Roboguru

Berikut ini adalah data hasil percobaan untuk menentukan hubungan antara konsentrasi awal pereaksi terhadap laju reaksi dari:   Laju reaksi diukur berdasarkan waktu yang diperlukan untuk mendapatkan sejumlah endapan belerang yang sama. Hasil yang diperoleh adalah sebagai berikut.   Tentukan orde reaksi terhadap  dan terhadap . Tuliskan persamaan (hukum) laju reaksinya.

Pertanyaan

Berikut ini adalah data hasil percobaan untuk menentukan hubungan antara konsentrasi awal pereaksi terhadap laju reaksi dari:

Na subscript 2 S subscript 2 O subscript 3 left parenthesis italic a italic q right parenthesis plus 2 H Cl left parenthesis italic a italic q right parenthesis yields 2 Na Cl left parenthesis italic a italic q right parenthesis plus H subscript 2 O open parentheses italic l close parentheses and S O subscript 2 open parentheses italic g close parentheses and S open parentheses italic s close parentheses 

Laju reaksi diukur berdasarkan waktu yang diperlukan untuk mendapatkan sejumlah endapan belerang yang sama. Hasil yang diperoleh adalah sebagai berikut.

 

  1. Tentukan orde reaksi terhadap Na subscript 2 S subscript 2 O subscript 3 dan terhadap H Cl.
  2. Tuliskan persamaan (hukum) laju reaksinya.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 

Na subscript 2 S subscript 2 O subscript 3 left parenthesis italic a italic q right parenthesis plus 2 H Cl left parenthesis italic a italic q right parenthesis yields 2 Na Cl left parenthesis italic a italic q right parenthesis plus H subscript 2 O open parentheses italic l close parentheses and S O subscript 2 open parentheses italic g close parentheses and S open parentheses italic s close parentheses 

mempunyai persamaan laju reaksi:

italic v equals italic k open square brackets Na subscript 2 S subscript 2 O subscript 3 close square brackets to the power of italic x open square brackets H Cl close square brackets to the power of italic y 

dengan:

k = tetapan laju reaksi
x = orde (tingkat atau pangkat) reaksi terhadap Na subscript 2 S subscript 2 O subscript 3 
y = orde reaksi terhadap H Cl 

a.   Orde reaksi Na subscript bold 2 S subscript bold 2 O subscript bold 3 dan H Cl 

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

  • Menghitung orde reaksi Na subscript bold 2 S subscript bold 2 O subscript bold 3 

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

    italic v subscript 2 over italic v subscript 3 equals fraction numerator italic k open square brackets Na S subscript 2 O subscript 3 close square brackets subscript 2 superscript italic x open square brackets H Cl close square brackets subscript 2 superscript italic y over denominator italic k open square brackets Na S subscript 2 O subscript 3 close square brackets subscript 3 superscript italic x open square brackets H Cl close square brackets subscript 3 superscript italic y end fraction fraction numerator begin display style bevelled 1 over italic t subscript 2 end style over denominator begin display style bevelled 1 over italic t subscript 3 end style end fraction equals fraction numerator up diagonal strike italic k left parenthesis 0 comma 05 right parenthesis to the power of italic x up diagonal strike left parenthesis 0 comma 4 right parenthesis to the power of italic y end strike over denominator up diagonal strike italic k left parenthesis 0 comma 10 right parenthesis to the power of italic x up diagonal strike left parenthesis 0 comma 4 right parenthesis to the power of italic y end strike end fraction fraction numerator begin display style bevelled 1 over 31 end style over denominator begin display style bevelled 1 over 15 end style end fraction equals fraction numerator up diagonal strike italic k left parenthesis 0 comma 05 right parenthesis to the power of italic x up diagonal strike left parenthesis 0 comma 4 right parenthesis to the power of italic y end strike over denominator up diagonal strike italic k left parenthesis 0 comma 10 right parenthesis to the power of italic x up diagonal strike left parenthesis 0 comma 4 right parenthesis to the power of italic y end strike end fraction 15 over 31 equals open parentheses 1 half close parentheses to the power of italic x 1 half equals open parentheses 1 half close parentheses to the power of italic x italic x equals 1 
     
