Roboguru

Data percobaan untuk reaksi:   2A(g)+B(g)+C(g)→hasilreaksi    adalah sebagai berikut.     Tentukan nilai x!

Pertanyaan

Data percobaan untuk reaksi:
 

2 A open parentheses italic g close parentheses space plus space B open parentheses italic g close parentheses space plus space C open parentheses italic g close parentheses space rightwards arrow space hasil space reaksi 
 

adalah sebagai berikut.
 


 

Tentukan nilai x!

Pembahasan Soal:

Percobaan 1 dan 2
 

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

Percobaan 2 dan 3
 

straight v subscript 2 over straight v subscript 3 equals fraction numerator k left square bracket straight A right square bracket subscript 2 superscript x left square bracket straight B right square bracket subscript 2 superscript y left square bracket straight C right square bracket subscript 2 superscript z over denominator k left square bracket straight A right square bracket subscript 3 superscript x left square bracket straight B right square bracket subscript 3 superscript y left square bracket straight C right square bracket subscript 3 superscript z end fraction fraction numerator 0 comma 02 over denominator 0 comma 04 end fraction equals fraction numerator up diagonal strike k up diagonal strike left square bracket 0 comma 2 right square bracket subscript blank superscript x end strike left square bracket 0 comma 1 right square bracket subscript blank superscript y up diagonal strike left square bracket 0 comma 1 right square bracket subscript 2 superscript z end strike over denominator up diagonal strike k up diagonal strike left square bracket 0 comma 2 right square bracket subscript blank superscript x end strike left square bracket 0 comma 2 right square bracket subscript blank superscript y up diagonal strike left square bracket 0 comma 1 right square bracket subscript 3 superscript z end strike end fraction fraction numerator 0 comma 02 over denominator 0 comma 04 end fraction equals open parentheses fraction numerator 0 comma 1 over denominator 0 comma 2 end fraction close parentheses to the power of straight y 1 half equals open parentheses 1 half close parentheses to the power of straight y straight y equals 1 
 

Percobaan 1 dan 4
 

straight v subscript 1 over straight v subscript 4 equals fraction numerator k left square bracket straight A right square bracket subscript 1 superscript x left square bracket straight B right square bracket subscript 1 superscript y left square bracket straight C right square bracket subscript 1 superscript z over denominator k left square bracket straight A right square bracket subscript 4 superscript x left square bracket straight B right square bracket subscript 4 superscript y left square bracket straight C right square bracket subscript 4 superscript z end fraction fraction numerator 0 comma 01 over denominator 0 comma 09 end fraction equals fraction numerator up diagonal strike k left square bracket 0 comma 1 right square bracket subscript blank superscript italic 1 left square bracket 0 comma 1 right square bracket subscript blank superscript italic 1 left square bracket 0 comma 1 right square bracket subscript 1 superscript z over denominator up diagonal strike k left square bracket 0 comma 3 right square bracket subscript blank superscript italic 1 left square bracket 0 comma 3 right square bracket subscript blank superscript italic 1 left square bracket 0 comma 3 right square bracket subscript 4 superscript z end fraction fraction numerator 0 comma 01 over denominator 0 comma 09 end fraction equals 1 third 1 third open parentheses fraction numerator 0 comma 1 over denominator 0 comma 3 end fraction close parentheses to the power of straight z 1 over 9 equals 1 over 9 open parentheses 1 third close parentheses to the power of straight z straight z equals 0 
 

Persamaan laju reaksi
 

v equals italic k open square brackets A close square brackets open square brackets B close square brackets 
 

Konstanta laju reaksi dan satuannya
 

v subscript 1 double bond k open square brackets A close square brackets open square brackets B close square brackets 0 comma 01 space M space s to the power of negative sign 1 end exponent equals k left parenthesis 0 comma 1 space M right parenthesis left parenthesis 0 comma 1 space M right parenthesis k equals fraction numerator 0 comma 01 space M space s to the power of negative sign 1 end exponent over denominator 0 comma 01 space M squared end fraction k equals 1 space M to the power of negative sign 1 end exponent space s to the power of negative sign 1 end exponent 
 

Nilai x
 

v subscript 5 double bond k open square brackets A close square brackets open square brackets B close square brackets italic x equals 1 left parenthesis 0 comma 5 right parenthesis left parenthesis 0 comma 4 right parenthesis italic x equals 0 comma 2 space M space s to the power of negative sign 1 end exponent 
 

Jadi, jawaban sesuai dengan uraian di atas. space 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Acfreelance

Terakhir diupdate 07 Oktober 2021

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Data reaksi: A(g)+2B(g)→AB2​(g) sebagai berikut.   Nilai x adalah ...

