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7/25/2019 Tugas Kinetika Dan Katalisis
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1. Per$%&aan 'ineti#a (ea#si 'iia :
Peran$an"an pa&ri# etil asetat dari asa asetat dan etan%l den"an #atalis asa sul)at
#apasitas 4!000 t%n per tahun
a. Persaaan (ea#siC H 3COO H (l )+C 2 H 5O H (l ) H 2 S04
⇔
C H 3COOC 2 H 5 (l)+ H 2 O(l)
asa asetat etan%l etil asetat uap air
&. '%ndisi Operasi
*ntu# pe&uatan etil asetat dai asa asetat dan etan%l en""una#an #atalis asa
sul)at &er%perasi pada suhu +0%, dan te#anan 1 at. Fase yan" ter-adi dala rea#si ini
adalah )ase $air dan erupa#an rea#si reersi&le.
$. /enis (ea#t%r an" Di"una#an
Pr%se pe&uatan etil asetat dila#u#an elalui rea#si esteri)i#asi yan" en""una#an
(ea#t%r alir an"#i erpen"adu# (A.
d. 'atalis
'atalis yan" di"una#an H2SO4 untu# eper$epat rea#si pe&entu#an etil asetat
e. in-auan 'ineti#a
Ditin-au dari se"i #ineti#a rea#si5 #e$epatan rea#si asa asetat dan etan%l en-adi etil
asetat adalah rea#si &er%rde dua. Persaaan la-u rea#si :
−r A=k 1C AA C E−k 1
k cC EA C W
k 1= (4,195C K +0,08815 )exp (−6500,1
T )Diana : r A : #e$epatan rea#si %l.63.deti# 61
# 1 : #%nstanta #e$epatan rea#si %l.61.deti# 61
# , : #%nstanta #eseti&an"an
, : #%nsentrasi %l.63
,' : persen %lue #atalis
: teperature '
AA : a$eti$ a$id
7 : ethan%l
7A : ethyl a$etate : 9ater
7/25/2019 Tugas Kinetika Dan Katalisis
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2. (ea#si : A+B→ Produk
i%le#uler eleenter C B0=2C A 0 ; tetap
,e# liitin" rea#tan
C A=C A 0
1 ; C B=
2C A0
1
Ma#a A liitin" rea#tan
Persaaan la-u rea#si :−r A=k C A C B
Nera$a %l :
¿−out +gen=acc
0−0+r A V =
d N A
dt
r A V =d N A
dt
r A=1
V
d N A
dt
r A=d C A
dt
St%i#i%etri :
C A=n A
V =
n A 0 (1− X A )V 0
=C A 0
(1− X A )
C A=nB
V =
nB0−2n A 0 X A
V 0=C
B0
−2C A0 X A=2C A0−C A 0 X A=C A0 (2− X A )
Den"an easu##an la-u rea#si dan st%i#i%etri :d C A
dt =−(k C A C B )
d [C A0 (1− X A )]dt
=−k C A0 (1− X A ) C A 0 (2− X A )
−C A0
d X A
dt =−k C A 0 (1− X A )C A0 (2− X A )
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d X A
dt =k C A0 (1− X A ) (2− X A )
∫0
X A d X A
(1− X A ) (2− X A )
=k C A0∫0
t
dt
Den"an inte"ral pe$ahan parsial :
∫0
X A d X A
(1− X A ) (2− X A )=
A
(1− X A )+
B
(2− X A )
A (2− X A )+B (1− X A )
(1− X A )( 2− X A ) =
2 A− A X A+B−B X A
(1− X A ) (2− X A )
'%e)isien X A0
→2 A+B=1
'%e)isien X A1
→− A−B=0→ B=− A
2 A+B=1
2 A+ (− A )=1→ A=1→ B=−1
Ma#a :
∫0
X A
( 1
(1− X A )− 1
(2− X A ) )d X A
Su&stitusi persaaan :
∫0
X A
( 1
(1− X A )−
1
(2− X A ) )d X A=k C A0∫0
t
dt
−ln (1− X A )+ln (2− X A )| X A0=k C A 0|t
0
ln (2− X A )2 (1− X A )
=k C A 0t
k = 1
C A0 t ln
(2− X A )2 (1− X A )
a. Penyelesaian den"an et%de inte"rasi
Den"an et%de least s<uares
n=25 a#a dapat dihitun" den"an :
7/25/2019 Tugas Kinetika Dan Katalisis
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k =
∑ [ t . ln (2− X A )2 (1− X A ) ]
∑C A0 t 2
Men"hitun" nilai >A pada asin" asin" 9a#tu : C A 1=C A 0(1− X A1)
140=200 (1− X A1 ) → X A1=0,3
C A 2=C A0(1− X A2)
100=200 (1− X A2 ) → X A2=0,5
C A 3=C A0(1− X A3)
75=200 (1− X A3 ) → X A3=0,625
C A 4=C A0(1− X A 4)
45=200 (1− X A4 ) → X A 4=0,775
C A 5=C A0(1− X A5)
30=200 (1− X A5 ) → X A5=0,85
N%%r 0 1 2 3 4 ! /ulah
t 0 20 40 ?0 120 180 420
,A 200 140 100 +! 4! 30 !@0
t2 0 400 1?00 3?00 14400 32400 !0800 X A 0 053 05! 05?2! 05++! 058! 350!
