Pada suhu , gas bromin dapat bereaksi dengan nitrogen monoksida menurut reaksi berikut: Dari reaksi tersebut, diperoleh data sebagai berikut: Tentukan persamaan laju reaksi dan tetapan laju reaksinya!

Pertanyaan

Pada suhu $273 degree C$ , gas bromin dapat bereaksi dengan nitrogen monoksida menurut reaksi berikut:

$2 N O open parentheses italic g close parentheses and Br subscript 2 open parentheses italic g close parentheses yields 2 N O Br open parentheses italic g close parentheses$

Dari reaksi tersebut, diperoleh data sebagai berikut:

Tentukan persamaan laju reaksi dan tetapan laju reaksinya! $space$

Pembahasan Soal:

Pertama menentukan orde reaksi NO dengan perbandingan data 4 banding 1 yaitu: $space$

$table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets N O close square brackets to the power of italic x open square brackets Br subscript 2 close square brackets to the power of italic y end cell equals V row cell open square brackets fraction numerator 0 comma 2 over denominator 0 comma 1 end fraction close square brackets to the power of x open square brackets fraction numerator 0 comma 05 over denominator 0 comma 05 end fraction close square brackets to the power of y end cell equals cell 24 over 6 end cell row cell open square brackets 2 close square brackets to the power of x end cell equals 4 row cell open square brackets 2 close square brackets to the power of x end cell equals cell 2 squared end cell row x equals 2 end table$

Sedangkan orde $Br subscript 2$ dapat ditentukan dengan perbandingan data 2 dan 1 yaitu:

$table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets N O close square brackets to the power of italic x open square brackets Br subscript 2 close square brackets to the power of italic y end cell equals V row cell open square brackets fraction numerator 0 comma 1 over denominator 0 comma 1 end fraction close square brackets to the power of x open square brackets fraction numerator 0 comma 10 over denominator 0 comma 05 end fraction close square brackets to the power of y end cell equals cell 12 over 6 end cell row cell open square brackets 2 close square brackets to the power of italic y end cell equals cell 2 to the power of 1 end cell row x equals 1 end table$

Dengan demikian maka persamaan lajunya adalah: $V double bond k open square brackets N O close square brackets squared open square brackets Br subscript 2 close square brackets$

Sedangkan tetapan laju reaksinya adalah:

$V double bond k open square brackets N O close square brackets squared open square brackets Br subscript 2 close square brackets k equals fraction numerator V over denominator open square brackets N O close square brackets squared open square brackets Br subscript 2 close square brackets end fraction k equals fraction numerator 6 space M space s to the power of negative sign 1 end exponent over denominator open square brackets 0 comma 1 M close square brackets squared open square brackets 0 comma 05 M close square brackets end fraction k equals fraction numerator 6 space M space s to the power of negative sign 1 end exponent over denominator 5 cross times 10 to the power of negative sign 4 end exponent space M cubed end fraction k equals 1 comma 2 cross times 10 to the power of 4 space M to the power of negative sign 2 end exponent space s to the power of negative sign 1 end exponent$

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

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

Berikut ini adalah data hasil percobaan laju reaksi dari: Reaksi tersebut mempunyai tetapan laju reaksi sebesar ....

Pembahasan Soal:

Orde reaksi menunjukkan besarnya pengaruh konsentrasi pereaksi terhadap laju reaksi. Penentuan orde reaksi dapat dihitung berdasarkan data eksperimen dan dinyatakan melalui persamaan laju reaksi.

Reaktan yang bereaksi adalah NO dan $H subscript 2$ , sehingga persamaan laju dapat dituliskan menjadi : $v double bond k open square brackets N O close square brackets to the power of m open square brackets H subscript 2 close square brackets to the power of n$ . Penentuan orde reaksi melalui data eksperimen dihitung berdasarkan perbandingan data percobaan yang sama.

Tentukan orde NO yaitu cari data $H subscript 2$ yang sama yaitu percobaan 3 dan 4.

