UNIVERSIDAD NACIONAL AUTONOMA DE MÉXICO

FACULTAD DE ESTUDIOS SUPERIORES ZARAGOZA

INGENIERIA QUÍMICA

TERMODINAMICA QUIMICA

“PROBLEMARIO”

CONTRERAS OROZCO RAFAEL

HERNÁNDEZ OROZCO EDUARDO

RAYÓN PICAZO GIZEH CECILIA

RUIZ AMADOR JOSÉ

PROBLEMARIO TERCERA UNIDAD TERMODINÁMICA QUÍMICA

1.- Los parámetros de la ecuación de Van Laar para el sistema binario acetato de metilo (1) y metanol (2) son A=0.4262 y B=0.4394 a la presión de 1 ATM. Muestra gráficamente el comportamiento, de la energía de Gibbs, de exceso de la solución a esa presión. Se sabe que esa solución forma azeótropo a la composición de x1= 0.675 ¿Cuál es el valor de G E a esas condiciones?

c1 acetato de metilo2 metanol

T 318.15R 8.314A 0.4262B 0.4394

X azt 0.675GE azt 0.0971

2.- Para una mezcla binaria, γ∞1= 3 y γ∞

2=2. Elabora graficas de las ecuaciones de Wilson y Margules sobre el mismo plano coordenado y compara las descripciones de que cada una realiza para esta mezcla binaria. Los parámetros de interacción binaria de Wilson son λ12= 0.3465 y λ21=0.9612.

Para Wilson:

3 2 0.3465

x1 x2 ḠE0 1 0

0.05 0.95 0.001460150.1 0.9 0.0067212

0.15 0.85 0.015124050.2 0.8 0.0260096

0.25 0.75 0.038718750.3 0.7 0.0525924

0.35 0.65 0.066971450.4 0.6 0.0811968

0.45 0.55 0.094609350.5 0.5 0.10655

0.55 0.45 0.116359650.6 0.4 0.1233792

0.65 0.35 0.126949550.7 0.3 0.1264116

0.75 0.25 0.121106250.8 0.2 0.1103744

0.85 0.15 0.093556950.9 0.1 0.0699948

0.95 0.05 0.039028851 0 0

0.9612

X1 X2 In 1 In 2 1 20 1 1.09867246 0 3.00018052 1

0.1 0.9 0.77583964 0.01624937 2.17241541 1.016382110.2 0.8 0.54573431 0.05631159 1.72587524 1.057927270.3 0.7 0.37776651 0.11186907 1.45902223 1.118366420.4 0.6 0.25393012 0.17818487 1.28908173 1.195046230.5 0.5 0.16286428 0.25236492 1.17687695 1.287065630.6 0.4 0.09700706 0.33254284 1.10186815 1.394509630.7 0.3 0.05110678 0.41746231 1.05243526 1.518104180.8 0.2 0.02138802 0.50624782 1.02161838 1.659054430.9 0.1 0.00505789 0.59827162 1.0050707 1.8189722

1 0 0 0.69307278 1 1.99985119

Para Margules:

T 318.15R 8.314 J/gmol

3

2

2905.93838 J/gmol A2 1833.44298 J/gmol

X1 X2 In 1 In 2 1 20 1 1.09861229 0 3 1

0.1 0.9 0.88987595 0.00693147 2.4348276 1.006955550.2 0.8 0.70311186 0.02772589 2.02002899 1.028113830.3 0.7 0.53832002 0.06238325 1.71312643 1.064370180.4 0.6 0.39550042 0.11090355 1.4851272 1.117287140.5 0.5 0.27465307 0.1732868 1.31607401 1.189207120.6 0.4 0.17577797 0.24953299 1.19217333 1.28342590.7 0.3 0.09887511 0.33964212 1.10392842 1.404444880.8 0.2 0.04394449 0.4436142 1.04492435 1.55832916

0.9 0.1 0.01098612 0.56144922 1.01104669 1.753211441 0 0 0.69314718 1 2

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10.9

1

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

1.9

2

2.1

2.2

2.3

2.4

2.5

2.6

2.7

2.8

2.9

3Grafica comparativa Margules vs Wilson

x1 vs gamma 1 (Wilson)x1 vs gamma2 (Wilson)x1 vs gamma2 (Margules)x1 vs gamma1 (Margules)

X

Y1-Y

2

5.- A continuación se presentan los datos de equilibrio liquido vapor experimentales para el sistema acetonitrilo (1) / benceno (2) a 45°C.

P(kPa) x1 y1 P(kPa) x1 y1

29.819 0.0000 0.0000 36.978 0.5458 0.509831.957 0.0455 0.1056 36.778 0.5946 0.537533.553 0.094 0.1818 35.792 0.7206 0.615735.285 0.1829 0.2783 34.372 0.8145 0.691336.457 0.2909 0.3607 32.331 0.8972 0.786936.996 0.398 0.4274 30.038 0.9573 0.891637.068 0.5069 0.4885 27.778 1.0000 1.0000

a) Encuentra los valores de los parámetros para la ecuación de margules de tercer orden en que

proporcionen el mejor ajuste de GE

RT a la información

b) Sobre el mismo plano coordenado dibuja las curvas experimentales y teóricas de: lnγ, vs x1,

GE

RT vs x1,

GE

RT x1 x2 vs x1.

c) Elabora un diagrama p-x-y que compare los datos experimentales con los datos calculados con la ecuación de Margules.

d) Determina la consistencia termodinámica termodinámica de acuerdo con el método de Smith – Van Ness – Abbott.

