Roboguru

Suatu reaksi berorde  terhadap salah satu pereaksi. Bagaimana perubahan laju reaksi tersebut apabila konsentrasi pereaksi tersebut diperbesar 4 kali?

Pertanyaan

Suatu reaksi berorde begin inline style 1 half end style terhadap salah satu pereaksi. Bagaimana perubahan laju reaksi tersebut apabila konsentrasi pereaksi tersebut diperbesar 4 kali?space

Pembahasan Soal:

Jika dimisalkan pereaksi tesebut adalah A dan pereaksi yang satunya adalah B, pereaksi yang diperbesar konsentrasi nya adalah A. Maka dapat ditentukan:
 

table attributes columnalign right center left columnspacing 0px end attributes row cell V subscript 2 over V subscript 1 end cell equals cell open square brackets fraction numerator 4 A over denominator A end fraction close square brackets to the power of begin inline style 1 half end style end exponent open square brackets B over B close square brackets to the power of italic x end cell row cell V subscript 2 over V subscript 1 end cell equals cell open square brackets 4 close square brackets to the power of begin inline style 1 half end style end exponent end cell row cell V subscript 2 end cell equals cell 2 cross times V subscript 1 end cell end table
 

Dengan demikian maka lajunya menjadi 2 kali laju semula.space

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

M. Pulungan

Mahasiswa/Alumni Universitas Negeri Medan

Terakhir diupdate 02 Mei 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

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. 

0

Roboguru

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.

0

Roboguru

Pada reaksi , kecepatan reaksinya adalah . Kecepatan reaksi ini tidak dipengaruhi oleh ....

Pembahasan Soal:

Laju reaksi dipengaruhi oleh beberapa faktor, diantaranya konsentrasi reaktan, luas permukaan, suhu, dan penambahan katalis.

Pada reaksi 2 A and B and 2 C yields D and 2 E, kecepatan reaksinya adalah V double bond k open parentheses A close parentheses point open parentheses B close parentheses squared. Hal tersebut berarti konsentrasi reaktan yang berpengaruh pada kecepatan reaksi hanya konsentrasi A dan B.

Jadi, kecepatan reaksi tersebut tidak dipengaruhi oleh konsentrasi C.

Jadi, jawaban yang tepat adalah C.space

0

Roboguru

Dalam suatu percobaan reaksi:   diperoleh data sebagai berikut.     Tentukan: a. orde reaksi terhadap A, b. orde reaksi terhadap B, c. orde reaksi terhadap C, d. orde total, e. rumus laju rea...

Pembahasan Soal:

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

Persamaan laju reaksi dapat ditulis berdasarkan reaksi diatas adalah:

r double bond k open square brackets A close square brackets to the power of x open square brackets B close square brackets to the power of y open square brackets C close square brackets to the power of z 

dimana x, y dan z 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 A, B dan orde C.

