Roboguru

Pada percobaan reaksi  diperoleh data-data percobaan sebagai berikut.   Tentukan: a. orde reaksi terhadap P

Pertanyaan

Pada percobaan reaksi P space plus space 2 Q yields PQ subscript 2 diperoleh data-data percobaan sebagai berikut.

 

Tentukan: a. orde reaksi terhadap P

Pembahasan Video:

Pembahasan Soal:

Persamaan laju reaksi adalah hubungan antara konsentrasi pereaksi dengan laju reaksi.

v space equals space k space open square brackets P close square brackets to the power of x space open square brackets Q close square brackets to the power of y 

Nilai v berbanding terbalik dengan waktu reaksi 

v space equals space 1 over t 

Bilangan pangkat pada persamaan di atas disebut sebagai orde reaksi atau tingkat reaksi pada reaksi yang bersangkutan. Jumlah bilangan pangkat konsentrasi pereaksi-pereaksi disebut sebagai orde reaksi total. Artinya, reaksi berorde x terhadap pereaksi A dan reaksi berorde y terhadap pereaksi B, orde reaksi total pada reaksi tersebut adalah (x + y). Faktor k yang terdapat pada persamaan tersebut disebut tetapan reaksi. 

Untuk dapat menentukan orde reaksi terhadap P, maka pilih konsentrasi Q yang tetap, yaitu percobaan 1 dan 2. 

italic v subscript 1 over italic v subscript italic 2 space equals space fraction numerator up diagonal strike italic k open square brackets P subscript 1 close square brackets to the power of x open square brackets Q subscript 1 close square brackets to the power of y over denominator up diagonal strike italic k open square brackets P subscript 2 close square brackets to the power of x open square brackets Q subscript 2 close square brackets to the power of italic y end fraction fraction numerator 1 over t subscript 1 over denominator 1 over t subscript 2 end fraction space equals space fraction numerator open square brackets P subscript 1 close square brackets to the power of x open square brackets Q subscript 1 close square brackets to the power of y over denominator open square brackets P subscript 2 close square brackets to the power of x open square brackets Q subscript 2 close square brackets to the power of italic y end fraction t subscript 2 over t subscript 1 space equals space fraction numerator left parenthesis 0 comma 1 right parenthesis to the power of x up diagonal strike left parenthesis 0 comma 1 right parenthesis to the power of y end strike over denominator left parenthesis 0 comma 2 right parenthesis to the power of x up diagonal strike left parenthesis 0 comma 1 right parenthesis to the power of italic y end strike end fraction 64 over 128 space equals space fraction numerator left parenthesis 0 comma 1 right parenthesis to the power of x up diagonal strike left parenthesis 0 comma 1 right parenthesis to the power of y end strike over denominator left parenthesis 0 comma 2 right parenthesis to the power of x up diagonal strike left parenthesis 0 comma 1 right parenthesis to the power of italic y end strike end fraction space space space space 1 half space equals space left parenthesis 1 half right parenthesis to the power of x space space space space space space space x space equals space 1 space 

Maka, orde reaksi terhadap P (x) adalah 1. 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

N. Puspita

Terakhir diupdate 13 September 2021

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Pertanyaan yang serupa

Dari percobaan pengukuran laju reaksi diperoleh data sebagai berikut.     Dari data tersebut dapat disimpulkan bahwa orde reaksi totalnya adalah ....

Pembahasan Soal:

Orde reaksi menunjukkan besarnya pengaruh konsentrasi 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. Misalkan pereaksi yang bereaksi adalah A dan B, maka persamaan laju reaksi ditulis sebagai berikut.

v double bond k open square brackets A close square brackets to the power of m open square brackets B close square brackets to the power of n 

Laju reaksi (v) berbanding terbalik terhadap waktu (t) dan dapat dinyatakan sebagai : v almost equal to 1 over t. Penentuan orde reaksi melalui data eksperimen dihitung berdasarkan perbandingan data percobaan yang sama.