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 italic A left parenthesis italic g right parenthesis plus 2 B open parentheses italic g close parentheses yields AB subscript 2 open parentheses italic g close parentheses mempunyai persamaan laju reaksi:

italic r equals italic k open square brackets italic A close square brackets to the power of italic x open square brackets B close square brackets to the power of italic y 

dengan:

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

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

Berikut langkah-langkah menentukan nilai x pada percobaan laju reaksi di atas:

  1. Menghitung orde reaksi

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

    italic r subscript 1 over italic r subscript 2 equals fraction numerator italic k left square bracket italic A right square bracket subscript 1 superscript italic x open square brackets B close square brackets subscript 1 superscript italic y over denominator italic k left square bracket italic A right square bracket subscript 2 superscript italic x open square brackets B close square brackets subscript 2 superscript italic y end fraction 5 over 80 equals fraction numerator up diagonal strike italic k left parenthesis 0 comma 5 right parenthesis 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 5 right parenthesis to the power of italic x end strike open parentheses 0 comma 4 close parentheses to the power of italic y end fraction 1 over 16 equals open parentheses 1 fourth close parentheses to the power of italic y italic y equals 2 
     
  2. Menghitung orde reaksi A 

    Kita pilih percobaan (2) dan (4)

    italic r subscript 2 over italic r subscript 4 equals fraction numerator italic k left square bracket italic A right square bracket subscript 2 superscript italic x open square brackets B close square brackets subscript 2 superscript italic y over denominator italic k left square bracket italic A right square bracket subscript 2 superscript italic x open square brackets B close square brackets subscript 2 superscript italic y end fraction 80 over 640 equals fraction numerator up diagonal strike italic k open parentheses 0 comma 5 close parentheses to the power of italic x open parentheses 0 comma 4 close parentheses squared over denominator up diagonal strike italic k open parentheses 1 comma 0 close parentheses to the power of italic x left parenthesis 0 comma 8 right parenthesis squared end fraction 1 over 8 equals open parentheses 1 half close parentheses to the power of italic x open parentheses 1 half close parentheses squared 1 over 8 equals open parentheses 1 half close parentheses to the power of italic x open parentheses 1 fourth close parentheses 1 over 8 cross times 4 over 1 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 
     
  3. Menghitung nilai x

    italic r subscript 2 over italic r subscript 3 equals fraction numerator italic k left square bracket italic A right square bracket subscript 2 superscript italic x open square brackets B close square brackets subscript 2 superscript italic y over denominator italic k left square bracket italic A right square bracket subscript 3 superscript italic x open square brackets B close square brackets subscript 3 superscript italic y end fraction italic r subscript 2 over italic r subscript 3 equals fraction numerator italic k left square bracket italic A right square bracket subscript italic 2 open square brackets B close square brackets subscript 2 superscript 2 over denominator italic k left square bracket italic A right square bracket subscript 3 open square brackets B close square brackets subscript 3 superscript 2 end fraction 80 over 32 equals fraction numerator up diagonal strike italic k open parentheses 0 comma 5 close parentheses up diagonal strike open parentheses 0 comma 4 close parentheses squared end strike over denominator up diagonal strike italic k open parentheses italic x close parentheses up diagonal strike left parenthesis 0 comma 4 right parenthesis squared end strike end fraction 5 over 2 equals fraction numerator 0 comma 5 over denominator italic x end fraction italic x equals fraction numerator 0 comma 5 cross times 2 over denominator 5 end fraction italic x equals 0 comma 2


Jadi, jawaban yang tepat adalah B.space space space

0

Roboguru

Pada reaksi: 2NO(g)+O2​(g)→N2​O4​(g) diperoleh data sebagai berikut. Hitung nilai tetapan laju reaksi dan satuannya!

