Chemical Kinetics - Result Question 32
####32. The bromination of acetone that occurs in acid solution is represented by this equation. [2008] $CH_3 COCH_3(aq)+Br_2(aq) \to$ $CH_3 COCH_2 Br(aq)+H^{+}(aq)+Br^{-}(aq)$ These kinetic data were obtained for given reaction concentrations.
\begin{tabular}{lll} Initial Concentrations, $\mathbf{M}$ & \\ {$[\mathbf{C H}_3 \mathbf{C O C H}_3]$} & {$[Br_2]$} & {$[\mathbf{H}^{+}]$} \\
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Solution:
- (a) Rewriting the given data for the reaction
$CH_3 COCH_3(aq)+Br_2(aq) \xrightarrow{H^{+}}$
$CH_3 COCH_2 Br(aq)+H^{+}(aq)+Br^{-}(aq)$
\begin{tabular}{ccccc} \hline \begin{tabular}{c} S. \\ No. \\ -ration of \\ $CH_3 COCH_3$ \\ in $M$ \end{tabular} & \begin{tabular}{c} Initial concentr \\ -ation of $Br_2$ \\ in $M$ \end{tabular} & \begin{tabular}{c} Initial concentr \\ -ation of $H^{+}$ \\ in $M$ \end{tabular} & \begin{tabular}{c} Rate of \\ disappearance \\ of $Br_2$ in $MS^{-1}$ \end{tabular} \\ \hline 1 & 0.30 & 0.05 & 0.05 & \begin{tabular}{c} i.e. $-\frac{d}{dt}[Br_2]$ or $\frac{dx}{dt}$ \end{tabular} \\ 2 & 0.30 & 0.10 & 0.05 & $5.7 \times 10^{-5}$ \\ 3 & 0.30 & 0.10 & 0.10 & $1.7 \times 10^{-5}$ \\ 4 & 0.40 & 0.05 & 0.20 & $3.1 \times 10^{-4}$ \\ \hline \end{tabular}
This reaction is autocatalyzed and involves complex calculation for concentration terms.
We can look at the above results in a simple way to find the dependence of reaction rate (i.e. rate of disappearance of $Br_2$ ).
From data (1) and (2) in which concentration of $CH_3 COCH_3$ and $H^{+}$remain unchanged and only the concentration of $Br_2$ is doubled, there is no change in rate of reaction. It means the rate of reaction is independent of concentration of $Br_2$. Again from (2) and (3) in which $(CH_3 CO CH_3)$ and $(Br_2)$ remain constant but $H^{+}$increases from $0.05 M$ to 0.10 i.e. doubled, the rate of reaction changes from $5.7 \times 10^{-5}$ to $1.2 \times 10^{-4}$ (or $12 \times 10^{-5}$ ), thus it also becomes almost doubled. It shows that rate of reaction is directly proportional to $[H^{+}]$. From (3) and (4), the rate should have doubled due to increase in conc of $[H^{+}]$from $0.10 M$ to $0.20 M$ but the rate has changed from $1.2 \times 10^{-4}$ to $3.1 \times 10^{-4}$. This is due to change in concentration of $CH_3 COCH_3$ from $0.30 M$ to $0.40 M$. Thus, the rate is directly proportional to $[CH_3 COCH_3]$.
rate $=k[CH_3 COCH_3]^{1}[Br_2]^{0}[H^{+}]^{1}$
$=k[CH_3 COCH_3][H^{+}]$.
