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Fisicoquímica II
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Problemario de Fisicoquímica II
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Fisicoquímica IIProblemario Unidad 3
1. En la descomposición en paralelo de A el producto deseado es C.
A → B υ B=1
A →C υC =2C A
A → D υ D =C 2A
Determinar la máxima concentración de C en un reactor discontinuo cuando CA0 = 4 mol/L
(CC = 1.6188 mol/L)
Solución:
C A 0=4 mol /L
dC B
=1
dt
−dC A
=1+2 C A +C 2A
dt
dC C
=2C A
dt
dC D
=C 2A
dt
dC C 2 CA
=
−dC A 1+ 2C A +C 2A
[ ]
CA
2CA 2
∫ dCC =−∫ 2
dC A = 2 ln ( C A +1 ) +
C A +1
C AB ( C A +1 )
[
C C = 2 ln ( 0+1 ) +
2
0+1 ][
− 2 ln ( 4+1 ) +
2
4 +1 ]
=−[ 2−3.6188 ] =1.6188 mol/ L
3. Se realizó un experimento de 17 minutos y se encontró un sistema de ecuaciones
paralelas al utilizar una concentración del reactivo de 1 mol/L. Usando los datos
, experimentales contenidos en la tabla, calcula las constantes de ambas reacciones y
determina cual es la reacción más rápida.
(k1 = 9.006x10-4 s-1; k2 = 1.8994x10-3 s-1)
A k1 B
→
A k2 C
→
[A] mol/L [B] mol/L [C] mol/L t (s) ln (CA)
1 0 0 0 0
0.997 0.000899 0.0019 1 -0.00300451
0.986 0.00447 0.00943 5 -0.01409892
0.972 0.00888 0.0187 10 -0.02839947
0.959 0.0132 0.0279 15 -0.0418642
0.946 0.0175 0.037 20 -0.05551271
0.932 0.0217 0.0459 25 -0.07042246
0.919 0.0259 0.0547 30 -0.08446916
0.907 0.03 0.0633 35 -0.09761283
0.894 0.0341 0.0719 40 -0.1120495
0.882 0.0381 0.0803 45 -0.12556322
0.869 0.042 0.0886 50 -0.14041215
0.857 0.0459 0.0969 55 -0.15431736
0.845 0.0497 0.105 60 -0.16841865
0.834 0.0535 0.113 65 -0.18152188
0.822 0.0572 0.121 70 -0.19601488
0.811 0.0609 0.129 75 -0.20948722
0.799 0.0645 0.136 80 -0.22439433
0.788 0.0681 0.144 85 -0.23825719
0.777 0.0716 0.151 90 -0.25231493
0.766 0.0751 0.158 95 -0.26657311
0.756 0.0785 0.166 100 -0.2797139
0.657 0.11 0.233 150 -0.42007126
0.571 0.138 0.291 200 -0.56036607
0.497 0.162 0.342 250 -0.69916525
0.432 0.183 0.386 300 -0.83932969
0.375 0.201 0.424 350 -0.98082925
0.326 0.217 0.457 400 -1.1208579
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