The following data were obtained at a certain temperature for the decomposition of `NH_3` in contact with catalyst.
`{:(P(mn), 25, 50),(t_(1//2),1.82,0.91):}`
The order of reaction is

Options

A. 2
B. 3
C. 1
D. 4

Answer

To determine the order of the reaction for the decomposition of ammonia (`NH3`) in contact with a catalyst, we will analyze the provided data regarding half-lives at different pressures.

### Step-by-Step Solution:

1. **Understand the Relationship**: The half-life of a reaction is related to the pressure and the order of the reaction. The relationship can be expressed as:
\[ t_{1/2} \propto \frac{1}{P^{n-1}} \]
where \( t_{1/2} \) is the half-life, \( P \) is the pressure, and \( n \) is the order of the reaction.

2. **Set Up the Equations**: From the data provided:
- For \( P = 25 \) mm, \( t_{1/2} = 1.82 \) min
- For \( P = 50 \) mm, \( t_{1/2} = 0.91 \) min

We can express this relationship for both conditions:
\[ t_{1/2,1} = K \cdot \frac{1}{P_1^{n-1}} \quad \text{(1)} \]
\[ t_{1/2,2} = K \cdot \frac{1}{P_2^{n-1}} \quad \text{(2)} \]

Substituting the values:
- From (1): \( 1.82 = K \cdot \frac{1}{25^{n-1}} \)
- From (2): \( 0.91 = K \cdot \frac{1}{50^{n-1}} \)

3. **Divide the Equations**: To eliminate \( K \), we can divide equation (1) by equation (2):
\[ \frac{1.82}{0.91} = \frac{K \cdot \frac{1}{25^{n-1}}}{K \cdot \frac{1}{50^{n-1}}} \]
This simplifies to:
\[ 2 = \frac{50^{n-1}}{25^{n-1}} \]

4. **Simplify the Right Side**: The expression \( \frac{50^{n-1}}{25^{n-1}} \) can be rewritten using the property of exponents:
\[ 2 = \left(\frac{50}{25}\right)^{n-1} = 2^{n-1} \]

5. **Solve for \( n \)**: Since the bases are the same, we can equate the exponents:
\[ 2^{1} = 2^{n-1} \]
Thus:
\[ 1 = n - 1 \implies n = 2 \]

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