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 and initial concentrations, we can follow these steps:

### Step 1: Understand the relationship between half-life and initial concentration
The half-life (\(t_{1/2}\)) of a reaction is related to the initial concentration (\(A_0\)) and the order of the reaction (\(n\)). For a reaction of order \(n\), the relationship can be expressed as:
\[ t_{1/2} \propto \frac{1}{A_0^{n-1}} \]
This means that the half-life is inversely proportional to the initial concentration raised to the power of \(n-1\).

### Step 2: Set up the equation using the given data
We are given two sets of data:
- For \(A_0 = 0.2 \, \text{mol L}^{-1}\), \(t_{1/2} = 100 \, \text{s}\)
- For \(A_0 = 0.4 \, \text{mol L}^{-1}\), \(t_{1/2} = 200 \, \text{s}\)

Using the relationship, we can write:
\[ \frac{t_{1/21}}{t_{1/22}} = \frac{A_0^{2-n}}{A_0^{1-n}} \]
Substituting the values:
\[ \frac{100}{200} = \frac{(0.4)^{2-n}}{(0.2)^{2-n}} \]

### Step 3: Simplify the equation
This simplifies to:
\[ \frac{1}{2} = \left(\frac{0.4}{0.2}\right)^{2-n} \]
Calculating \(\frac{0.4}{0.2} = 2\), we have:
\[ \frac{1}{2} = 2^{2-n} \]

### Step 4: Equate the powers
Since both sides are equal, we can equate the exponents:
\[ 2^{-1} = 2^{2-n} \]
This gives us:
\[ -1 = 2 - n \]

### Step 5: Solve for \(n\)
Rearranging the equation:
\[ n = 2 + 1 = 3 \]

### Conclusion
The order of the reaction is \(n = 3\).