$table attributes columnalign right center left columnspacing 0px end attributes row cell v subscript 3 over v subscript 4 end cell equals cell k subscript 3 over k subscript 4 fraction numerator open square brackets N O close square brackets subscript 3 superscript m over denominator open square brackets N O close square brackets subscript 4 superscript m end fraction fraction numerator open square brackets H subscript 2 close square brackets subscript 3 superscript n over denominator open square brackets H subscript 2 close square brackets subscript 4 superscript n end fraction end cell row cell fraction numerator 0 comma 5 over denominator 2 comma 0 end fraction end cell equals cell k subscript 3 over k subscript 4 open parentheses fraction numerator 0 comma 10 over denominator 0 comma 20 end fraction close parentheses to the power of m open parentheses fraction numerator 0 comma 25 over denominator 0 comma 25 end fraction close parentheses to the power of n end cell row cell 1 fourth end cell equals cell open parentheses 1 half close parentheses to the power of m end cell row cell open parentheses 1 half close parentheses squared end cell equals cell open parentheses 1 half close parentheses to the power of m end cell row m equals 2 end table$

Tentukan orde $H subscript 2$ yaitu cari data NO yang sama yaitu percobaan 3 dan 4.

$table attributes columnalign right center left columnspacing 0px end attributes row cell v subscript 1 over v subscript 2 end cell equals cell k subscript 1 over k subscript 2 fraction numerator open square brackets N O close square brackets subscript 1 superscript m over denominator open square brackets N O close square brackets subscript 2 superscript m end fraction fraction numerator open square brackets H subscript 2 close square brackets subscript 1 superscript n over denominator open square brackets H subscript 2 close square brackets subscript 2 superscript n end fraction end cell row cell fraction numerator 1 comma 6 over denominator 4 comma 8 end fraction end cell equals cell k subscript 3 over k subscript 4 open parentheses fraction numerator 0 comma 30 over denominator 0 comma 30 end fraction close parentheses to the power of m open parentheses fraction numerator 0 comma 05 over denominator 0 comma 15 end fraction close parentheses to the power of n end cell row cell 1 third end cell equals cell open parentheses 1 third close parentheses to the power of n end cell row cell open parentheses 1 third close parentheses to the power of 1 end cell equals cell open parentheses 1 half close parentheses to the power of n end cell row n equals 1 end table$

Tentukan persamaan laju reaksi dengan menyertakan orde reaksi masing-masing zat, dimana orde NO = 2 dan orde $H subscript 2$ = 1.

Tentukan tetapan laju reaksi (k). Penentuan tetapan laju reaksi dapat kita hitung berdasarkan salah satu data percobaan saja, misalnya percobaan 3. Oleh karena itu, kita memakai data percobaan 1 baik konsentrasi reaktan maupun laju reaksinya.

$table attributes columnalign right center left columnspacing 0px end attributes row v equals cell k open square brackets N O close square brackets squared open square brackets H subscript 2 close square brackets to the power of 1 end cell row cell 0 comma 5 end cell equals cell k open parentheses 0 comma 10 close parentheses squared open parentheses 0 comma 25 close parentheses end cell row cell 0 comma 5 end cell equals cell k open parentheses 0 comma 01 close parentheses open parentheses 0 comma 25 close parentheses end cell row k equals cell fraction numerator 0 comma 5 over denominator 0 comma 01 cross times 0 comma 25 end fraction end cell row k equals cell fraction numerator 0 comma 5 over denominator 0 comma 0025 end fraction end cell row k equals 200 end table$

Berdasarkan perhitungan diatas, maka persamaan laju reaksinya adalah $table attributes columnalign right center left columnspacing 0px end attributes row v equals cell k open square brackets N O close square brackets squared open square brackets H subscript 2 close square brackets to the power of 1 end cell end table$ dan tetapan laju reaksi sebesar 200.

Jadi, jawaban yang benar adalah E.

