)(ln

TTR

H

TP

TP Avap

A

A

)( 0* TPA = 1 atm, T0 = normal boiling point of A.

The same reasoning applies to substance B.

Once we have )(* TPA and )(* TPB , Raoult’s Law gives usxA as a unique function of T and P. This allows us toconstruct the liquid curve.

9

Composition of the vapor:

P

TPx

P

Py AAA

A

)(*

==

−

∆−

== Ab

vap

TTRT

H

A

A

A eP

TP

x

y ,

11* )(

This result is equivalent to

AliqAAvapA xRTyRT lnln *,

*, +=+ µµ

This last result is the starting point for the boiling pointelevation formula, except that there we assume yA = 1.

10

Recipe for construction of the phase diagram:

1. Calculate Tb,A and Tb,B at the pressure of thediagram.

2. Choose a temperature Tb,A < T < Tb,B

3. Calculate )(* TPA and ).(* TPB

4. )()(

)(**

*

TPTP

TPPx

BA

BA −

−=

5. (T)/P.P*AAA xy =

11

Cooling of a vapor mixture

Point a. Pure vapor, yA = zA

Point b. Vapor begins to condense at xA = zb’, yA = zA = zb

Point c. Comparable amounts of the two phases are

present. Lever rule: cc

cc

n

n

liquid

vapor

′′′

=

Point d. Vapor is almost all gone; xA = zd’ = zA, yA = zd”

Point e. Only liquid is present.

12

Distillation

Point a. Mixture starts to boil, with xA = zA, yA = zb

Points b-c. Vapor is condensed to form a liquid withxA= zb = zc

Point c. The liquid that was collected in the previousstep is boiled to form a vapor with xA=zd

Condensation of the last bit of vapor produces a liquidvery rich in either A (if Tb,A < Tb,B on the left) or B (ifTb,A > Tb,B on the right).

13

Non-ideal solutions

Left: Impossible phase diagram, because at Pmax, wherethe liquid of this composition just starts to boil, there is nocorresponding point on the vapor curve. (There is no tieline.)

The vapor should always lie below the liquid in apressure-composition diagram. At an extremum theymust touch.

Center: Vapor pressure reaches a maximum because ofrepulsion between A and B. This is an azeotrope, wherethe liquid and vapor have the same composition. (Note anerror in the drawing: The vapor curve does not have acusp, but rather is tangent to the liquid curve.)

Right: Vapor pressure reaches a minimum because ofattraction between A and B. This is also an azeotrope.

14

Distillation of non-ideal solutions

Left diagram: low boiling azeotrope. A solution ofcomposition za first boils at Ta. The vapor comes off withcomposition zb. It is condensed and then boils at Tc

The final vapor to come off has the azeotropiccomposition, and the remaining liquid is enriched in B (orA, depending on which side of the azeotrope the processstarted).

Right diagram: high boiling azeotrope. As before, za firstboils at Ta

The final vapor to come off is enriched in B (or A), andthe remaining liquid has the azeotropic composition.

15

Equilibrium between immiscible liquids

Upper CriticalTemperature

Lower CriticalTemperature

Double CriticalTemperature

The two phase region is always described by a tie line. Itis a “no man’s land.”

16

Boiling of immiscible liquids

Melting of immiscible solids

Mixture of non-reactivesolids.

Solids A and B react toform compound AB.

17

Lecture 38. Equilibrium Constants

Consider the following reaction:

H2 + Cl2 Ø 2HCl

Does it occur spontaneously? Let’s calculate DG for onemole of reaction.

0)( 20 =∆ HG f

0)( 20 =∆ ClG f

molkJHClG f /30.95)(0 −=∆

6.1900 −=∆ rG

It seems that the reaction occurs spontaneously.But what if we start with pure HCl. Shouldn’t some of itreact to form H2 and Cl2?