Por margules

P(Kpa) x₁ y₁ x₂ y₂ γ₁* γ₂* LNγ₁* LNγ₂* GE/RT* GE/RTX1X2*P°₂= 29.819 0 0 1 1 1 0 0 #¡DIV/0!

31.957 0.0455 0.1056 0.9545 0.8944 2.67003867 1.00421981 0.98209295 0.00421093 0.04870457 1.1214562833.553 0.094 0.1818 0.906 0.8182 2.33612678 1.01617746 0.84849434 0.016048 0.09429796 1.1072514135.285 0.1829 0.2783 0.8171 0.7217 1.93280771 1.0451498 0.65897372 0.04416022 0.15660961 1.047923336.457 0.2909 0.3607 0.7091 0.6393 1.62735527 1.1022626 0.48695617 0.09736498 0.21069705 1.0214268236.996 0.398 0.4274 0.602 0.5726 1.43022789 1.18009385 0.3578338 0.16559397 0.24210542 1.0104735537.068 0.5069 0.4885 0.4931 0.5115 1.28599849 1.28948625 0.25153545 0.25424388 0.25287098 1.0116765736.978 0.5458 0.5098 0.4542 0.4902 1.24339393 1.33837101 0.21784468 0.29145321 0.25127768 1.0136155136.778 0.5946 0.5375 0.4054 0.4625 1.19685269 1.40709372 0.17969535 0.34152639 0.24530165 1.0176345135.792 0.7106 0.6157 0.2894 0.3843 1.11642343 1.5939135 0.11013021 0.46619231 0.21317458 1.0366011734.372 0.8145 0.6913 0.1855 0.3087 1.05021761 1.91824663 0.04899739 0.65141155 0.16074522 1.0639055332.331 0.8972 0.7869 0.1028 0.2131 1.02081829 2.24758641 0.02060455 0.80985693 0.1017397 1.1030826930.038 0.9573 0.8916 0.0427 0.1084 1.00714509 2.55728626 0.00711969 0.93894664 0.0469087 1.14756537

P°₁= 27.778 1 1 0 0 1 #¡DIV/0! 0 0 #¡DIV/0!

DATOS EXPERIMENTALES

Datos Experimentales

γ i∗¿Y iP

x iPi°

GE∗¿

RT=(X ¿¿1 ln γ 1)+(X ¿¿2 ln γ 2)¿¿¿

GE∗¿

RT X1 X2=

GE∗¿

RTX 1X2

¿¿

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

LN GAMA1* Vs XPolynomial (LN GAMA1* Vs X)LN GAMA 2 Vs XLogarithmic (LN GAMA 2 Vs X)Logarithmic (LN GAMA 2 Vs X)Polynomial (LN GAMA 2 Vs X)

A) Parámetros de Margules del 3er. orden

ln γ1∞=1.04

ln γ2∞=1.0

RT ln γ1∞=A−B

RT ln γ2∞=A+B

RT ( ln γ1∞+ ln γ 2∞ )=2 A

R= 8.314 Jgmol/K

T=318.15 K

(8.314 JgmolK ) (3.18 .15K )(1.04+1.0)

2=A

A= 2698.00108 Jgmol

RT ln γ2∞=A+B

RT ln γ2∞−A=B

((8.314 JgmolK ) (3.18 .15K )∗1)−2698.00108Jgmol=B

B=-52.901982 J/gmol

A= 2698.00108 J/gmolKB= -52.901982 J/gmolK

A dilución infinita

LNγ₁= 1.04 A₁₂LNγ₂= 1 A₂₁

DATOS AJUSTADOS

GE/RT LNγ₁ γ₁ LNγ₂ γ₂ GE/RTX1X2*0 1.04 2.82921701 0 1 1.04

0.0450879 0.94419676 2.57074763 0.00222833 1.00223082 1.038180.08825034 0.84749675 2.33379746 0.00947643 1.00952148 1.036240.15433214 0.68458942 1.9829575 0.03563913 1.03628181 1.0326840.21212804 0.51123403 1.66734748 0.08942329 1.09354345 1.0283640.24536547 0.36536122 1.44103444 0.16603274 1.18061175 1.024080.25488245 0.24301339 1.2750857 0.26708368 1.30614974 1.0197240.25240625 0.20554177 1.22819028 0.30872205 1.36168384 1.0181680.24495972 0.16310537 1.17716072 0.36501547 1.44053629 1.0162160.20802822 0.0823413 1.08582634 0.51664302 1.67639058 1.0115760.15221084 0.03354449 1.03411345 0.67325526 1.96060923 1.007420.09261142 0.01023204 1.01028456 0.8115879 2.25148027 1.0041120.04094653 0.00175659 1.00175813 0.91955379 2.50817097 1.001708

0 0 1 1 2.71828183 1

GE

RT=x1 x2 [ ( A21 x1 )+( A12 x2 ) ]

GE

RT x1 x2=( A21 x1 )+( A12 x2 )

ln γ1=x22 [A12+2 ( A21+A12) x1 ]

ln γ1=x12 [A21+2 ( A12+A21) x2 ]

