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 between half-life and order of reaction:**
For a reaction of order \( n \), the half-life \( t_{1/2} \) is related to the initial concentration (or pressure in this case) \( P_0 \) by the formula:
\[ t_{1/2} \propto P_0^{(1-n)} \]
This means that the half-life depends on the initial concentration raised to the power of \( (1-n) \).

2. **Set up the ratios of half-lives:**
Given:
- \( t_{1/2,1} = 8 \) hours at \( P_0,1 = 100 \) mm
- \( t_{1/2,2} = 4 \) hours at \( P_0,2 = 200 \) mm

We can write the ratio of the two half-lives:
\[ \frac{t_{1/2,1}}{t_{1/2,2}} = \frac{P_0,1^{(1-n)}}{P_0,2^{(1-n)}} \]

3. **Substitute the known values:**
Substitute the values of half-lives and pressures into the equation:
\[ \frac{8}{4} = \frac{100^{(1-n)}}{200^{(1-n)}} \]
Simplifying the left side gives:
\[ 2 = \frac{100^{(1-n)}}{200^{(1-n)}} \]

4. **Rewrite the right side:**
The right side can be rewritten using the properties of exponents:
\[ 2 = \left(\frac{100}{200}\right)^{(1-n)} = \left(\frac{1}{2}\right)^{(1-n)} \]

5. **Equate and solve for \( n \):**
Now we have:
\[ 2 = \left(\frac{1}{2}\right)^{(1-n)} \]
This can be rewritten as:
\[ 2^1 = 2^{-(1-n)} \]
Therefore, we can equate the exponents:
\[ 1 = -(1-n) \]
Solving for \( n \):
\[ 1 = -1 + n \implies n = 2 \]

6. **Conclusion:**
The order of the reaction is \( n = 2 \).

### Final Answer:
The order of the reaction is 2.