Question

5. A mixture of ethanol (e) and toluene (t) in vapor-liquid equilibrium at 50 kPa behaves according to the following van Laar activity coefficient model: 1.757 1.757 Iny = and Iny, = = Does an azeotrope exist for this system at 50 kPa? If so, at what temperature and composition? If not, how do you know? Hint: because T is unknown, this one will be much easier to solve using an equation- solving program.

Answer 1

Let Ethanol and Toluene be represented as 1 and 2 respectively. Antoine's Constant for these components are: TC Component Ethanol (1) Toluene (2) A 8.1122 6.95805 1592.864 1346.773 226.184 219.693 And Antoine's equation is : ps = 10(A-THC where Ps is in mm Hg Tis in C Thus, the vapor pressures of both components are: Thus, PS - 10(8.1122 -T+226.187 P2 = 106.95805-T+219.693) Total Pressure (P) = 50 Kpa= 50 * 7.50062 mm Hg (1 Kpa = 7.50062 mm Hg) 375.031 mm Hg In(Yı) = 1.757 In(Y2) = Also, x2 Thus, ya Y = = = 1-xi exploit expl/ 7731-) = expl cxplan 1.757 Modified Raoults Law for non-ideal liquid and ideal vapor is given as : P* yi= Yi Xi* PS For all the components Thus, For ethanol, P *y Y1 XI * P.S For Toluene, p*y2=Y2 X2 * P2$ ---->(1) ----> (2)

For azeotrope, Xi = Yi At azeotrope, Thus, X X2 = = yi y2 Thus, equations (1) and (2) becomes P = Y1* PS P = Yz * P28 375.031 = 1.757 1 + 10(8.1122 - 7226.18 exp[1,157 z] * 10(8.1122 - T+226.184 --> (3) (1+_21)2 375.031 = exp[, 1,757 z] * 10(6.95805 - T+219.693) -> (4) 1) T *(1+ 10 +219.693) We have 2 equations - equation (3) and (4) and 2 unknowns - T and xi We can use a solver to solve these equations simultaneously On solving both equations, we get, T = 59.229 C 0.76324 Since the equations solved, we know that azeotrope exists at above conditions since we assumed that azeotrope exists to get the required equations. Thus, Azeotrope exists at 59.229 C Azeotropic compositions for both liquid and vapor are: X1 = y1 = 0.76324 1 -0.76324 = 0.23676