Roboguru

Percobaan kinetika reaksi: Menghasilkan data sebagai berikut:     Orde reaksi terhadap X adalah ....

Pertanyaan

Percobaan kinetika reaksi:

X and Y yields P and Q

Menghasilkan data sebagai berikut:
 


 

Orde reaksi terhadap X adalah ....space

  1. nolspace

  2. setengahspace

  3. satuspace

  4. duaspace

  5. tigaspace

Pembahasan Soal:

Persamaan laju reaksi adalah v double bond k open square brackets X close square brackets to the power of m open square brackets Y close square brackets to the power of n.

Orde reaksi terhadap X menggunakan data dari percobaan 1 dan 2 karena konsentrasi Y tetap.

Error converting from MathML to accessible text.

Berdasarkan perhitungan di atas, orde reaksi terhadap X 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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Pertanyaan yang serupa

Pengukuran laju reaksi biasanya ditentukan dengan menggunakan metode laju awal, yaitu laju reaksi berdasarkan konsentrasi awal zat-zat yang bereaksi. Untuk reaksi:   didapat data sebagai berikut.   ...

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 

0

Roboguru

Laju reaksi dari:     ditentukan dengan mengukur waktu yang diperlukan untuk membentuk jumlah tertentu endapan C.     Tentukan persamaan laju reaksinya! Berapakah waktu reaksi apabila konsent...

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

0

Roboguru

Pada percobaan reaksi gas nitrogen dan hidrogen dengan persamaan reaksi  didapatkan hasil percobaan sebagai berikut.    Persamaan laju reaksi dari data di atas adalah ….

Pembahasan Soal:

Laju reaksi menyatakan perubahan konsentrasi tiap satuan waktu. Persamaan laju reaksi tersebut adalah v double bond k open square brackets N subscript 2 close square brackets to the power of x open square brackets H subscript 2 close square brackets to the power of y. Untuk menentukan persamaan laju reaksi yang disertai dengan orde reaksi, langkah-langkahnya adalah sebagai berikut.

  • Menentukan orde reaksi terhadap N subscript 2
    Dengan menggunakan data konsentrasi H subscript 2 yang sama, yaitu data percobaan 1 dan 2.
    v2v20,00040,00082x====k[N2]1x[H2]1yk[N2]2x[H2]2y(0,002)x(0,004)x2x1 

    Jadi, orde raksi terhadap N subscript 2 adalah 1.
     
  • Menentukan orde reaksi terhadap H subscript 2
    Dengan menggunakan data konsentrasi N subscript 2 yang sama, yaitu data percobaan 2 dan 3.
    v2v30,00080,00324y====k[N2]3x[H2]3yk[N2]2x[H2]2y(0,002)y(0,008)y4y1 

    Jadi, orde reaksi terhadap H subscript 2 adalah 1.
     
  • Menuliskan persaaan laju reaksi
    Persamaan laju reaksi adalah v double bond k open square brackets N subscript 2 close square brackets open square brackets H subscript 2 close square brackets point 

Jadi, jawaban yang tepat adalah A.space 

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

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

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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