18

To determine when the reaction occurs spontaneously, wemust take the partial pressures into account:

+=∆

oCl

Clf P

PRTClG

2

2ln0)( 2

+−=∆

oHCl

HClf P

PRTHClG ln30.95)(

−

−

+−=∆o

Cl

o

HHClr P

PRT

P

PRT

P

PRTG 22 lnlnln6.190

2

0

At equilibrium, DGr = 0. That is,

=

=∆−=2222

22

2

ln)(

ln6.190ClH

HCl

o

Cl

o

H

oHCl

or PP

PRT

P

P

P

PP

P

RTG

0ln rP GKRT ∆−=

ln KP = 76.93 at 298 K

KP = 2.57 x 1033

19

So what happens if we start with pure HCl?

Suppose that initially HClP = 1 bar and 022

== ClH PP

At equilibrium, zPHCl 21−= and zPP ClH ==22

( ) 332

2

1057.221

xz

z =−

332 1057.2

1x

z≈

z=1.97 x 10-17

This reaction goes nearly to completion. But this won’tbe the case at higher T.

20

Another example:

N2O4 F 2NO2

=∆ 0fG 97.89 and 51.31 kJ/mol

30.4789.9731.5120 =−=∆ xGr

ln KP = -4.730/RT = 1.909

oON

NO

o

ON

o

NO

P PP

P

P

PP

P

K42

2

42

2

2

2

148.0 =

==

What is the composition of 2 bar of this material?

Let x be the mole fraction of NO2.

( )( ) oP PPx

xPK

−=

1

2

Noting that Po = 1 bar,

21

x2P2 = KPP - KPxP

P2 x2 + KPPx - KPP = 0

2

322

2

4

P

PKPKPKx PPP +±−

=

For P = 2 bar, x = 0.238. (Only the + sign is physicallymeaningful.) For P = 0.01 bar, x = 0.940

What is the composition of a very low pressure gas?

22

/41

P

KPPKPKx PPP ++−

=

( )P

PPPP

K

P

P

KPKPPKPKx

21

22//21

2

22

−=−++−≈

(Here we used the series (1+e)1/2 = 1 + e/2 -e2/8 + …)

For P = 0.01 bar bar, 966.0≈x

22

Suppose you somehow arranged to start with pure N2O4.What is the composition of an equilibrium mixture?

N2O4 NO2

Initial P0 0Final P0-z 2z

zP

zKP −

=0

2)2(

For KP = 0.148 and P0 = 1 bar, z = 0.175825.0175.01

42=−=ONP

349.0175.022

== xPNO

Ptot = 1.1744297.01744.1/349.0

2==NOx

23

The book defines the fraction of dissociation, a.

N2O4 NO2

Initial no.moles

N 0

Final no.moles

(1-a)n 2an

Final molefraction α

ααα

α+−=

+−−

1

1

21

1

αα

+1

2

2

2222

1

4

42

2

42

2

αα−

=== P

Px

Px

P

PK

ON

NO

ON

NOP

But P = (1+ a)P0

Substituting this result gives the same equation as before.This method is useful if the total pressure is specified.

How does KP change if P is increased?

How does the composition change if P is increased?

24

More complex stochiometry:

2H2S + CH4 F 4H2 + CS2

=

o

CH

o

SH

o

CS

o

H

P

P

P

P

P

P

P

P

P

K42

22

2

4

Set Po = 1 bar, so that

42

22

4

CHSH

CSHP PP

PPK =

Problem: Given that T = 700 C and P = 762 TorrInitial Final

H2S 11.02 mmol

CH4 5.48H2 0CS2 0 0.711

Find KP and DGro

25

Solution:

2H2S + CH4 F 4H2 + CS2

20

2

2

4

2

4

)(42

22

42

22

P

P

xx

xx

P

P

P

P

P

P

P

P

KCHSH

CSH

o

CH

o

SH

o

CS

o

H

P =

=

Set Po = 1 bar, so that

2PKK xP =

Problem: Given that T = 700 C and P = 762 Torr = 1.012 bar

Initial Final xH2S 11.02 mmol 9.598 0.536CH4 5.48 4.769 0.266H2 0 2.844 0.159CS2 0 0.711 0.040

Find KP and DGro

Kx = 3.34x10-4

KP = 3.43 x10-4

DGr0 = -RT lnKP = 94.5 kJ/mol

26

Lecture 39. Equilibrium Constants II

Derivation of KP for H2 + Cl2 F 2HCl

HClHClClClHHm nnnG µµµ 22222++=

HClHClClClHHm dndndndG µµµ 22222++=

Let x indicate the “extent” of reaction. It is a device forhandling the stochiometry of the reaction.

