How will the rate of reaction
`2SO_(2)(g) + O_(2)(g) rarr 2SO_(3)(g)` change if the volume of the reaction vessel is halved?

Options

A. (a) It will br `1//16th` of its initial value.
B. (b) it will be `1//4th` of its initial value.
C. It will be `8` times of its initial value.
D. It will be `4` times of its initial value.

Explanation

To determine how the rate of the reaction \( 2SO_2(g) + O_2(g) \rightarrow 2SO_3(g) \) changes when the volume of the reaction vessel is halved, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Rate Law**:
The rate of the reaction can be expressed using the rate law. For the given reaction, the rate law can be represented as:
\[ \text{Rate} = k [SO_2]^2 [O_2]^1 \]
where \( k \) is the rate constant, and \([SO_2]\) and \([O_2]\) are the concentrations of sulfur dioxide and oxygen, respectively.

2. **Effect of Volume on Concentration**:
Concentration is defined as the number of moles of a substance per unit volume. When the volume of the reaction vessel is halved, the concentration of each gas will double. This is because concentration is inversely proportional to volume:
\[ [\text{Concentration}] \propto \frac{1}{\text{Volume}} \]
Therefore, if the volume is halved, the new concentrations become:
\[ [SO_2]_{\text{new}} = 2[SO_2]_{\text{initial}} \]
\[ [O_2]_{\text{new}} = 2[O_2]_{\text{initial}} \]

3. **Substituting New Concentrations into the Rate Law**:
We can now substitute the new concentrations into the rate law:
\[ \text{Rate}_{\text{new}} = k (2[SO_2])^2 (2[O_2])^1 \]
Simplifying this gives:
\[ \text{Rate}_{\text{new}} = k \cdot 4[SO_2]^2 \cdot 2[O_2] = 8k [SO_2]^2 [O_2] \]

4. **Comparing New Rate with Initial Rate**:
The initial rate was:
\[ \text{Rate}_{\text{initial}} = k [SO_2]^2 [O_2] \]
Therefore, the new rate is:
\[ \text{Rate}_{\text{new}} = 8 \cdot \text{Rate}_{\text{initial}} \]

5. **Conclusion**:
Thus, when the volume of the reaction vessel is halved, the rate of the reaction increases by a factor of 8.

### Final Answer:
The rate of the reaction will become 8 times the initial rate.

---