For the reaction , `2SO_2+O_2hArr2SO_3, ` the rate of disappearance of `O_2` is `2xx10^(-4) molL^(-1)s^(-1)` . The rate of appearance of `SO_3` is

Options

A. `2xx10^(-4) molL^(-1)s^(-1)`
B. `4xx10^(-4) molL^(-1)s^(-1)`
C. `1xx10^(-4) molL^(-1)s^(-1)`
D. `6xx10^(-4) molL^(-1)s^(-1)`

Answer

To solve the problem, we need to analyze the given reaction and the rates of disappearance and appearance of the substances involved.

### Step-by-Step Solution:

1. **Write the balanced chemical equation**:
\[ 2SO_2 + O_2 \rightleftharpoons 2SO_3 \]

2. **Identify the rates of disappearance and appearance**:
- The rate of disappearance of \( O_2 \) is given as:
\[ \text{Rate of disappearance of } O_2 = -\frac{d[O_2]}{dt} = 2 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]
- We need to find the rate of appearance of \( SO_3 \):
\[ \text{Rate of appearance of } SO_3 = \frac{d[SO_3]}{dt} \]

3. **Relate the rates using stoichiometry**:
From the balanced equation, we see that:
- For every 1 mole of \( O_2 \) that disappears, 2 moles of \( SO_3 \) are formed.
- Therefore, the relationship between the rates can be expressed as:
\[ -\frac{1}{1} \frac{d[O_2]}{dt} = \frac{2}{1} \frac{d[SO_3]}{dt} \]

4. **Set up the equation**:
Substituting the known rate of disappearance of \( O_2 \):
\[ 2 \times 10^{-4} = 2 \frac{d[SO_3]}{dt} \]

5. **Solve for the rate of appearance of \( SO_3 \)**:
\[ \frac{d[SO_3]}{dt} = \frac{2 \times 10^{-4}}{2} = 1 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

6. **Final result**:
Therefore, the rate of appearance of \( SO_3 \) is:
\[ \frac{d[SO_3]}{dt} = 4 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

### Conclusion:
The rate of appearance of \( SO_3 \) is \( 4 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \).