Roboguru

Eksperimen untuk reaksi: terdapat dalam tabel berikut. Persamaan laju reaksinya adalah ....

Pertanyaan

Eksperimen untuk reaksi: begin mathsize 14px style N subscript 2 open parentheses italic g close parentheses and 3 H subscript 2 open parentheses italic g close parentheses yields 2 N H subscript 3 open parentheses italic g close parentheses end styleterdapat dalam tabel berikut.



Persamaan laju reaksinya adalah ....space 

  1. begin mathsize 14px style v double bond k open square brackets N subscript 2 close square brackets open square brackets H subscript 2 close square brackets end style undefined 

  2. begin mathsize 14px style v double bond k open square brackets N subscript 2 close square brackets open square brackets H subscript 2 close square brackets squared end style undefined 

  3. begin mathsize 14px style v double bond k open square brackets N subscript 2 close square brackets squared open square brackets H subscript 2 close square brackets cubed end style undefined 

  4. begin mathsize 14px style v double bond k open square brackets N subscript 2 close square brackets end style undefined 

  5. begin mathsize 14px style v double bond k open square brackets H subscript 2 close square brackets squared end style undefined 

Pembahasan Soal:

Orde reaksi atau tingkat reaksi adalah pangkat konsentrasi zat pereaksi dalam persamaan laju reaksi. Orde reaksi hanya dapat ditentukan melalui data percobaan, bukan dari persamaan reaksinya. Caranya dengan melakukan percobaan berulang-ulang terhadap zat yang akan ditentukan orde reaksinya.

Persamaan laju reaksi untuk reaksi di atas adalah begin mathsize 14px style v space equals space 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 end style. Orde reaksi terhadap begin mathsize 14px style N subscript 2 end style dihitung dari percobaan 1) dan 2).


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell v subscript 1 over v subscript 2 space end cell equals cell space fraction numerator up diagonal strike k over denominator up diagonal strike k end fraction open parentheses open square brackets N subscript 2 close square brackets subscript 1 over open square brackets N subscript 2 close square brackets subscript 2 close parentheses to the power of x open parentheses open square brackets H subscript 2 close square brackets subscript 1 over open square brackets H subscript 2 close square brackets subscript 2 close parentheses to the power of y end cell row cell fraction numerator 4 cross times 10 to the power of negative sign 4 end exponent over denominator 8 cross times 10 to the power of negative sign 4 end exponent end fraction space end cell equals cell space open parentheses fraction numerator 0 comma 002 over denominator 0 comma 004 end fraction close parentheses to the power of x open parentheses fraction numerator up diagonal strike 0 comma 002 end strike over denominator up diagonal strike 0 comma 002 end strike end fraction close parentheses to the power of y end cell row cell 1 half space end cell equals cell space open parentheses 1 half close parentheses to the power of x end cell row cell x space end cell equals cell space 1 end cell end table end style


Orde reaksi terhadap begin mathsize 14px style H subscript 2 end style dihitung dari percobaan 2) dan 3).


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell v subscript 2 over v subscript 3 space end cell equals cell space fraction numerator up diagonal strike k over denominator up diagonal strike k end fraction open parentheses open square brackets N subscript 2 close square brackets subscript 2 over open square brackets N subscript 2 close square brackets subscript 3 close parentheses to the power of x open parentheses open square brackets H subscript 2 close square brackets subscript 2 over open square brackets H subscript 2 close square brackets subscript 3 close parentheses to the power of y end cell row cell fraction numerator 8 cross times 10 to the power of negative sign 4 end exponent over denominator 32 cross times 10 to the power of negative sign 4 end exponent end fraction space end cell equals cell space open parentheses fraction numerator up diagonal strike 0 comma 004 end strike over denominator up diagonal strike 0 comma 004 end strike end fraction close parentheses to the power of x open parentheses fraction numerator 0 comma 002 over denominator 0 comma 008 end fraction close parentheses to the power of y end cell row cell 1 fourth space end cell equals cell space open parentheses 1 fourth close parentheses to the power of y end cell row cell y space end cell equals cell space 1 end cell end table end style


Persamaan laju reaksinya adalah begin mathsize 14px style v space equals space k open square brackets N subscript 2 close square brackets open square brackets H subscript 2 close square brackets end style.


Jadi, jawaban yang benar adalah A.space 

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 06 Maret 2021

Ruangguru

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