Roboguru

Pada reaksi  diperoleh data sebagai berikut.    Tentukan rumus laju reaksinya. Tentukan nilai tetapan laju reaksi (k).

Pertanyaan

Pada reaksi A and B and C yields diperoleh data sebagai berikut.

  

  1. Tentukan rumus laju reaksinya.
  2. Tentukan nilai tetapan laju reaksi (k).space space space

Pembahasan Soal:

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

Reaksi A and B and C yields mempunyai persamaan laju reaksi:

italic v equals italic k open square brackets A close square brackets to the power of italic x open square brackets B close square brackets to the power of italic y open square brackets C close square brackets to the power of italic z  

dengan:

k = tetapan laju reaksi
x = orde (tingkat atau pangkat) reaksi terhadap A
y = orde reaksi terhadap B
z = orde reaksi terhadap C

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

a.   Rumus laju reaksi

  • Menghitung orde reaksi B

    Untuk menghitung orde reaksi B, pilih 2 percobaan dimana A dan C mempunyai konsentrasi yang sama, yaitu percobaan (1) dan (2).

    italic v subscript 1 over italic v subscript 2 equals fraction numerator italic k open square brackets A close square brackets subscript 1 superscript italic x open square brackets B close square brackets subscript 1 superscript italic y begin italic style open square brackets C close square brackets end style subscript italic 1 superscript italic z over denominator italic k open square brackets A close square brackets subscript 2 superscript italic x open square brackets B close square brackets subscript 2 superscript italic y open square brackets C close square brackets subscript 2 superscript italic z end fraction 10 over 15 equals fraction numerator up diagonal strike italic k open parentheses 0 comma 1 close parentheses to the power of italic x end strike left parenthesis 0 comma 2 right parenthesis to the power of italic y up diagonal strike left parenthesis 0 comma 2 right parenthesis to the power of italic z end strike over denominator up diagonal strike italic k begin italic style left parenthesis straight 0 straight comma straight 1 right parenthesis end style to the power of italic x end strike left parenthesis 0 comma 3 right parenthesis to the power of italic y up diagonal strike left parenthesis 0 comma 2 right parenthesis to the power of italic z end strike end fraction 2 over 3 equals open parentheses 2 over 3 close parentheses to the power of italic y italic y equals 1  
     
  • Menghitung orde reaksi C 

    Untuk menghitung orde reaksi C, pilih 2 percobaan dimana A dan B mempunyai konsentrasi yang sama, yaitu percobaan (3) dan (4).

    italic v subscript 3 over italic v subscript 4 equals fraction numerator italic k open square brackets A close square brackets subscript 3 superscript italic x open square brackets B close square brackets subscript 3 superscript italic y open square brackets C close square brackets subscript 3 superscript italic z over denominator italic k open square brackets A close square brackets subscript 4 superscript italic x open square brackets B close square brackets subscript 4 superscript italic y open square brackets C close square brackets subscript 4 superscript italic z end fraction 80 over 160 equals fraction numerator up diagonal strike italic k open parentheses 0 comma 2 close parentheses to the power of italic x end strike up diagonal strike left parenthesis 0 comma 2 right parenthesis to the power of italic y end strike left parenthesis 0 comma 4 right parenthesis to the power of italic z over denominator up diagonal strike italic k left parenthesis 0 comma 2 right parenthesis to the power of italic x end strike up diagonal strike italic left parenthesis italic 0 italic comma italic 2 italic right parenthesis to the power of italic y end strike left parenthesis 0 comma 8 right parenthesis to the power of italic z end fraction 1 half equals open parentheses 1 half close parentheses to the power of italic z italic z equals 1 
     
  • Menghitung orde reaksi A

    Kita pilih percobaan (1) dan (3)