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell bold Orde bold space C space... space left parenthesis perc point 1 space dan space 2 comma space A space dan space B space tetap right parenthesis space end cell row cell k subscript 1 over k subscript 2 open square brackets A subscript 1 close square brackets to the power of x over open square brackets A subscript 2 close square brackets to the power of x open square brackets B subscript 1 close square brackets to the power of y over open square brackets B subscript 2 close square brackets to the power of y open square brackets C subscript 1 close square brackets to the power of z over open square brackets C subscript 2 close square brackets to the power of z end cell equals cell r subscript 1 over r subscript 2 end cell row cell open parentheses fraction numerator 1 comma 0 over denominator 1 comma 0 end fraction close parentheses to the power of x open parentheses fraction numerator 1 comma 0 over denominator 1 comma 0 end fraction close parentheses to the power of y open parentheses fraction numerator 1 comma 0 over denominator 0 comma 5 end fraction close parentheses to the power of z end cell equals cell fraction numerator 1 comma 0 cross times 10 to the power of negative sign 2 end exponent over denominator 2 comma 5 cross times 10 to the power of negative sign 3 end exponent end fraction end cell row cell open parentheses 2 close parentheses to the power of z end cell equals 4 row cell open parentheses 2 close parentheses to the power of z end cell equals cell open parentheses 2 close parentheses squared end cell row z equals 2 row blank blank blank row blank blank cell bold Orde bold space B space... space left parenthesis perc point 2 space dan space 3 comma space A space dan space C space tetap right parenthesis end cell row cell k subscript 2 over k subscript 3 open square brackets A subscript 2 close square brackets to the power of x over open square brackets A subscript 3 close square brackets to the power of x open square brackets B subscript 2 close square brackets to the power of y over open square brackets B subscript 3 close square brackets to the power of y open square brackets C subscript 2 close square brackets to the power of z over open square brackets C subscript 3 close square brackets to the power of z end cell equals cell r subscript 2 over r subscript 3 end cell row cell open parentheses fraction numerator 1 comma 0 over denominator 1 comma 0 end fraction close parentheses to the power of x open parentheses fraction numerator 1 comma 0 over denominator 2 comma 0 end fraction close parentheses to the power of y open parentheses fraction numerator 0 comma 5 over denominator 0 comma 5 end fraction close parentheses to the power of z end cell equals cell fraction numerator 2 comma 5 cross times 10 to the power of negative sign 3 end exponent over denominator 5 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 y end cell equals cell 1 half end cell row cell open parentheses 1 half close parentheses to the power of y end cell equals cell open parentheses 1 half close parentheses to the power of 1 end cell row y equals 1 row blank blank blank row blank blank cell bold Orde bold space italic A space... space left parenthesis perc point 3 space dan space 4 space serta space subs point space y space dan space z right parenthesis space end cell row cell k subscript 3 over k subscript 4 open square brackets A subscript 3 close square brackets to the power of x over open square brackets A subscript 4 close square brackets to the power of x open square brackets B subscript 3 close square brackets to the power of y over open square brackets B subscript 4 close square brackets to the power of y open square brackets C subscript 3 close square brackets to the power of z over open square brackets C subscript 4 close square brackets to the power of z end cell equals cell r subscript 3 over r subscript 4 end cell row cell open parentheses fraction numerator 1 comma 0 over denominator 2 comma 0 end fraction close parentheses to the power of x open parentheses fraction numerator 2 comma 0 over denominator 2 comma 0 end fraction close parentheses to the power of 1 open parentheses fraction numerator 0 comma 5 over denominator 1 comma 0 end fraction close parentheses squared end cell equals cell fraction numerator 5 comma 0 cross times 10 to the power of negative sign 3 end exponent over denominator 4 comma 0 cross times 10 to the power of negative sign 2 end exponent end fraction end cell row cell open parentheses 1 half close parentheses to the power of x cross times 1 cross times open parentheses 1 half close parentheses squared end cell equals cell 1 over 8 end cell row cell open parentheses 1 half close parentheses to the power of x end cell equals cell 1 over 8 colon open parentheses 1 half close parentheses squared end cell row cell open parentheses 1 half close parentheses to the power of x end cell equals cell 1 over 8 cross times 4 over 1 end cell row cell open parentheses 1 half close parentheses to the power of x end cell equals cell open parentheses 1 half close parentheses to the power of 1 end cell row x 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 A and orde space B and orde space C end cell row cell orde space total end cell equals cell 1 plus 1 plus 2 end cell row cell orde space total end cell equals 4 end table  

Langkah 3: Tentukan rumus laju reaksi. Substitusi nilai x = 1, y = 1, dan z = 2 ke dalam rumus laju reaksi.

r double bond k open square brackets A close square brackets to the power of x open square brackets B close square brackets to the power of y open square brackets C close square brackets to the power of z r double bond k open square brackets A close square brackets to the power of 1 open square brackets B close square brackets to the power of 1 open square brackets C close square brackets squared r double bond k open square brackets A close square brackets open square brackets B close square brackets open square brackets C close square brackets squared  

Dengan demikian, diperoleh hasil bahwa:

a. orde reaksi terhadap A = 1.

b. orde reaksi terhadap B = 1.

c. orde reaksi terhadap C = 2

d. orde total = 4

e. rumus laju reaksi, r double bond k open square brackets A close square brackets open square brackets B close square brackets open square brackets C close square brackets squared

Jadi, orde A, orde B, orde C, orde total, dan persamaan laju berturut-turut adalah 1, 1, 2, 4, dan r double bond k open square brackets A close square brackets open square brackets B close square brackets open square brackets C close square brackets squared.