Pembahasan Soal:

Berdasarkan reaksi tersebut, dapat disimpulkan persamaan laju reaksinya adalah begin mathsize 14px style v double bond k open square brackets N O close square brackets to the power of x open square brackets O subscript 2 close square brackets to the power of y end style, dimana x dan y merupakan orde reaksi. Maka langkah pertama adalah menentukan kedua orde tersebut, yaitu:

Orde begin mathsize 14px style N O end style dapat ditentukan menggunakan data 2 dan 3, sebagai berikut:

begin mathsize 14px style 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 space open square brackets N O close square brackets subscript 2 to the power of x space open square brackets O subscript 2 close square brackets subscript 2 to the power of y over denominator k space open square brackets N O close square brackets subscript 3 to the power of x space open square brackets O subscript 2 close square brackets subscript 3 to the power of y end fraction space end cell row cell fraction numerator 0 comma 02 over denominator 0 comma 08 end fraction end cell equals cell fraction numerator up diagonal strike k space left parenthesis 0 comma 1 right parenthesis to the power of x space left parenthesis 0 comma 2 right parenthesis to the power of y over denominator up diagonal strike k space left parenthesis 0 comma 2 right parenthesis to the power of x space left parenthesis 0 comma 2 right parenthesis to the power of y end fraction space end cell row cell 1 fourth end cell equals cell fraction numerator left parenthesis 0 comma 1 right parenthesis to the power of x space up diagonal strike left parenthesis 0 comma 2 right parenthesis to the power of y end strike over denominator left parenthesis 0 comma 2 right parenthesis to the power of x space up diagonal strike left parenthesis 0 comma 2 right parenthesis to the power of y end strike end fraction space end cell row cell 1 fourth end cell equals cell left parenthesis 1 half right parenthesis to the power of x space space end cell row x equals cell 2 space end cell end table end style

Orde begin mathsize 14px style O subscript 2 end style dapat ditentukan menggunakan data 1 dan 2, sebagai berikut:

undefined

Sehingga diperoleh  begin mathsize 14px style v double bond k open square brackets N O close square brackets squared open square brackets O subscript 2 close square brackets to the power of 1 end style

Selanjutnya menentukan konstanta/tetapan laju reaksi dengan satuannya menggunakan data 1 (diperbolehkan menggunakan data lainnya)

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row v equals cell k open square brackets N O close square brackets squared open square brackets O subscript 2 close square brackets to the power of 1 space end cell row cell 0 comma 01 space Ms to the power of negative sign 1 end exponent end cell equals cell k left parenthesis 0 comma 1 space M right parenthesis squared left parenthesis 0 comma 1 space M right parenthesis to the power of 1 end cell row cell space k end cell equals cell fraction numerator 0 comma 01 space Ms to the power of negative sign 1 end exponent over denominator left parenthesis 0 comma 1 space M right parenthesis cubed end fraction space end cell row k equals cell 10 space M to the power of negative sign 2 end exponent s to the power of negative sign 1 end exponent end cell end table end style


Jadi, nilai tetapan laju reaksi dan satuannya adalah 10 M-2 s-1.

0

Roboguru

Suatu percobaan reaksi: 2A+B→C diperoleh data sebagai berikut. Percobaan [A] M [B] M v (M s-1) 1) 0,1 0,1 0,01 2) 0,1 0,2 0,02 3) 0,2...

Pembahasan Soal:

a. Persamaan laju reaksi 

Menentukan orde A dengan percobaan 2 dan 3

 begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell v subscript 3 over v subscript 2 end cell equals cell fraction numerator k open square brackets A close square brackets subscript 3 superscript x open square brackets B close square brackets subscript 3 superscript y over denominator k open square brackets A close square brackets subscript 2 superscript x open square brackets B close square brackets subscript 2 superscript y end fraction end cell row cell fraction numerator 0 comma 08 over denominator 0 comma 02 end fraction end cell equals cell fraction numerator k open square brackets 0 comma 2 close square brackets to the power of x open square brackets 0 comma 2 close square brackets to the power of y over denominator k open square brackets 0 comma 1 close square brackets to the power of x open square brackets 0 comma 2 close square brackets to the power of y end fraction end cell row cell 4 space end cell equals cell space 2 to the power of x end cell row cell x space end cell equals cell space 2 end cell end table end style 

Menentukan orde B dengan percobaan 1 dan 2

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell v subscript 2 over v subscript 1 space end cell equals cell space fraction numerator k open square brackets A close square brackets subscript 2 superscript x open square brackets B close square brackets subscript 2 superscript y over denominator k open square brackets A close square brackets subscript 1 superscript x open square brackets B close square brackets subscript 1 superscript y end fraction end cell row cell fraction numerator 0 comma 02 over denominator 0 comma 01 end fraction space end cell equals cell space fraction numerator k open square brackets 0 comma 1 close square brackets to the power of x open square brackets 0 comma 2 close square brackets to the power of y over denominator k open square brackets 0 comma 1 close square brackets to the power of x open square brackets 0 comma 1 close square brackets to the power of y end fraction end cell row cell 2 space end cell equals cell space 2 to the power of y end cell row cell y space end cell equals cell space 1 end cell end table end style  