  • Tentukan orde A yaitu cari data B yang sama yaitu percobaan 2 dan 3.

table attributes columnalign right center left columnspacing 0px end attributes row cell v subscript 2 over v subscript 3 end cell equals cell k subscript 2 over k subscript 3 open square brackets A subscript 2 close square brackets to the power of m over open square brackets A subscript 3 close square brackets to the power of m open square brackets B subscript 2 close square brackets to the power of n over open square brackets B subscript 3 close square brackets to the power of n end cell row cell fraction numerator 1 over t subscript 2 over denominator 1 over t subscript 3 end fraction end cell equals cell k subscript 2 over k subscript 3 open square brackets A subscript 2 close square brackets to the power of m over open square brackets A subscript 3 close square brackets to the power of m open square brackets B subscript 2 close square brackets to the power of n over open square brackets B subscript 3 close square brackets to the power of n end cell row cell fraction numerator 1 fourth over denominator 1 fourth end fraction end cell equals cell k subscript 2 over k subscript 3 open parentheses 0 comma 1 close parentheses to the power of m over open parentheses 0 comma 2 close parentheses to the power of m open parentheses 0 comma 3 close parentheses to the power of n over open parentheses 0 comma 3 close parentheses to the power of n end cell row 1 equals cell open parentheses 1 half close parentheses to the power of m end cell row cell open parentheses 1 half close parentheses to the power of 0 end cell equals cell open parentheses 1 half close parentheses to the power of m end cell row m equals 0 end table  

  • Tentukan orde B yaitu cari data A 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 open square brackets A subscript 1 close square brackets to the power of m over open square brackets A subscript 2 close square brackets to the power of m open square brackets B subscript 1 close square brackets to the power of n over open square brackets B subscript 2 close square brackets to the power of n end cell row cell fraction numerator 1 over t subscript 1 over denominator 1 over t subscript 2 end fraction end cell equals cell k subscript 1 over k subscript 2 open square brackets A subscript 1 close square brackets to the power of m over open square brackets A subscript 2 close square brackets to the power of m open square brackets B subscript 1 close square brackets to the power of n over open square brackets B subscript 2 close square brackets to the power of n end cell row cell fraction numerator 1 over 36 over denominator 1 fourth end fraction end cell equals cell k subscript 1 over k subscript 2 open parentheses 0 comma 1 close parentheses to the power of m over open parentheses 0 comma 1 close parentheses to the power of m open parentheses 0 comma 1 close parentheses to the power of n over open parentheses 0 comma 3 close parentheses to the power of n end cell row cell 1 over 9 end cell equals cell open parentheses 1 third close parentheses to the power of n end cell row cell open parentheses 1 third close parentheses squared end cell equals cell open parentheses 1 third close parentheses to the power of n end cell row n equals 2 end table 

Berdasarkan perhitungan diatas, diperoleh bahwa orde A = m = 0, orde B = n = 2, sehingga orde total menjadi :

table attributes columnalign right center left columnspacing 0px end attributes row cell orde space total end cell equals cell orde space A and orde space B end cell row cell orde space total end cell equals cell 0 plus 2 end cell row cell orde space total end cell equals 2 end table 

Jadi, jawaban yang benar adalah C.

Roboguru

Laju awal serangkaian eksperimen pada reaksi  dan NO Tentukan orde  dan orde NO !

Pembahasan Soal:

menentukan orde terhadap begin mathsize 14px style O subscript 2 end style, maka pilih data NO yang tetap (reaksi 1 dan 2 )

Error converting from MathML to accessible text.  

menentukan orde terhadap NO, maka pilih data undefined yang tetap (reaksi 1 dan 3)

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator V subscript 1 space end subscript equals space K space open square brackets O subscript 2 close square brackets to the power of x open square brackets N O close square brackets to the power of y over denominator V subscript 3 equals space K space open square brackets O subscript 2 close square brackets to the power of x open square brackets N O close square brackets to the power of y end fraction end cell row blank blank cell fraction numerator 3 comma 21 space x space 10 to the power of negative sign 3 end exponent equals space up diagonal strike K space left square bracket up diagonal strike 1 comma 10 space x space 10 to the power of negative sign 2 end exponent right square bracket end strike to the power of x left square bracket 1 comma 30 space x space 10 to the power of negative sign 2 end exponent right square bracket to the power of y over denominator 12 comma 8 space space x space 10 to the power of negative sign 3 end exponent equals space up diagonal strike K space up diagonal strike left square bracket 1 comma 10 space x space 10 to the power of negative sign 2 end exponent right square bracket end strike to the power of x left square bracket 2 comma 60 space x space 10 to the power of negative sign 2 to the power of y end exponent end fraction end cell row cell 1 fourth end cell equals cell left square bracket 1 half right square bracket to the power of y end cell row y equals cell space 2 end cell end table end style  

jadi, orde begin mathsize 14px style O subscript 2 end style = 1 dan orde NO=2.

