Roboguru

Perhatikan data percobaan berikut:   Berdasarkan data di atas, rumus persamaan laju dan orde reaksi totalnya adalah ....

Pertanyaan

Perhatikan data percobaan berikut:

P left parenthesis italic a italic q right parenthesis plus Q left parenthesis italic a italic q right parenthesis yields R left parenthesis italic a italic q right parenthesis 

Berdasarkan data di atas, rumus persamaan laju dan orde reaksi totalnya adalah ....

  1. italic v equals italic k open square brackets P close square brackets semicolon space 2  

  2. italic v equals italic k open square brackets P close square brackets open square brackets Q close square brackets semicolon space 2 

  3. italic v equals italic k open square brackets P close square brackets open square brackets Q close square brackets squared semicolon space 3 

  4. italic v equals italic k open square brackets P close square brackets squared open square brackets Q close square brackets semicolon space 3 

  5. italic v equals italic k open square brackets P close square brackets squared open square brackets Q close square brackets squared semicolon space 4 

Pembahasan Soal:

Persamaan laju reaksi atau hukum laju merupakan sebuah persamaan yang memperlihatkan keterkaitan atau hubungan antara laju reaksi tertentu dengan konsentrasi pereaksinya. Persamaan laju reaksi tidak dapat ditentukan dari persamaan reaksi kimia tetapi harus ditentukan melalui eksperimen. 

Jika reaksi P left parenthesis italic a italic q right parenthesis plus Q left parenthesis italic a italic q right parenthesis yields R left parenthesis italic a italic q right parenthesis maka persamaan laju reaksinya dituliskan sebagai berikut.

italic v equals italic k open square brackets P close square brackets to the power of x open square brackets Q close square brackets to the power of y 

italic k merupakan tetapan laju reaksi dan x comma y merupakan orde reaksi. Dalam menentukan orde reaksi salah satu konsentrasi diubah dan konsentrasi zat lain dibuat tetap. 

  • Orde reaksi terhadap open square brackets P close square brackets 

table attributes columnalign right center left columnspacing 0px end attributes row cell italic v subscript 2 over italic v subscript 1 end cell equals cell italic k over italic k open parentheses open square brackets P close square brackets subscript 2 over open square brackets P close square brackets subscript 1 close parentheses to the power of italic x open parentheses open square brackets Q close square brackets subscript 2 over open square brackets Q close square brackets subscript 1 close parentheses to the power of italic y end cell row cell 40 over 20 end cell equals cell italic k over italic k open parentheses fraction numerator 0 comma 2 over denominator 0 comma 1 end fraction close parentheses to the power of italic x open parentheses fraction numerator 0 comma 1 over denominator 0 comma 1 end fraction close parentheses to the power of italic y end cell row 2 equals cell 2 to the power of x end cell row x equals 1 end table  

  • Orde reaksi terhadap open square brackets Q close square brackets 

table attributes columnalign right center left columnspacing 0px end attributes row cell italic v subscript 4 over italic v subscript 1 end cell equals cell italic k over italic k open parentheses open square brackets P close square brackets subscript 4 over open square brackets P close square brackets subscript 1 close parentheses to the power of italic x open parentheses open square brackets Q close square brackets subscript 4 over open square brackets Q close square brackets subscript 1 close parentheses to the power of italic y end cell row cell 180 over 20 end cell equals cell italic k over italic k open parentheses fraction numerator 0 comma 1 over denominator 0 comma 1 end fraction close parentheses to the power of italic x open parentheses fraction numerator 0 comma 3 over denominator 0 comma 1 end fraction close parentheses to the power of italic y end cell row 9 equals cell 3 to the power of italic y end cell row y equals 2 end table 

  • Orde reaksi total = 3
  • Persamaan laju reaksi : italic v equals italic k open square brackets P close square brackets open square brackets Q close square brackets squared 

Jadi, jawaban yang benar adalah C.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

S. Lubis

Mahasiswa/Alumni Universitas Sumatera Utara

Terakhir diupdate 30 Agustus 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Waktu-waktu yang terdaftar dalam tabel diukur pada  untuk reaksi yang menghasilkan  berikut.     Berapa waktu yang diharapkan pada percobaan 4?