B) Datos experimentales

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.2

0.4

0.6

0.8

1

1.2

LN GAMA 1 Vs X1GE/RTVs X1GE/RTX1X2 Vs X1LN GAMA 2 Vs X1

C) Diagrama P-xy

P=( x1γ 1P1° )+ (x2 γ2P2° )

DIAGRAMA P-XYP X₁ X₂ Y₁ Y₂

29.819 0 1 0 0.9315537131.774895 0.0455 0.9545 0.10976902 0.83629699

33.3670996 0.094 0.906 0.19604945 0.761423535.3237214 0.1829 0.8171 0.30616373 0.6658671836.5957984 0.2909 0.7091 0.39521342 0.5885905437.1247702 0.398 0.602 0.46066661 0.5317907737.1593565 0.5069 0.4931 0.51866454 0.4814606137.0632379 0.5458 0.4542 0.53932295 0.463533436.8570262 0.5946 0.4054 0.5662829 0.4401377635.8997882 0.7106 0.2894 0.64089482 0.3753901134.2419659 0.8145 0.1855 0.73348909 0.2950375132.0804101 0.8972 0.1028 0.8425315 0.2004113529.8322135 0.9573 0.0427 0.9585583 0.09972419

27.778 1 0 1 0

DIAGRAMA P-XY

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

5

10

15

20

25

30

35

40

P VS XP VS Y

X,Y

P (K

pa)

D) Consistencia Termodinámica

δ [ ln( γ1γ 2 )]=[ ln( γ1γ2 )−ln ( γ1¿

γ2¿ )] δ [ d( GE

RT )d x1

]=(GE

RT )−(GE

RT )¿

lδ(GE/RT)l lδ(LN(γ₁/γ₂)l0

-0.00361667 -0.03591359-0.00604762 0.00557399-0.00227748 0.03413680.00143098 0.032219550.00326005 0.007088660.00201147 -0.021361850.00112857 -0.02957176

-0.00034193 -0.04007906-0.00514637 -0.07823962-0.00853438 -0.03729661-0.00912828 -0.01210348-0.00596217 0.01402975

00.00237313 0.01153695

CONSISTENCIA TERMODINAMICA

PROMEDIO

6.- El siguiente es un conjunto de información del EVL para el sistema metanol(1)/agua(2) a 333.15 K:

P(kPa) x1 y1 P(kPa) x1 y119.953 0.0000 0.0000 60.614 0.5282 0.808539.223 0.1686 0.5714 63.998 0.6044 0.838342.984 0.2167 0.6268 67.924 0.6804 0.873348.852 0.3039 0.6943 70.229 0.7255 0.892252.784 0.3681 0.7345 72.832 0.7776 0.914156.652 0.4461 0.7742 84.562 1.0000 1.0000

a) Encuentra los valores de los parámetros para la ecuación de van Laar que proporcionen el

mejor ajuste de GE

RT a la información.

b) Sobre el mismo plano coordenado dibuja las curvas experimentales y teóricas de:

ln γi vs x1,GE

RTvs x1 y

GE

RT x1 x2vs x1.

c) Elabora un diagrama P−x− y que compare los datos experimentales con los calculados con la ecuación de van Laar.

d) Determina la consistencia termodinámica de acuerdo con el método de Smith – Van Ness – Abbott.

A) γ i∗¿Y iP

x iPi° ;GE∗¿

RT=(X ¿¿1 ln γ1)+(X ¿¿2 ln γ2);

GE∗¿

RT X1 X2=

GE∗¿

RTX1X2

¿¿¿¿¿

P(Kpa) x₁ y₁ x₂ y₂ γ₁* γ₂* LNγ₁* LNγ₂* GE/RT* GE/RTX1X2*P°₂= 19.953 0 0 1 1 1 0 0

39.223 0.1686 0.5714 0.8314 0.4286 1.57198438 1.01338565 0.45233876 0.01329686 0.08731932 0.6229350542.984 0.2167 0.6268 0.7833 0.3732 1.47028545 1.02638934 0.38545656 0.02604715 0.10393117 0.6122922748.852 0.3039 0.6943 0.6961 0.3057 1.31984693 1.07522153 0.27751577 0.07252671 0.13482289 0.6373254952.784 0.3681 0.7345 0.6319 0.2655 1.24552668 1.11150204 0.21955848 0.10571229 0.14761907 0.6346412656.652 0.4461 0.7742 0.5539 0.2258 1.16268199 1.15744301 0.15072939 0.14621327 0.14822791 0.5998827960.614 0.5282 0.8085 0.4718 0.1915 1.09718365 1.23303551 0.09274658 0.20947902 0.14782094 0.5931706363.998 0.6044 0.8383 0.3956 0.1617 1.04970234 1.31102791 0.04850664 0.27081149 0.13645044 0.5706820467.924 0.6804 0.8733 0.3196 0.1267 1.03097269 1.34953731 0.03050272 0.2997618 0.11655792 0.5360073170.229 0.7255 0.8922 0.2745 0.1078 1.02132989 1.38224394 0.02110559 0.32370822 0.10417001 0.5230737772.832 0.7776 0.9141 0.2224 0.0859 1.01247532 1.4098484 0.01239815 0.34348218 0.08603124 0.49746797

