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Kinetika Kimia Penurunan Rumus Order 2
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Kinetika Kimia Penurunan Rumus Order 2
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AllMaria Kerans
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Nama
: Aloysius K. Kerans
NIM
: 0806073154
Jurusan /Fakultas
: Kimia/ Sains dan Teknik
Reaksi Orde 2 Tipe 3
Reaksi :
A
− − − V=
=
+
2B
P ; Berlaku 2A=B, maka laju :
= K [A] [B]
Hasil Integrasi yang akan di peroleh, adalah : ln
= k t
Pembuktian integrasi adalah sebagai berikut :
= ∫ [] = ∫ [] [ A ] = [ B] = = + − = 2A[ ] BB = 2 A[] 2 [A]A] ] = [ ][] -2 d[A]
-2 2
2 ( A – A A0 )
- d[B]
-
B – B B0
2 A = 2A0 + B – B B0 A
, Selanjutnya
2A B
+ [B
[][[]] = [][]−[[]]+[] = [ ] [][]−[]+[] = = [][]−[]+[] [] []−[]+[] [ ] [ ] [ − + ] + [ ] = [][]−[]+[] [][]−[]+[] ==22[ [ ] –] 22[][ ] [[] ] [][] = 2[ ] [] 2[]
, subtitusi nilai [B] ke dalam persamaan awal.
1 1 1
Jadi :
[A](2m + n) = 0 (2m + n)
=0
n
=
2 = [] 2[ ] 1 = []−[] = 2 = 2 []−[] = []− [] ] ] [ 2 [ [] 2[][] [] 2[]2[ ]2[ ] [] = m
m
sehingga : n n n
masukan nilai m dan n ke dalam persamaan awal
[ ]
([]122[]) [] 2[ ]2[ []] [] = 1 | ]2[ ] n [ A ] l n 2[ [] | | (l )= [] 2[] [ ] [ ] 0 1 [ ] 2 2 [] [] 2[] [] ln [] = [ ] [ ] 1 l n [] 2[] [ ] []= [ ] [ ] 1 [] 2[] [] ln [ ]= ⁄ 1 [][] ⁄[ ] [] = [] 2[]1 [][] l n [] 2[][ ] [] [][] = ln [][] = [] 2[] ln [[]] ln [[AB]] =k t [B] 2[A] Pembuktian rumus integrasi di peroleh hasil yang sama, dan dapat di simpulkan rumus tersebut adalah “Benar” Keterangan :
[[]]
ln
Slope = k ( [B]0 - 2 [A]0 )
Y=
ln [[]]
x=t
Intersep = t
ln [[]]
Reaksi Orde 2 Tipe 4 Reaksi :
A
P ; berlaku A = P maka, Laju :
[ ] =[ ][ ], = 1 [ ] [ ] l n [] [] [] [] = = [] = [] = [ ][] [ ] = [], ]= ] [ [ ([]| )=[]| [] – [ ] = [] [] [] =[] [ ] [] [ ] = [ ][] [] = [ ][] [ ] = [ ][] [ ][ []] = ] [ [ ][] [ ] [] = 1 = [ ][] [ ] [] [ ] [] [ ] [] [ ] [ 1 [ ] ] [ ] = [ ][] [ ] [] [][] [ ] [] Hasil integrasi yang akan di peroleh, adalah :
V
Nilai [P] di subtitusikan ke dalam persamaan di atas :
1= [] [ ] [] [ ] 1=[ ] [] [] [ ]=0 =0 = [] [] = 1 = []+[] 1 1 1 = [ ][] [ ] [] [ ][] [] [] [][] [ ] [] [ ] [ ][] [] [] [][][ ] [ ] [] = [ ] [ ] 1 [ ][] [] = [] [] [ ] = [] 1 [] [ln[ ] 0 = [] 1 [] ln [ ] [ ] 1 [] [][] [ ] [] = [] [] [] [ ] [] = []−+ [] ln[] [ ] []| = []−+ [] ln [] +[[]]+ [] Subtitusikan nilai r dan s ke dalam persamaan awal :
Di integralkan masing-masing :
Penjumlahan hasil integrasi di atas :
] 1 [ ] 1 [ ] [ [] l n l n [] [] [] [] [] [] = [ ] 1 [ ] [ ] [] l n l n [] [] [] [] = [ ] 1 [ ] l n l n [] [] [] []= [ ] 1 [ ] l n l n [] [] [] []= [ ] 1 [ ] l n l n [] [] [] [] = [ ] ⁄ [ ] 1 l n [] [] [ ]⁄[] = [ ] [ ] 1 l n [] [] [][ ] = ;[ ]=[] ;[]=[] [ ] 1 [ ] l n [] [] [][] = ln [[ ]][[]] = [] [] ln ([[]] )ln [[]] = [] [] ln ([[]] )ln [[]] = [] [] ln ([[]] )= [] [] ln [[]]
Pembuktian rumus integrasi di peroleh hasil yang sama, dan dapat di simpulkan rumus tersebut adalah “Benar” Keterangan :
ln ([[]] )
Y= Slope= k ([A]0 + [P]0)
ln [[]]+−
x=t
Intersep = t
ln [[P]]
Reaksi Orde 3 Tipe 1 Reaksi
:
3A
= − [] = []3 Hasil integrasi
:
P, maka laju :
[] [] =2
Penurunan rumus untuk membuktikan hasil integrasi :
= [ ] = []3 [[] 3] = [[ ]3] = , [ ] []3 = 12 []−| = | 12 ([1] [1])= 1[] [1] =2 1[] =2 [1]
Pembuktian rumus integrasi di peroleh hasil yang sama, dan dapat di simpulkan rumus tersebut adalah “Benar” Keterangan :
[1]
Y= Slope= k
t
[]
x=t Intersep =
[]
Reaksi Orde 3 Tipe 2
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