Roboguru

Perhatikan reaksi bromometri dalam penentuan suatu kadar zat berikut:   Berdasarkan percobaan yang dilakukan diperoleh data sebagai berikut:   Tentukan persamaan laju reaksi diatas!

Pertanyaan

Perhatikan reaksi bromometri dalam penentuan suatu kadar zat berikut:

Br O subscript 3 to the power of minus sign left parenthesis italic a italic q right parenthesis plus 5 Br left parenthesis italic a italic q right parenthesis plus 6 H to the power of plus sign left parenthesis italic a italic q right parenthesis yields 3 Br subscript 2 left parenthesis italic a italic q right parenthesis plus 3 H subscript 2 O open parentheses italic l close parentheses 

Berdasarkan percobaan yang dilakukan diperoleh data sebagai berikut:

 

Tentukan persamaan laju reaksi diatas!space 

Pembahasan Soal:

Persamaan lau reaksi tersebut dapat ditenttukan menggunakan data percobaan yang telah diketahui. Pertama, menentukan orde reaksi  masing - masing pereaksi, yaitu sebagai berikut:

Menentukan orde reaksi Br O subscript 3 to the power of minus sign menggunakan data percobaan ke 1 dan 2.

space space space space space fraction numerator v 1 over denominator v 2 end fraction equals fraction numerator italic k left square bracket B italic r O subscript 3 to the power of minus sign right square bracket to the power of italic x left square bracket B italic r to the power of negative sign end exponent right square bracket to the power of italic y end exponent open square brackets H to the power of plus sign close square brackets to the power of italic z over denominator italic k left square bracket B italic r O subscript 3 to the power of minus sign right square bracket to the power of x left square bracket B italic r to the power of minus sign right square bracket to the power of y open square brackets H to the power of plus sign close square brackets to the power of z end fraction fraction numerator fraction numerator 1 over denominator 152 plus-or-minus 6 end fraction over denominator fraction numerator 1 over denominator 73 plus-or-minus 4 end fraction end fraction equals fraction numerator italic k italic left parenthesis italic 0 italic comma italic 4 italic right parenthesis to the power of italic x left parenthesis 0 comma 24 right parenthesis to the power of y left parenthesis 0 comma 01 right parenthesis to the power of italic z end exponent over denominator italic k italic left parenthesis italic 0 italic comma italic 8 italic right parenthesis to the power of x italic left parenthesis italic 0 italic comma italic 24 italic right parenthesis to the power of y italic left parenthesis italic 0 italic comma italic 01 italic right parenthesis to the power of z end fraction fraction numerator fraction numerator 1 over denominator 152 minus sign 6 end fraction over denominator 1 over 73 end fraction equals open parentheses 1 half close parentheses to the power of x space space space fraction numerator 1 over 146 over denominator 1 over 73 end fraction equals open parentheses 1 half close parentheses to the power of x space space space space space space 1 half equals open parentheses 1 half close parentheses to the power of x space space space space space space space space x equals 1

Menentukan orde reaksi Br to the power of minus sign dari data percobaan 1 dan 3.

space space space space space fraction numerator v 1 over denominator v 3 end fraction equals fraction numerator italic k left square bracket B italic r O subscript 3 to the power of minus sign right square bracket to the power of italic x left square bracket B italic r to the power of minus sign right square bracket to the power of italic y open square brackets H to the power of plus sign close square brackets to the power of italic z over denominator italic k left square bracket B italic r O subscript 3 to the power of minus sign right square bracket to the power of x left square bracket B italic r to the power of minus sign right square bracket to the power of y open square brackets H to the power of plus sign close square brackets to the power of z end fraction fraction numerator fraction numerator 1 over denominator 152 plus-or-minus 6 end fraction over denominator fraction numerator 1 over denominator 75 plus-or-minus 3 end fraction end fraction equals fraction numerator italic k italic left parenthesis italic 0 italic comma italic 4 italic right parenthesis to the power of italic x left parenthesis 0 comma 24 right parenthesis to the power of y left parenthesis 0 comma 01 right parenthesis to the power of italic z over denominator italic k italic left parenthesis italic 0 italic comma italic 8 italic right parenthesis to the power of x italic left parenthesis italic 0 italic comma italic 48 italic right parenthesis to the power of y italic left parenthesis italic 0 italic comma italic 01 italic right parenthesis to the power of z end fraction fraction numerator fraction numerator 1 over denominator 152 minus sign 6 end fraction over denominator fraction numerator 1 over denominator 75 minus sign 3 end fraction end fraction equals open parentheses 1 half close parentheses to the power of y space space space fraction numerator 1 over 146 over denominator 1 over 72 end fraction equals open parentheses 1 half close parentheses to the power of y space space space space space space 1 half equals open parentheses 1 half close parentheses to the power of y space space space space space space space space y equals 1 