     italic v subscript 1 over italic v subscript 3 equals fraction numerator italic k open square brackets A close square brackets subscript 1 superscript italic x open square brackets B close square brackets subscript 1 superscript italic y open square brackets C close square brackets subscript 1 superscript italic z over denominator italic k open square brackets A close square brackets subscript 3 superscript italic x open square brackets B close square brackets subscript 3 superscript italic y open square brackets C close square brackets subscript 3 superscript italic z end fraction 10 over 80 equals fraction numerator up diagonal strike italic k open parentheses 0 comma 1 close parentheses to the power of italic x up diagonal strike left parenthesis 0 comma 2 right parenthesis to the power of italic y end strike left parenthesis 0 comma 2 right parenthesis over denominator up diagonal strike italic k open parentheses 0 comma 2 close parentheses to the power of italic x up diagonal strike left parenthesis 0 comma 2 right parenthesis to the power of italic y end strike left parenthesis 0 comma 4 right parenthesis end fraction 1 over 8 equals open parentheses 1 half close parentheses to the power of italic x open parentheses 1 half close parentheses 1 over 8 cross times 2 over 1 equals open parentheses 1 half close parentheses to the power of italic x 1 fourth equals open parentheses 1 half close parentheses to the power of italic x italic x equals 2 
     
  • Persamaan laju reaksi

    italic v equals italic k open square brackets A close square brackets to the power of italic x open square brackets B close square brackets to the power of italic y open square brackets C close square brackets to the power of italic z italic v equals italic k open square brackets A close square brackets squared open square brackets B close square brackets open square brackets C close square brackets 


b.   Tetapan laju reaksi (k)

       Misal kita pilih percobaan nomor (1):

        italic v equals italic k open square brackets A close square brackets squared open square brackets B close square brackets open square brackets C close square brackets italic k equals fraction numerator italic v over denominator open square brackets A close square brackets squared open square brackets B close square brackets open square brackets C close square brackets end fraction italic k equals fraction numerator 10 space M space s to the power of negative sign 1 end exponent over denominator open parentheses 0 comma 1 space M close parentheses squared open parentheses 0 comma 2 space M close parentheses open parentheses 0 comma 2 space M close parentheses end fraction italic k equals 25.000 space M to the power of negative sign 3 end exponent space s to the power of negative sign 1 end exponent   


Jadi, persamaan laju reaksinya adalah italic v bold equals italic k bold left square bracket italic A bold right square bracket to the power of bold 2 bold open square brackets B bold close square brackets bold open square brackets C bold close square brackets dengan nilai tetapan laju reaksinya adalah bold 25 bold. bold 000 bold space italic M to the power of bold minus sign bold 3 end exponent bold space italic s to the power of bold minus sign bold 1 end exponent.space space space

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

S. Lubis

Mahasiswa/Alumni Universitas Sumatera Utara

Terakhir diupdate 30 Agustus 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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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. 

Roboguru

Dari suatu percobaan penyelidikan laju terhadap zat A diperoleh data sebagai berikut:   a. Buatlah grafik laju terhadap   b. Tulis orde reaksi terhadap A c. Tuliskan persamaan laju reaksinya d. T...

Pembahasan Soal:

a. Grafik laju terhadap open square brackets A close square brackets 

 

b. Orde reaksi terhadap A

V3V28×10216×1022x====k[A]3xk[A]2x(4×1048×104)x2x1

Oleh karena itu, orde reaksi terhadap A adalah 2.

c. Persamaan laju reaksi

v=k[A] 

Oleh karena itu, persamaan laju reaksi adalah italic v bold equals italic k begin bold style left square bracket italic A right square bracket end style to the power of bold 2.

d. Harga k

Percobaan No 3

v4×104k====k[A]2k(8×102)8×1024×104200s 

Jadi, orde terhadap A adalah 1 dengan persamaan laju reaksinya v=k[A]dan harga k adalah 200s.

Roboguru

Berikut diberikan data percobaan laju reaksi  pada beberapa kondisi.   Jika [Q] dan [T] masing-masing diubah menjadi 0,5 M, maka harga laju (v) reaksi saat itu adalah …..

Pembahasan Soal:

Rumus persamaan laju reaksi adalah v double bond k open square brackets Q close square brackets to the power of x open square brackets T close square brackets to the power of y.
Orde reaksi terhadap Q dapat dicari dengan memilih konsentrasi T yang tetap, yaitu percobaan 1 dan 2.


v subscript 1 over v subscript 2 equals fraction numerator k point open square brackets Q close square brackets subscript 1 superscript x. open square brackets T close square brackets subscript 1 superscript y over denominator k point open square brackets Q close square brackets subscript 2 superscript x. open square brackets T close square brackets subscript 2 superscript y end fraction fraction numerator 1 comma 25.10 to the power of negative sign 2 end exponent over denominator 5.10 to the power of negative sign 2 end exponent end fraction equals fraction numerator k point left parenthesis 0 comma 1 right parenthesis to the power of x left parenthesis 0 comma 1 right parenthesis to the power of y over denominator k point left parenthesis 0 comma 2 right parenthesis to the power of x left parenthesis 0 comma 1 right parenthesis to the power of y end fraction 1 fourth equals left parenthesis 1 half right parenthesis to the power of italic x space x equals 2 


Jadi, orde raksi terhadap Q adalah 2.

