Roboguru

Reaksi orde tiga mempunyai satuan tetapan laju reaksi ....

Pertanyaan

Reaksi orde tiga mempunyai satuan tetapan laju reaksi ....blank subscript blank

  1. detik to the power of negative sign 1 end exponent 

  2. mol to the power of negative sign 1 end exponent space dm to the power of negative sign 3 end exponent space detik to the power of negative sign 1 end exponent 

  3. mol to the power of negative sign 2 end exponent space dm to the power of 6 space detik to the power of negative sign 1 end exponent 

  4. mol squared space dm to the power of negative sign 6 end exponent space detik to the power of negative sign 1 end exponent 

  5. mol to the power of negative sign 3 end exponent space dm to the power of 9 space detik minus sign 1 

Pembahasan Soal:

Reaksi orde ketiga menunjukkan hubungan antara besarnya pengaruh konsentrasi reaktan terhadap laju reaksi. Laju reaksi berbanding lurus dengan pangkat tiga konsnetrasi reaktan. Pada orde ketiga, laju reaksi dirumuskan sebagai v double bond k open square brackets Reaktan close square brackets cubed . Tetapan laju reaksi (k) memiliki satuan yang berbeda-beda yang tergantung pada orde reaksi. Penentuan satuan k dapat diturunkan berdasarkan rumus laju reaksi berorde tiga yaitu sebagai berikut.

  • Laju reaksi memiliki satuan M space s to the power of negative sign 1 end exponent atau mol space dm to the power of negative sign 3 end exponent space detik to the power of negative sign 1 end exponent.
  • Konsentrasi reaktan memiliki satuan Molar atau mol space dm to the power of negative sign 3 end exponent.
  • Satuan tetapan laju reaksi adalah 

table attributes columnalign right center left columnspacing 0px end attributes row v equals cell k open square brackets Reaktan close square brackets cubed end cell row cell mol space dm to the power of negative sign 3 end exponent space detik to the power of negative sign 1 end exponent end cell equals cell k left parenthesis mol space dm to the power of negative sign 3 end exponent right parenthesis cubed end cell row cell mol space dm to the power of negative sign 3 end exponent space detik to the power of negative sign 1 end exponent end cell equals cell k cross times mol cubed space dm to the power of negative sign 9 end exponent end cell row k equals cell fraction numerator mol space dm to the power of negative sign 3 end exponent space detik to the power of negative sign 1 end exponent over denominator mol cubed space dm to the power of negative sign 9 end exponent end fraction end cell row k equals cell mol to the power of 1 minus sign 3 end exponent space dm to the power of left parenthesis minus sign 3 minus sign left parenthesis minus sign 9 right parenthesis right parenthesis end exponent detik to the power of negative sign 1 end exponent end cell row k equals cell mol to the power of negative sign 2 end exponent space dm to the power of 6 space detik to the power of negative sign 1 end exponent end cell end table   

Dengan demikian, satuan tetapan laju reaksi berorde tiga adalah mol to the power of negative sign 2 end exponent space dm to the power of 6 space detik to the power of negative sign 1 end exponent .

Jadi, jawaban yang benar adalah C.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

S. Lubis

Mahasiswa/Alumni Universitas Sumatera Utara

Terakhir diupdate 30 Agustus 2021

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Pertanyaan yang serupa

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

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

0

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

Gas nitrogen oksida dan gas bromin bereaksi pada  menurut persamaan reaksi berikut:     Laju reaksi diikuti dengan mengukur pertambahan konsentrasi NOBr dan diperoleh data sebagai berikut:     ...

Pembahasan Soal:

a. Orde reaksi terhadap NO, dengan perbandingan data 4 banding data 3 yaitu:
 

table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets N O close square brackets to the power of italic x open square brackets Br subscript 2 close square brackets to the power of italic y end cell equals v row cell open square brackets fraction numerator 0 comma 3 over denominator 0 comma 2 end fraction close square brackets to the power of x open square brackets fraction numerator 0 comma 1 over denominator 0 comma 1 end fraction close square brackets to the power of y end cell equals cell 108 over 48 end cell row cell open square brackets 3 over 2 close square brackets to the power of x end cell equals cell 9 over 4 end cell row cell open square brackets 3 over 2 close square brackets to the power of x end cell equals cell open parentheses 3 over 2 close parentheses squared end cell row x equals 2 end table
 

b. Orde reaksi terhadap Br subscript 2, dengan membandingkan data 2 dan 1, yaitu:
 

table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets N O close square brackets to the power of italic x open square brackets Br subscript 2 close square brackets to the power of italic y end cell equals v row cell open square brackets fraction numerator 0 comma 1 over denominator 0 comma 1 end fraction close square brackets to the power of x open square brackets fraction numerator 0 comma 2 over denominator 0 comma 1 end fraction close square brackets to the power of y end cell equals cell 24 over 12 end cell row cell open square brackets 2 close square brackets to the power of x end cell equals cell 2 to the power of 1 end cell row x equals 1 end table
 

c. Persamaan laju reaksi nya yaitu: V double bond k open square brackets N O close square brackets squared open square brackets Br subscript 2 close square brackets

d. Orde reaksi totalnya yaitu: x+y = 2 + 1 = 3

e. Jika menggunakan data nomor 1, maka nilai tetapan jenis(k) yaitu:

V double bond k open square brackets N O close square brackets squared open square brackets Br subscript 2 close square brackets 12 space M forward slash s double bond k open square brackets 0 comma 1 space M close square brackets squared open square brackets 0 comma 1 space M close square brackets k equals fraction numerator 12 space M forward slash s over denominator 0 comma 001 space M cubed end fraction k equals 12 cross times 10 cubed space M to the power of negative sign 2 end exponent middle dot s to the power of negative sign 1 end exponent
 

f. Laju reaksi saat konsentrasi pereaksi masing-masing 0,4 M, yaitu:space

V double bond k open square brackets N O close square brackets squared open square brackets Br subscript 2 close square brackets V equals 12 cross times 10 cubed open square brackets 0 comma 4 close square brackets squared open square brackets 0 comma 4 close square brackets V equals 12 cross times 10 cubed open square brackets 0 comma 16 close square brackets open square brackets 0 comma 4 close square brackets V equals 0 comma 768 cross times 10 cubed V equals 768 space M middle dot s to the power of negative sign 1 end exponent

Jadi orde reaksi, persamaan laju, dan konstata laju seperti diuraikan diatas

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 

0

Roboguru

Pada reaksi: , diperoleh data sebagai berikut.    Tentukan: a. orde reaksi terhadap P, b. orde reaksi terhadap Q, c. orde reaksi total, d. rumus laju reaksi, serta e. laju reaksi jika  dan  m...

Pembahasan Soal:

Hukum laju reaksi menyatakan sebuah persamaan yang memperlihatkan keterkaitan atau hubungan antara laju reaksi tertentu dengan konsentrasi pereaksinya.

Berdasarkan persamaan reaksi P open parentheses italic g close parentheses and Q open parentheses italic g close parentheses yields zat space hasil, maka persamaan laju reaksi dapat ditulis menjadi:

r double bond k open square brackets P close square brackets to the power of m open square brackets Q close square brackets to the power of n 

dimana m dan n adalah orde masing-masing reaktan.

Penentuan orde reaksi masing-masing reaktan dapat dicari berdasarkan data eksperimen dengan cara membandingkan data laju reaksi dan konsentrasi reaktan yang diketahui. Konsep yang perlu kita gunakan juga adalah hubungan waktu (t) terhadap laju reaksi (r).

r almost equal to 1 over t 

Langkah 1: Tentukan orde X dan orde Y.

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell bold Orde bold space P space... space perc point space left parenthesis 1 right parenthesis space dan space left parenthesis 2 right parenthesis end cell row cell k subscript 1 over k subscript 2 open square brackets P subscript 1 close square brackets to the power of m over open square brackets P subscript 2 close square brackets to the power of m open square brackets Q subscript 1 close square brackets to the power of n over open square brackets Q subscript 2 close square brackets to the power of n end cell equals cell r subscript 1 over r subscript 2 end cell row cell k subscript 1 over k subscript 2 open square brackets P subscript 1 close square brackets to the power of m over open square brackets P subscript 2 close square brackets to the power of m open square brackets Q subscript 1 close square brackets to the power of n over open square brackets Q subscript 2 close square brackets to the power of n end cell equals cell fraction numerator 1 over t subscript 1 over denominator 1 over t subscript 2 end fraction end cell row cell open parentheses fraction numerator 0 comma 1 over denominator 0 comma 2 end fraction close parentheses to the power of m open parentheses fraction numerator 0 comma 1 over denominator 0 comma 1 end fraction close parentheses end cell equals cell fraction numerator 1 over 80 over denominator 1 over 40 end fraction end cell row cell open parentheses 1 half close parentheses to the power of m end cell equals cell 40 over 80 end cell row cell open parentheses 1 half close parentheses to the power of m end cell equals cell open parentheses 1 half close parentheses to the power of 1 end cell row m equals 1 row blank blank blank row blank blank cell bold Orde bold space italic Q space... space perc point space left parenthesis 2 right parenthesis space dan space left parenthesis 3 right parenthesis space left parenthesis Substitusi space m equals 1 right parenthesis end cell row cell fraction numerator k subscript 2 over denominator k 3 end fraction open square brackets X subscript 2 close square brackets to the power of m over open square brackets X subscript 3 close square brackets to the power of m open square brackets Y subscript 2 close square brackets to the power of n over open square brackets Y subscript 3 close square brackets to the power of n end cell equals cell r subscript 2 over r subscript 3 end cell row cell fraction numerator k subscript 2 over denominator k 3 end fraction open square brackets P subscript 2 close square brackets to the power of m over open square brackets P subscript 3 close square brackets to the power of m open square brackets Q subscript 2 close square brackets to the power of n over open square brackets Q subscript 3 close square brackets to the power of n end cell equals cell fraction numerator 1 over t subscript 2 over denominator 1 over t subscript 3 end fraction end cell row cell open parentheses fraction numerator 0 comma 2 over denominator 0 comma 4 end fraction close parentheses to the power of 1 open parentheses fraction numerator 0 comma 1 over denominator 0 comma 2 end fraction close parentheses to the power of n end cell equals cell fraction numerator 1 over 40 over denominator 1 fifth end fraction end cell row cell open parentheses 1 half close parentheses open parentheses 1 half close parentheses to the power of n end cell equals cell 5 over 40 end cell row cell open parentheses 1 half close parentheses to the power of n end cell equals cell 5 over 40 division sign 1 half end cell row cell open parentheses 1 half close parentheses to the power of n end cell equals cell 5 over 40 cross times 2 over 1 end cell row cell open parentheses 1 half close parentheses to the power of n end cell equals cell 10 over 40 end cell row cell open parentheses 1 half close parentheses to the power of n end cell equals cell 1 fourth end cell row cell open parentheses 1 half close parentheses to the power of n end cell equals cell open parentheses 1 half close parentheses squared end cell row n equals 2 end table 

Langkah 2: Tentukan orde total.

table attributes columnalign right center left columnspacing 0px end attributes row cell orde space total end cell equals cell orde space P and orde space Q end cell row cell orde space total end cell equals cell 1 plus 2 end cell row cell orde space total end cell equals 3 end table  

Langkah 3: Tentukan rumus laju reaksi. Substitusi nilai m = 1 dan n = 2 ke dalam rumus laju reaksi.

table attributes columnalign right center left columnspacing 0px end attributes row r equals cell k open square brackets P close square brackets to the power of m open square brackets Q close square brackets to the power of n end cell row r equals cell k open square brackets P close square brackets to the power of 1 open square brackets Q close square brackets squared end cell row r equals cell k open square brackets P close square brackets open square brackets Q close square brackets squared end cell end table    

Langkah 4: Tentukan laju reaksi jika open square brackets P close square brackets dan open square brackets Q close square brackets masing-masing dinaikkan 3 kali.

table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets P close square brackets subscript akhir end cell equals cell 3 cross times open square brackets P close square brackets subscript awal end cell row cell open square brackets P close square brackets subscript akhir end cell equals cell 3 open square brackets P close square brackets end cell row cell open square brackets Q close square brackets subscript akhir end cell equals cell 3 cross times open square brackets Q close square brackets subscript awal end cell row cell open square brackets Q close square brackets subscript akhir end cell equals cell 3 open square brackets Q close square brackets end cell row cell laju space akhir space open parentheses r subscript 1 close parentheses end cell equals cell... ? end cell row blank blank blank row cell r subscript 1 over r end cell equals cell fraction numerator k open square brackets P close square brackets subscript akhir open square brackets Q close square brackets subscript akhir over denominator k open square brackets P close square brackets open square brackets Q close square brackets end fraction end cell row cell r subscript 1 over r end cell equals cell fraction numerator k cross times 3 open square brackets X close square brackets cross times 3 open square brackets Y close square brackets over denominator k cross times open square brackets X close square brackets cross times open square brackets Y close square brackets end fraction end cell row cell r subscript 1 over r end cell equals cell fraction numerator 3 cross times 3 over denominator 1 end fraction end cell row cell r subscript 1 over r end cell equals cell 9 over 1 end cell row cell r subscript 1 end cell equals cell 9 cross times r end cell end table   


Dengan demikian, diperoleh hasil bahwa:

a. orde reaksi terhadap P = 1.

b. orde reaksi terhadap Q = 2.

c. orde total = 3.

d. rumus laju reaksi, r double bond k open square brackets P close square brackets open square brackets Q close square brackets squared.

e. laju reaksi jika open square brackets P close square brackets dan open square brackets Q close square brackets masing-masing dinaikkan 3 kali menjadi 9 kali lebih cepat dibandingkan laju awal. 

Jadi, orde P, orde Q, orde total, rumus laju reaksi, dan perubahan laju jika konsentrasi dinaikkan 3 kali berturut-turut adalah 1, 2, 3, r double bond k open square brackets P close square brackets open square brackets Q close square brackets squared, dan 9 kali lebih cepat dibandingkan laju awal. 

0

Roboguru

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