Menentukan orde reaksi open square brackets H to the power of plus sign close square brackets menggunakan data perocabaan 2 dan 4.

 space space space space space fraction numerator v 2 over denominator v 4 end fraction equals fraction numerator italic k left square bracket B italic r O subscript 3 to the power of minus sign right square bracket to the power of italic x left square bracket B italic r to the power of minus sign right square bracket to the power of italic y open square brackets H to the power of plus sign close square brackets to the power of italic z over denominator italic k left square bracket B italic r O subscript 3 to the power of minus sign right square bracket to the power of x left square bracket B italic r to the power of minus sign right square bracket to the power of y open square brackets H to the power of plus sign close square brackets to the power of z end fraction fraction numerator fraction numerator 1 over denominator 75 plus-or-minus 4 end fraction over denominator fraction numerator 1 over denominator 19 plus-or-minus 4 end fraction end fraction equals fraction numerator italic k italic left parenthesis italic 0 italic comma italic 8 italic right parenthesis to the power of italic x left parenthesis 0 comma 24 right parenthesis to the power of y left parenthesis 0 comma 01 right parenthesis to the power of italic z over denominator italic k italic left parenthesis italic 0 italic comma italic 8 italic right parenthesis to the power of x italic left parenthesis italic 0 italic comma italic 24 italic right parenthesis to the power of y italic left parenthesis italic 0 italic comma italic 02 italic right parenthesis to the power of z end fraction space space space space fraction numerator 1 over 75 over denominator 1 over 15 end fraction equals open parentheses 1 half close parentheses to the power of z space space space space space space 1 fifth equals open parentheses 1 half close parentheses to the power of z space space space space space space space space z equals 2 comma 2 

Maka, persamaan laju reaksinya adalah v equals 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 v equals 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.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

I. Solichah

Terakhir diupdate 13 September 2021

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Pertanyaan yang serupa

Dari data eksperimen di atas, selesaikanlah permasalahan berikut. a. Tentukan orde reaksi total. b. Tentukan persamaan laju. c. Tentukan harga k. d. Berapa laju reaksinya jika [NO] = 0,2 M dan [Br...

Pembahasan Soal:

Jika diketahui t, maka nilai vnya adalah t1.

  1. Mencari orde total. 
    • Orde [NO], menggunakan percobaan 1&3
      (0,20,1)x(0,10,1)=9624(0,20,1)x(0,10,1)=9624(21)x=41(21)x=(21)2x=2  

      Maka, orde [NO] adalah 2. 
    • Mencari orde [Br2] menggunakan persamaan 1&3
      (0,10,1)x(0,20,1)y=9648(0,10,1)x(0,20,1)y=9624(21)y=21(21)y=(21)1y=1  

      Maka, orde [Br2] adalah 1. 
    • Orde total 
      ordetotal=x+yordetotal=2+1ordetotal=3 

      Jadi, orde totalnya adalah 3. 
  2. Persamaan laju reaksinya adalah:
    v=k[NO]2[Br2] 
  3. Menentukan nilai k menggunakan persamaan 1
    v=k[NO]2[Br2]t1=k[NO]2[Br2]961=k(0,1)2(0,1)k=96x1031k=10,4167 

    Jadi, nilai k adalah 10,4167.
     
  4. Berapa laju reaksinya jika [NO] = 0,2 M dan [Br2] = 0,3 M
    v=k[NO]2[Br2]v=10,4167(0,2)2(0,3)v=0,125 

    Jadi, laju reaksinya adalah 0,125.

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Laju reaksi terhadap:  ,  diketahui dengan mengukur jumlah mol  yang mengendap tiap liter per menit, dan diperoleh data sebagai berikut: Hitung laju reaksi jika konsentrasi awal  0,02 mol/L dan  0,...

Pembahasan Soal:

 

  • Menentukan orde begin mathsize 14px style Hg Cl subscript 2 end style dari percobaan 2 dan 3

               Error converting from MathML to accessible text. 

  • Menentukan orde begin mathsize 14px style C subscript 2 O subscript 4 superscript negative sign 2 end superscript end style dari percobaan 1 dan 2

                      Error converting from MathML to accessible text.

  • Menentukan nilai k

        Error converting from MathML to accessible text. 

