For the decompoistion of `HI` at `1000 K(2HI rarr H_(2)+I_(2))`, following data were obtained:
`|{:([HI] (M),"Rate of decomposition of HI" (mol L^(-1) s^(-1))),(0.1,2.75 xx 10^(-8)),(0.2,11 xx 10^(-8)),(0.3,24.75 xx 10^(-8)):}|`
The order of reaction is

Options

A. `1`
B. `2`
C. `0`
D. `1.5`

Explanation

To determine the order of the reaction for the decomposition of HI at 1000 K, we will analyze the provided experimental data and apply the method of rate laws.

### Step-by-Step Solution:

1. **Write the Rate Law Expression**:
The rate of the reaction can be expressed as:
\[ \text{Rate} = k \cdot [HI]^n \]
where \( k \) is the rate constant and \( n \) is the order of the reaction with respect to HI.

2. **Use Experimental Data**:
We have the following data:
- Experiment 1: \([HI] = 0.1 \, M\), Rate = \(2.75 \times 10^{-8} \, \text{mol L}^{-1} \text{s}^{-1}\)
- Experiment 2: \([HI] = 0.2 \, M\), Rate = \(11 \times 10^{-8} \, \text{mol L}^{-1} \text{s}^{-1}\)
- Experiment 3: \([HI] = 0.3 \, M\), Rate = \(24.75 \times 10^{-8} \, \text{mol L}^{-1} \text{s}^{-1}\)

3. **Set Up Ratios for Two Experiments**:
We will compare the first and second experiments to find \( n \).
\[ \frac{R_2}{R_1} = \frac{k \cdot [0.2]^n}{k \cdot [0.1]^n} \]
This simplifies to:
\[ \frac{R_2}{R_1} = \frac{[0.2]^n}{[0.1]^n} \]

4. **Substituting the Rates**:
Substitute the rates from the experiments into the equation:
\[ \frac{11 \times 10^{-8}}{2.75 \times 10^{-8}} = \frac{(0.2)^n}{(0.1)^n} \]
Simplifying the left side:
\[ 4 = \left(\frac{0.2}{0.1}\right)^n = (2)^n \]

5. **Solving for \( n \)**:
From the equation \( 4 = 2^n \), we can see that:
\[ n = 2 \]

6. **Conclusion**:
The order of the reaction with respect to HI is 2. Thus, the rate law can be expressed as:
\[ \text{Rate} = k \cdot [HI]^2 \]

### Final Answer:
The order of the reaction is **2**.