The reaction, `2SO_(2(g)) + O_(2(g))hArr2SO_(3(g))` is carried out in a 1 `dm^(3)` and 2 `dm^(3)`vessel separately. The ratio of the reaction velocity will be

Options

A. `1:8`
B. `1:4`
C. `4:1`
D. `8:1`

Explanation

To solve the problem, we need to determine the ratio of the reaction velocities (rates) of the reaction \(2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)\) when carried out in two different volumes: 1 dm³ and 2 dm³.

### Step-by-Step Solution:

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

2. **Calculate Concentrations**:
The concentration of a gas is given by:
\[ [C] = \frac{n}{V} \]
where \(n\) is the number of moles and \(V\) is the volume in dm³.

- For the 1 dm³ vessel:
\[ [SO_2] = \frac{n_{SO_2}}{1} = n_{SO_2} \]
\[ [O_2] = \frac{n_{O_2}}{1} = n_{O_2} \]

- For the 2 dm³ vessel:
\[ [SO_2] = \frac{n_{SO_2}}{2} \]
\[ [O_2] = \frac{n_{O_2}}{2} \]

3. **Write the Rate Expressions**:
- For the 1 dm³ vessel:
\[ R_1 = k [SO_2]^2 [O_2] = k (n_{SO_2})^2 (n_{O_2}) \]

- For the 2 dm³ vessel:
\[ R_2 = k \left(\frac{n_{SO_2}}{2}\right)^2 \left(\frac{n_{O_2}}{2}\right) = k \left(\frac{n_{SO_2}^2}{4}\right) \left(\frac{n_{O_2}}{2}\right) = k \frac{n_{SO_2}^2 n_{O_2}}{8} \]

4. **Calculate the Ratio of the Rates**:
To find the ratio \( \frac{R_1}{R_2} \):
\[ \frac{R_1}{R_2} = \frac{k (n_{SO_2})^2 (n_{O_2})}{k \frac{n_{SO_2}^2 n_{O_2}}{8}} = \frac{8(n_{SO_2})^2 (n_{O_2})}{(n_{SO_2})^2 (n_{O_2})} = 8 \]

5. **Final Ratio**:
Thus, the ratio of the reaction velocities when carried out in a 1 dm³ vessel to a 2 dm³ vessel is:
\[ R_1 : R_2 = 8 : 1 \]

### Conclusion:
The ratio of the reaction velocities is \(8:1\).