If the half - life of the first order reaction is 50 s, what be the value of its rate constant ?

Options

A. `1.38 xx10^(-2) s^(-1)`
B. `25 s^(-1)`
C. `34.66s^(-1)`
D. `1.38xx10^(-4)s^(-1)`

Answer

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

### Step-by-Step Solution:

1. **Identify the formula for half-life of a first-order reaction:**
The half-life (t_half) 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 half-life into the formula:**
We are given that the half-life \( t_{1/2} \) is 50 seconds. Substituting this value into the formula gives:
\[ 50 = \frac{0.693}{k} \]

3. **Rearrange the equation to solve for k:**
To find \( k \), we can rearrange the equation:
\[ k = \frac{0.693}{50} \]

4. **Calculate the value of k:**
Now, perform the calculation:
\[ k = \frac{0.693}{50} = 0.01386 \text{ s}^{-1} \]
This can also be expressed in scientific notation:
\[ k \approx 1.38 \times 10^{-2} \text{ s}^{-1} \]

5. **Final answer:**
The rate constant \( k \) for the first-order reaction is:
\[ k \approx 1.38 \times 10^{-2} \text{ s}^{-1} \]