Pertanyaan

Brominasi aseton memerlukan katalis asam dengan reaksi: CH 3 COCH 3 + Br 2 H + CH 3 COCH 2 Br + H + + Br − Data hasil percobaan reaksi di atas disajikan pada tabel berikut. Tentukan rumus laju reaksinya!

Brominasi aseton memerlukan katalis asam dengan reaksi:

$CH_{3} COCH_{3} + Br_{2} H^{+} CH_{3} COCH_{2} Br + H^{+} + Br^{-}$

Data hasil percobaan reaksi di atas disajikan pada tabel berikut.

Tentukan rumus laju reaksinya! $\;$

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Klaim

D. Hidayat

Master Teacher

Mahasiswa/Alumni Universitas Sebelas Maret

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Pembahasan

Orde reaksi terhadap Lihat data dan yang sama, yaitu pada percobaan 1 dan 5. Orde reaksi terhadap Lihat data dan yang sama, yaitu pada percobaan 1 dan 2. Orde reaksi terhadap Lihat data dan yang sama, yaitu pada percobaan 4 dan 5. Orde reaksi dibulatkan sehingga . Nilai K Gunakan data percobaan 2 k k k = = = [ CH 3 COCJH 3 ] [ H + ] v [ 0 , 3 M ] 2 [ 0 , 05 M ] 5 , 7 × 1 0 − 5 M / s 0 , 0038 M − 1 s − 1 Rumus laju reaksi

