If the rate constant of a first order reaction at a certain temperature is `1.5xx10^(-1)s^(-1)` and `t_(1)` and `t_(2)` are the respective times for `50%` and `75%` completion of the reaction, determine the ratio of `t_(2)" to "t_(1)`.

Explanation

To solve the problem of finding the ratio of \( t_2 \) to \( t_1 \) for a first-order reaction, we can use the integrated rate law for first-order reactions. The rate constant \( k \) is given as \( 1.5 \times 10^{-1} \, s^{-1} \).

### Step 1: Understand the Definitions
- \( t_1 \) is the time taken for 50% completion of the reaction.
- \( t_2 \) is the time taken for 75% completion of the reaction.

### Step 2: Use the Integrated Rate Law
For a first-order reaction, the integrated rate law is given by:
\[ t = \frac{2.303}{k} \log \left( \frac{A_0}{A} \right) \]
where:
- \( A_0 \) is the initial concentration,
- \( A \) is the concentration at time \( t \),
- \( k \) is the rate constant.

### Step 3: Calculate \( t_1 \) for 50% Completion
At 50% completion, \( A = \frac{A_0}{2} \):
\[ t_1 = \frac{2.303}{k} \log \left( \frac{A_0}{\frac{A_0}{2}} \right) = \frac{2.303}{k} \log (2) \]

### Step 4: Calculate \( t_2 \) for 75% Completion
At 75% completion, \( A = \frac{A_0}{4} \):
\[ t_2 = \frac{2.303}{k} \log \left( \frac{A_0}{\frac{A_0}{4}} \right) = \frac{2.303}{k} \log (4) \]

### Step 5: Find the Ratio \( \frac{t_2}{t_1} \)
Now, we can find the ratio of \( t_2 \) to \( t_1 \):
\[ \frac{t_2}{t_1} = \frac{\frac{2.303}{k} \log (4)}{\frac{2.303}{k} \log (2)} = \frac{\log (4)}{\log (2)} \]

### Step 6: Simplify the Ratio
Since \( \log (4) = \log (2^2) = 2 \log (2) \):
\[ \frac{t_2}{t_1} = \frac{2 \log (2)}{\log (2)} = 2 \]

### Final Answer
Thus, the ratio of \( t_2 \) to \( t_1 \) is:
\[ \frac{t_2}{t_1} = 2 \]