ξddnH −=2

ξddnCl −=2

ξddnHCl 2=

The chemical potentials are

+∆=

o

HomfH P

PRTHG 2

2ln)( 2,µ

+∆= o

ClomfCl P

PRTClG 2

2ln)( 2,µ

+∆=o

HClomfHCl P

PRTHClG ln)(,µ

27

Putting all the pieces together:

ξdP

PRTHClG

P

PRTClG

P

PRTHGdG

oHClo

mf

o

Clomfo

Homfm

}ln2)(2

ln)(ln)({

,

2,2,22

+∆+

−∆−

−∆−=

+∆=

22

2

lnClH

HClor

m

PP

PRTG

d

dG

ξ

At equilibrium,

0=ξd

dGm

orP GKRT ∆−=ln

=

22

2

ClH

HClP PP

PK

28

Note that KP is independent of pressure (for ideal gases).This does not mean that the equilibrium ratios of molefractions or concentrations are independent of P.

Let’s use 2H2S + CH4 as an example.

PxP CSCS 22= etc.

2

22

44

42

22

=

= ox

oCHoSH

oCSoH

P P

PK

P

Px

P

Px

P

Px

P

Px

K

In general,ν∆

=oxP P

PKK

29

For example, for N2O4F 2N2O

=oxP P

PKK

P

e

P

K

x

xK

RTGP

ON

ONx

omr /2 ,

42

2

∆−

−===

0lim420

=→ ON

Px

0lim2

=∞→ NO

Px

Increasing the pressure reduces the total number of moles,in accord with le Chatlier’s principle.

30

Temperature dependence of KP

RT

GK

omr

P,ln

∆−=

dT

TGd

RdT

Kd omrP

)/(1ln ,∆−=

Gibbs-Helmholtz equation:

2

,1ln

T

H

RdT

Kd omrP

∆=

2

,lnT

dT

R

HKd

omr

P

∆=

Van’t Hoff equation:

−

∆−=

12

,

1

2 11

)(

)(ln

TTR

H

TK

TK omr

P

P

31

Suppose that the reaction is exothermic, so that .0<∆ orH

For T2 > T1,

0)(

)(ln

1

2 <

TK

TK

P

P

In other words, increasing T shifts the reaction to the left.Otherwise, the reaction would “run away,” in accord withle Chatlier.

32

Example of H2 + Cl2 F 2HCl

62.184, −=∆ omrH kJ/mol

At 5000 K (assuming that DH is constant),

86.6298

1

5000

1

31451.8

620,18493.76)5000(ln =

−+=KK P

KP = 952

Now if we start off with 1 bar of pure HCl, theequilibrium mixture contains 0.032 bar of H2 and 0.032bar of Cl2.

33

How do real gases and condensed phases behave?

ioii aRT ln+= µµ

Gases: ai = fi/Po, ii

PPf =

→0lim

Solvent: aA = gA xA, AAx

xaA

=→1

lim

Solute: aB = PB/KB = gB xB, BBx

xaB

=→0

lim

We can also write BBB ba γ′=

Note:

BA

BB nn

nx

+=

AA

BB Mn

nb =

The activity of a pure solid or liquid is 1. Why? Becauseits activity doesn’t change, and no “correction term” isneeded.

34

Example: Liquid-vapor equilibrium:

CCl4(liq) F CCl4(vapor) at 298 K

vapor

ovapor

liq

vapor PPP

a

aK ===

1

/in bar

DGovap,m(298) = 4.46 kJ/mol

80.1298314.8

4460ln , −=−=

∆−=

xRT

GK

omvap

K = 0.165 fl vapor pressure = 124 Torr

Note: From the van’t Hoff equation,

−∆=

∆

bvap

vap

TTH

RT

TG 11)(0

0

This is the same result that we get from the Clapeyron-Clausius equation.

