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Rate Law from Initial Rate Data
Concepts tested here
- Method of initial rates
All Questions
2008 AIPMT 1 question
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The bromination of acetone that occurs in acid solution is represented by this equation.
$CH_3COCH_3(aq.) + Br_2(aq.) \rightarrow CH_3COCH_2Br(aq.) + H^+(aq.) + Br^-(aq.)$
These kinetic data were obtained for given reaction concentrations.
Initial concentrations (M) of $[CH_3COCH_3]$, $[Br_2]$, $[H^+]$ and the initial rate of disappearance of $Br_2$ (M s$^{-1}$):
Experiment 1: 0.30, 0.05, 0.05; rate $= 5.7 \times 10^{-5}$
Experiment 2: 0.30, 0.10, 0.05; rate $= 5.7 \times 10^{-5}$
Experiment 3: 0.30, 0.10, 0.10; rate $= 1.2 \times 10^{-4}$
Experiment 4: 0.40, 0.05, 0.20; rate $= 3.1 \times 10^{-4}$
Based on these data, the rate equation is :Let Rate $= k[CH_3COCH_3]^a[Br_2]^b[H^+]^c$.
Experiments 1 and 2: $[Br_2]$ is doubled (0.05 to 0.10) while the other two concentrations are constant, and the rate remains unaffected ($5.7 \times 10^{-5}$ in both).
So the order of reaction with respect to $Br_2$ is zero: $b = 0$.
Experiments 2 and 3: $[H^+]$ is doubled (0.05 to 0.10) and the rate becomes $\dfrac{1.2 \times 10^{-4}}{5.7 \times 10^{-5}} \approx 2$ times.
So the order with respect to $H^+$ is one: $c = 1$.
Experiments 1 and 4: $[CH_3COCH_3]$ becomes $\dfrac{0.40}{0.30} = \dfrac{4}{3}$ times and $[H^+]$ becomes 4 times; the rate becomes $\dfrac{3.1 \times 10^{-4}}{5.7 \times 10^{-5}} \approx 5.4$ times.
$\left(\dfrac{4}{3}\right)^a \times 4 = 5.4 \Rightarrow \left(\dfrac{4}{3}\right)^a = 1.35 \Rightarrow a = 1$
The rate law for the reaction is therefore: Rate $= k[CH_3COCH_3][H^+]$
This agrees with the mechanism: the slow step is the acid catalysed enolisation of acetone, and bromine reacts in a later fast step.