Pembahasan Soal:

Waktu reaksi berbanding terbalik dengan laju reaksi, jadi berdasarkan data tersebut, laju reaksi untuk masing masing percobaan yaitu:

v percobaan 1: 1 over 39 
v percobaan 2: 1 over 78 
v percobaan 3: 1 over 156 
v percobaan 4: ?

Untuk mengetahui waktu reaksi pada percobaan 4, maka harus dicari laju reaksinya terlebih dulu. Pertama, mencari orde reaksi yaitu sebagai berikut:

Menggunakan data percobaan 1 dan 2:
space space space space fraction numerator v 1 over denominator v 2 end fraction equals fraction numerator k left square bracket S subscript 2 O subscript 8 to the power of 2 minus sign end exponent right square bracket to the power of x left square bracket I to the power of negative sign end exponent right square bracket to the power of y end exponent over denominator k left square bracket S subscript 2 O subscript 8 to the power of 2 minus sign end exponent right square bracket to the power of x left square bracket I to the power of minus sign right square bracket to the power of y end fraction fraction numerator begin display style bevelled 1 over 39 end style over denominator 1 forward slash 78 end fraction equals fraction numerator k left square bracket 0 comma 0400 right square bracket to the power of x left square bracket 0 comma 0800 right square bracket to the power of y over denominator k left square bracket 0 comma 0400 right square bracket to the power of x left square bracket 0 comma 0400 right square bracket to the power of y end fraction space space space space space space space 2 equals 2 to the power of y space space space space space space space y equals 1 

Menggunakan data percobaan 1 dan 3:

fraction numerator v 1 over denominator v 2 end fraction equals fraction numerator k left square bracket S subscript 2 O subscript 8 to the power of 2 minus sign end exponent right square bracket to the power of x left square bracket I to the power of minus sign right square bracket to the power of y over denominator k left square bracket S subscript 2 O subscript 8 to the power of 2 minus sign end exponent right square bracket to the power of x left square bracket I to the power of minus sign right square bracket to the power of y end fraction fraction numerator begin display style bevelled 1 over 39 end style over denominator 1 forward slash 156 end fraction equals fraction numerator k left square bracket 0 comma 0400 right square bracket to the power of x left square bracket 0 comma 0800 right square bracket to the power of y over denominator k left square bracket 0 comma 0100 right square bracket to the power of x left square bracket 0 comma 0800 right square bracket to the power of y end fraction space space space space space space space space space space 4 equals 4 to the power of x space space space space space space space space space space x equals 1 

Jadi, orde reaksi terhadap S subscript 2 O subscript 8 to the power of 2 minus sign end exponent adalah 1 dan I to the power of minus sign adalah 1. Sehingga persamaan laju reaksinya adalah v double bond k left square bracket S subscript 2 O subscript 8 to the power of 2 minus sign end exponent right square bracket open square brackets I to the power of minus sign close square brackets 

Kemudian, mencari nilai k dengan memasukan salah satu data ke persamaan laju reaksi, yaitu sebagai berikut:

space space space space space space v double bond k left square bracket S subscript 2 O subscript 8 to the power of 2 minus sign end exponent right square bracket open square brackets I to the power of minus sign close square brackets bevelled 1 over 39 equals k left parenthesis 0 comma 0400 right parenthesis left parenthesis 0 comma 0800 right parenthesis space space space space space space k equals 8 comma 012  

Maka, laju reaksi percobaan 4 yaitu:

v double bond k left square bracket S subscript 2 O subscript 8 to the power of 2 minus sign end exponent right square bracket open square brackets I to the power of minus sign close square brackets space equals 8 comma 012 left parenthesis 0 comma 0200 right parenthesis left parenthesis 0 comma 0200 right parenthesis space equals 3 comma 21 cross times 10 to the power of negative sign 3 end exponent 

Waktu reaksi pada percobaan 4 yaitu:

t equals 1 over v space equals fraction numerator 1 over denominator 3 comma 21 cross times 10 to the power of negative sign 3 end exponent end fraction space equals 312 space detik 

Jadi, jawaban yang benar adalah C.

