$$\begin{array}{|c|c|}\hline \text { Time (Hours) } & {[\mathrm{A}] M} \\\ \hline 0 & {0.40} \\ \hline 1 & {0.20} \\ \hline 2 & {0.10} \\ \hline 3 & {0.05} \\ \hline\end{array}$$ Reactant A underwent a decomposition reaction. The concentration of A was measured periodically and recorded in the chart above. Based on the data in the chart, which of the following is the rate law for the reaction? (A) Rate \(=k[\mathrm{A}]\) (B) Rate \(=k[\mathrm{A}]^{2}\) (C) Rate \(=2 k[\mathrm{A}]\) (D) Rate \(=\frac{1}{2} k[\mathrm{A}]\)

Short Answer

Expert verified
The correct rate law for this reaction is Rate = k[A], which corresponds to Option (A).

Step by step solution

01

- Observing the change of concentration of A

Firstly, observe the concentration of A from the table. It appears to halve with each hour passed. This indicates the reaction is first order with respect to A, as the rate of reaction is proportional to the first power of the concentration of A.
02

- Analysis of Rate Law Options

Analyzing the given options, recognize that the constant k in the rate law doesn't affect the order of the reaction. It's the power to which the concentration is raised that determines whether the reaction is first order or not. The correct answer should have [A] to the power of 1 for a first order reaction.
03

- Matching the observation with the correct Rate Law

Based on the observations from Step 1 and Step 2, the rate law for this reaction should be Rate = k[A]. This matches with Option (A) from the given multiple choices. Therefore, option (A) is the correct answer.

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