`H_2+l_2 rarr` 2Hl (An elementary reaction) of second order
If the volume of the container containing the gaseous mixture is increased to two time, then final rate of the reaction.

Options

A. Become four times
B. Become `1/4` th of the original rate
C. Become 2 times
D. Become `1/2` of the original rate

Answer

To solve the problem, we will analyze the effect of increasing the volume of the container on the rate of the reaction \( H_2 + I_2 \rightarrow 2 HI \), which is a second-order reaction.

### Step-by-Step Solution:

1. **Identify the Reaction Order**:
The reaction \( H_2 + I_2 \rightarrow 2 HI \) is given as a second-order reaction. For a second-order reaction involving two reactants, the rate law can be expressed as:
\[ \text{Rate} = k [H_2]^m [I_2]^n \]
where \( m + n = 2 \). Since both reactants are involved, we can assume \( m = 1 \) and \( n = 1 \):
\[ \text{Rate} = k [H_2]^1 [I_2]^1 = k [H_2][I_2] \]

2. **Effect of Volume Increase**:
When the volume of the container is increased to two times, the concentration of each reactant will change. Concentration is defined as:
\[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume}} \]
If the volume is doubled, the concentration of each reactant will be halved:
\[ [H_2]_{\text{new}} = \frac{[H_2]}{2}, \quad [I_2]_{\text{new}} = \frac{[I_2]}{2} \]

3. **Calculate the New Rate**:
Substitute the new concentrations into the rate law:
\[ \text{New Rate} = k \left(\frac{[H_2]}{2}\right) \left(\frac{[I_2]}{2}\right) = k \frac{[H_2][I_2]}{4} \]
This shows that the new rate is:
\[ \text{New Rate} = \frac{1}{4} \times \text{Old Rate} \]

4. **Conclusion**:
Therefore, when the volume of the container is increased to two times, the final rate of the reaction will be \( \frac{1}{4} \) of the original rate.

### Final Answer:
The final rate of the reaction after increasing the volume is \( \frac{1}{4} \) of the original rate.

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