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 1: Understand the definition of half-life
The half-life of a reaction is the time required for the concentration of a reactant to decrease to half of its initial concentration.

### Step 2: Write the rate equation for a zero-order reaction
For a zero-order reaction, the rate of reaction is constant and is given by:
\[ \text{Rate} = -\frac{d[A]}{dt} = k \]
where \(k\) is the rate constant and \([A]\) is the concentration of the reactant.

### Step 3: Integrate the rate equation
For a zero-order reaction, we can express the concentration as:
\[ [A] = [A_0] - kt \]
where \([A_0]\) is the initial concentration.

### Step 4: Set up the equation for half-life
At half-life (\(t_{1/2}\)), the concentration \([A]\) will be half of the initial concentration:
\[ [A] = \frac{[A_0]}{2} \]
Substituting this into the integrated rate equation gives:
\[ \frac{[A_0]}{2} = [A_0] - kt_{1/2} \]

### Step 5: Solve for \(t_{1/2}\)
Rearranging the equation:
\[ kt_{1/2} = [A_0] - \frac{[A_0]}{2} \]
\[ kt_{1/2} = \frac{[A_0]}{2} \]
Now, solving for \(t_{1/2}\):
\[ t_{1/2} = \frac{[A_0]}{2k} \]

### Final Result
Thus, the half-life of a zero-order reaction is:
\[ t_{1/2} = \frac{[A_0]}{2k} \]