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
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