`A to B, K_(1) = 0.693sec^(-1)`
`C to D, K_(2) = 0.693 min^(-1)`. If `t_(1)` & `t_(2)` are half lives of two reactions, then

Options

A. `t_(1)=t_(2)`
B. `t_(1) = 60t_(2)`
C. `t_(2) = 60t_(1)`
D. `t_(2) = 2.303 t_(1)`

Explanation

To solve the problem, we need to find the relationship between the half-lives \( t_1 \) and \( t_2 \) of two first-order reactions given their rate constants \( K_1 \) and \( K_2 \).

### Step-by-Step Solution:

1. **Identify the half-life formula for first-order reactions**:
The half-life \( t_{1/2} \) for a first-order reaction is given by the formula:
\[ t_{1/2} = \frac{0.693}{K} \]

2. **Calculate \( t_1 \) for the reaction A to B**:
Given \( K_1 = 0.693 \, \text{sec}^{-1} \):
\[ t_1 = \frac{0.693}{K_1} = \frac{0.693}{0.693} = 1 \, \text{sec} \]

3. **Convert \( K_2 \) from minutes to seconds**:
Given \( K_2 = 0.693 \, \text{min}^{-1} \):
To convert \( K_2 \) to seconds, we use the conversion factor \( 1 \, \text{min} = 60 \, \text{sec} \):
\[ K_2 = \frac{0.693}{60} \, \text{sec}^{-1} \]

4. **Calculate \( t_2 \) for the reaction C to D**:
Using the half-life formula:
\[ t_2 = \frac{0.693}{K_2} = \frac{0.693}{\frac{0.693}{60}} = 60 \, \text{sec} \]

5. **Establish the relationship between \( t_1 \) and \( t_2 \)**:
Now we have:
\[ t_1 = 1 \, \text{sec} \quad \text{and} \quad t_2 = 60 \, \text{sec} \]
Therefore, the relationship can be expressed as:
\[ t_2 = 60 \cdot t_1 \]

### Final Answer:
The relationship between the half-lives is:
\[ t_2 = 60 \cdot t_1 \]