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Berikut ini adalah data hasil percobaan laju reaksi dari reaksi:     Reaksi tersebut mempunyai tetapan laju reaksi sebesar ....

Pertanyaan

Berikut ini adalah data hasil percobaan laju reaksi dari reaksi:

2 N O open parentheses italic g close parentheses and 2 H subscript 2 open parentheses italic g close parentheses yields N subscript 2 open parentheses italic g close parentheses and 2 H subscript 2 O open parentheses italic g close parentheses
 


 

Reaksi tersebut mempunyai tetapan laju reaksi sebesar ....space

  1. 0,2space

  2. 2space

  3. 20space

  4. 100space

  5. 200space

Pembahasan Video:

Pembahasan Soal:

Persamaan laju reaksi awal adalah r=k[NO]x[H2]y

  • Menentukan orde reaksi terhadap NO
    Dengan menggunakan data dari percobaan 3 dan 4.
    r3r4=k[NO]3x[H2]3yk[NO]4x[H2]4y0,52=k(0,1)x(0,25)yk(0,2)x(0,25)y4=(2)xx=2

    Berdasarkan perhitungan di atas, orde reaksi terhadap NO adalah 2.
     
  • Menentukan orde reaksi terhadap H subscript 2
    Dengan menggunakan data dari percobaan 1 dan 2.
    r1r2=k[NO]1x[H2]1yk[NO]2x[H2]2y1,64,8=k(0,3)x(0,05)yk(0,3)x(0,15)y3=(3)yy=1

    Berdasarkan perhitungan di atas, orde reaksi terhadap NO adalah 1.
     
  • Menentukan nilai k
    Persamaan laju reaksinya adalah r=k[NO]2[H2].
    Dengan menggunakan data percobaan 4, maka
    r2molL1s12molL1s12molL1s1kk======k[NO]2[H2]k(0,2molL1)2(0,25molL1)k(0,04mol2L2)(0,25molL1)k(0,01mol3L3)0,01mol3L32molL1s1200mol2L2s1

Dengan demikian, maka tetapan laju reaksi tersebut adalah 200mol2L2s1.

Jadi, jawaban yang benar adalah E.space

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

N. Puspita

Terakhir diupdate 21 September 2021

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Berikut ini diberikan data percobaan laju reaksi:   pada beberapa kondisi:     Jika  masing-masing diubah menjadi 0,5 M, maka harga laju reaksi  adalah ...

Pembahasan Soal:

Mencari orde reaksi Q

Dimisalkan, persamaan laju reaksi begin mathsize 14px style V equals italic k open square brackets Q close square brackets to the power of italic x open square brackets T close square brackets to the power of y end style. Cara mencari orde reaksi Q (x) adalah dengan memilih data konsentrasi T yang sama yaitu data (1) dan (2). 

begin mathsize 14px style V subscript left parenthesis 1 right parenthesis end subscript over V subscript left parenthesis 2 right parenthesis end subscript equals italic k subscript left parenthesis 1 right parenthesis end subscript over italic k subscript left parenthesis 2 right parenthesis end subscript cross times open square brackets italic Q subscript left parenthesis 1 right parenthesis end subscript over italic Q subscript left parenthesis 2 right parenthesis end subscript close square brackets to the power of italic x cross times open square brackets italic T subscript left parenthesis 1 right parenthesis end subscript over italic T subscript left parenthesis 2 right parenthesis end subscript close square brackets to the power of italic y end style, harga begin mathsize 14px style italic k subscript left parenthesis 1 right parenthesis end subscript equals italic k subscript left parenthesis 2 right parenthesis end subscript end style (karena suhu tetap) dan begin mathsize 14px style open square brackets T subscript left parenthesis 1 right parenthesis end subscript close square brackets equals open square brackets T subscript left parenthesis 2 right parenthesis end subscript close square brackets end style sehingga begin mathsize 14px style open square brackets T subscript left parenthesis 1 right parenthesis end subscript over T subscript left parenthesis 2 right parenthesis end subscript close square brackets to the power of italic y end style dapat dihilangkan (bernilai 1). 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 1 comma 25 cross times 10 to the power of negative sign 2 end exponent space Ms to the power of negative sign 1 end exponent over denominator 5 comma 00 cross times 10 to the power of negative sign 2 end exponent space Ms to the power of negative sign 1 end exponent end fraction end cell equals cell open square brackets fraction numerator 0 comma 1 over denominator 0 comma 2 end fraction close square brackets to the power of italic x end cell row cell 1 fourth end cell equals cell open square brackets 1 half close square brackets to the power of italic x end cell row italic x equals 2 end table end style 

