If initial concentration is reduced to its 1/4th in a zero order reaction, the time taken for half of the reaction to complete

Options

A. Remains same
B. Becomes 4 times
C. Becomes one-fourth
D. Doubles

Answer

To solve the problem, we need to understand the concept of half-life in a zero-order reaction and how the initial concentration affects it.

### Step-by-Step Solution:

1. **Understanding Zero-Order Reaction**:
In a zero-order reaction, the rate of reaction is constant and does not depend on the concentration of the reactants. The half-life (t_half) for a zero-order reaction can be expressed as:
\[ t_{1/2} = \frac{[A]_0}{2k} \]
where \([A]_0\) is the initial concentration and \(k\) is the rate constant.

2. **Initial Concentration Reduction**:
According to the problem, the initial concentration is reduced to its 1/4th. If the initial concentration is \([A]_0\), then the new concentration after reduction is:
\[ [A]_0' = \frac{[A]_0}{4} \]

3. **Calculating New Half-Life**:
Now we need to calculate the new half-life (\(t_{1/2}'\)) using the new concentration:
\[ t_{1/2}' = \frac{[A]_0'}{2k} = \frac{\frac{[A]_0}{4}}{2k} \]
Simplifying this gives:
\[ t_{1/2}' = \frac{[A]_0}{8k} \]

4. **Relating New Half-Life to Original Half-Life**:
From the original half-life formula, we know:
\[ t_{1/2} = \frac{[A]_0}{2k} \]
We can express the new half-life in terms of the original half-life:
\[ t_{1/2}' = \frac{1}{4} t_{1/2} \]
This shows that the new half-life is one-fourth of the original half-life.

### Final Answer:
The time taken for half of the reaction to complete when the initial concentration is reduced to its 1/4th in a zero-order reaction is one-fourth of the original half-life.

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