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

Answer

To determine the order of the reaction for the decomposition of HI at 1000 K, we will analyze the provided data and use the rate law.

### Step-by-Step Solution:

1. **Understand the Reaction**: The reaction is given as:
\[ 2 \text{HI} \rightarrow \text{H}_2 + \text{I}_2 \]
The rate of the reaction can be expressed as:
\[ \text{Rate} = k [\text{HI}]^x \]
where \( k \) is the rate constant and \( x \) is the order of the reaction.

2. **Gather the Data**: The data provided is:
- For \([HI] = 0.1 \, M\), Rate = \(2.75 \times 10^{-8} \, \text{mol L}^{-1} \text{s}^{-1}\) (let's call this \( R_1 \))
- For \([HI] = 0.2 \, M\), Rate = \(11 \times 10^{-8} \, \text{mol L}^{-1} \text{s}^{-1}\) (let's call this \( R_2 \))
- For \([HI] = 0.3 \, M\), Rate = \(24.75 \times 10^{-8} \, \text{mol L}^{-1} \text{s}^{-1}\) (let's call this \( R_3 \))

3. **Set Up the Rate Equations**:
- For \( R_1 \):
\[ R_1 = k (0.1)^x \]
Thus,
\[ 2.75 \times 10^{-8} = k (0.1)^x \quad \text{(1)} \]
- For \( R_2 \):
\[ R_2 = k (0.2)^x \]
Thus,
\[ 11 \times 10^{-8} = k (0.2)^x \quad \text{(2)} \]

4. **Divide Equations (1) and (2)**:
\[ \frac{R_1}{R_2} = \frac{k (0.1)^x}{k (0.2)^x} \]
This simplifies to:
\[ \frac{2.75 \times 10^{-8}}{11 \times 10^{-8}} = \frac{(0.1)^x}{(0.2)^x} \]
\[ \frac{2.75}{11} = \left(\frac{0.1}{0.2}\right)^x \]
\[ \frac{1}{4} = \left(\frac{1}{2}\right)^x \]

5. **Solve for \( x \)**:
Since \(\frac{1}{4} = \left(\frac{1}{2}\right)^2\), we can equate the exponents:
\[ x = 2 \]

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