If initial concentration is reduced to `1//4^(th)` in a zero order reaction , the time taken for half the reaction to complete

Options

A. Remains same
B. Becomes 4 time
C. Become one fourth
D. Doubles

Answer

To solve the problem, we need to understand the relationship between the initial concentration of a reactant in a zero-order reaction and the time taken for half of the reaction to complete.

### Step-by-Step Solution:

1. **Understanding Zero-Order Reactions**:
In a zero-order reaction, the rate of reaction is constant and does not depend on the concentration of the reactants. The rate law for a zero-order reaction can be expressed as:
\[ [A] = [A_0] - kt \]
where \([A]\) is the concentration at time \(t\), \([A_0]\) is the initial concentration, \(k\) is the rate constant, and \(t\) is time.

2. **Finding the Half-Life Expression**:
The half-life (\(t_{1/2}\)) for a zero-order reaction is given by:
\[ t_{1/2} = \frac{[A_0]}{2k} \]
This means that the half-life is directly proportional to the initial concentration \([A_0]\).

3. **Initial Concentration Reduction**:
According to the problem, the initial concentration is reduced to \(\frac{1}{4}\) of its original value. Thus, if the original concentration is \([A_0]\), the new concentration becomes:
\[ [A'] = \frac{[A_0]}{4} \]

4. **Calculating the New Half-Life**:
We can now calculate the new half-life (\(t'_{1/2}\)) using the new concentration:
\[ t'_{1/2} = \frac{[A']}{2k} = \frac{\frac{[A_0]}{4}}{2k} = \frac{[A_0]}{8k} \]

5. **Relating New Half-Life to Original Half-Life**:
The original half-life was:
\[ t_{1/2} = \frac{[A_0]}{2k} \]
Now, we can express the new half-life in terms of the original half-life:
\[ t'_{1/2} = \frac{[A_0]}{8k} = \frac{1}{4} \cdot \frac{[A_0]}{2k} = \frac{1}{4} t_{1/2} \]

6. **Conclusion**:
Therefore, if the initial concentration is reduced to \(\frac{1}{4}\), the time taken for half the reaction to complete is also reduced to \(\frac{1}{4}\) of the original half-life.

### Final Answer:
The time taken for half the reaction to complete when the initial concentration is reduced to \(\frac{1}{4}\) is \(\frac{1}{4} t_{1/2}\).