t . ln (2− X A )2 (1− X A ) 0 35883 1?521@ 3?53?8 12051+4 24158+2 4185!1?
[t . ln (2− X A )2 (1− X A ) ]1=t 1 . ln
(2− X A 1 )2 (1− X A 1)
=20.ln (2−0,3 )2 (1−0,3)
=3,883
[t . ln (2− X A )2 (1− X A ) ]2=t 2. ln
(2− X A 2 )2 (1− X A 2)
=40.ln (2−0,5 )2 (1−0,5 )
=16,219
[t . ln
(2− X A )
2 (1− X A ) ]3
=t 3 . ln (2− X A3 )
2 (1− X A 3 )=60.ln
(2−0,625 )
2 (1−0,625)
=36,368
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[t . ln (2− X A )2 (1− X A ) ]4=t 4 . ln
(2− X A4 )2 (1− X A 4 )
=120.ln (2−0,775 )2 (1−0,775 )
=120,174
[t . ln
(2− X A )2 (1− X A ) ]5
=t 5 . ln
(2− X A5 )2 (1− X A 5 )=180.ln
(2−0,85 )2 (1−0,85 )=241,872
∑ [ t . ln (2− X A )2 (1− X A ) ]=3,883+16,219+36,368+120,174+241,872=418,516
Ma#a :
k =
∑ [ t . ln (2− X A )2 (1− X A ) ]∑ t
2 =418,516
50800 =8,2385 x 10−3
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&. Penyelesaian den"an et%de di)erensial
−r A=k C An
d C A
dt =r A
−d C A
dt =k C A
n
−∆ C A
∆ t =k C A
n
Den"an linearisasi en-adi :
log(−∆ C A
∆ t )=log (k )+n log ( C A )
t ,A∆ t −∆ C A
−∆ C A
∆ t C A log(−∆ C A
∆ t ) log ( C A)
0 200
20 ?0 3 1+0 054++ 2523
20 140
20 40 2 120 05301 250+@
40 100
20 2! 152! 8+5! 050@+ 15@42
?0 +!
?0 30 05! ?0 605301 15++8120 4!
