Roboguru

Laju reaksi dari:     ditentukan dengan mengukur waktu yang diperlukan untuk membentuk jumlah tertentu endapan C.     Tentukan persamaan laju reaksinya! Berapakah waktu reaksi apabila konsentrasi A dan  masing-masing 0,3 M?

Pertanyaan

Laju reaksi dari:
 

2 A left parenthesis italic a italic q right parenthesis plus B subscript 2 left parenthesis italic a italic q right parenthesis yields C open parentheses italic s close parentheses and D left parenthesis italic a italic q right parenthesis
 

ditentukan dengan mengukur waktu yang diperlukan untuk membentuk jumlah tertentu endapan C.
 


 

  1. Tentukan persamaan laju reaksinya!
  2. Berapakah waktu reaksi apabila konsentrasi A dan B subscript 2 masing-masing 0,3 M?space

Pembahasan Soal:

a. pertama menentukan orde reaksi A dengan perbandingan data 2 dan 1 yaitu:
 

table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets A close square brackets to the power of x open square brackets B subscript 2 close square brackets to the power of y end cell equals V row cell open square brackets fraction numerator 0 comma 2 over denominator 0 comma 1 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 80 over 40 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
 

Menentukan orde B subscript 2 dengan membandingkan data 3 banding 2, yaitu:
 

table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets A close square brackets to the power of x open square brackets B subscript 2 close square brackets to the power of y end cell equals V row cell open square brackets fraction numerator 0 comma 2 over denominator 0 comma 2 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 40 over 10 end cell row cell open square brackets 2 close square brackets to the power of italic y end cell equals 4 row cell open square brackets 2 close square brackets to the power of y end cell equals cell 2 squared end cell row y equals 2 end table
 

Dengan demikian maka persamaan lajunya adalah: V double bond k open square brackets A close square brackets open square brackets B subscript 2 close square brackets squared

b. waktu jika konsentrasi pereaksi masing-masing adalah 0,3 M adalah:
 

table attributes columnalign right center left columnspacing 0px end attributes row cell V subscript dit over V subscript 1 end cell equals cell fraction numerator k open square brackets A close square brackets open square brackets B subscript 2 close square brackets squared over denominator k open square brackets A close square brackets open square brackets B subscript 2 close square brackets squared end fraction end cell row cell fraction numerator V subscript dit over denominator begin display style bevelled 1 over 80 end style end fraction end cell equals cell fraction numerator open square brackets 0 comma 3 close square brackets open square brackets 0 comma 3 close square brackets squared over denominator open square brackets 0 comma 1 close square brackets open square brackets 0 comma 1 close square brackets squared end fraction end cell row cell V subscript dit end cell equals cell fraction numerator open square brackets 0 comma 3 close square brackets open square brackets 0 comma 09 close square brackets over denominator open square brackets 0 comma 1 close square brackets open square brackets 0 comma 01 close square brackets end fraction cross times 1 over 80 end cell row cell V subscript dit end cell equals cell fraction numerator 0 comma 027 over denominator 0 comma 001 end fraction cross times 1 over 80 end cell row cell V subscript dit end cell equals cell 27 over 80 end cell row cell 1 over t end cell equals cell 27 over 80 end cell row t equals cell 80 over 27 end cell row t equals cell 2 comma 96 space s end cell end table
 

Dengan demikian maka waktu yang dibutuhkan adalah 2,96 s.space

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

M. Pulungan

Mahasiswa/Alumni Universitas Negeri Medan

Terakhir diupdate 02 Mei 2021

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Pertanyaan yang serupa

Reaksi gas bromin dengan gas nitrogen oksida sesuai dengan persamaan reaksi: Berdasarkan hasil dari percobaan yang diperoleh data sebagai berikut:     Tentukan orde reaksi terhadap NO!

