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? 
 

\([A]\) (mol/L)	\(t_{1/2}\) (min)
0.100	200
0.025	100

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. 

• A. A, B and D only
• B. A and C only
• C. A and B only
• D. C and D only

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, the half-life relates to the initial concentration by \( t_{1/2} \propto 1/[A_0] \).

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

Step 3: Statement B is correct due to the half-life's dependence on the initial concentration. Step 4: Statement D is accurate because for a second-order reaction, doubling \( [A_0] \) results in a doubling of the half-life. 

Final Conclusion: Option (1), which includes statements A, B, and D, is the correct answer.