?0 1! 052! 3+5! 605?02 15!+4
180 30
(∆ t )1=t 1−t 0=20−0=20
(∆ t )2=t 2−t 1=40−20=20
(∆ t )3=t 3−t 2=60−40=20
(∆ t )4= t 4−t 3=120−60=60
(∆ t )5=t 5−t 4=180−120=60
(−∆ C A )1=C A0−C A1=200−140=60
(−∆ C A )2=C A1−C A2=140−100=40
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(−∆ C A )3=C A2−C A3=100−75=25
(−∆ C A )4=C A 3−C A4=75−45=30
(−∆ C A )5=C A 4−C A5=45−30=15
(−∆ C A
∆ t )1
=(−∆ C A )1
( ∆t )1=
60
20=3
(−∆ C A
∆ t )2
=(−∆ C A )2
( ∆ t )2=
40
20=2
(−∆ C A
∆ t )3= (−∆ C A )3( ∆ t )3
=25
20=1,25
(−∆ C A
∆ t )4
=(−∆ C A )4
(∆ t )4=30
60=0,5
(−∆ C A
∆ t )5
=(−∆ C A )5
( ∆ t )5=15
60=0,25
( C A )1=1
2( C A 0+C A1 )=
1
2 (200+140 )=170
( C A )2=1
2( C A1+C A2 )=
1
2(140+100 )=120
( C A )3=1
2 (C A 2+C A3 )=
1
2 (100+75 )=87,5
( C A )4=1
2(C A3+C A4 )=
1
2 (75+45 )=60
( C A )5=1
2 (C A 4+C A5 )=
1
2( 45+30 )=37,5
[log
(
−∆ C A
∆t
)]1
= log (3)=0,477
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[ log(−∆ C A
∆t )]2
= log (2)=0,301
[log
(−∆ C A
∆t
)]3= log (1,25 )=0,097
[ log(−∆ C A
∆t )]4
=log (0,5)=−0,301
[ log(−∆ C A
∆t )]5
= log (0,25 )=−0,602
[ log ( C A ) ]1=log (170 )=2,23
[ log ( C A ) ]2=log (120 )=2,079
[ log ( C A ) ]3= log (87,5 )=1,942
[ log ( C A ) ]4=log (60 )=1,778
[ log ( C A ) ]5= log (37,5 )=1,574
Penentuan har"a #
[ log(−∆ C A
∆t )]1
= log ( k 1 )+n [log ( C A ) ]1
0,477=log
(k 1 )+(2 x2,23 )
lo g (k 1)=−3,983
k 1=10−3,983=1,04 x10
−4
[ log(−∆ C A
∆t )]2
= log ( k 2 )+n [log ( C A ) ]2
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0,301= log ( k 2 )+ (2 x 2,079)
log (k 2 )=−3,857
k 2=10−3,857
=1,39 x 10−4
[ log(−∆ C A
∆t )]3
= log ( k 3 )+n [ log ( C A ) ]3
0,097=log (k 3 )+ (2 x1,942)
log (k 3 )=−3,787
k 1=10−3,787=1,633 x 10
−4
[ log(−∆ C A
∆t )]4
=log (k 4 )+n [ log ( C A ) ]4
−0,301=log (k 4 )+(2 x1,778 )
log (k 4 )=−3,857
k 4=10−3,857=1,39 x10
−4
[ log(−∆ C A
∆t )]5
= log ( k 5 )+n [ log ( C A ) ]5
−0,602=log (k 5 )+ (2 x 1,574 )
log (k 5 )=−3,75
k 5=10−3,75=1,778 x10
−4
k =∑ k
∑
=k 1+k 2+k 3+k 4+k 5
5
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k =(1,04+1,39+1,633+1,39+1,778 ) x 10
−4
5
k =1,4462 x10−4
3. (ea#si : A →3 B
Persaaan la-u rea#si : −r A=k C A2
Nera$a %l :
¿−out +gen=acc
0−0+r A V =d N A
dt
r A V =d N A
dt
r A=1
V
d N A
dt
r A=d C A
dt
Den"an easu##an la-u rea#si :
d C A
dt =−(k C A
2)
∫C A 0
C A d C A
C A2 =−k ∫
0
t
dt
−1
C A |C
AC A0
=−kt
|t 0
1
C A−
1
C A 0
=k t → k =1
t ( 1C A−
1
C A0)
k = 1
!aktu
1
""ol/ #=( ""ol
# )−1
!aktu−1
Nilai # se$ara uu dapat dinyata#an dala :
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k =( ""ol
# )1−n
!aktu−1
(
""ol
#
)
1−n
=
(
""ol
#
)
−1
→ n=2
Ma#a5 ter&u#ti rea#si &er%rde 2
denti)i#asi asusi
Suhu tetap =0
e#anan tetap P=P0
Men"alai penurunan -ulah %l
Pen-a&aran : isal rea#si &erlan"sun" pada )asa "as
Persaaan "as ideal untu# #%ndisi t%tal : PV =nt $T
pada setiap saat t
P0V 0=nt 0 $ T 0 pada ula6ula
PV
P0 V 0=
n t $T
nt 0 $ T 0
V =V 0nt
nt 0
P0
P
T
T 0
V =V 0
nt 0−1
2 n A0 x A
nt 0
P0
P
T
T 0
V =V 0(1−1
2
n A0
nt 0
X A) P0
P
T
T 0
V =V 0(1−1
2 % A0 X A) P0
P
T
T 0
V =V 0 (1+& A X A ) P0
P
T
T 0
Analisis : #arena 0 B >A B 15 =05 dan P=P05 a#a nilai ; selalu tida# saa den"an
;05 atau ;C;0 siste aria&le density
Dala hal ini5 ɛA dapat &ernilai : p%siti)5 ne"ati)5 aupun n%l.
