ACTIVITY MODELS 1
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
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
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
Comparison with Equation of State Methods 5
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 1 1 2 2 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 = 0.6462, when = 0.6369 at P=66.9 mmhg ==> determine activity coefficient and relate to excess Gibbs energy 1 = y P 1 P =0.646266.9 1 0.636960.7 =1.118 = y P 2 2 x 2 P = 0.353866.9 2 0.363132.1 =2.031 G E 6 RT = ln 1 x 2 ln 2 =0.6369 ln1.1180.3631ln 2.031=0.328
plot to G E /RT vs 7
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
==> Dengan cara yg sama ==> Example (2): continued from example (1), at 30 o C show =0.1168 and y 1 =0.5316 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 = 0.328. 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 /[0.63690.3631]=1.42 =0.1168, we find : ln 1 = Ax 2 2 =1.420.8832 2 =1.107, 1 =3.03 ln 2 =A 2 =1.420.1168 2 =0.0194, 2 =1.02 9
Substituting into Modified Raoult's Law: y 1 P= 1 P 1 =0.11683.0360.7=21.48mmHg y 2 P= 2 P 2 =0.88321.0232.1=28.92mmHg P= y 1 P y 2 P=50.4 mmhg y 1 = y 1 P /P=21.48/50.4=0.426 10
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 2 21 1 12 2 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 2 1 2 A 21 ] T,P,n 2 =ln 1 n 1 n 2 2] 11
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
Example of Margules Equation Following is set of VLE data for the system methanol(1)/water(2) at 333.15 K: 13
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 =0.198 0.372x 2 x 2 G E RT = ln 1 x 2 ln 2 14
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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
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
Example of van Laar Equation Consider benzene(1) + ethanol(2) system which exhibits an azeotrope at 760 mmhg and 68.24 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 =6.87987-1196.76/(T+219.161) log P 2 =8.1122-1592.86/(T+226.18) 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=1.5094 where =0.552; x 2 =0.448 19
calculate A 12 and A 21 : A 12 =ln 1 [ 1 x 2 ln 2 ln 1]2 A 12 =1.3421; A 21 =1.8810 now consider =x 2 =0.5 ==> van Laar: 1 =1.579; =1.386 Problem ==> bubble temperature? [ A 21 =ln 1 x ln ]2 1 1 2 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
Wilson's Equation for Activity Coefficient G E RT = x ln x x x lnx x 1 1 2 12 2 2 1 21 ln 1 = ln x 2 12 x 2 ln 2 = ln 21 x 2 12 x 2 12 21 21 x 2 12 x 2 12 21 21 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
NRTL Equation for Activity Coefficient (3 parameters) G E x 2 RT = G 21 21 x 2 G 21 G 12 12 x 2 G 12 2[ ln 1 = x 2 21 ln 2 = 2[ 12 G 12 =exp 12 12 = b 12 RT G 21 x 2 G 212 G 12 12 x 2 G 12 2] G 12 x 2 G 122 G 21 21 21 = 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
For the infinite-dilution values of the activity coefficient: ln 1 = 21 12 exp 12 ln 2 = 12 21 exp 21 23
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
UNIQUAC for Binary Solution ==> ==> where ==> Q, R, v??? ln 1 =ln 1 1 1 5q 1 [ln 1 1 1 1 1 ] q 1[ 1 ln 1 2 21 1 1 2 21 2 12 1 12 2 ] ln 2 =ln 2 x 2 1 2 x 2 5q 2 [ln 2 2 1 2 2 ] q 2[ 1 ln 1 12 2 1 21 1 2 21 2 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 1 12 2] v k j Q k 25
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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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 28