P°₁= 84.562 1 1 0 0 1 0 0

DATOS EXPERIMENTALES

DONDE LN GAMA 1 = 68

DONDE LN GAMA 2 = 56

LNγ₁ 0.68 A₁₂LNγ₂ 0.56 A₂₁

R= 8.314 J/molKA= 1717.28164 J/gmolB= -166.188546 J/gmol

B)

lnγ1 = A12 ( 1+ A 12X 1A 21X 2

)-2 lnγ2 = A21 ( 1+ A 21X 2A 12X 1

)-2 ¿

RTX 1 X 2 =A12 A21

A12 X 1+A21 X 1)

ln γ1∞=1.04

ln γ2∞=1.0

RT ln γ1∞=A−B

RT ln γ2∞=A+B

RT ( ln γ1∞+ ln γ 2∞ )=2 A

RT ln γ2∞=A+B

GE GE/RTX1X2 LNγ₁ γ₁ LNγ₂ γ₂ GE/RT0 0.68 0.68 1.97387773 1 0

-264.293265 0.65628921 0.44476534 1.56012405 0.0218634 1.02210415 0.09199471-200.210604 0.64982492 0.38508682 1.46974192 0.03540972 1.03604411 0.110302

-148.63022 0.63842486 0.28814271 1.33394766 0.06721914 1.06952983 0.13505545-125.936003 0.63028409 0.22674602 1.2545112 0.09612056 1.10089178 0.14660559-104.623341 0.62066852 0.16374325 1.17791184 0.13689761 1.14671073 0.15336396-85.8266892 0.61085944 0.11065267 1.11700687 0.18590512 1.20430799 0.15222908-70.1901707 0.60202868 0.07253931 1.07523507 0.23642601 1.26671383 0.14394544-55.6432414 0.59347181 0.04403041 1.04501413 0.29116608 1.33798677 0.12905391-47.3525928 0.58850802 0.03108168 1.03156976 0.32553042 1.38476496 0.11720123-38.0124494 0.58287618 0.01937906 1.01956806 0.36684034 1.44316748 0.10080158

0 0.56 1 0.56 1.7506725 0

DATOS AJUSTADOS

C) P=( x1γ 1P1° )+ (x2 γ2P2° )

P X₁ Y₁ X₂19.953 0 0 1

39.1985359 0.1686 0.5714 0.831443.1249364 0.2167 0.6268 0.783349.1353124 0.3039 0.6943 0.696152.9298863 0.3681 0.7345 0.631957.1079047 0.4461 0.7742 0.553961.2289813 0.5282 0.8085 0.471864.9531703 0.6044 0.8383 0.395668.6582306 0.6804 0.8733 0.319670.8710212 0.7255 0.8922 0.274573.4462406 0.7776 0.9141 0.2224

84.562 1 1 0

DIAGRAMA P VS X,Y

D)

δ [ ln( γ1γ 2 )]=[ ln( γ1γ2 )−ln ( γ1¿

γ2¿ )]

δ [ d( GE

RT )d x1

]=(GE

RT )−(GE

RT )¿

δ(LN(γ₁/γ₂) δ(GE/RT)0

-0.01613996 0.00467539-0.00973231 0.006370830.01593451 0.000232570.01677927 -0.001013490.02232951 0.00513605

0.04148 0.004408140.05841815 0.0074950.02212341 0.012495990.00815389 0.01303121

-0.01637724 0.014770340

0.0119141 0.0056335

CONSISTENCIA TERMODINAMICA

Promedio

8.- Para el sistema acetona(1)/metanol(2) y con base en la ecuación modificada de Raoult junto con la ecuación de Wilson, realiza los siguientes cálculos:

a) Temperatura de burbuja a la presión de 101.325 kPa y x1=0.3.b) Temperatura de rocío a la presión de 101.325 kPa y y1=0.3.

c) Si existe un azeótropo a P=101.325 kPa, encontrar T az y x1az= y1

az.d) Fracción de vaporización, , a P=101.325 kPa y a la temperatura intermedia entre la

de burbuja y la de rocío, y composición de las fases en equilibrio, con z1=0.3.

𝑳𝑵𝜸₁= -ln (X1 +⋀12X2)+ β X2

𝑳𝑵𝜸2= -ln (X2 +⋀21X1)+ β BX1

β=Λ12

x1+Λ12 x2−

Λ21x1+ Λ21 x2

burbuja roció∑ ZK Z/K

PB PR Pprom φ

759.99998 760 759.99999 0.5

TB TB T

331.352748 333.147641 332.250194

4. Elabora un diagrama P−x− y a T=60 °C , para el sistema acetona(1)/agua(2), con base en la ecuación modificada de Raoult y en la ecuación NRTL.

P = ϒi*Xi*P°

P*yi = ϒi*Xi*P°

yi = ϒi*Xi*P°/P

T(K) 333.15componente A B C PvAcetona (1) 16.6513 2940.46 -35.93 860.995763Agua (2) 18.3036 3816.44 -46.13 149.429731

Datos sacados del smithg12 g21 α12 R cal/mol K631.05 1197.41 0.5343 1.987

τ12 τ21 G12 G210.9532923 1.808861 0.60088959 0.38042188

X1 X2 LNϒ1 LNϒ2 ϒ1 ϒ2 P1 P2 PTOTAL y1

0 1 2.38168442 0 10.8231182 1 0 149.429731 149.429731 0

0.1 0.9 1.58686955 0.0388967 4.88842201 1.03966308 420.891064 139.820918 560.711981 0.75063683

0.2 0.8 1.09164801 0.12447559 2.97917976 1.13255438 513.01223 135.389837 648.402066 0.79119462