Roboguru

Data eksperimen untuk reaksi disajikan dalam tabel berikut. Harga x adalah ....

Pembahasan Soal:

Persamaan laju reaksi adalah hasil kali dari tetapan laju reaksi dengan konsentrasi awal zat yang dipangkatkan dengan orde reaksi. Dengan orde reaksi pada percobaan diatas, yaitu:

Percobaan 1 dan 5

$\frac{v _{5}}{v _{1}} \frac{54}{6} 9 p = = = = \frac{k [ A ] _{5}^{p} [ B ] _{5}^{q}}{k [ A ] _{1}^{p} [ B ] _{1}^{q}} \frac{k [ 0 , 6 ] ^{p} [ 0 , 2 ] ^{q}}{k [ 0 , 2 ] ^{p} [ 0 , 2 ] ^{q}} 3^{p} 2$

Percobaan 1 dan 2

$\frac{v _{2}}{v _{1}} \frac{12}{6} 2 q = = = = \frac{k [ A ] _{2}^{p} [ B ] _{2}^{q}}{k [ A ] _{1}^{p} [ B ] _{1}^{q}} \frac{k [ 0 , 2 ] ^{p} [ 0 , 4 ] ^{q}}{k [ 0 , 2 ] ^{p} [ 0 , 2 ] ^{q}} 2^{q} 1$

Jadi, $orde A = p = 2$ dan $orde B = q = 1$

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

Mencari Nilai X
Nilai X dapat dicari dengan membanding percobaan 4 dengan percobaan 1, diperoleh:
$\frac{v _{4}}{v _{1}} \frac{X}{6} \frac{X}{6} X X = = = = = \frac{k [ A ] _{4}^{2} [ B ] _{4}}{k [ A ] _{1}^{2} [ B ] _{1}} \frac{k [ 0 , 4 ] ^{2} [ 0 , 2 ]}{k [ 0 , 2 ] ^{2} [ 0 , 2 ]} (2)^{2} (6) \times (4) 24 M s^{- 1}$

Jadi, jawaban yang benar adalah B.

Roboguru

Pada reaksi: diperoleh data basil percobaan sebagai berikut. Berdasarkan data tersebut, tentukan: a. orde reaksi terhadap dan ; b. persamaan laju reaksi; c. harga tetapan laju reaksi...

Pembahasan Soal:

a. Orde reaksi $begin mathsize 14px style S O subscript 2 end style$ dan $begin mathsize 14px style O subscript 2 end style$

Menentukan orde $begin mathsize 14px style S O subscript 2 end style$ dengan percobaan 1 dan 2

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell v subscript 2 over v subscript 1 end cell equals cell fraction numerator k open square brackets S O subscript 2 close square brackets subscript 2 superscript x open square brackets O subscript 2 close square brackets subscript 2 superscript y over denominator k open square brackets S O subscript 2 close square brackets subscript 1 superscript x open square brackets O subscript 2 close square brackets subscript 1 superscript y end fraction end cell row cell fraction numerator 1 comma 6 cross times 10 to the power of negative sign 6 end exponent over denominator 0 comma 4 cross times 10 to the power of negative sign 6 end exponent end fraction end cell equals cell fraction numerator k open square brackets 0 comma 008 close square brackets to the power of x open square brackets 0 comma 002 close square brackets to the power of y over denominator k open square brackets 0 comma 004 close square brackets to the power of x open square brackets 0 comma 002 close square brackets to the power of y end fraction end cell row cell 4 space end cell equals cell space 2 to the power of x end cell row cell x space end cell equals cell space 2 end cell end table end style$

Menentukan orde $begin mathsize 14px style O subscript 2 end style$ dengan percobaan 4 dan 5

