Roboguru

Pada reaksi , diperoleh data laju reaksi sebagai berikut. Jika , laju reaksi pada saat  mol/L dan  mol/L adalah ....

Pertanyaan

Pada reaksi 2 P and Q yields R, diperoleh data laju reaksi sebagai berikut.

Jika italic k equals 1 cross times 10 to the power of negative sign 5 end exponent, laju reaksi pada saat open square brackets P close square brackets equals 0 comma 02 mol/L dan open square brackets Q close square brackets equals 0 comma 6 mol/L adalah ....

  1. italic v equals 2 comma 0 cross times 10 to the power of negative sign 5 end exponent 

  2. italic v equals 2 comma 1 cross times 10 to the power of negative sign 6 end exponent 

  3. italic v equals 1 comma 2 cross times 10 to the power of negative sign 7 end exponent 

  4. italic v equals 7 comma 2 cross times 10 to the power of negative sign 8 end exponent 

  5. italic v equals 8 comma 6 cross times 10 to the power of negative sign 8 end exponent 

Pembahasan Soal:

Laju reaksi adalah laju berkurangnya konsentrasi reaktan per satuan waktu. Besarnya laju reaksi dapat dtentukan melalui persamaan laju reaksi atau hukum laju. Hukum laju merupakan sebuah persamaan yang memperlihatkan keterkaitan atau hubungan antara laju reaksi tertentu dengan konsentrasi pereaksinya.

Persamaan laju reaksi diperoleh berdasarkan eksperimen. Berdasarkan soal di atas, maka laju reaksi pada saat open square brackets P close square brackets equals 0 comma 02 mol/L dan open square brackets Q close square brackets equals 0 comma 6 mol/L adalah sebagai berikut.

  • Orde terhadap open square brackets P close square brackets 

table attributes columnalign right center left columnspacing 0px end attributes row cell t subscript 1 over t subscript 2 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 16 over 8 end cell equals cell open parentheses fraction numerator 0 comma 04 over denominator 0 comma 02 end fraction close parentheses to the power of italic x open parentheses fraction numerator 0 comma 6 over denominator 0 comma 6 end fraction close parentheses to the power of italic y end cell row 2 equals cell 2 to the power of italic x end cell row italic x equals 1 end table 

  • Orde terhadap open square brackets Q close square brackets 

table attributes columnalign right center left columnspacing 0px end attributes row cell t subscript 2 over t subscript 3 end cell equals cell italic k over italic k open parentheses open square brackets P close square brackets subscript 3 over open square brackets P close square brackets subscript 2 close parentheses to the power of italic x open parentheses open square brackets Q close square brackets subscript 3 over open square brackets Q close square brackets subscript 2 close parentheses to the power of italic y end cell row cell 8 over 4 end cell equals cell open parentheses fraction numerator 0 comma 02 over denominator 0 comma 02 end fraction close parentheses to the power of italic x open parentheses fraction numerator 0 comma 12 over denominator 0 comma 6 end fraction close parentheses to the power of italic y end cell row 2 equals cell 2 to the power of y end cell row y equals 1 end table 

  • Persamaan laju reaksi: italic v equals italic k open square brackets P close square brackets open square brackets Q close square brackets 
  • Laju reaksi saat laju reaksi pada saat open square brackets P close square brackets equals 0 comma 02 mol/L dan open square brackets Q close square brackets equals 0 comma 6 mol/L dengan nilai italic k equals 1 cross times 10 to the power of negative sign 5 end exponent:

 italic v equals italic k open square brackets P close square brackets open square brackets Q close square brackets equals 1 cross times 10 to the power of negative sign 5 end exponent left parenthesis 0 comma 02 right parenthesis left parenthesis 0 comma 6 right parenthesis equals 1 comma 2 cross times 10 to the power of negative sign 7 space 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

Gas karbon monoksida CO adalah gas beracun yang sangat berbahaya karena dapat bereaksi dengan hemoglobin (Hb) menggantikan . Laju reaksi antara hemoglobin (Hb) dan karbon monoksida (CO) dipelajari pad...

