Tugas Kinetika Dan Katalisis

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



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 )



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 :



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



[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



&. 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



(−∆ 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



[ 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



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



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 :



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



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



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

)



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



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



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 :



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



! !! 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



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



(∆ 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



( 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 #



[ 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



[ 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

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