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: 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.