0

Roboguru

Berikut data percobaan laju reaksi: : Jika [A] diubah menjadi 0,3 M dan [B] menjadi 0,4 M. maka nilai laju reaksi (v) pada saat itu adalah....M/det

Pembahasan Soal:

Untuk menentukan laju reaksi dengan konsentrasi tertentu maka ditentukan orde reaksi A dan B. Untuk menentukan orde reaksi A dapat digunakan data 1 dan 2.


table attributes columnalign right center left columnspacing 0px end attributes row cell v subscript 1 over v subscript 2 end cell equals cell fraction numerator italic k open square brackets A close square brackets to the power of x open square brackets B close square brackets to the power of y over denominator italic k open square brackets A close square brackets to the power of x open square brackets B close square brackets to the power of y end fraction end cell row cell fraction numerator 1 comma 25 cross times 10 to the power of negative sign 2 end exponent over denominator 5 cross times 10 to the power of negative sign 2 end exponent end fraction end cell equals cell fraction numerator italic k left square bracket 0 comma 1 right square bracket to the power of x left square bracket 0 comma 1 right square bracket to the power of y over denominator italic k left square bracket 0 comma 2 right square bracket to the power of x left square bracket 0 comma 1 right square bracket to the power of y end fraction end cell row cell 1 fourth end cell equals cell open parentheses 1 half close parentheses to the power of x end cell row x equals 2 end table


Orde reaksi dari A adalah 2, untuk menentukan orde reaksi B maka dapat digunakan data percobaan data 1 dan 3.


table attributes columnalign right center left columnspacing 0px end attributes row cell v subscript 1 over v subscript 3 end cell equals cell fraction numerator italic k open square brackets A close square brackets to the power of x open square brackets B close square brackets to the power of y over denominator italic k open square brackets A close square brackets to the power of x open square brackets B close square brackets to the power of y end fraction end cell row cell fraction numerator 1 comma 25 cross times 10 to the power of negative sign 2 end exponent over denominator 10 to the power of negative sign 1 end exponent end fraction end cell equals cell fraction numerator italic k left square bracket 0 comma 1 right square bracket to the power of x left square bracket 0 comma 1 right square bracket to the power of y over denominator italic k left square bracket 0 comma 1 right square bracket to the power of x left square bracket 0 comma 2 right square bracket to the power of y end fraction end cell row cell 1 over 8 end cell equals cell open parentheses 1 half close parentheses to the power of italic y end cell row y equals 3 end table


Untuk menentukan reaksi berdasarkan konsentrasi yang diinginkan maka dapat digunakan perbandingan data 1 dengan laju yang diharapkan dengan orde yang telah ditentukan.


table attributes columnalign right center left columnspacing 0px end attributes row cell v over v subscript 1 end cell equals cell fraction numerator italic k open square brackets A close square brackets to the power of x open square brackets B close square brackets to the power of y over denominator italic k open square brackets A close square brackets to the power of x open square brackets B close square brackets to the power of y end fraction end cell row cell fraction numerator italic v over denominator 1 comma 25 cross times 10 to the power of negative sign 2 end exponent end fraction end cell equals cell fraction numerator italic k left square bracket 0 comma 3 right square bracket squared left square bracket 0 comma 4 right square bracket cubed over denominator italic k left square bracket 0 comma 1 right square bracket squared left square bracket 0 comma 1 right square bracket cubed end fraction end cell row v equals cell fraction numerator 9 cross times 10 to the power of negative sign 2 end exponent cross times 64 cross times 10 to the power of negative sign 3 end exponent over denominator 10 to the power of negative sign 2 end exponent cross times 10 to the power of negative sign 3 end exponent end fraction cross times 1 comma 25 cross times 10 to the power of negative sign 2 end exponent end cell row v equals cell 7 comma 2 end cell end table


Oleh karena itu, laju reaksinya menjadi 7,2 M/det.

Jadi, jawaban yang benar adalah B.space 

0

Roboguru

Pengukuran laju reaksi biasanya ditentukan dengan menggunakan metode laju awal, yaitu laju reaksi berdasarkan konsentrasi awal zat-zat yang bereaksi. Untuk reaksi:   didapat data sebagai berikut.   ...

