TXY diagram for water-formic acid at 1 atm

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1 CHE 323, November 5, Chemical Engineering Thermodynamics. Tutorial problem 7. From the azeotrope database: formic acid water azeotrope forms an azeotrope at 1 atm, T=380.25K, and with a composition of mol fraction of formic acid (HCOOH). For this system: a) Find the Margules parameters for this system (A12, A21). Is the interaction between water and formic acid attractive or net repulsive? b) Find the Henry s constant for formic acid in water at 25 C. Compared with values in the literature (ranging from 0.8 to 6 kpa) c) Find the composition and pressure of the formic acid azeotrope at 80 C. d) Construct a T-X-Y diagram at 1 atm. For x1(formic acid)= 0.1, 0.25, 0.75, 0.9. Hint: this is a repeated exercise of bubble point temperature calculations at these compositions. Note: it is highly encouraged that you work in teams and each member of the team makes one of these different calculations, and share the results. P formic acid,kpa=exp( /(t C )) P water,kpa=exp( /(t C )) Then the table of compositions is Temperature X Y1 Temperature, Celsius TXY diagram for water-formic acid at 1 atm Molar fraction (x,y) of formic acid

2 Solution: a) At T C = ( ) = C, then P formic acid,kpa= kpa, P water,kpa= kpa. When one has an Azeotrope, xi=yi. The liquid-vapour equilibrium equation at low pressures (fugacity coefficients ~1, Poyinting effect is negligible) is: xi*pi *γi = yi*p Then, because of the azeotrope, γi = P/ Pi Then, γformic acid = kpa/115.42kpa = Then, γwater = kpa/133.55kpa = A12 = 2ln(γ2,az)/x1,az - ln(γ1,az)(x1,az x2,az)/x2,az 2 = 2*ln(0.759)/ ln(0.878)*( ( ))/( )^2 = A21 = 2ln(γ1,az)/x2,az - ln(γ2,az)(x2,az x1,az)/x1,az 2 = 2*ln(0.878)/( )-ln(0.759)*( )/(0.5741)^2 = In this case, the interactions are close to being symmetric, and the negative sign means they are attractive interactions. As the composition of the azeotrope is close to equimolar. b) To find Hformic acid = P formic acid*γformic acid, xformic acid=>0 γformic acid, xformic acid=>0 = = exp(1^2*( *( )*0)) = Hformic acid,25 C = 0.426*exp( /( )) = 3.85 kpa which is consistent with the range of value indicated in the literature. c) To find the composition and pressure of the formic acid azeotrope at 80 C: γi = P/ Pi Which means that γ1 P1 = γ2 P2 or that γ1 P1 - γ2 P2 = 0, then P1 *exp (x2 2 (A12 +2(A21-A12)x1)) - P2 * exp (x1 2 (A21 +2(A12-A21)x2)) = 0 At 25 C, P1 = exp( /( )) = and P2 = exp( /( )) = Considering f(x1) = 61.82*exp ((1-x1)^2*(A12 +2(A21-A12)*x1)) 47.9*exp (x1^2*(a21 +2(A12- A21)*(1-x1))) By trial and error we can find the value of x1 that makes this function equal to zero For x1 = 0.5 f(0.5) = 61.82*EXP((1-0.5)^2*( *( )*0.5)) *EXP(0.5^2*( *( )*(1-0.5))) = For x1 = 0.3 f(0.3) = 61.82*EXP((1-0.3)^2*( *( )*0.3)) *EXP(0.3^2*( *( )*(1-0.3))) = For x1 = 0.35 f(0.35) = 61.82*EXP((1-0.35)^2*( *( )*0.35)) *EXP(0.35^2*( *( )*(1-0.35))) = For x1 = 0.33 f(0.33) = 61.82*EXP((1-0.33)^2*( *( )*0.33)) *EXP(0.33^2*( *( )*(1-0.33))) = close enough to the zero. Then x1azeotrope=y1azeotrope = 0.33, P = 61.82*EXP((1-0.33)^2*( *( )*0.33)) = kpa. This means that in some situations we can change the composition of the azeotrope if we change the temperature/pressure of the system, but only if the ratio of vapour pressures changes with changes in temperature.