Pembahasan Soal:

a. Menentukan orde reaksi terhadap Hb dan CO

  • Pada perobaan 1 dan 2: open square brackets C O close square brackets space sama space dan space open square brackets Hb close square brackets space beda 
    table attributes columnalign right center left columnspacing 0px end attributes row cell v subscript 2 over v subscript 1 end cell equals cell fraction numerator k space open square brackets Hb close square brackets subscript 2 superscript x open square brackets C O close square brackets subscript 2 superscript y over denominator k space open square brackets Hb close square brackets subscript 1 superscript x open square brackets C O close square brackets subscript 1 superscript y end fraction end cell row cell fraction numerator 1 comma 24 over denominator 0 comma 619 end fraction end cell equals cell fraction numerator k space open square brackets 4 comma 42 close square brackets subscript 2 superscript x open square brackets 1 comma 00 close square brackets subscript 2 superscript y over denominator k space open square brackets 2 comma 21 close square brackets subscript 1 superscript x open square brackets 1 comma 00 close square brackets subscript 1 superscript y end fraction end cell row 2 equals cell open parentheses fraction numerator 4 comma 42 over denominator 2 comma 21 end fraction close parentheses to the power of italic x end cell row 2 equals cell open parentheses 2 close parentheses to the power of italic x end cell row x equals 1 end table 
    Jadi orde reaksi terhadap Hb adalah 1.
  • Pada perobaan 2 dan 3: open square brackets Hb close square brackets space sama space dan space open square brackets C O close square brackets space beda 
    table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator v subscript 3 over denominator v 2 end fraction end cell equals cell fraction numerator k space open square brackets Hb close square brackets subscript 3 superscript x open square brackets C O close square brackets subscript 3 superscript y over denominator k space open square brackets Hb close square brackets subscript 2 superscript x open square brackets C O close square brackets subscript 2 superscript y end fraction end cell row cell fraction numerator 3 comma 71 over denominator 1 comma 24 end fraction end cell equals cell fraction numerator k space open square brackets 4 comma 42 close square brackets subscript 3 superscript x open square brackets 3 comma 00 close square brackets subscript 3 superscript y over denominator k space open square brackets 4 comma 42 close square brackets subscript 2 superscript x open square brackets 1 comma 00 close square brackets subscript 2 superscript y end fraction end cell row 3 equals cell open parentheses fraction numerator 3 comma 00 over denominator 1 comma 00 end fraction close parentheses to the power of italic y end cell row 3 equals cell open parentheses 3 close parentheses to the power of y end cell row y equals 1 end table 
    Jadi orde reaksi terhadap CO adalah 1.

 

b. Menentukan hukum laju reaksi

Berdasarkan data pada soal (a) di dapat bahwa nilai x dan y adalah 1, maka hukum laju reaksinya adalah:

v double bond k open square brackets C O close square brackets open square brackets Hb close square brackets 

Jadi hukum laju reaksinya adalah italic v bold equals italic k begin bold style open square brackets C O close square brackets end style begin bold style open square brackets Hb close square brackets end style.

c. Menentukan nilai konstanta laju reaksinya

Untuk menentukan nilai konstanta laju reaksi kita tinggal memasukan salah satu data eksperimen ke hukum laju reaksi misal percobaan 1:

v double bond k open square brackets C O close square brackets open square brackets Hb close square brackets k equals fraction numerator v over denominator open square brackets C O close square brackets open square brackets Hb close square brackets end fraction k equals fraction numerator 0 comma 619 space μmol space L to the power of negative sign 1 end exponent space s to the power of negative sign 1 end exponent space over denominator open square brackets 2 comma 21 space space μmol space L to the power of negative sign 1 end exponent close square brackets open square brackets 1 comma 00 space space μmol space L to the power of negative sign 1 end exponent close square brackets end fraction k equals fraction numerator 0 comma 619 space μmol space L to the power of negative sign 1 end exponent space s to the power of negative sign 1 end exponent space over denominator open square brackets 2 comma 21 space space μmol space L to the power of negative sign 1 end exponent close square brackets open square brackets 1 comma 00 space space μmol space L to the power of negative sign 1 end exponent close square brackets end fraction k equals 0 comma 28 space L space μmol to the power of negative sign 1 end exponent space s to the power of negative sign 1 end exponent nstanta laju 
Jadi nilai konstanta laju reaksinya adalah bold 0 bold comma bold 28 bold space italic L bold space bold μmol to the power of bold minus sign bold 1 end exponent bold space italic s to the power of bold minus sign bold 1 end exponent.

