Roboguru

Diketahui reaksi: 2NO(g)+O2​(g)→NO2​(g). Jika diketahui persamaan laju reaksi percobaan tersebut v=k[NO]2[O2​] harga k=125, [NO]=2×10−3M dan [O2​]=3×10−3M. Laju reaksi untuk persamaan tersebut adalah ....

Pertanyaan

Diketahui reaksi: 2 N O open parentheses italic g close parentheses space plus space O subscript 2 open parentheses italic g close parentheses space rightwards arrow space N O subscript 2 open parentheses italic g close parentheses. Jika diketahui persamaan laju reaksi percobaan tersebut begin mathsize 14px style italic v equals italic k open square brackets N O close square brackets squared open square brackets O subscript 2 close square brackets end style harga begin mathsize 14px style italic k equals 125 end style, begin mathsize 14px style open square brackets N O close square brackets equals 2 cross times 10 to the power of negative sign 3 end exponent space M end style dan begin mathsize 14px style open square brackets O subscript 2 close square brackets equals 3 cross times 10 to the power of negative sign 3 end exponent space M end style. Laju reaksi untuk persamaan tersebut adalah ....

  1. begin mathsize 14px style 1 comma 5 cross times 10 to the power of negative 6 end exponent space straight M space straight s to the power of negative 1 end exponent end style 

  2. begin mathsize 14px style 3 cross times 10 to the power of negative 6 end exponent space straight M space straight s to the power of negative 1 end exponent end style

  3. begin mathsize 14px style 3 cross times 10 to the power of negative sign 3 space end exponent M space s to the power of negative sign 1 end exponent end style  

  4. begin mathsize 14px style 2 cross times 10 to the power of negative sign 3 end exponent space M space s to the power of negative sign 1 end exponent end style 

  5. begin mathsize 14px style 1 cross times 10 to the power of negative 2 end exponent space straight M space straight s to the power of negative 1 end exponent end style 

Pembahasan Video:

Pembahasan Soal:

Menentukan laju reaksi:

table attributes columnalign right center left columnspacing 0px end attributes row cell v space end cell equals cell space k open square brackets N O close square brackets squared open square brackets O subscript 2 close square brackets end cell row blank equals cell space 125 space open parentheses 2 cross times 10 to the power of negative sign 3 end exponent close parentheses squared open parentheses 3 cross times 10 to the power of negative sign 3 end exponent close parentheses end cell row blank equals cell space 1 comma 5 space cross times space 10 to the power of negative sign 6 end exponent space M space s to the power of negative sign 1 end exponent end cell end table  

Oleh karena itu, nilai laju reaksi adalah begin mathsize 14px style 1 comma 5 cross times 10 to the power of negative 6 end exponent space straight M space straight s to the power of negative 1 end exponent end style.

Jadi, jawaban yang benar adalah A.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Acfreelance

Terakhir diupdate 08 Oktober 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

Pertanyaan yang serupa

Pada suhu 27∘C terjadi reaksi sebagai berikut : 2NOCl(g)→2NO(g)+Cl2​(g) dengan data sebagai berikut : Tentukanlah : a. Orde reaksi b. Harga tetapan laju reaksi c. Ungkapan laju reaksi d. ...

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 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

Orde reaksi merupakan pangkat konsentrasi zat pereaksi dalam persamaan laju reaksi. Orde reaksi pada soal diatas dapat ditentukan dengan membandingkan data dari beberapa percobaan. Dimisalkan data yang dibandingkan adalah percobaan 1 dan 2, maka :


table attributes columnalign right center left columnspacing 0px end attributes row cell v subscript 1 over v subscript 2 end cell equals cell fraction numerator k open square brackets N O Cl close square brackets subscript 1 to the power of x over denominator k open square brackets N O Cl close square brackets subscript 2 to the power of x end fraction end cell row cell fraction numerator 3 comma 60 cross times 10 to the power of negative sign 9 end exponent over denominator 1 comma 44 cross times 10 to the power of negative sign 8 end exponent end fraction end cell equals cell left parenthesis fraction numerator 0 comma 30 over denominator 0 comma 60 end fraction right parenthesis to the power of x end cell row cell fraction numerator 3 comma 60 cross times 10 to the power of negative sign 9 end exponent over denominator 14 comma 4 cross times 10 to the power of negative sign 9 end exponent end fraction end cell equals cell left parenthesis 1 half right parenthesis to the power of x end cell row cell 1 fourth end cell equals cell left parenthesis 1 half right parenthesis to the power of x end cell row cell left parenthesis 1 half right parenthesis squared end cell equals cell left parenthesis 1 half right parenthesis to the power of x end cell row x equals 2 end table


Jadi, orde reaksinya adalah 2.