0.3 0.7 0.76222049 0.23309875 2.14302952 1.26250615 553.541802 132.059168 685.60097 0.80738188

0.4 0.6 0.53077367 0.3569314 1.70024724 1.42893784 585.562267 128.115479 713.677745 0.82048554

0.5 0.5 0.36088706 0.49542139 1.43460142 1.64118968 617.592873 122.621266 740.214139 0.83434352

0.6 0.4 0.23248724 0.65209773 1.26173435 1.91956333 651.808757 114.735933 766.54469 0.85032062

0.7 0.3 0.13482305 0.83350955 1.14433427 2.30138141 689.686872 103.168441 792.855314 0.86987734

0.8 0.2 0.06309147 1.04920846 1.06512426 2.85539006 733.653978 85.3360337 818.990012 0.89580333

0.9 0.1 0.01692505 1.31242427 1.01706909 3.71516937 788.122962 55.5156759 843.638638 0.93419496

1 0 0 1.6414226 1 5.1625085 860.995763 0 860.995763 1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

100

200

300

400

500

600

700

800

900

1000

Series2

P vs XY

XY

P

11.- Elabora un diagrama T−x− y a P=101.325 kPa, para el sistema acetona(1)/agua(2), con base en la ecuación modificada de Raoult y en la ecuación NRTL.

P = Ƴi*Xi*P°

yi = (P°/P)*Xi

P(mmHg) 760

A B CAcetona (1) 16.6513 2940.46 -35.93Agua (2) 18.3036 3816.44 -46.13

g12 g21 α12 R cal/mol K631.05 1197.41 0.5343 1.987

X1 X2 Tn f(T) f`(t) Tn+1 error

0 1 373.156672 0.000163131 27.1254232 373.156666 -6.014E-060.1 0.9 366.101725 1.20512E-05 25.949404 366.101725 -4.6441E-070.2 0.8 359.855671 4.8846E-05 25.2854206 359.855669 -1.9318E-060.3 0.7 354.342946 -3.598E-05 24.956882 354.342948 1.4417E-060.4 0.6 349.48397 8.31307E-05 24.8536257 349.483967 -3.3448E-060.5 0.5 345.177081 7.56971E-05 24.8909459 345.177078 -3.0411E-060.6 0.4 341.354863 0.000475477 25.0286723 341.354844 -1.8997E-050.7 0.3 337.907391 4.02948E-05 25.2080777 337.907389 -1.5985E-060.8 0.2 334.813459 6.85278E-05 25.4336598 334.813456 -2.6944E-060.9 0.1 332.011216 0.000179645 25.6831962 332.011209 -6.9946E-061 0 329.448615 1.38699E-05 25.9396191 329.448614 -5.347E-07

τ12 τ21 G12 G21 LNƳ1 LNƳ2

0.85108844 1.61493037 0.63461513 0.42195471 2.15504398 0

0.8674893 1.64605082 0.62907831 0.4149966 1.48317216 0.03329809

0.88254641 1.6746215 0.62403767 0.40870965 1.03912922 0.11028107

0.89627671 1.70067459 0.61947642 0.40305976 0.73130577 0.21192972

0.9087379 1.72431955 0.61536563 0.39799974 0.50932623 0.33077842

0.92007653 1.74583446 0.61164888 0.39345076 0.34434555 0.46529577

0.93037881 1.76538291 0.6082913 0.38936266 0.21960136 0.61748448

0.93987092 1.78339409 0.60521408 0.38563365 0.12563813 0.79192955

0.94855605 1.79987401 0.60241211 0.38225296 0.05784237 0.99559776

0.95656205 1.81506532 0.59984073 0.37916288 0.01522955 1.23812297

0.96400263 1.82918372 0.5974608 0.37631344 0 1.53251592

Ƴ1 Ƴ2 P°1 y18.62826961 1 2784.33493 04.40690292 1.03385868 2311.03227 0.304083192.82675447 1.11659186 1946.37448 0.512203812.07779196 1.23606101 1663.30592 0.656568131.66416955 1.3920513 1441.52477 0.758697251.41106614 1.59248513 1265.03024 0.832256731.2455801 1.85425775 1123.1118 0.886667211.13387177 2.20765209 1006.21243 0.926774611.05954796 2.7063416 909.732872 0.957613551.01534611 3.44913326 828.849377 0.981532161 4.62981039 760.010541 1.00001387

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1305

315

325

335

345

355

365

375

Series2

T-XY

XY

T

12.- Los coeficientes de actividad a dilución infinita para la acetona (1) y el cloroformo (2) en las soluciones que éstos forman a 1 atm, son 0.37 y 0.46, respectivamente. La mezcla forma un azeótropo con x1=0.345 a 64.5 °C. Hallar los valores de los coeficientes de actividad a x1=0.25 ,0.50 ,75, empleando estas dos series de datos junto con:

a) La ecuación de Margules de tercer orden.b) La ecuación de Wilson.c) La ecuación NRTL con α 12=0.30.