d. Menentukan laju awal eksperimen

Untuk menentukan laju awal eksperimen kita tinggal memasukan data-datanya ke rumus karena konstanta laju reaksinya sudah ada pada jawaban (c).

table attributes columnalign right center left columnspacing 0px end attributes row cell v subscript 0 end cell equals cell k open square brackets C O close square brackets subscript 0 open square brackets Hb close square brackets subscript 0 end cell row cell v subscript 0 end cell equals cell 0 comma 28 space L space μmol to the power of negative sign 1 end exponent space s to the power of negative sign 1 end exponent cross times open square brackets 2 comma 24 space μmo space L to the power of negative sign 1 end exponent close square brackets subscript 0 cross times open square brackets 3 comma 36 space μmo space L to the power of negative sign 1 end exponent close square brackets subscript 0 end cell row cell v subscript 0 end cell equals cell 2 comma 26 space μmol space L to the power of negative sign 1 end exponent space s to the power of negative sign 1 end exponent end cell end table  

Jadi nilai laju awal eksperimen adalah Error converting from MathML to accessible text..space 

0

Roboguru

Beberapa reaksi gas terjadi dalam mesin mobil dan sistem pembuangan. Salah satunya adalah reaksi berikut:   Gunakan data berikut untuk menentukan orde reaksi dan hukum laju reaksi bagi reaksi terseb...

Pembahasan Soal:

Laju reaksi adalah laju pengurangan konsentrasi pereaksi atau penambahan konsentrasi hasil reaksi per satuanspacewaktu.

  • Menghitung orde reaksi N O subscript 2, maka cari open square brackets C O close square brackets yang sama (Percobaan 1 danspace2)

table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets open square brackets N O subscript 2 close square brackets subscript 2 over open square brackets N O subscript 2 close square brackets subscript 1 close square brackets to the power of m end cell equals cell v subscript 2 over v subscript 1 end cell row cell open square brackets fraction numerator 0 comma 4 over denominator 0 comma 1 end fraction close square brackets to the power of m end cell equals cell fraction numerator 0 comma 0800 over denominator 0 comma 0050 end fraction end cell row cell open square brackets 4 close square brackets to the power of m end cell equals 16 row m equals 2 end table
 

  • Menghitung orde reaksi C O, maka cari open square brackets N O subscript 2 close square brackets yang sama (Percobaan 1 danspace3)

table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets open square brackets C O close square brackets subscript 3 over open square brackets C O close square brackets subscript 1 close square brackets to the power of n end cell equals cell v subscript 3 over v subscript 1 end cell row cell open square brackets fraction numerator 0 comma 2 over denominator 0 comma 1 end fraction close square brackets to the power of n end cell equals cell fraction numerator 0 comma 0050 over denominator 0 comma 0050 end fraction end cell row cell open square brackets 2 close square brackets to the power of n end cell equals 1 row n equals 0 end table
 

  • Menentukan hukum lajuspacereaksi

table attributes columnalign right center left columnspacing 0px end attributes row v equals cell k space open square brackets N O subscript 2 close square brackets to the power of m space open square brackets C O close square brackets to the power of n end cell row v equals cell k space open square brackets N O subscript 2 close square brackets squared space open square brackets C O close square brackets to the power of 0 end cell row v equals cell k space open square brackets N O subscript 2 close square brackets squared end cell end table


Jadi,

  • Orde reaksi terhadap N O subscript 2 adalah 2,
  • Orde reaksi terhadap C O adalah 0,
  • Persamaan laju reaksinya adalah Error converting from MathML to accessible text..

