In the 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 from the given half-lives at different pressures, we can follow these steps:

### Step 1: Understand the relationship between half-life and order of reaction
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 is given by the formula:
\[ t_{1/2} \propto \frac{1}{A^{(n-1)}} \]
where \( A \) is the concentration (or pressure in this case).

### Step 2: Set up the ratios of half-lives and pressures
From the problem, we have:
- At 50 torr, \( t_{1/2} = 3.64 \) hours
- At 100 torr, \( t_{1/2} = 1.82 \) hours

We can write the relationship as:
\[ \frac{t_{1/2,1}}{t_{1/2,2}} = \frac{A_2^{(n-1)}}{A_1^{(n-1)}} \]
Substituting the values:
\[ \frac{3.64}{1.82} = \frac{100^{(n-1)}}{50^{(n-1)}} \]

### Step 3: Simplify the ratio of half-lives
Calculating the left side:
\[ \frac{3.64}{1.82} = 2 \]

### Step 4: Express the pressure ratio
Now we can express the right side:
\[ 2 = \frac{100^{(n-1)}}{50^{(n-1)}} \]
This simplifies to:
\[ 2 = \left(\frac{100}{50}\right)^{(n-1)} = 2^{(n-1)} \]

### Step 5: Solve for \( n \)
Taking logarithm on both sides or equating the exponents:
\[ 2 = 2^{(n-1)} \]
This implies:
\[ n - 1 = 1 \implies n = 2 \]

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