Modified Raoult's Law and Excess Gibbs Free Energy

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Steven Russell
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1 ACTIVITY MODELS 1

2 Modified Raoult's Law and Excess Gibbs Free Energy Equilibrium criteria: f V i = L f i For vapor phase: f V i = y i i P For liquid phase, we may use an activity coefficient ( i ), giving: f L L i = x i i f i Poynting method is used to calculate the pure component liquid phase fugacities: Combining: y i i P=x i i i P i exp V L i P P i RT f L i = i P i exp V L i P P i RT 2

3 K-ratio ==> Ki=yi/xi At low pressure: Poynting Correction and ratio of for the component approach unity ==> MODIFIED RAOULT'S LAW K i = L i P i P [ i exp V i L P P i /RT i ] K i = L i P i P atau y i P= x i i P i i / i 3

4 Excess Gibbs free energy: G E =G-G is dst diperoleh: Activity Coefficient and excess Gibbs free energy are coupled. Excess Gibbs energy is zero for an ideal solution Activity Coefficients as derivatives: G E =RT i G E n i T,P,n j i =RT ln i x i ln i 4

5 Comparison with Equation of State Methods 5

6 Determination of G E from Experimental Data Modified Raoult's Law: x i P i Excess Gibbs energy from activity coefficient (for binary system): G E RT =x ln x ln i = y i P plot G E /RT vs Example (1): system of 2-propanol (1) + water (2) from citation data: T=30 o C, P 1 = 60.7 mmhg, P 2 = 32.1 mmhg, y 1 = , when = at P=66.9 mmhg ==> determine activity coefficient and relate to excess Gibbs energy 1 = y P 1 P = =1.118 = y P 2 2 x 2 P = =2.031 G E 6 RT = ln 1 x 2 ln 2 = ln ln 2.031=0.328

7 plot to G E /RT vs 7

8 One-Parameter MARGULES Equation the simplest expression for Gibbs excess energy function G E RT = Ax 2 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 ==> G E RT = An 2 n 1 n 1 RT G E n i T,P,n j i =An 2[ 1 n n 1 n 2] = A n 2 n [ 1 n 1 n ] =Ax 21 8

9 ==> Dengan cara yg sama ==> Example (2): continued from example (1), at 30 o C show = and y 1 = at P=60.3 mmhg. What are the pressure and vapor phase compositions predicted by the one-parameter Margules Equation. at =0.6369, we found G E /RT = Fitting the Margules equation: G E ln 1 = Ax 2 2 at the new composition: ln 2 =A 2 RT = A x 2 A=0.328 /[ ]=1.42 =0.1168, we find : ln 1 = Ax 2 2 = =1.107, 1 =3.03 ln 2 =A 2 = =0.0194, 2 =1.02 9

Substituting into Modified Raoult's Law: y 1 P= 1 P 1 =0.11683.0360.7=21.

10 Substituting into Modified Raoult's Law: y 1 P= 1 P 1 = =21.48mmHg y 2 P= 2 P 2 = =28.92mmHg P= y 1 P y 2 P=50.4 mmhg y 1 = y 1 P /P=21.48/50.4=

11 Two-Parameters MARGULES Equation requires differentiation of ng E /RT with respect to a mole number the equation is multiplied by n (mole numbers) where =n 1 /(n 1 +n 2 ), and x 2 =n 2 /(n 1 +n 2 ): ng E RT = A n A n n 1 n differentiation with respect to n 1 : G E x 2 RT = A 21 A 12 x 2 alternatively G E ln 1 =n 2[ A 21 n 1 A 12 n 2 n 1 n 2 2 Reconversion of n i to x i : RT =A 21 A 12 x 2 x 2 ng E /RT n 1 1 n 1 n 2 2 2n 1 n 1 n 2 3 n 1 A 21 ln 1 = x 2[A 21 A 12 x A 21 ] T,P,n 2 =ln 1 n 1 n 2 2] 11

12 note: x 2 =1-, so: ln 1 = x 2 2[ A 12 2 A 21 A 12 ] Margules Equation ln 2 = 2[ A 21 2A 12 A 21 x 2 ] For the limiting conditions of infinite dilution: ln 1 = A 12 =0 andd ln 2 = A 21 x 2 =0 12

13 Example of Margules Equation Following is set of VLE data for the system methanol(1)/water(2) at K: 13

14 calculate i ==> modified Raoult's law calculate ln i i = y i P x i P i plot ln i vs x i A 12 and A 21 are values of the intercepts at =0 and =1 of the straight line drawn to represent the G E / x 2 RT: ln 1 = A 12 =0 andd ln 2 = A 21 x 2 =0 we get: A 12 =0.372 and A 21 =0.198 calculate G E /RT and G E / x 2 RT We have equation: or ==> G E RT = x 2 x 2 G E RT = ln 1 x 2 ln 2 14