Orde reaksi terhadap T dapat dicari dengan memilih konsentrasi Q yang tetap, yaitu percobaan 1 dan 3.


v subscript 1 over v subscript 3 equals fraction numerator k point open square brackets Q close square brackets subscript 1 superscript x. open square brackets T close square brackets subscript 1 superscript y over denominator k point open square brackets Q close square brackets subscript 3 superscript x. open square brackets T close square brackets subscript 3 superscript y end fraction fraction numerator 1 comma 25.10 to the power of negative sign 2 end exponent over denominator 10 to the power of negative sign 1 end exponent end fraction equals fraction numerator k point left parenthesis 0 comma 1 right parenthesis to the power of x left parenthesis 0 comma 1 right parenthesis to the power of y over denominator k point left parenthesis 0 comma 1 right parenthesis to the power of x left parenthesis 0 comma 2 right parenthesis to the power of y end fraction 1 over 8 equals left parenthesis 1 half right parenthesis to the power of y space y equals 3 


Jadi, orde reaksi terhadap T adalah 3.

Rumus persamaan laju reaksi adalah  v double bond k open square brackets Q close square brackets squared open square brackets T close square brackets cubed. 

Penentuan harga k, pilih salah satu percobaan, misalnya percobaan 2.


v subscript 2 double bond k open square brackets Q close square brackets subscript 2 superscript 2 open square brackets T close square brackets subscript 2 superscript 3 space 5.10 to the power of negative sign 2 end exponent equals k left square bracket 0 comma 2 right square bracket squared left square bracket 0 comma 1 right square bracket cubed space k equals 1.250 space M to the power of negative sign 4 end exponent det to the power of negative sign 1 end exponent 


Jika [Q] dan [T] masing-masing diubah menjadi 0,5 M, maka harga laju (v) reaksi, sebagai berikut:


v double bond k open square brackets Q close square brackets squared open square brackets T close square brackets cubed space v equals 1.250 cross times left parenthesis 0 comma 5 right parenthesis squared left parenthesis 0 comma 5 right parenthesis cubed space v equals 39 space M forward slash det   


Jadi, jawaban yang paling tepat adalah E.space 

Roboguru

Reaksi berikut ini berlangsung pada temperatur . Laju reaksinya diamati dengan menggunakan metode laju awal, yang dinyatakan sebagai laju berkurangnya tekanan parsial reaktan (pereaksi). Reaksi diamat...

Pembahasan Soal:

Laju reaksi adalah laju pengurangan konsentrasi pereaksi atau penamahan konsentrasi hasil reaksi per satuan waktu.

  • Menghitung orde reaksi terhadap N O left parenthesis P subscript H subscript 2 end subscript equals 400 space mmHg right parenthesis (Percobaan 2 danspace3)

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses open parentheses P subscript NO close parentheses subscript 2 over open parentheses P subscript NO close parentheses subscript 3 close parentheses to the power of m end cell equals cell v subscript 2 over v subscript 3 end cell row cell open parentheses 300 over 152 close parentheses to the power of m end cell equals cell fraction numerator 0 comma 515 over denominator 0 comma 125 end fraction end cell row cell 2 to the power of m end cell equals 4 row m equals 2 end table
 

  • Menghitung orde reaksi terhadap H subscript 2 left parenthesis P subscript NO equals 400 space mmHg right parenthesis (Percobaan 2 danspace3)

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses open parentheses P subscript H subscript 2 end subscript close parentheses subscript 1 over open parentheses P subscript H subscript 2 end subscript close parentheses subscript 3 close parentheses to the power of n end cell equals cell v subscript 1 over v subscript 3 end cell row cell open parentheses 289 over 147 close parentheses to the power of n end cell equals cell fraction numerator 0 comma 800 over denominator 0 comma 395 end fraction end cell row cell 2 to the power of n end cell equals 2 row n equals 1 end table
 