Pembahasan Soal:

Rumus persamaan laju reaksi adalah v double bond k open square brackets N O close square brackets to the power of x open square brackets Br subscript 2 close square brackets to the power of y

Menentukan orde reaksi terhadap NO menggunakan data percobaan 1 dan 3 karena konsentrasi Br tetap.

begin inline style v subscript 1 over v subscript 3 end style equals begin inline style fraction numerator k open square brackets N O close square brackets subscript 1 to the power of x open square brackets Br subscript 2 close square brackets subscript 1 to the power of y over denominator k open square brackets N O close square brackets subscript 3 to the power of x open square brackets Br subscript 2 close square brackets subscript 3 to the power of y end fraction end style begin inline style 6 over 24 end style begin inline style equals end style begin inline style fraction numerator up diagonal strike k left parenthesis 0 comma 1 right parenthesis to the power of x up diagonal strike left parenthesis 0 comma 5 right parenthesis to the power of y end strike over denominator up diagonal strike k left parenthesis 0 comma 2 right parenthesis to the power of x up diagonal strike left parenthesis 0 comma 5 right parenthesis to the power of y end strike end fraction end style begin inline style 1 fourth end style equals open parentheses begin inline style 1 half end style close parentheses to the power of x x equals 2

Berdasarkan perhitungan di atas, orde reaksi terhadap NO adalah 2.

Jadi, jawaban yang benar adalah E.space

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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

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Reaksi antara ion bromat dan ion bromida dalam larutan asam dinyatakan dengan persamaan reaksi:   Campuran reaksi dibuat melalui pencampuran larutan-Iarutan induk dengan volume dan Iaju awal berkurang...

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

Untuk persamaan reaksi :   Diperoleh data hasil eksperimen pada suhu tetap sebagai berikut.       Tentukan persamaan laju reaksi?

Pembahasan Soal:

Laju reaksi adalah laju berkurangnya konsentrasi reaktan atau laju bertambahnya konsentrasi produk per satuan waktu. Laju reaksi dinyatakan dalam persamaan laju sebagai berikut :

v=k[X]a[Y]b

Untuk menentukan persamaan laju harus menentukan orde reaksi terlebih dahulu. Penentuan orde reaksi adalah dengan menggunakan 2 data percobaan yang memuat konsentrasi dan laju reaksi, dimana konsentrasi salah satunya bernilai sama. Namun karena pada soal diketahui adalah waktu, maka laju (v) menjadi v=t1

Orde total 
Menentukan orde reaksi Y
Menggunakan data percobaan 1 dan 2, karena konsentrasi X sama
Error converting from MathML to accessible text. 


Menentukan orde reaksi X
Menggunakan data percobaan 1 dan 3, karena konsentrasi Y sama
Error converting from MathML to accessible text. 


Orde total=1+1=2

Harga k
Error converting from MathML to accessible text.

 

Persamaan laju reaksi
v=33,33M1s1[X][Y] 
 

Jadi, persamaan laju reaksinya adalah v=33,33M1s1[X][Y] 

0

Roboguru

Berikut ini diberikan data percobaan laju reaksi:   pada beberapa kondisi:     Jika  masing-masing diubah menjadi 0,5 M, maka harga laju reaksi  adalah ...

Pembahasan Soal:

Mencari orde reaksi Q

Dimisalkan, persamaan laju reaksi begin mathsize 14px style V equals italic k open square brackets Q close square brackets to the power of italic x open square brackets T close square brackets to the power of y end style. Cara mencari orde reaksi Q (x) adalah dengan memilih data konsentrasi T yang sama yaitu data (1) dan (2). 

begin mathsize 14px style V subscript left parenthesis 1 right parenthesis end subscript over V subscript left parenthesis 2 right parenthesis end subscript equals italic k subscript left parenthesis 1 right parenthesis end subscript over italic k subscript left parenthesis 2 right parenthesis end subscript cross times open square brackets italic Q subscript left parenthesis 1 right parenthesis end subscript over italic Q subscript left parenthesis 2 right parenthesis end subscript close square brackets to the power of italic x cross times open square brackets italic T subscript left parenthesis 1 right parenthesis end subscript over italic T subscript left parenthesis 2 right parenthesis end subscript close square brackets to the power of italic y end style, harga begin mathsize 14px style italic k subscript left parenthesis 1 right parenthesis end subscript equals italic k subscript left parenthesis 2 right parenthesis end subscript end style (karena suhu tetap) dan begin mathsize 14px style open square brackets T subscript left parenthesis 1 right parenthesis end subscript close square brackets equals open square brackets T subscript left parenthesis 2 right parenthesis end subscript close square brackets end style sehingga begin mathsize 14px style open square brackets T subscript left parenthesis 1 right parenthesis end subscript over T subscript left parenthesis 2 right parenthesis end subscript close square brackets to the power of italic y end style dapat dihilangkan (bernilai 1). 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 1 comma 25 cross times 10 to the power of negative sign 2 end exponent space Ms to the power of negative sign 1 end exponent over denominator 5 comma 00 cross times 10 to the power of negative sign 2 end exponent space Ms to the power of negative sign 1 end exponent end fraction end cell equals cell open square brackets fraction numerator 0 comma 1 over denominator 0 comma 2 end fraction close square brackets to the power of italic x end cell row cell 1 fourth end cell equals cell open square brackets 1 half close square brackets to the power of italic x end cell row italic x equals 2 end table end style 

Jadi orde reaksi terhadap Q adalah 2.
 

