The half-life periof for catalystic decompoistion of `AB_(3)` at `50 mm` is found to be `4` hr and at `100 mm` it is `2.0 hr`. The order of reaction is

Options

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

Explanation

To determine the order of the reaction for the catalytic decomposition of \( AB_3 \) based on the given half-life periods at different concentrations, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Half-Life in Relation to Reaction Order**:
The half-life (\( t_{1/2} \)) of a reaction is related to the concentration of the reactants and the order of the reaction. For a general nth order reaction, the half-life is given by:
\[ t_{1/2} \propto [A]^{(1-n)} \]
where \( [A] \) is the concentration of the reactant.

2. **Setting Up the Given Data**:
We have two half-life values:
- At \( [A] = 50 \, \text{mm} \), \( t_{1/2} = 4 \, \text{hr} \)
- At \( [A] = 100 \, \text{mm} \), \( t_{1/2} = 2 \, \text{hr} \)

3. **Formulating the Ratio of Half-Lives**:
We can express the ratio of the two half-lives in terms of their concentrations:
\[ \frac{t_{1/2,1}}{t_{1/2,2}} = \frac{[A]_1^{(1-n)}}{[A]_2^{(1-n)}} \]
Substituting the known values:
\[ \frac{4 \, \text{hr}}{2 \, \text{hr}} = \frac{50^{(1-n)}}{100^{(1-n)}} \]
Simplifying the left side:
\[ 2 = \frac{50^{(1-n)}}{100^{(1-n)}} \]

4. **Simplifying the Right Side**:
The right side can be simplified further:
\[ 2 = \left(\frac{50}{100}\right)^{(1-n)} = \left(\frac{1}{2}\right)^{(1-n)} \]

5. **Equating the Powers**:
Now we can equate the powers:
\[ 2 = 2^{(n-1)} \]
This implies:
\[ n - 1 = 1 \quad \Rightarrow \quad n = 2 \]

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

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