  • Menghitung orde reaksi HCl

    Untuk menghitung orde reaksi H Cl, pilih 2 percobaan dimana Na subscript 2 S subscript 2 O subscript 3 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 Na S subscript 2 O subscript 3 close square brackets subscript 1 superscript italic x open square brackets H Cl close square brackets subscript 1 superscript italic y over denominator italic k open square brackets Na S subscript 2 O subscript 3 close square brackets subscript 2 superscript italic x open square brackets H Cl close square brackets subscript 2 superscript italic y end fraction fraction numerator begin display style bevelled 1 over italic t subscript 1 end style over denominator begin display style bevelled 1 over italic t subscript 2 end style end fraction equals fraction numerator up diagonal strike italic k left parenthesis 0 comma 05 right parenthesis to the power of italic x end strike left parenthesis 0 comma 2 right parenthesis to the power of italic y over denominator up diagonal strike italic k left parenthesis 0 comma 05 right parenthesis to the power of italic x end strike left parenthesis 0 comma 4 right parenthesis to the power of italic y end fraction fraction numerator begin display style bevelled 1 over 30 end style over denominator begin display style bevelled 1 over 31 end style end fraction equals open parentheses 1 half close parentheses to the power of italic y 31 over 30 equals open parentheses 1 half close parentheses to the power of italic y 1 equals open parentheses 1 half close parentheses to the power of italic y italic y equals 0 


Jadi, orde reaksi terhadap Na subscript bold 2 S subscript bold 2 O subscript bold 3 adalah 1 dan orde reaksi terhadap HCl adalah 0.


b.   Persamaan (hukum) laju reaksi

       italic v equals italic k open square brackets Na subscript 2 S subscript 2 O subscript 3 close square brackets to the power of italic x open square brackets H Cl close square brackets to the power of italic y italic v equals italic k open square brackets Na subscript 2 S subscript 2 O subscript 3 close square brackets to the power of 1 open square brackets H Cl close square brackets to the power of 0 italic v equals italic k open square brackets Na subscript 2 S subscript 2 O subscript 3 close square brackets 


Jadi, persamaan laju reaksinya adalah italic v bold equals italic k bold open square brackets Na subscript bold 2 S subscript bold 2 O subscript bold 3 bold close square brackets.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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Pertanyaan yang serupa

Berikut adalah tabel data laju reaksi :     Rumus laju reaksinya adalah ....

Pembahasan Soal:

Orde reaksi menunjukkan besarnya pengaruh kosentrasi pereaksi terhadap laju reaksi. Tingkat reaksi total merupakan penjumlahan orde semua pereaksi. Penentuan orde reaksi dapat dihitung berdasarkan data eksperimen dan dinyatakan melalui persamaan laju reaksi. Persamaan laju yang mungkin untuk reaksi diatas adalah : v double bond k open square brackets N O close square brackets to the power of m open square brackets Br subscript 2 close square brackets to the power of n. Penentuan orde reaksi melalui data eksperimen dihitung berdasarkan perbandingan data percobaan yang sama.

  • Tentukan orde NO yaitu cari data Br subscript 2 yang sama yaitu percobaan 4 dan 5.

table attributes columnalign right center left columnspacing 0px end attributes row cell v subscript 4 over v subscript 5 end cell equals cell k subscript 4 over k subscript 5 fraction numerator open square brackets N O close square brackets subscript 4 superscript m over denominator open square brackets N O close square brackets subscript 5 superscript m end fraction fraction numerator open square brackets Br subscript 2 close square brackets subscript 4 superscript n over denominator open square brackets Br subscript 2 close square brackets subscript 5 superscript n end fraction end cell row cell 24 over 54 end cell equals cell k subscript 4 over k subscript 5 open parentheses fraction numerator 0 comma 20 over denominator 0 comma 30 end fraction close parentheses to the power of m open parentheses fraction numerator 0 comma 05 over denominator 0 comma 05 end fraction close parentheses to the power of n end cell row blank blank blank row cell 4 over 9 end cell equals cell open parentheses 2 over 3 close parentheses to the power of m end cell row cell open parentheses 2 over 3 close parentheses squared end cell equals cell open parentheses 2 over 3 close parentheses to the power of blank to the power of m end exponent end cell row m equals 2 end table    

  • Tentukan orde Br subscript 2 yaitu cari data NO yang sama yaitu percobaan 1 dan 2.

table attributes columnalign right center left columnspacing 0px end attributes row cell v subscript 1 over v subscript 2 end cell equals cell k subscript 1 over k subscript 2 fraction numerator open square brackets N O close square brackets subscript 1 superscript m over denominator open square brackets N O close square brackets subscript 2 superscript m end fraction fraction numerator open square brackets Br subscript 2 close square brackets subscript 1 superscript n over denominator open square brackets Br subscript 2 close square brackets subscript 2 superscript n end fraction end cell row cell 6 over 12 end cell equals cell k subscript 1 over k subscript 2 open parentheses fraction numerator 0 comma 10 over denominator 0 comma 10 end fraction close parentheses to the power of m open parentheses fraction numerator 0 comma 05 over denominator 0 comma 10 end fraction close parentheses to the power of n end cell row blank blank blank row cell 1 half end cell equals cell open parentheses 1 half close parentheses to the power of n end cell row cell open parentheses 1 half close parentheses to the power of 1 end cell equals cell open parentheses 1 half close parentheses to the power of n end cell row n equals 1 end table 

