Question

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?

\[ \begin{array}{|c|c|} \hline \textbf{[A] (mol/L)} & \textbf{t$_{1/2}$ (min)} \\ \hline 0.100 & 200 \\ 0.025 & 100 \\ \hline \end{array} \]

• 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.

Hint:

For reaction rate problems, use the relationship between concentration and half-life to determine the reaction order.

Answer 1

The Correct Option is A

Step 1: Determine the reaction order from the provided data. For a first-order reaction, half-life is proportional to the inverse of the initial concentration, \( t_{1/2} \propto 1/[A_0] \).

Step 2: Statement A is accurate because \( t_{1/2} \propto \frac{1}{\sqrt{[A_0]}} \) signifies a fractional order reaction.

Step 3: Statement B is correct as the half-life's dependency on initial concentration is established.

Step 4: Statement D is correct; in a second-order reaction, doubling \( [A_0] \) results in a doubling of the half-life.

Final Conclusion: Option (1) is the correct choice, encompassing statements A, B, and D.