b. Harga tetapan laju reaksi

Harga tetapan laju reaksi (k) dapat ditentukan dengan cara mensubstitusikan nilai x atau orde ke dalam salah satu data percobaan. Misal, nilai x dimasukkan ke dalam data percobaan ketiga maka :


table attributes columnalign right center left columnspacing 0px end attributes row v equals cell k open square brackets N O Cl close square brackets squared end cell row cell 3 comma 24 cross times 10 to the power of negative sign 8 end exponent end cell equals cell k left parenthesis 0 comma 90 right parenthesis squared end cell row cell 3 comma 24 cross times 10 to the power of negative sign 8 end exponent end cell equals cell k left parenthesis 8 comma 1 cross times 10 to the power of negative sign 1 end exponent right parenthesis end cell row k equals cell fraction numerator 3 comma 24 cross times 10 to the power of negative sign 8 end exponent over denominator 8 comma 1 cross times 10 to the power of negative sign 1 end exponent end fraction end cell row k equals cell 0 comma 4 cross times 10 to the power of negative sign 7 end exponent end cell row k equals cell 4 cross times 10 to the power of negative sign 8 end exponent end cell end table


Jadi, harga tetapan laju reaksinya adalah 4 cross times 10 to the power of negative sign 8 end exponent.

c. Ungkapan laju reaksi dapat ditentukan dengan cara memasukkan nilai orde (x) dan harga tetapan laju reaksi (k) yang telah dihitung ke dalam persamaan laju reaksi.

Jadi, ungkapan laju reaksinya yaitu v equals 4 cross times 10 to the power of negative sign 8 end exponent space open square brackets N O Cl close square brackets squared.

d. Laju reaksi sesaat jika diketahui open square brackets N O Cl close square brackets equals 1 comma 20 space bevelled M over liter

Laju reaksinya dapat ditentukan dengan cara mensubstitusikan konsentrasi NOCl tersebut ke dalam persamaan laju reaksi, maka :


table attributes columnalign right center left columnspacing 0px end attributes row v equals cell 4 cross times 10 to the power of negative sign 8 end exponent open square brackets N O Cl close square brackets squared end cell row blank equals cell 4 cross times 10 to the power of negative sign 8 end exponent left parenthesis 1 comma 20 right parenthesis squared end cell row blank equals cell 4 cross times 10 to the power of negative sign 8 end exponent middle dot 1 comma 44 end cell row blank equals cell 5 comma 76 cross times 10 to the power of negative sign 8 end exponent space bevelled M over detik end cell end table


Jadi, laju reaksi sesaat jika diketahui open square brackets N O Cl close square brackets equals 1 comma 20 space bevelled M over liter adalah Error converting from MathML to accessible text..space 

0

Roboguru

Persamaan laju reaksi pada reaksi: NH4+​(aq)+NO2−​(aq)→N2​(g)+2H2​O(l) adalah V=k[NH4+​][NO2−​]. Pada suhu 25∘C nilai k=4,0×10−4. Hitung laju reaksinya jika [NH4+​]=0,50M dan [NO2−​]=0,04M!

Pembahasan Soal:

Persamaan laju reaksi: begin mathsize 14px style V equals italic k open square brackets N H subscript 4 to the power of plus close square brackets open square brackets N O subscript 2 to the power of minus sign close square brackets end style dan nilai begin mathsize 14px style k equals 4 comma 0 cross times 10 to the power of negative 4 end exponent end style

Sehingga nilai laju reaksi untuk begin mathsize 14px style open square brackets N H subscript 4 to the power of plus close square brackets equals 0 comma 50 space M end style dan begin mathsize 14px style open square brackets N O subscript 2 to the power of minus sign close square brackets equals 0 comma 04 space M end style adalah:

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 N H subscript 4 to the power of plus close square brackets open square brackets N O subscript 2 to the power of minus sign close square brackets end cell row blank equals cell 4 comma 0 cross times 10 to the power of negative sign 4 end exponent space left parenthesis 0 comma 50 right parenthesis space left parenthesis 0 comma 04 right parenthesis end cell row blank equals cell 8 comma 0 cross times 10 to the power of negative sign 6 end exponent space mol space L to the power of negative sign 1 end exponent space s to the power of negative sign 1 end exponent end cell end table end style 

Jadi, jawaban yang benar adalah begin mathsize 14px style 8 comma 0 cross times 10 to the power of negative sign 6 end exponent space mol space L to the power of negative sign 1 end exponent space s to the power of negative sign 1 end exponent end style.

0

Roboguru

Untuk reaksi berikut: 2ICl(g)+H2​(g)→2HCl(g)+I2​(g)  Pada suhu tertentu tetapan laju reaksinya adalah 1,63×10−6L/mol.s. Berapa orde reaksi keseluruhannya?