Berdasarkan perhitungan diatas, diperoleh bahwa orde NO = 2, orde Br subscript 2 = 1, sehingga persamaan laju reaksi dapat dinyatakan sebagai berikut.

v double bond k open square brackets N O close square brackets squared open square brackets Br subscript 2 close square brackets 

Jadi, jawaban yang benar adalah C.

Roboguru

Pada reaksi:  diperoleh data percobaan sebagai berikut:   Tentukan: orde reaksi total, rumus laju reaksi, nilai tetapan laju reaksi (k) dan satuannya, nilai x

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 X left parenthesis italic a italic q right parenthesis plus Y left parenthesis italic a italic q right parenthesis yields Z left parenthesis italic a italic q right parenthesis mempunyai persamaan laju reaksi:

italic v equals italic k open square brackets X close square brackets to the power of italic m open square brackets Y close square brackets to the power of italic n 

dengan:

k = tetapan laju reaksi
m = orde (tingkat atau pangkat) reaksi terhadap X
n = orde reaksi terhadap Y

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

  • Menghitung orde reaksi X

    Untuk menghitung orde reaksi X, pilih 2 percobaan dimana Y 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 X close square brackets subscript 1 superscript italic m open square brackets Y close square brackets subscript 1 superscript italic n over denominator italic k open square brackets X close square brackets subscript 2 superscript italic m open square brackets Y close square brackets subscript 2 superscript italic n end fraction fraction numerator 6 comma 0 cross times 10 to the power of negative sign 3 end exponent over denominator 2 comma 4 cross times 10 to the power of negative sign 2 end exponent end fraction equals fraction numerator up diagonal strike italic k begin italic style left parenthesis straight 0 straight comma straight 2 right parenthesis end style to the power of italic m up diagonal strike begin italic style left parenthesis straight 0 straight comma straight 5 right parenthesis end style to the power of italic n end strike over denominator up diagonal strike italic k left parenthesis 0 comma 4 right parenthesis to the power of italic m up diagonal strike italic left parenthesis italic 0 italic comma italic 5 italic right parenthesis to the power of italic n end strike end fraction 1 fourth equals open parentheses 1 half close parentheses to the power of italic m m equals 2 
     
  • Menghitung orde reaksi Y 

    Untuk menghitung orde reaksi Y, pilih 2 percobaan dimana X mempunyai konsentrasi yang sama, yaitu percobaan (1) dan (3).
    italic v subscript 1 over italic v subscript 3 equals fraction numerator italic k open square brackets X close square brackets subscript 1 superscript italic m open square brackets Y close square brackets subscript 1 superscript italic n over denominator italic k open square brackets X close square brackets subscript 3 superscript italic m open square brackets Y close square brackets subscript 3 superscript italic n end fraction fraction numerator 6 comma 0 cross times 10 to the power of negative sign 3 end exponent over denominator 1 comma 20 cross times 10 to the power of negative sign 3 end exponent end fraction equals fraction numerator up diagonal strike italic k italic left parenthesis italic 0 italic comma italic 2 italic right parenthesis to the power of italic m end strike left parenthesis 0 comma 5 right parenthesis to the power of italic n over denominator up diagonal strike italic k italic left parenthesis italic 0 italic comma italic 2 italic right parenthesis to the power of italic m end strike left parenthesis 0 comma 1 right parenthesis to the power of italic n end fraction 5 over 1 equals open parentheses 5 over 1 close parentheses to the power of italic n n equals 1 
     
  • Menghitung orde reaksi total

    Orde space reaksi space total double bond m and n Orde space reaksi space total equals 2 plus 1 Orde space reaksi space total equals 3 


Jadi, orde reaksi totalnya adalah 3.


b.   Rumus laju reaksi

       italic v equals italic k open square brackets X close square brackets to the power of italic m open square brackets Y close square brackets to the power of italic n italic v equals italic k open square brackets X close square brackets squared open square brackets Y close square brackets 


Jadi, rumus laju reaksinya adalah italic v bold equals italic k bold left square bracket italic X bold right square bracket to the power of bold 2 bold open square brackets Y bold close square brackets.