  • Menghitung orde total reaksi

m and n equals 2 plus 1 equals 3
 

  • Menghitung tetapan laju reaksi tersebut pada 826 degree C dimasukkan 0,1 mol gas H subscript 2 dan 0,2 mol NO

table attributes columnalign right center left columnspacing 0px end attributes row cell P subscript H subscript 2 end subscript end cell equals cell fraction numerator mol space H subscript 2 over denominator mol space H subscript 2 and mol space N O end fraction cross times P space total end cell row blank equals cell fraction numerator 0 comma 1 space mol over denominator 0 comma 1 space mol plus 0 comma 2 space mol end fraction cross times left parenthesis 400 space mmHg plus 400 space mmHg right parenthesis end cell row blank equals cell fraction numerator 0 comma 1 space mol over denominator 0 comma 3 space mol end fraction cross times 800 space mmHg end cell row blank equals cell 267 space mmHg end cell end table


table attributes columnalign right center left columnspacing 0px end attributes row cell P subscript NO end cell equals cell fraction numerator mol space N O over denominator mol space H subscript 2 and mol space N O end fraction cross times P space total end cell row blank equals cell fraction numerator 0 comma 2 space mol over denominator 0 comma 1 space mol plus 0 comma 2 space mol end fraction cross times left parenthesis 400 space mmHg plus 400 space mmHg right parenthesis end cell row blank equals cell fraction numerator 0 comma 2 space mol over denominator 0 comma 3 space mol end fraction cross times 800 space mmHg end cell row blank equals cell 533 space mmHg end cell end table
 

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses fraction numerator open parentheses P subscript NO close parentheses over denominator open parentheses P subscript NO close parentheses subscript 3 end fraction close parentheses to the power of m space open parentheses fraction numerator open parentheses P subscript H subscript 2 end subscript close parentheses over denominator open parentheses P subscript H subscript 2 end subscript close parentheses subscript 3 end fraction close parentheses to the power of n end cell equals cell v over v subscript 3 end cell row cell open parentheses 533 over 152 close parentheses squared space open parentheses 267 over 147 close parentheses to the power of 1 end cell equals cell fraction numerator v over denominator 0 comma 395 end fraction end cell row cell open parentheses 12 comma 25 close parentheses space open parentheses 1 comma 82 close parentheses to the power of 1 end cell equals cell fraction numerator v over denominator 0 comma 395 end fraction end cell row cell 22 comma 295 end cell equals cell fraction numerator v over denominator 0 comma 395 end fraction end cell row v equals cell 8 comma 806 space mmHg end cell end table


table attributes columnalign right center left columnspacing 0px end attributes row v equals cell k space open parentheses P subscript NO close parentheses squared space open parentheses P subscript H subscript 2 end subscript close parentheses end cell row cell 8 comma 806 end cell equals cell k space open parentheses 533 close parentheses squared space open parentheses 267 close parentheses end cell row cell 8 comma 806 end cell equals cell k space left parenthesis 75.851 comma 763 right parenthesis end cell row k equals cell 1 comma 2 cross times 10 to the power of negative sign 4 end exponent end cell end table
 

  • Menentukan tekanan total gas pada awal reaksi

table attributes columnalign right center left columnspacing 0px end attributes row cell P subscript total end cell equals cell P subscript NO and P subscript H subscript 2 end subscript end cell row blank equals cell 400 space mmHg plus 400 space mmHg end cell row blank equals cell 800 space mmHg end cell end table
 

  • Menentukan laju awal reaksi dengan komposisi tersebut

table attributes columnalign right center left columnspacing 0px end attributes row v equals cell k space open parentheses P subscript NO close parentheses squared space open parentheses P subscript H subscript 2 end subscript close parentheses end cell row blank equals cell 1 comma 2 cross times 10 to the power of negative sign 4 end exponent space open parentheses 400 close parentheses squared space open parentheses 400 close parentheses end cell row blank equals cell 76 comma 8 space mmHg forward slash s end cell end table

Jadi, jawaban untuk pertanyaan a-e seperti penjelasan di atas.

Roboguru

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

Pembahasan Soal:

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

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

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

dengan:

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

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

a.   Orde reaksi terhadap masing-masing pereaksi

  • Orde reaksi N O  

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

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

     
  • Orde reaksi O subscript bold 2 

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

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


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


b.   Rumus laju reaksi

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


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


c.   Nilai tetapan laju reaksi (k)

      Misal kita ambil percobaan nomor (1)

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


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

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