0

Roboguru

Berdasarkan data berikut:     Bagaimana persamaan laju untuk reaksi:

Pembahasan Soal:

Misal laju reaksi diibaratkan: v double bond 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 

  • Pada percobaan 2 dan 3  open square brackets A close square brackets sama, open square brackets B close square brackets beda
    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 italic k open square brackets A close square brackets subscript 2 superscript italic x open square brackets B close square brackets subscript 2 superscript italic y over denominator italic k open square brackets A close square brackets subscript 3 superscript italic x open square brackets B close square brackets subscript 3 superscript italic y end fraction end cell row cell fraction numerator 4 x over denominator 16 x end fraction end cell equals cell fraction numerator italic k open square brackets 0 comma 30 space mol space L to the power of negative sign 1 end exponent close square brackets subscript 2 superscript italic x open square brackets 0 comma 20 space mol space L to the power of negative sign 1 end exponent close square brackets subscript 2 superscript italic y over denominator italic k open square brackets 0 comma 30 space mol space L to the power of negative sign 1 end exponent close square brackets subscript 3 superscript italic x open square brackets 0 comma 40 space mol space L to the power of negative sign 1 end exponent close square brackets subscript 3 superscript italic y end fraction end cell row cell 1 fourth end cell equals cell fraction numerator open square brackets 0 comma 20 space mol space L to the power of negative sign 1 end exponent close square brackets subscript 2 superscript italic y over denominator open square brackets 0 comma 40 space mol space L to the power of negative sign 1 end exponent close square brackets subscript 3 superscript italic y end fraction end cell row cell 1 fourth end cell equals cell open parentheses 1 half close parentheses to the power of y 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 y end cell row y equals 2 row blank blank blank end table 
    Jadi orde reaksi terhadap open square brackets B close square brackets adalah 2.
  • Untuk mengetahui orde open square brackets A close square brackets kita masukan data-data percobaan 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 subscript 1 superscript italic x open square brackets B close square brackets subscript 1 superscript italic y over denominator italic k open square brackets A close square brackets subscript 2 superscript italic x open square brackets B close square brackets subscript 2 superscript italic y end fraction end cell row cell fraction numerator x over denominator 4 x end fraction end cell equals cell fraction numerator italic k open square brackets 0 comma 25 space mol space L to the power of negative sign 1 end exponent close square brackets subscript 1 superscript italic x open square brackets 0 comma 20 space mol space L to the power of negative sign 1 end exponent close square brackets subscript 1 superscript 2 over denominator italic k open square brackets 0 comma 30 space mol space L to the power of negative sign 1 end exponent close square brackets subscript 2 superscript italic x open square brackets 0 comma 40 space mol space L to the power of negative sign 1 end exponent close square brackets subscript 2 superscript 2 end fraction end cell row cell 1 fourth end cell equals cell open parentheses 5 over 6 close parentheses to the power of x open parentheses 1 half close parentheses squared end cell row cell 1 fourth end cell equals cell open parentheses 5 over 6 close parentheses to the power of x open parentheses 1 fourth close parentheses end cell row 1 equals cell open parentheses 5 over 6 close parentheses to the power of x end cell row x equals 0 row blank blank blank end table 
    Jadi orde reaksi terhadap open square brackets A close square brackets adalah 0.
  • Pada perhitungan di atas kita telah menemukan nilai x=0, dan y=2, maka persamaan laju reaksinya adalah:
    italic v bold equals italic k begin bold style left square bracket italic A right square bracket end style to the power of bold 0 begin bold style open square brackets B close square brackets end style to the power of bold 2 italic v bold equals italic k begin bold style open square brackets B close square brackets end style to the power of bold 2 

Oleh karena itu, jawaban yang benar adalah B.space 

0

Roboguru

Berikut data percobaan laju reaksi.           Laju reaksi jika [CO] = 0,3 M dan [] = 0,2 M adalah ....

