For the reaction `N_(2)O_(5) rarr 2NO_(2) + (1)/(2) O_(2)`, the rate of disappearance of `N_(2)O_(5)` is `6.25 xx 10^(-3) "mol L"^(-1) s^(-1)`. The rate of formation of `NO_(2)` and `O_(2)` will be respectively.

Options

A. `6.25xx10^(-3)molL^(-1)S^(-1)and 6.25xx10^(-3)molL^(-1)S^(-1)`
B. `1.25xx10^(-2)molL^(-1)S^(-1)and 3.125xx10^(-3)molL^(-1)S^(-1)`
C. `6.25xx10^(-3)molL^(-1)S^(-1)and 3.125xx10^(-3)molL^(-1)S^(-1)`
D. `1.25xx10^(-2)molL^(-1)S^(-1)and 6.25xx10^(-3)molL^(-1)S^(-1)`

Explanation

To solve the problem, we need to determine the rates of formation of \( \text{NO}_2 \) and \( \text{O}_2 \) based on the given rate of disappearance of \( \text{N}_2\text{O}_5 \).

### Step-by-Step Solution:

1. **Write the balanced chemical equation:**
\[ \text{N}_2\text{O}_5 \rightarrow 2\text{NO}_2 + \frac{1}{2}\text{O}_2 \]

2. **Identify the stoichiometric relationships:**
From the balanced equation, we see that:
- 1 mole of \( \text{N}_2\text{O}_5 \) produces 2 moles of \( \text{NO}_2 \).
- 1 mole of \( \text{N}_2\text{O}_5 \) produces 0.5 moles of \( \text{O}_2 \).

3. **Given rate of disappearance of \( \text{N}_2\text{O}_5 \):**
The rate of disappearance of \( \text{N}_2\text{O}_5 \) is given as:
\[ -\frac{d[\text{N}_2\text{O}_5]}{dt} = 6.25 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} \]

4. **Relate the rate of formation of \( \text{NO}_2 \):**
According to the stoichiometry of the reaction:
\[ -\frac{d[\text{N}_2\text{O}_5]}{dt} = \frac{1}{2} \frac{d[\text{NO}_2]}{dt} \]
Rearranging gives:
\[ \frac{d[\text{NO}_2]}{dt} = -2 \frac{d[\text{N}_2\text{O}_5]}{dt} \]

5. **Substituting the value:**
\[ \frac{d[\text{NO}_2]}{dt} = -2 \times (-6.25 \times 10^{-3}) = 1.25 \times 10^{-2} \, \text{mol L}^{-1} \text{s}^{-1} \]

6. **Relate the rate of formation of \( \text{O}_2 \):**
Similarly, for \( \text{O}_2 \):
\[ -\frac{d[\text{N}_2\text{O}_5]}{dt} = \frac{1}{2} \frac{d[\text{O}_2]}{dt} \]
Rearranging gives:
\[ \frac{d[\text{O}_2]}{dt} = -2 \times \frac{d[\text{N}_2\text{O}_5]}{dt} \]

7. **Substituting the value:**
\[ \frac{d[\text{O}_2]}{dt} = \frac{1}{2} \times 6.25 \times 10^{-3} = 3.125 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} \]

### Final Results:
- The rate of formation of \( \text{NO}_2 \) is \( 1.25 \times 10^{-2} \, \text{mol L}^{-1} \text{s}^{-1} \).
- The rate of formation of \( \text{O}_2 \) is \( 3.125 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} \).