For a given reaction \( R \rightarrow P \), \( t_{1/2} \) is related to \([A_0]\) as given in the table. Given: \( \log 2 = 0.30 \). Which of the following is true?
\[
\begin{array}{|c|c|}
\hline
\textbf{[A] (mol/L)} & \textbf{t$_{1/2}$ (min)} \\
\hline
0.100 & 200 \\
0.025 & 100 \\
\hline
\end{array}
\]
\[ \begin{array}{|c|c|} \hline \textbf{[A] (mol/L)} & \textbf{t$_{1/2}$ (min)} \\ \hline 0.100 & 200 \\ 0.025 & 100 \\ \hline \end{array} \]
Show Hint
- A. The order of the reaction is \( \frac{1}{2} \).
- B. If \( [A_0] \) is 1 M, then \( t_{1/2} \) is \( 200/\sqrt{10} \) min.
- C. The order of the reaction changes to 1 if the concentration of reactant changes from 0.100 M to 0.500 M.
- D. \( t_{1/2} \) is 800 min for \( [A_0] = 1.6 \) M.
The Correct Option is A
Solution and Explanation
Step 1: From the given data, calculate the order of the reaction. The relationship between half-life and concentration is given by the formula \( t_{1/2} \propto 1/[A_0] \) for a first-order reaction.
Step 2: Statement A is correct as \( t_{1/2} \propto \frac{1}{\sqrt{[A_0]}} \), indicating a fractional order reaction.
Step 3: Statement B is correct because the half-life depends on the initial concentration.
Step 4: Statement D is correct because doubling \( [A_0] \) doubles the half-life for a second-order reaction.
Final Conclusion: The correct answer is Option (1), A, B and D Only.
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