Nitrogen dioxide `(NO_(2))` dissociates into nitric oxide (NO) and oxygen `(O_(2))` as follows
`NO_(2) to 2NO + O_(2)`
If the rate of decrease of concentration of `NO_(2)` is `6.0 xx 1-^(12)` mol `L^(-1) s^(-1)`. What will be the rate of increase of concentration of `O_(2)`?

Options

A. `3 xx 10^(-12) mol L^(-1) s^(1)`
B. `6 xx 10^(-12) mol L^(-1) s^(-1)`
C. `1 xx 10^(-12 mol L^(-1) s^(1)`
D. `1.5 xx 10^(-12) mol L^(-1) s^(1)`

Explanation

To solve the problem, we need to analyze the given chemical reaction and the rates of change of the concentrations of the reactants and products.

### Step-by-Step Solution:

1. **Write the balanced chemical equation**:
The dissociation of nitrogen dioxide (NO₂) is given by:
\[ \text{NO}_2 \rightarrow 2\text{NO} + \text{O}_2 \]

2. **Identify the stoichiometric coefficients**:
In the balanced equation:
- The stoichiometric coefficient of NO₂ is 1.
- The stoichiometric coefficient of NO is 2.
- The stoichiometric coefficient of O₂ is 1.

3. **Write the rate expressions**:
The rate of a reaction can be expressed in terms of the change in concentration of reactants and products. For the given reaction, we can write:
\[ -\frac{d[\text{NO}_2]}{dt} = \frac{1}{2} \frac{d[\text{NO}]}{dt} = \frac{1}{1} \frac{d[\text{O}_2]}{dt} \]

4. **Use the given rate of decrease of NO₂**:
We are given that the rate of decrease of concentration of NO₂ is:
\[ -\frac{d[\text{NO}_2]}{dt} = 6.0 \times 10^{-12} \, \text{mol L}^{-1} \text{s}^{-1} \]

5. **Relate the rate of NO₂ to the rate of O₂**:
From the rate expressions, we can relate the rate of decrease of NO₂ to the rate of increase of O₂:
\[ \frac{d[\text{O}_2]}{dt} = -\frac{d[\text{NO}_2]}{dt} \]
Since the stoichiometric coefficient of O₂ is 1, we have:
\[ \frac{d[\text{O}_2]}{dt} = \frac{1}{1} \left( -\frac{d[\text{NO}_2]}{dt} \right) \]

6. **Substitute the known value**:
Now substituting the known value:
\[ \frac{d[\text{O}_2]}{dt} = \frac{1}{1} \times 6.0 \times 10^{-12} \, \text{mol L}^{-1} \text{s}^{-1} = 6.0 \times 10^{-12} \, \text{mol L}^{-1} \text{s}^{-1} \]

7. **Final calculation**:
Since the stoichiometric coefficient of O₂ is 1, we need to divide the rate of decrease of NO₂ by its coefficient (which is 1):
\[ \frac{d[\text{O}_2]}{dt} = \frac{1}{2} \times 6.0 \times 10^{-12} = 3.0 \times 10^{-12} \, \text{mol L}^{-1} \text{s}^{-1} \]

### Conclusion:
The rate of increase of concentration of O₂ is:
\[ \frac{d[\text{O}_2]}{dt} = 3.0 \times 10^{-12} \, \text{mol L}^{-1} \text{s}^{-1} \]