A first order reaction is 50 % completed in 30 minutes . The rate constant of the reaction is

Options

A. `2.31"min"^(-1) `
B. `2.31xx10 ^(2) "min"^(-1) `
C. `2.31xx10^(2)"min"^(-1)`
D. `2.31xx10^(-1)"min"^(-1)`

Explanation

To find the rate constant \( k \) for a first-order reaction that is 50% completed in 30 minutes, we can use the relationship between the half-life of a first-order reaction and the rate constant.

### Step-by-Step Solution:

1. **Understanding Half-Life**: For a first-order reaction, the half-life (\( t_{1/2} \)) is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
where \( k \) is the rate constant.

2. **Given Data**: The problem states that the reaction is 50% completed in 30 minutes. Therefore, we can directly substitute \( t_{1/2} \) with 30 minutes:
\[ t_{1/2} = 30 \text{ minutes} \]

3. **Setting Up the Equation**: Substitute \( t_{1/2} \) into the half-life formula:
\[ 30 = \frac{0.693}{k} \]

4. **Rearranging the Equation**: To find \( k \), rearrange the equation:
\[ k = \frac{0.693}{30} \]

5. **Calculating \( k \)**: Now, perform the calculation:
\[ k = \frac{0.693}{30} \approx 0.0231 \text{ min}^{-1} \]

6. **Final Answer**: The rate constant \( k \) for the reaction is approximately:
\[ k \approx 2.31 \times 10^{-2} \text{ min}^{-1} \]