3 To calculate the bubble point, all liquid, zi=xi, then we need to assure that Σ yi =1 But y i = x i γ i P i /P that incorporating into the bubble point condition leads to P = Σ x i γ i P i For x1= 0.1 For x1 = 0.1, γ1 = exp (x2 2 (A12 +2(A21-A12)x1)), γ1 = exp(0.9^2*( *( )*0.1)) = For x2 = 0.9, γ2 = exp (x1 2 (A21 +2(A12-A21)x2)), γ2 = exp(0.1^2*( *( )*0.9)) = Then, 0.1*0.511*exp( /(T C ))+0.9*0.991*exp( /(T C )) Pcalculated = 0.1*0.511*exp( /( ))+0.9*0.991*exp( /( )) = kpa Pcalculated = 0.1*0.511*exp( /( ))+0.9*0.991*exp( /( )) = Pcalculated = 0.1*0.511*exp( /( ))+0.9*0.991*exp( /( )) = Pcalculated = 0.1*0.511*exp( /( ))+0.9*0.991*exp( /( )) = kpa close enough to kpa (P=1 atm) In this case, y1 = 0.1*0.511*exp( /( ))/101.3 = For x1= 0.25 For x1 = 0.25, γ1 = exp (x2 2 (A12 +2(A21-A12)x1)), γ1 = exp(0.75^2*( *( )*0.25)) = For x2 = 0.75, γ2 = exp (x1 2 (A21 +2(A12-A21)x2)), γ2 = exp(0.25^2*( *( )*0.75)) = Then, 0.25*0.64*exp( /(T C ))+0.75*0.945*exp( /(T C )) Pcalculated = 0.25*0.64*exp( /( ))+0.75*0.945*exp( /( )) = Pcalculated = 0.25*0.64*exp( /( ))+0.75*0.945*exp( /( )) = kpa Pcalculated = 0.25*0.64*exp( /( ))+0.75*0.945*exp( /( )) = Pcalculated = 0.25*0.64*exp( /( ))+0.75*0.945*exp( /( )) = kpa close enough to kpa (P=1 atm) In this case, y1 = 0.25*0.640*exp( /( ))/101.3 = For x1= 0.75

4 For x1 = 0.75, γ1 = exp (x2 2 (A12 +2(A21-A12)x1)), γ1 = exp(0.25^2*( *( )*0.75)) = For x2 = 0.25, γ2 = exp (x1 2 (A21 +2(A12-A21)x2)), γ2 = exp(0.75^2*( *( )*0.25)) = Then, 0.75*0.959*exp( /(T C ))+0.25*0.640*exp( /(T C )) Pcalculated = 0.75*0.959*exp( /( ))+0.25*0.640*exp( /( )) = Pcalculated = 0.75*0.959*exp( /( ))+0.25*0.640*exp( /( )) = kpa Pcalculated = 0.75*0.959*exp( /( ))+0.25*0.640*exp( /( )) = Pcalculated = 0.75*0.959*exp( /( ))+0.25*0.640*exp( /( )) = kpa close enough to kpa. In this case, y1 = 0.75*0.959*exp( /( ))/101.3 = For x1= 0.9 For x1 = 0.9, γ1 = exp (x2 2 (A12 +2(A21-A12)x1)), γ1 = exp(0.1^2*( *( )*0.9)) = For x2 = 0.1, γ2 = exp (x1 2 (A21 +2(A12-A21)x2)), γ2 = exp(0.9^2*( *( )*0.1)) = Then, 0.9*0.994*exp( /(T C ))+0.1*0.541*exp( /(T C )) Pcalculated = 0.9*0.994*exp( /( ))+0.1*0.541*exp( /( )) = kpa Pcalculated = 0.9*0.994*exp( /( ))+0.1*0.541*exp( /( )) = Pcalculated = 0.9*0.994*exp( /( ))+0.1*0.541*exp( /( )) = 98.9 Pcalculated = 0.9*0.994*exp( /( ))+0.1*0.541*exp( /( )) = kpa close enough to kpa. In this case, y1 =0.9*0.994*exp( /( ))/101.3 = Then the table of compositions is Temperature = X Y

5 The T-X-Y diagram is Temperature, Celsius TXY diagram for water-formic acid at 1 atm T-X T-Y Molar fraction (x,y) of formic acid

CH2351 Chemical Engineering Thermodynamics II Unit I, II Phase Equilibria. Dr. M. Subramanian

CH2351 Chemical Engineering Thermodynamics II Unit I, II Phase Equilibria Dr. M. Subramanian Associate Professor Department of Chemical Engineering Sri Sivasubramaniya Nadar College of Engineering Kalavakkam