c.   Nilai tetapan laju reaksi 

Misal kita pilih percobaan nomor (1):

        italic v equals italic k open square brackets X close square brackets squared open square brackets Y close square brackets italic k equals fraction numerator italic v over denominator open square brackets X close square brackets squared open square brackets Y close square brackets end fraction italic k equals fraction numerator 6 comma 0 cross times 10 to the power of negative sign 3 end exponent space mol space L to the power of negative sign 1 end exponent space detik to the power of negative sign 1 end exponent over denominator open parentheses 0 comma 2 space mol space L to the power of negative sign 1 end exponent close parentheses squared open parentheses 0 comma 5 space mol space L to the power of negative sign 1 end exponent close parentheses end fraction italic k equals fraction numerator 6 comma 0 cross times 10 to the power of negative sign 3 end exponent space mol space L to the power of negative sign 1 end exponent space detik to the power of negative sign 1 end exponent over denominator 2 comma 0 cross times 10 to the power of negative sign 2 end exponent space mol cubed space L to the power of negative sign 3 end exponent end fraction italic k equals 0 comma 3 space mol to the power of negative sign 2 end exponent space L to the power of 2 space end exponent space detik to the power of negative sign 1 end exponent  


Jadi, nilai tetapan laju reaksinya adalah bold 0 bold comma bold 3 bold space bold mol to the power of bold minus sign bold 2 end exponent bold space italic L to the power of bold 2 bold space end exponent bold space bold detik to the power of bold minus sign bold 1 end exponent.


d.   nilai x

        italic v equals italic k open square brackets X close square brackets squared open square brackets Y close square brackets italic x equals open parentheses 0 comma 3 space mol to the power of negative sign 2 end exponent space L squared space detik to the power of negative sign 1 end exponent close parentheses open parentheses 0 comma 3 space mol space L to the power of negative sign 1 end exponent close parentheses squared open parentheses 0 comma 4 space mol space L to the power of negative sign 1 end exponent close parentheses italic x equals 1 comma 08 cross times 10 to the power of negative sign 2 end exponent space mol space L to the power of negative sign 1 end exponent space detik to the power of negative sign 1 end exponent 


Jadi, nilai x pada percobaan adalah bold 1 bold comma bold 08 bold cross times bold 10 to the power of bold minus sign bold 2 end exponent bold space bold mol bold space italic L to the power of bold minus sign bold 1 end exponent bold space bold detik to the power of bold minus sign bold 1 end exponent.space space space

Roboguru

Data hasil eksperimen dari reaksi itu adalah sebagai berikut: Tentukan: a. Tingkat reaksi terhadap  b. Tingkat reaksi terhadap   c. Tingkat reaksi total d. Persamaan laju reaksinya

Pembahasan Soal:

persamaan laju reaksi mula - mula :

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 

pembahasan a:

Untuk menentukan tingkat reaksi terhadap NO, kita gunakan konsentrasi mol Bryang sama, yaitu percobaan 1 dan 4. Sehingga diperoleh:

fraction numerator V subscript 1 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 over denominator V subscript 4 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 end fraction fraction numerator 6 equals k left square bracket 0 comma 1 right square bracket to the power of X up diagonal strike open square brackets Br subscript 2 close square brackets to the power of y end strike over denominator 24 equals k left square bracket 0 comma 2 right square bracket to the power of X up diagonal strike open square brackets Br subscript 2 close square brackets to the power of y end strike end fraction open parentheses 1 fourth close parentheses equals open parentheses 1 half close parentheses to the power of italic y x space space equals space 2   

pembahasan b :

Untuk menentukan tingkat reaksi terhadap Brdigunakan konsentrasi mol [NO] yang sama yaitu percobaan 1 dan 2.

fraction numerator V subscript 1 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 over denominator V subscript 2 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 end fraction fraction numerator V subscript 1 equals up diagonal strike k open square brackets N O close square brackets to the power of X end strike left square bracket 0 comma 05 right square bracket to the power of y over denominator V subscript 2 equals up diagonal strike k open square brackets N O close square brackets to the power of X end strike left square bracket 0 comma 10 right square bracket to the power of y end fraction open parentheses 1 half close parentheses equals open parentheses 1 half close parentheses to the power of italic y italic y space space equals space 1 

pembahasan c :

Tingkat reaksi orde total adalah x + y = 2+1 = 3

pembahasan d :

Persamaan laju reaksinya yaitu :

V double bond k open square brackets N O close square brackets squared open square brackets Br subscript 2 close square brackets to the power of 1 

Jadi, jawaban benar seperti penjabaran 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

Roboguru

Laju reaksi:  ditentukan dari percobaan berikut.   Berdasarkan data tersebut, tentukan: a. orde reaksi terhadap X, b. orde reaksi terhadap Y, c. orde total, d. rumus laju reaksi, serta e. laj...

