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 1: Write the relationship between half-life and initial concentration (or pressure)
The half-life (\(t_{1/2}\)) of a reaction is related to the initial concentration (\(A_0\)) and the order of the reaction (\(n\)) by the formula:
\[ t_{1/2} \propto \frac{1}{A_0^{(n-1)}} \]
This means that the half-life is inversely proportional to the initial concentration raised to the power of \(n-1\).

### Step 2: Set up the ratio of half-lives
Given:
- At 50 torr, \(t_{1/2} = 3.64\) hours
- At 100 torr, \(t_{1/2} = 1.82\) hours

We can set up the ratio of the half-lives:
\[ \frac{t_{1/2,1}}{t_{1/2,2}} = \frac{3.64}{1.82} \]
And the corresponding pressures:
\[ \frac{A_{0,1}}{A_{0,2}} = \frac{50}{100} = \frac{1}{2} \]

### Step 3: Substitute into the half-life relationship
Substituting into the relationship gives us:
\[ \frac{3.64}{1.82} = \left(\frac{1}{2}\right)^{(1-n)} \]

### Step 4: Calculate the left side
Calculating the left side:
\[ \frac{3.64}{1.82} = 2 \]

### Step 5: Set up the equation
Now we have:
\[ 2 = \left(\frac{1}{2}\right)^{(1-n)} \]

### Step 6: Rewrite the equation
Rewriting the equation:
\[ 2 = 2^{-1+n} \]

### Step 7: Equate the exponents
Since the bases are the same, we can equate the exponents:
\[ 1 = -1 + n \]

### Step 8: Solve for \(n\)
Solving for \(n\):
\[ n = 2 \]

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

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