Roboguru

Senyawa  jika dipanaskan dapat terdekomposisi menjadi  dan  Dalam suatu percobaan,  0,003  dipanaskan pada suhu  selama 2,1 menit. Ternyata, konsentrasinya berkurang menjadi 0,0013 . Jika laju reaksi dekomposisi tersebut adalah reaksi orde satu, hitung nilai tetapan jenis reaksinya!

Pertanyaan

Senyawa N subscript 2 O subscript 5 jika dipanaskan dapat terdekomposisi menjadi N O subscript 2 dan O subscript 2 Dalam suatu percobaan, N subscript 2 O subscript 5 0,003 mol space L to the power of negative sign 1 end exponent dipanaskan pada suhu 65 degree C selama 2,1 menit. Ternyata, konsentrasinya berkurang menjadi 0,0013 mol space L to the power of negative sign 1 end exponent. Jika laju reaksi dekomposisi tersebut adalah reaksi orde satu, hitung nilai tetapan jenis reaksinya!space

Pembahasan Soal:

Berdasarkan data di soal dapat ditentukan laju reaksinya yaitu:
 

table attributes columnalign right center left columnspacing 0px end attributes row V equals cell fraction numerator increment M over denominator increment t end fraction end cell row blank equals cell fraction numerator left parenthesis 0 comma 003 minus sign 0 comma 0013 right parenthesis space mol space L to the power of negative sign 1 end exponent over denominator 2 comma 1 space menit end fraction end cell row blank equals cell fraction numerator 0 comma 0017 space mol space L to the power of negative sign 1 end exponent over denominator 2 comma 1 space menit end fraction end cell row blank equals cell 8 comma 09 cross times 10 to the power of negative sign 4 end exponent space M space menit to the power of negative sign 1 end exponent end cell end table
 

Dengan demikian maka nilai tetapan jenis reaksinya adalah:space
 

V double bond k open square brackets N subscript 2 O subscript 5 close square brackets 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 8 comma 09 cross times 10 to the power of negative sign 4 end exponent space M space menit to the power of negative sign 1 end exponent over denominator 3 cross times 10 to the power of negative sign 3 end exponent space M end fraction k equals 2 comma 69 cross times 10 to the power of negative sign 1 end exponent space menit 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

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Pertanyaan yang serupa

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.

0

Roboguru

Data eksperimen dari reaksi:    adalah sebagai berikut.   Nilai k 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 H subscript 2 open parentheses italic g close parentheses and Cl subscript 2 open parentheses italic g close parentheses yields 2 H Cl open parentheses italic g close parentheses mempunyai persamaan laju reaksi:

italic r equals italic k open square brackets H subscript 2 close square brackets to the power of italic x open square brackets Cl subscript 2 close square brackets to the power of italic y 

dengan:

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

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

Berikut langkah-langkah menentukan nilai k pada persamaan reaksi di atas:

  1. Menghitung orde reaksi H subscript bold 2 

    Untuk menghitung orde reaksi H subscript 2, pilih 2 percobaan dimana Cl subscript 2 mempunyai konsentrasi yang sama, yaitu percobaan (1) dan (3).

    italic r subscript 1 over italic r subscript 3 equals fraction numerator italic k open square brackets H subscript 2 close square brackets subscript 1 superscript italic x open square brackets Cl subscript 2 close square brackets subscript 1 superscript italic y over denominator italic k open square brackets H subscript 2 close square brackets subscript 3 superscript italic x open square brackets Cl subscript 2 close square brackets subscript 3 superscript italic y end fraction fraction numerator 1 comma 2 cross times 10 to the power of negative sign 4 end exponent over denominator 2 comma 4 cross times 10 to the power of negative sign 4 end exponent end fraction equals fraction numerator up diagonal strike italic k open parentheses 3 cross times 10 to the power of negative sign 2 end exponent close parentheses to the power of italic x up diagonal strike left parenthesis 2 cross times 10 to the power of negative sign 2 end exponent right parenthesis to the power of italic y end strike over denominator up diagonal strike italic k open parentheses 6 cross times 10 to the power of negative sign 2 end exponent close parentheses to the power of italic x up diagonal strike open parentheses 2 cross times 10 to the power of negative sign 2 end exponent close parentheses to the power of italic y end strike end fraction 1 half equals open parentheses 1 half close parentheses to the power of italic x italic x equals 1 
     
