Roboguru

Suatu reaksi : diketahui data sebagai berikut : Tentukanlah : a. Orde reaksi b. Ungkapan laju reaksinya c. Harga tetapan lajunya

Pertanyaan

Suatu reaksi :


2 space N O open parentheses g close parentheses space plus space 2 space H subscript 2 open parentheses g close parentheses space rightwards arrow space 2 space N subscript 2 open parentheses g close parentheses space plus space 2 space H subscript 2 O open parentheses l close parentheses


diketahui data sebagai berikut :



Tentukanlah :

a. Orde reaksi

b. Ungkapan laju reaksinya

c. Harga tetapan lajunya space 

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

TIPS! Untuk menentukan orde reaksi dari NO maka gunakan data percobaan yang konsentrasi H subscript 2 nya sama, dan begitupun sebaliknya.

  • Orde reaksi dari NO, gunakan data percobaan 3 dan 4.


table attributes columnalign right center left columnspacing 0px end attributes row v equals cell 1 over t end cell row blank blank blank row cell v subscript 3 over v subscript 4 end cell equals cell fraction numerator k open square brackets N O close square brackets subscript 3 to the power of x middle dot open square brackets H subscript 2 close square brackets subscript 3 to the power of y over denominator k open square brackets N O close square brackets subscript 4 to the power of x middle dot open square brackets H subscript 2 close square brackets subscript 4 to the power of y end fraction end cell row cell fraction numerator begin display style 1 over t subscript 3 end style over denominator begin display style 1 over t subscript 4 end style end fraction end cell equals cell fraction numerator k open square brackets N O close square brackets subscript 3 to the power of x middle dot open square brackets H subscript 2 close square brackets subscript 3 to the power of y over denominator k open square brackets N O close square brackets subscript 4 to the power of x middle dot open square brackets H subscript 2 close square brackets subscript 4 to the power of y end fraction end cell row cell t subscript 4 over t subscript 3 end cell equals cell fraction numerator k left parenthesis 0 comma 2 right parenthesis to the power of x middle dot left parenthesis 0 comma 1 right parenthesis to the power of y over denominator k left parenthesis 0 comma 3 right parenthesis to the power of x middle dot left parenthesis 0 comma 1 right parenthesis to the power of y end exponent end fraction end cell row cell 8 over 12 end cell equals cell left parenthesis 2 over 3 right parenthesis to the power of x fraction numerator 2 over denominator 3 end fraction equals left parenthesis 2 over 3 right parenthesis to the power of x end cell row x equals 1 end table

 

  • Orde reaksi H subscript 2, gunakan data percobaan 1 dan 2.


table attributes columnalign right center left columnspacing 0px end attributes row v equals cell 1 over t end cell row blank blank blank row cell v subscript 1 over v subscript 2 end cell equals cell fraction numerator k open square brackets N O close square brackets subscript 1 to the power of x middle dot open square brackets H subscript 2 close square brackets subscript 1 to the power of y over denominator k open square brackets N O close square brackets subscript 2 to the power of x middle dot open square brackets H subscript 2 close square brackets subscript 2 to the power of y end fraction end cell row cell fraction numerator begin display style 1 over t subscript 1 end style over denominator begin display style 1 over t subscript 2 end style end fraction end cell equals cell fraction numerator k open square brackets N O close square brackets subscript 1 to the power of x middle dot open square brackets H subscript 2 close square brackets subscript 1 to the power of y over denominator k open square brackets N O close square brackets subscript 2 to the power of x middle dot open square brackets H subscript 2 close square brackets subscript 2 to the power of y end fraction end cell row cell t subscript 2 over t subscript 1 end cell equals cell fraction numerator k left parenthesis 0 comma 1 right parenthesis to the power of x middle dot left parenthesis 0 comma 1 right parenthesis to the power of y over denominator k left parenthesis 0 comma 1 right parenthesis to the power of x middle dot left parenthesis 0 comma 2 right parenthesis to the power of y end fraction end cell row cell 6 over 24 end cell equals cell left parenthesis 1 half right parenthesis to the power of y end cell row cell 1 fourth end cell equals cell left parenthesis 1 half right parenthesis to the power of y end cell row cell left parenthesis 1 half right parenthesis squared end cell equals cell left parenthesis fraction numerator 1 over denominator 2 end fraction right parenthesis blank to the power of y end cell row y equals 2 end table


Dengan demikian, orde reaksi NO adalah 1 dan orde reaksi H subscript 2 adalah 2. Sementara orde reaksi totalnya adalah 1 + 2 = 3.

b. Ungkapan laju reaksi

Ungkapan laju reaksi dapat ditentukan dengan cara memasukkan nilai orde reaksi (x dan y) ke dalam persamaan laju reaksi, maka :


table attributes columnalign right center left columnspacing 0px end attributes row v equals cell k open square brackets N O close square brackets to the power of x middle dot open square brackets H subscript 2 close square brackets to the power of y end cell row v equals cell k space open square brackets N O close square brackets space open square brackets H subscript 2 close square brackets squared end cell end table


Jadi, ungkapan laju reaksinya adalah Error converting from MathML to accessible text..

c.  Harga tetapan laju reaksi

Harga tetapan laju reaksi dapat ditentukan dengan cara mensubstitusikan rumus persamaan laju reaksi yang telah diketahui ke dalam salah satu data percobaan. Misal, disubstitusi ke dalam percobaan 1 , maka :


table attributes columnalign right center left columnspacing 0px end attributes row v equals cell k open square brackets N O close square brackets space open square brackets H subscript 2 close square brackets squared end cell row cell 1 over t end cell equals cell k open square brackets N O close square brackets space open square brackets H subscript 2 close square brackets squared end cell row cell 1 over 24 end cell equals cell k space left parenthesis 0 comma 1 right parenthesis left parenthesis 0 comma 1 right parenthesis squared end cell row cell 1 over 24 end cell equals cell k middle dot 10 to the power of negative sign 3 end exponent end cell row k equals cell fraction numerator 1 over denominator 24 cross times 10 to the power of negative sign 3 end exponent end fraction end cell row k equals cell 41 comma 67 end cell end table


Jadi, harga tetapan lajunya adalah 41,67.space​​​​​​​ 

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 10 Juni 2021

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