The half life period for catalytic decomposition of `AB_(3)` at 50 mm is found to be 4 hrs and at 100 mm it in 2 hrs. The order of reaction is :

Options

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

Answer

To determine the order of the reaction based on the half-life periods provided, we can follow these steps:

### Step 1: Understand the relationship between half-life and concentration
The half-life period (t_half) of a reaction is related to the initial concentration (A0) and the order of the reaction (N). The relationship can be expressed as:
\[ t_{1/2} \propto [A_0]^{(1 - N)} \]

### Step 2: Set up the equations for the two given conditions
We have two conditions given:
1. When \( A_0 = 50 \, \text{mm} \), \( t_{1/2} = 4 \, \text{hours} \)
2. When \( A_0 = 100 \, \text{mm} \), \( t_{1/2} = 2 \, \text{hours} \)

Using the proportionality, we can write:
\[ t_{1/2,1} \propto [50]^{(1 - N)} \]
\[ t_{1/2,2} \propto [100]^{(1 - N)} \]

### Step 3: Formulate the ratio of the two half-lives
Taking the ratio of the two half-lives, we have:
\[ \frac{t_{1/2,1}}{t_{1/2,2}} = \frac{4}{2} = 2 \]
This can be expressed in terms of concentrations:
\[ \frac{[50]^{(1 - N)}}{[100]^{(1 - N)}} = 2 \]

### Step 4: Simplify the equation
This simplifies to:
\[ \left(\frac{50}{100}\right)^{(1 - N)} = 2 \]
\[ \left(\frac{1}{2}\right)^{(1 - N)} = 2 \]

### Step 5: Solve for N
Taking logarithms or recognizing that \( \left(\frac{1}{2}\right)^{-1} = 2 \), we can equate the exponents:
\[ 1 - N = -1 \]
Thus:
\[ N = 2 \]

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

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