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 step by step, we need to follow the principles of zero-order reactions and the half-life formula. Here’s how we can approach the solution:

### Step 1: Understand the half-life of a zero-order reaction
For a zero-order reaction, the half-life (T_half) 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)
From the problem, we know:
- \( T_{1/2} = 1 \, \text{h} \)
- \( A_0 = 2.0 \, \text{mol L}^{-1} \)

Using the half-life formula:
\[ 1 \, \text{h} = \frac{2.0 \, \text{mol L}^{-1}}{2K} \]

Rearranging to find \( K \):
\[ K = \frac{2.0 \, \text{mol L}^{-1}}{2 \times 1 \, \text{h}} = 1.0 \, \text{mol L}^{-1} \text{h}^{-1} \]

### Step 3: Use the zero-order reaction formula to find time (T)
For a zero-order reaction, the change in concentration over time can be expressed as:
\[ K = \frac{A_0 - A}{T} \]
Where:
- \( A_0 \) is the initial concentration
- \( A \) is the final concentration
- \( T \) is the time taken

We need to find the time taken for the concentration to change from \( 0.50 \, \text{mol L}^{-1} \) to \( 0.25 \, \text{mol L}^{-1} \).

Substituting the values:
- \( A_0 = 0.50 \, \text{mol L}^{-1} \)
- \( A = 0.25 \, \text{mol L}^{-1} \)

The equation becomes:
\[ 1.0 = \frac{0.50 - 0.25}{T} \]

### Step 4: Solve for time (T)
Rearranging the equation gives:
\[ T = \frac{0.50 - 0.25}{1.0} = \frac{0.25}{1.0} = 0.25 \, \text{h} \]

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

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