  2. Menghitung orde reaksi Cl subscript bold 2  

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

    italic r subscript 1 over italic r subscript 2 equals fraction numerator italic k open square brackets H subscript 2 close square brackets subscript 1 superscript italic x open square brackets Cl subscript 2 close square brackets subscript 1 superscript italic y over denominator italic k open square brackets H subscript 2 close square brackets subscript 2 superscript italic x open square brackets Cl subscript 2 close square brackets subscript 2 superscript italic y end fraction fraction numerator 1 comma 2 cross times 10 to the power of negative sign 4 end exponent over denominator 4 comma 8 cross times 10 to the power of negative sign 4 end exponent end fraction equals fraction numerator up diagonal strike italic k left parenthesis 3 cross times 10 to the power of negative sign 2 end exponent right parenthesis to the power of italic x end strike open parentheses 2 cross times 10 to the power of negative sign 2 end exponent close parentheses to the power of italic y over denominator up diagonal strike italic k left parenthesis 3 cross times 10 to the power of negative sign 2 end exponent right parenthesis to the power of italic x end strike open parentheses 4 cross times 10 to the power of negative sign 2 end exponent close parentheses to the power of italic y end fraction 1 fourth equals open parentheses 1 half close parentheses to the power of italic y italic y equals 2 
     
  3. Menentukan persamaan laju reaksi

    italic r equals italic k open square brackets H subscript 2 close square brackets to the power of italic x open square brackets Cl subscript 2 close square brackets to the power of italic y italic r equals italic k open square brackets H subscript 2 close square brackets open square brackets Cl subscript 2 close square brackets to the power of italic 2  
     
  4. Menghitung nilai k

    Misal kita gunakan persamaan laju reaksi dari percobaan (1)

    italic r equals italic k open square brackets H subscript 2 close square brackets open square brackets Cl subscript 2 close square brackets squared italic k equals fraction numerator italic r over denominator open square brackets H subscript 2 close square brackets open square brackets Cl subscript 2 close square brackets squared end fraction italic k equals fraction numerator 1 comma 2 cross times 10 to the power of negative sign 4 end exponent over denominator open parentheses 3 cross times 10 to the power of negative sign 2 end exponent close parentheses open parentheses 2 cross times 10 to the power of negative sign 2 end exponent close parentheses squared end fraction italic k equals fraction numerator 1 comma 2 cross times 10 to the power of negative sign 4 end exponent over denominator open parentheses 3 cross times 10 to the power of negative sign 2 end exponent close parentheses open parentheses 4 cross times 10 to the power of negative sign 4 end exponent close parentheses end fraction italic k equals fraction numerator 1 comma 2 cross times 10 to the power of negative sign 4 end exponent over denominator 1 comma 2 cross times 10 to the power of negative sign 5 end exponent end fraction italic k equals 10 


Jadi, jawaban yang tepat adalah B.space space space

0

Roboguru

Brominasi aseton memerlukan katalis asam dengan reaksi:   Data hasil percobaan reaksi di atas disajikan pada tabel berikut:   Tentukan rumus laju reaksinya. Tentukan nilai tetapan laju reaksi (...

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. Jika terdapat katalis dalam reaksi, maka konsentrasi katalis diikutsertakan dalam persamaan laju reaksi.

Reaksi C H subscript 3 C O C H subscript 3 and Br subscript 2 yields with H to the power of plus sign on top C H subscript 3 C O C H subscript 2 Br and H to the power of plus sign and Br to the power of minus sign mempunyai persamaan laju reaksi:

italic v equals italic k open square brackets C H subscript 3 C O C H subscript 3 close square brackets to the power of italic x open square brackets Br subscript 2 close square brackets to the power of italic y open square brackets H to the power of plus sign close square brackets to the power of italic z 

dengan:

k = tetapan laju reaksi
x = orde (tingkat atau pangkat) reaksi terhadap C H subscript 3 C O C H subscript 3 
y = orde reaksi terhadap Br subscript 2 
z = orde reaksi terhadap H to the power of plus sign 

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

a.   Rumus laju reaksi

  • Menghitung orde reaksi C H subscript bold 3 C O C H subscript bold 3 

    Untuk menghitung orde reaksi C H subscript 3 C O C H subscript 3, pilih 2 percobaan dimana Br subscript 2 dan H to the power of plus sign  mempunyai konsentrasi yang sama, yaitu percobaan (1) dan (5).

