For the reaction `2NO_(2)rarrN_(2)O_(3)+O_(2)`, rate expression is as follows `:`
`-(d[NO_(2)])/(dt)=K[NO_(2)]^(n)`, where `K=3xx10^(-3)mol^(-1)L sec^(-1)`
If rate of formation of oxygen is `1.5xx10^(-4)mol L^(-1)sec^(-1)` then the molar concentration of `NO_(2)`in mole `L^(-1)` is `:`

Options

A. `1.5xx10^(-4)`
B. `0.0151`
C. `0.214`
D. `0.316`

Explanation

To solve the problem, we need to determine the molar concentration of \( NO_2 \) given the rate of formation of oxygen and the rate expression for the reaction. Let's break down the steps:

### Step 1: Write the Rate Expression
The rate expression for the reaction \( 2NO_2 \rightarrow N_2O_3 + O_2 \) is given as:
\[ -\frac{d[NO_2]}{dt} = K[NO_2]^n \]
where \( K = 3 \times 10^{-3} \, \text{mol}^{-1} \text{L sec}^{-1} \).

### Step 2: Relate the Rate of Formation of Oxygen to the Rate of Consumption of \( NO_2 \)
From the stoichiometry of the reaction, we know that:
- For every 2 moles of \( NO_2 \) that react, 1 mole of \( O_2 \) is formed.
- Therefore, the rate of formation of \( O_2 \) is related to the rate of consumption of \( NO_2 \) by the equation:
\[ \text{Rate of formation of } O_2 = \frac{1}{2} \left(-\frac{d[NO_2]}{dt}\right) \]

### Step 3: Substitute the Given Rate of Formation of Oxygen
We are given that the rate of formation of \( O_2 \) is:
\[ \frac{d[O_2]}{dt} = 1.5 \times 10^{-4} \, \text{mol L}^{-1} \text{sec}^{-1} \]
Thus, we can express the rate of consumption of \( NO_2 \):
\[ 1.5 \times 10^{-4} = \frac{1}{2} \left(-\frac{d[NO_2]}{dt}\right) \]
This implies:
\[ -\frac{d[NO_2]}{dt} = 3.0 \times 10^{-4} \, \text{mol L}^{-1} \text{sec}^{-1} \]

### Step 4: Substitute into the Rate Expression
Now we can substitute this into the rate expression:
\[ 3.0 \times 10^{-4} = K[NO_2]^n \]
Substituting \( K \) and \( n \):
\[ 3.0 \times 10^{-4} = (3 \times 10^{-3})[NO_2]^2 \]

### Step 5: Solve for \( [NO_2]^2 \)
Rearranging the equation gives:
\[ [NO_2]^2 = \frac{3.0 \times 10^{-4}}{3 \times 10^{-3}} = \frac{3.0}{3} \times 10^{-4 + 3} = 1.0 \times 10^{-1} \]

### Step 6: Calculate \( [NO_2] \)
Taking the square root of both sides:
\[ [NO_2] = \sqrt{1.0 \times 10^{-1}} = 0.316227766 \approx 0.316 \, \text{mol L}^{-1} \]

### Conclusion
Thus, the molar concentration of \( NO_2 \) is approximately \( 0.316 \, \text{mol L}^{-1} \).