The time for half-life period of a certain reaction, `A rarr` products is `1 h`. When the initial concentration of the reactant `'A'` is `2.0 "mol" L^(-1)`, how much time does it take for its concentration to come from `0.50` to `0.25 "mol" L^(-1)`, if it is zero order reaction ?

Options

A. 4 h
B. `0.5 h`
C. `0.25 h`
D. 1 h

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Understand the half-life of a zero-order reaction
For a zero-order reaction, the half-life \( t_{1/2} \) is given by the formula:
\[ t_{1/2} = \frac{[A]_0}{2K} \]
where \([A]_0\) is the initial concentration and \(K\) is the rate constant.

### Step 2: Calculate the rate constant \( K \)
We know from the question that:
- The half-life \( t_{1/2} = 1 \) hour
- The initial concentration \([A]_0 = 2.0 \, \text{mol L}^{-1}\)

Using the half-life formula:
\[ 1 = \frac{2.0}{2K} \]
Rearranging gives:
\[ 2K = 2.0 \implies K = 1.0 \, \text{mol L}^{-1} \text{hour}^{-1} \]

### Step 3: Use the zero-order reaction formula to find the time
For a zero-order reaction, the time \( t \) taken for the concentration to change from \([A]_0\) to \([A]\) is given by:
\[ t = \frac{[A]_0 - [A]}{K} \]
Here, we need to find the time taken for the concentration to go from \(0.50 \, \text{mol L}^{-1}\) to \(0.25 \, \text{mol L}^{-1}\).

Substituting the values:
- Initial concentration \([A]_0 = 0.50 \, \text{mol L}^{-1}\)
- Final concentration \([A] = 0.25 \, \text{mol L}^{-1}\)
- Rate constant \( K = 1.0 \, \text{mol L}^{-1} \text{hour}^{-1} \)

Now substituting these values into the formula:
\[ t = \frac{0.50 - 0.25}{1.0} = \frac{0.25}{1.0} = 0.25 \, \text{hours} \]

### Conclusion
The time taken for the concentration of \( A \) to decrease from \( 0.50 \, \text{mol L}^{-1} \) to \( 0.25 \, \text{mol L}^{-1} \) is \( 0.25 \, \text{hours} \).

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