The half -lie period for the decomposition of `AB_(2)(g)` at `100mm` pressure is `10min` and at `200mm` pressure is 5 min, the order of reaction is

Explanation

To determine the order of the reaction for the decomposition of \( AB_2(g) \) based on the given half-life periods at different pressures, we can follow these steps:

### Step-by-Step Solution

1. **Identify the Given Data:**
- Half-life at \( P_1 = 100 \, \text{mm} \) is \( T_{1/2} = 10 \, \text{min} \)
- Half-life at \( P_2 = 200 \, \text{mm} \) is \( T_{1/2} = 5 \, \text{min} \)

2. **Use the Relationship for Half-Life:**
The half-life \( T_{1/2} \) for a reaction of order \( n \) is given by the formula:
\[ T_{1/2} \propto \frac{1}{P^{n-1}} \]
This means that:
\[ T_{1/2} \cdot P^{n-1} = \text{constant} \]

3. **Set Up the Ratio of Half-Lives:**
From the relationship, we can write:
\[ \frac{T_{1/2,1}}{T_{1/2,2}} = \frac{P_2^{n-1}}{P_1^{n-1}} \]

4. **Substitute the Known Values:**
Substitute \( T_{1/2,1} = 10 \, \text{min} \), \( T_{1/2,2} = 5 \, \text{min} \), \( P_1 = 100 \, \text{mm} \), and \( P_2 = 200 \, \text{mm} \):
\[ \frac{10}{5} = \frac{200^{n-1}}{100^{n-1}} \]

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

6. **Solve for \( n \):**
Since \( 2 = 2^{n-1} \), we can equate the exponents:
\[ n - 1 = 1 \]
Therefore:
\[ n = 2 \]

### Conclusion
The order of the reaction is \( n = 2 \).

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