The reaction `A to C+D` was found to be second order in A. The rate constant for the reaction was determined to be `2.42 `L/mol/s . If the initial concentration is `0.5` mole/L what is the value of `t_(1/2)`.

Explanation

To find the half-life (\( t_{1/2} \)) of a second-order reaction, we can use the formula for the half-life of a second-order reaction:

\[ t_{1/2} = \frac{1}{k \cdot C_0} \]

Where:
- \( k \) is the rate constant,
- \( C_0 \) is the initial concentration.

### Step-by-Step Solution:

1. **Identify the given values**:
- Rate constant (\( k \)) = 2.42 L/mol/s
- Initial concentration (\( C_0 \)) = 0.5 mol/L

2. **Substitute the values into the half-life formula**:
\[ t_{1/2} = \frac{1}{k \cdot C_0} = \frac{1}{2.42 \, \text{L/mol/s} \cdot 0.5 \, \text{mol/L}} \]

3. **Calculate the denominator**:
\[ k \cdot C_0 = 2.42 \cdot 0.5 = 1.21 \, \text{L/s} \]

4. **Calculate \( t_{1/2} \)**:
\[ t_{1/2} = \frac{1}{1.21} \approx 0.826 \, \text{s} \]

5. **Round the answer**:
\[ t_{1/2} \approx 0.83 \, \text{s} \]

### Final Answer:
The half-life (\( t_{1/2} \)) of the reaction is approximately **0.83 seconds**.