The half life period for catalytic decomposition of `XY_3` at 100 mm is found to be 8 hrs and at 200 mm it is 4 hrs. The order of reaction is

Options

A. 3
B. 2
C. 4
D. None of these

Answer

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

### Step-by-Step Solution:

1. **Understand the relationship for half-life in nth order reactions**:
The half-life \( t_{1/2} \) for an nth order reaction is given by the relationship:
\[ t_{1/2} \propto \frac{1}{P_0^{(n-1)}} \]
where \( P_0 \) is the initial pressure and \( n \) is the order of the reaction.

2. **Set up the known values**:
From the problem, we have:
- At \( P_0 = 100 \, \text{mm} \), \( t_{1/2} = 8 \, \text{hrs} \)
- At \( P_0 = 200 \, \text{mm} \), \( t_{1/2} = 4 \, \text{hrs} \)

3. **Formulate the ratio of half-lives**:
We can write the ratio of the half-lives as:
\[ \frac{t_{1/2,1}}{t_{1/2,2}} = \frac{P_{0,2}^{(n-1)}}{P_{0,1}^{(n-1)}} \]
Substituting the values:
\[ \frac{8}{4} = \frac{200^{(n-1)}}{100^{(n-1)}} \]

4. **Simplify the equation**:
This simplifies to:
\[ 2 = \left(\frac{200}{100}\right)^{(n-1)} = 2^{(n-1)} \]

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

6. **Solve for n**:
Rearranging gives:
\[ n = 2 \]

### Conclusion:
The order of the reaction is **2**.