`5A rarr` Product In above reaction, half-life period is directly proportional to initial concentration of reactant. The initial rate of reaction is `400" mol lit"^(–1)" min"^(–1)` . Calculate the half-life period (in sec) when initial concentration of reactant is `200" mol lit"^(–1)`.

Answer

To solve the problem, we need to calculate the half-life period of a zero-order reaction given the initial concentration of the reactant and the initial rate of the reaction.

### Step-by-Step Solution:

1. **Identify the Reaction Order**:
The problem states that the half-life period is directly proportional to the initial concentration of the reactant. This characteristic is typical of a zero-order reaction.

2. **Write the Half-Life Formula for Zero-Order Reactions**:
The half-life (\( T_{1/2} \)) for a zero-order reaction is given by the formula:
\[ T_{1/2} = \frac{C_0}{2k} \]
where \( C_0 \) is the initial concentration and \( k \) is the rate constant.

3. **Determine the Rate Constant \( k \)**:
The initial rate of the reaction is given as \( 400 \, \text{mol L}^{-1} \text{min}^{-1} \). For a zero-order reaction, the rate is equal to the rate constant \( k \):
\[ k = 400 \, \text{mol L}^{-1} \text{min}^{-1} \]

4. **Substitute the Values into the Half-Life Formula**:
We are given \( C_0 = 200 \, \text{mol L}^{-1} \). Now we can substitute \( C_0 \) and \( k \) into the half-life formula:
\[ T_{1/2} = \frac{200}{2 \times 400} \]

5. **Calculate the Half-Life**:
Simplifying the equation:
\[ T_{1/2} = \frac{200}{800} = 0.25 \, \text{minutes} \]

6. **Convert Half-Life from Minutes to Seconds**:
Since we need the half-life in seconds, we convert minutes to seconds:
\[ T_{1/2} = 0.25 \, \text{minutes} \times 60 \, \text{seconds/minute} = 15 \, \text{seconds} \]

### Final Answer:
The half-life period when the initial concentration of the reactant is \( 200 \, \text{mol L}^{-1} \) is **15 seconds**.