A) Por Margules de 3er orden

A dilución infinita

RT ln γ∞1= A-B

RT ln γ∞2= A+B

γ∞1= 0.37

γ∞2= 0.46

RT= (1.987 cal/gmol K) (337.75K)RT= 670.9105 cal/gmol

RT (-0.9942)= A-BRT (-0.7765)= A+B

Resolviendo por eliminación suma y resta

670.9105 (-0.9942)= A-B670.9105 (-0.7765)= A+B

-607.0192= A-B-520.9620=A+B-1187.9812= 2ª

A= -1187.9812/2A= -593.9906 cal/gmol

Sustituyendo A para obtener BB= -520.9620+593.9906B= 73.0286 cal/gmol

R= 1.987 CAL/gmol*K

X1+X2=1

X2=1-X1

ECUACION DE MARGULES

RTLN γ1= (A+3B ) x22−4 Bx2

3

RTLN γ2= (A−3 B ) x12−4 Bx1

3

componente X1 X1 azeotropo X1 X1 X1 X1acetona (1) 0.25 0.35 0.5 0.75 0 1cloroformo (2) 0.75 0.65 0.5 0.25 1 0RTLnɣ1 -334.119713 -238.6191951 -130.2405 -27.9958375 -667.0192 0RTLnɣ2 -46.2529875 -87.0774541 -166.7548 -334.119713 0 -520.962ɣ1 0.60773927 0.700707533 0.82355497 0.959130511 0.37001938 1ɣ2 0.93338211 0.878279885 0.77993088 0.607739266 1 0.46001327

B) Utilizando Wilson

A dilución infinita

1) ln γ∞1= -ln λ12 1- λ21

2) ln γ∞2= -ln λ21 1- λ12

1) ln (0.37)= -ln λ12 1- λ21

2) ln (0.46)= -ln λ 21 1- λ12

1) -0.9942= -ln λ12 1- λ21

2) -0.7765= -ln λ21 1- λ12

(−1.9942=−ln λ121−λ21−1.7765=−ln λ211−λ12 )−1

1) 1.9942= -ln λ12 1- λ21

2) 1.7765= -ln λ21 1- λ12

Despejando a λ21

λ21= 1.9942 - ln λ12

Sustituyendo

1.7765= ln(1.9942 - ln λ12)+ λ12

Programamos en Excel la ecuación para encontrar el valor de λ12

λ12= 1.167

Sustituyendo λ12

λ21= 1.9942 – ln (1.167)

λ21= 1.8397

ECUACION DE WILSON

𝑳𝑵𝜸₁= -ln (X1 +⋀12X2)+ β X2

𝑳𝑵𝜸2= -ln (X2 +⋀21X1)+ β BX1

β=Λ12

x1+Λ12 x2−

Λ21x1+ Λ21 x2

componente X1 X1 azeotropo X1 X1 X1 X1acetona (1) 0.25 0.35 0.5 0.75 0 1cloroformo (2) 0.75 0.65 0.5 0.25 1 0ln(ɣ1) -0.48055869 -0.342970747 -0.18951413 -0.04304593 -0.99413635 0ln(ɣ2) -0.06970722 -0.128470492 -0.24123364 -0.48201016 0 -0.77660251ɣ1 0.61843778 0.709658971 0.82736102 0.957867397 0.37004289 1ɣ2 0.93266684 0.879439512 0.78565804 0.617540786 1 0.45996609

C) Para NRTL con α12= 0.30

1) ln γ∞1= Ϯ21 + Ϯ12 G12

2) ln γ∞2= Ϯ12 + Ϯ21 G21

Como G12= exp (-α12 Ϯ12)

G21= exp (-α12 Ϯ21)

Sustituyendo G12 y G21 en las ecuaciones 1 y 2 respectivamente

1) ln γ∞1= Ϯ21 + Ϯ12 (exp (-α12 Ϯ12))

2) ln γ∞2= Ϯ12 + Ϯ21 (exp (-α12 Ϯ21))

Sustituir α12

1) ln (0.37)= Ϯ21 + Ϯ12 (exp ((-0.30) Ϯ12))2) ln (0.46)= Ϯ12 + Ϯ21 (exp ((-0.30) Ϯ21))

Despejar Ϯ21 de la ecuación 1

Ϯ21= -0.9942 - Ϯ12 (exp ((-0.30) Ϯ12))

Sustituir Ϯ21 en la ecuación 2

-0.7765= Ϯ12 + [-0.9942 - Ϯ12 (exp ((-0.30) Ϯ12))] [exp (-0.30 {-0.9942- Ϯ12 (exp ((-0.30) Ϯ12)}]

En Excel se programa la ecuación para encontrar el valor Ϯ12

Ϯ12= -1.2279

Sustituir Ϯ12= -1.2279 en Ϯ21 despejada

Ϯ21= -0.9942 – [-1.2279 (exp (-0.30* -1.2279))]

Ϯ21= 0.7806

ECUACION NRTL

Ln ϒ1 = X22[τ 21( G 21

X 1+X 2G 21 )2

+ τ 12G12(X 2+X 1G 12 )2 ]

Ln ϒ2 = X12[τ 12( G12

X 2+X 1G 12 )2

+ τ 21G21(X1+X 2G 21 )2 ]

α12= 0.3τ12= -1.2279τ21= 0.7806

componente X1 X1 azeotropo X1 X1 X1 X1acetona (1) 0.25 0.35 0.5 0.75 0 1cloroformo (2) 0.75 0.65 0.5 0.25 1 0ln(ɣ1) -0.74728957 -0.55132513 -0.28171637 -0.04383536 -0.99417809 0ln(ɣ2) -0.06570266 -0.112074611 -0.19378264 -0.37055008 0 -0.61027415ɣ1 0.4736486 0.576185783 0.75448765 0.957111527 0.37002745 1ɣ2 0.93640926 0.893977554 0.82383695 0.690354476 1 0.54320193

G12= 1.445376729G21= 0.791219384

13.- A la presión atmosférica y temperatura de 71.8 °C, el acetato de etilo y el alcohol etílico forman una mezcla azeotrópica que contiene 53.9 mol % del componente más volátil. Estimar:

a) Los valores de las constantes A y B de la siguiente forma de la ecuación de van Laar

Los parámetros A y B no son constantes.

