ISOTERMAS DE ADSORCION
lolitaaasfrs19 de Agosto de 2014
543 Palabras (3 Páginas)384 Visitas
Haga una gráfica de la cantidad adsorbida en función de la concentración de las soluciones de ácido acético y compruebe si cumple con la ecuación de Henry. Calcule la constante k.
NaOH+ 〖CH〗_3 COOH 〖CH〗_3 COONa + H_2 O
Solución
Calculo de moles de 〖CH〗_3 COOH iniciales
Calculo de moles de 〖CH〗_3 COOH finales
Calculo de no. De moles de 〖CH〗_3 COOH por cada gramo de carbón activado.
(X/g)
B C_(〖CH〗_3 COOH INCIAL )=(5ml )(1M)/((250ml) )=0.02 M〖CH〗_3 COOH
((0.02 mol )/(1 l))(0.005 l〖CH〗_3 COOH)
= 0.0004 mol 〖CH〗_3 COOH C_(〖CH〗_3 COOH FINAL)=((0.1M)(1.1 ml))/((5 ml) )=0.02 M
((0.02 mol)/(1 l ))(0.025l )=0.0005 mol 〖CH〗_3 COOH
0.0004 mol-0.0005 mol= -0.0001 mol 〖CH〗_3 COOH
((-0.0001〖CH〗_3 COOH)/(1 g de Carbon activado))=
-0.0001 mol 〖CH〗_3 COOH/ g Car. Activado.
C C_(〖CH〗_3 COOH INCIAL )=(5ml )(1M)/((100ml) )=0.05 M〖CH〗_3 COOH
((0.05 mol )/(1 l))(0.005 l〖CH〗_3 COOH)
= 0.00025 mol 〖CH〗_3 COOH C_(〖CH〗_3 COOH FINAL)=((0.1M)(1.6 ml))/((5 ml) )=0.032 M
((0.032 mol)/(1 l ))(0.025l )=0.0008 mol 〖CH〗_3 COOH
0.00025 mol-0.0008 mol= -0.0005 mol 〖CH〗_3 COOH
((-0.0005〖CH〗_3 COOH)/(1 g de Carbon activado))=
-0.0005 mol 〖CH〗_3 COOH/ g Car. Activado.
D C_(〖CH〗_3 COOH INCIAL )=(10ml )(1M)/((100ml) )=0.1 M〖CH〗_3 COOH
((0.1 mol )/(1 l))(0.01 l〖CH〗_3 COOH)
= 0.001 mol 〖CH〗_3 COOH C_(〖CH〗_3 COOH FINAL)=((0.1M)(3.5 ml))/((5 ml) )=0.07 M
((0.07 mol)/(1 l ))(0.025l )=0.00175 mol 〖CH〗_3 COOH
0.001 mol-0.00175 mol= -0.0007mol 〖CH〗_3 COOH
((-0.0007〖CH〗_3 COOH)/(1 g de Carbon activado))=
-0.0007 mol 〖CH〗_3 COOH/ g Car. Activado.
E C_(〖CH〗_3 COOH INCIAL )=(20ml )(1M)/((100ml) )=0.2 M〖CH〗_3 COOH
((0.2 mol )/(1 l))(0.02 l〖CH〗_3 COOH)
= 0.004 mol 〖CH〗_3 COOH C_(〖CH〗_3 COOH FINAL)=((0.1M)(7.6 ml))/((5 ml) )=0.152 M
((0.152 mol)/(1 l ))(0.025l )=0.00312 mol 〖CH〗_3 COOH
0.004 mol-0.00312 mol= 0.00088mol 〖CH〗_3 COOH
((0.00088〖CH〗_3 COOH)/(1 g de Carbon activado))=
0.00088 mol 〖CH〗_3 COOH/ g Car. Activado.
F C_(〖CH〗_3 COOH INCIAL )=(25ml )(1M)/((100ml) )=0.25 M〖CH〗_3 COOH
((0.25 mol )/(1 l))(0.025 l〖CH〗_3 COOH)
= 0.00625 mol 〖CH〗_3 COOH C_(〖CH〗_3 COOH FINAL)=((0.1M)(10.15 ml))/((5 ml) )=0.203 M
((0.203 mol)/(1 l ))(0.025l )=0.00507 mol 〖CH〗_3 COOH
0.00625 mol-0.00507 mol= 0.00118mol 〖CH〗_3 COOH
((0.00118〖CH〗_3 COOH)/(1 g de Carbon activado))=
0.00118 mol 〖CH〗_3 COOH/ g Car. Activado.
G C_(〖CH〗_3 COOH INCIAL )=(50ml )(1M)/((100ml) )=0.5 M〖CH〗_3 COOH
((0.5 mol )/(1 l))(0.05 l〖CH〗_3 COOH)
= 0.025 mol 〖CH〗_3 COOH C_(〖CH〗_3 COOH FINAL)=((0.1M)(20.2 ml))/((5 ml) )=0.404 M
((0.404 mol)/(1 l ))(0.025l )=0.0101 mol 〖CH〗_3 COOH
0.025 mol-0.0101 mol= 0.0149mol 〖CH〗_3 COOH
((0.0149〖CH〗_3 COOH)/(1 g de Carbon activado))=
0.0149 mol 〖CH〗_3 COOH/ g Car. Activado.
H C_(〖CH〗_3 COOH INCIAL )=(75ml )(1M)/((100ml) )=0.75 M〖CH〗_3 COOH
((0.75 mol )/(1 l))(0.075 l〖CH〗_3 COOH)
= 0.0562 mol 〖CH〗_3 COOH C_(〖CH〗_3 COOH FINAL)=((0.1M)(31.55 ml))/((5 ml) )=0.631 M
((0.631 mol)/(1 l ))(0.025l )=0.0157 mol 〖CH〗_3 COOH
0.0562 mol-0.0157 mol= 0.0405mol 〖CH〗_3 COOH
((0.0405〖CH〗_3 COOH)/(1 g de Carbon
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