Roboguru

Pada suhu , gas bromin dapat bereaksi dengan nitrogen monoksida menurut reaksi:     Dari reaksi tersebut, diperoleh data sebagai berikut:     Tentukan persamaan laju reaksi dan konstanta laju reaksinya!

Pertanyaan

Pada suhu 273 degree C, gas bromin dapat bereaksi dengan nitrogen monoksida menurut reaksi:
 

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

Dari reaksi tersebut, diperoleh data sebagai berikut:
 


 

Tentukan persamaan laju reaksi dan konstanta laju reaksinya!space

Pembahasan Soal:

Menentukan orde reaksinya dapat ditentukan dengan perbandingan data 3 banding 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 3 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 fraction numerator 0 comma 012 over denominator 0 comma 001 end fraction end cell row cell open square brackets 3 close square brackets to the power of x open square brackets 2 close square brackets to the power of y end cell equals 12 row cell open square brackets 3 close square brackets to the power of x open square brackets 2 close square brackets to the power of y end cell equals cell open square brackets 3 close square brackets to the power of 1 open square brackets 2 close square brackets squared end cell row cell maka colon space x end cell equals cell 1 space dan space y equals 2 end cell end table
 

Dengan demikian maka persamaan lajunya adalah: V double bond k open square brackets N O close square brackets open square brackets Br subscript 2 close square brackets squared

Sedangkan Konstanta laju reaksinya adalah:
 

table attributes columnalign right center left columnspacing 0px end attributes row V equals cell k open square brackets N O close square brackets open square brackets Br subscript 2 close square brackets squared end cell row k equals cell fraction numerator V over denominator open square brackets N O close square brackets open square brackets Br subscript 2 close square brackets squared end fraction end cell row k equals cell fraction numerator 0 comma 001 space M space s to the power of negative sign 1 end exponent over denominator open square brackets 0 comma 1 M close square brackets open square brackets 0 comma 1 M close square brackets squared end fraction end cell row k equals cell fraction numerator 0 comma 001 space M space s to the power of negative sign 1 end exponent over denominator 0 comma 001 space M cubed end fraction end cell row k equals cell 1 space M to the power of negative sign 2 end exponent space s to the power of negative sign 1 end exponent end cell end table

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

M. Pulungan

Mahasiswa/Alumni Universitas Negeri Medan

Terakhir diupdate 02 Mei 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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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

Laju reaksi terhadap   diketahui dengan mengukur jumlah mol H yang mengendap tiap liter per menit, dan diperoleh data sebagai berikut: Tentukanlah: a. Orde reaksi terhadap  dan   b. Orde total p...

Pembahasan Soal:

Orde reaksi menampilkan hubungan antara perubahan konsentrasi dengan perubahan laju reaksi. Oleh karena itu, orde reaksi yang dihasilkan adalah

a. Orde reaksi

Menentukan orde reaksi terhadap begin mathsize 14px style Hg Cl subscript 2 end style, yang digunakan percobaan 2 dan 3 dimana konsentrasi begin mathsize 14px style C subscript 2 O subscript 4 to the power of 2 minus sign end exponent end style tetap:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell v subscript 1 over v subscript 2 end cell equals cell italic k over italic k open parentheses open square brackets Hg Cl subscript 2 close square brackets subscript 2 over open square brackets Hg Cl subscript 2 close square brackets subscript 3 close parentheses to the power of x open parentheses fraction numerator left square bracket C subscript 2 O subscript 4 to the power of negative sign 2 end exponent right square bracket 2 over denominator left square bracket C subscript 2 O subscript 4 to the power of negative sign 2 end exponent right square bracket 3 end fraction close parentheses to the power of y end cell row cell fraction numerator 7 comma 1 middle dot 10 to the power of negative sign 5 end exponent over denominator 3 comma 5 middle dot 10 to the power of negative sign 5 end exponent end fraction end cell equals cell open parentheses fraction numerator 0 comma 105 over denominator 0 comma 052 end fraction close parentheses to the power of x space space open parentheses fraction numerator 0 comma 30 over denominator 0 comma 30 end fraction close parentheses to the power of y end cell row 2 equals cell 2 to the power of italic x end cell row italic x equals 1 end table end style 

Diperoleh orde reaksi terhadap begin mathsize 14px style open square brackets Hg Cl subscript 2 close square brackets end style sebesar x=1.