Mencari orde reaksi T

Untuk menentukan orde reaksi T (y)  dipilih data konsentrasi Q yang sama yaitu data (1) dan (3).

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell V subscript left parenthesis 1 right parenthesis end subscript over V subscript left parenthesis 3 right parenthesis end subscript end cell equals cell italic k subscript left parenthesis 1 right parenthesis end subscript over italic k subscript left parenthesis 3 right parenthesis end subscript cross times open square brackets italic Q subscript left parenthesis 1 right parenthesis end subscript over italic Q subscript left parenthesis 3 right parenthesis end subscript close square brackets to the power of italic x cross times open square brackets italic T subscript left parenthesis 1 right parenthesis end subscript over italic T subscript left parenthesis 3 right parenthesis end subscript right square bracket close square brackets to the power of italic y end cell row cell fraction numerator 1 comma 25 cross times 10 to the power of negative sign 2 end exponent space Ms to the power of negative sign 1 end exponent over denominator 1 x 10 to the power of negative sign 1 end exponent space Ms to the power of negative sign 1 end exponent end fraction end cell equals cell fraction numerator 0 comma 1 over denominator 0 comma 1 end fraction cross times open square brackets 1 half close square brackets to the power of y end cell row cell 1 over 8 end cell equals cell open square brackets 1 half close square brackets to the power of y end cell row y equals 3 end table end style   

Jadi orde reaksi terhadap T adalah 3.
 

Mencari nilai k

Pesamaan laju reaksi pembentukan begin mathsize 14px style T subscript 2 Q end style dapat dituliskan sebagai berikut :

begin mathsize 14px style V double bond k open square brackets Q close square brackets squared open square brackets T close square brackets cubed end style 

Untuk menentukan harga konstanta laju reaksi k maka dapat memasukan data percobaan di atas misalkan pada data percobaan (1), maka diperoleh:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row V equals cell italic k open square brackets Q close square brackets squared open square brackets T close square brackets cubed end cell row cell 1 comma 25 cross times 10 to the power of negative sign 2 end exponent space Ms to the power of negative sign 1 end exponent end cell equals cell italic k cross times left parenthesis 10 to the power of negative sign 1 end exponent right parenthesis squared cross times left parenthesis 10 to the power of negative sign 1 end exponent right parenthesis cubed end cell row italic k equals cell fraction numerator 1 comma 25 cross times 10 to the power of negative sign 2 end exponent space Ms to the power of negative sign 1 end exponent over denominator 10 to the power of negative sign 5 end exponent end fraction end cell row italic k equals cell 1 comma 25 cross times 10 cubed end cell end table end style 

Jadi harga konstanta laju reaksi adalah begin mathsize 14px style bold 1 bold comma bold 25 bold cross times bold 10 to the power of bold 3 end style.
 

Menentukan laju reaksi pada Q 0,5 M dan T 0,5 M  

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row V equals cell italic k open square brackets Q close square brackets squared open square brackets T close square brackets cubed end cell row V equals cell 1 comma 25 cross times 10 cubed cross times open square brackets 5 cross times 10 to the power of negative sign 1 end exponent space M close square brackets squared cross times left square bracket 5 cross times 10 to the power of negative sign 1 end exponent space M right square bracket cubed space end cell row V equals cell 1 comma 25 cross times 10 cubed cross times 5 cross times 10 to the power of negative sign 2 end exponent cross times 5 cross times 10 to the power of negative sign 3 end exponent end cell row V equals cell 3 comma 125 cross times 10 to the power of negative sign 1 end exponent space Ms to the power of negative sign 1 end exponent end cell end table end style 

Jadi laju reaksi pada Q 0,5 M dan T 0,5 M adalah begin mathsize 14px style bold 3 bold comma bold 125 bold cross times bold 10 to the power of bold minus sign bold 1 end exponent bold space bold Ms to the power of bold minus sign bold 1 end exponent end style.space 

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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