Question

The bromination of acetone that occurs in acid solution is represented by this equation.$C{{H}_{3}}COC{{H}_{3}}(aq.)+B{{r}_{2}}(aq.)\to C{{H}_{3}}COC{{H}_{2}}Br(aq.)+{{H}^{+}}(aq.)+B{{r}^{-}}(aq.)$
These kinetic data were obtained from given reaction concentrations.
Initial concentrations, (M)
\[\begin{align}
  & [C{{H}_{3}}COC{{H}_{3}}]\,\,\,\,\,\,\,\,\,[B{{r}_{2}}]\,\,\,\,\,\,\,\,\,[{{H}^{+}}] \\
 & 0.30\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.05\,\,\,\,\,\,\,\,\,\,\,\,0.05 \\
 & 0.30\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.10\,\,\,\,\,\,\,\,\,\,\,\,0.05 \\
 & 0.30\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.10\,\,\,\,\,\,\,\,\,\,\,\,0.10 \\
 & 0.40\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.05\,\,\,\,\,\,\,\,\,\,\,\,0.20 \\
\end{align}\]
Corresponding initial rate of disappearance of$B{{r}_{2}}$, $M{{s}^{-1}}$
\[\begin{align}
  & 5.7\times {{10}^{-5}} \\
 & 5.7\times {{10}^{-5}} \\
 & 1.2\times {{10}^{-4}} \\
 & 3.1\times {{10}^{-4}} \\
\end{align}\]
Based on these data, the rate equation is:
A.$rate=k[C{{H}_{3}}COC{{H}_{3}}][B{{r}_{2}}]{{[{{H}^{+}}]}^{2}}$
B. $rate=k[C{{H}_{3}}COC{{H}_{3}}][B{{r}_{2}}][{{H}^{+}}]$
C.$rate=k[C{{H}_{3}}COC{{H}_{3}}]{{[{{H}^{+}}]}^{2}}$
D. $rate=k[CH=COC{{H}_{3}}][B{{r}_{2}}]$

Answer 1

Hint: Rate law gives the rate equation of any reaction which is the rate constant multiplied by the concentrations of reactants raised to their stoichiometric values in a balanced equation.

Complete step by step answer:
We have been given a reaction, where bromination occurs of acetone, which forms, 1-bromo-propanone. We are only given with the concentration data of the reactants, and their corresponding rates of disappearance of bromine. We have to find its rate of equation.
As we know, the rate of any equation is the concentration of the compounds involved. The disappearance rates of bromine, corresponds to the initial concentrations as,
\[\begin{align}
  & \,\,\,\,\,[C{{H}_{3}}COC{{H}_{3}}]\,\,\,\,\,\,\,\,\,[B{{r}_{2}}]\,\,\,\,\,\,\,\,\,[{{H}^{+}}] \\
 & 1.\,\,\,\,\,0.30\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.05\,\,\,\,\,\,\,\,\,\,\,\,0.05 \\
 & 2.\,\,\,\,0.30\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.10\,\,\,\,\,\,\,\,\,\,\,\,0.05 \\
 & 3.\,\,\,\,0.30\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.10\,\,\,\,\,\,\,\,\,\,\,\,0.10 \\
 & 4.\,\,\,\,0.40\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.05\,\,\,\,\,\,\,\,\,\,\,\,0.20 \\
\end{align}\] \[\begin{align}
  & disappearance\,rate\,of\,B{{r}_{2}} \\
 & 5.7\times {{10}^{-5}} \\
 & 5.7\times {{10}^{-5}} \\
 & 1.2\times {{10}^{-4}} \\
 & 3.1\times {{10}^{-4}} \\
\end{align}\]
On comparing these values, it can be inferred that, rate of disappearance of bromine is the rate of reaction. Comparing 1 and 2, there is no change in the rates, due to the same values, when the bromine concentration is doubled.
Comparing the values of 3 and 4, we have the concentrations as, $[C{{H}_{3}}COC{{H}_{3}}]\to \dfrac{3}{4}[C{{H}_{3}}COC{{H}_{3}}]$and, $[{{H}^{+}}]\to 2[{{H}^{+}}]$. This means that the rate becomes $\dfrac{8}{3}$ times the initial rate. So, the rate equation will consist of concentrations of acetone and hydrogen ions.
Hence the rate of reaction will be$rate=k[C{{H}_{3}}COC{{H}_{3}}]{{[{{H}^{+}}]}^{2}}$.

So,option C is correct.

Note: For equilibrium the rate of any reaction is the concentration of products upon the concentration of reactants raised to their stoichiometric values in a balanced equation. While, for the rate law of any reaction it is the rate constant multiplied by concentrations of reactants.