Pembahasan Soal:

Berdasarkan percobaan di atas, kita tentukan dahulu orde reaksi terhadap begin mathsize 14px style C O end style dan begin mathsize 14px style O subscript 2 end style.
Menentukan orde CO (cari dua data begin mathsize 14px style open square brackets O subscript 2 close square brackets end style yang sama, yaitu data 1 dan 3

begin mathsize 14px style open parentheses italic r subscript 1 over italic r subscript 3 close parentheses equals italic k over italic k open parentheses open square brackets C O close square brackets subscript 1 over open square brackets C O close square brackets subscript 3 close parentheses to the power of x open parentheses open square brackets O subscript 2 close square brackets subscript 1 over open square brackets O subscript 2 close square brackets subscript 3 close parentheses to the power of y end style

(karena k dan undefined sama, maka dapat dicoret)

begin mathsize 14px style open parentheses fraction numerator x over denominator 4 x end fraction close parentheses equals open parentheses fraction numerator 0 comma 2 over denominator 0 comma 4 end fraction close parentheses to the power of x open parentheses 1 fourth close parentheses equals open parentheses 1 half close parentheses to the power of x open parentheses 1 half close parentheses squared equals open parentheses 1 half close parentheses to the power of x x equals 2 end style


Menentukan orde undefined (cari dua data begin mathsize 14px style open square brackets C O close square brackets end style yang sama, yaitu data 1 dan 2

begin mathsize 14px style open parentheses italic r subscript 1 over italic r subscript 2 close parentheses equals italic k over italic k open parentheses open square brackets C O close square brackets subscript 1 over open square brackets C O close square brackets subscript 2 close parentheses to the power of x open parentheses open square brackets O subscript 2 close square brackets subscript 1 over open square brackets O subscript 2 close square brackets subscript 2 close parentheses to the power of y end style

(karena k dan undefined sama, maka dapat dicoret)

begin mathsize 14px style open parentheses fraction numerator x over denominator 3 x end fraction close parentheses equals open parentheses fraction numerator 0 comma 1 over denominator 0 comma 3 end fraction close parentheses to the power of y open parentheses 1 third close parentheses equals open parentheses 1 third close parentheses to the power of y y equals 1 end style

Maka, persamaan lajunya sebagai berikut :

begin mathsize 14px style italic r italic equals italic k open square brackets C O close square brackets squared open square brackets O subscript 2 close square brackets end style

Adapun laju reaksi jika begin mathsize 14px style open square brackets C O close square brackets equals 0 comma 3 M end style dan begin mathsize 14px style open square brackets O subscript 2 close square brackets equals 0 comma 2 M end style adalah begin mathsize 14px style italic r italic equals italic k left parenthesis 0 comma 3 right parenthesis squared left parenthesis 0 comma 2 right parenthesis end style


Jadi, jawaban yang benar adalah B.

0

Roboguru

Reaksi brominasi aseton berlangsung sebagai berikut.   Untuk menentukan laju reaksinya, diperoleh data (konsentrasi mula-mula, M) sebagai berikut.     Persamaan laju brominasi berdasarkan data di...

Pembahasan Soal:

Misal laju reaksinya adalah: v double bond k open square brackets C H subscript 3 C O C H subscript 3 close square brackets to the power of x open square brackets Br subscript 2 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 z 