a. Den"an et%de "ra)i# pe&andin"
Orde 1
Persaaan a-u (ea#si :−r A=k C A
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d C A
dt =−k C A
∫C A 0
C A d C A
C A=−k ∫
0
t
dt
lnC A| C AC A0
=−kt |t 0
ln( C A
C A0)=k t → k =
1
t ln( C A
C A 0)
0 50 100 150 200 250
0
0.5
1
1.5
2
2.5
k orde 1
t
ln (Ca/Ca0)
Orde 2
Persaaan a-u (ea#si : −r A=k C A2
d C A
dt =−(k C A2)
∫C A 0
C A d C A
C A2 =−k ∫
0
t
dt
−1
C A |C A
C A0
=−kt |t 0
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1
C A−
1
C A 0
=k t→k =1
t ( 1C A−
1
C A0)
0 50 100 150 200 250
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
k orde 2
t
((1/Ca)-(1/Ca0))
Orde 3
Persaaan a-u (ea#si : −r A=k C A3
d C A
dt =−(k C A3
)
∫C A 0
C A d C A
C A3 =−k ∫
0
t
dt
−1
C A2| C A
C A 0
=−kt |t 0
1
C A2−
1
C A 0
2=k t→k =1
t ( 1
C A2−
1
C A 0
2
)
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0 50 100 150 200 250
0
0
0
0
0
0.01
k orde 3
t
(1/Ca^2)-(1/Ca0^2)
Analisis: dari #eti"a "ra)i# terse&ut diatas5 terlihat &ah9a:
Era)i# pe&andin" pada te&a#an %rde 1 enun-u##an #elen"#un"an ne"ati) dan %rde
3 enun-u##an #elen"#un"an p%siti)5 yan" &erarti &ah9a %rde se&enarnya tida# le&ih
&esar dari %rde 3 dan tida# le&ih #e$il dari %rde 1.
Era)i# pe&andin" pada te&a#an %rde 2 enun-u##an trendline "aris lurus5 diana
trendline "aris yan" ter&entu# ele9ati seua titi# data5 yan" &erarti &ah9area#si ini
dapat dian""ap en"i#uti %rde 2.
&. Met%de Merata6(ata#an har"a #
Met%de Sh%rt nteral
t 0 10 3! @0 1!0 210
,A 120 81 4@ 2+ 20 1!
# %rde62 1064 6 45012 3544@ 3518@ 25++8 25++8
k 1=1
t ( 1C A1
− 1
C A 0)= 1
10 ( 181− 1
120 )=4,012 x10−4
k 2=1
t
( 1
C A2
− 1
C A 0
)=
1
35
( 1
49−
1
120
)=3,449 x10
−4
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k 3=1
t ( 1C A1
− 1
C A 0)= 1
90 ( 127− 1
120 )=3,189 x10−4
k 4=1
t
(
1
C A1
− 1
C A0
)=
1
150
(
1
20
− 1
120
)=2,778 x 10
−4
k 5=1
t ( 1C A1
− 1
C A 0)= 1
210 ( 115− 1
120 )=2,778 x10−4
k =∑ k
∑ =
k 1+k 2+k 3+k 4+k 5
5
k =( 4,012+3,449+3,189+2,778+2,778 ) x10−4
5
k =3,2412 x 10−4
Met%de %n" nteral
t 0 10 3! @0 1!0 210
,A 120 81 4@ 2+ 20 1!
# %rde62 1064 6 45012 3522! 35023 251?1 25++8
k 1= 1
t 1−0 ( 1C A 1
− 1
C A0)= 1
10 ( 181− 1
120 )=4,012 x10−4
k 2= 1
t 2−t 1 ( 1
C A 2
− 1
C A1)= 1
35 ( 149− 1
81 )=3,225 x 10−4
k 3= 1
t 3−t 2 ( 1
C A3
− 1
C A2)= 1
90 ( 127− 1
49 )=3,023 x10−4
k 4= 1
t 4−t 3 ( 1
C A4
− 1
C A3)= 1
150 ( 120− 1
27 )=2,161 x10−4
k 5= 1
t 5−t 4 ( 1
C A 5
− 1
C A4)= 1
210 ( 115− 1
20 )=2,778 x 10−4
k =∑ k
∑ =
k 1+k 2+k 3+k 4+k 5
5
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k =( 4,012+3,225+3,023+2,161+2,778 ) x 10
−4
5
k =3,0398 x 10−4
$. Met%de east S<uares
n=25 a#a dapat dihitun" den"an :
k =
∑ [ t ( 1C A−
1
C A 0)]
∑ t 2
N%%r 0 1 2 3 4 ! /ulaht 0 10 3! @0 1!0 210 4@!