    v subscript 1 over v subscript 5 equals fraction numerator italic k open square brackets C H subscript 3 C O C H subscript 3 close square brackets subscript 1 superscript italic x open square brackets Br subscript 2 close square brackets subscript 1 superscript italic y open square brackets H to the power of plus sign close square brackets subscript 1 superscript italic z over denominator italic k open square brackets C H subscript 3 C O C H subscript 3 close square brackets subscript 5 superscript italic x open square brackets Br subscript 2 close square brackets subscript 5 superscript italic y open square brackets H to the power of plus sign close square brackets subscript 5 superscript italic z end fraction fraction numerator 5 comma 7 cross times 10 to the power of negative sign 5 end exponent over denominator 7 comma 6 cross times 10 to the power of negative sign 5 end exponent end fraction equals fraction numerator up diagonal strike italic k open parentheses 0 comma 30 close parentheses to the power of italic x up diagonal strike open parentheses 0 comma 05 close parentheses to the power of italic y end strike up diagonal strike open parentheses 0 comma 05 close parentheses to the power of italic z end strike over denominator up diagonal strike italic k begin italic style left parenthesis straight 0 straight comma straight 40 right parenthesis end style to the power of italic x up diagonal strike begin italic style left parenthesis straight 0 straight comma straight 05 right parenthesis end style to the power of italic y end strike up diagonal strike begin italic style left parenthesis straight 0 straight comma straight 05 right parenthesis end style to the power of italic z end strike end fraction 3 over 4 equals open parentheses 3 over 4 close parentheses to the power of italic x italic x equals 1 
     
  • Menghitung orde reaksi Br subscript bold 2 

    Untuk menghitung orde reaksi Br subscript 2, pilih 2 percobaan dimana C H subscript 3 C O C H subscript 3 dan H to the power of plus sign mempunyai konsentrasi yang sama, yaitu percobaan (1) dan (2).

    v subscript 1 over v subscript 2 equals fraction numerator italic k open square brackets C H subscript 3 C O C H subscript 3 close square brackets subscript 1 superscript italic x open square brackets Br subscript 2 close square brackets subscript 1 superscript italic y open square brackets H to the power of plus sign close square brackets subscript 1 superscript italic z over denominator italic k open square brackets C H subscript 3 C O C H subscript 3 close square brackets subscript 2 superscript italic x open square brackets Br subscript 2 close square brackets subscript 2 superscript italic y open square brackets H to the power of plus sign close square brackets subscript 2 superscript italic z end fraction fraction numerator 5 comma 7 cross times 10 to the power of negative sign 5 end exponent over denominator 5 comma 7 cross times 10 to the power of negative sign 5 end exponent end fraction equals fraction numerator up diagonal strike italic k open parentheses 0 comma 30 close parentheses to the power of italic x end strike open parentheses 0 comma 05 close parentheses to the power of italic y up diagonal strike open parentheses 0 comma 05 close parentheses to the power of italic z end strike over denominator up diagonal strike italic k left parenthesis 0 comma 30 right parenthesis to the power of italic x end strike open parentheses 0 comma 10 close parentheses to the power of italic y up diagonal strike left parenthesis 0 comma 05 right parenthesis to the power of italic z end strike end fraction 1 equals open parentheses 1 half close parentheses to the power of italic y italic y equals 0 
     
  • Menghitung orde reaksi H to the power of bold plus sign 

    Untuk menghitung orde reaksi H to the power of plus sign, pilih 2 percobaan dimana C H subscript 3 C O C H subscript 3 dan Br subscript 2 mempunyai konsentrasi yang sama, yaitu percobaan (4) dan (5).

