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SOAL LATIHAN MEKANISME REAKSI

Nama

: Tiara Ria Eva Veronika

NIM

: 160332605810

Offering

:H

1. The condensation reaction of acetone, (CH3)2CO (propanon), in aqueous solution is catalyzed by bases, B, which react reversibly with acetone to form the carbanion C3H5O-. The carbanion then reacts with a molecule of acetone to give the product . a simplified version of the mechanism is (1) AH + B → BH+ + A(2) A- + BH+ → AH + B (3) A- + HA → product Where AH stands for acetone and A- its carbanion. Use the steady-state approximation to find the concentration of the carbanion and derive the rate equation for the formation of the product Jawab : (1) AH + B → BH+ + A(2) A- + BH+ → AH + B (3) A- + HA → product r=

𝑑 [𝑃] 𝑑𝑡

= 𝑘3 [𝐴− ][𝐻𝐴]

…………………………

(1)

𝑑 [𝐴− ] = 𝑘1 [𝐴𝐻][𝐵] − 𝑘2 [𝐴− ][𝐵𝐻 + ] − 𝑘3 [𝐴− ][𝐻𝐴] = 0 𝑑𝑡 Catatan : tanda (-) apabila A- berada pada reaktan, sedangkan tanda (+) apabila A- berada pada produk Mencari [A-] 𝑘1 [𝐴𝐻][𝐵] − 𝑘2 [𝐴− ][𝐵𝐻 + ] − 𝑘3 [𝐴− ][𝐻𝐴] = 0 [A-] = 𝑘

𝑘1 [𝐴𝐻][𝐵] ………………………………. (2) + ]+ 𝑘 [𝐻𝐴] [𝐵𝐻 2 3

Mencari turunan persamaan laju pembentukan produk Persamaan 2 disubstitusi kedalam persamaan 1 𝑑 [𝑃] = 𝑘3 [𝐴− ][𝐻𝐴] 𝑑𝑡

𝑑 [𝑃] 𝑘1 [𝐴𝐻][𝐵] = 𝑘3 ( ) [𝐻𝐴] 𝑑𝑡 𝑘2 [𝐵𝐻 + ] + 𝑘3 [𝐻𝐴] Sehingga turunan persamaan laju pembentukan produk adalah : 𝑑 [𝑃] 𝑘3 𝑘1 [𝐴𝐻]2 [𝐵] = 𝑑𝑡 𝑘2 [𝐵𝐻 + ] + 𝑘3 [𝐻𝐴] 2. Consider the following chain mechanism : (1) AH → A. + H. (2) A. → B. + C (3) AH + B. →A. + D (4) A. + B. → P Identify the initiation, propagation, and termination steps, and use the steady-state approximation to deduce that the decomposition of AH is first order in AH. Jawab : Yang termasuk inisiasi, propagasi, dan terminasi adalah: (1) AH → A. + H.

inisiasi

(2) A. → B. + C

propagasi

(3) AH + B. →A. + D (4) A. + B. → P 𝑑[𝐴𝐻] 𝑑𝑡

terminasi

= −𝑘1 [𝐴𝐻] − 𝑘3 [𝐴𝐻][B. ] ……………………………………………………(1)

𝑑[B. ] = 0 = 𝑘2 [A. ] − 𝑘3 [𝐴𝐻][𝐵. ] − 𝑘4 [A. ][B. ] 𝑑𝑡 𝑘3 [𝐴𝐻][𝐵. ] + 𝑘4 [A. ][B. ] = 𝑘2 [A. ] [B. ](𝑘3 [𝐴𝐻] + 𝑘4 [A. ]) = 𝑘2 [A. ] [B. ] =

𝑘2 [A.] 𝑘3 [𝐴𝐻]+ 𝑘4 [A.]