Menentukan orde reaksi terhadap [undefined], yang digunakan percobaan 1 dan 2 dimana konsentrasi undefined tetap:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell v subscript 1 over v subscript 2 end cell equals cell italic k over italic k open parentheses open square brackets Hg Cl subscript 2 close square brackets subscript 1 over open square brackets Hg Cl subscript 2 close square brackets subscript 2 close parentheses to the power of x open parentheses left square bracket C subscript 2 O subscript 4 to the power of negative sign 2 end exponent right square bracket subscript 1 over left square bracket C subscript 2 O subscript 4 to the power of negative sign 2 end exponent right square bracket subscript 2 close parentheses to the power of y end cell row cell fraction numerator 7 comma 1 middle dot 10 to the power of negative sign 5 end exponent over denominator 1 comma 8 middle dot 10 to the power of negative sign 5 end exponent end fraction end cell equals cell open parentheses fraction numerator 0 comma 105 over denominator 0 comma 105 end fraction close parentheses to the power of x space space open parentheses fraction numerator 0 comma 30 over denominator 0 comma 15 end fraction close parentheses to the power of y end cell row 4 equals cell 2 to the power of italic y end cell row y equals 2 end table end style  


b. Orde keseluruhan (orde total atau orde reaksi) adalah

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


c. Berdasarkan hasil orde reaksi, maka persamaan laju reaksinya:

begin mathsize 14px style v double bond k open square brackets Hg Cl subscript 2 close square brackets open square brackets C subscript 2 O subscript 4 to the power of 2 minus sign end exponent close square brackets squared end style 


d. Untuk menentukan konstanta laju reaksi, data dalam percobaan 1 disubtitusikan ke dalam persamaan laju reaksi.


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row v equals cell k open square brackets Hg Cl subscript 2 close square brackets open square brackets C subscript 2 O subscript 4 to the power of 2 minus sign end exponent close square brackets squared end cell row cell 1 comma 8 middle dot 10 to the power of negative sign 5 end exponent end cell equals cell k open square brackets 0 comma 105 space M close square brackets open square brackets 0 comma 15 space M close square brackets squared end cell row k equals cell fraction numerator 1 comma 8 middle dot 10 to the power of negative sign 5 end exponent space over denominator 2 comma 3 middle dot 10 to the power of negative sign 3 end exponent space end fraction end cell row blank equals cell 7 comma 82 middle dot 10 to the power of negative sign 3 end exponent space M squared space menit to the power of negative sign 1 end exponent end cell row blank equals cell 1 comma 3 middle dot 10 to the power of negative sign 4 end exponent space M squared space detik to the power of negative sign 1 end exponent end cell end table end style 


Jadi, orde laju reaksi, orde total reaksi, persamaan laju reaksi, dan konstanta laju reaksinya adalah sesuai penjelasan di atas.

0

Roboguru

Pada reaksi:   diperoleh data basil percobaan sebagai berikut.       Berdasarkan data tersebut, tentukan: a. orde reaksi terhadap  dan ; b. persamaan laju reaksi; c. harga tetapan laju reaksi...

Pembahasan Soal:

a. Orde reaksi begin mathsize 14px style S O subscript 2 end style dan begin mathsize 14px style O subscript 2 end style

Menentukan orde begin mathsize 14px style S O subscript 2 end style dengan percobaan 1 dan 2

begin mathsize 14px style 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 open square brackets S O subscript 2 close square brackets subscript 2 superscript x open square brackets O subscript 2 close square brackets subscript 2 superscript y over denominator k open square brackets S O subscript 2 close square brackets subscript 1 superscript x open square brackets O subscript 2 close square brackets subscript 1 superscript y end fraction end cell row cell fraction numerator 1 comma 6 cross times 10 to the power of negative sign 6 end exponent over denominator 0 comma 4 cross times 10 to the power of negative sign 6 end exponent end fraction end cell equals cell fraction numerator k open square brackets 0 comma 008 close square brackets to the power of x open square brackets 0 comma 002 close square brackets to the power of y over denominator k open square brackets 0 comma 004 close square brackets to the power of x open square brackets 0 comma 002 close square brackets to the power of y end fraction end cell row cell 4 space end cell equals cell space 2 to the power of x end cell row cell x space end cell equals cell space 2 end cell end table end style 