A+B¿1A21

A-B¿1A12

ln γ1=A12(1+A12 X1A21 X2

)−2

ln γ2=A21(1+A21 X2A12 X1

)−2

15.- Una mezcla equimolar de acetona (1), alcohol etílico (2) y éter etílico (3) se encuentra a 55 °C. El comportamiento de la fase vapor puede tomarse como ideal y el de la fase líquida se puede describir con la ecuación de Wilson. Determinar las presiones de punto de burbuja y de punto de rocío y las correspondientes composiciones de las fases en equilibrio. La ecuación de Wilson para calcular ln γi en sistemas multicomponente es

X interacion ᴧ ij ᴧ ji ᴧ jj ᴧ ii0.33333333 12 0.30771 1.20101 0 00.33333333 13 0.4988 0.86494 0 00.33333333 23 0.29207 0.77045 0 0

1.31365127 1 0 0.80438423 0.528889130.4198379 1 0.38153278 0 0.47111087

0.60673098 1 0.61846722 0.19561577 0

0.98037791 ɣ1= 2.66546336 10.56719426 ɣ2= 1.76331271 10.79264798 ɣ3= 2.20923872 1

A B C T P PB16.6513 2940.46 -35.93 328.15 726.917385 1909.2436318.9119 3803.98 -41.68 328.15 279.52519916.0828 2511.29 -41.95 328.15 1492.491

Ki Xi F pr103.378982 0.18129811 1.000524326.2981247 0.71269013175.925521 0.10653607

P R18.7424138

Los coeficientes están dados en la siguiente tabla

Desarrollo

lnɣ 1=−ln (X1 ᴧ11+X2 ᴧ 12+X3 ᴧ 13)+1−(( X1 ᴧ 11X1 ᴧ 11+X 2 ᴧ12+X3 ᴧ13 )+( X2 ᴧ21

X1 ᴧ 21+X 2 ᴧ22+X3 ᴧ23 )+( X3 ᴧ 11X1 ᴧ 31+X 2 ᴧ32+X3 ᴧ33 ))

lnɣ 2=−ln (X 1 ᴧ21+X2 ᴧ22+X3 ᴧ23 )+1−(( X1 ᴧ21X 1 ᴧ11+X2 ᴧ 12+X3 ᴧ 13 )+( X2 ᴧ22

X1 ᴧ21+X2 ᴧ 22+X3 ᴧ 23 )+( X3 ᴧ32X1 ᴧ31+X2 ᴧ 32+X3 ᴧ 33 ))

lnɣ 2=−ln (X 1 ᴧ31+X2 ᴧ32+X3 ᴧ33 )+1−(( X1 ᴧ31X1 ᴧ11+X2 ᴧ 12+X3 ᴧ 13)+( X2 ᴧ 32

X1 ᴧ 21+X2 ᴧ 22+X3 ᴧ 23 )+( X3 ᴧ 33X1 ᴧ31+X2 ᴧ 32+X3 ᴧ 33 ))

−ln ( X1 ᴧ 11+X2 ᴧ12+X3 ᴧ 13 )=1.313651271

−ln ( X1 ᴧ 21+X 2 ᴧ22+X3 ᴧ23 )=0.419837904

−ln ( X1 ᴧ 11+X2 ᴧ12+X3 ᴧ 13 )=0.606730981

(( X1 ᴧ 11X1 ᴧ 11+X2 ᴧ12+X3 ᴧ13 )+( X2 ᴧ21

X1 ᴧ21+X2 ᴧ22+X3 ᴧ23 )+( X 3 ᴧ11X 1 ᴧ31+X2 ᴧ32+X3 ᴧ33 ))=0+0.804384226+0.528889133

(( X 1 ᴧ21X1 ᴧ 11+X2 ᴧ12+X3 ᴧ13 )+( X2 ᴧ22

X1 ᴧ21+X2 ᴧ22+X3 ᴧ23 )+( X3 ᴧ32X 1 ᴧ31+X2 ᴧ32+X3 ᴧ33 ))0.381532777+0+0.471110867

(( X 1 ᴧ31X1 ᴧ 11+X2 ᴧ12+X3 ᴧ13 )+( X2 ᴧ32

X1 ᴧ21+X2 ᴧ22+X3 ᴧ23 )+( X3 ᴧ33X 1 ᴧ31+X2 ᴧ32+X3 ᴧ33 ))=0.618467223+0.195615774+0

lnɣ 1=0.98037791

lnɣ 2=0.56719426

lnɣ 3=0.79264798

ɣ1=2.66546336

ɣ2=1.76331271

ɣ3=2.20923872

Componente A B C T°K P°i

acetona 16.6513 2940.46 -35.93 328.15 726.917385etanol 18.9119 3803.98 -41.68 328.15 279.525199eter etilico 16.0828 2511.29 -41.95 328.15 1492.491

