`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 need to analyze how the rate of the reaction changes when the volume of the container is increased. Here are the steps to arrive at the solution:

### Step-by-Step Solution:

1. **Identify the Reaction and Its Order**:
The reaction given is:
\[ H_2 + I_2 \rightarrow 2HI \]
This is an elementary reaction of second order.

2. **Write the Rate Law Expression**:
For a second-order reaction involving two reactants, the rate law can be expressed as:
\[ \text{Rate} = k [H_2][I_2] \]
where \( k \) is the rate constant, and \([H_2]\) and \([I_2]\) are the concentrations of the reactants.

3. **Understand the Effect of Volume Change**:
When the volume of the container is increased to twice its original volume (from \( V \) to \( 2V \)), the concentration of each reactant will change. Concentration is defined as:
\[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume}} \]
Therefore, if the volume doubles, the concentration of each reactant will be halved.

4. **Calculate New Concentrations**:
Let the initial concentrations of \( H_2 \) and \( I_2 \) be \([H_2] = C\) and \([I_2] = C\). After doubling the volume:
\[ [H_2]_{\text{new}} = \frac{C}{2} \quad \text{and} \quad [I_2]_{\text{new}} = \frac{C}{2} \]

5. **Write the New Rate Law**:
Substitute the new concentrations into the rate law:
\[ \text{New Rate} = k \left(\frac{C}{2}\right) \left(\frac{C}{2}\right) = k \left(\frac{C^2}{4}\right) = \frac{1}{4} k C^2 \]

6. **Compare New Rate with Original Rate**:
The original rate was:
\[ \text{Original Rate} = k C^2 \]
Therefore, the new rate is:
\[ \text{New Rate} = \frac{1}{4} \times \text{Original Rate} \]

7. **Conclusion**:
The final rate of the reaction becomes \( \frac{1}{4} \) of the original rate.

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

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