Roboguru

Diketahui reaksi: Menurut data percobaan pada suhu tetap:   Reaksi di atas merupakan reaksi orde ....

Pertanyaan

Diketahui reaksi:

2 Fe to the power of 3 plus sign left parenthesis italic a italic q right parenthesis plus 3 S to the power of 2 minus sign left parenthesis italic a italic q right parenthesis yields S open parentheses italic s close parentheses and 2 Fe S open parentheses italic s close parentheses

Menurut data percobaan pada suhu tetap:
 


Reaksi di atas merupakan reaksi orde ....space

  1. 1space

  2. 2space

  3. 3space

  4. 4space

  5. 5space

Pembahasan Soal:

Rumus persamaan laju reaksi adalah v double bond k open square brackets Fe to the power of 2 plus sign close square brackets to the power of x open square brackets S to the power of 2 minus sign close square brackets to the power of y

  • Orde reaksi terdap Fe to the power of 2 plus sign menggunakan data dari percobaan 1 dan 2 karena konsentrasi S to the power of 2 minus sign tetap.

begin inline style v subscript 1 over v subscript 2 end style equals begin inline style fraction numerator k open square brackets Fe to the power of 2 plus sign close square brackets subscript 1 to the power of x open square brackets S to the power of 2 minus sign close square brackets subscript 1 to the power of y over denominator k open square brackets Fe to the power of 2 plus sign close square brackets subscript 2 to the power of x open square brackets S to the power of 2 minus sign close square brackets subscript 2 to the power of y end fraction end style begin inline style 2 over 8 end style equals 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 1 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 1 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 italic x equals 2

Berdasarkan perhitungan di atas, orde reaksi terhadap Fe to the power of 2 plus sign adalah 2.

  • Orde reaksi terhadap S to the power of 2 minus sign menggunakan data dari percobaan 2 dan 3 karena konsentrasi Fe to the power of 2 plus sign tetap.

v subscript 2 over v subscript 3 equals begin inline style fraction numerator k open square brackets Fe to the power of 2 plus sign close square brackets subscript 2 to the power of x open square brackets S to the power of 2 minus sign close square brackets subscript 2 to the power of y over denominator k open square brackets Fe to the power of 2 plus sign close square brackets subscript 3 to the power of x open square brackets S to the power of 2 minus sign close square brackets subscript 3 to the power of y end fraction end style begin inline style 8 over 16 end style equals begin inline style fraction numerator up diagonal strike k left parenthesis 0 comma 2 right parenthesis to the power of x end strike left parenthesis 0 comma 1 right parenthesis to the power of y over denominator up diagonal strike k left parenthesis 0 comma 2 right parenthesis to the power of x end strike left parenthesis 0 comma 2 right parenthesis to the power of y 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 y y equals 1

Berdasarkan perhitungan di atas, orde reaksi terhadap S to the power of 2 minus sign adalah 1.

Orde reaksi total adalah 2 plus 1 equals 3.

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

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Data eksperimen untuk reaksi adalah sebagai berikut:     Dari data tersebut dapat disimpulkan bahwa persamaan laju reaksinya adalah ....

Pembahasan Soal:

Rumus persamaan laju reaksi adalah v double bond k open square brackets A close square brackets to the power of x open square brackets B close square brackets to the power of y

Untuk menentukan orde reaksi terhadap A menggunakan data dari percobaan 1 dan 4 karena konsentrasi B tetap

begin inline style v subscript 1 over v subscript 4 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 4 to the power of x open square brackets B close square brackets subscript 4 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 1 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 1 right parenthesis to the power of y end strike end fraction space 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 A adalah 2.

Untuk menentukan orde reaksi terhadap B menggunakan data dari percobaan 1 dan 2 karena konsentrasi A tetap.

v subscript 1 over v subscript 2 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 6 over 12 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 end strike left parenthesis 0 comma 1 right parenthesis to the power of y over denominator up diagonal strike k left parenthesis 0 comma 1 right parenthesis to the power of x end strike left parenthesis 0 comma 2 right parenthesis to the power of y end fraction space 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 y y equals 1

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

Persamaan dari laju reaksi tersebut adalah v double bond k open square brackets A close square brackets squared open square brackets B close square brackets

Jadi, jawaban yang benar adalah E.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 

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Roboguru

Reaksi orde tiga mempunyai satuan tetapan laju reaksi ....

Pembahasan Soal:

Reaksi orde ketiga menunjukkan hubungan antara besarnya pengaruh konsentrasi reaktan terhadap laju reaksi. Laju reaksi berbanding lurus dengan pangkat tiga konsnetrasi reaktan. Pada orde ketiga, laju reaksi dirumuskan sebagai v double bond k open square brackets Reaktan close square brackets cubed . Tetapan laju reaksi (k) memiliki satuan yang berbeda-beda yang tergantung pada orde reaksi. Penentuan satuan k dapat diturunkan berdasarkan rumus laju reaksi berorde tiga yaitu sebagai berikut.