 

Roboguru

Pada reaksi: , diperoleh data sebagai berikut.    Tentukan: a. orde reaksi terhadap P, b. orde reaksi terhadap Q, c. orde reaksi total, d. rumus laju reaksi, serta e. laju reaksi jika  dan  m...

Pembahasan Soal:

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

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

r double bond k open square brackets P close square brackets to the power of m open square brackets Q close square brackets to the power of n 

dimana m dan n adalah orde masing-masing reaktan.

Penentuan orde reaksi masing-masing reaktan dapat dicari berdasarkan data eksperimen dengan cara membandingkan data laju reaksi dan konsentrasi reaktan yang diketahui. Konsep yang perlu kita gunakan juga adalah hubungan waktu (t) terhadap laju reaksi (r).

r almost equal to 1 over t 

Langkah 1: Tentukan orde X dan orde Y.

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell bold Orde bold space P space... space perc point space left parenthesis 1 right parenthesis space dan space left parenthesis 2 right parenthesis end cell row cell k subscript 1 over k subscript 2 open square brackets P subscript 1 close square brackets to the power of m over open square brackets P subscript 2 close square brackets to the power of m open square brackets Q subscript 1 close square brackets to the power of n over open square brackets Q subscript 2 close square brackets to the power of n end cell equals cell r subscript 1 over r subscript 2 end cell row cell k subscript 1 over k subscript 2 open square brackets P subscript 1 close square brackets to the power of m over open square brackets P subscript 2 close square brackets to the power of m open square brackets Q subscript 1 close square brackets to the power of n over open square brackets Q subscript 2 close square brackets to the power of n end cell equals cell fraction numerator 1 over t subscript 1 over denominator 1 over t subscript 2 end fraction end cell row cell open parentheses fraction numerator 0 comma 1 over denominator 0 comma 2 end fraction close parentheses to the power of m open parentheses fraction numerator 0 comma 1 over denominator 0 comma 1 end fraction close parentheses end cell equals cell fraction numerator 1 over 80 over denominator 1 over 40 end fraction end cell row cell open parentheses 1 half close parentheses to the power of m end cell equals cell 40 over 80 end cell row cell open parentheses 1 half close parentheses to the power of m end cell equals cell open parentheses 1 half close parentheses to the power of 1 end cell row m equals 1 row blank blank blank row blank blank cell bold Orde bold space italic Q space... space perc point space left parenthesis 2 right parenthesis space dan space left parenthesis 3 right parenthesis space left parenthesis Substitusi space m equals 1 right parenthesis end cell row cell fraction numerator k subscript 2 over denominator k 3 end fraction open square brackets X subscript 2 close square brackets to the power of m over open square brackets X subscript 3 close square brackets to the power of m open square brackets Y subscript 2 close square brackets to the power of n over open square brackets Y subscript 3 close square brackets to the power of n end cell equals cell r subscript 2 over r subscript 3 end cell row cell fraction numerator k subscript 2 over denominator k 3 end fraction open square brackets P subscript 2 close square brackets to the power of m over open square brackets P subscript 3 close square brackets to the power of m open square brackets Q subscript 2 close square brackets to the power of n over open square brackets Q subscript 3 close square brackets to the power of n end cell equals cell fraction numerator 1 over t subscript 2 over denominator 1 over t subscript 3 end fraction end cell row cell open parentheses fraction numerator 0 comma 2 over denominator 0 comma 4 end fraction close parentheses to the power of 1 open parentheses fraction numerator 0 comma 1 over denominator 0 comma 2 end fraction close parentheses to the power of n end cell equals cell fraction numerator 1 over 40 over denominator 1 fifth end fraction end cell row cell open parentheses 1 half close parentheses open parentheses 1 half close parentheses to the power of n end cell equals cell 5 over 40 end cell row cell open parentheses 1 half close parentheses to the power of n end cell equals cell 5 over 40 division sign 1 half end cell row cell open parentheses 1 half close parentheses to the power of n end cell equals cell 5 over 40 cross times 2 over 1 end cell row cell open parentheses 1 half close parentheses to the power of n end cell equals cell 10 over 40 end cell row cell open parentheses 1 half close parentheses to the power of n end cell equals cell 1 fourth end cell row cell open parentheses 1 half close parentheses to the power of n end cell equals cell open parentheses 1 half close parentheses squared end cell row n equals 2 end table 