  • Menghitung laju reaksi jika konsentrasi begin mathsize 14px style Hg Cl subscript 2 space 0 comma 02 space mol forward slash L end style dan begin mathsize 14px style C subscript 2 O subscript 4 to the power of 2 minus sign end exponent space 0 comma 22 space mol forward slash L end style

          Error converting from MathML to accessible text. 

 

Jadi, didapatkan nilai laju reaksinya adalah begin mathsize 14px style 7 comma 37 space cross times space 10 to the power of negative sign 6 end exponent space M forward slash menit end style.

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Data eksperimen untuk reaksi  sebagai berikut:   Persamaan laju reaksi di atas adalah ...

Pembahasan Soal:

Persamaan laju reaksi di atas secara umum adalah sebagai berikut: begin mathsize 14px style v double bond k middle dot open square brackets A close square brackets to the power of m middle dot open square brackets B close square brackets to the power of n end style, dimana m merupakan orde reaksi pereaksi A dan n merupakan orde reaksi dari pereaksi B. Nilai orde reaksi m dan n dapat dihitung dengan cara berikut.space


Orde A (m):

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses fraction numerator open square brackets A subscript 1 close square brackets over denominator open square brackets A subscript 4 close square brackets end fraction close parentheses to the power of m end cell equals cell V subscript 1 over V subscript 4 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 end cell equals cell 6 over 24 end cell row cell open parentheses 1 half close parentheses to the power of m end cell equals cell 1 fourth end cell row m equals 2 end table end style  

Orde B (n):

begin mathsize 14px style open parentheses fraction numerator open square brackets B subscript 1 close square brackets over denominator open square brackets B subscript 2 close square brackets end fraction close parentheses to the power of n equals V subscript 1 over V subscript 2 open parentheses fraction numerator 0 comma 1 over denominator 0 comma 2 end fraction close parentheses to the power of n equals 6 over 12 open parentheses 1 half close parentheses to the power of n equals 1 half n equals 1 end style 

Maka persamaan laju reaksinya 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 to the power of m middle dot open square brackets B close square brackets to the power of n end cell end table end style atau begin mathsize 14px style v double bond k middle dot open square brackets A close square brackets squared middle dot open square brackets B close square brackets end style.space

Jadi, jawaban yang tepat adalah A.space

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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

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Data reaksi antara NO dan  membentuk NOCI pada suhu 295 K adalah sebagai berikut:   Maka persamaan laju reaksi di atas adalah ....

Pembahasan Soal:

Laju reaksi adalah perubahan konsentrasi reaktan atau produk per satuan waktu .Besaran laju reaksi dilihat dari ukuran cepat lambatnya suatu reaksi kimia.

Untuk menentukan persamaan laju reaksi tersebut menggunakan data percobaan yang telah diketahui, langkakh - langkah yaitu sebagai berikut:

Mencari orde reaksi terhadap NO, menggunakan data ke 1 dan 3.

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator v 3 over denominator v 1 end fraction end cell equals cell fraction numerator k open square brackets N O close square brackets subscript 3 superscript blank to the power of x open square brackets Cl subscript 2 close square brackets subscript 3 to the power of y over denominator k open square brackets N O close square brackets subscript 1 to the power of x open square brackets Cl subscript 2 close square brackets subscript 1 to the power of y end fraction end cell row cell fraction numerator 9 cross times 10 to the power of negative sign 3 end exponent over denominator 1 cross times 10 to the power of negative sign 3 end exponent end fraction end cell equals cell fraction numerator k left parenthesis 0 comma 150 right parenthesis to the power of x left parenthesis 0 comma 050 right parenthesis to the power of y over denominator k left parenthesis 0 comma 050 right parenthesis to the power of x left parenthesis 0 comma 050 right parenthesis to the power of y end fraction end cell row cell space 9 end cell equals cell 3 to the power of x end cell row cell 3 squared end cell equals cell space 3 to the power of x end cell row x equals 2 end table  

Jadi, orde reaksi NO adalah 2.

Mencari orde raksi terhadap Cl subscript 2 menggunakan data 1 dan 2.

v1v21×1033×1033y====k[NO]1x[Cl2]1yk[NO]2x[Cl2]2yk(0,050)x(0,050)yk(0,050)x(0,150)y3y1 

Jadi, orde reaksi Cl subscript 2 adalah 1.

Berdasarkan orde reaksi tersebut, persamaan lau reaksinya yaitu:

v double bond k open square brackets N O close square brackets squared open square brackets Cl subscript 2 close square brackets 

Jadi, persamaan laju reaksi tersebut adalah v double bond k open square brackets N O close square brackets squared open square brackets Cl subscript 2 close square brackets.

0

Roboguru

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