For the reaction `2NO(g) + H_(2)(g) rarr N_(2)O(g) + H_(2)O(g)`, at `900 K` following data are observed.
`{:("Initial pressure of NO (atm)","Initial pressure of" H_(2)o (atm),"Initial rate of pressure decrease" (atm min^(-1))),(0.150,0.40,0.020),(0.075,0.40,0.005),(0.150,0.20,0.010):}`
Find the order of reaction.

Options

A. `3`
B. `2`
C. `1`
D. `0`

Explanation

To find the order of the reaction for the given reaction \(2NO(g) + H_2(g) \rightarrow N_2O(g) + H_2O(g)\) at \(900 K\), we will use the initial rate method based on the provided experimental data.

### Step-by-Step Solution:

1. **Write the Rate Law Expression:**
The rate law for the reaction can be expressed as:
\[ \text{Rate} = k [NO]^x [H_2]^y \]
where \(x\) is the order with respect to \(NO\) and \(y\) is the order with respect to \(H_2\).

2. **Use the First Two Experiments to Find \(x\):**
From the data:
- Experiment 1: \(P_{NO} = 0.150 \, \text{atm}, P_{H_2} = 0.40 \, \text{atm}, \text{Rate} = 0.020 \, \text{atm min}^{-1}\)
- Experiment 2: \(P_{NO} = 0.075 \, \text{atm}, P_{H_2} = 0.40 \, \text{atm}, \text{Rate} = 0.005 \, \text{atm min}^{-1}\)

Set up the ratio of the rates:
\[ \frac{0.020}{0.005} = \frac{k (0.150)^x (0.40)^y}{k (0.075)^x (0.40)^y} \]
This simplifies to:
\[ 4 = \left(\frac{0.150}{0.075}\right)^x \]
\[ 4 = 2^x \]
Solving for \(x\):
\[ x = 2 \]

3. **Use the First and Third Experiments to Find \(y\):**
From the data:
- Experiment 1: \(P_{NO} = 0.150 \, \text{atm}, P_{H_2} = 0.40 \, \text{atm}, \text{Rate} = 0.020 \, \text{atm min}^{-1}\)
- Experiment 3: \(P_{NO} = 0.150 \, \text{atm}, P_{H_2} = 0.20 \, \text{atm}, \text{Rate} = 0.010 \, \text{atm min}^{-1}\)

Set up the ratio of the rates:
\[ \frac{0.020}{0.010} = \frac{k (0.150)^x (0.40)^y}{k (0.150)^x (0.20)^y} \]
This simplifies to:
\[ 2 = \left(\frac{0.40}{0.20}\right)^y \]
\[ 2 = 2^y \]
Solving for \(y\):
\[ y = 1 \]

4. **Calculate the Total Order of the Reaction:**
The total order of the reaction is given by:
\[ \text{Total Order} = x + y = 2 + 1 = 3 \]

### Final Answer:
The order of the reaction is **3**.