  • Pada bercobaan 1 dan 2: open square brackets C H subscript 3 C O C H subscript 3 close square brackets space open square brackets H to the power of plus sign close square brackets  sama dan open square brackets Br subscript 2 close square brackets beda 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 C H subscript 3 C O C H subscript 3 close square brackets subscript 1 superscript x open square brackets Br subscript 2 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 C H subscript 3 C O C H subscript 3 close square brackets subscript 2 superscript x open square brackets Br subscript 2 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 fraction numerator 0 comma 57 over denominator 0 comma 57 end fraction end cell equals cell fraction numerator k open square brackets 0 comma 30 close square brackets subscript 1 superscript x open square brackets 0 comma 05 close square brackets subscript 1 superscript y open square brackets 0 comma 05 close square brackets subscript 1 superscript z over denominator k open square brackets 0 comma 30 close square brackets subscript 2 superscript x open square brackets 0 comma 10 close square brackets subscript 2 superscript y open square brackets 0 comma 05 close square brackets subscript 2 superscript z end fraction end cell row cell fraction numerator 0 comma 57 over denominator 0 comma 57 end fraction end cell equals cell open parentheses fraction numerator 0 comma 05 over denominator 0 comma 10 end fraction close parentheses to the power of y end cell row 1 equals cell open parentheses 1 half close parentheses to the power of y end cell row y equals 0 end table  
    Jadi orde reaksi terhadap open square brackets Br subscript 2 close square brackets adalah 0.
  • Pada bercobaan 2 dan 3 open square brackets C H subscript 3 C O C H subscript 3 close square brackets space open square brackets Br subscript 2 close square brackets sama dan open square brackets H to the power of plus sign close square brackets beda 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 C H subscript 3 C O C H subscript 3 close square brackets subscript 2 superscript x open square brackets Br subscript 2 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 C H subscript 3 C O C H subscript 3 close square brackets subscript 3 superscript x open square brackets Br subscript 2 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 fraction numerator 0 comma 57 over denominator 1 comma 2 end fraction end cell equals cell fraction numerator k open square brackets 0 comma 30 close square brackets subscript 1 superscript x open square brackets 0 comma 10 close square brackets subscript 1 superscript y open square brackets 0 comma 05 close square brackets subscript 1 superscript z over denominator k open square brackets 0 comma 30 close square brackets subscript 2 superscript x open square brackets 0 comma 10 close square brackets subscript 2 superscript y open square brackets 0 comma 10 close square brackets subscript 2 superscript z end fraction end cell row cell fraction numerator 0 comma 57 over denominator 1 comma 2 end fraction end cell equals cell open parentheses fraction numerator 0 comma 05 over denominator 0 comma 10 end fraction close parentheses to the power of italic z end cell row cell 1 half end cell equals cell open parentheses 1 half close parentheses to the power of 1 end cell row z equals 1 end table  
    Jadi orde reaksi terhadap open square brackets H to the power of plus sign close square brackets adalah 1.
  • Untuk mendapatkan orde reaksi terhadap open square brackets C H subscript 3 C O C H subscript 3 close square brackets kita tinggal memasukan ke persamaan 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 k open square brackets C H subscript 3 C O C H subscript 3 close square brackets subscript 1 superscript x open square brackets Br subscript 2 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 C H subscript 3 C O C H subscript 3 close square brackets subscript 2 superscript x open square brackets Br subscript 2 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 y equals 0 row z equals 1 row cell fraction numerator 0 comma 57 over denominator 0 comma 57 end fraction end cell equals cell fraction numerator k open square brackets 0 comma 30 close square brackets subscript 1 superscript x open square brackets 0 comma 05 close square brackets subscript 1 superscript 0 open square brackets 0 comma 05 close square brackets subscript 1 superscript 1 over denominator k open square brackets 0 comma 30 close square brackets subscript 2 superscript x open square brackets 0 comma 10 close square brackets subscript 2 superscript 0 open square brackets 0 comma 05 close square brackets subscript 2 superscript 1 end fraction end cell row cell fraction numerator 0 comma 57 over denominator 0 comma 57 end fraction end cell equals cell open parentheses fraction numerator 0 comma 30 over denominator 0 comma 30 end fraction close parentheses to the power of x end cell row 1 equals cell open parentheses 1 close parentheses to the power of x end cell row x equals 1 end table 
    Jadi orde reaksi terhadap adalah 1
  • Pada perhitungan di atas kita telah menemukan nilai x=1, y=0 dan z=1, maka persamaan laju reaksinya adalah:
    v double bond k open square brackets C H subscript 3 C O C H subscript 3 close square brackets to the power of 1 open square brackets Br subscript 2 close square brackets to the power of 0 open square brackets H to the power of plus sign close square brackets to the power of 1 
    italic v bold equals italic k begin bold style open square brackets C H subscript 3 C O C H subscript 3 close square brackets end style to the power of bold 1 begin bold style open square brackets H to the power of plus sign close square brackets end style to the power of bold 1 

Oleh karena itu, jawaban yang benar adalah B.space 

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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