Upload
camarada-rojo
View
28
Download
0
Tags:
Embed Size (px)
DESCRIPTION
simple proof of gibbs phase rule thermodynamics
Citation preview
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