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Pertanyaan

Diketahui persamaan reaksi antara K, L dan M sebagai berikut : K + L + M → N Kemudian, didapatkan data percobaan sebagai berikut : Percobaan [ K ] (M) [ L ] (M) [ M ] (M) v ( M/det) 1 2 3 4 0,1 0,2 0,2 0,3 0,2 0,2 0,4 0,8 0,3 0,3 0,3 0,9 0,001 0,001 0,002 0,036 Berdasarkan data percobaan, persamaan laju reaksi tersebut adalah…

Diketahui persamaan reaksi antara K, L dan M sebagai berikut :

K + L + M N

Kemudian, didapatkan data percobaan sebagai berikut :

 

Percobaan

[ K ]

(M)

[ L ]

(M)

[ M ]

(M)

v

( M/det)

1

2

3

4

0,1

0,2

0,2

0,3

0,2

0,2

0,4

0,8

0,3

0,3

0,3

0,9

0,001

0,001

0,002

0,036

 

Berdasarkan data percobaan, persamaan laju reaksi tersebut adalah…

  1. v = k [K] [L] [M]2

  2. v = k [K] [L]2 [M]2

  3. v = k [K]2 [L] [M]2

  4. v = k [K] [M]2

  5. v = k [L] [M]2

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I. Solichah

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah E.

jawaban yang tepat adalah E.

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Pembahasan

Menghitung orde [K] (pilih data [L] dan [M] yang sama yaitu 1 dan 2) : Menghitung orde [L] (pilih data [ K ] dan [M] yang sama yaitu 2 dan 3) : Menghitung orde [M] (pilih data 3 dan 4) : Persamaan laju reaksinya : v = k [ K ] x [ L ] y [ M ] z v = k [ K ] 0 [ L ] 1 [ M ] 2 v = k [ L ] [ M ] 2 Jadi, jawaban yang tepat adalah E.

Menghitung orde [K] (pilih data [L] dan [M] yang sama yaitu 1 dan 2) :

v subscript 1 over v subscript 2 equals left parenthesis fraction numerator k left square bracket K right square bracket subscript 1 to the power of x over denominator k left square bracket K right square bracket subscript 2 to the power of x end fraction right parenthesis fraction numerator left square bracket L right square bracket subscript 1 to the power of y over denominator left square bracket L right square bracket subscript 2 to the power of y end fraction fraction numerator left square bracket M right square bracket subscript 1 to the power of z over denominator left square bracket M right square bracket subscript 2 to the power of z end fraction fraction numerator 0 comma 001 over denominator 0 comma 001 end fraction equals fraction numerator k left square bracket 0 comma 1 right square bracket to the power of x over denominator k left square bracket 0 comma 2 right square bracket to the power of x end fraction fraction numerator left square bracket 0 comma 2 right square bracket to the power of y over denominator left square bracket 0 comma 2 right square bracket to the power of y end fraction fraction numerator left square bracket 0 comma 3 right square bracket to the power of z over denominator left square bracket 0 comma 3 right square bracket to the power of z end fraction 1 over 1 equals 1 to the power of x over 2 to the power of x x space space equals space space 0 

Menghitung orde [L] (pilih data [K] dan [M] yang sama yaitu 2 dan 3) :

v subscript 2 over v subscript 3 equals fraction numerator k left square bracket K right square bracket subscript 2 to the power of x over denominator k left square bracket K right square bracket subscript 3 to the power of x end fraction fraction numerator left square bracket L right square bracket subscript 2 to the power of y over denominator left square bracket L right square bracket subscript 3 to the power of y end fraction fraction numerator left square bracket M right square bracket subscript 2 to the power of z over denominator left square bracket M right square bracket subscript 3 to the power of z end fraction fraction numerator 0 comma 001 over denominator 0 comma 002 end fraction equals fraction numerator horizontal strike k left square bracket 0 comma 2 right square bracket to the power of x end strike over denominator horizontal strike k left square bracket 0 comma 2 right square bracket to the power of x end strike end fraction fraction numerator left square bracket 0 comma 2 right square bracket to the power of y over denominator left square bracket 0 comma 4 right square bracket to the power of y end fraction fraction numerator horizontal strike left square bracket 0 comma 3 right square bracket to the power of z end strike over denominator horizontal strike left square bracket 0 comma 3 right square bracket to the power of z end strike end fraction 1 half equals 1 to the power of y over 2 to the power of y y space space equals space space 1 

 Menghitung orde [M] (pilih data 3 dan 4) :

v subscript 3 over v subscript 4 equals fraction numerator k left square bracket K right square bracket subscript 3 to the power of x over denominator k left square bracket K right square bracket subscript 4 to the power of x end fraction fraction numerator left square bracket L right square bracket subscript 3 to the power of y over denominator left square bracket L right square bracket subscript 4 to the power of y end fraction fraction numerator left square bracket M right square bracket subscript 3 to the power of z over denominator left square bracket M right square bracket subscript 4 to the power of z end fraction fraction numerator 0 comma 002 over denominator 0 comma 036 end fraction equals fraction numerator k left square bracket 0 comma 2 right square bracket to the power of 0 over denominator k left square bracket 0 comma 3 right square bracket to the power of 0 end fraction fraction numerator left square bracket 0 comma 4 right square bracket to the power of 1 over denominator left square bracket 0 comma 8 right square bracket to the power of 1 end fraction fraction numerator left square bracket 0 comma 3 right square bracket to the power of z over denominator left square bracket 0 comma 9 right square bracket to the power of z end fraction 1 over 18 equals 1 half 1 to the power of z over 3 to the power of z 2 over 18 equals 1 to the power of z over 3 to the power of z 1 over 9 equals 1 to the power of z over 3 to the power of z z space space equals space space 2 

Persamaan laju reaksinya :

v = k [K]x [L]y [M]z

v = k [K]0 [L]1 [M]2

v = k [L] [M]2

Jadi, jawaban yang tepat adalah E.

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