Jadi, orde reaksi A = 2 dan orde reaksi B = 1

Persamaan laju reaksi: begin mathsize 14px style v space equals space k open square brackets A close square brackets squared open square brackets B close square brackets end style 

b. Harga tetapan laju reaksi

Percobaan 1

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell v space end cell equals cell space k open square brackets A close square brackets squared open square brackets B close square brackets end cell row cell 0 comma 01 space end cell equals cell space k left parenthesis 0 comma 1 right parenthesis squared left parenthesis 0 comma 1 right parenthesis end cell row cell k space end cell equals cell space 10 end cell end table end style 

c.  Harga Y

Percobaan 4

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell v space end cell equals cell space k space open square brackets A close square brackets squared open square brackets B close square brackets end cell row cell Y space end cell equals cell space 10 left parenthesis 0 comma 2 right parenthesis squared left parenthesis 0 comma 3 right parenthesis end cell row cell Y space end cell equals cell space 0 comma 12 space M space s to the power of negative sign 1 end exponent end cell end table end style 

Jadi, harga persamaan laju reaksi adalah begin mathsize 14px style v space equals space k open square brackets A close square brackets squared open square brackets B close square brackets end styleharga tetapan laju reaksi adalah 10 dan harga Y adalah begin mathsize 14px style 0 comma 12 space M space s to the power of negative sign 1 end exponent end style

0

Roboguru

Untuk reaksi ; 4NH3​(g)+5O2​(g)→4NO(g)+6H2​O(g) diketahui data sebagai berikut : Tentukanlah : a. Orde reaksi b. Harga tetapan lajunya c. Ungkapan laju reaksinya

Pembahasan Soal:

Persamaan umum laju reaksi dapat dituliskan sebagai berikut.


v double bond k open square brackets pereaksi close square brackets to the power of x  Keterangan space colon v double bond laju space reaksi k double bond tetapan space laju space reaksi x double bond orde space reaksi


Soal diatas dapat diselesaikan dengan cara yaitu :

a. Orde reaksi

TIPS! Jika dalam data percobaan tidak ada konsentrasi yang sama untuk bisa dicoret, maka untuk memudahkan perhitungan gunakan data laju reaksi yang saling berkelipatan sehingga bisa diselesaikan dengan metode aljabar yaitu :

  • Buat persamaan matematika dari percobaan 4 dan 1


table attributes columnalign right center left columnspacing 0px end attributes row v equals cell k open square brackets N H subscript 3 close square brackets to the power of x middle dot open square brackets O subscript 2 close square brackets to the power of y end cell row blank blank blank row cell v subscript 4 over v subscript 1 end cell equals cell fraction numerator k open square brackets N H subscript 3 close square brackets subscript 4 to the power of x middle dot open square brackets O subscript 2 close square brackets subscript 4 to the power of y over denominator k open square brackets N H subscript 3 close square brackets subscript 1 to the power of x middle dot open square brackets O subscript 2 close square brackets subscript 1 to the power of y end fraction end cell row cell fraction numerator 2 comma 4 cross times 10 to the power of negative sign 4 end exponent over denominator 3 comma 0 cross times 10 to the power of negative sign 5 end exponent end fraction end cell equals cell fraction numerator k space left parenthesis 0 comma 04 right parenthesis to the power of x space left parenthesis 0 comma 06 right parenthesis to the power of y over denominator k space left parenthesis 0 comma 01 right parenthesis to the power of x space left parenthesis 0 comma 03 right parenthesis to the power of y end fraction end cell row cell fraction numerator 2 comma 4 cross times 10 to the power of negative sign 4 end exponent over denominator 0 comma 3 cross times 10 to the power of negative sign 4 end exponent end fraction end cell equals cell 4 to the power of x middle dot 2 to the power of y end cell row 8 equals cell 4 to the power of x middle dot 2 to the power of y end cell row cell 2 cubed end cell equals cell left parenthesis 2 right parenthesis to the power of 2 x end exponent middle dot 2 to the power of y end cell row blank blank cell maka colon end cell row cell 2 x and y end cell equals cell 3 space...............1 right parenthesis end cell end table

 