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Roboguru

Data reaksi:  adalah sebagai berikut.    Rumus laju reaksi adalah ...

Pembahasan Soal:

Persamaan laju reaksi tidak dapat diturunkan dari stoikiometri reaksi tetapi ditentukan melalui percobaan.

Reaksi N subscript 2 open parentheses g close parentheses and H subscript 2 open parentheses g close parentheses yields 2 N H subscript 3 open parentheses g close parentheses mempunyai persamaan laju reaksi:

r double bond k open square brackets N subscript 2 close square brackets to the power of x open square brackets H subscript 2 close square brackets to the power of y space 

dengan:

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

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

Berikut langkah-langkah menentukan persamaan laju reaksi di atas:

Menghitung orde reaksi x terhadap N subscript bold 2 
Ditentukan dengan membandingkan dua percobaan dengan konsentrasi H subscript 2 sama, yaitu percobaan (1) dan (2).
table attributes columnalign right center left columnspacing 0px end attributes row cell r subscript 2 over r subscript 1 end cell equals cell fraction numerator k open square brackets N subscript 2 close square brackets subscript 2 superscript space space x end superscript space space open square brackets H subscript 2 close square brackets subscript 2 to the power of y over denominator k open square brackets N subscript 2 close square brackets subscript 1 superscript space space x end superscript space space open square brackets H subscript 2 close square brackets subscript 1 superscript space space y end superscript end fraction end cell row cell fraction numerator 0 comma 8 over denominator 0 comma 4 end fraction end cell equals cell fraction numerator k left parenthesis 0 comma 02 right parenthesis to the power of x left parenthesis 0 comma 02 right parenthesis to the power of y over denominator k left parenthesis 0 comma 01 right parenthesis to the power of x left parenthesis 0 comma 02 right parenthesis to the power of y end fraction end cell row 2 equals cell 2 to the power of x end cell row 1 equals x end table 
 

Menghitung orde reaksi y terhadap H subscript bold 2  
Pada percobaan di atas tidak ada dua percobaan yang memiliki konsentrasi H subscript 2 sama, sehingga untuk membandingkannya dipilih yaitu percobaan (2) dan (3), tetapi bisa juga membandingkan dengan percobaan lain.
table attributes columnalign right center left columnspacing 0px end attributes row cell r subscript 3 over r subscript 2 end cell equals cell fraction numerator k open square brackets N subscript 2 close square brackets subscript 3 superscript space space x end superscript space space open square brackets H subscript 2 close square brackets subscript 3 to the power of y over denominator k open square brackets N subscript 2 close square brackets subscript 2 superscript space space x end superscript space space open square brackets H subscript 2 close square brackets subscript 2 superscript space space y end superscript end fraction end cell row cell fraction numerator 10 comma 8 over denominator 0 comma 8 end fraction end cell equals cell fraction numerator k left parenthesis 0 comma 03 right parenthesis to the power of 1 left parenthesis 0 comma 06 right parenthesis to the power of y over denominator k left parenthesis 0 comma 02 right parenthesis to the power of 1 left parenthesis 0 comma 02 right parenthesis to the power of y end fraction end cell row cell 13 comma 5 end cell equals cell 1 comma 5 cross times 3 to the power of italic y end cell row cell fraction numerator 13 comma 5 over denominator 1 comma 5 end fraction end cell equals cell 3 to the power of italic y end cell row 9 equals cell 3 to the power of italic y end cell row cell 3 squared end cell equals cell 3 to the power of italic y end cell row 2 equals y end table 

Menuliskan persamaan laju reaksi

Orde reaksi  x terhadap N subscript 2  = 1
orde reaksi y terhadap H subscript 2 = 2
r double bond k open square brackets N subscript 2 close square brackets to the power of x open square brackets H subscript 2 close square brackets to the power of y r double bond k open square brackets N subscript 2 close square brackets open square brackets H subscript 2 close square brackets squared 

Jadi, jawaban yang tepat adalah Bspace space space

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

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