Langkah 2: Tentukan orde total.

table attributes columnalign right center left columnspacing 0px end attributes row cell orde space total end cell equals cell orde space P and orde space Q end cell row cell orde space total end cell equals cell 1 plus 2 end cell row cell orde space total end cell equals 3 end table  

Langkah 3: Tentukan rumus laju reaksi. Substitusi nilai m = 1 dan n = 2 ke dalam rumus laju reaksi.

table attributes columnalign right center left columnspacing 0px end attributes row r equals cell k open square brackets P close square brackets to the power of m open square brackets Q close square brackets to the power of n end cell row r equals cell k open square brackets P close square brackets to the power of 1 open square brackets Q close square brackets squared end cell row r equals cell k open square brackets P close square brackets open square brackets Q close square brackets squared end cell end table    

Langkah 4: Tentukan laju reaksi jika open square brackets P close square brackets dan open square brackets Q close square brackets masing-masing dinaikkan 3 kali.

table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets P close square brackets subscript akhir end cell equals cell 3 cross times open square brackets P close square brackets subscript awal end cell row cell open square brackets P close square brackets subscript akhir end cell equals cell 3 open square brackets P close square brackets end cell row cell open square brackets Q close square brackets subscript akhir end cell equals cell 3 cross times open square brackets Q close square brackets subscript awal end cell row cell open square brackets Q close square brackets subscript akhir end cell equals cell 3 open square brackets Q close square brackets end cell row cell laju space akhir space open parentheses r subscript 1 close parentheses end cell equals cell... ? end cell row blank blank blank row cell r subscript 1 over r end cell equals cell fraction numerator k open square brackets P close square brackets subscript akhir open square brackets Q close square brackets subscript akhir over denominator k open square brackets P close square brackets open square brackets Q close square brackets end fraction end cell row cell r subscript 1 over r end cell equals cell fraction numerator k cross times 3 open square brackets X close square brackets cross times 3 open square brackets Y close square brackets over denominator k cross times open square brackets X close square brackets cross times open square brackets Y close square brackets end fraction end cell row cell r subscript 1 over r end cell equals cell fraction numerator 3 cross times 3 over denominator 1 end fraction end cell row cell r subscript 1 over r end cell equals cell 9 over 1 end cell row cell r subscript 1 end cell equals cell 9 cross times r end cell end table   


Dengan demikian, diperoleh hasil bahwa:

a. orde reaksi terhadap P = 1.

b. orde reaksi terhadap Q = 2.

c. orde total = 3.

d. rumus laju reaksi, r double bond k open square brackets P close square brackets open square brackets Q close square brackets squared.

e. laju reaksi jika open square brackets P close square brackets dan open square brackets Q close square brackets masing-masing dinaikkan 3 kali menjadi 9 kali lebih cepat dibandingkan laju awal. 

Jadi, orde P, orde Q, orde total, rumus laju reaksi, dan perubahan laju jika konsentrasi dinaikkan 3 kali berturut-turut adalah 1, 2, 3, r double bond k open square brackets P close square brackets open square brackets Q close square brackets squared, dan 9 kali lebih cepat dibandingkan laju awal. 

Roboguru

Pada suhu 273 , gas bromin dapat bereaksi dengan gas nitrogen monoksida menurut persamaan reaksi   Dari reaksi tersebut diperoleh data berikut. Hitung orde reaksi tersebut.

Pembahasan Soal:

Orde reaksi adalah bilangan yang menyatakan seberapa besar pengaruh konsentrasi terhadap laju reaksi. Orde reaksi dapat ditentukan dengan cara membandingkan dua data, dan salah satu datanya harus sama.