    v subscript 4 over v subscript 5 equals fraction numerator italic k open square brackets C H subscript 3 C O C H subscript 3 close square brackets subscript 4 superscript italic x open square brackets Br subscript 2 close square brackets subscript 4 superscript italic y open square brackets H to the power of plus sign close square brackets subscript 4 superscript italic z over denominator italic k open square brackets C H subscript 3 C O C H subscript 3 close square brackets subscript 5 superscript italic x open square brackets Br subscript 2 close square brackets subscript 5 superscript italic y open square brackets H to the power of plus sign close square brackets subscript 5 superscript italic z end fraction fraction numerator 3 comma 1 cross times 10 to the power of negative sign 4 end exponent over denominator 7 comma 6 cross times 10 to the power of negative sign 5 end exponent end fraction equals fraction numerator up diagonal strike italic k open parentheses 0 comma 40 close parentheses to the power of italic x end strike up diagonal strike open parentheses 0 comma 05 close parentheses to the power of italic y end strike open parentheses 0 comma 20 close parentheses to the power of italic z over denominator up diagonal strike italic k left parenthesis 0 comma 40 right parenthesis to the power of italic x end strike up diagonal strike left parenthesis 0 comma 05 right parenthesis to the power of italic y end strike open parentheses 0 comma 05 close parentheses to the power of italic z end fraction 4 over 1 equals open parentheses 4 over 1 close parentheses to the power of italic z italic z equals 1 
     
  • Persamaan laju reaksi

    italic v equals italic k open square brackets C H subscript 3 C O C H subscript 3 close square brackets to the power of italic x open square brackets Br subscript 2 close square brackets to the power of italic y open square brackets H to the power of plus sign close square brackets to the power of italic z italic v equals italic k open square brackets C H subscript 3 C O C H subscript 3 close square brackets to the power of 1 open square brackets Br subscript 2 close square brackets to the power of 0 open square brackets H to the power of plus sign close square brackets to the power of 1 italic v equals italic k open square brackets C H subscript 3 C O C H subscript 3 close square brackets open square brackets H to the power of plus sign close square brackets   
     

b.   Nilai tetapan laju reaksi (k)

      Misal kita ambil percobaan nomor (1)

      italic v equals italic k open square brackets C H subscript 3 C O C H subscript 3 close square brackets open square brackets H to the power of plus sign close square brackets italic k equals fraction numerator italic v over denominator open square brackets C H subscript 3 C O C H subscript 3 close square brackets open square brackets H to the power of plus sign close square brackets end fraction italic k equals fraction numerator 5 comma 7 cross times 10 to the power of negative sign 5 end exponent space M space detik to the power of negative sign 1 end exponent over denominator open parentheses 0 comma 30 space M close parentheses open parentheses 0 comma 05 space M close parentheses end fraction italic k equals 0 comma 0038 space M to the power of negative sign 1 end exponent space detik to the power of negative sign 1 end exponent 


Jadi, rumus laju reaksi untuk reaksi C H subscript bold 3 C O C H subscript bold 3 bold and Br subscript bold 2 bold yields with H to the power of bold plus sign on top C H subscript bold 3 C O C H subscript bold 2 Br bold and H to the power of bold plus sign bold and Br to the power of bold minus sign adalah italic v bold equals italic k bold open square brackets C H subscript bold 3 C O C H subscript bold 3 bold close square brackets bold open square brackets H to the power of bold plus sign bold close square brackets dengan nilai italic k bold equals bold 0 bold comma bold 0038 bold space italic M 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

0

Roboguru

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

Pembahasan Soal:

Persamaan laju reaksi awal adalah r=k[NO]x[H2]y

  • Menentukan orde reaksi terhadap NO
    Dengan menggunakan data dari percobaan 3 dan 4.
    r3r4=k[NO]3x[H2]3yk[NO]4x[H2]4y0,52=k(0,1)x(0,25)yk(0,2)x(0,25)y4=(2)xx=2

    Berdasarkan perhitungan di atas, orde reaksi terhadap NO adalah 2.
     
  • Menentukan orde reaksi terhadap H subscript 2
    Dengan menggunakan data dari percobaan 1 dan 2.
    r1r2=k[NO]1x[H2]1yk[NO]2x[H2]2y1,64,8=k(0,3)x(0,05)yk(0,3)x(0,15)y3=(3)yy=1

    Berdasarkan perhitungan di atas, orde reaksi terhadap NO adalah 1.
     
  • Menentukan nilai k
    Persamaan laju reaksinya adalah r=k[NO]2[H2].
    Dengan menggunakan data percobaan 4, maka
    r2molL1s12molL1s12molL1s1kk======k[NO]2[H2]k(0,2molL1)2(0,25molL1)k(0,04mol2L2)(0,25molL1)k(0,01mol3L3)0,01mol3L32molL1s1200mol2L2s1

Dengan demikian, maka tetapan laju reaksi tersebut adalah 200mol2L2s1.

Jadi, jawaban yang benar adalah E.space

0

Roboguru

Pada reaksi:  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

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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