Pembahasan Soal:

a. Menentukan orde reaksi pada begin bold style left square bracket Br O subscript 3 to the power of minus sign right square bracket end style bold comma bold space begin bold style open square brackets Br to the power of minus sign close square brackets end style bold comma bold space begin bold style open square brackets H to the power of plus sign close square brackets end style dan orde reaksi total

  • Menentukan orde reaksi bold left square bracket Br O subscript bold 3 to the power of bold minus sign bold right square bracket 
    Untuk menentukan orde reaksinya kita bisa menggunakan percobaan 1 dan 2: bold left square bracket Br O subscript bold 3 to the power of bold minus sign bold right square bracket space beda space dan bold space bold open square brackets Br to the power of bold minus sign bold close square brackets bold comma bold space bold open square brackets H to the power of bold plus sign bold close square brackets space sama maka:
    table attributes columnalign right center left columnspacing 0px end attributes row cell v subscript 1 over v subscript 2 end cell equals cell fraction numerator k open square brackets Br O subscript 3 to the power of minus sign close square brackets subscript 1 superscript x open square brackets Br to the power of minus sign close square brackets subscript 1 superscript y open square brackets H to the power of plus sign close square brackets subscript 1 superscript z over denominator k open square brackets Br O subscript 3 to the power of minus sign close square brackets subscript 2 superscript x open square brackets Br to the power of minus sign close square brackets subscript 2 superscript y open square brackets H to the power of plus sign close square brackets subscript 2 superscript z end fraction end cell row cell 1 half end cell equals cell fraction numerator k open square brackets 0 comma 1 space M close square brackets subscript 1 superscript x open square brackets 0 comma 1 space M close square brackets subscript 1 superscript y open square brackets 0 comma 1 space M close square brackets subscript 1 superscript z over denominator k open square brackets 0 comma 2 space M close square brackets subscript 2 superscript x open square brackets 0 comma 1 space M close square brackets subscript 2 superscript y open square brackets 0 comma 1 space M close square brackets subscript 2 superscript z end fraction end cell row cell 1 half end cell equals cell open parentheses fraction numerator 0 comma 1 space M over denominator 0 comma 2 space M end fraction close parentheses to the power of x end cell row cell 1 half end cell equals cell open parentheses 1 half close parentheses to the power of x end cell row x equals 1 end table 
    Jadi orde reaksi terhadap bold left square bracket Br O subscript bold 3 to the power of bold minus sign bold right square bracket adalah 1.
  • Menentukan orde reaksi bold open square brackets Br to the power of bold minus sign bold close square brackets 
    Untuk menentukan orde reaksinya kita bisa menggunakan percobaan 2 dan 3: bold open square brackets Br to the power of bold minus sign bold close square brackets bold space space beda space dan space bold left square bracket Br O subscript bold 3 to the power of bold minus sign bold right square bracket bold comma bold space bold open square brackets H to the power of bold plus sign bold close square brackets space sama maka:
     table attributes columnalign right center left columnspacing 0px end attributes row cell v subscript 2 over v subscript 3 end cell equals cell fraction numerator k open square brackets Br O subscript 3 to the power of minus sign close square brackets subscript 2 superscript x open square brackets Br to the power of minus sign close square brackets subscript 2 superscript y open square brackets H to the power of plus sign close square brackets subscript 2 superscript z over denominator k open square brackets Br O subscript 3 to the power of minus sign close square brackets subscript 3 superscript x open square brackets Br to the power of minus sign close square brackets subscript 3 superscript y open square brackets H to the power of plus sign close square brackets subscript 3 superscript z end fraction end cell row cell 2 over 4 end cell equals cell fraction numerator k open square brackets 0 comma 2 space M close square brackets subscript 2 superscript x open square brackets 0 comma 1 space M close square brackets subscript 2 superscript y open square brackets 0 comma 1 space M close square brackets subscript 2 superscript z over denominator k open square brackets 0 comma 2 space M close square brackets subscript 3 superscript x open square brackets 0 comma 2 space M close square brackets subscript 3 superscript y open square brackets 0 comma 1 space M close square brackets subscript 3 superscript z end fraction end cell row cell 1 half end cell equals cell open parentheses fraction numerator 0 comma 1 space M over denominator 0 comma 2 space M end fraction close parentheses to the power of italic y end cell row cell 1 half end cell equals cell open parentheses 1 half close parentheses to the power of italic y end cell row y equals 1 end table 
    Jadi orde reaksi terhadap bold open square brackets Br to the power of bold minus sign bold close square brackets adalah 1.
  • Menentukan orde reaksi bold open square brackets H to the power of plus sign bold close square brackets 
    Untuk menentukan orde reaksinya kita bisa menggunakan percobaan 1 dan 4: bold open square brackets H to the power of bold plus sign bold close square brackets space beda space dan space bold left square bracket Br O subscript bold 3 to the power of bold minus sign bold right square bracket bold comma bold open square brackets Br to the power of bold minus sign bold close square brackets bold space sama maka:
    table attributes columnalign right center left columnspacing 0px end attributes row cell v subscript 1 over v subscript 4 end cell equals cell fraction numerator k open square brackets Br O subscript 3 to the power of minus sign close square brackets subscript 1 superscript x open square brackets Br to the power of minus sign close square brackets subscript 1 superscript y open square brackets H to the power of plus sign close square brackets subscript 1 superscript z over denominator k open square brackets Br O subscript 3 to the power of minus sign close square brackets subscript 4 superscript x open square brackets Br to the power of minus sign close square brackets subscript 4 superscript y open square brackets H to the power of plus sign close square brackets subscript 4 superscript z end fraction end cell row cell 1 fourth end cell equals cell fraction numerator k open square brackets 0 comma 1 space M close square brackets subscript 1 superscript x open square brackets 0 comma 1 space M close square brackets subscript 1 superscript y open square brackets 0 comma 1 space M close square brackets subscript 1 superscript z over denominator k open square brackets 0 comma 1 space M close square brackets subscript 4 superscript x open square brackets 0 comma 1 space M close square brackets subscript 4 superscript y open square brackets 0 comma 2 space M close square brackets subscript 4 superscript z end fraction end cell row cell 1 fourth end cell equals cell open parentheses fraction numerator 0 comma 1 space M over denominator 0 comma 2 space M end fraction close parentheses to the power of italic z end cell row cell open parentheses 1 half close parentheses squared end cell equals cell open parentheses 1 half close parentheses to the power of italic z end cell row z equals 2 end table  
    Jadi orde reaksi terhadap bold open square brackets H to the power of plus sign bold close square brackets adalah 2.
  • Menentukan orde reaksi total
    orde space reaksi space total double bond x and y and z orde space reaksi space total equals 1 plus 1 plus 2 orde space reaksi space total equals 4 
    Jadi orde reaksi total adalah 4.space 

