Seorang siswa sedang melakukan percobaan pengaruh perubahan konsentrasi terhadap laju reaksi dengan data hasil percobaan sebagai berikut.       Berdasarkan hasil percobaan di atas, berikut ini pernyataan yang benar adalah ...  Orde reaksi terhadap masing-masing reaktan secara berurutan adalah 1, 0, dan 1. Persamaan laju reaksinya adalah v=k [BrO3−​] [H+]  Jika konsentrasi tiap reaktan adalah 0,5 M, laju reaksinya menjadi 9×10−4 M/s. Tetapan laju reaksinya memiliki satuan Ms.

Pertanyaan

Seorang siswa sedang melakukan percobaan pengaruh perubahan konsentrasi terhadap laju reaksi dengan data hasil percobaan sebagai berikut.
 

begin mathsize 14px style BrO subscript 3 to the power of minus open parentheses a q close parentheses plus 5 Br to the power of minus open parentheses a q close parentheses plus 6 straight H to the power of plus open parentheses a q close parentheses rightwards arrow 3 Br subscript 2 open parentheses l close parentheses plus 3 straight H subscript 2 straight O left parenthesis l right parenthesis end style 
 


Berdasarkan hasil percobaan di atas, berikut ini pernyataan yang benar adalah ...space 

  1. Orde reaksi terhadap masing-masing reaktan secara berurutan adalah 1, 0, dan 1.
  2. Persamaan laju reaksinya adalah begin mathsize 14px style straight v equals straight k blank left square bracket BrO subscript 3 to the power of minus right square bracket blank left square bracket straight H to the power of plus right square bracket end style 
  3. Jika konsentrasi tiap reaktan adalah 0,5 M, laju reaksinya menjadi begin mathsize 14px style 9 cross times 10 to the power of negative 4 end exponent end style M/s.
  4. Tetapan laju reaksinya memiliki satuan Ms.
  1. 1, 2, dan 3 SAJA yang benar.space 

  2. 1 dan 3 SAJA yang benar.space 

  3. 2 dan 4 SAJA yang benar.space 

  4. HANYA 4 yang benar.space 

  5. SEMUA pilihan benar.space 

M. Rizki

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah A.undefined 

Pembahasan

Persamaan laju reaksi:
 

begin mathsize 14px style straight v equals straight k blank left square bracket BrO subscript 3 to the power of minus right square bracket to the power of straight x blank blank left square bracket Br to the power of minus right square bracket to the power of straight y blank blank left square bracket straight H to the power of plus right square bracket to the power of straight z end style 


Menentukan orde begin mathsize 14px style begin bold style left square bracket BrO subscript 3 to the power of minus right square bracket end style end style

Orde reaksi terhadap begin mathsize 14px style open square brackets BrO subscript 3 to the power of minus close square brackets end style dapat ditentukan dari konsentrasi begin mathsize 14px style blank left square bracket Br to the power of minus right square bracket end style dan begin mathsize 14px style left square bracket straight H to the power of plus right square bracket end style yang sama, yaitu pada percobaan 1 dan 4.
 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell straight v subscript 1 over straight v subscript 4 end cell equals cell fraction numerator straight k blank left square bracket BrO subscript 3 to the power of minus right square bracket subscript 1 to the power of straight x blank left square bracket Br to the power of minus right square bracket subscript 1 to the power of straight y blank left square bracket straight H to the power of plus right square bracket subscript 1 to the power of straight z over denominator straight k blank left square bracket BrO subscript 3 to the power of minus right square bracket subscript 4 to the power of straight x blank left square bracket Br to the power of minus right square bracket subscript 4 to the power of straight y blank left square bracket straight H to the power of plus right square bracket subscript 4 to the power of straight z end fraction end cell row cell fraction numerator 5 , 4 cross times 10 to the power of negative 5 end exponent over denominator 7 , 2 cross times 10 to the power of negative 5 end exponent end fraction end cell equals cell fraction numerator straight k blank open square brackets 0 , 30 close square brackets to the power of straight x blank open square brackets 0 , 05 close square brackets to the power of straight y blank open square brackets 0 , 05 close square brackets to the power of straight z over denominator straight k blank open square brackets 0 , 40 close square brackets to the power of straight x blank open square brackets 0 , 05 close square brackets to the power of straight y blank open square brackets 0 , 05 close square brackets to the power of straight z end fraction end cell row cell fraction numerator 5 , 4 over denominator 7 , 2 end fraction end cell equals cell open square brackets 0 , 30 close square brackets to the power of straight x over open square brackets 0 , 40 close square brackets to the power of straight x end cell row cell 3 over 4 end cell equals cell open parentheses 3 over 4 close parentheses to the power of straight x end cell row straight x equals 1 end table end style 

 

Menentukan orde begin mathsize 14px style begin bold style left square bracket Br to the power of minus right square bracket end style end style.

Orde reaksi terhadap begin mathsize 14px style left square bracket Br to the power of minus right square bracket end style dapat ditentukan dari konsentrasi begin mathsize 14px style open square brackets BrO subscript 3 to the power of minus close square brackets space d a n space left square bracket straight H to the power of plus right square bracket end style yang sama, yaitu pada percobaan 1 dan 2.