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell v subscript 4 over v subscript 5 space end cell equals cell space fraction numerator k open square brackets S O subscript 2 close square brackets subscript 4 superscript x open square brackets O subscript 2 close square brackets subscript 4 superscript y over denominator k open square brackets S O subscript 2 close square brackets subscript 5 superscript x open square brackets O subscript 2 close square brackets subscript 5 superscript y end fraction end cell row cell fraction numerator 9 comma 6 cross times 10 to the power of negative sign 6 end exponent over denominator 2 comma 4 cross times 10 to the power of negative sign 6 end exponent end fraction space end cell equals cell space fraction numerator k open square brackets 0 comma 008 close square brackets to the power of x open square brackets 0 comma 016 close square brackets to the power of y over denominator k open square brackets 0 comma 008 close square brackets to the power of x open square brackets 0 comma 004 close square brackets to the power of y end fraction end cell row cell 4 space end cell equals cell space 4 to the power of y end cell row cell y space end cell equals cell space 1 end cell end table end style$

Jadi, orde reaksi $begin mathsize 14px style S O subscript 2 end style$ dan orde reaksi $begin mathsize 14px style O subscript 2 end style$ berturut-turut adalah 2 dan 1.

b. Persamaan laju reaksi: $begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell v space end cell equals cell space k open square brackets S O subscript 2 close square brackets to the power of x open square brackets O subscript 2 close square brackets to the power of y end cell row cell v space end cell equals cell space k open square brackets S O subscript 2 close square brackets squared open square brackets O subscript 2 close square brackets end cell end table end style$

Jadi, persamaan laju reaksi yang benar adalah $begin mathsize 14px style v space equals space k open square brackets S O subscript 2 close square brackets squared open square brackets O subscript 2 close square brackets end style$ .

c. Harga tetapan laju reaksi

Percobaan 1

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell v space end cell equals cell space k open square brackets S O subscript 2 close square brackets squared open square brackets O subscript 2 close square brackets end cell row cell 0 comma 4 cross times 10 to the power of negative sign 6 end exponent space end cell equals cell space k left parenthesis 0 comma 004 right parenthesis squared left parenthesis 0 comma 002 right parenthesis end cell row cell k space end cell equals cell 12 comma 5 end cell end table end style$

Jadi, harga tetapan laju reaksi yang benar adalah 12,5.

Roboguru

Terdapat suatu reaksi antara senyawa A dan senyawa B menghasilkan AB sesuai persamaan berikut. Data percobaan yang dihasilkan adalah sebagai berikut. Dari data tersebut, nilai x sebesar .....

Pembahasan Soal:

Ingat! Persamaan laju adalah $v double bond k space open square brackets reaktan close square brackets to the power of orde$ . Persamaan reaksi

$A and B yields AB$

memiliki persamaan laju

$v double bond k space open square brackets A close square brackets to the power of x open square brackets B close square brackets to the power of y$

Perlu dilakukan penentuan orde reaksi dari masing-masing reaktan.

Penentuan orde [A] menggunakan data [B] yang sama, yaitu data percobaan 4 dan 5.

$Error converting from MathML to accessible text.$

Penentuan orde [B] menggunakan data [A] yang sama, yaitu percobaan 1 dan 5.

$Error converting from MathML to accessible text.$

Berdasarkan orde reaksi yang diperoleh, persamaan laju reaksinya adalah sebagai berikut.

$table attributes columnalign right center left columnspacing 0px end attributes row v equals cell k space open square brackets A close square brackets to the power of x open square brackets B close square brackets to the power of y end cell row v equals cell k space open square brackets A close square brackets to the power of 0 open square brackets B close square brackets to the power of 1 end cell end table$

Penentuan konstanta laju reaksi bisa mencoba salah satu percobaan, misalkan percobaan 1.

^{$Error converting from MathML to accessible text.$}

Selanjutnya x pada percobaan 2 dapat dihitung sebagai berikut.

