For zero order reaction , `t_(1//2)` will be (`A_0` is the initial concentration , K is rate constant)

Options

A. 0.693/K
B. 2.303/K
C. `A_0/(2K)`
D. `("In"A_0)/2`

Answer

To find the half-life (\(t_{1/2}\)) of a zero-order reaction, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Zero-Order Reaction**:
- A zero-order reaction is characterized by a rate that is independent of the concentration of the reactant. The rate law can be expressed as:
\[ \text{Rate} = k \]
- Here, \(k\) is the rate constant.

2. **Write the Rate Equation**:
- For a zero-order reaction, the rate of reaction can also be expressed as:
\[ -\frac{d[A]}{dt} = k \]
- This indicates that the change in concentration of \(A\) over time is constant.

3. **Integrate the Rate Equation**:
- Rearranging the equation gives:
\[ d[A] = -k \, dt \]
- Integrating both sides from \(t = 0\) to \(t = t\) and from \([A]_0\) to \([A]_t\):
\[ \int_{[A]_0}^{[A]_t} d[A] = -k \int_{0}^{t} dt \]
- This results in:
\[ [A]_t - [A]_0 = -kt \]
- Rearranging gives:
\[ [A]_t = [A]_0 - kt \]

4. **Determine the Half-Life**:
- The half-life (\(t_{1/2}\)) is defined as the time required for the concentration of the reactant to decrease to half of its initial concentration:
\[ [A]_{t_{1/2}} = \frac{[A]_0}{2} \]
- Substituting this into the equation:
\[ \frac{[A]_0}{2} = [A]_0 - kt_{1/2} \]
- Rearranging gives:
\[ kt_{1/2} = [A]_0 - \frac{[A]_0}{2} = \frac{[A]_0}{2} \]

5. **Solve for \(t_{1/2}\)**:
- Dividing both sides by \(k\):
\[ t_{1/2} = \frac{[A]_0}{2k} \]

### Final Result:
Thus, the expression for the half-life of a zero-order reaction is:
\[ t_{1/2} = \frac{[A]_0}{2k} \]