Menentukan orde begin mathsize 14px style O subscript 2 end style dengan percobaan 4 dan 5

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell v subscript 4 over v subscript 5 space end cell equals cell space fraction numerator k open square brackets S O subscript 2 close square brackets subscript 4 superscript x open square brackets O subscript 2 close square brackets subscript 4 superscript y over denominator k open square brackets S O subscript 2 close square brackets subscript 5 superscript x open square brackets O subscript 2 close square brackets subscript 5 superscript y end fraction end cell row cell fraction numerator 9 comma 6 cross times 10 to the power of negative sign 6 end exponent over denominator 2 comma 4 cross times 10 to the power of negative sign 6 end exponent end fraction space end cell equals cell space fraction numerator k open square brackets 0 comma 008 close square brackets to the power of x open square brackets 0 comma 016 close square brackets to the power of y over denominator k open square brackets 0 comma 008 close square brackets to the power of x open square brackets 0 comma 004 close square brackets to the power of y end fraction end cell row cell 4 space end cell equals cell space 4 to the power of y end cell row cell y space end cell equals cell space 1 end cell end table end style 

Jadi, orde reaksi begin mathsize 14px style S O subscript 2 end style dan orde reaksi begin mathsize 14px style O subscript 2 end style berturut-turut adalah 2 dan 1.

b. Persamaan laju reaksi: begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell v space end cell equals cell space k open square brackets S O subscript 2 close square brackets to the power of x open square brackets O subscript 2 close square brackets to the power of y end cell row cell v space end cell equals cell space k open square brackets S O subscript 2 close square brackets squared open square brackets O subscript 2 close square brackets end cell end table end style    

Jadi, persamaan laju reaksi yang benar adalah begin mathsize 14px style v space equals space k open square brackets S O subscript 2 close square brackets squared open square brackets O subscript 2 close square brackets end style.

c. Harga tetapan laju reaksi

Percobaan 1

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell v space end cell equals cell space k open square brackets S O subscript 2 close square brackets squared open square brackets O subscript 2 close square brackets end cell row cell 0 comma 4 cross times 10 to the power of negative sign 6 end exponent space end cell equals cell space k left parenthesis 0 comma 004 right parenthesis squared left parenthesis 0 comma 002 right parenthesis end cell row cell k space end cell equals cell 12 comma 5 end cell end table end style 

Jadi, harga tetapan laju reaksi yang benar adalah 12,5.

0

Roboguru

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

Brominasi aseton memerlukan katalis asam dengan reaksi:   Data hasil percobaan reaksi di atas disajikan pada tabel berikut. Tentukan nilai tetapan laju reaksi (k)!

Pembahasan Soal:

Langkah-langkah untuk menentukan nilai konstanta laju reaksi adalah:

1. Menuliskan persamaan laju reaksi

 begin mathsize 14px style v double bond k middle dot open square brackets C H subscript 3 C O C H subscript 3 close square brackets to the power of x middle dot open square brackets Br subscript 2 close square brackets to the power of y middle dot open square brackets H to the power of plus sign close square brackets to the power of z end style

2. Mencari nilai x dengan melihat data begin mathsize 14px style open square brackets Br subscript 2 close square brackets end style dan begin mathsize 14px style open square brackets H to the power of plus sign close square brackets end style yang sama yaitu data 1 dan 5, lalu kita masukan data 1 dan 5 dalam perhitungan yaitu:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses fraction numerator 0 comma 30 over denominator 0 comma 40 end fraction close parentheses to the power of italic x end cell equals cell open parentheses fraction numerator 5 comma 7 middle dot down diagonal strike 10 to the power of negative sign 5 end exponent end strike over denominator 7 comma 6 middle dot down diagonal strike 10 to the power of negative sign 5 end exponent end strike end fraction close parentheses end cell row cell open parentheses 3 over 4 close parentheses to the power of italic x end cell equals cell open parentheses 3 over 4 close parentheses end cell row cell open parentheses 3 over 4 close parentheses to the power of italic x end cell equals cell open parentheses 3 over 4 close parentheses to the power of 1 end cell row x equals 1 end table end style 

