What will be the half-life of the first order reaction for which the value of rate constant is `200 s^(-1)`?

Options

A. `3.46 xx 10^(-2) x`
B. `3.46 xx 10^(-3) s`
C. `4.26 xx 10^(-2) s`
D. `4.26 xx 10^(-3) s`

Explanation

To find the half-life of a first-order reaction with a given rate constant, we can follow these steps:

### Step-by-Step Solution:

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

2. **Substitute the given value of the rate constant**:
In this case, the rate constant \( k \) is given as \( 200 \, \text{s}^{-1} \). We will substitute this value into the formula:
\[ t_{1/2} = \frac{0.693}{200} \]

3. **Calculate the half-life**:
Now, perform the division:
\[ t_{1/2} = \frac{0.693}{200} = 0.003465 \, \text{s} \]

4. **Express the result in scientific notation**:
To express \( 0.003465 \) in scientific notation:
\[ t_{1/2} = 3.465 \times 10^{-3} \, \text{s} \]

5. **Final answer**:
Therefore, the half-life of the first-order reaction is:
\[ t_{1/2} = 3.465 \times 10^{-3} \, \text{s} \]