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`

Answer

To solve the problem, we need to determine the ratio of the reaction velocities for the given reaction in two different volumes (1 dm³ and 2 dm³). The reaction is:

\[ 2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \]

### Step-by-Step Solution:

1. **Identify the Rate Expression**:
The rate of the reaction can be expressed as:
\[ \text{Rate} = k [SO_2]^2 [O_2]^1 \]
where \( k \) is the rate constant.

2. **Calculate Concentrations in the 1 dm³ Vessel**:
For the 1 dm³ vessel, let’s assume the initial concentrations of \( SO_2 \) and \( O_2 \) are \( [SO_2]_1 \) and \( [O_2]_1 \) respectively. The rate \( R_1 \) can be expressed as:
\[ R_1 = k [SO_2]_1^2 [O_2]_1 \]

3. **Calculate Concentrations in the 2 dm³ Vessel**:
For the 2 dm³ vessel, the concentrations will be half of those in the 1 dm³ vessel due to the increased volume. Therefore, we have:
\[ [SO_2]_2 = \frac{[SO_2]_1}{2}, \quad [O_2]_2 = \frac{[O_2]_1}{2} \]
The rate \( R_2 \) can be expressed as:
\[ R_2 = k \left(\frac{[SO_2]_1}{2}\right)^2 \left(\frac{[O_2]_1}{2}\right) \]

4. **Substituting the Concentrations into the Rate Expression**:
Substitute the concentrations into the rate expression for \( R_2 \):
\[ R_2 = k \left(\frac{[SO_2]_1^2}{4}\right) \left(\frac{[O_2]_1}{2}\right) = k \frac{[SO_2]_1^2 [O_2]_1}{8} \]

5. **Finding the Ratio of Reaction Velocities**:
Now, we can find the ratio of the reaction velocities \( \frac{R_1}{R_2} \):
\[ \frac{R_1}{R_2} = \frac{k [SO_2]_1^2 [O_2]_1}{k \frac{[SO_2]_1^2 [O_2]_1}{8}} = \frac{1}{\frac{1}{8}} = 8 \]

6. **Final Ratio**:
Therefore, the ratio of the reaction velocities \( R_1 : R_2 \) is:
\[ R_1 : R_2 = 8 : 1 \]

### Conclusion:
The ratio of the reaction velocity in the 1 dm³ vessel to that in the 2 dm³ vessel is \( 8 : 1 \).