3. Mencari nilai y dengan melihat data begin mathsize 14px style open square brackets C H subscript 3 C O C H subscript 3 close square brackets end style dan undefined yang sama yaitu data 1 dan 2, lalu kita masukan data 1 dan 2 dalam perhitungan yaitu:

 begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses fraction numerator 0 comma 05 over denominator 0 comma 10 end fraction close parentheses to the power of italic y end cell equals cell open parentheses fraction numerator down diagonal strike 5 comma 7 middle dot 10 to the power of negative sign 5 end exponent end strike over denominator down diagonal strike 5 comma 7 middle dot 10 to the power of negative sign 5 end exponent end strike end fraction close parentheses end cell row cell open parentheses 1 half close parentheses to the power of italic y end cell equals 1 row y equals 0 end table end style  

4. Mencari nilai z dengan melihat data begin mathsize 14px style open square brackets C H subscript 3 C O C H subscript 3 close square brackets end style dan undefined yang sama yaitu data 1 dan 3, lalu kita masukan data 1 dan 3 dalam perhitungan yaitu:

 begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row 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 equals cell open parentheses fraction numerator 5 comma 7 middle dot 10 to the power of negative sign 5 end exponent over denominator 1 comma 2 middle dot 10 to the power of negative sign 4 end exponent end fraction close parentheses end cell row cell open parentheses 1 half close parentheses to the power of italic z end cell equals cell open parentheses 4 comma 75 middle dot 10 to the power of negative sign 1 end exponent close parentheses end cell row cell open parentheses 1 half close parentheses to the power of italic z end cell equals cell open parentheses 0 comma 475 close parentheses equals 0 comma 5 end cell row cell open parentheses 1 half close parentheses to the power of italic z end cell equals cell open parentheses 1 half close parentheses end cell row cell open parentheses 1 half close parentheses to the power of italic z end cell equals cell open parentheses 1 half close parentheses to the power of 1 end cell row z equals 1 end table end style 

5. Setelah itu kita susun persamaan yang baru yaitu:

begin mathsize 14px style v double bond k middle dot open square brackets C H subscript 3 C O C H subscript 3 close square brackets to the power of 1 middle dot open square brackets Br subscript 2 close square brackets to the power of 0 middle dot open square brackets H to the power of plus sign close square brackets to the power of 1 space v double bond k middle dot open square brackets C H subscript 3 C O C H subscript 3 close square brackets to the power of 1 middle dot open square brackets H to the power of plus sign close square brackets to the power of 1 italic v bold equals italic k bold middle dot bold open square brackets C H subscript bold 3 C O C H subscript bold 3 bold close square brackets bold middle dot bold open square brackets H to the power of bold plus sign bold close square brackets end style 

6. Lalu kita mencari nilai k dengan memasukkan salah satu data ke dalam persamaan, misal kita memasukan data nomor 1 yaitu:

 begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 5 comma 7 middle dot 10 to the power of negative sign 5 end exponent end cell equals cell k middle dot left parenthesis 0 comma 30 right parenthesis middle dot left parenthesis 0 comma 05 right parenthesis end cell row cell 5 comma 7 middle dot 10 to the power of negative sign 5 end exponent end cell equals cell k middle dot left parenthesis 0 comma 015 right parenthesis end cell row cell fraction numerator 5 comma 7 middle dot 10 to the power of negative sign 5 end exponent over denominator 0 comma 015 end fraction end cell equals k row cell fraction numerator 5 comma 7 middle dot 10 to the power of negative sign 5 end exponent over denominator 1 comma 5 middle dot 10 to the power of negative sign 2 end exponent end fraction end cell equals k row k equals cell 3 comma 8 middle dot 10 to the power of negative sign 3 end exponent end cell end table end style        

Jadi, nilai konstanta laju reaksi atau k adalah begin mathsize 14px style bold 3 bold comma bold 8 bold middle dot bold 10 to the power of bold minus sign bold 3 end exponent end style

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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