,A 120 81 4@ 2+ 20 1! 312
t2 0 100 122! 8100 22!00 44100 +?02!
t ( 1C A−
1
C A 0) 0 0504 05282 154@? 15@44 35! +52?2
[t ( 1C A−
1
C A0 )]1=t 1( 1C A1
− 1
C A0 )=10( 181−
1
120 )=0,04
[t ( 1C A−
1
C A0 )]2=t 2( 1C A2
− 1
C A 0)=35( 149−
1
120 )=0,282
[t ( 1C A−
1
C A0 )]3=t 3( 1C A3
− 1
C A 0)=90( 127− 1
120 )=1,496
[t
(
1
C A−
1
C A0
)]4
=t 4
(
1
C A 4
− 1
C A0
)=150
(
1
20
− 1
120
)=1,944
[t ( 1C A−
1
C A0 )]5=t 5( 1C A5
− 1
C A 0)=210 ( 115−
1
120 )=3,5
∑ [ t ( 1C A−
1
C A0 )]=0,04+0,282+1,496+1,944+3,5=7,262
Ma#a :
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k =
∑ [ t ( 1C A−
1
C A 0)]
∑ t 2
= 7,262
76025=9,55 x10
−4
d. Met%de Di)erensial
Met%deDi)erensial Earis Sin""un"
N% t enit ,A %lGliter
1 0 120
2 10 81
3 3! 4@
4 @0 2+
! 1!0 20
? 210 1!
0 50 100 150 200 250
0
20
40
60
80
100
120
140
Grafk Garis Singgung
t (menit)
CA (mmol/liter)
Pe&a$aa
n Data #e6
'%%rdinat titi#
sin""un" t5 ,aSl%pe "aris sin""un" ,a 6
dCa
dt
1 3 104120−80
10−0 104 1152
2 18 ?888−60
22−8 ?8 2
3 8 !?68−46
37−16 !? 1504+?
4 41 4?56−36
60−28 4? 05?2!
65432
1
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! !! 3848−26
86−35 38 054314
? @0 2830−20
150−77 28 0513?@
(ea#si: A 2
−r A=k C An
d C A
dt =r A
−d C A
dt =k C A
n
inearisasi terhadap persaaan diatas5 en"hasil#an:
ln(−dC A
dt )=ln k +n lnC A identi# den"an persaaan linear:' =a0+a1 x
Ma#a:
C A −dC A
dt
X =ln C A❑' = ln
−dC A
dt
>2 >
104 1152 45?444 2541!@ 215!+04 +50?03
?8 2 4521@! 05?@31 1+58042 25@24!
!? 1504+? 4502!3 0504?! 1?52030 0518+2
4? 05?2! 3.828? 6054+ 145?!82 615+@@4
38 054314 35?3+? 605840+ 1352321 6350!81
28 0513?@ 353322 615@88! 115103! 6?.?2?0
Σ 235?8+? 60510++ @45!+14 615311!
Ma#a:
n a0+ ( X a1= ( '
( X a0+ ( X 2
a1= ( X'
6a0+23,6876 a1=−0,1077
23,6876a0+94,5714 a1=−1,3115
Sehin""a5 den"an et%de su&titusi diper%leh:a0=−8,3944
dana1=2,0105
Den"an dei#ian:
a1=n=2,0105 → n )2
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a0=ln k =−8,3944 →k =exp (−8,3944 )=2,2613*10−4
lter "ol−1
"ent −1
/adi rea#si ini eenuhi #ineti#a &er%erde 25 den"an # = 252?13 *
10−4
lter"ol *"e nt
e. Met%de Di)erensial
Met%de Di)erensial Sederhana
−r A=k C An
d C A
dt =r A
−d C A
dt =k C A
n
−∆ C A
∆ t =k C A
n
Den"an linearisasi en-adi :
log
(
−∆ C A
∆ t
)= log (k )+n log ( C A )
t ,A∆ t −∆ C A
−∆ C A
∆ t C A log(−∆ C A
∆ t ) log ( C A)
0 120
10 3@ 35@ 1005! 05!@1 25002
10 81
2! 32 1528 ?! 0510+ 15813
3! 4@
!! 22 054 38 6053@8 15!+@
@0 2+
?0 + 0511+ 235! 605@32 153+1
1!0 20
?0 ! 05083 1+5! 615081 1.243
210 1!