  • Orde NO

    Membandingkan data 1 dan 4

    table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses open square brackets N O close square brackets subscript 4 over open square brackets N O close square brackets subscript 1 close parentheses to the power of x end cell equals cell v subscript 4 over v subscript 1 end cell row cell open parentheses fraction numerator 0 comma 2 over denominator 0 comma 1 end fraction close parentheses to the power of x end cell equals cell 24 over 6 end cell row cell 2 to the power of x end cell equals 4 row x equals 2 end table 

    Oleh karena itu, orde reaksi NO adalah 2.
     
  • Orde Br subscript 2 

    Membandingkan data 2 dan 3

    table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses open square brackets Br subscript 2 close square brackets subscript 3 over open square brackets Br subscript 2 close square brackets subscript 2 close parentheses to the power of y end cell equals cell v subscript 3 over v subscript 2 end cell row cell open parentheses fraction numerator 0 comma 2 over denominator 0 comma 1 end fraction close parentheses to the power of y end cell equals cell 24 over 12 end cell row cell 2 to the power of y end cell equals 2 row y equals 1 end table 

    Oleh karena itu, orde reaksi Br subscript 2 adalah 1.

Jadi, orde reaksi NO adalah 2 dan orde reaksi Br subscript 2 adalah 1.space 

Roboguru

Dari reaksi:  diperoleh data eksperimen pada suhu tetap:   Dari data tersebut, orde reaksi totalnya adalah ....

Pembahasan Soal:

Orde reaksi menunjukkan besarnya pengaruh konsentrasi 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 reaksi berdasarkan data percobaan dituliskan menjadi : v double bond k open square brackets Fe to the power of 3 plus sign close square brackets to the power of m open square brackets S to the power of 2 minus sign close square brackets to the power of n 

  • Tentukan orde Fe to the power of 3 plus sign yaitu cari data S to the power of 2 minus sign 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 Fe to the power of 3 plus sign close square brackets subscript 1 superscript m over denominator open square brackets Fe to the power of 3 plus sign close square brackets subscript 2 superscript m end fraction fraction numerator open square brackets S to the power of 2 minus sign close square brackets subscript blank to the power of 1 end subscript superscript n over denominator open square brackets S to the power of 2 minus sign close square brackets subscript blank squared end subscript superscript n end fraction end cell row cell 2 over 8 end cell equals cell k subscript 1 over k subscript 2 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 to the power of n end cell row cell 1 fourth end cell equals cell open parentheses 1 half close parentheses to the power of blank to the power of m end exponent end cell row cell open parentheses 1 half close parentheses squared end cell equals cell open parentheses 1 half close parentheses to the power of m end cell row m equals 2 end table 

  • Tentukan orde S to the power of 2 minus sign yaitu cari data Fe to the power of 3 plus sign yang sama yaitu percobaan 2 dan 3.

table attributes columnalign right center left columnspacing 0px end attributes row cell v subscript 2 over v subscript 3 end cell equals cell k subscript 2 over k subscript 3 fraction numerator open square brackets Fe to the power of 3 plus sign close square brackets subscript 1 superscript m over denominator open square brackets Fe to the power of 3 plus sign close square brackets subscript 2 superscript m end fraction fraction numerator open square brackets S to the power of 2 minus sign close square brackets subscript blank to the power of 1 end subscript superscript n over denominator open square brackets S to the power of 2 minus sign close square brackets subscript blank squared end subscript superscript n end fraction end cell row cell 8 over 16 end cell equals cell k subscript 2 over k subscript 3 open parentheses fraction numerator 0 comma 2 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 2 end fraction close parentheses to the power of n end cell row cell 1 half end cell equals cell open parentheses 1 half close parentheses to the power of blank to the power of n end exponent 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 Fe to the power of 3 plus sign = 2, orde S to the power of 2 minus sign = 1, sehingga orde total menjadi :

table attributes columnalign right center left columnspacing 0px end attributes row cell orde space total end cell equals cell orde space Fe to the power of 3 plus sign and orde space S to the power of 2 minus sign end cell row cell orde space total end cell equals cell 2 plus 1 end cell row cell orde space total end cell equals 3 end table  

Jadi, jawaban yang benar adalah A.

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