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 relate the rate of disappearance of \( O_2 \) to the rate of appearance of \( SO_3 \) using the stoichiometry of the reaction:

### Step-by-Step Solution:

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

2. **Identify the Rates:**
According to the stoichiometry of the reaction:
- The rate of disappearance of \( O_2 \) is related to the rate of appearance of \( SO_3 \).
- For every 1 mole of \( O_2 \) that disappears, 2 moles of \( SO_3 \) are produced.

3. **Given Rate of Disappearance of \( O_2 \):**
The rate of disappearance of \( O_2 \) is given as:
\[ -\frac{d[O_2]}{dt} = 2 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

4. **Relate the Rates Using Stoichiometry:**
From the balanced equation, we can write:
\[ -\frac{1}{1} \frac{d[O_2]}{dt} = \frac{2}{2} \frac{d[SO_3]}{dt} \]
This simplifies to:
\[ \frac{d[SO_3]}{dt} = -\frac{1}{2} \frac{d[O_2]}{dt} \]

5. **Substitute the Given Rate:**
Substitute the rate of disappearance of \( O_2 \) into the equation:
\[ \frac{d[SO_3]}{dt} = -\frac{1}{2} \left(-2 \times 10^{-4}\right) \]
This simplifies to:
\[ \frac{d[SO_3]}{dt} = \frac{1}{2} \times 2 \times 10^{-4} = 1 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

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