Jadi orde reaksi terhadap Q adalah 2.
 

Mencari orde reaksi T

Untuk menentukan orde reaksi T (y)  dipilih data konsentrasi Q yang sama yaitu data (1) dan (3).

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell V subscript left parenthesis 1 right parenthesis end subscript over V subscript left parenthesis 3 right parenthesis end subscript end cell equals cell italic k subscript left parenthesis 1 right parenthesis end subscript over italic k subscript left parenthesis 3 right parenthesis end subscript cross times open square brackets italic Q subscript left parenthesis 1 right parenthesis end subscript over italic Q subscript left parenthesis 3 right parenthesis end subscript close square brackets to the power of italic x cross times open square brackets italic T subscript left parenthesis 1 right parenthesis end subscript over italic T subscript left parenthesis 3 right parenthesis end subscript right square bracket close square brackets to the power of italic y end cell row cell fraction numerator 1 comma 25 cross times 10 to the power of negative sign 2 end exponent space Ms to the power of negative sign 1 end exponent over denominator 1 x 10 to the power of negative sign 1 end exponent space Ms to the power of negative sign 1 end exponent end fraction end cell equals cell fraction numerator 0 comma 1 over denominator 0 comma 1 end fraction cross times open square brackets 1 half close square brackets to the power of y end cell row cell 1 over 8 end cell equals cell open square brackets 1 half close square brackets to the power of y end cell row y equals 3 end table end style   

Jadi orde reaksi terhadap T adalah 3.
 

Mencari nilai k

Pesamaan laju reaksi pembentukan begin mathsize 14px style T subscript 2 Q end style dapat dituliskan sebagai berikut :

begin mathsize 14px style V double bond k open square brackets Q close square brackets squared open square brackets T close square brackets cubed end style 

Untuk menentukan harga konstanta laju reaksi k maka dapat memasukan data percobaan di atas misalkan pada data percobaan (1), maka diperoleh:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row V equals cell italic k open square brackets Q close square brackets squared open square brackets T close square brackets cubed end cell row cell 1 comma 25 cross times 10 to the power of negative sign 2 end exponent space Ms to the power of negative sign 1 end exponent end cell equals cell italic k cross times left parenthesis 10 to the power of negative sign 1 end exponent right parenthesis squared cross times left parenthesis 10 to the power of negative sign 1 end exponent right parenthesis cubed end cell row italic k equals cell fraction numerator 1 comma 25 cross times 10 to the power of negative sign 2 end exponent space Ms to the power of negative sign 1 end exponent over denominator 10 to the power of negative sign 5 end exponent end fraction end cell row italic k equals cell 1 comma 25 cross times 10 cubed end cell end table end style 

Jadi harga konstanta laju reaksi adalah begin mathsize 14px style bold 1 bold comma bold 25 bold cross times bold 10 to the power of bold 3 end style.
 

Menentukan laju reaksi pada Q 0,5 M dan T 0,5 M  

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row V equals cell italic k open square brackets Q close square brackets squared open square brackets T close square brackets cubed end cell row V equals cell 1 comma 25 cross times 10 cubed cross times open square brackets 5 cross times 10 to the power of negative sign 1 end exponent space M close square brackets squared cross times left square bracket 5 cross times 10 to the power of negative sign 1 end exponent space M right square bracket cubed space end cell row V equals cell 1 comma 25 cross times 10 cubed cross times 5 cross times 10 to the power of negative sign 2 end exponent cross times 5 cross times 10 to the power of negative sign 3 end exponent end cell row V equals cell 3 comma 125 cross times 10 to the power of negative sign 1 end exponent space Ms to the power of negative sign 1 end exponent end cell end table end style 

Jadi laju reaksi pada Q 0,5 M dan T 0,5 M adalah begin mathsize 14px style bold 3 bold comma bold 125 bold cross times bold 10 to the power of bold minus sign bold 1 end exponent bold space bold Ms to the power of bold minus sign bold 1 end exponent end style.space 

0

Roboguru

Data eksperimen untuk reaksi :     Tentukan: a. orde reaksi total b. persamaan laju reaksi c. nilai konstanta (k) d. tentukan laju reaksi jika konsentrasi gas A = 2 dan B = 4

Pembahasan Soal:

a. Orde reaksi total

    Berikut adalah langkah-langkah mencari orde reaksi total:

  • Mencari orde reaksi A menggunakan data percobaan 1 dan 4

    Error converting from MathML to accessible text.  
     
  • Mencari orde reaksi B menggunakan data percobaan 2 dan 3

    begin mathsize 14px style v subscript 3 over v subscript 2 equals fraction numerator k space open square brackets A close square brackets subscript 3 superscript x space open square brackets B close square brackets subscript 3 superscript y over denominator k space open square brackets A close square brackets subscript 2 superscript x space open square brackets B close square brackets subscript 2 superscript y end fraction 18 over 12 equals fraction numerator up diagonal strike k middle dot up diagonal strike left parenthesis 0 comma 1 right parenthesis to the power of x end strike middle dot left parenthesis 0 comma 3 right parenthesis to the power of y over denominator up diagonal strike k middle dot up diagonal strike left parenthesis 0 comma 1 right parenthesis to the power of x end strike middle dot left parenthesis 0 comma 2 right parenthesis to the power of y end fraction 3 over 2 equals open parentheses 3 over 2 close parentheses to the power of y sehingga comma space y equals 1 end style 
     

   Jadi, orde reaksi total = 3.


b. Persamaan laju reaksi 

    Setelah menemukan orde reaksi dari A dan B, maka persamaan reaksinya adalah v = k [A]2 [B] 


c. Nilai konstanta

    Untuk mencari nilai konstanta dapat menggunakan salah satu data dari percobaan. Berikut adalah perhitungannya:


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row v equals cell k middle dot open square brackets A close square brackets squared middle dot open square brackets B close square brackets end cell row 6 equals cell k middle dot open parentheses 0 comma 1 close parentheses squared middle dot 0 comma 1 end cell row 6 equals cell k middle dot 0 comma 001 end cell row 6000 equals k end table end style 
 

d. Nilai laju reaksi jika konsentrasi gas A = 2 dan B = 4

   Setelah menemukan orde reaksi, persamaan laju reaksi dan nilai konstanta laju reaksi, maka nilai laju reaksi pada konsentrasi gas A = 2 dan B = 4 adalah


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row v equals cell k middle dot open square brackets A close square brackets squared middle dot open square brackets B close square brackets end cell row blank equals cell 6000 middle dot left parenthesis 2 right parenthesis squared middle dot 4 end cell row blank equals cell 96000 space M space det to the power of negative sign 1 end exponent end cell end table end style 


Jadi, orde reaksi total = 3, persamaan laju reaksi = v = k [A]2 [B], nilai konstanta = 6000, dan nilai laju reaksi pada konsentrasi gas A = 2 dan B = 4 adalah 96000 M det-1.space 

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Roboguru

Reaksi antara ion bromat dan ion bromida dalam larutan asam dinyatakan dengan persamaan reaksi:   Campuran reaksi dibuat melalui pencampuran larutan-Iarutan induk dengan volume dan Iaju awal berkurang...

Pembahasan Soal:

Volume total semua percobaan adalah 3 mL. v subscript 1 over v subscript 2 equals fraction numerator italic k space open square brackets Br to the power of minus sign close square brackets subscript 1 to the power of a open square brackets Br O subscript 3 to the power of minus sign close square brackets subscript 1 to the power of b open square brackets H subscript 3 O to the power of plus sign close square brackets subscript 1 to the power of c over denominator italic k space open square brackets Br to the power of minus sign close square brackets subscript 2 to the power of a open square brackets Br O subscript 3 to the power of minus sign close square brackets subscript 2 to the power of b open square brackets H subscript 3 O to the power of plus sign close square brackets subscript 2 to the power of c end fraction v subscript 1 over v subscript 2 equals fraction numerator italic k space open parentheses begin display style fraction numerator horizontal strike n subscript Br to the power of minus sign end subscript end strike cross times V subscript Br to the power of minus sign end subscript over denominator horizontal strike V subscript tot end strike end fraction end style close parentheses subscript 1 to the power of a open parentheses fraction numerator horizontal strike n subscript Br O subscript 3 to the power of minus sign end subscript end strike cross times V subscript Br O subscript 3 to the power of minus sign end subscript over denominator horizontal strike V subscript tot end strike end fraction close parentheses subscript 1 to the power of b open parentheses fraction numerator horizontal strike n subscript H subscript 3 O to the power of plus sign to the power of minus sign end subscript end strike cross times V subscript H subscript 3 O to the power of plus sign end subscript over denominator horizontal strike V subscript tot end strike end fraction close parentheses subscript 1 to the power of c over denominator italic k space open parentheses fraction numerator horizontal strike n subscript Br to the power of minus sign end subscript end strike cross times V subscript Br to the power of minus sign end subscript over denominator horizontal strike V subscript tot end strike end fraction close parentheses subscript 2 to the power of a open parentheses fraction numerator horizontal strike n subscript Br O subscript 3 to the power of minus sign end subscript end strike cross times V subscript Br O subscript 3 to the power of minus sign end subscript over denominator horizontal strike V subscript tot end strike end fraction close parentheses subscript 2 to the power of b open parentheses fraction numerator horizontal strike n subscript H subscript 3 O to the power of plus sign to the power of minus sign end subscript end strike cross times V subscript H subscript 3 O to the power of plus sign end subscript over denominator horizontal strike V subscript tot end strike end fraction close parentheses subscript 2 to the power of c end fraction v subscript 1 over v subscript 2 equals fraction numerator open parentheses V subscript Br to the power of minus sign end subscript close parentheses subscript 1 to the power of a open parentheses V subscript Br O subscript 3 to the power of minus sign end subscript close parentheses subscript 1 to the power of b open parentheses V subscript H subscript 3 O to the power of plus sign end subscript close parentheses subscript 1 to the power of c over denominator open parentheses V subscript Br to the power of minus sign end subscript close parentheses subscript 2 to the power of a open parentheses V subscript Br O subscript 3 to the power of minus sign end subscript close parentheses subscript 2 to the power of b open parentheses V subscript H subscript 3 O to the power of plus sign end subscript close parentheses subscript 2 to the power of c end fraction fraction numerator 5 comma 63 cross times 10 to the power of negative sign 6 end exponent over denominator 1 comma 09 cross times 10 to the power of negative sign 5 end exponent end fraction equals fraction numerator open parentheses 0 comma 1 close parentheses to the power of a horizontal strike open parentheses 0 comma 5 close parentheses to the power of b open parentheses 1 close parentheses to the power of c end strike over denominator open parentheses 0 comma 2 close parentheses to the power of a horizontal strike open parentheses 0 comma 5 close parentheses to the power of b open parentheses 1 close parentheses to the power of c end strike end fraction 1 half equals open parentheses 1 half close parentheses to the power of a a equals 1

v subscript 1 over v subscript 3 equals fraction numerator open parentheses V subscript Br to the power of minus sign end subscript close parentheses subscript 1 to the power of a open parentheses V subscript Br O subscript 3 to the power of minus sign end subscript close parentheses subscript 1 to the power of b open parentheses V subscript H subscript 3 O to the power of plus sign end subscript close parentheses subscript 1 to the power of c over denominator open parentheses V subscript Br to the power of minus sign end subscript close parentheses subscript 3 to the power of a open parentheses V subscript Br O subscript 3 to the power of minus sign end subscript close parentheses subscript 3 to the power of b open parentheses V subscript H subscript 3 O to the power of plus sign end subscript close parentheses subscript 3 to the power of c end fraction fraction numerator 5 comma 63 cross times 10 to the power of negative sign 6 end exponent over denominator 1 comma 13 cross times 10 to the power of negative sign 5 end exponent end fraction equals fraction numerator open parentheses 0 comma 1 close parentheses to the power of a open parentheses 0 comma 5 close parentheses to the power of b open parentheses 1 close parentheses to the power of c over denominator open parentheses 0 comma 1 close parentheses to the power of a open parentheses 1 close parentheses to the power of b open parentheses 1 close parentheses to the power of c end fraction 1 half equals open parentheses 1 half close parentheses to the power of italic b b equals 1

v subscript 3 over v subscript 4 equals fraction numerator open parentheses V subscript Br to the power of minus sign end subscript close parentheses subscript 3 to the power of a open parentheses V subscript Br O subscript 3 to the power of minus sign end subscript close parentheses subscript 3 to the power of b open parentheses V subscript H subscript 3 O to the power of plus sign end subscript close parentheses subscript 3 to the power of c over denominator open parentheses V subscript Br to the power of minus sign end subscript close parentheses subscript 4 to the power of a open parentheses V subscript Br O subscript 3 to the power of minus sign end subscript close parentheses subscript 4 to the power of b open parentheses V subscript H subscript 3 O to the power of plus sign end subscript close parentheses subscript 4 to the power of c end fraction fraction numerator 1 comma 13 cross times 10 to the power of negative sign 5 end exponent over denominator 5 comma 50 cross times 10 to the power of negative sign 6 end exponent end fraction equals fraction numerator open parentheses 0 comma 1 close parentheses open parentheses 1 close parentheses open parentheses 1 close parentheses to the power of c over denominator open parentheses 0 comma 2 close parentheses open parentheses 0 comma 5 close parentheses open parentheses 0 comma 7 close parentheses to the power of c end fraction fraction numerator 1 over denominator 0 comma 49 end fraction equals open parentheses fraction numerator 1 over denominator 0 comma 7 end fraction close parentheses to the power of italic c c equals 2

Sehingga hukum laju reaksinya adalah v space equals k space open square brackets Br to the power of minus sign close square brackets open square brackets Br O subscript 3 to the power of minus sign close square brackets open square brackets H subscript 3 O to the power of plus sign close square brackets squared. Tetapan laju dari reaksi tersebut adalah 

 v space equals k space open square brackets Br to the power of minus sign close square brackets open square brackets Br O subscript 3 to the power of minus sign close square brackets open square brackets H subscript 3 O to the power of plus sign close square brackets squared 5 comma 63 cross times 10 to the power of negative sign 6 end exponent equals k space open parentheses fraction numerator 0 comma 1 cross times 1 comma 37 over denominator 3 end fraction close parentheses open parentheses fraction numerator 0 comma 5 cross times 7 comma 1 cross times 10 to the power of negative sign 3 end exponent over denominator 3 end fraction close parentheses open parentheses fraction numerator 1 cross times 0 comma 573 over denominator 3 end fraction close parentheses squared k space equals space 2 comma 83 space M to the power of negative sign 3 end exponent s to the power of negative sign 1 end exponent

Dengan demikian, maka jawaban yang tepat adalah sesuai penjelasan di atas.

0

Roboguru

Untuk reaksi , diperoleh data percobaan sebagai berikut: Tentukan: Orde terhadap zat  Orde terhadap zat  Orde total Persamaan laju reaksi Tetapan laju reaksinya () Tentukan laju reaksi pe...

Pembahasan Soal:

1. Untuk mencari orde reaksi pada begin mathsize 14px style P end style adalah dengan memperhatikan variabel kontrol (konsentrasi begin mathsize 14px style Q end style) pada percobaan.

Menggunakan percobaan pertama dan kedua:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses begin inline style open square brackets P close square brackets subscript 1 over open square brackets P close square brackets subscript 2 end style close parentheses to the power of x end cell equals cell begin inline style v subscript 1 over v subscript 2 end style end cell row cell open parentheses begin inline style fraction numerator 0 comma 1 over denominator 0 comma 2 end fraction end style close parentheses to the power of x end cell equals cell begin inline style 6 over 12 end style end cell row cell open parentheses begin inline style 1 half end style close parentheses to the power of x end cell equals cell begin inline style 1 half end style end cell row x equals 1 end table end style


2. Untuk mencari orde reaksi pada begin mathsize 14px style Q end style adalah dengan memperhatikan variabel kontrol (konsentrasi begin mathsize 14px style P end style) pada percobaan.

Menggunakan percobaan kedua dan ketiga:


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses begin inline style open square brackets Q close square brackets subscript 2 over open square brackets Q close square brackets subscript 3 end style close parentheses to the power of y end cell equals cell begin inline style v subscript 2 over v subscript 3 end style end cell row cell open parentheses begin inline style fraction numerator 0 comma 1 over denominator 0 comma 2 end fraction end style close parentheses to the power of y end cell equals cell begin inline style 12 over 48 end style end cell row cell open parentheses begin inline style 1 half end style close parentheses to the power of y end cell equals cell begin inline style 1 fourth end style end cell row y equals 2 end table end style


3. Orde total:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell x and y end cell equals cell 1 plus 2 end cell row blank equals 3 end table end style


4. Persamaan laju reaksi:

begin mathsize 14px style v double bond k cross times open square brackets P close square brackets to the power of 1 cross times open square brackets Q close square brackets squared end style


5. Nilai k

Mencari nilai k dapat dilakukan dengan mensubstitusikan salah satu percobaan ke dalam persamaan laju reaksi: (misal menggunakan percobaan ke 1)


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row v equals cell k cross times open square brackets P close square brackets to the power of 1 cross times open square brackets Q close square brackets squared end cell row 6 equals cell k cross times left parenthesis 0 comma 1 right parenthesis to the power of 1 cross times left parenthesis 0 comma 1 right parenthesis squared end cell row 6 equals cell k cross times 10 to the power of negative sign 3 end exponent end cell row k equals cell 6 cross times 10 cubed end cell end table end style


5. v pada percobaan keempat:


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row v equals cell k cross times open square brackets P close square brackets to the power of 1 cross times open square brackets Q close square brackets squared end cell row blank equals cell 6 cross times 10 cubed cross times left parenthesis 0 comma 3 right parenthesis to the power of 1 cross times left parenthesis 0 comma 3 right parenthesis squared end cell row blank equals cell 162 space M forward slash det end cell end table end style

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Roboguru

Pada reaksi:  diperoleh data sebagai berikut:   Tentukan orde reaksi terhadap masing-masing pereaksi. Tentukan rumus laju reaksinya. Hitung nilai tetapan laju reaksi dan satuannya.

Pembahasan Soal:

Salah satu faktor yang mempengaruhi laju reaksi adalah konsentrasi reaktan. Hubungan antara laju reaksi dengan konsentrasi reaktan dirumuskan dalam suatu persamaan laju reaksi. 

Reaksi 2 N O open parentheses italic g close parentheses and O subscript 2 open parentheses italic g close parentheses yields N subscript 2 O subscript 4 open parentheses italic g close parentheses mempunyai persamaan laju reaksi:

italic v equals italic k open square brackets N O close square brackets to the power of italic x open square brackets O subscript 2 close square brackets to the power of italic y 

dengan:

k = tetapan laju reaksi
x = orde (tingkat atau pangkat) reaksi terhadap N O  
y = orde reaksi terhadap O subscript 2  

Orde reaksi ditentukan melalui percobaan, tidak berkaitan dengan koefisien reaksi.

a.   Orde reaksi terhadap masing-masing pereaksi

  • Orde reaksi N O  

    Untuk menghitung orde reaksi N O, pilih 2 percobaan dimana O subscript 2 mempunyai konsentrasi yang sama, yaitu percobaan (2) dan (3).

    v subscript 2 over v subscript 3 equals fraction numerator italic k open square brackets N O close square brackets subscript 2 superscript italic x open square brackets O subscript 2 close square brackets subscript 2 superscript italic y over denominator italic k open square brackets N O close square brackets subscript 3 superscript italic x open square brackets O subscript 2 close square brackets subscript 3 superscript italic y end fraction fraction numerator 0 comma 02 over denominator 0 comma 08 end fraction equals fraction numerator up diagonal strike italic k open parentheses 0 comma 1 close parentheses to the power of italic x up diagonal strike open parentheses 0 comma 2 close parentheses to the power of italic y end strike over denominator up diagonal strike italic k left parenthesis 0 comma 2 right parenthesis to the power of italic x up diagonal strike left parenthesis 0 comma 2 right parenthesis to the power of italic y end strike end fraction 1 fourth equals open parentheses 1 over 1 close parentheses to the power of italic x italic x equals 2  

     
  • Orde reaksi O subscript bold 2 

    Untuk menghitung orde reaksi O subscript 2, pilih 2 percobaan dimana N O mempunyai konsentrasi yang sama, yaitu percobaan (1) dan (2).

    v subscript 1 over v subscript 2 equals fraction numerator italic k open square brackets N O close square brackets subscript 1 superscript italic x open square brackets O subscript 2 close square brackets subscript 1 superscript italic y over denominator italic k open square brackets N O close square brackets subscript 2 superscript italic x open square brackets O subscript 2 close square brackets subscript 2 superscript italic y end fraction fraction numerator 0 comma 01 over denominator 0 comma 02 end fraction equals fraction numerator up diagonal strike italic k open parentheses 0 comma 1 close parentheses to the power of italic x end strike open parentheses 0 comma 1 close parentheses to the power of italic y over denominator up diagonal strike italic k left parenthesis 0 comma 1 right parenthesis to the power of italic x end strike open parentheses 0 comma 2 close parentheses to the power of italic y end fraction 1 half equals open parentheses 1 half close parentheses to the power of italic y italic y equals 1 


Jadi, orde reaksi terhadap N O bold thin space bold dan bold space O subscript bold 2 berturut-turut adalah 2 dan 1. 


b.   Rumus laju reaksi

      italic v equals italic k open square brackets N O close square brackets to the power of italic x open square brackets O subscript 2 close square brackets to the power of italic y italic v equals italic k open square brackets N O close square brackets squared open square brackets O subscript 2 close square brackets 


Jadi, persamaan laju reaksinya adalah italic v bold equals italic k begin bold style open square brackets N O close square brackets end style to the power of bold 2 begin bold style open square brackets O subscript 2 close square brackets end style.


c.   Nilai tetapan laju reaksi (k)

      Misal kita ambil percobaan nomor (1)

      italic v equals italic k open square brackets N O close square brackets squared open square brackets O subscript 2 close square brackets italic k equals fraction numerator italic v over denominator open square brackets N O close square brackets squared open square brackets O subscript 2 close square brackets end fraction italic k equals fraction numerator 0 comma 01 space M space detik to the power of negative sign 1 end exponent over denominator open parentheses 0 comma 1 space M close parentheses squared open parentheses 0 comma 1 space M close parentheses end fraction space italic k equals 10 space M to the power of negative sign 2 end exponent space detik to the power of negative sign 1 end exponent 


Jadi, nilai tetapan lajunya adalah bold 10 bold space italic M to the power of bold minus sign bold 2 end exponent bold space bold detik to the power of bold minus sign bold 1 end exponent.space space space

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Roboguru

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