Las gamas en el liquido se consideran como ideal por lo cual es 1

PB=∑ X i ɣ iPi0

PB=(0.333∗1∗726.917 )+ (0.333∗1∗279 )+(0.333∗1∗1492 )=1909.24

La fase vapor se calcula de a cuerdo con los resultados de las gamas optenidas anterior mente por medio del siguiente método

ɣ=cte K i=ɣ iPi

0

P X i=P/K i f (PR )=∑ X i cuando fpr=1 se para con el metodo

ɣ Ki Xi F pr2.66546336 103.378982 0.18129811 1.00052431.76331271 26.2981247 0.712690132.20923872 175.925521 0.10653607Como las gamas ya estaban dadas lo que se varia es la presión para determinar la presión de rocio y esta fue de 18.7424mmHg

16.- las composiciones de las fases liquido y vapor que coexisten en equilibrio de etanol (1) y tolueno (2) a 55° son x1=0.7186 y 0.7431, a P=307.81 mmHg, según reportaron Kretschmer y Wiebe en 1949. J. Amer. Chem. Soc. 71:1793. Estimar la presiondel punto de burbuja a 55°C y x1=0.1, aplicando la ecuación de Van Laar.

Etanol(1) Tolueno(2)

Ant A= 18.9119 16.0137Ant B= 3803.98 3096.52Ant C= -41.68 -53.67

A12=A’ 1.0659

B12=B' 578

T= 55 °CT= 328.15 K

x1= 0.7186x2= 0.2814

In 1= 1.05593129 1= 2.87465105In 2= 0.012698442= 1.01277941

P°(1) 279.525199 mmHgP°(2) 113.5558926 mmHg

PB=609.784970

5 mmHg

17.- Modelar el comportamiento del sistema etanol (1)- Tolueno (2) a 55°C utilizando la ecuación UNIQUAC y los parámetros r1=2.1005, r2=3.9228, q2=1.972, q2=2.968, α12=-76.1573 K y α21=438.005 K.

T= 328.15 °K1 2

etanol toluenor 2.1055 3.9228q 1.972 2.968q' 0.96 1

l1 -0.438l2 1.8512

x1 x2 φ1* θ1 θ'1 lnγ1 lnγ1 γ1 φ2* θ2 θ'2 lnγ2 lnγ2 γ2 Ge comb Ge res GE0 1 0 0 0 #¡DIV/0! 1.03060714 #¡DIV/0! 1 1 1 0 0 1 #¡DIV/0! 0 #¡DIV/0!

0.1686 0.8314 0.09816032 0.11873942 0.16295497 0.04464397 0.50038424 1.72465704 0.90183968 0.88126058 0.83704503 0.000585859 0.04353517 1.04510883 0.00801406 0.21597856 0.223992620.2167 0.7833 0.1292896 0.15527126 0.20985099 0.042581431 0.40864685 1.57023969 0.8707104 0.84472874 0.79014901 0.001079929 0.06534821 1.06868417 0.0100733 0.24885034 0.258923650.3039 0.6961 0.18984043 0.22484797 0.29533408 0.038371494 0.28123173 1.37658146 0.81015957 0.77515203 0.70466592 0.002575975 0.10980897 1.11894351 0.01345423 0.28530437 0.29875860.3681 0.6319 0.23818987 0.2790424 0.35865686 0.034899462 0.21131648 1.27917578 0.76181013 0.7209576 0.64134314 0.004340686 0.14506291 1.16114153 0.01558937 0.29636589 0.311955260.4461 0.5539 0.30180997 0.3485813 0.43603664 0.030293134 0.14635811 1.19321488 0.69819003 0.6514187 0.56396336 0.007522152 0.18946862 1.2177328 0.01768029 0.29509475 0.312775030.5282 0.4718 0.37534993 0.42655512 0.51801666 0.025058501 0.09598035 1.12866876 0.62465007 0.57344488 0.48198334 0.012525298 0.23711571 1.28356455 0.01914534 0.27922411 0.298369450.6044 0.3956 0.45055709 0.50374807 0.59459903 0.019957135 0.06182805 1.08522266 0.54944291 0.49625193 0.40540097 0.019222246 0.28155422 1.35090734 0.01966641 0.25336596 0.273032370.6804 0.3196 0.53329016 0.58583423 0.67145841 0.01479819 0.03717335 1.05334576 0.46670984 0.41416577 0.32854159 0.028542971 0.32569102 1.42508861 0.01919102 0.21857605 0.237767070.7255 0.2745 0.58653442 0.63716252 0.71729593 0.011795099 0.02617444 1.03869959 0.41346558 0.36283748 0.28270407 0.035665407 0.35168599 1.47307404 0.0183475 0.19423108 0.212578580.7776 0.2224 0.65237199 0.69907451 0.7704604 0.008481699 0.0163109 1.02510249 0.34762801 0.30092549 0.2295396 0.045721812 0.38146798 1.53294357 0.0167639 0.16305827 0.17982217

1 0 1 1 1 0 0 1 0 0 0 #¡DIV/0! 0.50470072 #¡DIV/0! #¡DIV/0! 0 #¡DIV/0!

a12 -76.1573a21 438.005

τ12= 1.26122158τ21= 0.26321853

0 0.2 0.4 0.6 0.8 1 1.20

2

4

6

8

10

12