The half life period for a reaction at initial concentration of 0.2 and 0.4 mol L`^(-1) are 100 s and 200 s respectively. The order of the reaction is

Options

A. 0
B. 1
C. 2
D. 3

Answer

To determine the order of the reaction based on the given half-life periods at different initial concentrations, we will follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between half-life and concentration**:
The half-life \( t_{1/2} \) of a reaction is related to the initial concentration \( [A]_0 \) and the order of the reaction \( n \) by the formula:
\[ t_{1/2} \propto [A]_0^{1-n} \]
This means that the half-life is directly proportional to the initial concentration raised to the power of \( 1-n \).

2. **Set up the ratio of half-lives**:
Given:
- \( t_{1/2,1} = 100 \, \text{s} \) at \( [A]_0 = 0.2 \, \text{mol L}^{-1} \)
- \( t_{1/2,2} = 200 \, \text{s} \) at \( [A]_0 = 0.4 \, \text{mol L}^{-1} \)

We can write the equations for the half-lives:
\[ t_{1/2,1} = k [0.2]^{1-n} \]
\[ t_{1/2,2} = k [0.4]^{1-n} \]

3. **Form the ratio of the two half-lives**:
\[ \frac{t_{1/2,1}}{t_{1/2,2}} = \frac{k [0.2]^{1-n}}{k [0.4]^{1-n}} \]
Simplifying this gives:
\[ \frac{100}{200} = \frac{[0.2]^{1-n}}{[0.4]^{1-n}} \]
\[ \frac{1}{2} = \left(\frac{0.2}{0.4}\right)^{1-n} \]

4. **Calculate the ratio of concentrations**:
\[ \frac{0.2}{0.4} = \frac{1}{2} \]
Therefore, we have:
\[ \frac{1}{2} = \left(\frac{1}{2}\right)^{1-n} \]

5. **Equate the exponents**:
Since the bases are the same, we can equate the exponents:
\[ 1 = 1 - n \]

6. **Solve for \( n \)**:
Rearranging gives:
\[ n = 1 - 1 = 0 \]

### Conclusion:
The order of the reaction is \( n = 0 \).