Examen Final FQ 2016 II

August 1, 2022 | Author: Anonymous | Category: N/A
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4.ara una determinada reaccin la constante de velocidad es 2.$%& * 10 -4 se'-1 a 3&$ºC / la ener'a es 211.3 211.3 kJ * mol-1 mol-1 . Calcul Calcular ar el valor valor de la constante constante de v velocidad elocidad a 400º C. (R = 8.314 J * º-1*mol !

k1= 1=

0.0002$%& se se'-1 +48 º

2= k2= ,,

 E

∗¿

(

211300 J J))º*mol

+&3 º

[ ]   R

1

T 2

 Ln

*=



 1

 Ln

(

 K  1

 K 2

)

 Ln

T 1

 0.  0 . 00 0002 0259 597 7  K 2

(

 K 

1

 K 2

 Ln

(

 K 2

[ ]  

1

T 2

)=211300 /

  8.314  1

1

−648

[  ] 673

 0.0002597

)= E */

)=− )= −1. 45693245

0.0002$%&)k2 = 0.232%4%&++ k2 = 0.0002$%&)0.232%4%& 0.0002$%&)0.232%4%&++ ++ k2= 0.0011 0.001 1148 seg-1

R



 1

T 1

 

  e activacin   1" ( 2 #tos"

 

SOLUCION PROB.1 773ºK

!kN"%2$

Po= 7.33

TIEMPO !%&$

L !Po"P#$ P'

0 2 + 10

!kN"%2$ &.33 8.$3 10.$ 11.%

P#(=P'-Po

P#=Po-P#(

0 1.2 3.1& 4.$&

P#=2Po-P' &.33 +.13 4.1+ 2.&+

0.0000 0.1&88 0.$++$ 0.%&+&

RR R5

L!Po"P#$= K'

= 0.0%&8&8$ &8$& min-1 r 1= 0.%%%&1

7)3ºK

Po= 7.33

TIEMPO !%&$

P'

!kN"%2$ L !Po"P#$

P#(=P'-Po

P#=Po-P#(

0 2

!kN"%2$ &.33 %.+&

0 2.34

P#=2Po-P' &.33 4.%%

0.0000 0.384$

+ 10

12.3 13.+

4.%& +.2&

2.3+ 1.0+

1.1333 1.%33&

RR R5

0.1%2&&%8 min-1 0.%%%8%

L!Po"P#$= K'

= r 1=

813ºK

Po= 7.33

TIEMPO !%&$

P'

0 2 +

!kN"%2$ &.33 11.2 13.%

!kN"%2$ L !Po"P#$

P#(=P'-Po

P#=Po-P#(

0 3.8& +.$&

P#=2Po-P' &.33 3.4+ 0.&+

0.0000 0.&$0& 2.2++4

RR R5

= 0.3&&%0$$ 0$$3 min-1 r 1= 1.00000

L!Po"P#$= K'

k=6e-*)R ln k = ln 6 7 (-*)R" (1)" k

* k

0.0%&% 0.1%28 0.3&&%

-2.3240 -1.+4+2 -0.%&31

T !ºK$ &&3 &%3 813

1"T 0.0012%4 0.0012+1 0.001230

  -E+"R = -21220.,804)8 E+= n a= 2.123737,43

-42038.1,8 CL"MOL

 

r=

-0.))))2

E+=

-17,428.74 / /""MOL

 

813

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