In te decomposition of Ammonia it was found that at 50 torr pressure `t_((1//2))` was 3.64 hour while at 100 torr `t_((1//2))` was 1.82 hours. Then order of reaction would be:

Explanation

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

### Step-by-Step Solution:

1. **Understanding the Relationship**: The half-life (\(t_{1/2}\)) of a reaction is related to the concentration of the reactant and the order of the reaction. For a reaction of order \(n\), the half-life can be expressed as:
\[ t_{1/2} \propto \frac{1}{A^{(n-1)}} \]
where \(A\) is the concentration (or pressure in this case).

2. **Setting Up the Ratios**: We have two pressures and their corresponding half-lives:
- At \(50 \, \text{torr}\), \(t_{1/2} = 3.64 \, \text{hours}\)
- At \(100 \, \text{torr}\), \(t_{1/2} = 1.82 \, \text{hours}\)

We can set up the following ratio based on the half-life formula:
\[ \frac{t_{1/2,1}}{t_{1/2,2}} = \left(\frac{A_2}{A_1}\right)^{(1-n)} \]
Substituting the values:
\[ \frac{3.64}{1.82} = \left(\frac{100}{50}\right)^{(1-n)} \]

3. **Calculating the Left Side**: Calculate the left side of the equation:
\[ \frac{3.64}{1.82} = 2 \]

4. **Calculating the Right Side**: The right side simplifies to:
\[ \left(\frac{100}{50}\right)^{(1-n)} = 2^{(1-n)} \]

5. **Setting Up the Equation**: Now we equate both sides:
\[ 2 = 2^{(1-n)} \]

6. **Solving for \(n\)**: Since the bases are the same, we can set the exponents equal to each other:
\[ 1 = 1 - n \]
Rearranging gives:
\[ n = 2 \]

### Conclusion:
The order of the reaction is \(n = 2\), indicating that it is a second-order reaction.

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