Metode analisis kecepatan

4 metode

• 1. Integral

• 2. Differensial

• 3. Laju awal

• 4. Waktu paruh

Reaksi Orde Satu

• Pada reaksi orde satu, kecepatan reaksi berbanding lurus dengan konsentrasi reaktan.

• Persamaan laju reaksi orde satu dinyatakan sebagai :

- = k1 [A]

- = k1 dt

ln = k1 (t – t0)

Bila t = 0 A = A0

ln [A] = ln [A0] - k1 t [A] = [A0] e-k

1t

dt

dA

][

]0[

A

A

][A

dA

Tetapan laju (k1) dapat dihitung dari grafik ln [A] terhadap t, dengan –k1 sebagai gradiennya.

ln [A]

ln [A]0

gradien = -k1

t

Grafik ln [A] terhadap t untuk reaksi orde satu

Waktu paruh (t1/2) adalah waktu yang dibutuhkan agar konsentrasi reaktan hanya tinggal setengahnya. Pada reaksi orde satu, waktu paruh dinyatakan sebagai

k1 = ln

k1 =

t1/2

1

2/1

1

2/1

693,0

t

. Reaksi Orde Dua • Persamaan laju reaksi untuk orde dua

dinyatakan sebagai :

- = k2 [A]2

- = k2 t

- = k2 (t – t0)

dt

dA

2][A

dA

][

1

A ]0[

1

A

Tetapan laju (k2) dapat dihitung dari grafik 1/A terhadap t dengan k2 sebagai gradiennya.

ln 1/[A]

ln 1/[A]0

gradien = -k2

t

Grafik ln 1/[A] terhadap t untuk reaksi orde dua

Waktu paruh untuk reaksi orde dua dinyatakan sebagai

t1/2 = ]0[2

1

Ak

Menentukan Orde Reaksi dari Persamaan Laju Integral

00

11lnln

Akt

AAktA

tt

Waktu Paruh Reaksi

Reaction Order Using the Integrated Rate Equations

Order in [A] Rate Law Integrated Form,

y = mx + b

Straight Line Plot

Half-Life t1/2

zeroth

order

(n = 0)

rate = k [A]o= k [A]t = - k t +[A]o [A]t vs. t

(slope = - k)

first

order

(n = 1)

rate = k [A]1 ln[A]t = - k t + ln[A]o

ln[A]t vs. t

(slope = - k)

second

order

(n = 2)

rate = k [A]2

vs. t

(slope = k)

1

[A]tk t

1

[A]0

1

[A]t

t1/ 2 [A]0

2k

t1/ 2 ln 2

k

0.693

k

t1/ 2 1

k[A]0

These types of plots are usually used to determine the values k for runs at

various temperatures and then used to determine the activation energy. •

Zero Order First Order Second Order

                                                                                              

                                                                                              

                                                                                   

                                                                                              

                                                                                   

                                                                                                                                                           

                                                                                                                                                                                

                                                                                                                                                                

Differential Method

                                                   

                                                                                 

                                                                              

• Taking the natural log of

• The reaction order can be found from a ln-ln plot of:

time (s) 0 t1 t2 t3

concentration (mol/dm3) CAo CA1 CA2 CA3

Three Ways to Determine (-dCA/dt) from Concentration-Time Data (Graphical,    Polynomial, Finite Difference, Non-Linear Least Squares Analysis)

However, we are usually given concentration as a function of time from batch reactor experiments.

Graphical

This method accentuates measurement error!

Polynomial (using Polymath)

• CA = ao + a1t + a2t2 + a3t3 +a4t4

•

Finite Difference

METHODA LAJU AWAL

Disini, diukur ro untuk beberapa run percobaan.

Variasi konsentrasi awal dari satu reaktan pada waktu tertentu

Misal diukur rAo dua konsentrasi awal dari A , yaitu

CAo,1 dan CAo,2 dengan CBo, CCo, - - - dibuat tetap.

Karena hanya Cao yang berubah sedangkan yang lainnya

dijaga tetap, misal untuk persamaan laju reaksi :

- rA = k CAa CB

b CCc - - - ,

laju reaksi untuk run 1 dan 2 : rAo,2 CAo,2

--------- = --------- rAo,1 CAo,1

a . . . (121)

maka data a dapat ditentukan..

Laju reaksi awal - rAo dapat dicari dari data differensial pada waktu = 0

Orde reaksi b , c , - - - dapat dicari dengan acara yang sama.

Contoh 1. The initial-rate method

Kelarutan dolumite, calcium magnesium carbonate dalam

HCl adalah

4HCl + CaMg(CO3)2 Mg+2 + Ca+2 + 4Cl- + 2 CO3 + 2H2O

Tentukan orde reaksi HCl dari data seperti pada Fig.1

Solution

- rA = k CAa CB

b

- rAo = k CAoa Cbo

b

CBo,1 = CBo,2 and CAo,1 CAo,2

- rAo,2 CAo,2

----------- = --------- - rAo,1 CAo,1

a

Run 1 Run 2

•

•

••

•

4N HCl

• •

••

•1 N HCl

4.000

3.999

3.998

3.997

3.996 0 2 4 6 8

t (min)

CH

Cl

1.0000

0.9990

0.9980

0.9970

0.9960

t (min)

0 2 4 6 8

CH

Cl

Figure 1

- dCA 3.9982 - 4.000-rAO,2 = ------- = -------------------------- = 3.6 x 10 –4 mole/lt min dt 5 - 0

- dCA 0.9987 - 1.0000-rAo,1 = -------- = ---------------------------- = 2.2 x 10 –4 mole/lt.min dt 6 - 0

Jadi :

- rAo,2 CAo,2

----------- = ----------

- rAO,1 CAo,1

a

3.6 x 10 –4 4 ------------- = ------ 2,2 x 10 – 4 1

a

a = 0.44.

Orde reaksi b dicari dengan cara yang sama dimana hanya (CBo) yang berubah

• EXAMPLE: Given the following experimental kinetic data, determine the kinetic rate expression and overall order of reaction.

• 2HgCl2 + C2O42- ------ Hg2Cl2 + 2CO2 + 2Cl-

• Our tentative rate expression for this reaction is: rate = k[HgCl2]x[C2O4

2-]y

• We need to determine what the exponents (orders) x and y are on the reactant concentrations. Then we can solve for the rate constant k.

• Step #1: find two experiments where the concentration of one of the reactants stays the same. Since the rate constant k is a constant and the concentration of one of the components is not changing (also a constant for this comparison), we only have to worry about the one reactant that is changing.

Exp # [HgCl2] [C2O42-] Initial Rate

1 0.10 M 0.15 M 2 x 10-5

2 0.10 M 0.30 M 8 x 10-5

3 0.05 M 0.30 M 4 x 10-5

We can then set up the following proportionality:

Rate(exp #2)

8 x 10 -5

2 x 10 -5

Rate(exp #1)=

=

=

=[ ]exp#2A y [ ]exp#2A

[ ]exp#1A y [ ]exp#1A

y

y

y

0.30

0.15

4 (2)

If y o u ca n ’t so lv e th is b y in sp e ctio n , th enta k e th e o f ea c h sid e o f th e e q u a tio n :lo g

log(4) = log(2)y

R e a rr a n g e a n d so lv e fo r :y

y = log(4) 0.602

log(2) 0.301= = 2

Step #2: Repeat for the other reactant concentration terms in the rate equation:

Rate(exp #2)

8 x 10 -5

4 x 10 -5

Rate(exp #3)=

=

=

[ ]exp#2A

[ ]exp#3A

x

x

x

0.10

0.05

2 (2) x = 1

Now that we have solved for the orders of the kinetic rate expression, we can write out the rate equation:

rate = k[HgCl2][C2O42-]2

• The overall order of the rate expression is the sum of the individual orders (x + y): 1 + 2 = 3. So this is called a third order rxn or rate law.

• Note that the orders DO NOT correspond to the coefficients on the chemical equation. Many times they do, but just as many times they won’t. There is NO FORMAL CONNECTION!!

Step # 3: Calculate the rate constant (if asked for). We now just plug in the experimental data from any one of the experiments and rearrange and solve for k, the rate constant. For this example we will use the data from

experiment # 1.

rate 2 = [HgCl ] [C O - ]2 2 42k

[HgCl ] [C O - ]2 2 42 2k =

k =

k =

rate

2 10x -5

2 10x -5

2.25 10x -3

k = 8.88 10x -3

(0.10 M) (0.15 M) 2

M sec -1

M sec -1

M 3

M sec-2 -1

• Consider a reaction A + B + C-> Products (this, too, is not an elementary reaction). Several initial conditions of this reaction are investigated and the following data are obtained:

Run #

[A]0 [B]0 [C]0 v0

1 0.151 M 0.213 M 0.398 M 0.480 M/s

2 0.251 M 0.105 M 0.325 M 0.356 M/s

3 0.151 M 0.213 M 0.525 M 1.102 M/s

4 0.151 M 0.250 M 0.480 M 0.988 M/s

rate = k [A]a [B]b [C]c

The ratio of the initial rates of runs 1 and 3 is then:

• Note that we have chosen this ratio so that many terms on the right hand side cancel, i.e. k and the concentrations of species A and B. The ratio reduces to:

• Note that none of the numbers in this equation have any units anymore because we have divided them out in a dimensionless ratio. Now, how to solve for c? Take the natural log (i.e. ln) of both sides of the equation

• Atau

• To continue, we pick another pair of runs that have a change in the initial concetration of C and just one other reactant. Let us choose the runs 1 and 4. The ratio of the initial rates of these runs is:

• Note that everthing cancels, is known, or is the order b. So:

• We can determine b as before by taking the ln of both sides of this equation:

•

• and b = 1.00. The reaction is First Order in B! Now that we know the order of Reactants A and B, we can use another pair of experimental runs to determine the remaining unknown order. We must use a pair of runs where the initial concentration of A changes, so we pick runs 1 and 2. As before, the ratio of initial rates is

• The rate constant cancels as before and we can solve for a:

• and, as usual, take the ln of both sides:

which results in a = 2.00 and the reaction is Second Order in A.

The rate constant for the reaction may now be evaluated from any of

the experimental runs:

Atau

k = 1.57 x 103    l5 / mol5.s

Problem: Consider the following reaction and kinetic data.

What is the kinetic rate expression? • Ni(CO)4 + CNCH3 Ni(CO)3(CNCH3) + CO

Exp # [Ni(CO)4] [CNCH3] Initial Rate (Msec-1)

1 0.04 0.06 0.2 x 10-5

2 0.04 0.12 0.2 x 10-5

3 0.08 0.06 0.4 x 10-5

4 0.08 0.12 0.4 x 10-5

Problem: Consider the following reaction and kinetic data. What

is the rate constant? • CO(g) + H2O(g) CO2(g) + H2(g)

Exp # [CO] [H2O] Initial Rate (Msec-1)

1 1.5 1.5 0,2

2 1.5 4.5 0,6

3 3.0 4.5 1,2

4 1.5 0.75 0,1

5 3.0 3.0 0,8

problemDari percobaan reaksi A + B AB, diperoleh data sebagai berikut

Perc [A] M [B] M V M/s

1

2

3

4

1,3.10-2

6,5.10-3

3,9.10-2

1,3.10-2

2,1.10-2

1,05.10-2

4,2.10-2

1,05.10-2

1,4.10-1

3,5.10-2

8,4.10-1

7.10-2

TentukanA. Orde reaksi untuk A dan BB. Persamaan laju reaksiC. Harga tetapan laju reaksiD. Laju reaksi jika konsentrasi A 0,026 M dan konsentrasi B

0,021 M

11

22

33

44

55

The following data were obtained for the reaction A + B + C products:

Experi-ment

[A]0 [B]0 [C]0 Initial rate, v0

(mol L-1 s-1) 1 1.25 x 10-3 M 1.25 x 10-3 M 1.25 x 10-3 M 0.0087

2 2.50 x 10-3 M 1.25 x 10-3 M 1.25 x 10-3 M 0.0174

3 1.25 x 10-3 M 3.02 x 10-3 M 1.25 x 10-3 M 0.0508

4 1.25 x 10-3 M 3.02 x 10-3 M 3.75 x 10-3 M 0.457

5 3.01 x 10-3 M 1.00 x 10-3 M 1.15 x 10-3 M ?

(a) Write the rate law for the reaction. (b) What is the overall order of the reaction? (c) Determine the value of the rate constant.

[Hint: Use your rate law from (a) and appropriate data.](d) Use the data to predict the reaction rate for experiment 5.

Penentuan Hukum Laju Cara Waktu paruh

[ ]ndA

k Adt

Untuk reaksi orde n:

0

[ ]

[ ] 0[ ] [ ]

A tn n

AA d A k dt

10[ ] [ ]

1

nA Ak t

n

1

0(1 )[ ]nx n A

10

0

[ ]1 [ ] ( 1) . 1

[ ]nA

A n kt ut nA

Untuk [A] = ½ [A]0 dan t = t1/2

1

1/ 2 10

2 1. 1

( 1)[ ]

n

nt ut n

n A k

Persamaan laju reaksi :Persamaan laju reaksi :

- 1 dNA mole A

- rA = ------ -------- ---------------- V dt (volume)(waktu)

Bentuk lain :

- 1 dNA mole A

- rA = ------ -------- ------------------------------------------ S dt (luas permukaan solid)(waktu)

- 1 dNA mole A

- rA = ------ -------- ----------------------------- W dt (berat katalis)(waktu)

Reaktor batch (phase gas) :

volume reaktor konstan tetapi terjadi perubahan

jumlah mole selama reaksi ( tekanan total akan

berubah dengan perubahan waktu ) , atau

aA + bB r R + s S . . . (1)

n = ( r + s ) - ( a + b )

Untuk reaktan A : pA = CART = pAo + ------- ( o - ) a

n

Untuk produk R : pR = CR RT = pRO - ------- ( o - )

. . . (2)

r

n

Reaktor batch :

Phase gas : konstan tekanan maupun temperatur ,

tetapi terjadi perubahan jumlah volume atau densiti

selama reaksi yaitu :

V = Vo (1 + A.XA) . . . (3)

dimana A. Adalah fraksi perubahan volume, atau

VA habis bereaksi - VA mula-2 VXA=1 - VXA=0

A. = -------------------------------- = ----------------------

V A mula-2 VXA=0

7

- Untuk sistem volume & densiti konstan :

dV = 0 dan NA = CA.V

- rA = ---- --------- = ---- -----------------

- 1 d(CA.V) -1 V dCA + CA dV

V dt V dt

= 0

or

- rA = ----- ------- = --------- -1 dNA - dCA

V dt dt

12

KONSENTRASI dan TEKANAN PARSIAL

Seperti pada Pers diatas:

- rA = - (1/V)(dNA/dt) = kCCAa CB

b . . . . . (4)

Hubungan antara konsentrasi dengan tekanan parsial

dari hukum gas ideal : pA = CA RT

Persamaan laju reaksi menjadi :

- rA = - ----- ------- = kC CAa CB

b

17

1 dNA

V dt

- rA = ------------- = kC ( ------- )(a+b) pAa pB

b

- rA* = -------- = kC( RT )1 - (a+b) pAa pB

b

= kP pAa pB

b (unit tekanan) / (unit waktu)

. . . (5)

d(pA/RT) 1

dt RT

dpA

dt

18

- rA = -------- = kC pAa pB

b mol/unit volume.wkt

dapat dinyatakan sebagai :

- dCA

dt

Untuk volume konstan :

19

Dimana laju reaksi spesifik atau konstanta laju

reaksi adalah

kP = kC (RT) 1 - n ; n = a + b

Untuk reaksi fasa gas komposisi dalam persamaan

laju reaksi umumnya dinyatakan dalam bentuk

tekanan parsial dari komponen yang bereaksi.

Pada berbagai situasi biasanya lebih baik bila

konsentrasi diubah ke tekanan parsial sehingga lebih

mudah dalam evaluasinya.

. . . (16)

. . . (17)

- rA = kC. CAa CB

b - rA = kP pAa pB

b

(mol/lt)1-n /jam

mol/lit.jam (mol/lt)n

mol/lit.atm n. jam

mol/lit. jam (atm)n

kP = kC (RT)–n mol/lit.atmn.jam

Contoh 1. Mengubah bentuk tekanan ke konsentrasi

Pada 610oK dan 1 atm, reaksi bolak balik elementer diketahui

sebagai berikut :

2 A B,

- rA = k1 pA2 – k2

* pB mol/lit.jam ( k2*selerusnya ditulis k2)

Ubah laju reaksi ini ke unit konsentrasi & tekanan

a. - rA’ = k1 CA2 – k2 CB mol/lit.jam

b. - rA* = k1”CA2 – k2” CB atm/jam

Data diketahui : 1. k1 ; 2. KP

3. T ; 4.

k1 = 10 -3 mol/lt.atm2 jam

Kp = 0,5 atm -1

1

2

Solution :

a.) diketahui - rA = k1 pA2 – k2 pB (mole/lit.jam)

diingini bentuk - rA = k1* CA2 – k2* CB (mole/liter.jam)

jadi disini bentuk dari – rA tetap sebagai – dCA/dt mole/liter.jam

atau yang akan diubah hanya bentuk pA dan pB ke CA dan CB

Diketahui : pA = CA.RT

mol mol mol

- rA = (10-3 ------------- )( pA2 atm2) – ( k2 --------------)(pB atm) ---------

lit.atm2.jam lit.atm.jam lit.jam

mole

k1 pA2 = (10-3 -------------- ) { ( CART)2 } (atm)2

lit atm2 jam

mole liter.atm

CA.RT = ----------- ---------------- oK = atm

liter mole. oK

Jadi :

k1.pA2 = ( 10-3 mole/lt.atm2.jam ) [ CA

2 ( mole/lt)2] [ (0,082 )(610) ]2

(liter.atm/mole)2

= (10 –3)(0,082)2(610)2 CA2 mole/liter.jam

= 2,5 CA2 mole/liter.jam

k2.pB = ( 2 x 10-3 ------------ ) ( pB atm )

k2 pB = ( 2 x 10 –3 )( CB RT)

= ( 2 x 10 –3 )(RT) CB

= k2” CB

k2” = (2 x 10 –3 ------------ )(0,082 ----------- )( 610 K)

= 0,1 jam -1

mol

lit atm jam

lit.atm

mol.oK

mol

lit.atm.jam

25

KP = 0,05 atm –1 = ----- or k2 = ----- = ------------------------- 10-3 mol/lit.atm2.jam

0,5 atm -1

k1 k1

k2 KP

Persamaan yang diingini menjadi :

- rA’ = 2,5 CA2 – 0,1CB mol/lt.jam

b. Diingini : - rA* dalam atm/jam

dari :

dCA

- rA = ------- = k1.pA2 - k2 pB mol/lit.jam

dt

dCA 1 dpA

- rA = ------- = ----- --------- ( mol/lit.jam)

dt RT dt

- rA = -------- = ----- --------- = k1.pA2 – k2.pB mole/lit.jam

- rA* = ------- = k1 (RT) pA2 - k2 (RT) pB atm/jam

- dCA

dt

-1 d pA

RT dt

- dPA

dt

k1RTpA2 = (10-3 mol/lit.atm2.jam)(0,082 lit.atm/mol. K) (610 K) pA

2 atm2

= 0,05 . pA2 atm/jam

k2 RT pB = (2 x 10 –3mol/lit.atm.jam )(0,082 lit.atm/mol. K)(610 K) pB

= 10 –1 pB atm/jam

Persamaan laju reaksi menjadi :

- rA* = 0,05 pA2 - 0,1 pB atm/jam

24

Contoh 2. Mengubah tekanan ke konsentrasi

Reaksi dengan persamaan stoikiometeri : A R (aq)

dengan persamaan laju reaksi : - rA = 0,5 CA2 mol/lit.det

terjadi pada 100oC & 1 atm.

Ubah bentuk ini ke bentuk unit konsentrasi dan tekanan

a). - rA* = kP pA2 mol/lit.det

b). - rA” = kP pA2 atm/ det

Solution :

a), - rA = - rA* mol/lit.det or 0,5 CA2 = kP pA

2 mol/lit.det

(0,5)(pA/RT)2 = kP pA2 mol/lit.det

{ 0,5 / (RT)2 } pA2 = kP pA

2 mol/lit.det

- rA = (0,5)/{(0,082 lit.atm/mol. K)(373 K)}2 pA2 atm2

-rA* = 5x10 - 4 pA2 = kP pA

2

kP = 5 x 10-4 mol/lit.atm2 det

25

b). - rA’ = -dpA/dt atm/det

- rA = - dCA/dt = 0,5 CA2

- rA = - d(pA/RT)/dt = 0,5 (pA/RT)2

- rA = - (RT) –1 dpA/dt = 0,5 (RT) –2 pA2

- rA’ = - dpA/dt = { 0,5 (RT) 1 – 2 } pA2 atm/det

- rA’ = - dpA/dt = [0,5/{(0,082)(373)} –1] pA2 atm/det

- rA’ = - dpA/dt = 0,016 pA2 atm/det

where kP’ = 0,016 (atm.det) -1

89

Contoh 10 : Order of reaction

Hinshelwood dan Askey mempelajari reaksi phase gas

dari dekomposisi dimethylether dengan mengukur

kenaikan tekanan :

Reaksi

(CH3)2O CH4 + H2 + CO

Data diperoleh pada 504oC dan tekanan awal ether 312 mm, adalah :

Kenaikan tekanan,mm : 96 176 250 467 619

Waktu, sec : 390 777 1197 3155

Hitunglah orde reaksi dan konstanta laju reaksi k.