Orde reaksi terhadap
Lihat data  dan  yang sama, yaitu pada percobaan 1 dan 5.
$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell italic v subscript 1 over italic v subscript 5 end cell equals cell fraction numerator italic k open parentheses open square brackets C H subscript 3 C O C H subscript 3 close square brackets close parentheses subscript left parenthesis 1 right parenthesis end subscript superscript italic x space open parentheses open square brackets Br subscript 2 close square brackets close parentheses subscript left parenthesis 1 right parenthesis end subscript superscript italic y space open parentheses open square brackets H to the power of plus sign close square brackets close parentheses subscript left parenthesis 1 right parenthesis end subscript superscript italic z over denominator italic k open parentheses open square brackets C H subscript 3 C O C H subscript 3 close square brackets close parentheses subscript left parenthesis 5 right parenthesis end subscript superscript italic x space open parentheses open square brackets Br subscript 2 close square brackets close parentheses subscript left parenthesis 5 right parenthesis end subscript superscript italic y space open parentheses open square brackets H to the power of plus sign close square brackets close parentheses subscript left parenthesis 5 right parenthesis end subscript superscript italic z end fraction end cell row cell fraction numerator 5 comma 7 cross times 10 to the power of negative sign 5 end exponent over denominator 7 comma 6 cross times 10 to the power of negative sign 5 end exponent end fraction end cell equals cell open parentheses fraction numerator 0 comma 30 over denominator 0 comma 40 end fraction close parentheses to the power of italic x open parentheses fraction numerator 0 comma 05 over denominator 0 comma 05 end fraction close parentheses to the power of italic y open parentheses fraction numerator 0 comma 05 over denominator 0 comma 05 end fraction close parentheses to the power of italic z end cell row cell 3 over 4 end cell equals cell open parentheses 3 over 4 close parentheses to the power of italic x end cell row italic x equals 1 end table end style$
Orde reaksi terhadap
Lihat data  dan  yang sama, yaitu pada percobaan 1 dan 2.
$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell v subscript 1 over v subscript 2 end cell equals cell fraction numerator italic k open parentheses open square brackets C H subscript 3 C O C H subscript 3 close square brackets close parentheses subscript left parenthesis 1 right parenthesis end subscript superscript italic x space open parentheses open square brackets Br subscript 2 close square brackets close parentheses subscript left parenthesis 1 right parenthesis end subscript superscript italic y space open parentheses open square brackets H to the power of plus sign close square brackets close parentheses subscript left parenthesis 1 right parenthesis end subscript superscript italic z over denominator italic k open parentheses open square brackets C H subscript 3 C O C H subscript 3 close square brackets close parentheses subscript left parenthesis 2 right parenthesis end subscript superscript italic x space open parentheses open square brackets Br subscript 2 close square brackets close parentheses subscript left parenthesis 2 right parenthesis end subscript superscript italic y space open parentheses open square brackets H to the power of plus sign close square brackets close parentheses subscript left parenthesis 2 right parenthesis end subscript superscript italic z end fraction end cell row cell fraction numerator 5 comma 7 cross times 10 to the power of negative sign 5 end exponent over denominator 5 comma 7 cross times 10 to the power of negative sign 5 end exponent end fraction end cell equals cell open parentheses fraction numerator 0 comma 30 over denominator 0 comma 30 end fraction close parentheses to the power of italic x open parentheses fraction numerator 0 comma 05 over denominator 0 comma 10 end fraction close parentheses to the power of italic y open parentheses fraction numerator 0 comma 05 over denominator 0 comma 05 end fraction close parentheses to the power of italic z end cell row 1 equals cell open parentheses 1 half close parentheses to the power of italic y end cell row italic y equals 0 end table end style$
Orde reaksi terhadap
Lihat data dan yang sama, yaitu pada percobaan 4 dan 5.
$begin mathsize 14px style 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 italic k open parentheses open square brackets C H subscript 3 C O C H subscript 3 close square brackets close parentheses subscript left parenthesis 4 right parenthesis end subscript superscript italic x space open parentheses open square brackets Br subscript 2 close square brackets close parentheses subscript left parenthesis 4 right parenthesis end subscript superscript italic y space open parentheses open square brackets H to the power of plus sign close square brackets close parentheses subscript left parenthesis 4 right parenthesis end subscript superscript italic z over denominator italic k open parentheses open square brackets C H subscript 3 C O C H subscript 3 close square brackets close parentheses subscript left parenthesis 5 right parenthesis end subscript superscript italic x space open parentheses open square brackets Br subscript 2 close square brackets close parentheses subscript left parenthesis 5 right parenthesis end subscript superscript italic y space open parentheses open square brackets H to the power of plus sign close square brackets close parentheses subscript left parenthesis 5 right parenthesis end subscript superscript italic z end fraction end cell row cell fraction numerator 3 comma 1 cross times 10 to the power of negative sign 4 end exponent over denominator 7 comma 6 cross times 10 to the power of negative sign 5 end exponent end fraction end cell equals cell open parentheses fraction numerator 0 comma 40 over denominator 0 comma 40 end fraction close parentheses to the power of italic x open parentheses fraction numerator 0 comma 05 over denominator 0 comma 05 end fraction close parentheses to the power of italic y open parentheses fraction numerator 0 comma 20 over denominator 0 comma 05 end fraction close parentheses to the power of italic z end cell row cell 4 comma 0789 end cell equals cell open parentheses 4 close parentheses to the power of italic z end cell row italic z equals cell fraction numerator log space 4 comma 0789 over denominator log space 4 end fraction end cell row italic z equals cell 1 comma 0141 end cell end table end style$
Orde reaksi dibulatkan sehingga .
Nilai K
Gunakan data percobaan 2
$k k k = = = \frac{v}{[ CH _{3} COCJH _{3} ] [ H ^{+} ]} \frac{5 , 7 \times 1 0 ^{- 5} M / s}{[ 0 , 3 M ] ^{2} [ 0 , 05 M ]} 0, 0038 M^{- 1} s^{- 1}$
Rumus laju reaksi
$\begin{array}{rcl}v&=&k\;{\lbrack{\mathrm{CH}}_3{\mathrm{COCH}}_3\rbrack}^{\mathrm x}\;{\lbrack{\mathrm{Br}}_2\rbrack}^{\mathrm y}\;{\lbrack\mathrm H^+\rbrack}^{\mathrm z}\\v&=&0,038\mathit\;{\lbrack{\mathrm{CH}}_3{\mathrm{COCH}}_3\rbrack}^1{\;\lbrack{\mathrm{Br}}_2\rbrack}^0\;{\lbrack\mathrm H^+\rbrack}^1\\{\color[rgb]{0.0, 0.5, 0.0}v}&{\color[rgb]{0.0, 0.5, 0.0}=}&{\color[rgb]{0.0, 0.5, 0.0}0}{\color[rgb]{0.0, 0.5, 0.0},}{\color[rgb]{0.0, 0.5, 0.0}038}{\color[rgb]{0.0, 0.5, 0.0}\mathit\;}{\color[rgb]{0.0, 0.5, 0.0}\left[{\mathrm{CH}}_3{\mathrm{COCH}}_3\right]}{\color[rgb]{0.0, 0.5, 0.0}\;}{\color[rgb]{0.0, 0.5, 0.0}\left[\mathrm H^+\right]}\end{array}$

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