Roboguru

Pengukuran laju reaksi biasanya ditentukan dengan menggunakan metode laju awal, yaitu laju reaksi berdasarkan konsentrasi awal zat-zat yang bereaksi. Untuk reaksi: BrO3−​+5Br−+6H+→3Br2​+3H2​O  didapat data sebagai berikut.     a. Tentukan orde reaksi pada [BrO3−​],[Br−],[H+] dan orde reaksi total. b. Tentukanlah persamaan laju reaksinya. c. Tentukan laju relatif reaksi pada kondisi kosentrasi  berturut-turut 0,3 M; 0,2 M, dan 0,1 M.

Pertanyaan

Pengukuran laju reaksi biasanya ditentukan dengan menggunakan metode laju awal, yaitu laju reaksi berdasarkan konsentrasi awal zat-zat yang bereaksi.

Untuk reaksi: Br O subscript 3 to the power of minus sign plus 5 Br to the power of minus sign and 6 H to the power of plus sign yields 3 Br subscript 2 and 3 H subscript 2 O 

didapat data sebagai berikut.
 


 

a. Tentukan orde reaksi pada open square brackets Br O subscript 3 to the power of minus sign close square brackets comma space open square brackets Br to the power of minus sign close square brackets comma space open square brackets H to the power of plus sign close square brackets dan orde reaksi total.

b. Tentukanlah persamaan laju reaksinya.

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

Pembahasan Soal:

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

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

 

b. Menentukan persamaan laju reaksinya

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

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

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

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

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

S. Lubis

Mahasiswa/Alumni Universitas Sumatera Utara

Terakhir diupdate 07 Oktober 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

Pertanyaan yang serupa

Pada reaksi: P(g)+Q(g)→zathasil, 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 r...

0

Roboguru

Pada reaksi: 2X(aq)+Y(aq)→Z(aq) diperoleh data percobaan sebagai berikut:   Tentukan: orde reaksi total, rumus laju reaksi, nilai tetapan laju reaksi (k) dan satuannya, nilai x

2

Roboguru

Laju reaksi: X(g)+Y(g)→zathasil ditentukan dari percobaan berikut.   Berdasarkan data tersebut, tentukan: a. orde reaksi terhadap X, b. orde reaksi terhadap Y, c. orde total, d. rumus laju rea...

0

Roboguru

Pada reaksi: 2NO(g)+O2​(g)→N2​O4​(g) diperoleh data sebagai berikut:   Tentukan orde reaksi terhadap masing-masing pereaksi. Tentukan rumus laju reaksinya. Hitung nilai tetapan laju reaksi dan s...

0

Roboguru

Gas nitrogen oksida dan gas bromin bereaksi pada 0∘C menurut persamaan reaksi berikut:   2NO(g)+Br2​(g)→2NOBr(g)   Laju reaksi diikuti dengan mengukur pertambahan konsentrasi NOBr dan diperoleh data...

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ikuti Kami

©2021 Ruangguru. All Rights Reserved