Solution :

• Reaksi : A R + S + T

• Assumsi : reaksi orde satu

90

- dCA/dt = k.CA

atau - dpA/dt = k.(RT)1-1 pA

- ln (pA/pAO) = k.t

aDari Eq. (2) pA = pAo - ----- ( - O ) n

reactant A murni atau pAO = O

Jadi a

pA = O - ------ ( - O ) n

1 pA = 312 - ---- ( )

3-1

pA = 312 – ½ ( )

Pada t = 390 sec, pA = 312 – ½ (96) = 264 mm

Begitu juga untuk data yang lain :

91

Diperoleh pada berbagai t.

t pA pA/pAo - ln (pA/pAo)

390 96 264 0.843

777 176 226 0.724

1197 250 187 0.599

3155 467 78,5 0.251

Apabila di plot - ln (pA/pAO) terhadap waktu (t),

menunjukkan garis lurus maka reaksi mengikuti reaksi orde

satu

slope >> k = sec -1

92

k1

k2

Contoh 11. Reaksi bolak balik

Nitrogen dan oxygen bereaksi membentuk nitrous oxide pada

2700 K pada batch reaktor volume tetap ( 0.4 dm3 ) pada

tekanan 20 atm. Feed terdiri dari 77% N2, 15% O2, dab 8% gas,

lain yang dapat dianggap sebagai inerts. Pada temperatur ini

konstanta kesetimbangan Ke adalah 0.01.

Apabila konversi 80% dari konversi kesetimbangan terjadi

pada 151 det, maka tentukan konstanta laju reaksi ke kanan.

Solution

Reaksi : N2 + O2 2 NO

atau A + B 2 R

Keadaan setimbang :

( - rA )e = 0 = k1 CACB – ½ k2CR2

93

- rAe = 0 = k1CACB – k2’ CR2 . . . (i)

CR

0 = k1 CACB - ------ . . . (ii) Ke

2

Batch reaktor volume tetap : A = 0, and

CA = CAo (1 – XA) . . . (iii)

CB = CAO ( M – XA) . . . (iv)

CR = CRo + 2CAoXA = 2CAo + XA . . . (v)

Dari Eq (ii) dan Eq(v) diperoleh

4Xe2 = Ke(1 – Xe)(M – Xe) . . . (vi)

4 ----- - 1 Xe

2 + (1 + M) Xe - M = 0 . . . (vii) Ke

M = CBo/CAo = 0.15/0.77 = 0.195

Substitusi Ke = 0.01 dan M dalam Eq (vii), maka :

94

399 Xe2 + 1.195 Xe - 0.195 = 0

Atau Xe = 0.02

Untuk 80% dari konversi kesetimbangan :

XA = (0,8)(0.02 ) = 0.016

Laju reaksi :

- rA = -------- = k1 CACB – k2’ CR2

CAo dXA

------------- = k1CAo2(1-XA)(M-XA) – (k1/Ke)(CAoXA)2

dt

1 dXA

t = -------- ------------------------------------ k1CAo (1 – XA)(M – XA) – 4X2/Ke

yA o (0.77)(20 atm) CAo = --------- = ---------------------------------------- RT (0.082 atm lit/mol.K)(2700 K)

- dCA

dt

XA

0

95

1 dX k1 = --------------------------------------- ------------------------------------ (151 x 10-6 s)(0.0696 mol/lit) 0.195 – 1.195XA – 399 XA

2

Dengan hukum Simpson dengan h = 0.008 to obtain

1 (X) = --------------------------------- 0.195 – 0.195 X – 399 X2

h f(X)dX = ---- [ f(Xo) + 4 f(X1) + f (X2) ]

3

ketika Xo = 0 - - - f(Xo) = 5.128 X1 = 0.008 - - - f(X1) = 6.25

X2 = 0.016 - - - f(X2) = 13.56

0.016

0

0.008k1 = 9.51 x 104 -------- 5.128 + 4(6.25) + 13.56 dm3/mol.sec 3

= 1.11 x 104 dm3/mol.sec