35

Lecture 40. Equilibrium of Elecytrolytes

Solubility Product

For a saturated solution,

AgCl(s) Ø Ag+(aq) + Cl-(aq)

2)( o

ClAgSP b

bbK

−+ −+=γγ

Given that K=1.8x10-10 and −+ =ClAg

bb , and assuming thatg+g- = 1,

SPo

Ag Kb

b=

+

51035.1 −=+ xbAg mol/Kg

36

Debye-H�ckel limiting law:

IAzz ||log 10` −+± −=γ

I = ionic strength of the solution (dimensionless)A = 0.50926 for water at 25 C.

∑=i

oii bbzI )/(

2

1 2

zi = charge number of ion i

Here, z+ = 1, z-

= -1, I = b @ 1.35x10-5

log10 g≤ = -0.00187

g≤

= 0.9957

51035.1 −=+ xbAg mol/Kg

37

Example: Determine the solubility of Ni3(PO4)2, giventhat KSP = 4.74x10-32. Assume first that g

≤= 1.

One mole of electrolyte produces 3 moles of Ni+ andtwo moles of PO4

2+. Let b = the molality of theNi3(PO4)2 that gets dissolved. The activity of thedissolved Ni2+ is 3g+b, and the activity of thedissolved PO4

2- is 3g-b.

523

10823

=

= −+ oooSP b

b

b

b

b

bK γγ

b = 2.13x10-7 mol/kg

Now calculate the ionic strength.z+ = 2, z

-= 3I = 0.5{3b(2)2 + 2b(3)2} = 3.196x10-6

005463.0)3)(2(50926.0log −=−=± Iγ

235−+± = γγγ g

≤= 0.9875

5

5108

= ± oSP b

bK γ

b = 2.16x10-7 mol/kg

38

Ionization of water

Water:2H2O F H3O

+ + OH-

Note: pH meters are calibrated for activity.

WKOH =+][ 3

KW(250C) = 10-13.997@ 10-14

[H3O+] = 10-7 M

pH = -log10[H3O+] = 7

What is Kw at 00C?

We need DG(00C) for acid dissociation. Must correct forDCP.

dTT

H

T

Gd

2

∆−=

∆

2332

2

3 ][]][[)]([

)()( +−+−+

=== OHOHOHOHa

OHaOHaKw

39

( )00 )()( TTCTHTH P −∆+∆=∆

−−∆+

−∆+∆=∆

00

000

0

ln1)()()(T

TTTTC

T

TTHTG

T

TTG P

DG(250C) = 79.868 kj/mol

DH(250C) = 56.563 kj/mol

DCP(250C) = 19.7 j/mol/deg

DG(00C) = 78.127 kj/mol

Kw = 1.198x10-15

pH = 7.47

40

Lecture 42. Acid-Base Equilibria

Strong acid:HA + H2O Ø H3O

+ + A-

[H3O+] = [HA]0 >> 10-7

pH = - log[HA]0

Strong base:[OH-] >> [H3O

+]

[H3O+] = KW/[OH-]

pH = pKW + log[B]

Weak acid:HA + H2O F H3O

+ + A-

0

233

][

][

][

]][[

HA

OH

HA

AOHKa

+−+

≈=

703 10][][ −+ >>= HAKOH a

pH = ½ pKa - ½ logA0

41

Weak base:

A- + H2O Ø H3O+ + OH-

][

][

][

]][[ 2

−

−

−

−

≈=A

OH

A

OHHAKb

KaKb = Kw

awb KKBBKOH /][ 00 ==−

03 /]/[][ BKKOHKOH aww == −+

pH = ½ pKw + ½ pKa - log B0

42

Problem 9.16: Acid rain. Calculate the pH of rainwatercaused by dissolved CO2, given:

atmxPCO4106.3

2

−=KH = 1.25 x 106 Torr

pKa =6.37

The reaction involved is

H2CO3(aq) + H2O(liq) F H2CO3-(aq) + H3O

+(aq)

][][

][]][[

32

23

32

332

COHOH

COHOHCOH

aK++−

≈=

pKa = 2pH + log[H2CO3]

Assume that [H2CO3] = [CO2] in the liquid phase.

015.18/1000

][

][

][ 2

2

2

2

2

22

2

2

CO

OH

CO

n

n

nn

nx

OH

CO

OHCO

COCO ==≈

+=

018015.0/][22 COxCO =

43

We calculate 2COx using Henry’s law.

