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ACTIVITY MODELS

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Modified Raoult's Law and Excess Gibbs Free Energy

Equilibrium criteria: For vapor phase: For liquid phase, we may use an activity coefficient

(i), giving: Poynting method is used to calculate the pure

component liquid phase fugacities: Combining:

f iV= f i

L

f iV= yi

i P

f iL= xii f i

L

f iL=i

sat Pisat exp V i

LP−Pisat

RT yii P=xiii

sat Pisat expV i

LP−Pisat

RT

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K-ratio ==> Ki=yi/xi

At low pressure: Poynting Correction and ratio of for the component approach unity ==> MODIFIED RAOULT'S LAW

Ki=i

L Pisat

P [isat exp V i

LP−Pisat /RT

i]

isat / i

Ki=i

L Pisat

Patau yi P= xii Pi

sat

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Excess Gibbs free energy: GE=G-Gis

dst diperoleh: Activity Coefficient and excess Gibbs free energy

are coupled. Excess Gibbs energy is zero for an ideal solution Activity Coefficients as derivatives:

GE=RT∑i

xi lni

∂GE

∂ ni T , P , n j≠i

=RT ln i

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Comparison with Equation of State Methods

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Determination of GE from Experimental Data Modified Raoult's Law: Excess Gibbs energy from activity coefficient (for

binary system):

plot GE/RT vs x1

Example (1): system of 2-propanol (1) + water (2) from citation data: T=30 oC, P1

sat = 60.7 mmHg, P2sat = 32.1

mmHg, y1 = 0.6462, when x1 = 0.6369 at P=66.9 mmHg

==> determine activity coefficient and relate to excess Gibbs energy

i=yi P

xi Pisat

GE

RT=x1 ln1x2 ln2

1=y1 P

x1 P1sat=

0.646266.90.636960.7

=1.118 2=y2 P

x2 P2sat=

0.353866.90.363132.1

=2.031

GE

RT=x1 ln1x2 ln2=0.6369 ln 1.1180.3631 ln 2.031=0.328

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plot to GE/RT vs x1

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One-Parameter MARGULES Equation the simplest expression for Gibbs excess energy

function

Parameter A is a constant which is not associated with the other uses of the variable (equation of state parameters, Helmholtz energy, Antoine coefficients).

Example: derive the expression for the activity coefficients for the one-parameter Margules equation

==>

GE

RT=A x1 x2

GE

RT=A n2

n1

n

1RT ∂G E

∂ ni T , P , n j≠i

=An2[ 1n−

n1

n2 ]=An2

n [1−n1

n ]=Ax21− x1

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==> Dengan cara yg sama ==> Example (2): continued from example (1), at 30 oC

show x1=0.1168 and y1=0.5316 at P=60.3 mmHg. What are the pressure and vapor phase compositions predicted by the one-parameter Margules Equation.

at x1=0.6369, we found GE/RT = 0.328. Fitting the Margules equation:

at the new composition:

ln1=Ax22

ln2=Ax12

GE

RT=A x1 x2 A=0.328 /[0.63690.3631]=1.42

x1=0.1168,we find : ln1=Ax22=1.420.88322=1.107,1=3.03

ln2=Ax12=1.42 0.11682=0.0194,2=1.02

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Substituting into Modified Raoult's Law:

y1 P=x11 P1sat=0.11683.0360.7=21.48mmHg

y2 P=x12 P2sat=0.88321.0232.1=28.92mmHg

P= y1 P y2 P=50.4mmHgy1= y1 P /P=21.48/50.4=0.426

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Two-Parameters MARGULES Equation

requires differentiation of nGE/RT with respect to a mole number

the equation is multiplied by n (mole numbers) where x1=n1/(n1+n2), and x2=n2/(n1+n2):

differentiation with respect to n1:

Reconversion of ni to xi:

GE

x1 x2 RT=A21 x1A12 x2 alternatively G E

RT=A21 x1A12 x2x1 x2

nGE

RT= A21n1A12 n2

n1n2

n1n22

∂ nGE /RT∂ n1 T , P , n2

= ln 1

ln1=n2[A21n1A12 n2 1n1n2

2−2n1

n1n23 n1 A21

n1n22 ]

ln1= x2 [A21 x1A12 x21−2x1A21 x1]

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note: x2=1-x1, so:

For the limiting conditions of infinite dilution:

ln1= x22 [ A122 A21−A12x1 ]

ln2=x12 [ A212A12−A21 x2 ]Margules Equation

ln1∞=A12x1=0 andd ln2

∞=A21x2=0

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Example of Margules Equation Following is set of VLE data for the system

methanol(1)/water(2) at 333.15 K:

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calculate i ==> modified Raoult's law

calculate ln i

plot ln i vs xi

A12 and A21 are values of the intercepts at x1=0 and x1=1 of the straight line drawn to represent the GE/x1x2RT:

we get: A12=0.372 and A21=0.198

calculate GE/RT and GE/x1x2RT We have equation: or ==>

i=yi P

xi Pisat

ln1∞=A12x1=0 andd ln2

∞=A21x2=0

G E

RT=0.198 x10.372x2x1 x2

G E

RT=x1 ln1x2 ln2

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16

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Use 1 and 2 to calculate VLE predictions:

Another method: by fitting ln i vs xi for the following Margules equation: or to obtain A12 and A21

i=yi P

xi Pisat P=x11 P1

satx22 P2sat

y1=x11 P1

sat

x11 P1sat x22 P2

sat

ln1= x22 [ A122 A21−A12x1 ] ln2=x1

2 [ A212A12−A21 x2 ]

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van Laar Equation for Activity Coefficient Model

by infinite solutions method:

or rearranging to:

ln1=A12

[1 A12 x1

A21 x2 ]2

van Laar Equation ln2=A21

[1 A21 x1

A12 x2 ]2

ln1∞=A12x1=0 andd ln2

∞=A21x2=0

A12=ln1[1 x2 ln2

x1 ln1 ]2

A21=ln 2[1 x1 ln1

x2 ln2 ]2

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Example of van Laar Equation Consider benzene(1) + ethanol(2) system which exhibits an

azeotrope at 760 mmHg and 68.24 oC containing 44.8 mole% ethanol. Calculate the composition of the vapor in equilibrum with an equimolar liquid solution at 760 mmHg given the Antoine constants:log P1

sat=6.87987-1196.76/(T+219.161)log P2

sat=8.1122-1592.86/(T+226.18) Solution: at T=68.24 oC, P1

sat=519.7 mmHg; P2sat=503.5 mmHg

at azeotrope: x1=y1 ==> 1=P/P1sat; =P/P2

sat

1=760/519.7=1.4624; =760/503.5=1.5094

where x1=0.552; x2=0.448

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calculate A12 and A21:

A12=1.3421; A21=1.8810

now consider x1=x2=0.5 ==> van Laar:

1=1.579; =1.386 Problem ==> bubble temperature? Guess T=60oC ==> P1

sat=391.6 mmHg; P2sat=351.8 mmHg

yi=xi Pisat/P ==>y1=0.407; y2=0.321; yi=0.728 ==> T is too low

at T=68.24oC, P1sat=519.7 mmHg; P2

sat=503.5 mmHg

yi=xi Pisat/P ==>y1=0.540; y2=0.459; yi=0.999 ==> T is ok

(Tazeotrope)

A12=ln1[1 x2 ln2

x1 ln1 ]2

A21=ln 2[1 x1 ln1

x2 ln2 ]2

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Wilson's Equation for Activity Coefficient

Vi, Vj : molar volume of pure liquid i, j, at temperature T

ln1=−ln x1x212x2 12

x1 x212−

21

x121x2

Wilson's Equation

ln2=−ln x121x2 x1 12

x1x212−

21

x121 x2 12=

V 2

V 1exp−A12

RT and. 21=V 1

V 2exp−A21

RT

GE

RT=−x1ln x1x212−x2 lnx2 x121

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NRTL Equation for Activity Coefficient (3 parameters)

, b12, and b21 are parameters specific to a particular pair of

species, and are independent of composition and temperature NRTL: Non Random Two Liquid

ln1= x22[21 G 21

x1 x2 G 212

G1212

x2x1G122 ]

NRTL EquationGE

x1 x2 RT=

G 2121

x1x2G 21

G 1212

x2x1 G12

ln2=x12[12 G 12

x2x1 G12 2

G 2121

x1x2 G212 ]

G12=exp−12 G21=exp−21

12=b12

RT21=

b21

RT

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For the infinite-dilution values of the activity coefficient: ln1

∞=2112 exp−12

ln2∞=1221 exp−21

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UNIQUAC Equation for Activity Coefficient For multicomponent solution:

UNIQUAC equation requires two adjustable parameters characterized from experimental data for each binary system ==>

GE

RT=∑

jx j ln j/ x j−5∑

jq j x jln j/ j−∑

jq j x j ln∑i

i ijln k=ln k

COMB lnkRES

lnkCOMB=ln k /xk 1−k / xk −5qk [ lnk /k 1−k /k ]

ln kRES=qk[1−ln∑i

iik −∑j

jkj

∑iiij ]

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UNIQUAC for Binary Solution

==>

==>

where ==>

Q, R, v ???

ln 1=ln 1

x11−

1

x1−5q1[ ln

1

11−

1

1]

q1[1−ln 1221−1

1221−2

12

1122 ]ln2=ln

2

x21−

2

x2−5q2[ ln

2

21−

2

2]

q2[1−ln 1122−121

1221−

2

1122 ] j≡

x j r j

∑i

xi ri

j≡x j q j

∑i

xi qi

r j=∑k

vk j Rk ; q j=∑

kvk jQk

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Rk parameter ==> group volume

Qk parameter ==> group surface area

Molecule size (rj) and molecule shape (qj) may be calculated by multiplying the group parameters by the number of times each group appears in the molecule,

and summing all the groups in the molecule where vk

(j) is the number of groups of the kth type in the jth molecule.

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Technique for Fitting Model to Experimental data Tools in Matlab: lsqcurvefit

%Assume you determined xdata and ydata experimentally

xdata = [0.9 1.5 13.8 19.8 24.1 28.2 35.2 60.3 74.6 81.3];

ydata = [455.2 428.6 124.1 67.3 43.2 28.1 13.1 -0.4 -1.3 -1.5];

x0 = [100; -1] % Starting guess

[x,resnorm] = lsqcurvefit(@myfun,x0,xdata,ydata)

function F = myfun(x,xdata)

F = x(1)*exp(x(2)*xdata);

Result: x =

498.8309 -0.1013

resnorm =

9.5049

or using GUI of Matlab: >>cftool