$table attributes columnalign right center left columnspacing 0px end attributes row v equals cell k space open square brackets A close square brackets to the power of 0 open square brackets B close square brackets to the power of 1 end cell row v equals cell 5 cross times 10 to the power of negative sign 3 end exponent space det to the power of negative sign 1 end exponent space left square bracket 1 M right square bracket to the power of 0 space left square bracket 0 comma 6 M right square bracket to the power of 1 end cell row v equals cell 5 cross times 10 to the power of negative sign 3 end exponent space det to the power of negative sign 1 end exponent space left square bracket 1 right square bracket space left square bracket 0 comma 6 M right square bracket end cell row v equals cell 3 cross times 10 to the power of negative sign 3 end exponent space M space det to the power of negative sign 1 end exponent end cell end table$

Jadi, jawaban yang benar adalah opsi C.

Roboguru

Untuk reaksi , diperoleh data percobaan sebagai berikut: Tentukan: Orde terhadap zat Orde terhadap zat Orde total Persamaan laju reaksi Tetapan laju reaksinya () Tentukan laju reaksi pe...

Pembahasan Soal:

1. Untuk mencari orde reaksi pada $begin mathsize 14px style P end style$ adalah dengan memperhatikan variabel kontrol (konsentrasi $begin mathsize 14px style Q end style$ ) pada percobaan.

Menggunakan percobaan pertama dan kedua:

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses begin inline style open square brackets P close square brackets subscript 1 over open square brackets P close square brackets subscript 2 end style close parentheses to the power of x end cell equals cell begin inline style v subscript 1 over v subscript 2 end style end cell row cell open parentheses begin inline style fraction numerator 0 comma 1 over denominator 0 comma 2 end fraction end style close parentheses to the power of x end cell equals cell begin inline style 6 over 12 end style end cell row cell open parentheses begin inline style 1 half end style close parentheses to the power of x end cell equals cell begin inline style 1 half end style end cell row x equals 1 end table end style$

2. Untuk mencari orde reaksi pada $begin mathsize 14px style Q end style$ adalah dengan memperhatikan variabel kontrol (konsentrasi $begin mathsize 14px style P end style$ ) pada percobaan.

Menggunakan percobaan kedua dan ketiga:

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses begin inline style open square brackets Q close square brackets subscript 2 over open square brackets Q close square brackets subscript 3 end style close parentheses to the power of y end cell equals cell begin inline style v subscript 2 over v subscript 3 end style end cell row cell open parentheses begin inline style fraction numerator 0 comma 1 over denominator 0 comma 2 end fraction end style close parentheses to the power of y end cell equals cell begin inline style 12 over 48 end style end cell row cell open parentheses begin inline style 1 half end style close parentheses to the power of y end cell equals cell begin inline style 1 fourth end style end cell row y equals 2 end table end style$

3. Orde total:

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell x and y end cell equals cell 1 plus 2 end cell row blank equals 3 end table end style$

4. Persamaan laju reaksi:

$begin mathsize 14px style v double bond k cross times open square brackets P close square brackets to the power of 1 cross times open square brackets Q close square brackets squared end style$

5. Nilai k

Mencari nilai k dapat dilakukan dengan mensubstitusikan salah satu percobaan ke dalam persamaan laju reaksi: (misal menggunakan percobaan ke 1)

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row v equals cell k cross times open square brackets P close square brackets to the power of 1 cross times open square brackets Q close square brackets squared end cell row 6 equals cell k cross times left parenthesis 0 comma 1 right parenthesis to the power of 1 cross times left parenthesis 0 comma 1 right parenthesis squared end cell row 6 equals cell k cross times 10 to the power of negative sign 3 end exponent end cell row k equals cell 6 cross times 10 cubed end cell end table end style$

5. v pada percobaan keempat:

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row v equals cell k cross times open square brackets P close square brackets to the power of 1 cross times open square brackets Q close square brackets squared end cell row blank equals cell 6 cross times 10 cubed cross times left parenthesis 0 comma 3 right parenthesis to the power of 1 cross times left parenthesis 0 comma 3 right parenthesis squared end cell row blank equals cell 162 space M forward slash det end cell end table end style$

Roboguru