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17 Use 1 and 2 to calculate VLE predictions: i = y i P P=x 1 1 P 1 x 2 2 P 2 x i P i x y 1 = 1 1 P 1 1 P 1 x 2 2 P 2 Another method: by fitting ln i vs x i for the following Margules equation: ln 1 = x 2 2[ A 12 2 A 21 A 12 ] ln 2 = 2[ A 21 2A 12 A 21 x 2 ] or to obtain A 12 and A 21 17

18 van Laar Equation for Activity Coefficient Model van Laar Equation ln 2 = by infinite solutions method: ln 1 = A 12 =0 andd ln 2 = A 21 x 2 =0 or rearranging to: [ A 12 =ln 1 1 x ln 2 2 ln 1]2 ln 1 = A 12 [ A 1 12 A 21 x 2]2 A 21 [ A 1 x 21 1 A 12 x 2]2 [ A 21 =ln 1 ln ]2 1 2 x 2 ln 2 18

19 Example of van Laar Equation Consider benzene(1) + ethanol(2) system which exhibits an azeotrope at 760 mmhg and o C 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 P 1 = /(T ) log P 2 = /(T ) Solution: at T=68.24 o C, P 1 =519.7 mmhg; P 2 =503.5 mmhg at azeotrope: =y 1 ==> 1 =P/P 1 ; =P/P 2 1 =760/519.7=1.4624; =760/503.5= where =0.552; x 2 =

20 calculate A 12 and A 21 : A 12 =ln 1 [ 1 x 2 ln 2 ln 1]2 A 12 =1.3421; A 21 = now consider =x 2 =0.5 ==> van Laar: 1 =1.579; =1.386 Problem ==> bubble temperature? [ A 21 =ln 1 x ln ] x 2 ln 2 Guess T=60 o C ==> P 1 =391.6 mmhg; P 2 =351.8 mmhg y i =x i P i /P ==>y 1 =0.407; y 2 =0.321; y i =0.728 ==> T is too low at T=68.24 o C, P 1 =519.7 mmhg; P 2 =503.5 mmhg y i =x i P i /P ==>y 1 =0.540; y 2 =0.459; y i =0.999 ==> T is ok (T azeotrope ) 20

21 Wilson's Equation for Activity Coefficient G E RT = x ln x x x lnx x ln 1 = ln x 2 12 x 2 ln 2 = ln 21 x 2 12 x x 2 12 x x 2 12 = V 2 V 1 exp A 12 and. RT 21 = V 1 V 2 exp A 21 RT Wilson's Equation V i, V j : molar volume of pure liquid i, j, at temperature T 21

22 NRTL Equation for Activity Coefficient (3 parameters) G E x 2 RT = G x 2 G 21 G x 2 G 12 2[ ln 1 = x 2 21 ln 2 = 2[ 12 G 12 =exp = b 12 RT G 21 x 2 G 212 G x 2 G 12 2] G 12 x 2 G 122 G = b 21 RT x 2 G 21 2] G 21 =exp 21 NRTL Equation, b 12, and b 21 are parameters specific to a particular pair of species, and are independent of composition and temperature NRTL: Non Random Two Liquid 22

23 For the infinite-dilution values of the activity coefficient: ln 1 = exp 12 ln 2 = exp 21 23

24 UNIQUAC Equation for Activity Coefficient For multicomponent solution: G E RT = x j ln j / x j 5 j j ln k =ln COMB RES k ln k q j x j ln j / j j q j x j ln i i ij ln COMB k =ln k[ k /x k 1 k / x k 5q k [ln k / k 1 k / k ] ln RES k =q 1 ln i ik j kj i j i i ij] UNIQUAC equation requires two adjustable parameters characterized from experimental data for each binary system ==> 24

25 UNIQUAC for Binary Solution ==> ==> where ==> Q, R, v??? ln 1 =ln q 1 [ln ] q 1[ 1 ln ] ln 2 =ln 2 x x 2 5q 2 [ln ] q 2[ 1 ln r j = k j x j r j x i r i i j x j q j x i q i i v k j R k ; q j = k ] v k j Q k 25

26 R k parameter ==> group volume Q k parameter ==> group surface area Molecule size (r j ) and molecule shape (q j ) 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 v k (j) is the number of groups of the k th type in the j th molecule. 26

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28 Technique for Fitting Model to Experimental data Tools in Matlab: lsqcurvefit %Assume you determined xdata and ydata experimentally xdata = [ ]; ydata = [ ]; 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 = resnorm = or using GUI of Matlab: >>cftool 28

Note: items marked with * you should be able to perform on a closed book exam. Chapter 10 Learning Objective Checklist

Note: items marked with * you should be able to perform on a closed book exam. Chapter 10 Learning Objective Checklist Sections 10.1-10.13 find pure component properties on a binary P-x-y or T-x-y diagram.*