Roboguru

Tabel berikut memberikan informasi reaksi. Orde reaksi terhadap A adalah ....

Pertanyaan

Tabel berikut memberikan informasi reaksi.

A and B yields C and D

Orde reaksi terhadap A adalah ....space

  1. 0space

  2. 0,5space

  3. 1space

  4. 2space

  5. 3space

Pembahasan Soal:

Rumus persamaan laju reaksi adalah v double bond k open square brackets A close square brackets open square brackets B close square brackets

Orde reaksi terhadap A menggunakan data dari percobaan 1 dan 2 karena konsentrasi B tetap.

begin inline style v subscript 1 over v subscript 2 end style equals begin inline style fraction numerator k open square brackets A close square brackets subscript 1 to the power of x open square brackets B close square brackets subscript 1 to the power of y over denominator k open square brackets A close square brackets subscript 2 to the power of x open square brackets B close square brackets subscript 2 to the power of y end fraction end style begin inline style fraction numerator 0 comma 1 over denominator 0 comma 2 end fraction end style equals begin inline style fraction numerator up diagonal strike k left parenthesis 0 comma 2 right parenthesis to the power of x up diagonal strike left parenthesis 0 comma 1 right parenthesis to the power of y end strike over denominator up diagonal strike k left parenthesis 0 comma 4 right parenthesis to the power of x up diagonal strike left parenthesis 0 comma 1 right parenthesis to the power of y end strike end fraction end style begin inline style 1 half end style equals open parentheses begin inline style 1 half end style close parentheses to the power of x italic x equals 1

Berdasarkan perhitungan di atas, orde reaksi terhadap A adalah 1.

Jadi, jawaban yang tepat adalah C.space

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

B. Rohmawati

Mahasiswa/Alumni Universitas Negeri Semarang

Terakhir diupdate 13 September 2021

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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

Pada reaksi:  diperoleh data sebagai berikut:   Tentukan orde reaksi terhadap masing-masing pereaksi. Tentukan rumus laju reaksinya. Hitung nilai tetapan laju reaksi dan satuannya.

Pembahasan Soal:

Salah satu faktor yang mempengaruhi laju reaksi adalah konsentrasi reaktan. Hubungan antara laju reaksi dengan konsentrasi reaktan dirumuskan dalam suatu persamaan laju reaksi. 

Reaksi 2 N O open parentheses italic g close parentheses and O subscript 2 open parentheses italic g close parentheses yields N subscript 2 O subscript 4 open parentheses italic g close parentheses mempunyai persamaan laju reaksi:

italic v equals italic k open square brackets N O close square brackets to the power of italic x open square brackets O subscript 2 close square brackets to the power of italic y 

dengan:

k = tetapan laju reaksi
x = orde (tingkat atau pangkat) reaksi terhadap N O  
y = orde reaksi terhadap O subscript 2  

Orde reaksi ditentukan melalui percobaan, tidak berkaitan dengan koefisien reaksi.

a.   Orde reaksi terhadap masing-masing pereaksi

  • Orde reaksi N O  

    Untuk menghitung orde reaksi N O, pilih 2 percobaan dimana O subscript 2 mempunyai konsentrasi yang sama, yaitu percobaan (2) dan (3).

    v subscript 2 over v subscript 3 equals fraction numerator italic k open square brackets N O close square brackets subscript 2 superscript italic x open square brackets O subscript 2 close square brackets subscript 2 superscript italic y over denominator italic k open square brackets N O close square brackets subscript 3 superscript italic x open square brackets O subscript 2 close square brackets subscript 3 superscript italic y end fraction fraction numerator 0 comma 02 over denominator 0 comma 08 end fraction equals fraction numerator up diagonal strike italic k open parentheses 0 comma 1 close parentheses to the power of italic x up diagonal strike open parentheses 0 comma 2 close parentheses to the power of italic y end strike over denominator up diagonal strike italic k left parenthesis 0 comma 2 right parenthesis to the power of italic x up diagonal strike left parenthesis 0 comma 2 right parenthesis to the power of italic y end strike end fraction 1 fourth equals open parentheses 1 over 1 close parentheses to the power of italic x italic x equals 2  

     
  • Orde reaksi O subscript bold 2 

    Untuk menghitung orde reaksi O subscript 2, pilih 2 percobaan dimana N O mempunyai konsentrasi yang sama, yaitu percobaan (1) dan (2).

    v subscript 1 over v subscript 2 equals fraction numerator italic k open square brackets N O close square brackets subscript 1 superscript italic x open square brackets O subscript 2 close square brackets subscript 1 superscript italic y over denominator italic k open square brackets N O close square brackets subscript 2 superscript italic x open square brackets O subscript 2 close square brackets subscript 2 superscript italic y end fraction fraction numerator 0 comma 01 over denominator 0 comma 02 end fraction equals fraction numerator up diagonal strike italic k open parentheses 0 comma 1 close parentheses to the power of italic x end strike open parentheses 0 comma 1 close parentheses to the power of italic y over denominator up diagonal strike italic k left parenthesis 0 comma 1 right parenthesis to the power of italic x end strike open parentheses 0 comma 2 close parentheses to the power of italic y end fraction 1 half equals open parentheses 1 half close parentheses to the power of italic y italic y equals 1 


Jadi, orde reaksi terhadap N O bold thin space bold dan bold space O subscript bold 2 berturut-turut adalah 2 dan 1. 


b.   Rumus laju reaksi

      italic v equals italic k open square brackets N O close square brackets to the power of italic x open square brackets O subscript 2 close square brackets to the power of italic y italic v equals italic k open square brackets N O close square brackets squared open square brackets O subscript 2 close square brackets 


Jadi, persamaan laju reaksinya adalah italic v bold equals italic k begin bold style open square brackets N O close square brackets end style to the power of bold 2 begin bold style open square brackets O subscript 2 close square brackets end style.


c.   Nilai tetapan laju reaksi (k)

      Misal kita ambil percobaan nomor (1)

      italic v equals italic k open square brackets N O close square brackets squared open square brackets O subscript 2 close square brackets italic k equals fraction numerator italic v over denominator open square brackets N O close square brackets squared open square brackets O subscript 2 close square brackets end fraction italic k equals fraction numerator 0 comma 01 space M space detik to the power of negative sign 1 end exponent over denominator open parentheses 0 comma 1 space M close parentheses squared open parentheses 0 comma 1 space M close parentheses end fraction space italic k equals 10 space M to the power of negative sign 2 end exponent space detik to the power of negative sign 1 end exponent 


Jadi, nilai tetapan lajunya adalah bold 10 bold space italic M to the power of bold minus sign bold 2 end exponent bold space bold detik to the power of bold minus sign bold 1 end exponent.space space space

0

Roboguru

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

Pembahasan Soal:

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

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

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

dimana m dan n adalah orde masing-masing reaktan.

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

r almost equal to 1 over t 

Langkah 1: Tentukan orde X dan orde Y.

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

Langkah 2: Tentukan orde total.

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

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

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

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

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


Dengan demikian, diperoleh hasil bahwa:

a. orde reaksi terhadap P = 1.

b. orde reaksi terhadap Q = 2.

c. orde total = 3.

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

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

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

0

Roboguru

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

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

Pembahasan Soal:

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

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

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

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

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

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

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

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

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

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

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

0

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