Roboguru

Data percobaan laju reaksi untuk:   Berdasarkan data di atas, maka rumus laju reaksi yang benar adalah ….

Pertanyaan

Data percobaan laju reaksi untuk:


2 N O subscript open parentheses g close parentheses end subscript and 2 H subscript 2 open parentheses g close parentheses end subscript space space N subscript 2 subscript open parentheses g close parentheses end subscript plus 2 H subscript 2 O subscript open parentheses g close parentheses end subscript 



Berdasarkan data di atas, maka rumus laju reaksi yang benar adalah ….space 

  1. v double bond k open parentheses N O close parentheses open parentheses H subscript 2 close parentheses 

  2. v double bond k open parentheses N O close parentheses open parentheses H subscript 2 close parentheses squared 

  3. v double bond k open parentheses N O close parentheses squared open parentheses H subscript 2 close parentheses 

  4. v double bond k open parentheses N O close parentheses squared 

  5. v double bond k open parentheses H subscript 2 close parentheses squared 

Pembahasan Soal:

Pada soal tersebut persamaan reaksi yang benar adalah 2 N O open parentheses g close parentheses and 2 H subscript 2 open parentheses g close parentheses yields N subscript 2 open parentheses g close parentheses and 2 H subscript 2 O open parentheses g close parentheses.

Untuk menentukan persamaan laju reaksi pada reaksi tersebut maka perlu dintentukan orde reaksi dari pereaksinya yaitu NO dan H subscript 2 

Untuk menentukan orde reaksi H subscript 2 maka dapat digunakan data percobaan 2 dan 1


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 space open square brackets N O close square brackets to the power of m space open square brackets H subscript 2 close square brackets to the power of n over denominator k space open square brackets N O close square brackets to the power of m space open square brackets H subscript 2 close square brackets to the power of n end fraction end cell row cell fraction numerator 64 cross times 10 to the power of negative sign 7 end exponent over denominator 32 cross times 10 to the power of negative sign 7 end exponent end fraction end cell equals cell fraction numerator k space left square bracket 4 cross times 10 to the power of negative sign 3 end exponent right square bracket to the power of m space left square bracket 3 comma 0 cross times 10 to the power of negative sign 3 end exponent right square bracket to the power of n over denominator k space left square bracket 4 cross times 10 to the power of negative sign 3 end exponent right square bracket to the power of m space left square bracket 1 comma 5 cross times 10 to the power of negative sign 3 end exponent right square bracket to the power of n end fraction end cell row 2 equals cell 2 to the power of n end cell row n equals 1 end table


Orde reaksi untuk H subscript 2 adalah 1 untuk menentukan orde reaksi dari NO maka dapat digunakan data 4 dan 5


table attributes columnalign right center left columnspacing 0px end attributes row cell V subscript 4 over V subscript 5 end cell equals cell fraction numerator k space open square brackets N O close square brackets to the power of m space open square brackets H subscript 2 close square brackets to the power of n over denominator k space open square brackets N O close square brackets to the power of m space open square brackets H subscript 2 close square brackets to the power of n end fraction end cell row cell fraction numerator 32 cross times 10 to the power of negative sign 7 end exponent over denominator 7 comma 9 cross times 10 to the power of negative sign 7 end exponent end fraction end cell equals cell fraction numerator k space left square bracket 2 cross times 10 to the power of negative sign 3 end exponent right square bracket to the power of m space left square bracket 6 comma 0 cross times 10 to the power of negative sign 3 end exponent right square bracket to the power of n over denominator k space left square bracket 1 cross times 10 to the power of negative sign 3 end exponent right square bracket to the power of m space left square bracket 6 comma 0 cross times 10 to the power of negative sign 3 end exponent right square bracket to the power of n end fraction end cell row 4 equals cell 2 to the power of m end cell row m equals 2 end table

 

Oleh karena itu, persamaan laju yang benar adalah v double bond k open parentheses N O close parentheses squared open parentheses H subscript 2 close parentheses

Jadi, jawaban yang benar adalah C.space 

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 03 Juni 2021

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