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)

( M/det)

0,1

0,2

0,3

0,2

0,4

0,8

0,3

0,9

0,001

0,002

0,036

Berdasarkan data percobaan, persamaan laju reaksi tersebut adalah…

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

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I. Solichah

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah E.

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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