Pembahasan Soal:

Hukum laju reaksi menyatakan sebuah persamaan yang memperlihatkan keterkaitan atau hubungan antara laju reaksi tertentu dengan konsentrasi pereaksinya.

Berdasarkan persamaan reaksi X open parentheses italic g close parentheses and Y open parentheses italic g close parentheses yields zat space hasil, maka persamaan laju reaksi dapat ditulis menjadi:

r 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

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.

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 Y 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 X subscript 1 close square brackets to the power of m over open square brackets X subscript 2 close square brackets to the power of m open square brackets Y subscript 1 close square brackets to the power of n over open square brackets Y 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 open parentheses fraction numerator 0 comma 01 over denominator 0 comma 01 end fraction close parentheses open parentheses fraction numerator 0 comma 01 over denominator 0 comma 02 end fraction close parentheses to the power of n end cell equals cell fraction numerator 1 cross times 10 to the power of negative sign 3 end exponent over denominator 2 cross times 10 to the power of negative sign 3 end exponent end fraction 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 to the power of 1 end cell row n equals 1 row blank blank blank row blank blank cell bold Orde bold space italic X space... space perc point space left parenthesis 2 right parenthesis space dan space left parenthesis 3 right parenthesis space left parenthesis Substitusi space n 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 open parentheses fraction numerator 0 comma 01 over denominator 0 comma 02 end fraction close parentheses to the power of m open parentheses fraction numerator 0 comma 02 over denominator 0 comma 04 end fraction close parentheses to the power of 1 end cell equals cell fraction numerator 2 cross times 10 to the power of negative sign 3 end exponent over denominator 8 cross times 10 to the power of negative sign 3 end exponent end fraction end cell row cell open parentheses 1 half close parentheses to the power of m open parentheses 1 half close parentheses end cell equals cell 1 fourth end cell row cell open parentheses 1 half close parentheses to the power of m end cell equals cell 1 fourth division sign 1 half end cell row cell open parentheses 1 half close parentheses to the power of m end cell equals cell 1 fourth cross times 2 over 1 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 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 X and orde space Y end cell row cell orde space total end cell equals cell 1 plus 1 end cell row cell orde space total end cell equals 2 end table 

Langkah 3: Tentukan rumus laju reaksi. Substitusi nilai m = 1 dan n = 1 ke dalam rumus laju reaksi.

table attributes columnalign right center left columnspacing 0px end attributes row r equals cell 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 end cell row r equals cell k open square brackets X close square brackets to the power of 1 open square brackets Y close square brackets to the power of 1 end cell row r equals cell k open square brackets X close square brackets open square brackets Y close square brackets end cell end table 

Langkah 4: Tentukan laju reaksi jika open square brackets X close square brackets dan open square brackets Y close square brackets masing-masing dinaikkan 8 kali.

open square brackets X close square brackets subscript akhir equals 8 cross times open square brackets X close square brackets subscript awal open square brackets X close square brackets subscript akhir equals 8 open square brackets X close square brackets open square brackets Y close square brackets subscript akhir equals 8 cross times open square brackets Y close square brackets subscript awal open square brackets Y close square brackets subscript akhir equals 8 open square brackets Y close square brackets laju space akhir space open parentheses r subscript 1 close parentheses equals... ? r subscript 1 over r equals fraction numerator k open square brackets X close square brackets subscript akhir open square brackets Y close square brackets subscript akhir over denominator k open square brackets X close square brackets open square brackets Y close square brackets end fraction r subscript 1 over r equals fraction numerator k cross times 8 open square brackets X close square brackets cross times 8 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 r subscript 1 over r equals 64 over 1 r subscript 1 equals 64 cross times r 


Dengan demikian, diperoleh hasil bahwa:

a. orde reaksi terhadap X = 1.

b. orde reaksi terhadap Y = 1.

c. orde total = 2

d. rumus laju reaksi, r double bond k open square brackets X close square brackets open square brackets Y close square brackets

e. laju reaksi jika open square brackets X close square brackets dan open square brackets Y close square brackets masing-masing dinaikkan 8 kali, r = 64 kali lebih cepat dibandingkan laju awal. 

Jadi, orde X, orde Y, orde total, persamaan laju, dan perubahan laju jika dinaikkan 8 kali berturut-turut adalah 1, 1, 2, r double bond k open square brackets X close square brackets open square brackets Y close square brackets, dan 64 kali lebih cepat dibandingkan laju awal. 

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