Pembahasan Soal:

Bentuk umum hubungan orde reaksi total dengan satuan k:

table attributes columnalign right center left columnspacing 0px end attributes row k equals cell M to the power of 1 minus sign n end exponent space s to the power of negative sign 1 end exponent end cell row k equals cell left parenthesis mol space L to the power of negative sign 1 end exponent right parenthesis to the power of 1 minus sign n end exponent space s to the power of negative sign 1 end exponent end cell row L equals liter row n equals cell orde space reaksi space total end cell row s equals detik end table 

Pada soal diketahui  satuan k double bond L forward slash mol point s atau dapat diubah menjadi L point mol to the power of negative sign 1 end exponent space s to the power of negative sign 1 end exponent
space space space space space space space space space space space space space space space space space space k equals left parenthesis mol space L to the power of negative sign 1 end exponent right parenthesis to the power of 1 minus sign n end exponent space s to the power of negative sign 1 end exponent L point mol to the power of negative sign 1 end exponent space s to the power of negative sign 1 end exponent equals left parenthesis mol space L to the power of negative sign 1 end exponent right parenthesis to the power of 1 minus sign n end exponent space s to the power of negative sign 1 end exponent space space space space space space L point mol to the power of negative sign 1 end exponent equals left parenthesis mol space L to the power of negative sign 1 end exponent right parenthesis to the power of 1 minus sign n end exponent left parenthesis mol point L to the power of negative sign 1 end exponent right parenthesis to the power of negative sign 1 end exponent equals left parenthesis mol space L to the power of negative sign 1 end exponent right parenthesis to the power of 1 minus sign n end exponent space space space space space space space space space space space space space minus sign 1 equals 1 minus sign n space space space space space space space space space space space space space space space space space n equals 1 plus 1 space space space space space space space space space space space space space space space space space n equals 2      

Jadi orde reaksi total adalah 2.

Oleh karena itu, jawaban yang benar adalah C.space 

1

Roboguru

Diketahui reaksi A → B dengan laju reaksi  pada saat konsentrasi A sebesar . Hitunglah tetapan laju reaksinya jika reaksi orde 3 terhadap A!

Pembahasan Soal:

Reaksi tersebut memiliki order reaksi terhadap A sebesar 3, maka persamaan laju reaksinya adalah begin mathsize 14px style v double bond k open square brackets A close square brackets cubed end style. Dari persamaan tersebut, perhitungan tetapan laju reaksi adalah sebagai berikut 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row v equals cell k open square brackets A close square brackets cubed end cell row cell space k end cell equals cell v over open square brackets A close square brackets cubed end cell end table end style 

 begin mathsize 14px style k equals fraction numerator 1 comma 6 cross times 10 to the power of negative sign 2 end exponent space mol space dm to the power of negative sign 3 end exponent space detik to the power of negative sign 1 end exponent over denominator left parenthesis 0 comma 35 space mol space dm to the power of negative sign 3 end exponent right parenthesis cubed end fraction k equals fraction numerator 1 comma 6 cross times 10 to the power of negative sign 2 end exponent space mol space dm to the power of negative sign 3 end exponent space detik to the power of negative sign 1 end exponent over denominator 0 comma 043 space mol cubed space dm to the power of negative sign 9 end exponent end fraction k equals 0 comma 372 space mol to the power of negative sign 2 end exponent space dm to the power of 6 space detik to the power of negative sign 1 end exponent end style 

 

Jadi, tetapan laju reaksinya adalah undefined

0

Roboguru

Gas dinitrogen pentaoksida N2​O5​, terurai membentuk nitrogen dioksida dan gas oksigen dengan data laju awal pada suhu 25∘C diberikan dalam tabel berikut:     a. Tuliskan persamaan untuk reaksi ini...

Pembahasan Soal:

a. Menuliskan persamaan reaksi

    2 N subscript 2 O subscript 5 yields 4 N O subscript 2 and O subscript 2 

b. Menentukan hukum laju reaksi dan tetapan laju reaksi

    Menentukan hukum laju reaksi

    table attributes columnalign right center left columnspacing 0px end attributes row cell misal space v end cell equals cell italic k open square brackets N subscript 2 O subscript 5 close square brackets to the power of italic x italic comma italic space maka colon end cell row cell v subscript 1 over v subscript 2 end cell equals cell fraction numerator open square brackets N subscript 2 O subscript 5 close square brackets subscript 1 superscript x over denominator open square brackets N subscript 2 O subscript 5 close square brackets subscript 2 superscript x end fraction end cell row cell fraction numerator 7 comma 98 cross times 10 to the power of negative sign 4 end exponent space M forward slash menit over denominator 3 comma 42 cross times 10 to the power of negative sign 4 end exponent space M forward slash menit end fraction end cell equals cell fraction numerator open square brackets 0 comma 350 space M close square brackets subscript 1 superscript x over denominator open square brackets 0 comma 150 space M close square brackets subscript 2 superscript x end fraction end cell row cell 2 comma 33 end cell equals cell open square brackets 2 comma 33 close square brackets to the power of x end cell row x equals 1 end table 

   Bila diperiksa dengan data lain didapatkan juga x=1, dengan demikian hukum laju reaksinya adalah v equals italic k open square brackets N subscript 2 O subscript 5 close square brackets.

     Menentukan tetapan laju reaksi

     k equals fraction numerator v over denominator open square brackets N subscript 2 O subscript 5 close square brackets end fraction k equals fraction numerator 1 comma 48 cross times 10 to the power of negative sign 3 end exponent over denominator 0 comma 650 space M end fraction k equals 2 comma 28 cross times 10 to the power of negative sign 3 end exponent space menit to the power of negative sign 1 end exponent 

    Jadi jawaban yang benar adalah sesuai uraian di atas.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?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ikuti Kami

©2021 Ruangguru. All Rights Reserved