……………………………………………………(2)

𝑑[A. ] = 0 = 𝑘1 [𝐴𝐻] − 𝑘2 [A. ] + 𝑘3 [𝐴𝐻][𝐵. ] − 𝑘4 [A. ][B. ] 𝑑𝑡 𝑘2 [A. ] + 𝑘4 [A. ][B. ] = 𝑘1 [𝐴𝐻] + 𝑘3 [𝐴𝐻][𝐵. ] [A. ] = (𝑘2 + 𝑘4 [B. ]) = 𝑘1 [𝐴𝐻] + 𝑘3 [𝐴𝐻][𝐵. ] [A. ] =

𝑘1 [𝐴𝐻]+ 𝑘3 [𝐴𝐻][𝐵.] 𝑘2 +𝑘4 [B.]

……………………………………………………(3)

Disubstitusi persamaan 3 ke persamaan 2 [B. ] =

𝑘2 [A. ] 𝑘3 [𝐴𝐻] + 𝑘4 [A. ]

[B. ] = 𝑘2

𝑘1 [𝐴𝐻]+ 𝑘3 [𝐴𝐻][𝐵.] 𝑘2 +𝑘4 [B.] 𝑘 [𝐴𝐻]+ 𝑘 [𝐴𝐻][𝐵.] 𝑘3 [𝐴𝐻]+ 𝑘4 [ 1 𝑘 +𝑘 3[B.] ] 2 4

𝑘2 𝑘1 [𝐴𝐻] + 𝑘2 𝑘3 [𝐴𝐻][𝐵. ] 𝑘2 + 𝑘4 [B. ] [B. ] = 𝑘 [𝐴𝐻] + 𝑘3 [𝐴𝐻][𝐵. ] 𝑘3 [𝐴𝐻] + 𝑘4 [ 1 ] 𝑘2 + 𝑘4 [B. ] 𝑘2 𝑘1 [𝐴𝐻] + 𝑘3 𝑘2 [𝐴𝐻][𝐵. ] ) 𝑘2 + 𝑘4 [B. ] [B. ] = 𝑘 𝑘 [𝐴𝐻] + 𝑘2 𝑘4 [𝐴𝐻][𝐵. ] + 𝑘1 𝑘4 [𝐴𝐻] + 𝑘4 𝑘3 [𝐴𝐻][𝐵. ] ( 3 2 ) 𝑘2 + 𝑘4 [B. ] (

[B. ] =

𝑘1 𝑘2 [𝐴𝐻]+𝑘2 𝑘3 [𝐴𝐻][𝐵.] 𝑘2 +𝑘4 [B.]

[B. ] =

𝑘1 𝑘2 [𝐴𝐻] + 𝑘2 𝑘3 [𝐴𝐻][𝐵. ] 𝑘2 𝑘3 [𝐴𝐻] + 𝑘3 𝑘4 [𝐴𝐻][𝐵. ] + 𝑘1 𝑘4 [𝐴𝐻] + 𝑘4 𝑘3 [𝐴𝐻][𝐵. ]

x𝑘

𝑘2 +𝑘4 [B.] 2 𝑘3 [𝐴𝐻]+𝑘3 𝑘4 [𝐴𝐻][𝐵.]+𝑘1 𝑘4 [𝐴𝐻]+𝑘4 𝑘3 [𝐴𝐻][𝐵.]

𝑘2 𝑘3 [𝐵. ][𝐴𝐻] + 𝑘3 𝑘4 [𝐴𝐻][𝐵. ]2 + 𝑘1 𝑘4 [𝐴𝐻][𝐵. ] + 𝑘3 𝑘4 [𝐴𝐻][𝐵. ]2 = 𝑘1 𝑘2 [𝐴𝐻] + 𝑘2 𝑘3 [𝐴𝐻][𝐵. ] 𝑘2 𝑘3 [𝐵. ][𝐴𝐻] − 𝑘2 𝑘3 [𝐴𝐻][𝐵. ] + 𝑘3 𝑘4 [𝐵. ]2 [𝐴𝐻] + 𝑘1 𝑘4 [𝐴𝐻][𝐵. ] + 𝑘3 𝑘4 [𝐴𝐻][𝐵. ]2 = 𝑘1 𝑘2 [𝐴𝐻] 2𝑘3 𝑘4 [𝐵. ]2 [𝐴𝐻] + 𝑘1 𝑘4 [𝐴𝐻][𝐵. ] = 𝑘1 𝑘2 [𝐴𝐻]

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