  • Buat persamaan matematika dari percobaan 5 dan 1


table attributes columnalign right center left columnspacing 0px end attributes row cell v subscript 5 over v subscript 1 end cell equals cell fraction numerator k open square brackets N H subscript 3 close square brackets subscript 5 to the power of x middle dot open square brackets O subscript 2 close square brackets subscript 5 to the power of y over denominator k open square brackets N H subscript 3 close square brackets subscript 1 to the power of x middle dot open square brackets O subscript 2 close square brackets subscript 1 to the power of y end fraction end cell row cell fraction numerator 2 comma 7 cross times 10 to the power of negative sign 4 end exponent over denominator 3 comma 0 cross times 10 to the power of negative sign 5 end exponent end fraction end cell equals cell fraction numerator k space left parenthesis 0 comma 03 right parenthesis to the power of x space left parenthesis 0 comma 09 right parenthesis to the power of y over denominator k space left parenthesis 0 comma 01 right parenthesis to the power of x space left parenthesis 0 comma 03 right parenthesis to the power of y end fraction end cell row cell fraction numerator 2 comma 7 cross times 10 to the power of negative sign 4 end exponent over denominator 0 comma 3 cross times 10 to the power of negative sign 4 end exponent end fraction end cell equals cell 3 to the power of x middle dot 3 to the power of y end cell row 9 equals cell 3 to the power of x middle dot 3 to the power of y end cell row cell 3 squared end cell equals cell 3 blank to the power of x middle dot 3 blank to the power of y end cell row blank blank cell maka colon end cell row cell x and y end cell equals 2 row x equals cell 2 minus sign y space space...........2 right parenthesis end cell end table

 

  • Substitusikan persamaan 2) ke dalam persamaan 1)


table attributes columnalign right center left columnspacing 0px end attributes row x equals cell 2 minus sign y space rightwards arrow space 2 x and y equals 3 end cell row blank blank blank row cell 2 left parenthesis 2 minus sign y right parenthesis plus y end cell equals 3 row cell 4 minus sign 2 y and y end cell equals 3 row cell 4 minus sign y end cell equals 3 row y equals cell 4 minus sign 3 end cell row y equals 1 row blank blank cell maka colon end cell row x equals cell 2 minus sign y end cell row x equals cell 2 minus sign 2 end cell row x equals 1 end table


Dengan demikian, orde reaksi terhadap open square brackets N H subscript 3 close square brackets adalah 1, dan orde reaksi terhadap open square brackets O subscript 2 close square brackets juga 1. Orde reaksi total = 1 + 1 = 2.

b. Ungkapan laju reaksi

Ungkapan laju reaksi dapat ditentukan dengan cara memasukkan nilai orde reaksi (x dan y) ke dalam persamaan laju reaksi, maka :


table attributes columnalign right center left columnspacing 0px end attributes row v equals cell k open square brackets N H subscript 3 close square brackets to the power of x middle dot open square brackets O subscript 2 close square brackets to the power of y end cell row v equals cell k open square brackets N H subscript 3 close square brackets space open square brackets O subscript 2 close square brackets end cell end table


Jadi, ungkapan laju reaksinya adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank v end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank k end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open square brackets N H subscript 3 close square brackets end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open square brackets O subscript 2 close square brackets end cell end table.

c.  Harga tetapan laju reaksi

Harga tetapan laju reaksi dapat ditentukan dengan cara mensubstitusikan rumus persamaan laju reaksi yang telah diketahui ke dalam salah satu data percobaan. Misal, disubstitusi ke dalam percobaan 1 , maka :


table attributes columnalign right center left columnspacing 0px end attributes row v equals cell k open square brackets N H subscript 3 close square brackets middle dot open square brackets O subscript 2 close square brackets end cell row cell 3 comma 0 cross times 10 to the power of negative sign 5 end exponent end cell equals cell k space left parenthesis 0 comma 01 right parenthesis space left parenthesis 0 comma 03 right parenthesis end cell row k equals cell fraction numerator 3 comma 0 cross times 10 to the power of negative sign 5 end exponent over denominator 10 to the power of negative sign 2 end exponent middle dot 3 comma 0 cross times 10 to the power of negative sign 2 end exponent end fraction end cell row k equals cell 10 to the power of negative sign 1 end exponent end cell row k equals cell 0 comma 1 end cell end table


Jadi, harga tetapan lajunya adalah 0,1.space

0

Roboguru

Pada reaksi: 2X(aq)+Y(aq)→Z(aq) 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

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