The half-life period of a first order reaction is 100 sec. The rate constant of the reaction is

Options

A. `6.93 xx 10^(-2) sec^(-1)`
B. `6.93 xx 10^(-4) sec^(-1)`
C. `0.693 sec^(-1)`
D. `69.3 sec^(-1)`s

Answer

To find the rate constant (k) of a first-order reaction given its half-life (t½), we can use the formula that relates the two:

### Step-by-Step Solution:

1. **Understand the relationship between half-life and rate constant**:
The half-life (t½) of a first-order reaction is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
where \( k \) is the rate constant.

2. **Rearrange the formula to solve for k**:
To find the rate constant \( k \), we can rearrange the formula:
\[ k = \frac{0.693}{t_{1/2}} \]

3. **Substitute the given half-life into the equation**:
We are given that the half-life \( t_{1/2} \) is 100 seconds. Substituting this value into the equation gives:
\[ k = \frac{0.693}{100 \text{ sec}} \]

4. **Calculate the value of k**:
Performing the calculation:
\[ k = 0.00693 \text{ sec}^{-1} \]
or in scientific notation:
\[ k = 6.93 \times 10^{-3} \text{ sec}^{-1} \]

5. **Final answer**:
Therefore, the rate constant \( k \) for the reaction is:
\[ k = 6.93 \times 10^{-3} \text{ sec}^{-1} \]