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell straight v subscript 1 over straight v subscript 2 end cell equals cell fraction numerator straight k blank left square bracket BrO subscript 3 to the power of minus right square bracket subscript 1 to the power of straight x blank left square bracket Br to the power of minus right square bracket subscript 1 to the power of straight y blank left square bracket straight H to the power of plus right square bracket subscript 1 to the power of straight z over denominator straight k blank left square bracket BrO subscript 3 to the power of minus right square bracket subscript 2 to the power of straight x blank left square bracket Br to the power of minus right square bracket subscript 2 to the power of straight y blank left square bracket straight H to the power of plus right square bracket subscript 2 to the power of straight z end fraction end cell row cell fraction numerator 5 , 4 cross times 10 to the power of negative 5 end exponent over denominator 5 , 4 cross times 10 to the power of negative 5 end exponent end fraction end cell equals cell fraction numerator straight k blank open square brackets 0 , 30 close square brackets to the power of straight x blank open square brackets 0 , 05 close square brackets to the power of straight y blank open square brackets 0 , 05 close square brackets to the power of straight z over denominator straight k blank open square brackets 0 , 30 close square brackets to the power of straight x blank open square brackets 0 , 10 close square brackets to the power of straight y blank open square brackets 0 , 05 close square brackets to the power of straight z end fraction end cell row 1 equals cell open square brackets 0 , 05 close square brackets to the power of straight y over open square brackets 0 , 10 close square brackets to the power of straight y end cell row 1 equals cell open parentheses 1 half close parentheses to the power of straight y end cell row straight y equals 0 end table end style 

 

Menentukan orde begin mathsize 14px style begin bold style left square bracket H to the power of plus right square bracket end style end style.

Orde reaksi terhadap begin mathsize 14px style open square brackets straight H to the power of plus close square brackets end style dapat ditentukan dari konsentrasi begin mathsize 14px style open square brackets BrO subscript 3 to the power of minus close square brackets space dan space left square bracket Br to the power of minus right square bracket end style yang sama, yaitu pada percobaan 1 dan 3.
 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell straight v subscript 1 over straight v subscript 3 end cell equals cell fraction numerator straight k blank left square bracket BrO subscript 3 to the power of minus right square bracket subscript 1 to the power of straight x blank left square bracket Br to the power of minus right square bracket subscript 1 to the power of straight y blank left square bracket straight H to the power of plus right square bracket subscript 1 to the power of straight z over denominator straight k blank left square bracket BrO subscript 3 to the power of minus right square bracket subscript 3 to the power of straight x blank left square bracket Br to the power of minus right square bracket subscript 3 to the power of straight y blank left square bracket straight H to the power of plus right square bracket subscript 3 to the power of straight z end fraction end cell row cell fraction numerator 5 , 4 cross times 10 to the power of negative 5 end exponent over denominator 2 , 7 cross times 10 to the power of negative 4 end exponent end fraction end cell equals cell fraction numerator straight k blank open square brackets 0 , 30 close square brackets to the power of straight x blank open square brackets 0 , 05 close square brackets to the power of straight y blank open square brackets 0 , 05 close square brackets to the power of straight z over denominator straight k blank open square brackets 0 , 30 close square brackets to the power of straight x blank open square brackets 0 , 05 close square brackets to the power of straight y blank open square brackets 0 , 15 close square brackets to the power of straight z end fraction end cell row cell 2 cross times 10 to the power of negative 1 end exponent end cell equals cell open square brackets 0 , 05 close square brackets to the power of straight z over open square brackets 0 , 25 close square brackets to the power of straight z end cell row cell 2 over 10 end cell equals cell open parentheses 5 over 25 close parentheses to the power of straight z end cell row cell 2 over 10 end cell equals cell open parentheses 5 over 25 close parentheses to the power of straight z end cell row straight z equals 1 row blank blank blank end table end style 


Orde x, y, dan z secara berurutan adalah 1, 0, dan 1. (Pernyataan 1 benar)

 

Menentukan persamaan laju.
 

undefined 


(Pernyataan 2 benar)

 

Menentukan tetapan laju reaksi.

Tetapan laju reaksi dapat dihitung menggunakan data percobaan mana pun. Kita gunakan data percobaan 1.


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell straight v subscript 1 end cell equals cell straight k blank left square bracket BrO subscript 3 to the power of minus right square bracket subscript 1 blank left square bracket straight H to the power of plus right square bracket subscript 1 end cell row cell 5 , 4 cross times 10 to the power of negative 5 end exponent blank straight M divided by straight s end cell equals cell straight k blank open parentheses 0 , 30 blank straight M close parentheses left parenthesis 0 , 05 blank straight M right parenthesis end cell row cell 5 , 4 cross times 10 to the power of negative 5 end exponent blank straight M divided by straight s end cell equals cell straight k blank left parenthesis 1 , 5 cross times 10 to the power of negative 2 end exponent blank straight M squared right parenthesis end cell row straight k equals cell fraction numerator 5 , 4 cross times 10 to the power of negative 5 end exponent blank straight M divided by straight s over denominator 1 , 5 cross times 10 to the power of negative 2 end exponent blank straight M squared end fraction end cell row straight k equals cell 3 , 6 cross times 10 to the power of negative 3 end exponent blank straight M to the power of negative 1 end exponent straight s to the power of negative 1 end exponent end cell row blank blank blank row blank blank blank end table end style 


Tetapan laju reaksi memiliki satuan undefined. (Pernyataan 4 salah)


Jika konsentrasi tiap reaktan adalah 0,5 M, laju reaksinya dapat dihitung sebagai berikut.


begin mathsize 14px style straight v equals straight k blank left square bracket BrO subscript 3 to the power of minus right square bracket blank left square bracket straight H to the power of plus right square bracket straight v equals 3 , 6 cross times 10 to the power of negative 3 end exponent blank straight M to the power of negative 1 end exponent straight s to the power of negative 1 end exponent cross times 0 , 5 blank straight M cross times 0 , 5 blank straight M straight v equals 9 cross times 10 to the power of negative 4 end exponent blank straight M divided by straight s end style 


(Pernyataan 3 benar)


Jadi, jawaban yang benar adalah A.undefined 

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