22 COHCO xKP =

76

4

102.21025.1

760106.32

2

−−

=== xx

xx

K

Px

H

COCO

Mxx

CO 57

2 102.1018015.0

102.2][ −

−

==

44

Exact solution: 4 equations with 4 unknowns

x = [H3O+], y = [OH-], z = [A-], A = [AH]

Kw = xy

Ka = xz/A

x = y + z

A = A0 - z

Solution:y = x - z

Kw = x(x-z) fl z = x - Kw/x

wa

w

w

wa KKxxA

xKx

x

KxA

KxK

+−−

=−−

−=

20

3

0

2

waaaw KKxKxAKxKx +−=− 20

3

pH= ½ pKa - ½ log(1.2x10-5) = 5.645

45

Titration of a Weak Acid with a Strong Base

Chemical equations:

BOH Ø B+ + OH-

OH- + HA Ø H2O + A-

HA + H2O F H3O+ + A-

A- + H2O F HA + OH-

Unknowns: HA, H3O+, A-, B+, OH-

Algebraic equations:

][

]][[ 3

HA

AOHK a

−+

=

][]][[

−

−

=A

OHHAKb

(note: KaKb = Kw)

H3O+ + B+ = A- + OH-

[HA] = [HA]0 + [A-]

46

[B+] = [BOH]0 = S

Initial point:pH = ½ pKa - ½ log A0

Before the equivalence point:

OH- + HA Ø H2O + A-

HA + H2O F H3O+ + A-

[A-] = S

[HA] = A0 - S

(S is the concentration of salt produced byBOH + HAØ H2O + B+A-)

SA

SOH

HA

AOHKa −

== +−+

03

3 ][][

]][[

pH = pKa - log((A0 - S)/S)

47

At the equivalence point:

HA has been all converted, and the relevant chemicalequation is A- + H2O F HA + OH-

S

OH

A

OHHAKb

2][][

]][[ −

−

−

==

2/1)]/([][ awb KKSSKOH ==−

pOH = ½ pKw - ½ pKa - ½ log S = pKw - pH

pH = ½ pKa + ½ pKw + ½ log S

(Exercise: Work out the corresponding equation fortitration of a weak base by a strong acid.)

End point:

[H3O+] = Kw/[OH-] = Kw/(B0-A0)

pH = pKw + log(B0-A0)

48

Lecture 43. Standard States

The choice of standard state is arbitrary. It affects thevalue of K, but not what you actually observe in the lab.

Gas: 1 barLiquid or solid: the pure materialSolute, including H+: 1 MH+ in biochemical reactions: 10-7 M (i.e., pH 7)

Suppose you lived on a planet where the atmosphericpressure is l bar. You my choose to use as your standarda pressure of l bar. How does this affect m and KP?

λµλµµ lnln RTP

PRT o

o

ooo +=

+=′

49

For the conditions of your experiment,

′

+′=

+

+−′=

+−′=

+=

oo

oo

o

o

oo

oo

P

PRT

RTP

PRTRT

P

P

P

PRTRT

P

PRT

ln

lnlnln

lnln

ln

µ

λλ

λµ

λλ

λµ

µµ

For the reaction aA + bB F cC + dD ,

λνλ lnln)( ,,, RTGRTbadcGG omr

omr

omr ∆+∆=−−++∆=′∆

νλ

λλ

λλ ∆−=

=′ Pb

oB

a

oA

d

oD

c

oC

P K

P

P

P

P

P

P

P

P

K

A reaction that is at equilibrium at Po will not be atequilibrium at .oP′\

This idea applies also to ionic equilibria.Normally we choose for the standard state co = 1M.But for biochemical systems, we choose for H+

Mc o 710−=′

50

[ ]pHRT

c

HRTaRTHH o

o ⋅−=

+=+=

+++ 302.2ln0ln)()( µµ

[ ]RTH

c

HRTH

o302.2)(

10ln)(

7−=

=′ +

−

++ µµ

For a reaction A + nH+Ø P

RTGG omr

omr 303.27,, ⋅+∆=′∆ ν