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 dissociation reaction of nitrogen dioxide (NO₂) into nitric oxide (NO) and oxygen (O₂):

\[ \text{NO}_2 \rightarrow 2\text{NO} + \text{O}_2 \]

### Step 1: Write the Rate Expression
The rate of the reaction can be expressed in terms of the change in concentration of the reactants and products. For the given reaction, the rate of decrease of NO₂ is related to the rate of increase of NO and O₂.

The rate of decrease of NO₂ can be expressed as:
\[ -\frac{d[\text{NO}_2]}{dt} \]

The rate of increase of NO and O₂ can be expressed as:
\[ \frac{d[\text{NO}]}{dt} = 2 \cdot \frac{d[\text{NO}_2]}{dt} \]
\[ \frac{d[\text{O}_2]}{dt} = \frac{d[\text{NO}_2]}{dt} \]

### Step 2: Relate the Rates
From the stoichiometry of the reaction, we can see that for every 1 mole of NO₂ that decomposes, 1 mole of O₂ is produced and 2 moles of NO are produced. Therefore, the relationship between the rates is:
\[ -\frac{1}{1} \frac{d[\text{NO}_2]}{dt} = \frac{1}{2} \frac{d[\text{NO}]}{dt} = \frac{1}{1} \frac{d[\text{O}_2]}{dt} \]

### Step 3: Substitute the Given Rate
We are given that the rate of decrease of NO₂ is:
\[ -\frac{d[\text{NO}_2]}{dt} = 6.0 \times 10^{-12} \, \text{mol L}^{-1} \text{s}^{-1} \]

### Step 4: Calculate the Rate of Increase of O₂
Using the relationship from Step 2, we can find the rate of increase of O₂:
\[ \frac{d[\text{O}_2]}{dt} = -\frac{d[\text{NO}_2]}{dt} \]

Substituting the value:
\[ \frac{d[\text{O}_2]}{dt} = \frac{6.0 \times 10^{-12}}{1} = 6.0 \times 10^{-12} \, \text{mol L}^{-1} \text{s}^{-1} \]

### Step 5: Final Calculation for O₂
Since the stoichiometry shows that for every 1 mole of NO₂ that decomposes, 1 mole of O₂ is produced, the rate of increase of O₂ is:
\[ \frac{d[\text{O}_2]}{dt} = \frac{1}{2} \times 6.0 \times 10^{-12} = 3.0 \times 10^{-12} \, \text{mol L}^{-1} \text{s}^{-1} \]

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