  • Laju reaksi memiliki satuan M space s to the power of negative sign 1 end exponent atau mol space dm to the power of negative sign 3 end exponent space detik to the power of negative sign 1 end exponent.
  • Konsentrasi reaktan memiliki satuan Molar atau mol space dm to the power of negative sign 3 end exponent.
  • Satuan tetapan laju reaksi adalah 

table attributes columnalign right center left columnspacing 0px end attributes row v equals cell k open square brackets Reaktan close square brackets cubed end cell row cell mol space dm to the power of negative sign 3 end exponent space detik to the power of negative sign 1 end exponent end cell equals cell k left parenthesis mol space dm to the power of negative sign 3 end exponent right parenthesis cubed end cell row cell mol space dm to the power of negative sign 3 end exponent space detik to the power of negative sign 1 end exponent end cell equals cell k cross times mol cubed space dm to the power of negative sign 9 end exponent end cell row k equals cell fraction numerator mol space dm to the power of negative sign 3 end exponent space detik to the power of negative sign 1 end exponent over denominator mol cubed space dm to the power of negative sign 9 end exponent end fraction end cell row k equals cell mol to the power of 1 minus sign 3 end exponent space dm to the power of left parenthesis minus sign 3 minus sign left parenthesis minus sign 9 right parenthesis right parenthesis end exponent detik to the power of negative sign 1 end exponent end cell row k equals cell mol to the power of negative sign 2 end exponent space dm to the power of 6 space detik to the power of negative sign 1 end exponent end cell end table   

Dengan demikian, satuan tetapan laju reaksi berorde tiga adalah mol to the power of negative sign 2 end exponent space dm to the power of 6 space detik to the power of negative sign 1 end exponent .

Jadi, jawaban yang benar adalah C.

0

Roboguru

Data percobaan laju reaksi diperoleh dari reaksi: , sebagai berikut: Percobaan [A] molar [B] molar Laju reaksi (molar/detik) 1 0,01 0,20 0,02 2 0,02 ...

Pembahasan Soal:

Persamaan laju reaksi merupakan persamaan yang menunjukkan keterkaitan atau hubungan antara laju reaksi tertentu dengan konsentrasi pereaksinya. Persamaan reaksi diperoleh melalui percobaan bukan melalui persamaan reaksi kimia. Pada persamaan laju reaksi perlu ditentukan terlebih dahulu orde reaksinya, yaitu bilangan pangkat konsentrasi. Orde reaksi diperoleh dengan cara mengubah konsentrasi salah satu pereaksi dan membuat konsentrasi zat lain tetap. 

  • Orde terhadap [A]

table attributes columnalign right center left columnspacing 0px end attributes row cell italic v subscript 2 over italic v subscript 1 end cell equals cell italic k over italic k open parentheses open square brackets A close square brackets subscript 2 over open square brackets A close square brackets subscript 1 close parentheses to the power of italic x open parentheses open square brackets B close square brackets subscript 2 over open square brackets B close square brackets subscript 1 close parentheses to the power of italic y end cell row cell fraction numerator 0 comma 08 over denominator 0 comma 02 end fraction end cell equals cell open parentheses fraction numerator 0 comma 02 over denominator 0 comma 01 end fraction close parentheses to the power of italic x open parentheses fraction numerator 0 comma 20 over denominator 0 comma 20 end fraction close parentheses to the power of italic y end cell row 4 equals cell 2 to the power of x end cell row x equals 2 row blank blank blank end table   

  • Orde terhadap [B]

table attributes columnalign right center left columnspacing 0px end attributes row cell italic v subscript 4 over italic v subscript 3 end cell equals cell italic k over italic k open parentheses open square brackets A close square brackets subscript 4 over open square brackets A close square brackets subscript 3 close parentheses to the power of italic x open parentheses open square brackets B close square brackets subscript 4 over open square brackets B close square brackets subscript 3 close parentheses to the power of italic y end cell row cell fraction numerator 0 comma 36 over denominator 0 comma 18 end fraction end cell equals cell open parentheses fraction numerator 0 comma 03 over denominator 0 comma 03 end fraction close parentheses to the power of italic x open parentheses fraction numerator 0 comma 40 over denominator 0 comma 20 end fraction close parentheses to the power of italic y end cell row 2 equals cell 2 to the power of italic y end cell row y equals 1 row blank blank blank end table  

  • Persamaan laju reaksi: italic v equals italic k open square brackets A close square brackets squared open square brackets B close square brackets 

Jadi, jawaban yang benar adalah A.

0

Roboguru

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

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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