 

b. Menentukan persamaan laju reaksinya

v double bond k left square bracket Br O subscript 3 to the power of minus sign right square bracket to the power of italic x open square brackets Br to the power of minus sign close square brackets to the power of y open square brackets H to the power of plus sign close square brackets to the power of italic z x equals 1 comma space y equals 1 comma space dan space z equals 2 v double bond k left square bracket Br O subscript 3 to the power of minus sign right square bracket open square brackets Br to the power of minus sign close square brackets open square brackets H to the power of plus sign close square brackets squared 

Jadi persamaan laju reaksinya adalah italic v bold equals italic k bold left square bracket Br O subscript bold 3 to the power of bold minus sign bold right square bracket bold open square brackets Br to the power of bold minus sign bold close square brackets bold open square brackets H to the power of bold plus sign bold close square brackets to the power of bold 2  
 

c. Menentukan laju relatif reaksi pada kondisi kosentrasi begin bold style left square bracket Br O subscript 3 to the power of minus sign right square bracket end style bold comma bold space begin bold style open square brackets Br to the power of minus sign close square brackets end style bold comma bold space begin bold style open square brackets H to the power of plus sign close square brackets end style  berturut-turut 0,3 M; 0,2 M, dan 0,1 M.space 

  • Menentukan tetapan laju reaksi
    v double bond k left square bracket Br O subscript 3 to the power of minus sign right square bracket open square brackets Br to the power of minus sign close square brackets open square brackets H to the power of plus sign close square brackets squared k equals fraction numerator v over denominator left square bracket Br O subscript 3 to the power of minus sign right square bracket open square brackets Br to the power of minus sign close square brackets open square brackets H to the power of plus sign close square brackets squared end fraction comma space masukan space ke space pecobaan space 4 k equals fraction numerator 4 space M forward slash detik over denominator left square bracket 0 comma 1 space M right square bracket left square bracket 0 comma 1 M right square bracket left square bracket 0 comma 2 space M right square bracket squared end fraction k equals fraction numerator 4 space M space detik to the power of negative sign 1 end exponent over denominator 4 cross times 10 to the power of negative sign 4 end exponent space M to the power of 4 end fraction k equals 10 to the power of 4 space M to the power of negative sign 3 end exponent space detik to the power of negative sign 1 end exponent 
  • Menentukan laju reaksi
    v double bond k left square bracket Br O subscript 3 to the power of minus sign right square bracket open square brackets Br to the power of minus sign close square brackets open square brackets H to the power of plus sign close square brackets squared k equals 10 to the power of 4 space M to the power of negative sign 3 end exponent space detik to the power of negative sign 1 end exponent v equals 10 to the power of 4 space M to the power of negative sign 3 end exponent space detik to the power of negative sign 1 end exponent left square bracket 3 cross times 10 to the power of negative sign 1 end exponent space M right square bracket left square bracket 2 cross times 10 to the power of negative sign 1 end exponent space M right square bracket left square bracket 1 cross times 10 to the power of negative sign 1 end exponent space M right square bracket squared v equals 6 cross times 10 to the power of 4 cross times 10 to the power of negative sign 4 end exponent space M to the power of negative sign 3 end exponent space M to the power of 4 space detik to the power of negative sign 1 end exponent v equals 6 space M space detik to the power of negative sign 1 end exponent 
    Jadi laju reaksinya adalah 6 M/detik.
    space 

0

Roboguru

Berikut ini diberikan data percobaan laju reaksi:  pada beberapa kondisi.   Jika [Q] dan [T] masing-masing diubah menjadi 0,5 M, harga laju (v) reaksi saat itu adalah ... M/det.

Pembahasan Soal:

Untuk mengetahui laju reaksi tersebut, maka dicari orde reaksi dan konstanta laju reaksi terlebih dulu agar diperoleh persamaan laju reaksinya.

Mencari orde reaksi Q menggunakan data 1 dan 2:

space space space space space space space space space space space space space space space fraction numerator v 1 over denominator v 2 end fraction equals fraction numerator k open square brackets Q close square brackets to the power of x open square brackets T close square brackets to the power of y over denominator k open square brackets Q close square brackets to the power of x open square brackets T close square brackets to the power of y end fraction fraction numerator 1 comma 25 cross times 10 to the power of negative sign 2 end exponent over denominator 5 cross times 10 to the power of negative sign 2 end exponent end fraction equals fraction numerator k left square bracket 0 comma 1 right square bracket to the power of x left square bracket 0 comma 1 right square bracket to the power of y over denominator k left square bracket 0 comma 2 right square bracket to the power of x left square bracket 0 comma 1 right square bracket to the power of y end fraction space space space space space space space space space space space space space space space space 1 fourth equals open parentheses 1 half close parentheses to the power of x space space space space space space space space space space space space space space space space space space x equals 2 

Mencari orde reaksi T menggunakan data 1 dan 3:

space space space space space space space space space space space space space space space fraction numerator v 1 over denominator v 2 end fraction equals fraction numerator k open square brackets Q close square brackets to the power of x open square brackets T close square brackets to the power of y over denominator k open square brackets Q close square brackets to the power of x open square brackets T close square brackets to the power of y end fraction fraction numerator 1 comma 25 cross times 10 to the power of negative sign 2 end exponent over denominator 10 to the power of negative sign 4 end exponent end fraction equals fraction numerator k left square bracket 0 comma 1 right square bracket to the power of x left square bracket 0 comma 1 right square bracket to the power of y over denominator k left square bracket 0 comma 1 right square bracket to the power of x left square bracket 0 comma 2 right square bracket to the power of y end fraction space space space space space space space space space space space space space space space space 1 over 8 equals open parentheses 1 half close parentheses to the power of y space space space space space space space space space space space space space space space space space space y equals 3 

Jadi, orde raksi terhadap Q adalah 2 dan orde reaksi terhadap T adalah 3. Sehingga, persamaan laju reaksinya adalah v double bond k open square brackets Q close square brackets squared open square brackets T close square brackets cubed.