(∆ t )1=t 1−t 0=10−0=10
(∆ t )2=t 2−t 1=35−10=25
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(∆ t )3=t 3−t 2=90−35=55
(∆ t )4= t 4−t 3=150−90=60
(∆ t )5=t 5−t 4=210−150=60
(−∆ C A )1=C A0−C A1=120−81=39
(−∆ C A )2=C A1−C A2=81−49=32
(−∆ C A )3=C A2−C A3=49−27=22
(−∆ C A )4=C A 3−C A4=27−20=7
(−∆ C A )5=C A 4−C A5=20−15=5
(−∆ C A
∆ t )1
=(−∆ C A )1
( ∆t )1=
39
10=3,9
(−∆ C A
∆ t )2=(−∆ C A )2
( ∆ t )2 =32
25=1,28
(−∆ C A
∆ t )3
=(−∆ C A )3
( ∆ t )3=22
55=0,4
(−∆ C A
∆ t )4
=(−∆ C A )4
(∆ t )4= 7
60=0,117
(−∆ C A
∆ t )5=(−∆ C A )5
( ∆ t )5 = 560=0,083
( C A )1=1
2( C A 0+C A1 )=
1
2 (120+81 )=100,5
( C A )2=1
2( C A1+C A2 )=
1
2(81+49 )=65
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( C A )3=1
2 (C A 2+C A3 )=
1
2 (49+27 )=38
( C A )4=1
2(C A3+C A4 )=
1
2 (27+20 )=23,5
( C A )5=1
2 (C A 4+C A5 )=
1
2(20+15 )=17,5
[ log(−∆ C A
∆t )]1
= log (3,9)=0,591
[log
(−∆ C
A
∆t )]2=log (1,28 )=0,107
[ log(−∆ C A
∆t )]3
= log (0,4 )=−0,398
[ log(−∆ C A
∆t )]4
=log (0,117 )=−0,932
[ log(−∆ C A
∆t )]5= log (0,083 )=−1,081
[ log ( C A ) ]1=log (100,5 )=2,002
[ log ( C A) ]2= log (65)=1,813
[ log ( C A ) ]3= log (38 )=1,579
[ log ( C A) ]4=log (23,5 )=1,371
[ log ( C A ) ]5= log (17,5 )=1,243
Penentuan har"a #
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[ log(−∆ C A
∆t )]1
= log ( k 1 )+n [log ( C A ) ]1
0,591= log ( k 1 )+ (2 x 2,002)
log (k 1 )=−3,414
k 1=10−3,414=3,855 x10
−4
[ log(−∆ C A
∆t )]2
= log ( k 2 )+n [log ( C A ) ]2
0,107=log (k 2 )+ (2 x1,813)
log (k 2 )=−5,519
k 2=10−3,519=3,028 x 10
−4
[log
(−∆ C A
∆t
)]3
= log ( k 3 )+n
[log ( C A )
]3
−0,398= log ( k 3 )+(2 x 1,579 )
log (k 3 )=−3,556
k 1=10−3,556=2,779 x 10
−4
[ log(−∆ C A
∆t )]4=log (k 4 )+n [ log ( C A ) ]4
−0,933= log ( k 4 )+(2 x1,371 )
log (k 4 )=−3,675
k 4=10−3,675=2,113 x 10
−4
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[ log(−∆ C A
∆t )]5
= log ( k 5 )+n [ log ( C A ) ]5
−1,081=log (k 5 )+(2 x1,243 )
log (k 5 )=−3,567
k 5=10−3,567=2,710 x10
−4
k =∑ k
∑ =
k 1+k 2+k 3+k 4+k 5
5
´k =
(3,855+3,028+2,779+2,113+2,710) x10−4
5
k =2,897 x10−4