Kemudian, mencari nilai konstanta laju reaksi dengan memasukan salah satu data ke persamaan laju reaksi, yaitu sebagai berikut:

space space space space space space space space space space space space space space space space v double bond k open square brackets Q close square brackets squared open square brackets T close square brackets cubed space 1 comma 25 cross times 10 to the power of negative sign 2 end exponent equals k left square bracket 0 comma 1 right square bracket squared left square bracket 0 comma 1 right square bracket cubed space 1 comma 25 cross times 10 to the power of negative sign 2 end exponent equals k point 10 to the power of negative sign 5 end exponent space space space space space space space space space space space space space space space space space k equals fraction numerator 1 comma 25 cross times 10 to the power of negative sign 2 end exponent over denominator 10 to the power of negative sign 5 end exponent end fraction space space space space space space space space space space space space space space space space space space space equals space 1.250   

Laju reaksi untuk [Q] dan [T] adalah 0,5 M, yaitu sebagai berikut:

v double bond k open square brackets Q close square brackets squared left square bracket T cubed right square bracket space space equals 1.250 left parenthesis 0 comma 5 right parenthesis squared left parenthesis 0 comma 5 right parenthesis cubed space space space space space equals 39 comma 0 space M forward slash det  

Jadi, jawaban yang benar adalah E.

0

Roboguru

Data percobaan laju reaksi diperoleh dari reaksi: , sebagai berikut: Percobaan [A] molar [B] molar Laju reaksi (molar/detik) 1 0,01 0,20 0,02 2 0,02 ...

Pembahasan Soal:

Persamaan laju reaksi merupakan persamaan yang menunjukkan keterkaitan atau hubungan antara laju reaksi tertentu dengan konsentrasi pereaksinya. Persamaan reaksi diperoleh melalui percobaan bukan melalui persamaan reaksi kimia. Pada persamaan laju reaksi perlu ditentukan terlebih dahulu orde reaksinya, yaitu bilangan pangkat konsentrasi. Orde reaksi diperoleh dengan cara mengubah konsentrasi salah satu pereaksi dan membuat konsentrasi zat lain tetap. 

  • Orde terhadap [A]

table attributes columnalign right center left columnspacing 0px end attributes row cell italic v subscript 2 over italic v subscript 1 end cell equals cell italic k over italic k open parentheses open square brackets A close square brackets subscript 2 over open square brackets A close square brackets subscript 1 close parentheses to the power of italic x open parentheses open square brackets B close square brackets subscript 2 over open square brackets B close square brackets subscript 1 close parentheses to the power of italic y end cell row cell fraction numerator 0 comma 08 over denominator 0 comma 02 end fraction end cell equals cell open parentheses fraction numerator 0 comma 02 over denominator 0 comma 01 end fraction close parentheses to the power of italic x open parentheses fraction numerator 0 comma 20 over denominator 0 comma 20 end fraction close parentheses to the power of italic y end cell row 4 equals cell 2 to the power of x end cell row x equals 2 row blank blank blank end table   

  • Orde terhadap [B]

table attributes columnalign right center left columnspacing 0px end attributes row cell italic v subscript 4 over italic v subscript 3 end cell equals cell italic k over italic k open parentheses open square brackets A close square brackets subscript 4 over open square brackets A close square brackets subscript 3 close parentheses to the power of italic x open parentheses open square brackets B close square brackets subscript 4 over open square brackets B close square brackets subscript 3 close parentheses to the power of italic y end cell row cell fraction numerator 0 comma 36 over denominator 0 comma 18 end fraction end cell equals cell open parentheses fraction numerator 0 comma 03 over denominator 0 comma 03 end fraction close parentheses to the power of italic x open parentheses fraction numerator 0 comma 40 over denominator 0 comma 20 end fraction close parentheses to the power of italic y end cell row 2 equals cell 2 to the power of italic y end cell row y equals 1 row blank blank blank end table  

  • Persamaan laju reaksi: italic v equals italic k open square brackets A close square brackets squared open square brackets B close square brackets 

Jadi, jawaban yang benar adalah A.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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